Applied Calculus For Business Gordon .pdf

APPLIED CALCULUSFOR BUSINESS, ECONOMICS, AND FINANCE Warren B. Gordon / Walter O. Wang / April Allen Materowski Baruch College City University of New York Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Cover designed by Joshua Gordon. Photograph of the TI-89 Platinum Graphing Calculator reproduced by permission of Texas Instruments. Copyright © 2007, 2006 by Pearson Custom Publishing All rights reserved. Permission in writing must be obtained from the publisher before any part of this work may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying and recording, or by any information storage or retrieval system. All trademarks, service marks, registered trademarks, and registered service marks are the property of their respective owners and are used herein for identi cation purposes only. Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 ISBN 0-536-46018-3 2007360639 CS Please visit our web site at www.pearsoncustom.com PEARSON CUSTOM PUBLISHING 501 Boylston Street, Suite 900, Boston, MA 02116 A Pearson Education Company Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Preface Applied Calculus for Business, Economics and Finance is a combination of the authors two texts Precalculus and Elements of Calculus and Applied Calculus. This single text may be used to cover the content of an applied calculus course for non-science majors in a variety of ways. This text continues the approach used in its precursor texts, that is, the integration of precalculus with the calculus as well as the integration of technology. Most sections of this text conclude with Calculator Tips, illustrating how the calculator may be used to enhance the understanding and appreciation of the topics considered in that section. While the text illustrates the TI 89 calculator, many of the calculator examples and exercises may be worked with most graphing calculators. As most students are now required to have a basic knowledge of spreadsheets, we utilize them several times in the text. They make the calculations involved in Newton s method painless, they allow us to easily illustrate the use of left, right and midpoint variations in computing Riemann sums, and even suggest the notion of speed of convergence. Many of the graphs and tables found in this text were created using MAPLE®, the TI 89 calculator and Excel®. The exponential and logarithmic functions are introduced in this text, assuming no prior knowledge. There are abundant examples illustrating the importance of these functions, perhaps more than time may allow and therefore selection is left to the instructor. In an introductory applied calculus text we believe it is inappropriate to give a detailed treatment of multivariable functions. We make no attempt to develop the methods used to sketch surfaces, but instead, examine functions of two variables using level curves. Our objective in this chapter is for students to understand the notion of a partial derivative, and the optimization of multivariable functions, including the method of Lagrange multipliers, an important tool in Economics and Finance. The chapter concludes with an examination of double integrals and their application to areas, volumes and probability. To learn mathematics, students must work out exercises and we have included many in this text. The text includes answers to both the odd and even exercises. For many students, applied calculus is a terminal course. Most of the exercises were written with the goal of reinforcing the concepts and skills introduced in the section. We also include, in most exercise sets, problems which will allow the stronger students to stretch their understanding of the calculus. These exercises can be assigned by the instructor as extra credit or honors problems. As there are a significant number of students majoring in quantitative disciplines who will continue with their study of mathematics, they may find some of these exercises useful in easing their transition to the next course. On the publisher s web page http://www.pearsoncustom.com/gordon_appliedcalcbus/ there is located a link to the text videos. These videos, presented by two of the authors, correspond to the text material, section by section. We have found that the videos are very useful to students who miss a class, or just need another detailed look at the material. The videos allow students to go over the section material at their own speed. There is more material in this text than can be included in a one semester course, and as a result, there is some flexibility in the inclusion of topics. Chapter 0 is provided for students who need an algebra refresher. It has been our experience that students should be referred to this material as needed (or to the corresponding videos). The content of this iii Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. iv * ** Preface chapter is aimed directly at the algebraic skills needed for calculus. Appendix A, Matrices and Linear Systems, is generally not found in most applied calculus texts. We include it as many business schools require their students to be familiar with matrices and their applications. No other chapter in the text uses this material, and it may be included or omitted as required by the needs of the course. Trigonometric functions do not appear in this text. While we have been tempted to include them, we realize that periodic functions arise mostly in the study of physics, infrequently, if ever, in business applications. A supplement can be made available to those who wish to briefly examine the calculus of these functions. In the inside cover of this text are images of the TI 89 and TI 89 Titanium calculators. Note that in addition to the symbol on each key there are two additional symbols associated with each key; on the TI 89, either in orange, accessed by first pressing the orange 2nd key and then the key with the associated symbol, or in green, accessed by first pressing the green diamond key and then the key with the associated symbol. We shall indicate the green diamond by the symbol * in this text. The TI 89 Titanium model has the * key in light green and the 2nd key in blue. Both of these keys are found in the upper left portion of the keypad on each of the calculators. Also note the purple alpha key which is needed for the space and and insertions. We gratefully acknowledge the help of many of our colleagues at Baruch College who have provided suggestions and offered constructive criticism of this text, in particular, we thank Sherman Wong who is our local Maple guru. We thank Delia Uherec, Acquisitions Editor at Pearson Publication for her assistance with publication of this text, and Curt Hinrichs and Ann Day, Editors at Thompson Brooks/Cole Publishing, for allowing us to use material written by the first author from the text Succeeding in Applied Calculus: Algebra Essentials, and its website. Warren B. Gordon Walter O. Wang April Allen Materowski April, 2006 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Contents Preface iii Chapter 0 Reviewing the Basics 2 3 5 5 6 9 10 12 14 15 17 18 21 22 24 25 26 27 28 29 32 32 38 39 0.1 Solving Linear Equations Addition and Multiplication Properties Linear Equations with Fractions Linear Equations with Decimals Solving for a Particular Variable Applications of Linear Equations Calculator Tips Exercises Isolation of Squared Term Isolation of Squared Binomial Term Calculator Tips Exercises Completion of the Square Calculator Tips Exercises 0.2 Solving Equations of the Form ax2 - b = 0 0.3 Completing the Square 0.4 The Quadratic Formula and Applications Quadratic Formula Clearing Fractions Applications Equations Reducible to Quadratics Calculator Tips Exercises Sign Analysis Interval Notation Calculator Tips Exercises 0.5 Solving Non-Linear Inequalities Chapter 1 1.1 The Line Functions and their Applications 42 43 45 46 47 49 v Two Dimensional Coordinate System Horizontal and Vertical Lines The Slope Intercept Form Graphing The Point-Slope Equation The Slope Formula Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. vi * ** Contents The General Linear Equation An Economic Application Calculator Tips Exercises 51 53 56 58 60 61 62 64 66 67 68 69 70 70 71 75 79 80 82 83 84 88 90 91 91 94 94 97 99 102 103 104 105 109 111 113 114 115 116 117 119 124 124 126 1.2 Basic Notions of Functions Definition of a Function Functional Notation Difference Quotient Domain and Range Independent and Dependent Variables Vertical Line Test Combining Functions Composition Decomposition Functions of Several Variables Calculator Tips Exercises Break-Even Analysis Depreciation Piecewise Linear Graphs Calculator Tips Exercise 1.3 Applications of Linear Functions 1.4 Quadratic Functions Parabolas Scaling Vertical Translation Axis of a Parabola Horizontal Translation Locating the Vertex Graphing a Parabola in the form y = ax2 + bx + c Applications to Optimization Calculator Tips Exercise Definition of a Circle Equation of a Circle Graphing a Circle Tangent Line The Ellipse Calculator Tips Exercises Supply Function Demand Function Market Equilibrium Revenue, Cost and Profit Functions Marginal Functions Calculator Tips Exercises 1.5 The Circle 1.6 Economic Functions Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Contents * ** vii 1.7 More on Functions Using the Zeros Even Functions Symmetry About the y-axis Odd Functions Symmetry About the Origin Rational Functions Vertical Asymptotes Horizontal Asymptotes Translations Calculator Tips Exercises 127 130 131 132 135 136 144 147 149 151 153 155 156 156 161 162 1.8 Regression Scatter Plot Line of Best Fit Linear Regression Correlation Coefficient Non-Linear Regression Calculator Tips Exercises Chapter Summary Chapter 2 An Introduction to Calculus 169 170 171 171 174 175 176 177 179 180 181 182 184 185 186 188 190 191 193 196 196 198 199 2.1 Slope of a Curve Slope of a Tangent Line The Slope as a Limit Slope of a Curve Equation of a Tangent Line A Place Where No Tangent Exists The Derivative Calculator Tips Exercises Derivative of a Linear Function The Simple Power Rule The Constant Multiplier Rule The Sum Rule Calculator Tips Exercises The Limit Limits by Substitution One Sided Limits Jumps and Holes Continuity Removable Discontinuities Differentiability and Continuity Calculator Tips Exercises 2.2 Derivatives Rules 1 2.3 Limits and Continuity Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. viii * ** Contents 2.4 Limits at Infinity, Infinite Limits and Asymptotes Limits at Infinity Dominant Terms Horizontal asymptotes Infinite limits Vertical asymptotes Calculator Tips Exercises The Product Rule The Quotient Rule Calculator Tips Exercises 202 203 205 207 207 208 209 210 211 214 214 217 218 221 221 222 224 225 227 229 229 232 232 234 235 237 237 239 242 242 243 245 246 246 247 248 249 252 252 253 2.5 Derivative Rules 2 2.6 The Chain Rule The Chain Rule The General Power Rule Calculator Tips Exercises Marginal Functions Average Cost Velocity Average and Instantaneous Rates of Change Calculator Tips Exercises Finding a Tangent Line Finding the Derivative Calculator Tips Exercises Vertical Angles Parallel Lines Similarity Congruence Midpoint Formula Exercises 2.7 Marginal Functions and Rates of Change 2.8 Implicit Differentiation 2.9 Elements of Geometry 2.10 Related Rates A Geometric Example An Ecological Example An Economic Example Using Similarity Exercises Newton s Method Calculator Tips Exercises Chapter Review 2.11 Newton s Method Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Contents * ** ix Chapter 3 Applications of the Derivative 256 257 258 259 261 264 265 266 267 269 274 274 276 277 278 279 281 281 288 289 290 294 295 295 298 299 300 302 303 306 309 310 312 314 315 315 317 319 3.1 Extrema of a Function Continuity Maximum and Minimum Values Extreme Value Theorem Relative Maxima and Minima Critical Numbers and Critical Points Calculator Tips Exercises Increasing and Decreasing Functions The First Derivative Test Sign Diagrams Calculator Tips Exercises The Second Derivative Higher Order Derivatives Velocity and Acceleration Concavity The Second Derivative Test for Concavity The Second Derivative Test for Relative Extrema Implicit Differentiation and Curve Sketching Calculator Tips Exercises Area and Perimeter Optimization Procedure Volume Distance and Velocity Calculator Tips Exercises 3.2 The First Derivative Test 3.3 Concavity and the Second Derivative 3.4 Applications I Geometric Optimization Problems 3.5 Applications II Business and Economic Optimization Problems Price, Demand and Revenue Cost and Average Cost Elasticity of Demand Exercises 3.6 Linearization and Differentials Linearization Differentials The Differential Approximation Differentiable Functions Differential Formulas Exercises Chapter Review Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. x * ** Contents Chapter 4 Exponential and Logarithmic Functions 322 322 323 325 329 331 332 332 335 336 339 340 342 343 343 346 349 350 353 353 354 355 357 359 359 362 363 364 364 367 368 370 371 373 375 376 377 378 380 382 383 384 384 4.1 Inverse Functions One-to-One Function Horizontal Line Test Increasing and Decreasing Functions Inverse Function Composition Property Derivative of the Inverse Calculator Tips Exercises Exponential Expressions The Graph of y = f1x2 = bx Solving Special Exponential Equations Finding the Exponential Function Growth and Decay Rates Power Function Calculator Tips Exercises Continuous Compounding of Interest The Constant e Calculator Tips Exercises The Simple Exponential Rule The Generalized Exponential Rule Exponential Domination Calculator Tips Exercises 4.2 Exponential Functions 4.3 The Number e 4.4 The Derivative of the Exponential Function 4.5 Logarithmic Functions Definition of a Logarithm Base 10 and e pH of a Solution Graphing Logarithmic Functions The Simple Logarithmic Rule The Generalized Logarithmic Rule Calculator Tips Exercises Multiplicative and Division Properties Exponential Property Derivatives Using the Properties Logarithmic Equations Exponential Equations Change of Base Derivatives in Different Bases Logarithmic Differentiation Calculator Tips Exercises 4.6 Properties of Logarithmic Functions Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Contents * ** xi 4.7 Applications of Exponential and Logarithmic Functions Exponential Growth Population Growth Continuous Compounding Radioactive Decay Carbon Dating Logistic Growth Richter Scale Calculator Tips Exercises Chapter Review 387 387 388 388 389 390 392 393 394 395 Chapter 5 Integration and its Applications 398 399 399 401 402 403 404 405 407 410 411 412 413 414 415 418 419 421 421 423 424 425 426 433 433 435 436 436 439 440 444 445 5.1 Antidifferentiation Integration Antiderivative Integration Theorems Simple Power Rule Simple Logarithmic Rule Simple Exponential Rule Calculator Tips Exercises 5.2 Applications of Antidifferentiation Particular Solutions Equations of Motion Marginal Functions Separable Differential Equations Calculator Tips Exercises Reversing the Chain Rule Generalized Power Rule Generalized Logarithmic Rule Generalized Exponential Rule Calculator Tips Exercises Areas by Rectangles Left Endpoints Right Endpoints Midpoints Calculator Tips Exercises 5.3 The Substitution Method 5.4 Approximation of Areas 5.5 Sigma Notation and Areas Sigma Notation Linearity Property Summation Formulas Riemann Sums Areas by Riemann Sums Calculator Tips Exercises Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. xii * ** Contents 5.6 The Definite Integral Definite Integral Fundamental Theorem of Integral Calculus Basic Properties Calculator Tips Exercises Substitution Odd and Even Functions Average Value Derivative of an Integral Calculator Tips Exercises 447 448 450 456 459 462 464 466 468 470 471 472 479 481 483 484 485 486 489 492 494 494 5.7 Substitution and Properties of Definite Integrals 5.8 Applications of the Definite Integral Area Between Curves Consumer and Producer Surplus Continuous Income Flow Probability Calculator Tips Exercises Substitution Integration by Parts Tabular Integration Exercises Chapter Review 5.9 Two Integration Techniques Chapter 6 An Introduction to Functions of Several Variables 6.1 Functions of Several Variables Functions of Several Variables Difference Quotients Three-Dimensional Coordinate System Surfaces Calculator Tips Exercises 498 499 499 500 501 503 504 507 509 511 513 516 518 519 6.2 Partial Derivatives Partial Derivative Visualization of the Partial Derivative Level Curves Contours Cobb-Douglas Production Function Utility Functions and Indifference Curves Higher Order Partial Derivatives Calculator Tips Exercises Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 0 Reviewing the Basics T his chapter reviews some of the basic concepts from elementary and intermediate algebra. Each section begins with a pretest which you may take to see if you remember the material. If you do, then the section may be skipped, if you cannot solve most of the problems in the pretest, then you should review the entire section. Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 2 * ** Section 0.1 Solving Linear Equations 0.1 » » » » » » Solving Linear Equations Addition and Multiplication Properties Linear Equations with Fractions Linear Equations with Decimals Solving for a Particular Variable Applications of Linear Equations Calculator Tips Pretest 0.1 - Time 10 minutes Each question is worth one point. Solve for the unknown: 1. x + 2 = 3 3. 3y - 2 = 7 5. 2z - 5 = 5z - 3 7. 9. 3 5x 2. x + 5 = 2 4. - 2x + 4 = - 10 6. 0.2x + 3.212 - 5x2 = .5 8. 3 4w = 12 = 4 3 - 1 4 = 2 3w + 7 2 2x 3x - 2 10. Solve for y: 2x - 3y = 5 In this section we review the method of solving a linear equation. The objective of this section is to remind you how to solve an equation of the form ax + b = c for x (we assume, of course, that a Z 0). The basic idea of solving an equation is to isolate the unknown on one side of the equation. The solution is then the number on the other side. Usually, two steps are required. First isolate the term containing the unknown, then isolate the unknown itself. This often takes the application of two properties: the first is the addition property which states that the same term may be added (or subtracted) to each side of an equation. For example if A + D = E then A + D - D = E - D or A = E - D. This property may be rephrased as follows: an expression may moved from one side of an equation to the other provided its sign is reversed. Notice we move D from the left side of the equation A + D = E to the right side by changing its sign and writing A = E - D. Sometimes the procedure is called transposition. Once the unknown term is isolated, we use the multiplication property to solve for it; this property states that two sides of an equation may be multiplied by the same (non-zero) expression. For example, if A = B then AC = BC or if AB = D, then if we multiply this equation by 1/B (or equivalently divide by B) we have that A = D/B. We illustrate how these properties are used in the following examples. Example 1 Solve the following equation for x: x + 2 = 5. Solution We isolate the x term by transposing the 2 to the right hand side of the equation and write x = 5 - 2, or x = 3. Addition and Multiplication Properties Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 2 to the right hand side of the equation and write 3x = 7 + 2 or 3x = 9 we next solve for x by multiplying both sides of the equation by the reciprocal of 3.Section 0. Published by Pearson Learning Solutions. which is 3/2. Thus. we need only multiply both sides of the equation by the reciprocal of 2/3. we may multiply every term on each side of the equation by the LCD (lowest common denominator) to clear the fractions. Economics.2 = 7 Solution We isolate the x term by transposing the . and April Allen Materowski. by Warren B. 2 x = 5 3 3 2 3 a b x = a b5 2 3 2 15 x = 2 Method II Recall that whenever an equation contains fraction. . In this example. our solution checks the original equation Example 2 Solve the following equation for x: 3x .2 = 7 7 = 7 which checks our solution. Walter O. we could have equivalently divided each term by 3. 1/3 yielding. This is performed by substituting our result for x wherever it appears in the original problem. ? 3 + 2 = 5 5 = 5 Thus. Inc.1 Solving Linear Equations * ** 3 It is important to check the result. and Finance. Gordon. ? 3132 . In Linear Equations with Fractions Applied Calculus for Business. 11/323x = 11/329 or x = 3 We check our result. instead of multiplying each term on both sides of the equation by 1/3. Note: in the last example. Copyright © 2007 by Pearson Education. Example 3 2 Solve the following equation for x: x = 5 3 Solution Method I Since the x term is already isolated. Wang.2 = 7 ? 9 . w + for w.2y = 4 or 3y = 4 and finally.7 = 2y . Copyright © 2007 by Pearson Education. and Finance. 4 6 2 3 Solution We multiply every term on both sides of the equation by the LCD which is 12 to obtain 2 3 5 1 12 # w + 12 # = 12 # a . by Warren B.9 8w + 2w = 21 10w = 21 w = 21/10 Applied Calculus for Business. we multiply both sides by 1/3 (or equivalently divide both sides by 3) to obtain y = 4/3 Example 5 2 3 1 5 Solve the equation w + = . Solution We need to isolate the y term. Example 4 Solve the equation 5y . Economics. Wang. and April Allen Materowski.7 from the left to the right side of the equation by changing its sign. Inc. Walter O.11 for y. First we transpose the . Published by Pearson Learning Solutions. Sometimes one method may be preferred to another as it results in less work in obtaining the solution. 2 x = 5 3 2 3# x = 3#5 3 2x = 15 1 1 a b 2x = a b 15 2 2 15 x = 2 We leave the checking of the solution in this and the rest of the examples as exercises for you.4 * ** Section 0.1 Solving Linear Equations this example.2w + 30 8w = . As we see from the last example.11 + 7 5y = 2y . . there is often more than one way to solve an equation.2w + 30 .w b + 12 # 3 4 6 2 8w + 9 = .4 We next transpose the 2y term to the left hand side by changing its sign to obtain 5y . Gordon. we need only multiply each term on either side of the equation by the LCD which is 3 and then complete the solution as follows. so we have 5y = 2y . 3 24 . Observe that the smallest decimal involves hundredths (0.2x2 = 100 # 0.6m2 = 9 . isolate the term containing the desired variable. Copyright © 2007 by Pearson Education.5x + 100 # 0. Published by Pearson Learning Solutions.42 + 315 .19 + 16m .28/5. Economics.417 . Gordon.2x2 = 0. Inc. 6m . to obtain x = or x = 8 4 y. We isolate the y term by transposing the x term to the right hand side of the equation to obtain Applied Calculus for Business. and Finance.4y 3 Solving for a Particular Variable Linear Equations with Decimals (b)We next solve the original equation for y.4y Now we divide each side of the equation by 3.12m + 3 = .19 + 16m .28 + 16m .) Sometimes.04 Solution Just as with fractions.5x + 0. but since the original problem involved decimals. so we multiply each term on each side of the equation by 100 to obtain 100 # 0.04 50x + 2013 .1 Solving Linear Equations * ** 5 Example 6 Solve the equation 312m . and April Allen Materowski.12m = . combine like terms. we may do the same thing with decimals.6 (Note that we could have written the answer as . we left the answer in decimal form. and transpose to isolate m and then solve.40x = 4 10x = .4m2 for m. Solution (a) As before we isolate the x term by transposing the y term to obtain 3x = 24 . Solution We first distribute.04).12m = .18m = 9 .2x2 = 4 50x + 60 .12 + 15 .56 10 = . Walter O.56 x = . Wang.213 . . by Warren B.22 .3 .22 + 16m .22 m = 22/28 = 11/14 Example 7 Solve the equation 0.Section 0. where they are cleared by multiplying by the LCD. Example 8 Solve the equation 3x + 4y = 24 for (a) x (b) y.12m .5.16m = . Our tactics are still the same. more than one variable may appear in an equation and we are asked to solve for one of them in terms of the others.213 .28m = . The longer piece is 3 feet longer than twice the shorter piece. so we must have 2127 . What are their present ages? Solution Let Mark s present age = m. Ten years from now. The longer piece = 3 + 2x (Note that is translates into = .6 * ** Section 0. Inc. Example 10 The sum of Susan and Mark s present ages is 27 years.m + 102 respectively. twice Susan s age (then) will exceed Mark s age (then) by 28 years.) The sum of the shorter piece and the longer piece is the total length which is 24 feet (see Figure 1). The next few examples illustrate how we translate an applied problem into a mathematical equation and then solve the equation. Published by Pearson Learning Solutions. and April Allen Materowski. Copyright © 2007 by Pearson Education. therefore.m.x 4 4 It is important that we be able to use mathematics as means by which we may solve applied problems. Solving.m + 102 = 1m + 102 + 28 twice Susan s age is 28 years more that Mark s age Note that twice Susan s age exceeds Marks age by 28. Find the length of the longer piece.x + 6 4 3 24 . At that time.3 3x = 21 x = 7 Thus. Wang.1 Solving Linear Equations 4y = 24 . Gordon. we have Susan s present age = 27 . Application of Linear Equations Example 9 A 24 foot rope is cut into two pieces. . Let us translate the statement the longer piece is 3 feet longer than twice the shorter piece. we need to add 28 to Mark.3x = 6 . more than into + . since the sum of their present ages is 27. Economics. by Warren B. we have e !!!! 24 !!!! e !!!! e ! !!!!! x 3 + 2x Figure 1 x + 13 + 2x2 = 24 3x + 3 = 24 3x = 24 . In 10 years their ages will be 1m + 102 and 127 . twice Susan s age (two times her age) will be 28 years more than Mark. and Finance. Walter O.3x Dividing each side of the equation by 4 gives y = often. so for them to be equal. we have Applied Calculus for Business. the shorter piece is 7 feet long and the longer piece is 3 + 2172 = 17 feet long. Solution Let x = the length of the shorter piece. this last equation is written as 3 y = . .3m = .000 less than twice Bob s salary = 2b . together they earn $100. If the perimeter of the room is to be 50 feet. Inc.Section 0. Economics. and Finance. From the perimeter relationship. so we have Al s salary + Bob s Salary = 100000 or 2b . Walter O.12 = 15 years. w l = 2w + 7 Figure 2: Porch Applied Calculus for Business. Due to the remaining lot size.000. Gordon.2m = m + 38 .36 m = 12 Thus. representing the porch. and April Allen Materowski. Al s salary is $8. Copyright © 2007 by Pearson Education.8000 + b = 100000 or 3b = 108000 or Bob s salary is b = 108000/3 = $36. Mark s present age is 12 years.8000 = 72000 .000 less than twice Bob s salary. we could subtract Bob s salary from $100.000 and Al s salary is 2b . How much does each earn? Solution Let b = Bob s salary. the width of the porch is to be 6 feet. (Alternately.m2 = m + 38 74 . we have that 50 = 2l + 2w = 212w + 72 + 2w or 50 = 4w + 14 + 2w 36 = 6w 6 = w Thus. Published by Pearson Learning Solutions. and the length 2162 + 7 = 19 feet. the length of the proposed porch must be 7 feet more than twice the width.1 Solving Linear Equations * ** 7 2137 . and Susan s is 27 .000. by Warren B.000 to compute Al s salary.000.8000 = $64. Solution We recall that the perimeter of a rectangle of length l and width w is P = 2l + 2w. determine the dimensions of the room. why?) Exercise 12 The Smiths want to add a rectangular porch to their home. Wang.8000 Together they earn 100. Exercise 11 Al earns $8.8000 = 21360002 . We labeled the above rectangle. in Figure 2 with all the pertinent information. ) If we solve this equation for r we have that r = d/t. how long does it take them to be 385 miles apart? Solution Let t be the time they need to travel to be 385 miles apart. Recall that if two fractions are equal. Gordon. . and Finance. I = 500. Walter O. or if solve this equation for t. the relationship between the Principal (P . Published by Pearson Learning Solutions. Using I = Prt. The next example illustrates a simple application of this result.1 Solving Linear Equations Recall that when objects (cars.033333 L 3. planes. the distance he travels in this time is 50t. Depending on what needs to be solved for. If John averages 50 mph and Isabel averages 60 mph. and April Allen Materowski.8 * ** Section 0. we have 500 = 3000r152 or 500 = 15000r or r = 500/15000 = 0. Similarly. that is. we have that t = d/r. bicycles.the money invested). the distance Isabel travels in this time is 60t. Inc.) are moving at a constant rate (speed). Note that the sum of John and Isabel s distances must be 385 miles (see Figure 3). if Applied Calculus for Business. people. rate times time is equal to distance. the distance (d) and the time (t) is given by the formula rt = d (that is. Wang. Example 13 John and Isabel leave the parking lot and travel in opposite directions. in 31*2 hours they will be 385 miles apart. e !!!!!385 miles!!!!! e !!!!! e !!!!!!!! 50t 60t John s distance Isabel s distance Figure 3 When money is invested and earns simple interest. and the time in years the money is invested (t) and the interest earned (I) is I = Prt Example 14 If Carl wishes to earn $500 in interest by investing $3000 for five years.33% We conclude this section with one more application which further reviews the concept of solving an equation using the method of cross multiplication. we use one of these equivalent forms to solve motion problems. the annual interest rate (r). and need to find the annual interest r. by Warren B. the relationship among the rate (r). Copyright © 2007 by Pearson Education. what simple interest rate will be required? Solution We are given P = 3000. Economics. t = 5. Since John s average rate is 50 mph. etc. so we have the equation 50t + 60t = 385 or 110t = 385 t = 385/110 = 31*2 hours Thus. where eqn stands for the equation that is to be solved and variable represents the unknown you are solving for. and April Allen Materowski. and Finance.Section 0. the original fraction was 3 3 = 2132 + 1 7 The calculator may be used to solve most equations for an unknown. Solution Let x be the numerator of the original fraction. If the numerator is doubled and the denominator is increased by two the resulting fraction is 2/3. scrolling down to solve and then press enter. We illustrate in Figure 4. Copyright © 2007 by Pearson Education. Gordon. we have 312x2 = 212x + 32 or 6x = 4x + 6 or 2x = 6 or x = 3 Therefore. For example. solve(eqn. as seen in Figure 5. Wang. by Warren B. Published by Pearson Learning Solutions. it follows that AD = BC Example 15 The denominator of a fraction is one more than twice the numerator. Therefore the original fraction is represented by . You may obtain this command by pressing the Catalog key and then either scrolling down to s or pressing the letter s (the alpha key followed by s). the solve command is used. variable).1 Solving Linear Equations * ** 9 A C = B D Then by multiplying each side of the equation by the common denominator BD. . Calculator Tips Applied Calculus for Business.81/20. Economics. If the numerator is dou2x + 1 2x bled and the denominator is increased by two. we have 2x 2 = 2x + 3 3 cross multiplying. then the denominator is x 1 + 2x. Pressing enter produces the solution x = . then the new fraction is .= 2x + 4 for x. 3 5 Note the syntax. Since this 2x + 3 new fraction is 2/3. Walter O. Inc. suppose we wish to solve the equation 2 7 x . Determine the original fraction. 4w .35 + 0. and April Allen Materowski.5x + 312x .92 = 12x 13.0.3 = w + 5 6. 18.3t2 11.62 = 19 + 417 .4 = 4 y + 10 + 3 19. 9 . x . For x: 5x . 17. .0. 0. 0.24x = 0.10 * ** Section 0. x + 11 = 5 3. Published by Pearson Learning Solutions.5r = 4r + 11 8.2y = 19 23.92 = 7t . 5n .7 = 5w .2p2 10.4r .215 . For x: ax + by = c 25.2x . 0.7 5. 13 . 3 5y = 9 4 5 2z = 3 3 1 4w + 2 = w .6n 12. 15.4 = 3m + 12 7. and Finance. 16.3 = 219n . by Warren B. For C: F = 9/5C + 32 = 8 Applied Calculus for Business. For y: ax + by = c 26. 1. 22.3w2 9. 2 3x 14. Wang.319 . Walter O. 314w . Copyright © 2007 by Pearson Education. 5x = 0 4.3x2 21.1 In Exercise 1 21. solve for the indicated variable.72 + 14 .2r2 = 100 In Exercises 22 29.7 = 9 2. 2 + 315t . 3p + 215p . 7m . solve for the unknown.1 Solving Linear Equations Figure 4: Using the Calculator to Solve an Equation Figure 5 EXERCISE SET 0.2 2 1 1 3 4w .2x2 = 0. Gordon.315 . For y: 5x .2y = 19 24. 2w . Economics. Inc.4 20. and then check your solution.52140 .52 = 719 .3w + 4 = 6w 3 3 1 7 5 y . and April Allen Materowski. The numerator of a fraction is three less than the denominator. If the equal sides are 6 centimeters less than twice the unequal side. If one angle is 15 degrees more than twice the other. 1 . In how many hours will they be 4400 miles apart? 40.1. Walter O. How long before the cars are 60 miles apart? 1 Solving Linear Equations * ** 11 39. find the three integers. 4 3x 1 6x = 20 2 3 = 3 4x + 2 = 3 4 8. One train leaves Boston traveling towards New York at an average speed of 80 mph. James has a collection of nickels and dimes which total to $5. 3 . If the perimeter is 36 feet. If the numerator is increased by 1 and the denominator is decreased by 2 the resulting fraction is 1*2. For S: 1/R = 1/S + 1/W 30.52 . The sum of three consecutive integers is 96. In five years.15. 3x 2 . 6. What annual simple interest does a $10. If one angle is 30 degrees less than six times the other. Mike is 5 years older then Joan. What are their present ages? 32. Economics. one flying north at 500 mph and the other south at 600 mph. 5x .000 investment earn if it accumulates to $11. If he has 5 fewer nickels than dimes. The length of a rectangle is two feet less than three times its width. what are the dimensions of the rectangle? 34. 8. Determine the original fraction. $6. If his grades on the first three exams were 82.1 27.7y = 26 5. Today. what is the length of the two equal sides? 35. . Copyright © 2007 by Pearson Education.x = 2 2. the sum of their ages then will be 41. (a) How long will it take them to meet.115x .42 = 2 10. For h: V = pr2h 28. Find the three integers. (b) How far has each train traveled when they meet? 38. An isosceles triangle has a perimeter of 33 centimeters.Section 0. Boston and New York are approximately 280 miles apart. 88 and 78. The sum of two consecutive odd integers is 15 more than the next odd integer. A nine foot rope is cut into two pieces so that one piece is twice the other. Solve for y: 3x . Gordon. Solve for the unknown: 1. the sum of their ages will be 57. to what does it accumulate? 46. Two angles are said to be complementary if the sum of the measures of their angles is 90 degrees. 5 . Tripling the numerator and quadrupling the denominator results in a fraction equal to 3/5. and Finance. For b: A = *2h1a + b2 29.3z2 = 519 + 2z2 7. Maria s age is 5 years less than twice Louie s age.2 = 8 3. In ten years.412 . what grade did he get on the fourth exam? 33.4y = 9 Applied Calculus for Business. 44. Alphonse took four examinations and his average on these four exams was 83. The denominator of a fraction is 1 less than three times the numerator. 36. Determine the original fraction. what is the measure of the smaller angle? 41. 45.200 after three years? Posttest 0.000 to earn a total interest of $600 if the annual simple interest rate is 4%? 47. Two angles are said to be supplementary if the sum of the measures of their angles is 180 degrees.5x 4. by Warren B.000 is invested for two years earning annual simple interest at 4%.213 + 2x2 . Published by Pearson Learning Solutions. How long is the larger piece? 9. Two planes leave JFK airport at the same time. One travels at 60 mph and the other at 45 mph. 37. Inc. How long does it take $5. Another train leaves New York at the same time traveling toward Boston at an average speed of 60 mph. Wang. what is the measure of the larger angle? 42. Two cars leave a city traveling in the same direction.Time 10 Minutes Each question is worth one point. What are their present ages? 31. 312z . how many dimes has he? 43. Copyright © 2007 by Pearson Education. 2x . 1x . You may not be able to (easily) factor every quadratic equation.3 or x = 3.12 = 0 6. remembering that x could also be negative. 16x2 . we shall examine other techniques for solving quadratic equations which will easily yield their solutions.12 = 0 may be factored as 12x .2 » » » Solving Equations of the Form ax 2 * b + 0 Isolation of Squared Term Isolation of Squared Binomial Term Calculator Tips Pretest 0. so the two solutions to the given quadratic are x = . then either A = 0 or B = 0.321x + 32 = 0 Isolation of Squared Term and we immediately find that x = . therefore. We recall that if AB = 0. Inc.20x = 0 4.b = 0. 31x .522 + 81 = 0 You should recall how to solve quadratic equations by factoring.3213x + 42 = 0. the first equation yields x = 3/2 and the second x = . Consider the equation x2 . and April Allen Materowski. given the equation ax 2 . 3 More generally. 3 2 2 x = 0 4 3 2. 1.Time 10 minutes Solve each of the following quadratic equations for the unknown.49 = 0 5.2 Solving Equations of the Form ax2 . so we have x1 = 29 = 3 or x2 = . or written more succinctly as x = .9 = 0. 4x 2 + 32 = 0 8. We write.2 . . Walter O.4/3. 5x 2 . Economics. Published by Pearson Learning Solutions.4/3 and x = 3/2. also called their roots.522 . so we have the two possibilities.522 . we can write Applied Calculus for Business. for example the equation 6x2 . or more succinctly x = . Gordon. We can also solve this equation by isolating the x 2 term. 9x2 .29 = 3. 4x2 = 5x 3. 12x . 29 = .3 = 0 or 3x + 4 = 0.1213x + 22 = 24 7. This problem is most easily solved by factoring.14 = 1 4 9.18 = 0 10. 3. 1x . Consider the following. by Warren B.x .b = 0 0. Wang. and Finance. 3 12x . x2 = 9 We now take the square root of each side.12 * ** Section 0. Rewrite the equation with the x2 term isolated on one side of the equation. 2 1 12 # x2 . if fractions appear in the quadratic equation. Walter O. Economics.12 # = 12 # 0 3 4 2 8x . A8 A8 2 4 Applied Calculus for Business.25 = 0. the first thing we should do is to clear any fractions. We multiply every term on each side of the equation by the LCD which is 12. Example 1 Solve the quadratic equation 4x 2 . we have x2 = 8 x = .= 0 4 3 Solution We begin by first clearing fractions. first multiply every term on each side of the equation by the LCD. Copyright © 2007 by Pearson Education. Solution Isolating the x2 term. = .b = 0 * ** 13 ax2 = b b x2 = a b x = .Section 0. 2 22 Example 3 2 1 Solve the equation x 2 . = . we would have to simplify the radical expressions. .3 = 0 8x2 = 3 3 x2 = 8 3 3#2 26 x = . Published by Pearson Learning Solutions. the solutions could be complex numbers.8 = 0. Inc. and Finance. Wang. Gordon. That is. A4 5 x = . Aa Of course.2 Solving Equations of the Form ax2 . One general remark: for simplicity in carrying out the algebraic manipulations. we have 4x2 = 25 25 x2 = 4 25 x = . and sometimes. and April Allen Materowski. This results in an equivalent quadratic equation all of whose terms are integers. 2 Example 2 Solve the equation x 2 . 28 x = . Solution Isolating the x2 term. We illustrate the various cases in the examples that follow. by Warren B. = . .522 = 4 We next take the square roots. Economics.b = 0 Example 4 Solve the equation x 2 + 8 = 0. i 24 22 = . Solution Isolating the x2 term. but some expression squared. 2 .. = 4 4 4 2 2 2 Applied Calculus for Business. In such a case.3 . 212 = . Wang. we could have easily solved the problem by factoring. Solution We first isolate the squared term by rewriting the expression as 1z . and April Allen Materowski. However.6 . Copyright © 2007 by Pearson Education.2 Solving Equations of the Form ax2 . and Finance.2 z = 5 .5 = . 14y + 622 = 12 4y + 6 = . Walter O. If we break it into two fractions..14 * ** Section 0. that could not be done so easily on the next example. z . Example 5 Solve the quadratic equation 1z . In the study of calculus we will be mostly concerned with solutions which are real numbers.522 . Gordon. Published by Pearson Learning Solutions. we have y = 6 2 23 3 23 . we isolate the squared expression and then take the square roots. Inc. the radicand was negative resulting in imaginary solutions. 2 23 4y = . 2 23 .10z + 21 = 0.6 .8 = . as illustrated in the next two examples. 2 23 y = 4 We can reduce this expression by either breaking it into two fractions or by factoring. 2 23 = . we have x2 = . by Warren B. i 28 = . 2 z1 = 5 . A slight variation on the above method of solution occurs when the left hand side is not x2.2 or z2 = 5 + 2 z1 = 3 or z2 = 7 We remark that if we actually squared the expression and rewrote the quadratic as z2 .5 = .12 = 0 Solution. Example 6 Solve the quadratic equation 14y + 622 .4 = 0.6 . 24 z .8 x = . 23 . 2 22i Isolation of Squared Binomial Term Note that in the last example. ) The calculator produces the two solutions.2. the object of the next section.634 and y2 L . that is of the form a . .C = 0 may be solved by isolating the squared binomial term and then take the square roots. If we need a numer2 2 ical approximation. by Warren B. and will yield the real solution(s).3 . we must first learn how to complete the square. see Figure 2. Consider the problem Calculator Tips Figure 1: Using solve on a Quadratic Equation (Note that ¿ is used for exponentiation. linear or not. Note that the two solutions to the quadratic equation in the previous examples are conjugates of each other. 2 23 . 23 = = 4 2 4 Thus.2b and a + 2b.3 . Walter O. This command will work on most equations. Applied Calculus for Business. Any quadratic equation written in the form 1x + B22 . For approximate solutions you need only press (the green diamond and then Enter) Enter. Wang. Published by Pearson Learning Solutions. we can use a calculator to find y1 L . Inc. To do so. We now make one very important observation. Gordon. we have the two solutions y1 = . Try solving this last example by multiplying out the expression and factoring. It is precisely this observation that will enable us to solve any quadratic equation. we could factor the expression and write y = 2 A .0. You will quickly see that it is not a simple matter to find the factors since they are irrational numbers. What we shall do is to learn to rewrite any given quadratic equation in the above form. Copyright © 2007 by Pearson Education.3 .Section 0.366. Economics.3 + 23 . and Finance. This will be true in general.6 .23 or y2 = .b = 0 * ** 15 Alternately. You may recall in the last section we introduced the solve command to solve a linear equation and quadratic. as we shall see.2 Solving Equations of the Form ax2 . 23 B . and April Allen Materowski. see Figure 3. . meaning there are no real solutions. Gordon. see Figure 4. Copyright © 2007 by Pearson Education. Walter O. Economics. as in Example 4. Wang. by Warren B. Figure 3: Using solve When There Is No Real Solution The complex solution may be obtained by using csolve in place of solve.2 Solving Equations of the Form ax2 . and April Allen Materowski. and Finance.16 * ** Section 0. Figure 4: Using csolve Applied Calculus for Business. Published by Pearson Learning Solutions. Inc.b = 0 Figure 2: Obtaining Approximate Answers with the Calculator Note that if the solutions are not real. the solve command results in the answer false. 81 = 0 5.2 Solving Equations of the Form ax2 . 1x + 1212x . 55. Gordon. 8x . 16x . 7x2 = 3x 3.522 + 18 = 0 49.Section 0. y + 49 = 0 19.9 43. .4 = 3 1 2 8x + 3 = 0 2 . 56. 14x .24 = 0 41. 4x 2 .24 = 0 10.322 = 49 37.222 + 25 = 0 46. Wang. 54. 16x2 .32 + 24 = 9 3 15 3 2 81x + 72 + 4 = 2 3 15 3 2 8 12r + 72 + 2 = 4 2 1 2 513x .40 = 24 9.18x = 0 4.722 = 18 39.4 = 0 4. 1x . 41x . Published by Pearson Learning Solutions.19 = 1 3 9. 13x .22 = 9 32. y2 . 4 2 3 x = 0 5 10 2. x2 + 25 = 0 16. x2 = 36 3.322 . 16y .45 26. x + 4 = 0 18. 1x + 322 . 2 2 3x 3 2 4y 5 2 6r 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 29. 1x . 57. 5y2 + 12 = 3y2 . Economics.32 = 12 34. y . 13x . z2 . 6x 2 + 120 = 0 8. x + 7 = 16 7.4)2 = 64 38. x = 25 2. and April Allen Materowski. 12r . x2 + 16 = 7 20. y2 .45 = 0 6. by Warren B.24 25. 30.7 = 2x + 20 10.8 = 0 + 24 = 0 2 3 = 3 4 Posttest 0.Time 10 minutes Solve each of the following quadratic equations for the unknown. 14x .b = 0 * ** 17 EXERCISE SET 0. 1x . 5 15x + 422 . Inc. 3x + 24 = 0 23. 20x2 . 1x + 322 = 25 35. 1x + 322 = . 2 2 3 1x + 32 = 12 3 2 2 13y .422 + 16 = 0 44. 1x . and Finance.2 Find the unknown in each of the following exercises.5214x + 52 = .52 + 2 = 3 5 2 1 5 6w .36 17. 2y = 48 15. 17x + 20 = 11x + 74 11. 27. 1.321x + 22 = 14 33.49 = 0 5. 3z . 1y . Walter O.96 = 0 14. 15x + 422 = 20 40.222 + 9 = 41 42.422 + 3 = 39 36. Copyright © 2007 by Pearson Education. x2 = 18 12.522 + 48 = 0 Applied Calculus for Business. 1. 51. 12y . x + 20 = 0 22.18 = 0 4 2 5 15w + 22 + 25 = 0 3 2 412z . 52. 12y .9 7.25 = 39 6. x2 . 53. 14y + 722 + 36 = 0 47. 5x .322 + 24 = 0 48.2 . r + 21 = 5 21. 6y2 + 9 = 4y2 . (3z .422 + 48 = 0 50.4 = 21 8.52 + 3 = 2 1 2 2 5 12z . 6w2 + 96 = 0 24. 28.24 = 0 13. w = .522 + 12 = 0 45. 31.22 . x2 + 5x . x2 .422 . take one-half the coefficient of the x term.25/4 The expression may be rewritten as x 2 + 5x = A x + 1*2152 B 2 . .16 The expression may be rewritten as x2 . and Finance.A 1*2152 B 2 Do you see the pattern in each of the examples? We spell it out more generally. and April Allen Materowski. The question that naturally arises is given any quadratic equation. Walter O. Let us make some observations on several expressions which are identities: x 2 + 6x = 1x + 322 . Leave the answer in simplest form. 3.8x = A x + 1*21 .3 Completing the Square 0.A 1*2162 B 2 x 2 .8x so that it is a perfect square? 2.5 = 0 5.12x . (the sign is part of the coefficient) add it to x.82 B 2 x 2 + 5x = 1x + 5/222 .10x . by Warren B. then subtract the square of one-half the coefficient of the x term from the previous squared sum. 3x 2 .82 B 2 . x 2 .8x = 1x . What number should be added to x 2 + 3x so that it is a perfect square? Complete the square to solve each of the following quadratic equations.3 = 0 4.4 = 0 In the preceding section we observed that any quadratic equation written in the form 1x + B22 . how do we rewrite it so that the unknown appears within a square as in (1)? The procedure by which this is accomplished is called completion of the square.3 . Economics.A 1*21 . square the sum. Bx + a x . 2 2 1 1 Bb * a Bb 2 2 (2) Applied Calculus for Business. 1.3 » » Completing the Square Completion of the Square Calculator Tips Pretest 0. Inc. The sum x 2 + Bx may be written as a perfect square as follows: In words. Wang. What number should be added to x2 .C = 0 (1) Completion of the Square may be solved by isolating the squared term and then take square roots. Copyright © 2007 by Pearson Education. Published by Pearson Learning Solutions. and in addition to the solution of quadratic equations is useful in numerous applications.9 The expression may be rewritten as x2 + 6x = A x + 1*2162 B 2 .18 * ** Section 0. Gordon.Time 10 minutes Each question is worth two points. 522 = 1x . We isolate the square term and solve.5 = .Section 0.522 .10x = 1x .10x + 10 = 0 We isolate the two terms involving x. . Solution The coefficient of the x term is .25.3/222 . one-half this number is . by Warren B. Walter O.3.9/4 It is just one more step to see how rewriting a quadratic may be used to solve any quadratic equation.522 = 15 x .25 = .522 .5.3x = 1x .10x in the perfect square form as given on the right-hand side of (2).1 . so we have x2 .1622 or x 2 + 12x = 1x + 622 .3 Completing the Square * ** 19 Of course. Example 1 Rewrite the expression x2 + 12x in the perfect square form as given on the right-hand side of (2). This identity is useful not only in solving quadratic equations. Published by Pearson Learning Solutions. 215 x = 5 . 215 Applied Calculus for Business.522 . Copyright © 2007 by Pearson Education. and Finance. one-half this number is .36 Example 2 Rewrite the expression x2 . Economics. Inc. The quadratic equation may now be written as 1x .522 . x2 .3/222 = 1x . but in other applications as well. so we have x2 . you may verify that the identity is true by simply multiplying out the right hand side of (2). Solution The coefficient of the x term is .10 The problem is now in the form we studied in the previous section. therefore 1*2 1122 = 6. and April Allen Materowski.10x = .10x = 1x .25 Example 3 Rewrite the expression x2 .10. Gordon.1 . Solution The coefficient of x is 12. as you shall see later in this text as well as in calculus. and comparing it to the left hand side.10 We rewrite the left-hand side of this equation as in Example 2 as x 2 . 1x . so we have x2 + 12x = 1x + 622 . Wang.3/2.3x in the perfect square form as given on the right-hand side of (2).3/222 . Consider the quadratic equation x2 . Economics.4 and next rewrite the left-hand side as x 2 . Solution Multiplying each term on both sides of the equation by the 1/4.422 . Walter O. 2 23 x = 4 .25/4 = 0 rewriting.4 = . Therefore.3 Completing the Square We illustrate how the method of completion of the square is used to solve quadratic equations in the following examples. the coefficient of the x 2 term has been 1.4 1x .49/4 therefore we have. the reciprocal of 4. we have 1x .49/4 = 6 or 1x .7x .= .7/222 = 73/4 7 73 273 x .7/222 .7x = 6 x2 .82 B 2 = 1x . we have x 2 + 3x = 25/4 Applied Calculus for Business. by Warren B.16. = 2 2 2 In all the examples considered so far.1 . 1x .A 1*21 .6 = 0 Solution We rewrite the equation as x2 .8x = A x + 1*21 . Published by Pearson Learning Solutions. 212 = . Example 4 Solve the quadratic equation x2 . 2 23 Example 5 Solve the quadratic equation x2 .7/222 = 49/4 + 6 1x . . We illustrate in the following examples. Inc.20 * ** Section 0. and April Allen Materowski. multiplying each term on both sides of the equation by the reciprocal of the coefficient reduces the problem to an equivalent one with coefficient 1.7/222 .82 B 2 .7x = 1x .25 = 0. Gordon. and Finance.8x = . 273 x = . When this is not the case.8x + 4 = 0 Solution We rewrite the equation as x 2 . gives x2 + 3x . Wang.7/222 = 1x .422 .7/222 . 2 A4 2 7 273 7 .16 = . Example 6 Solve the quadratic equation 4x2 + 12x .422 = 12 x . = . Copyright © 2007 by Pearson Education. 42 B 2 . in the previous example. 234 x = . and Finance. We illustrate these tips along with the screen shots on other examples at the end of the next section.25/2 or 1x .17/2 x . 234i i = 2 2 The use of completion of the square in this section was to solve a quadratic equation.42 B 2 = .13/222 = 25/4 1x + 3/222 . We first multiply each term on each side of the equation by the reciprocal of the coefficient of the x2 term. we would use a calculator to approximate the square roots.4. For example.i = . However. 1*2.4x = .8x + 25 = 0 Solution.Section 0. for example. Example 7 Solve the quadratic equation 2x 2 . we have 1x + 3/222 . you would enter csolve 12x 2 .9/4 = 25/4 1x + 3/222 = 25/4 + 9/4 1x + 3/222 = 34/4 3 234 = . 234/4 = . Copyright © 2007 by Pearson Education. we have A x + 1*21 .A 1*21 .3 Completing the Square * ** 21 completing the square and solving. if we want to solve the quadratic in Example 6. solve 12x 2 . = 2 2 2 x + We remark.222 = . indicating there is no real solution. Inc. Walter O. the roots to the nearest onethousandth are 1. it tells us there is no solution.8x + 25 = 0. x) the calculator will give us the two solutions. we enter solve(4x2 + 12x .4x + 25/2 = 0 or x2 . giving x2 . x2 the calculator responds with false. use the function csolve which will solve for complex (as well as real) solutions. if you try to solve a problem with complex solutions with the solve function. Wang. to solve this problem for the complex solutions.i = . A 34 234 17 17 17 # 2 = . Calculator Tips Applied Calculus for Business. if you entered the last example.i = . you should be warned.i A4 2 2 A2 A2 2 x = 2 .2 = . and April Allen Materowski.8x + 25 = 0. 2 2 3 234 . Published by Pearson Learning Solutions. 234 4 .416 and . Thus.222 = 4 .4 = . we may use the solve function to solve equations having real solutions. that is.3 . Gordon. by Warren B. x2 the calculator then provides the solution obtained above.25 = 0. As we illustrated in the last section.25/2 1x .416.25/2 1x .222 .25/2 completing the square. Instead. Economics. that if approximate answers are needed.. . 9 = 0 5. 1. 12x . 3 .421x + 12 = 6 26. 2x2 + x . x2 . The area of a triangle is 20 square feet. 1 2 2x 3 2 5x Solve each of the quadratics by first completing the square. x . x . 52. x2 + 6x 3. Copyright © 2007 by Pearson Education. by Warren B. Gordon.5x = 5 22. x + 12x 7.C. x2 . 13.6x 4. x2 . 1*2 x2 + 5x + 2 = 0 47.1 = 0 23. x + 5x 11. also give the solutions to the nearest onethousandth. When the roots are irrational.3 . 4x 2 .4x = 8 36.7x 12.8 = 0 Applied Calculus for Business.22 * ** Section 0. 6x2 + 23x = . .10 30. x2 + 6x = 3 18.6x + 23 = 0 32. and April Allen Materowski. 1x .321x + 52 = 20 27. x2 + 10x + 50 = 0 34.48 33. h.16t2 . x2 .3 Complete the square in each of the following by putting in the form 1x + B22 .4 = 0 4. What number should be added to x 2 + 5x so that it is a perfect square? Complete the square to solve each of the following quadratic equations. and Finance.3213x + 52 = . x2 .8x 5. 12x2 . find the dimensions of the plot.3x 9. x2 + 6x + 3 = 0 15. x2 + 5x = 5 21. 1x + 321x + 22 = . x . A rock is thrown down from the ledge of a mountain 200 feet above the ground with an initial velocity of 48 feet per second. x2 . 53.12x = . x2 . 3. x2 .3x + 3 = 0 38. x2 + 3x .3213x + 12 = .5x 10. x2 .4 = 0 20. x2 .4x2 = 8x 50.12x . 6x2 + 5x . Give your answers rounded to the nearest tenth of a meter.8x = 4 19. Economics.10x 6. 2x2 + 5x + 10 = 0 45.20 44.48t + 200. If the height is three feet less than the base.3x .3 4x = .3 2x = 5 6 7 10 49.6 = 0 41. x2 + 5x + 10 = 0 39. A farmer wants to set aside a rectangular plot of land to contain 100 square meters.7x = 12 43. x2 .20 31. find the length of the height and base of the triangle. x2 + 6x + 21 = 0 37. x2 + 3x . how long does it take for the rock to hit the ground? Give your answer to the nearest one-hundredth of a second.Time 10 minutes Each question is worth two points. If the width of the plot is 10 meters less than the length.10 = 0 16. x2 + 2x .12x so that it is a perfect square? 2. Inc. x .16x .4 . 1. where t is the time in seconds. x2 . 9 . 12x . 1x . Walter O.6 = 0 42. x2 + 3x 8. 2x2 + 8x + 1 = 0 40.10 = 0 14. Posttest 0. If the height.3 Completing the Square EXERCISE SET 0. Leave the answer in simplest form.321x + 52 = 12 29. x2 + 4x 2.2x . 1x . Published by Pearson Learning Solutions. x 2 + 2x + 10 = 0 25.5x + 8 = 0 46.3x2 + 5x = 8 + 2x2 51.8x + 36 = 0 35.6x + 3 = 0 17. Wang. What number should be added to x2 . x2 . 48. is given by the equation h = . x2 .1 = 0 24. x + 9x 2 2 2 2 2 2 2 28.12x . 4x2 . obtaining x2 + we rewrite this equation as x2 + we next complete the square and obtain ax + b 2 b 2 c b .280 feet? We saw in the last section that any quadratic equation could be solved by the method of completion of the square. . 3x2 . The vertical height h of a rocket measured in feet at time t measured in seconds is given by the equation h = . To the nearest thousandth of a second. by Warren B. Economics. which means that a Z 0.4ac b = = = a 2a 4a 2 4a 2 4a 2 4a 2 Applied Calculus for Business. We proceed by solving (1) using completion of the square.Time 10 minutes Solve each of the following using the quadratic formula.4 » » » » » The Quadratic Formula and Applications Quadratic Formula Clearing Fractions Applications Equations Reducible to Quadratics Calculator Tips Pretest 0. Gordon.6x + 4 = 0 = 2 3x + 1 6 5. Walter O. squaring and combining fractions.Section 0. and April Allen Materowski. Inc. we have ax + b 2 b2 c b2 4ac b2 . That means we will have a formula for the solutions to any quadratic equation.16t2 + 3200t. Wang.4 The Quadratic Formula and Applications * ** 23 0. 2x2 . and Finance. b and c.1 4. 3 2 4x 2. Suppose we apply this method to the general quadratic equation ax2 + bx + c = 0 (1) What should then happen is that our solution should depend on a. how long does it take the rocket to reach a height of 5.4 . Leave answers in simplest radical form.4x . 4x . Copyright © 2007 by Pearson Education. We multiply every term on each side of (1) by 1/a.a b = a 2a 2a b c x = a a b c x + = 0 a a Transposing. Published by Pearson Learning Solutions.5 = 0 3. We assume that (1) is indeed a quadratic.x 2 = . Each question is worth two points. 1. 5 .6 = 0.24 * ** Section 0. Clear all fractions so the coefficients are all integers. Rewrite the equation so that the coefficient of the squared term is positive. we call one solution x1 and the second solution x2. symbol is a shorthand that tells us first use the + sign to write one solution.5 . Economics. and Step 3 is accomplished by multiplying each term in the equation by the least common denominator. Gordon. We illustrate the use of this formula on the following exercises. we have x + Solving for x.2212x + 32 = 0 yields x = 2/3 and x = .4ac b . Wang. 3. Note that the preceding example could have easily been solved by factoring as 6x 2 + 5x . Copyright © 2007 by Pearson Education. by Warren B. Sometimes. We identify the coefficients. Example 1 Use the quadratic formula to solve the equation 6x2 + 5x = 6. . and April Allen Materowski. 2a 2a b2 . and then use the .5 + 132/12 = 8/12 = 2/3 and x2 = 1 . Given any quadratic equation. some suggestions that will make the use of the formula more convenient. We remind you that there are two solutions.5 .6 = 13x . 2 2a 2a B 4a x + (2) Quadratic Formula The solution given in (2) is the solution to the general quadratic equation and is known as the quadratic formula.41621 . if required: 1.3/2. 21522 .6. Rewrite the equation so that the terms are in descending powers. perform the following steps. Applied Calculus for Business.4 The Quadratic Formula and Applications taking square roots.132/12 = . b = 5 and c = . and Finance. we have as our two solutions (roots). We next substitute into the quadratic formula to obtain x = .4ac = = . x1 = 1 . First. These remarks will be illustrated in the examples that follow. we find x = or writing as a single fraction we have * b_ 2b 2 * 4ac 2a b 2b2 .4ac . Walter O.sign to write the other solution. 13 = = = 2162 12 12 12 therefore. a = 6.5 . 2.1. 2169 . Inc. Published by Pearson Learning Solutions. Solution We first rewrite the equation as 6x2 + 5x . we call the solutions the roots.18/12 = .3/2. 225 + 144 . We remark that Step 2 is easily accomplished by multiplying each term in the equation by . 2b2 . as the . Sometimes.62 .5 . 9. and c = 2. Walter O. 217 = = 2182 16 16 Thus.9x + 2 = 0.922 . b = .2 + 2 27 . We set a = 8.4 . 21 .Section 0. We set a = 2. 2 27 B .1 .4182122 9 .2 . we have .4 .2 . We rewrite this equation as 2z2 .8 = 0. b = 4.305 or x2 = 0. and c = .4 .82 . Copyright © 2007 by Pearson Education.217 9 + 217 and x2 = 16 16 (b) Using 217 L 4. (b) Approximate the roots to the 4 6 3 nearest thousandth Solution (a)We first clear fractions by multiplying each term by the least common denominator 12. b = . Solution We first apply Step 1 and rewrite the equation as 3x 2 + 4x . We have. 216 # 7 .64 9 . .4 The Quadratic Formula and Applications * ** 25 Example 2 Use the quadratic formula to solve the equation 4x = . We now identify a = 3.3x2 + 8.41321 . we would compute them using a calculator to approximate the square root.2 27 and x2 = 3 3 If we needed numerical solutions. we have irrational conjugate expressions as the two roots (solutions) x1 = . and Finance. 2 27 = = = 6 6 6 3 thus. Economics. Inc.8.1231056 we obtain. 2112 = = 2132 6 6 x = We next simplify this expression. x = 2 A .4 and c = 5.820. Substituting into the quadratic equation.92 . 21422 . remembering what we learned about the simplification of radicals.2 . Wang. by Warren B. This gives x = . 216 + 96 .4z + 5 = 0.4 . x1 = 0.4 . Example 3 2 3 1 (a) Solve the quadratic equation x 2 . 4 27 . Gordon. the two roots are x1 = 9 . 281 . 8 2 4 Solution We clear fractions by multiplying each term by the LCD which is 8. gives Clearing Fractions Applied Calculus for Business. Example 4 1 5 1 Solve the quadratic equation z2 + = z. Published by Pearson Learning Solutions. and April Allen Materowski. giving 2z2 + 5 = 4z. Substitution into the quadratic formula. to three decimal places.x + = 0. We obtain 8x2 . Why are there two solutions? When the rocket is going upward. 21 . (a) How long does it take the rocket to reach an altitude of 4700 feet? (b) How long does it take for the rocket to return to the ground? Solution (a) We are asked to find t when h = 4700 feet.903 seconds.25 seconds.25002 yielding t = 0 or t = 2500/16 = 156. . 2 26i = = = 4 4 4 4 z = 2 . there are two complex conjugate roots. and Finance. However. 225002 . Applied Calculus for Business. Wang.16t2 + 2500t.42 . i 224 4 . i 24 26 4 . and April Allen Materowski. z1 = Applications In applications involving quadratic equations.4 # 16 # 4700 32 2500 . When t = 0.4122152 4 . by Warren B.26 * ** Section 0. Inc. Consider the following examples. it reaches a height of 4700 feet in approximately 1. where t is the time from firing in seconds. so we need to solve the equation 0 = . Published by Pearson Learning Solutions. Gordon. (b) When the rocket returns to the ground its altitude h = 0. Copyright © 2007 by Pearson Education. we sometimes find that both solutions make sense and sometimes we find that one of the solutions needs to be rejected as it makes no physical sense. so the time it takes for the rocket to return to the ground is 156.347 seconds after launch it will again be at this position but coming down. this rocket will achieve its maximum altitude and then begin to come down. we find the two solutions are t1 L 154. Example 5 The altitude h of a rocket fired vertically upward is given by the equation h = .t116t . Economics.16t2 + 2500t We rewrite this equation as 16t2 . 26i B 4 .422 .24 = = = 2122 4 4 2 A 2 . 216 . 2 .1 . the rocket is being launched. At approximately 154. 2 2 Thus.4 The Quadratic Formula and Applications z = .2500t + 4700 = 0 Using the quadratic formula.25 seconds.903 seconds.40 4 . 26i 2 2 . we have t = 2500 . 25949200 t = 32 Using a calculator. Walter O.347 seconds or t2 L 1.26i 2 + 26i and z2 = .16t2 + 2500t This equation may be solved by factoring 0 = . Therefore we need to solve the equation 4700 = . Economics. find (to the nearest integer) the value of x which produces a profit of $1. find the length of the other leg of the triangle to the nearest thousandth of an inch. and April Allen Materowski. Exercise 8 Solve for x: (note that x Z .2000. 2160 L = . x + 8 x .8. so we solve . x = . x = 23 or 43. Wang. Copyright © 2007 by Pearson Education. see Figure 1. by Warren B. it must be a positive number. therefore. 12.4 The Quadratic Formula and Applications * ** 27 Exercise 6 One leg of a right triangle is 6 inches and the hypothenuse is 14 inches. as the next two examples illustrate.Section 0. Solution Let x represent the length of the unknown leg.2000 = 0 or 3x 2 . The other leg of the triangle is therefore approximately 12. x = 12 or 54. and Finance. we need to perform some algebraic simplifications on an equation before we recognize it as a quadratic. .200x + 2000 = 0 20 A 5 .200x + 3000 = 0 We may solve this equation using the quadratic formula. We find (verify) x = 10 A 10 . Published by Pearson Learning Solutions. we have x + 6 = 14 isolating the unknown. 3 Sometimes.000. 2. Gordon. we find that x2 = 160. we find x = . Solution (a) We need to solve the equation .649 Since x represents a length. we reject the negative solution. Walter O. (b) A break even situation arises when the profit is zero. 2 2 2 x 6 Figure 1 14 Exercise 7 (a) If the profit P (in dollars) earned by selling x units of some item is given by the equation P = . (b) Find the quantity (to the nearest integer) which produces a break even situation 1Profit = 02. or to the nearest integer. Inc. Using Pythagoras theorem.649 inches.3x 2 + 200x .2000 = 1000 or equivalently 3x2 .2 Equations Reducible to Quadratics Applied Calculus for Business. 210 B Solving. Why?) 24 20 + = 12.3x2 + 200x . 210 B 3 Or to the nearest integer.3x2 + 200x . as illustrated by the next example.4213x + 192 = 0 Therefore x1 = . Since u = x8. Sometimes.. Wang. when u = . and Finance. 3 6 3 3 Calculator Tips The solve command will work well on all the problems considered in this section. Exercise 9 3 3 Solve the equation 8x 4 . we have x = 6561/256. 3 8 3 8 8 However. Consider the last example.22. we need to find x. Copyright © 2007 by Pearson Education.) appears on the input line to indicate you need to scroll to see the rest of the equation.27 = 0 which is a quadratic in the variable u and factors as 18u .22 + 201x + 82 = 121x + 821x . we keep the base and multiply the exponents. Published by Pearson Learning Solutions. Thus.22 Multiplying out and collecting similar terms. a trinomial may be at first glance appear not to be a quadratic equation. 8x 3/4 .304 = 0 or 413x 2 + 7x .1. we have x = 127/828/3 = 1127/821/328 = 13/228 = 6561/256.e.128/3 = 1. Economics. by Warren B.19u . we enter the equation into the calculator as illustrated in Figure 1 and press enter.762 = 41x . so the ellipsis symbol (. but by a simple substitution it may be transformed into a trinomial which is a quadratic.. i.27 = 0.2721u + 12 = 0 Thus u = 27/8 or u = . x = 1 does not check (it is extraneous that is.27 = 0.28 * ** Section 0. Applied Calculus for Business.) We may rewrite the original equation as 8u2 .4 The Quadratic Formula and Applications Solution We multiply each term on both sides of the equation by the LCD which is 1x + 821x . Walter O. u = x8 then u2 = A x 8 B 2 = x 8 = x 4 (Recall that when raising a power to a power.19x 3/8 . Solution The key to this problem and problems similar to it is that the exponent of the highest power term in the trinomial is twice the exponential of the next highest power term.1. We leave it as an exercise for you to check the solutions. yielding 241x . .. and April Allen Materowski. thus. You ll note that the equation is too big for the screen. the black arrow appears on the screen to indicate the entire image is not there and you need to scroll to see the remainder of the image. Inc.19x 8 . Similarly. we obtain 12x 2 + 28x . Therefore. we have x = 1 . however.19/3 and x2 = 4. we have 1u23 = A x 8 B 3 or x = u3. if we let the smallest power be called u. Gordon. it does not check the original equation). when u = 27/8. x2 + 6x + 21 = 0 39.Section 0.4 The Quadratic Formula and Applications * ** 29 Figure 1: Solving 8x4 . x2 .19x8 . 1 2 2x In each of the following. and c. x2 .8x . x2 .421x + 12 = 6 28. x2 . x2 .4215x .12 = 0 4. and then determine the coefficients a. with a 7 0. x2 + 5x + 10 = 0 41. 4x2 = 9x 12. x2 + 6x = 3 21.321x + 52 = 12 31.4 30. Do not solve the equation. 2x2 = 10 9.10 = 0 19. 4x = 9x2 + 4 7.4x = 8 38. 12x .3 4x = 5 6 Applied Calculus for Business.2213x + 42 = 5 13. x2 .3x . and April Allen Materowski. 3x2 + 7x .1 = 0 26. Write the solutions in its simplest form.6 = 0 43.48 35. x2 + 6x + 3 = 0 18. solve the given quadratic equation exactly using the quadratic formula.10 = 0 5.8x = 4 . 1x . 1.20 33. 12x2 . Gordon.4 In exercises 1 15 rewrite the equation in the form ax2 + bx + c = 0. by Warren B. 1x .20 46.7212x .8x + 36 = 0 37. x2 + 10x + 50 = 0 36. x2 . x + 4x . 3 .27x + 10 .321x + 52 = 20 29. 1x . 12x .32 = 3x12x .4 = 0 23. 2x2 + x .27 = 0 with the Calculator 3 3 EXERCISE SET 0. 5x2 . 1x . 6x2 + 5x .11 8. x2 .6x + 23 = 0 34.2x = 8x + 5x2 .8 = 0 3. Inc. 1*2 x2 + 5x + 2 = 0 49. and Finance.10 32. x2 . x2 + 2x + 10 = 0 27. 3x = 7x 2 11. 413x . x2 . determine all irrational solutions to the nearest thousandth. x2 + 5x = 5 24. Walter O. 5x2 = . Wang. x2 .5x + 8 = 0 48. b. 6x2 + 23x = .11 10. 4x2 .5x = 5 25. 2x2 + 5x + 10 = 0 47.5214x + 32 = 7 14. 7x + 3x2 .3213x + 52 = .42 2 22.11 = 0 2. . Using a calculator.3213x + 12 = . 16. 1x .7x = 12 45. x2 + 2x . Published by Pearson Learning Solutions.12x .6x + 3 = 0 20.12x = . . 1x + 321x + 22 = .10 = 0 17.2x .3x + 3 = 0 40. Copyright © 2007 by Pearson Education.5x2 = 0 6. 2x2 + 8x + 1 = 0 42.62 = 25 15.6 = 0 44. Economics. 12x . 30 50. 18 9 12 + = x . 6x6 . h is given by the equation h = . Wang. how long. The hypothenuse and one leg of a right triangle are 18 and 12 inches respectively. is the hypothenuse? 56.15000.2x 2 = . Leave answers in simplest radical form. If the rectangular rug she purchases has an area of 216 square feet and is placed an equal distance from each wall (a) how wide is the uniform strip of uncovered flooring? (b) what are the dimensions of the rug? Solve each of the following equations for the real values of x. what was his speed each way? 63. they can build a computer in 2 hours and 24 minutes.5x + 3 = 0 = 1 10 x + 1 2 5. Given the quadratic equation ax2 + bx + c = 0 and cx2 + bx + a = 0. 3 . The sum of the squares of three consecutive odd integers is 515.16t2 + 80t + 75. 67.4ac. If the legs of an isosceles right triangle are each 12 inches. measured in seconds. 4x 2 .2 2x + 2 69. 68. how wide is the strip? 60. by Warren B. Inc. to the nearest one-thousandth of an inch is the other leg? 57. If a rectangular strip of grass of uniform width is to go around the pool.36 = 0 70. is given by the equation s = . A ball is thrown vertically upward from a ledge of a building 200 feet above the ground. Now factor the left-hand-side of this equation and complete the proof. If the car moving north traveled 12 miles more than the one going east. (a) How many items should be produced to break even? (b) How many items should be produced to maximize profit? (c) What is the maximum profit? 65.17x3 + 12 = 0 71. 1. (a) how long does it take the ball to reach a height of 90 feet? (b) How long before the ball is back to its original position (at 75 feet)? (c) How long before the ball hits the ground? Give each answer to the nearest one-thousandth of a second. where t is measured in seconds.35x 3 + 216 = 0 72. Published by Pearson Learning Solutions.3x . x4 + 5x2 . and the area of this strip is 624 square feet. 66.16t2 + 86t + 200. measured in feet.4 . how many items should be produced to (a) yield a profit of $2. If the combined time for both trips was 22 hours.3x2 + 5x = 8 + 2x2 53. .4ac and add b2 to each side giving 4a2x2 + 4abx + b2 = b2 . The profit in dollars in producing x-items of some commodity is given by the equation P = . 2 2 5x 2. 4x2 . Working together. Walter O. * ** Section 0. How long does it take the ball to hit the ground? Applied Calculus for Business. Some time later they are 125 miles apart. 54. On his return his average speed is 2 miles per hour more than when going. find the integers.4 4. how many far (to the nearest mile) did each car travel? 61. x3 . Rewrite the equation as 4a2x 2 + 4abx = . Its height above the ground s. and April Allen Materowski. A ball is thrown vertically upward from the ledge of a building 75 feet above ground. 55.20x2 + 1300x . 9 .2x2 + 400x . to the nearest one-thousandth of an inch. How long before the object is (a) 100 feet above the ground? (b) 50 feet above the ground? (c) on the ground? Give each answer to the nearest one-thousandth of a second. Copyright © 2007 by Pearson Education. 74.3x + = 2 3x .4 7 10 The Quadratic Formula and Applications 62. The ball s height h in feet above the ground at time t in seconds is given by the equation h = .2 = 0 3. Economics. 2 1 Posttest 0. Each question is worth two points. To the nearest integer. An object is dropped from a helicopter hovering at 250 feet above the ground.4 x + 4 2 18 8 = 1 2x + 3 x + 5 4x 16 . in terms of its time of flight t. Mary can build a computer in two hours less time than Tim. 59.Time 10 minutes Solve each of the following using the quadratic formula. The objects height in feet.3 2x = 2 51.000? (b) break even? 3 2 5x . Gordon. John bikes a distance of 120 miles and then returns over the same route. and multiply each term of the equation by 4a.3 = 0 73. Here is another proof of the quadratic formula. How long does it take Mary alone to build a computer? 64. and Finance.16t2 + 250.4x = 8x 52. one goes north and the other goes east. 2x4 . Barbara wants to purchase an area rug for her dining room whose dimensions are 20 feet by 24 feet. Begin with ax2 + bx + c = 0. The sum of the squares of three consecutive integers is 110. prove that the roots of one equation are the reciprocals of the roots of the other equation. A rectangular swimming pool is 30 feet by 40 feet. This gives 4a2x 2 + 4abx + 4ac = 0. Two cars leave an intersection at the same time.15000. how long. The profit in dollars in producing x-items of some commodity is given by the equation P = . x .4x2 . 58. find the integers. 2x 1. Thus.5 2x 31x 2 . . and April Allen Materowski. we need only choose any test value for x in each of these regions and examine the sign of 3x .2 = 3112 .2 = + Thus. This inequality is completely solved.22 The only time the term 3x . Published by Pearson Learning Solutions. Therefore. Therefore. Wang. Let us now re-examine the same problem and suggest another approach which will work on any inequality which may be factored and is to be less than or greater than zero. to solve the inequality 3x .2 at the test value. we wish to find all values of x for which this inequality is a true statement. We first solve 3x .2 7 0. x . for any x-value greater than 2/3. Sign of 3x-2 I 0 II !!!!!! e !!!!!! value of x 2/3 x Figure 1: Examining the Sign of 13x . Consider Table 1 where we examine the sign at a test value in each region. We recall that the rules for manipulating an inequality are the same as those for solving an equation with one exception. the inequality is true. we could write 3x 7 2 or x 7 2/3. Walter O.22 3102 .2 = 0.3 4.22 Region I II Test Value 0 1 sign of 13x .5 » » » Solving Non-Linear Inequalities Sign Analysis Interval Notation Calculator Tips Pretest 0. by Warren B.16 0 Given the linear inequality. 7 2 0 3x + 4 x2 . in each of the Regions I or II the sign 3x .2 7 0. 3x2 + 10x Ú 8 2. 5. Applied Calculus for Business.2 must always be the same. Inc.5 Solving Non-Linear Inequalities * ** 31 0. Gordon. For any other value of x it is either positive (greater than 0) or negative (less than 0). and Finance.Section 0. that is. we see that in region II the expression 13x . Copyright © 2007 by Pearson Education.2 is zero is when x = 2/3. We draw the number line and indicate this value on the line as indicated in Figure 1.Time 15 minutes Each question is worth two points. This is of course yields x = 2/3.22 is positive therefore our solution is x 7 2/3.12x2 + 18x 6 0 3. Solve the given inequality 3 . 3x . namely when an inequality is multiplied or divided by a negative number the inequality symbol reverses. Table 1: Sign of 13x . 2x3 .3x + 22 2x . Economics.5 . 5 Solving Non-Linear Inequalities We now generalize the second method and consider the following non-linear inequality in factored form 1x . and Finance.10 6 0. when x = . this is shorthand for the interval . only positive or only negative in each of the three indicated regions.3x . [a. Table 2: Sign of 1x . This is done in Figure 3. and April Allen Materowski. Gordon. by Warren B. Published by Pearson Learning Solutions. All that we Interval Notation Applied Calculus for Business.2. That means for any other value of x the product of these factors must always be positive or negative. b] is shorthand for the interval a x b. indicating the sign of the product in each of the three regions. that is.2 6 x 6 5 and Region III to x 7 5. .2 6 x 6 5.2 or x = 5. Region II corresponds to . Copyright © 2007 by Pearson Education. we put a zero in the diagram.821 . 52. sign of (x .2. Walter O. + 0 -2 0 5 + x Figure 3: Sign Diagram for 1x . or some other equivalent formulation could have been given.2 6 x 6 5.521x + 22 1 . Region I corresponds to the region in which x 6 .42132 = 112182 = + We can now redraw the sign diagram in Figure 2. This is important as we shall see. (How would you indicate a in interval x 6 b notation?) Notice that the given inequality could have been written as x 2 . because in that region the product is negative. We do this in Table 2. find those values of x for which the product of the two factors is negative. Since the product is only zero at the indicated two x-values. More generally. that is. b) is shorthand for the interval a 6 x 6 b.521x + 22 Note that at x = . That is to indicate that the product of the two factors is zero at each of those x values.12 = + 1 . b] is shorthand for a 6 x Note that the bracket is used to include the endpoint and b. Sometimes the answer is written as 1 . we included letters for the three regions they created. Observe that the only time the product on the left-hand-side is zero is when either factor is zero. that is.3x 6 10.521x + 22 6 0 is any x in the interval .521x + 22 This tells us that the solution to the inequality 1x . the parenthesis is used to exclude it. Economics. We first observe that the right hand side of the inequality is the number zero.5)(x + 2) value of x I 0 -2 II 0 5 III x Sign Analysis Figure 2: Examining the Sign of 1x . Wang. A simple way to determine this is as follows: draw the number line and indicate these two values for x on the number line as indicated in Figure 2.2 and x = 5. Inc.521x + 22 6 0 Our problem is to find those value of x which satisfy the inequality. it will be of the same sign. Moreover. All we need to do is to select any x-value in each of these regions and determine the sign of the product.521x + 22 Region I II III Test value x = -3 x = 1 x = 6 Sign of Product 1x . (a. or 2 x .32 * ** Section 0. and (a. Section 0. we rewrite the inequality so that zero is on one side. We have 2x3 .22 or finally. or x . Draw a number line indicating those x-values (in increasing order) found in Step 2. The above suggests a general procedure for the solution of an inequality. Example 1 Determine the values of x satisfying the inequality (a) 2x 3 . Wang. Rewrite the inequality so that the expression involving x is on one side of the inequality symbol and the number 0 is on the other side of the inequality symbol. Read the solution from the sign diagram.5 Solving Non-Linear Inequalities * ** 33 need do is rewrite the expression so that one side of the inequality is the number zero. Solve for those x-value at which both the numerator = 0 and the denominator = 0. Factor the left hand side. Applied Calculus for Business. Published by Pearson Learning Solutions. 3. Economics.8x 2 Ú . Choose any x-value in each of the regions separated by the values found in Step 2. Solving an Inequality 1. I 0 0 II 0 2 III x Figure 4: Examining the Sign of 2x1x2 . . Solution (a) First. 5. to obtain 2x1x2 . Walter O.4x + 42 or 2x1x . Copyright © 2007 by Pearson Education.4x + 42 We choose values in each of the regions and determine the sign of the expression at these test values.8x 2 + 8x 0 . by Warren B. 2. and Finance.8x. They are indicated in Table 3. and then proceed as above.221x . Compute the sign of the expression at each of these values and enter them on the sign diagram.2 = 0. Inc. Put the number 0 in the sign diagram above each of the zeros of the numerator and ND for each zero of the denominator. yielding x = 0 or x = 2.8x 2 (b) 2x3 .222 0 0 0 The left hand side will be zero when 2x = 0. 4. Gordon. 2x1x .8x We want to determine when the expression on the left will be negative or zero. and April Allen Materowski. We illustrate the procedure on the following examples. We indicate the number line in Figure 4. .121 . because then we would be dividing by zero.322 = 21121 .34 * ** Section 0. as [0.3. q 2.q . (Note that when x = 0 or 2. Inc. 02 or x = 2.5 Find those values of x so that 0 x + 3 Solution The problem reduces to determining when the quotient on the left of the inequality is negative or zero. the expression is zero. Economics.122 = + 21321122 = + We now can complete the sign diagram as given in Figure 5. See Figure 6. Be careful. and when they are different it will be negative) we proceed in almost the identical way as above. Since the rules of signs for quotients is the same as for products. I ND -3 II 0 5/2 III x x . Using the interval notation we may write the solution as 1 . and Finance. (The symbol . The numerator will be zero when x = 5/2 (verify!).3. Copyright © 2007 by Pearson Education.5 Solving Non-Linear Inequalities Table 3: Sign of 2x1x .) Example 2 2x . Walter O. Published by Pearson Learning Solutions. Its solution is read immediately from Figure 4. (when the factors have the same sign the quotient will be positive.8x 2 + 8x Ú 0. They are indicated in Table 4. and the denominator will be zero when x = . Applied Calculus for Business. We allow x = 2 because the inequality is 2x3 .222 21 . ) (b) this reduces to solving the inequality 2x 3 .222 The expression will be negative or zero when x 0 or x = 2.222 Region I II III Test value x = -1 x = 1 x = 3 Sign of Product 2x1x . elsewhere in the interval it is positive.5 Figure 6: Examining the Sign of 2 x + 3 We next choose and test points in each of the regions.3. by Warren B. We indicate this on the number line by writing ND (not defined) above . namely the interval x Ú 0.8x 2 + 8x 0. and April Allen Materowski.q is read minus infinity. 0 + 0 2 0 + x Figure 5: Sign Diagram for 2x1x . or in interval notation. The factors that determine the sign are the numerator and denominator. allowing the expression to equal to zero. Gordon. Wang. x is not allowed to be . .3 6 x 5/2 or in interval notation. We have .13 Sign of x 2x + 5 x = -3 x = 0 x = 14 -/. Inc.5 Figure 7: Sign Diagram for 2x x + 3 We may now read our solution directly from the sign diagram.8 . and Finance.8 2x + 5 x .8 Solution We first rewrite the inequality so that one side is zero.13 Figure 8: Examining the Sign x 2x + 5 To determine the sign in each region.) We now can complete the sign diagram as given in Figure 7. . 2x + 5 We combine into one fraction to obtain 3x . by Warren B. The numerator is zero when x = 13. Walter O. 5/2].13 = = 7 0 2x + 5 2x + 5 2x + 5 2x + 5 The method of solution is now similar to the previous example. namely . Note that x Z . Gordon.5 Table 4: Sign of 2xx + 3 Solving Non-Linear Inequalities * ** 35 Region I II III Test value .5 .8 Solve the inequality 7 1. Economics.3. Copyright © 2007 by Pearson Education. + ND -3 0 5/2 + x .3. 2x + 5 3x .5/3 = . I ND -5/2 II 0 III 13 x .Section 0.= + .12x + 52 3x .13 Table 5: Sign of x 2x + 5 Region I II III Test value . 1 . Example 3 3x .13/ ./ + = 3/7 = + / + = + (Notice that all we really need is the sign not the actual value of the expression at the test values.= + -/+ = +/+ = + Applied Calculus for Business. Wang. and April Allen Materowski.1 7 0.1 = . Published by Pearson Learning Solutions.5/2. and is zero at 5/2./ . but x = 5/2 since the expression is not defined at . and the denominator is zero (the fraction is not defined) at x = .3. we test points as indicated in Table 5. We begin the sign analysis with Figure 8.5 Sign of 2xx + 3 x = -4 x = 0 x = 4 . 32 2 We need seven test points to determine the sign in each region. Gordon. We remark that some texts use the set theoretic symbol ´ instead of writing the word or.5/22 or 113. by Warren B. Walter O.92 = 4x 51x + 221x . and even number means it is positive.3 or 3.4. in place of 1 . if an inequality has no solution we indicate this by writing Ø. we may write the solution as 1 . and Finance. Also.22 4x 4x1x 2 .32 4x51x + 221x . q 2. Economics. Be careful when working with exponents.22 1x + 421x + 3221x . + ND -5/2 0 + 13 x . Thus. 1 . . 4x41x 3 . In determining the sign of the expression. we have as our completed sign diagram.13 This expression x 2x + 5 is positive when x 6 . an odd number means the expression is negative. All we need do is count the total number of negative signs.4x2 1x2 + 7x + 1221x 2 . and April Allen Materowski. or 2.q .5/22 or 113.36 * ** Section 0.22 1x + 421x + 32 1x . Solution We must first factor the given expression. q 2. q 2 they would write 1 . the procedure for finding the sign of an expression generalizes to any number of factors. which represents the empty set.5/22 ´ 113. . . Wang.321x + 32 1x + 421x + 3221x .q .42 = 1x + 321x + 421x .2. Published by Pearson Learning Solutions.q 6 x 6 . Our next example illustrates that as long as an expression is written in factored form.92 Ú 0. . Using interval notation.25 counts as five Applied Calculus for Business. For example.5 Solving Non-Linear Inequalities Thus. . and the denominator will be zero (the expression will not be defined) when x = .q . Inc. Example 4 Solve the inequality 4x41x3 .4x2 1x2 + 7x + 1221x 2 . .13 Figure 9: Sign Diagram for x 2x + 5 . rather than actually evaluating it for each test value. We indicate these point on the number line in Figure 10. Copyright © 2007 by Pearson Education.5/2 or x 7 13 (which may also be written as . I ND II ND III 0 IV 0 V 0 VI ND VII -4 -3 -2 0 2 3 x Figure 10: Examining the Sign of 4x51x + 221x . we must solve the inequality Ú 0 The numerator will be zero when x = 0.32 Thus. we will instead indicate the sign of each factor for each test value.5/2 or 13 6 x 6 q . the expression that needed to be analyzed was easily factorable. we can sometimes perform the analysis needed to determine the sign of a given expression.2/1 + 21 + 221 .32 or 1 . and x2 = . Economics.3.32 2 in each of the seven regions.22 1x + 421x + 32 1x .32 2 -5 .221 . Solution The zeros of the quadratic x 2 + 2x .732. Table 6 summarizes the sign of the expression 4x51x + 221x .2 = 41 + 251 + 21 + 2/1 + 21 + 221 + 2 = + The information obtained from Table 5 is now placed into our sign diagram in Figure 11.2 or 0 x 2 or x 7 3.2. .3 or -3 6 x .2 = 41 .2/1 + 21 + 221 .5 -1 1 2.3.2 = + 41 + 251 + 21 + 2/1 + 21 + 221 .2/1 . We begin our sign diagram in Figure 12.2 = + 41 . and April Allen Materowski. Wang. and Finance.22 counts as two.5 4 41 .251 .4 6 x 6 .21 .2 are found by the quadratic formula to be x1 = . we can write it as 1 .23 L . We now proceed in the usual way. Example 5 Solve the inequality x2 + 2x . Inc.2] or [0.4.21 .32 We see that the expression will be zero or positive when .2 Applied Calculus for Business.21 .32 2 Test value Sign of 4x 51x + 221x .2 6 0.251 .2 = 41 + 251 + 21 .22 1x + 421x + 32 1x .732 x2 * 0.251 + 21 . I 0 II 0 III x x1-*2.1 + 23 L 0. Gordon.5 Solving Non-Linear Inequalities * ** 37 negative signs.22 1x + 421x + 32 1x .221 . by Warren B. q 2.Section 0. Published by Pearson Learning Solutions.2/1 + 21 + 221 .1 . Table 6: Sign of Region I II III IV V VI VII 4x 51x + 221x .22 1x + 421x + 32 1x . 1 . 2] or 13.2/1 + 21 .21 .5 .251 .732 Figure 12: Examining the Sign of x2 + 2x .2 = + 41 . Copyright © 2007 by Pearson Education. .2. Even when this is not the case. Walter O. ND + ND -4 -3 0 -2 0 0 + 0 0 2 2 ND 3 + x Figure 11: Sign of 4x51x + 221x . In each of the examples we considered so far. The next example illustrates. .732. Using the interval notation. Walter O. That means there is no real number at which the quadratic is zero. The next example illustrates this. the empty set.2 Applied Calculus for Business. Table 7: Sign of x 2 + 2x .2 = + We now complete the sign diagram as given in Figure 13.5 Solving Non-Linear Inequalities We next test a point in each region. The calculator cannot directly solve an inequality. the same analysis tell us that there is no solution.q 6 x 6 q . We see that for this value of x the quadratic is positive.1 . Suppose we have a quadratic whose zeros are complex numbers. and April Allen Materowski.732 Figure 13: Sign of x2 + 2x .2 at the test points x = . therefore the sign of the quadratic is always positive or always negative. Solution Since x2 + 9 has no real zeros (verify!). therefore it is everywhere positive. Note for the inequality x2 + 9 6 0. Consider the determination of the sign of the expression x2 + 2x .3. and Finance. We choose x = 0. Published by Pearson Learning Solutions.2 x = -3 x = 0 x = 1 9 .2 Therefore. or the solution is ¤.6 . Inc.23 6 x 6 . . we see that x2 + 2x . Gordon.2 = + -2 = 1 + 2 .38 * ** Section 0. by Warren B.1 + 23.2 Region I II III Test value Sign of x 2 + 2x . we need only determine its sign at any convenient x-value. 0 Calculator Tips Figure 14: Entering y11x2 = x2 + 2x . so we have as our solution the interval . We need only test any particular x-value to determine the sign of the quadratic. Example 6 Solve the inequality x2 + 9 7 0. in particular the determination of the sign at the test points. + 0 x1-*2.732 0 + x x2 * 0. Copyright © 2007 by Pearson Education. it can be used to perform some of the analysis for us. Economics. Wang. as indicated in Table 7.2 6 0 when . 2212x + 32 6 28 36.25 Ú 0 26. 53. 10x2 + 7x .5215x + 42 6 .42 1x . by Warren B. 12x + 3213x .12 2 2 3 x x .3 12x .2x + 5 6 0 66. which means in Region II the expression is negative.x .x3 3 3 2 54. 13x .3 7 0 2.2 examined in Example 5 is now in the calculator s memory and it is named y1(x).52 Ú 0 12. 55. 40. 47.7212x + 5 5 6 2 3 0 0 7 0 923 6 1x .5 Solving Non-Linear Inequalities * ** 39 and 1. which means in this Region III the expression is positive.5213x .5215x + 42 Ú . 1.321x + 42 Ú 0 4.221x + 32 13x + 42 12x . 43. 46.72 7 0 12x .32 3 6 2 Ú 2 Ú 4 6 4 0 48. x12x .3 1x .221x + 52 Ú 0 17.42 12x .321x + 52 6 0 9.25 20 11 7 0 7 4 3x 21x .x . 13x .1 6 2 1x .Section 0.42 1x .4x212 . 16.32 15. .2.221x + 32 0 0 4 37.25 4 7 1 1 20 11 0 22. 25x .2x21x + 42 7 0 10. x2 .3 7x x + 3 7x x + 3 3 2 2 2 0 0 7 0 0 0 41.12 6 0 25.5213x . Walter O.321x + 42 6 0 5.1 Ú 0 63. 13x . 10x + 7x . solve the given inequality.42 7 0 21.3 x 2 . 14x . Inc. The expression x 2 + 2x . x2 . x2 + 7x + 12 Ú 0 19. 6x . 52.25 6 0 27.22 2 Ú 0 6 0 x12x . 1x + 221x . To compute the value of the expression at 1.2. see Figure 14.42 10 x . 1x + 121x . 25x . Economics. Copyright © 2007 by Pearson Education. 2x .7x + 10 6 0 20. 15x .x 7 0 31. 1x .72 7. x3 . and y1102 = .41 7 0 64. x 2 + 4 7 5 33. x2 . we type in y1(1) and the calculator gives 1.3 10 x .9 39.32 1x .12 x2 . x2 . 13x + 5214x . Wang. 12x + 321x . and April Allen Materowski. 24x2 + 10x . x2 . x .16 2 x .2x . 3x21x . 44. 51.2x . 6. 13x + 5214x .3x2 Ú 0 11. 14x .42 x12x .321x + 52 Ú 0 8. 6x2 . and for y1(x) we enter x 2 + 2x . x + x .221x + 32 13x + 42 12x .8 7 0 61. 42.521x + 12 18.9 32. x2 .4x . 56.52 6 0 13. similarly y11 . x2 .4x .16x Ú 0 28. x2 . Gordon. 13 .2212x + 32 Ú 28 60.3 0 3.12 7 x .32 Ú 11 34. 12x + 321x .121x + 22 x .12 Ú 0 24. and Finance. 1x + 2221x . 2x .16x 6 0 29.5 In each of the following exercises.12 7 0 23.22 1x .32 6 11 35. x2 .121x + 22 x .12 x . EXERCISE SET 0.1 Ú 2 x 3 x .32 = 1 which means in Region I the expression is also positive. 58. Published by Pearson Learning Solutions. 49. 50.8 0 30. x12x . x2 . x1x . We then return to the home screen by pressing the HOME key.2x + 5 Ú 0 0 Applied Calculus for Business.32 Ú 0 14. 15x .7212x + 92 1x . 1x .3 1x + 221x .9 38.22 12x + 12 5x 214x .16 2 x . 45.1 6 0 62. 57. 15 .41 65.22 12x + 12 0 7 59. 24x2 + 10x . We proceed as follows: we go to the Y = screen by pressing * F1.12 5x 14x . Walter O.5 8x + 12 0 Applied Calculus for Business. Economics. 2 . Published by Pearson Learning Solutions. 24x4 . 10x2 + 11x Ú 6 4.9x2 0 3. . and April Allen Materowski.7x 7 -1 4 + 5x 5.38x 2 + 20x 25 . Solve the given inequality 1.Time 15 minutes Each question is worth two points. by Warren B. Inc. 2x . Copyright © 2007 by Pearson Education. Wang.5 Solving Non-Linear Inequalities POSTTEST 0. 2.5 .42x3 + 18x 2 6 0 12x 3 . Gordon. and Finance.40 * ** Section 0. 1 Functions and Their Applications This chapter examines the basic notions of a function. It begins with the linear function, and then the quadratic functions the parabola. The functional concept needed for the study of calculus are included as well as relevant applications to economics and statistics. Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 42 * ** Section 1.1 The Line 1.1 » » » » » » » » » The Line Two Dimensional Coordinate System Horizontal and Vertical Lines The Slope Intercept Form Graphing The Point-Slope Equation The Slope Formula The General Linear Equation An Economic Application Calculator Tips PRETEST 1.1- Time 15 minutes Each question is worth one point. 1. Determine the equation of the line whose slope is 2 and whose y-intercept is (0,5). For questions 2 4, use the line 5x - 2y = 9. 2. 3. 4. 5. 6. 7. 8. 9. Determine the slope of the line. Determine its y-intercept. Determine its x- intercept. Find the equation of the line with slope 3 passing through the point 12, - 52. Find the equation of the line passing through (5, 2) and 15, - 72. Find the equation of the line passing through (5, 2) and 1 - 3, 22. Find the equation of the line passing through (4, 2) and (6, 7). Find the equation of the line parallel to the line 3x - 2y = 7 and passing through the point (6, 5). Two Dimensional Coordinate System You have learned about lines at several different times in your mathematical education. We give a complete development of the line and slope in this section along with some applications. We believe a detailed review of this topic is useful as some of the basic notions of calculus generalize ideas examined in this section. Once again, we provide a short pretest for those of you who believe you remember the topic well. The box above indicates the major points of discussion of this section. Coordinate geometry is one of the most useful tools in gaining a visual understanding of functions. With coordinate geometry, algebraic formulas may be translated into graphs. In many cases, having the graph is the end of the problem. As you know from everyday experience, a picture may be far more informative than a collection of data. In other cases, the picture may reveal the solution to a problem that might otherwise appear to be too difficult to attack. Although none of this should be new to you, let us review the ideas and thereby fix what will become our standard notation and terminology. We remark that our approach is probably not the one you first learned, but as a second time through the material, it will most quickly obtain the results needed to understand linear functions. We begin with a Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 1.1 The Line * ** 43 two-dimensional universe consisting of all ordered pairs of real numbers, usually denoted by R2. Examples of points in this universe are (1,4), (0.5 , 6), A 0, 22 B , and 1 - p, 3.12. Because the two real numbers are ordered, (2,7) and (7,2) are different points. When coordinate axes are introduced in a plane, every ordered pair is associated with a point in the plane; and, conversely, every point in the plane has attached to it a unique ordered pair of coordinates. Let us briefly explain how this is done. First, as in Figure 1, a pair of number lines are drawn at right angles to one another, intersecting at the point zero on each line. The horizontal line is called the x-axis and the vertical line is called the y-axis. Construct a vertical line through any point in the plane. At the point where this line crosses the x-axis is a number called the x-coordinate (or abscissa) of the point. Now, construct a horizontal line through the point. At the point where this line crosses the y-axis is a number called the y-coordinate (or ordinate) of the point. If the point is called P, and its coordinates are (x, y), then we may refer to it as P(x, y). In Figure 1, the points P(2, 9), Q1 - 4, - 52, and R1 - 5, 42, S14, - 32 are shown. Notice that every point on the x-axis has ordinate 0 and every point on the y-axis has abscissa 0. Horizontal and Vertical Lines P(2,9) R(-5,4) (0,0) S(4,-3) Q(-4,-5) Figure 1: The Two Dimensional Coordinate System The point where the axes intersect, called the origin, has coordinates (0,0). If we have a relation between two quantities x and y, then we may plot all the points whose coordinates satisfy the rule. The resulting picture is the graph of the relationship. Of course, there may be an infinite number of points, so that actually plotting them all is impossible. On the other hand, in most cases of practical interest, the graph assumes an easily observable pattern which we can visualize without seeing every point. It is also possible to go the other way. That is, we could have a verbal description of the geometric shape that we want as a graph and then try to find the algebraic relationship between the coordinates, that would produce it. For example, the vertical line crossing the x-axis at 3 would consist of all points with x-coordinate, 3. Thus, it would be described by the rule x = 3 and y = anything. Since the last restriction is no restriction at all, we shall simply refer to this graph as the line x = 3. In general, any vertical line will be represented by the equation Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 44 * ** Section 1.1 The Line x = a, where a is some constant. If a is positive, then the line lies to the right of the y-axis. If a is negative, the line lies to the left of the y-axis. Of course, x = 0 is the y-axis. See Figure 2. We note that the graphs in Figure 2 cannot be graphs of functions of x, since there are many y values corresponding to the value x = 3. In a similar manner, the equation y = b has as its graph a horizontal straight line which crosses the y-axis at (0, b). If b is positive the line lies above the x-axis; if b is negative the line lies below the x-axis; and, naturally, y = 0 is the x-axis. See Figure 3 where we graph the lines y = 4 and y = - 3. The graph defined by y = b is a very simple one, and this relation is called the constant function. (We shall discuss functions in detail later in this chapter.) y x = -1 x=3 x Figure 2: The lines x = - 1 and x = 3 y=4 y=-3 Figure 3: The lines y = 4 and y = - 3 Let us now see what kind of function gives rise to any other straight line graph. To begin, we consider a line passing through the origin and making an angle u with the positive x-axis (see Figure 4). Let P(x, y) be any point on the line, other than the origin. For any choice of (x, y), the triangles formed by the given line, the vertical line through P and the x-axis are similar. Therefore, from the figure, it is clear that the ratio y/x is the same for every point P. That is, y/x is a constant. We will denote the constant by m, and call this the slope of the line. That is, the coordinates of every point on this line except the origin must satisfy the condition that y/x = m, which is a constant. Multiplying by x, we obtain y = mx and in this form we can also allow 1x, y2 = 10, 02, so that y = mx Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 1.1 y = mx P(x, y) y The Line * ** 45 * x Figure 4: The line y = mx is the equation of a line of slope m passing through the origin. Indeed, every non-vertical line passing through the origin is the graph of y = mx for some value of m. Notice that for m = 0, this equation reduces to just y = 0, which we saw above is an equation for the x-axis. y = mx (0, b) y = mx + b Figure 5 Now consider the line parallel to the line y = mx that cuts the y-axis at (0,b). If b is positive, then this line lies above the original line y = mx (as drawn in Figure 5) and for each x, the y-coordinate of every point on the new line is just b more than the value y = mx on the first line. Therefore, we conclude that every point on this line satisfies the equation y = mx + b. Of course, if b is negative, the line y = mx + b will lie below the original line, but in all other respects the analysis will be the same. Note that parallel lines have the same slope. Technically, the y-intercept is the point where the line crosses the y-axis and it has coordinates (0, b). In practice, the number b will be referred to as the y-intercept. Thus, we conclude that the equation of any non-vertical straight line may be written in the slope intercept form which follows. The Slope Intercept Form Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 46 * ** Section 1.1 The Line The Slope Intercept Form of a Line An equation for any non-vertical line is y * mx + b where m and b are constants, m is called the slope of the line, and b is called the y-intercept. Drawing the graph of a straight line is particularly simple, since we know that two points determine a line. Thus, we need locate only two points on the line and lay a straight edge across them. Let us look at some examples. Graphing Example 1. Plot the graphs of (a) y = 2x - 1 (b) y = - 1*2x + 4 (c) y = - 3 (d) x = 3 Solution. In (a) we have a line of slope 2 and y-intercept - 1. That is, the line crosses the y-axis at 10, - 12. In order to draw its graph we need only one additional point. So, we substitute any convenient value for x into the equation, say, x = 1. When x = 1, y = 2112 - 1 = 1. That is, the line passes through 10, - 12 and (1, 1). We plot these two points and draw the line as shown in Figure 6(a). In (b), the slope is - 1*2 and the y-intercept is 4. This line passes through (0, 4). Picking x = 2 for convenience, we get a second point y = - 1*2122 + 4 = 3. Thus, for a second point we have (2, 3). The graph is shown in Figure 6(b). We see no x term in equation (c). However, we recognize this as being of the form y = constant, which is a horizontal line consisting of all points with y-coordinate - 3. Its graph is Figure 6(c). We note that this line could be thought of as y = 0x + 1 - 32, that is, a line of zero slope, parallel to the x-axis, with y-intercept - 3. Finally, in (d), there is no y term and we recognize this line as a special case that cannot be put into slope intercept form. Its graph is the vertical line shown in Figure 6(d). y = 2x -1 y = * *+ x + 4 Figure 6(a): y = 2x - 1 Figure 6(b): y = - 1*2x + 4 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 1.1 The Line * ** 47 x=3 y = -3 Figure 6(c): y = - 3 Figure 6(d): x = 3 Example 2 Find an equation of a line having slope 3, and which passes through the point 1 - 1, 42. Solution. form Since this is not a vertical line, its equation can be written in slope intercept y = 3x + b, where b is to be determined. Since the coordinates of the point 1 - 1, 42 must satisfy the equation, we have 4 = 31 - 12 + b 4 = -3 + b Thus, b = 7, and the equation of the line is y = 3x + 7 In the many applications, we shall frequently encounter examples in which one knows the slope of a line and one point on the line as in Example 2. Therefore, it will be convenient to have a simple formula into which such data can be substituted to find the equation of the line directly. Let us suppose that we know the slope, m, and one point 1x1, y12 on a line. As in the previous example, we know that the equation must be y = mx + b, where b is to be determined. Substituting the known point 1x1, y12, yields y1 = mx1 + b We solve for b, obtaining y1 - mx1 = b We substitute this for b into y = mx + b, to obtain y = mx + 1y1 - mx12 = mx + y1 - mx1 Now we subtract y1 from both sides of the equation, giving y - y1 = mx - mx1. The Point-Slope Equation Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 48 * ** Section 1.1 The Line Finally, we factor the m from both terms on the right hand side of the equation, to get y - y1 = m1x - x12. This is the so-called point-slope equation of a straight line. The Point-Slope Equation of a Line If you are given the slope of a line m, and know the coordinates of one point 1x1, y12 on the line then you may determine an equation for the line by simply substituting the information into the point-slope equation y + y1 * m1x + x 12 Of the various forms of the equation of the straight line, the Point-Slope form is the most useful for calculus. When simplifying the expression, we either write the final expression in the form y = mx + b or in the form Ax + By = C where A, B, and C are integers. Example 3 Using the point-slope equation, rework Example 2. Solution. In Example 2, we were asked to find an equation for a line of slope 3, that passes through 1 - 1, 42. The given information exactly suits the point-slope formula that was just derived. Therefore, we substitute directly m = 3, x1 = - 1, y1 = 4 to obtain y - 4 = 31x - 1 - 122 y - 4 = 31x + 12 y - 4 = 3x + 3, and adding 4 to both sides, we obtain y = 3x + 7 the same form as before. Here is another frequently encountered type of problem. Example 4 (a) Find an equation for the line that passes through (2, 9) and 1 - 5, - 62. (b) Identify its slope and y-intercept. Solution. (a) Although we know more than one point, we cannot use the point-slope formula yet, since we do not know the slope. However, since both points satisfy the same equation, y = mx + b, we substitute each set of coordinates to get two equations 9 = 2m + b - 6 = - 5m + b Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 1.1 The Line * ** 49 If we subtract the lower equation from the upper, we will eliminate b, thus, 9 - 1 - 62 = 2m - 1 - 5m2 15 = 7m m = 15/7 Now we may use the point-slope equation using either of the given points for 1x1, y12. Let us use the first point (no reason to burden ourselves with extra negative signs): y - 9 = 115/721x - 22 Multiply through by 7 to simplify, yielding 7y - 63 = 151x - 22 7y - 63 = 15x - 30 7y = 15x + 33 (b) Notice that the equation we just obtained looks nice but it is not in the form y = mx + b (or, if you prefer, y = f1x2). This is not unusual. Since we are asked to identify the slope and y-intercept of this line we must divide the equation by 7 to get y = 15 33 x + 7 7 Thus, the slope is 15/7 and the y-intercept is 33/7. It is not uncommon to encounter cases in which two points are known, and an equation of the line determined by them is needed. What would help is a simple way to find the slope; then you can use the point-slope formula as in the last example. Therefore, let us suppose that we know the coordinates of two points on a non-vertical line. Call them P1x1, y12 and Q1x2, y22. Proceeding as in the preceding example, we realize that both points satisfy the equation y = mx + b So y2 = mx2 + b and y1 = mx1 + b subtracting the lower equation from the upper, we have y2 - y1 = mx2 - mx1 = m1x2 - x12. Since the line is not vertical, no two points on it have the same x-coordinate. Therefore, x1 Z x2 and x2 - x1 Z 0. Thus, dividing by 1x2 - x12, we obtain the slope formula. The Slope Formula The Slope Formula Let 1x1, y12 and 1x2, y22 be any two points on a line for which x1 Z x2, then the slope of the line is given by m = y2 - y1 x2 - x1 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 50 * ** Section 1.1 The Line This result is known as the slope formula. Do we have to worry about the possibility that x2 - x1 = 0? Not really. If the difference is zero, this means that x2 = x1. If the two points are different points and x2 = x1 then our line passes through two points with the same x-coordinate. That is, the line is vertical. The slope of a vertical line is undefined in any event, so we can safely use the above formula, and if the denominator is zero, we will know that the desired line is vertical. Note that it is correct to write m = but m Z y2 - y1 y1 - y2 and m Z x 1 - x2 x2 - x 1 1Why?2 y1 - y2 x 1 - x2 Example 5 Using the slope formula, rework Example 4. Solution. We know the two points, 1x1, y12 = 12, 92 and 1x2, y22 = 1 - 5, - 62. Substituting into the slope formula, m = -6 - 9 - 15 15 = = -5 - 2 -7 7 of course, now that we have m, we may finish the problem exactly as in Example 4. The slope formula also gives us a good indication of what the slope really represents. Call the two given points, P1x1, y12 and Q1x2, y22 We can always think of Q as lying to the right of P (see Figure 7). That is, always identify x2 and x1, so that x2 7 x1; then x2 - x1 is always positive. Thus, in the slope formula, the denominator is always positive. If Q is higher than P, the numerator is also positive; hence, m is positive. So, if the line is rising as you go from left to right, the slope is positive; otherwise it is negative. A zero slope indicates that the numerator is zero, which means that y2 = y1 and the line is horizontal. P(x1, y1) Q(x2, y2) P(x1, y1) Q(x2, y2) Figure 7(a) Q is higher than P, m 7 0 Figure 7(a) Q is lower than P, m 6 0 The numerator is the change in y as you go from the first point to the second, and the denominator is the change in x. Thus, m is sometimes referred to colloquially as the rise over the run; that is, the vertical change divided by the horizontal change. It now becomes clear that if Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 1.1 The Line * ** 51 the magnitude of the slope is large, then the change in y is large for a relatively small change in x. That is, steep lines have large slopes. Here, large negative m means m is large. (Remember, m means the absolute value or magnitude of m. So, for example, 6 = 6 and - 6 = 6.) Incidentally, the division between large and small in this context is 1. A line of slope 1 makes a 45* angle with the x-axis ( - 1 means the line makes an angle of 45* but measured from the negative portion of the x-axis.) Figure 8 shows several examples to give you an idea of how steep, lines of different slopes are. m3 m2 m1 Figure 8: Lines with Slopes m1 6 m2 6 m3 In summary, we have the following cases summarized in Table 1. Table 1: The Relationship Between a Line and its Slope Slope Property y decreases (gets smaller) as we move from left to right, i.e. the line slopes downhill. y increases (gets larger) as we move from left to right, i.e. the line slopes uphill. line is horizontal (parallel to x-axis). line is vertical. m 6 0 m 7 0 m = 0 undefined We shall see, when we study the calculus, how these observations generalize to nonlinear graphs. In short, we have seen equations of a straight line may take any of three forms: 1. x = a, a vertical straight line. 2. y = b, a horizontal straight line. 3. y = mx + b, any other straight line. Of course, the second form of a line is only a special case of Form 3 with m = 0. All this can be summarized in the following: Theorem: Every equation of the form Ax + By * C, A, B, and C constants, A and B not both zero, is an equation of a straight line. Accordingly, every such equation is called a linear equation. The General Linear Equation Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 52 * ** Section 1.1 The Line This may be verified as follows. Since A and B cannot both be zero, first consider the case where B = 0, and A is nonzero. The equation is now Ax = C, and we can divide by A yielding x = C/A. That is, x = constant, which is a vertical line. On the other hand, if B Z 0, then we can solve for B, By = - Ax + C Now divide by B, y = - 1A/B2x + 1C/B2 which is of the form y = mx + b. Note that the slope of the line Ax + By = C is m = - A/B. Example 6 Find the slope and y-intercept of the line whose equation is 3x + 4y = 8. Also find the x-intercept (the point where the line crosses the x-axis) and plot the line. Solution. The equation is 3x + 4y = 8, which we solve for y in order to get it into the usual form: 4y = - 3x + 8 Dividing by 4, 3 y = - x + 2 4 Now, by inspection, we see that the slope is - 3/4 and the y-intercept is (0, 2). To find the x-intercept, we are really asking to find the value of x for which y = 0. In other words, we substitute y = 0 into the given equation and solve for x. 3x + 4102 = 8 3x = 8 x = 8/3 The x-intercept is (8/3, 0). In general, unless a line passes through the origin, (or is horizontal or vertical) the easiest way to draw its graph is to plot the intercepts. Thus, in Figure 9 we show the intercepts and the line. 3x + 4y = 8 Figure 9: The Line 3x + 4y = 8 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 1.1 The Line * ** 53 Example 7 Determine an equation for the line parallel to the line 3x - 2y = 8 and passing through the point 11, - 22. Solution. As above, we put the given line in the usual form by solving for y. We find that y = 3/2x - 4. Since the required line is parallel to the given line, its slope must also be 3/2. Thus, the required line has slope m = 3/2, passes through the point 11, - 22, and by the point-slope formula its equation is y - 1 - 22 = 3 2 1x - 12 Simplifying, we find that the slope-intercept equation of the required line is y = 3/2x - 7/2. We can also write an equation for the line in the form 3x - 2y = 7. An Economic Application The slope is defined as the change in y-values divided by a corresponding change in x-values. Thus, slope is essentially the average rate of change of y with respect to x. It is precisely this interpretation of slope that is essential to our understanding many applications, especially in Economics and Finance. The next example illustrates such an application. Example 8 When a wholesaler sold CD players at $60 per player, weekly sales averaged 150 players. For each $5 drop in the wholesale price the average number of players sold increased by 15. (a) Describe the relationship between the wholesale price and average weekly sales and (b) what is the average weekly sales if the wholesaler charges $42 per unit? Solution. We let x represent the average weekly sales, and y the wholesale price. (a) We first plot some points to see if we notice a pattern. When y = 60, x = 150, that is, (150, 60) is our first point. If we decrease the price, y by 5, then weekly sales, x increases by 15, so we have the point (165, 55), if we drop y again by 5, then x increases by another 15, and we have as our third point, (180, 50). These points are plotted in Figure 9. Note that as we move the sales right 15 units to the right (increase), the price falls 5 dollars. Thus, we go from the point (150, 60) to the point (165, 55). Once again, as we move from the point (165, 55) and move right 15 units, we then fall another 5 units to the point (180, 50) and this trend continues. This ratio of the price to the weekly sales, or equivalently, the average rate of change of price with respect to weekly sales is the slope of the line connecting the points. Thus, we see a linear relationship between the variables y and x, the price and sales. The rate of change of price with respect to sales is 5/15 = 1/3. Since sales are increasing as the price falls, we have a negative rate of change, that is, the slope is - 1/3. We can now write the equation that represents this relationship. We have m = - 1/3 and choosing the point (150, 60) we have y - 60 = - 1/31x - 1502 or simplifying, we have 3y + x = 330 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 54 * ** Section 1.1 The Line Since it makes sense to think of the price as driving the demand for the CDs, we solve for x as a function of y. This gives x = - 3y + 330 (b) When the wholesale price y = 42, we have x = - 31422 + 330 = 204. Thus, at a price of $42 per unit, the average sales are 204. 15 5 15 5 Figure 10: Price Plotted versus Average Weekly Sales We have already seen that parallel lines have the same slope, and conversely that two lines with the same slope are parallel. We now investigate the relationship between the slope of perpendicular lines, that is, lines that intersect at an angle of 90 degrees right angles. Let us first dispense with the case where one line is vertical, then any line perpendicular to it must be horizontal. Thus, we assume in what follows that neither line is vertical. We remind you that the formula for the distance d between the points 1x1, y12 and 1x2, y22 is given by d = 21x2 - x122 + 1y2 - y122 Without loss of generality, assume the two lines intersect at the origin. (Or equivalently, think of the origin as being at the intersection of the two lines.) Let m1 and m2 be the slope of these two lines, then their equations are y = m1x and y = m2x (why?). Choose points A and B on each of these lines with x-coordinate 1, then the corresponding y-coordinates are m1 and m2. Consider Figure 11. Let us first assume the lines intersect at right angles at the origin. The triangle shown with sides a, b, and c is therefore a right triangle, and Pythagoras theorem is applicable. By the distance formula, we have a = 211 - 022 + 1m1 - 022 = 21 + m1 2 b = 211 - 022 + 1m2 - 022 = 21 + m2 2 and the vertical line connecting A to B has distance c = m1 - m2 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Walter O. by working backwards. Wang. Published by Pearson Learning Solutions. .2m1m2 + m2 2 or 2 = .2m1m2 or m1m2 = . we have shown that if two lines are perpendicular. or equivalently one is the negative reciprocal of the other.1/3. the product of their slopes is . m2 = .2/5 therefore its negative reciprocal is m2 = 5/2.1 Thus.1. beginning with two lines whose slope product m1m2 = .1 The Line * ** 55 y = m1x m1 a c b m2 y = m2x Figure 11: Slopes and Perpendicular Lines by Pythagoras theorem a 2 + b2 = c 2 we have 1 + m1 2 + 1 + m2 2 = 1m1 . Example 9 Determine the slope of the line perpendicular to the line (a) y = 3x (b) y = .1. that is. Gordon. we obtain the first statement which is Pythagoras theorem.Section 1. Solution. by Warren B. Economics. m2 = . Applied Calculus for Business. (b) m1 = .1/m1 We also observe that each step in the above argument is reversible. that is. implying the triangle is a right triangle.2/5x + 7 (c) 5x + 2y = 11 (d) x = 2 (e) y = 1. and April Allen Materowski. Copyright © 2007 by Pearson Education. (a) m1 = 3 therefore the slope of the perpendicular line is its negative reciprocal. and Finance. therefore the lines intersect at right angles. Inc.m222 or 2 + m1 2 + m2 2 = m1 2 . 2 key to the left of the Enter key on the bottom of the calculator.1 The Line (c) rewriting the line in the form y = .5 Calculator Tips The calculator may draw graphs for us in regions called windows. Gordon. then we enter . Be careful. by rewriting it as y = 3/2x . we only need y1. any line perpendicular to it is vertical and vertical lines have no slope.2/3x .5/2x + 11/2. Suppose you want to draw the line y = .5/3 or 2x + 3y = . Inc. You will notice that on top of the F1 button of the calculator. the negative sign in front of the 2 is input by pressing the 1 . The slope of the line perpendicular to it is the negative reciprocal so we have m2 = .2/3. Solution. you press the green diamond and then the F1 key1 F12. For a single line. We illustrate with lines.2y = 12. and so on. (The calculator distinguishes between the operations subtraction and negation. By the point slope formula. and April Allen Materowski. If we do not set the window. Horizontal lines have 0 slope.2x + 7. Walter O. therefore any line perpendicular to it is horizontal. Wang.2y = 12. we see immediately that its slope m1 = 3/2.32 = . Economics.5/2 therefore the slope of the line perpendicular to it is 2/5.2x + 7.56 * ** Section 1. (d) x = 2 is a vertical line. the calculator chooses the default window which may or may not be what we desire.22 or y = . in green is Y = . and Finance. .1 . Example 10 Determine the equation of the line passing through the point 12.6. Copyright © 2007 by Pearson Education.) Figure 12: The Y = Screen Applied Calculus for Business. the equation of the required line is y .32 and perpendicular to the line 3x . by Warren B. we see its slope m1 = . You now see the screen indicated in Figure 12. We write the equation in the form y = mx + b. Published by Pearson Learning Solutions. y1. We first find the slope of the line 3x . y2. It has on separate lines. To get the screen associated with Y = .2/31x . y3. . (e) The line y = 1 is a horizontal line. 2x + 7 You can see what the default window is by pressing the window key 1* F22. Gordon. Figure 15: The Window Screen Applied Calculus for Business. Published by Pearson Learning Solutions.2x + 7 Stored in the calculator s memory is the equation y = . Walter O. Copyright © 2007 by Pearson Education. Economics. Figure 14: y = . The calculator calls it y1. see Figure 15. Figure 13: y1 = . . Inc.1 The Line * ** 57 We now have Figure 13. it gives each one a separate name.Section 1. Wang. and April Allen Materowski. see Figure 14. If you now press the graph button 1*F32 it draws the line in the default window. by Warren B.2x + 7. and Finance. in the event that you store other equations in its memory. (b) Find an equation for the vertical line passing through the x-intercept of the line in 41(b). 42. the line y = 7 would be entered as y1 = 0x + 7. the highest is xmax which is 10. 72. 43. Once again.2. Exercise 4 25. (b) Find the equation of the line parallel to the line given in (a) and four units above it. (0. Exercise 9 30.4) and 1 . (0. 82.2. Exercise 16 37. (b) Find the equation of the line parallel to the line given in (a) and passing through 1 . (2. Exercise 14 35. One note of caution. Plot the line.2 = 0 10. experiment. Exercise 18 39.3x + 4. Exercise 11 32. Published by Pearson Learning Solutions. If you change any of the setting indicated above.7y + 11. 2x + 4y . Experiment.1 The Line xmin. xscl = 1 means each x tick mark is 1. Copyright © 2007 by Pearson Education. Exercise 2 23. the largest (ymax) is 10. 02. Note the zoom button is F2. Plot the line on the same set of axes.2). change some of them and then press the graph key to redraw the line. Exercise 17 38. Exercise 8 29. 1*2. 1/4. Exercise 1 22. and April Allen Materowski. 0. A vertical line passing through (12. EXERCISE SET 1.62.1. (a) Find the slope of the line whose equation is 3x + 7y + 42 = 0. 4y = 5x + 12 9. In exercises 21 40 plot the lines found in 21. Applied Calculus for Business.4y = 35 In exercises 11 15 give an equation of the line with the given slope and passing through the given point. and 4 units from it. Exercise 3 24. 0) 3.3. Economics. and Finance.10. 44.3. each y tick mark is 1. . 0) 14. A line with intercepts (0. find the slope. 1 . 11. 1. it is sometimes useful to enter the equation of the horizontal line say y = a as y1 = 0x + a. you change the window.1. Exercise 6 27. (3. (a) Find the slope of the line whose equation is 2x .4y = 12 7. 2. 16 . 5) 12. 5. . Exercise 10 31. 10. Find the equations of two lines parallel to the line y = . 6. . 02 13. determine the equation of the line whose slope and y-intercept are given. 4) 2. 82. (0. x-intercept and y-intercept of the given line. Gordon. . 3x . Exercise 12 33. the lowest x-value is . this button allows you to change the window more quickly by zooming in and out on the graph.10. 0. 1. Walter O. Exercise 15 36. A horizontal line passing through 12. 19. the lowest y-value (ymin) in the window is . . 0) In exercises 16 20 find equation for the given line. xres is the resolution. Wang.9 = 0 8.5y = 6. A vertical line passing through 1 . Find the x and y-intercepts of the line. Inc.1 In Exercises 1 5.32 15. 16. by Warren B. 3) In exercises 6 10.22 4.3. 45. (a) Find an equation for the horizontal line passing through the y-intercept of the line in 42(a). 20. Plot the line. Exercise 7 28. (b) Find an equation for the vertical line passing through the x-intercept of the line in 42(b). 1*2. Horizontal lines all have zero slope. 4. A horizontal line passing through 1 . we usually leave this alone.1. Exercise 20 41.58 * ** Section 1. Thus.1. 17. Exercise 19 40. 1/6. Plot the line on the same set of axes.3. 1 . 18. Find the x and y-intercepts of the line. (0. (0. Exercise 13 34. 14. (a) Find an equation for the horizontal line passing through the y-intercept of the line in 41(a). Exercise 5 26. The last item. Similarly. 1/4). (a) Find the length of the portion of the line 5x + 12y = 84 that is cut off by the two axes.1. Applied Calculus for Business. In 1990 the Massachusetts Non-Resident State Income Tax calls for a tax of 5% on earned income and 10% on unearned income. Exercise 58.42 (c) A 1*2. Exercise 48. Determine an equation for the line parallel to y = 3x . Plot each of the following lines on the same set of axes. 4) and 1 . and 6 units from it. (b) Repeat for Ax + By + C = 0 for A. Show if ab Z 0. . 49. Given the two parallel lines y = mx + b and y = mx + B. . 1 . (a) y = 2x (b) y = 21x .h2 + b related? (m. In general. 2) and (8. (0. 16) and 112. 10. 48. 90. 62 and (0. Exercise 55. For each $2 increase in price the producer will supply an additional 6 radios. (a) Plot these points. A 1*2. derive the distance formula. Exercise 54. determine the perpendicular distance between these two lines. 0). Copyright © 2007 by Pearson Education. (b) Do they lie on the same line? (c) How can you tell without plotting? 79.42. Walter O. Find an equation of a line whose y-intercept is 4 and such that the area of the triangle formed by the line and the two axes is 20 square units. 52. how are the lines y = mx + b and y . 89. 02. 3) and 1 . 56. Gordon. 4) and (2. (a) y = 2x (b) y . B.32. 67. 10. How many radios are supplied if their per unit price is $72? 82.92 and (12. Suppose a person has total income of $40. The area of a triangle formed by a line and the two axes is 40 and the slope of the line is . 0) and (0. 2/32 and (0. Determine an equation for the line (a) parallel (b) perpendicular to 3x + 7y = 11 and passing through the point 11.32 (c) y = 21x + 32 (d) How are these lines related? 83.52 and 12. . 68. .7. 5).42. What is the average monthly sales if the price is $400 per color television? 81. 02.32 (d) y . (4. Exercise 57. 0).4 = 2x (c) y + 4 = 2x (d) How are these lines related? 84. 16 B and A 1*2.52. Find the area of the triangle formed by the line.92 and (1/2.72.732. by Warren B. . Using the previous exercise. Plot each of the following lines on the same set of axes. (0. . Find her tax. 50. (b) Repeat for Ax + By + C = 0 for A. a b 87.4 = 21x + 32 (e) y + 4 = 21x + 32 (f) How are these lines related? 85. and April Allen Materowski. 11.1. (a) y = 2x (b) y . Determine an equation for the line (a) parallel (b) perpendicular to 2x . 2/52 and (0.52. Find an equation for the line. a producer will supply 100 radios each month for sale. The Line * ** 59 75. determine the equation of the line with intercepts (a) (3. 88. 61. then the line with intercepts (a. When the price for a color television is $240. Exercise 56. Exercise 51. (a) Plot the points 1 . 62.4).7 and passing through the point 11. 55.5y = 9 and passing through the point 1 .1/42. Find the equations of two lines parallel to x = 2. When the price is $50 per radio. 54. . Exercise 49. (12. 11.1/3. .0). A 1*2. For each $10 increase in price. 51. 73. 63. . 1 . In Exercises 47 58 find the slope of the line passing through each pair of points 47. 4). Find the point on the line y = 2x + 3 that is equidistant from the points 1 . (0. Plot each of the following lines on the same set of axes.000 of which amount x is earned. Exercise 50.1 46. . the average monthly sales fall by 20 units. 2/3). 6) (b) (2. (b) Obtain the right triangle formed by drawing a horizontal line from A and a vertical line through B. Let A1x1. (0. Wang.Section 1.4 = 21x . y22 be in any two points in the plane. 0). the x-axis and the y-axis. (a) Plot the line 4x + 6y + 12 = 0. y12 and B1x2. . b) has the equay x tion + = 1. What are the coordinates of the point at which these two lines intersect? (c) Using the theorem of Pythagoras. 74. 80.3). . 53.2 B and 11/4. and k are constants. 60.73 B . as a function of x.) 86.5. . 70. h. .2. 12. 58. and Finance. (Two possible answers. 78.k = m1x . 72.5. . the average monthly sales for this item at a department store is 450. Inc.) 77. Exercise 53 66.) 76.1/4. 71.(Two possible answers. In Exercise 59 70 find the equations of the lines and plot the lines from 59. . 64.22 and 11. Published by Pearson Learning Solutions.32 (c) y + 4 = 21x . b. C positive. (1. 0 B . 0). t.6. 69. 57. B. C positive. Exercise 47. Economics. Exercise 52 65. Time 15 minutes Each question is worth one point. Find the equation of the line slope 3/2 passing through the point passing through 1 . Published by Pearson Learning Solutions. 52. taxicab fare is a function of the distance traveled. Gordon. The more you earn.1. the more you (usually) pay in taxes. Determine its x. . 19. and April Allen Materowski. 8. 1 . 72. use the line 2x + 7y = 12. we may now formulate the definition of a function. for example.2 Basic Notions of Functions » » » » » » » » » » » Definition of a Function Functional Notation Difference Quotient Domain and Range Independent and Dependent Variables Vertical Line Test Combining Functions Composition Decomposition Functions of Several Variables Calculator Tips Definition of a Function The concept of a function comes up regularly in everyday usage. by Warren B. 52. 1. 62 and (3. 4. 3. Determine the equation of the line whose slope is 3 and whose y-intercept is 10.32. 2. to represent functions. f Applied Calculus for Business. trips of the same distance should cost the same. Determine the slope of the line. 12. With these examples in mind. but any letter may be chosen). Find the equation of the line parallel to 2x + 5y = 9 and passing through the point 110. Wang.8. taxicab fare as a function of distance means to each trip we associate a fare. A function is a set of ordered pairs obtained by some rule so that to each first element in the ordered pair there corresponds a unique second element in the ordered pair. a child s height is usually a function of age. your tax rate is a function of your total earnings. . Let us look at a few of these examples to understand what we mean by a function.60 * ** Section 1.82. Find the equation of the line passing passing through (3. or h. Determine its y-intercept. two people with the same income (and with all other deductions being equal) should pay the same taxes. Find the equation of the line with 6. 1. Observe the definition indicates that a function is a set. 4) and 1 . through (5. Inc. Walter O.8.2.2 Basic Notions of Functions POSTTEST 1. 6). g. f = 511.32.intercept. For questions 2 4. and Finance. Find the equation of the line 7. . Copyright © 2007 by Pearson Education. 32. 5. 3) and 15. 1126 is an example of a function (not a very interesting one) that we named f (we often use the letters f. Economics. 9. For example. and longer trips should cost more than shorter ones. . we could instead describe this function by its graph. Notice that each first element (in this example 1. We know that y = 2x + 1 (remember y is the same as f(x). it is the equation which defines the functional relationship between x and its corresponding y-value.2 Basic Notions of Functions * ** 61 contains three members. Note that in this example x could be any real number. when we write y = f1x2. we indicate the graph of f in Figure 1. For example if x = 5 then the corresponding y-value is 2152 + 1 = 11.3) and (1.3). Published by Pearson Learning Solutions. 9) corresponds to a unique second element. That means that the function has an infinite number of elements. it would be wrong to say the graph of the function f1x2 = 2x + 1. Note that any non-vertical line defines a function. Wang. or we would write f152 = 11. We could not have (1. or it maybe more complex. 11).Section 1.3x + 4. then the second element is found by multiplication of the first element by 2 then adding 1. we might say a function is defined by the equation f1x2 = 2x + 1. Thus. Economics. Often the rule by which we obtain the correspondence between x and y is given by an equation of this form. by Warren B. and Finance. y = f1x2 is not a function. Inc. This is sometimes called functional notation. we mean that y is the value associated with the number x. as we shall see. 5 corresponds 2 and 11 corresponds to 9. Copyright © 2007 by Pearson Education. (2. instead of saying 3 corresponds to 1. The equation can be a simple one. 5). The function is the set of all ordered pairs obtained through the use of this equation. For example.12 (d) f(2) (e) f(2x). Since it is impossible to list the elements of the function. (9. In general. f122 = 5 f192 = 11. Determine (a) f(0) (b) f(1) (c) f1 .7) belonging to the function since we would have two different second elements (3 and 7) corresponding to the same first element (1). so each point on the line is an element of f. Walter O. 2. (1. This means if x is any first element in an ordered pair. Function Notation f(x) = 2x + 1 Figure 1: The Graph of f1x2 = 2x + 1 Note that the caption in Figure1 states the graph of f1x2 = 2x + 1. (Why did we say non-vertical?) Example 1 Given the function defined by the equation y = f1x2 = x 2 . Gordon. . Applied Calculus for Business. that is. and April Allen Materowski. we prefer to write this correspondence as f112 = 3. y * f(x)) is the equation of a line. 3x .12 + 4 = 8 (d) f122 = 1222 . One of the most important uses of functional notation in the study of the calculus is the evaluation of the difference quotient. You will verify this remark in the exercises (Exercise 23).3122 + 4 = 2 (e) f12x2 = 12x22 . Economics.c 2 d B A a 2b + c 2 d B = a2b .3h = = 2x + h . a function defined by an equation whose graph is a non-vertical line. (a) f102 = 1022 . Example 2 Compute the difference quotient for the function defined by the equation f1x2 = x2 . The product of two conjugates A a 2b . Applied Calculus for Business.3x + 4. say f1x2 = mx + b. f1x + h2 . It is precisely this quotient that generalizes the concept of slope which is fundamental to the development of the calculus. and a non-zero number h. its conjugate is a 2b . given the expression a 2b + c 2 d. We have.c 2 d.312x2 + 4 = 4x 2 . Gordon. the difference quotient will turn out to be precisely the slope of the line.12 = 1 . We first compute f1x + h2.3h + 4 . f1x + h2 = 1x + h22 . therefore.1x2 . we replace x everywhere it appears with the given x-value.31 . .3h therefore. we replaced x by 2x wherever x appeared. Given a function defined by the equation y = f1x2.3x + 42 = 2xh + h2 . then the difference quotient is defined as f1x + h2 . Wang. Published by Pearson Learning Solutions.6x + 4 Difference Quotient Note that to determine the y-value at any x-value. We have f1x2 = x2 .32 2xh + h2 .31x + h2 + 4 = x 2 + 2xh + h2 . to find f(2x) in the preceding example.3x . and Finance. Recall.62 * ** Section 1. Copyright © 2007 by Pearson Education.2 Basic Notions of Functions Solution. Inc. This means we replace x by x + h. by Warren B.c2d contains no radicals. The trick was to multiply the binomial to be rationalized by its conjugate expression.f1x2 h12x + h . Solution. Thus.3102 + 4 = 4 (b) f112 = 1122 .f1x2 h (1) We remark that for a linear function. The basic idea was to eliminate the square roots appearing in a binomial expression. everywhere it appears in the equation.f1x2 = x2 + 2xh + h2 .3 = h h h We momentarily digress and remind you about the rationalization of denominators and numerators that you learned about in algebra.3x + 4. Walter O.3h + 4 and f1x + h2 .122 . and April Allen Materowski.3112 + 4 = 2 (c) f1 . that is. We illustrate the computation of the difference quotient in the next example. Wang. Solution. as the next example illustrates Example 3 29 + h . and Finance. Economics.3 B A 29 + h + 3 B 29 + h . Gordon. in problems similar to the last one. Thus. in this last example. that after performing rationalizations like these. Inc. when we study the calculus. the difference quotient is Applied Calculus for Business. we would like to determine what the rationalized expression equals if we allow h to be 0. You will note that evaluation before the rationalization produces 0/0 an expression which is called an indeterminate form. Sometimes.2 Basic Notions of Functions * ** 63 Consider the expression (for h Z 0) h 2 . A 29 + h . the result should not contain any radicals in the numerator. Copyright © 2007 by Pearson Education.Section 1. we obtain the result 1/6. after rationalization. determine the value of the expression for h = 0.3 9 + h . We have h 2 . .24 + h B A 2 + 24 + h B h A 2 + 24 + h B h = = - = . Example 4 Determine the difference quotient for f1x2 = 22x + 1.22x + 1 therefore. However. and April Allen Materowski.3 Rationalize the numerator of the expression for h Z 0.24 + h h A 2 + 24 + h B 4 . Walter O. Published by Pearson Learning Solutions.24 + h This last expression is algebraically equivalent to the original expression when h is not zero.14 + h2 = h A 2 + 24 + h B A 2 . if we evaluate the expression with the numerator rationalized with h = 0. h cancelled in both the numerator and denominator.9 = = h h A 29 + h + 3 B h A 29 + h + 3 B = h h A 29 + h + 3 B = 1 A 29 + h + 3 B We shall see. by Warren B. h Solution. After the rationalization process is completed.24 + h and let us rationalize the denominator. Note the cancellation of h in both the numerator and denominator of the last expression.f1x2 = 22x + 2h + 1 .2 . f1x + h2 = 221x + h2 + 1 = 22x + 2h + 1 subtracting f(x) from this expression yields f1x + h2 . we have to rationalize the numerator in order to perform the necessary evaluation. When using a calculator to compute the difference quotient. 1126. We illustrate in the next few examples. Inc. The set of all first elements in the ordered pairs contained within a function is called the domain of the function. Thus this function has as its domain D = 51.12x + 12 h A 22x + 2h + 1 + 22x + 1 B = 2h h A 22x + 2h + 1 + 22x + 1 B 2 = = A 22x + 2h + 1 + 22x + 1 B Notice the cancellation of the h in the numerator and denominator. and Finance.2x + 5 if 1 x 4.3 and 3. 5. In those texts. by Warren B. When the correspondence which determines the function is given by an equation. 32. we have 2 22x + 1 + 22x + 1 = 2 2 22x + 1 = 1 22x + 1 Some texts use the symbol ¢ x in place of h.64 * ** Section 1. If we now set h = 0. the domain is determined by examination of the permissible values of the (independent) variable. 2. We also assume that only real numbers are allowed in the domain. . Solution. and 9 and three second elements 3.32. Example 5 Determine the domain and range of the function defined by the equation y = f1x2 = . 5. 11. 96 and its range R = 53. 3) and 14. Applied Calculus for Business. namely the set of all x such that 1 x 4.3 y 3. we need only determine the range. Walter O. this function has three first elements 1.f1x2 22x + 2h + 1 . therefore the range is the set of all y-values with . its interpretation is the same. . Published by Pearson Learning Solutions. Note that the graph of this function is a line segment connecting the two endpoints (1. 52. and the set of all second elements is called the range of the function. and April Allen Materowski. Economics. The range is often obtained from the sketch of the graph. 116. f1x + h2 . 12. y can assume all values between .f1x2 ¢x In does not matter which symbol is used. 19. Gordon.22x + 1 # 22x + 2h + 1 + 22x + 1 = B R B R= h h 22x + 2h + 1 + 22x + 1 12x + 2h + 12 . 2. Wang.2 Basic Notions of Functions f1x + h2 . Therefore.22x + 1 = h h We rationalize the numerator by multiplying numerator and denominator by the conjugate of the numerator. the form using h is easier to enter. the difference quotient is written as Domain and Range f1x + ¢ x2 . Note that the domain is specified in this example. Copyright © 2007 by Pearson Education.f1x2 22x + 2h + 1 . In the example in which f = 511. We sketch the graph of this linear function in Figure 2. solving the inequality 2x . Sometimes this is written in interval notation as [3/2.3 Ú 0 for x.2 27 L 2. Inc. Published by Pearson Learning Solutions.2 Basic Notions of Functions * ** 65 (1.7 25 L 2.2x + 5.3 x 3/2 2 3 4 5 6 y = f1x2 = 22x . by Warren B. we obtain that the domain is x Ú 3/2. To determine the range we will sketch a graph of the function.3 0 1 23 L 1. Of course. q 2. Thus. y Ú 0 or in interval notation [0. and Finance. Copyright © 2007 by Pearson Education. -3) Figure 2: The Graph of y = f1x2 = . Economics. we have as the range. . 0) and the graph rises and will assume all positive y-values. Gordon. Therefore.6 3 Using these points. we draw the graph in Figure 3. To determine the range recall that we must deal only with real numbers. Table 1: Points to Plot the Graph of y = f1x2 = 22x . Solution.Section 1. 1 x 4 Example 6 Determine the domain and range of the function defined by the equation y = f1x2 = 22x . We do this by plotting some points as indicated in Table 1.3 Ú 0 (if it were not.3.3) y = f(x) = x+5 (4. we begin with x = 3/2. The next example illustrates another way the domain may be restricted. we must require that 2x . Wang. Walter O. and April Allen Materowski. then the radicand would be negative resulting in imaginary values for y). Note that the lowest y-value on the graph is at the point (3/2. Applied Calculus for Business. Thus. q 2. For example velocity as a function of time is often written as v = f1t2. the amount paid in taxes (the dependent variable) depends upon the income earned (the independent variable). y. Sometimes.3 Example 7 Determine the domain of the function defined by the equation f1x2 = x . by Warren B. Independent and Dependent Variables Note that we did not find the range of the function defined in the previous example. Often we use the letter x to indicate the independent variable and f(x) or y to indicate the dependent variable. This function is an example of a rational function and will be considered in detail later in this chapter.7x + 12 Solution. or 1x . Walter O. x and produces an output.7x + 12 = 0. the domain consists of all x-values except 3 or 4. In the illustrations above. The denominator will be zero when x 2 . Published by Pearson Learning Solutions. it is convenient to visualize a functions as a machine as indicated in Figure 4.42 = 0. and April Allen Materowski. x+ + y = f(x) f Figure 4: A Function Viewed as a Machine Applied Calculus for Business. Inc. you will learn how to easily sketch its graph and (sometimes) obtain its range from examination of its graph. Wang. Gordon. . Usually. and Finance.321x . we call the elements in the domain the independent variables and the elements in the range the dependent variables. The terminology certainly makes sense. Thus. At that time. or when x = 3 or 4. where v represents velocity and t time. Economics.2 Basic Notions of Functions Figure 3: The Graph of the Function Defined by y = f1x2 = 22x . This machine accepts an input. x 2 . but any symbol may be used.66 * ** Section 1. All x-values are allowable with the exception of those which make the denominator zero. Copyright © 2007 by Pearson Education. the fare (the dependent variable) is dependent on the ride s length (the independent variable). the length of the ride determined the taxicab fare. Economics. there corresponds a unique output (the dependent variable). Consider the graph in Figure 5. and April Allen Materowski. The set of all possible inputs accepted by the machine is its domain. which of these represent a function? Figure 6a Figure 6b Figure 6c Applied Calculus for Business.Section 1.2 Basic Notions of Functions * ** 67 The machine performs an operation on the input x producing the output y. A graph represents a function if and only if any vertical line intersects the graph in at most one point. (This gives the correspondence between x and y. by Warren B. there are two different y values. . This is true for all graphs which represent functions. when x = 10. .1 and 5 (violating the condition that each x corresponds to a unique y). Published by Pearson Learning Solutions. Inc. Copyright © 2007 by Pearson Education. Wang. and the set of all outputs produced by the machine is its range. you would expect that its output should be replicated by using the same input (independent variable). as shown. Remember. that is. Walter O.) If a machine is reliable. Gordon. Note that a vertical line passing through this graph will intersect the graph in more than one point. to each input. Note in Figure 1 that if we were to draw a vertical line anywhere on the graph it would intersect the graph (which represents the function) exactly once. -1) Figure 5: A Graph That Does Not Represent a Function Example 8 Consider the graphs in Figure 6. to each x-value (first element) there corresponds and unique y-value (second element). and Finance. therefore this graph cannot represent a function. 5) (10. Vertical Line Test x =10 (10. 68 * ** Section 1. If we draw a vertical line through any point (except x = 2) in Figure 6c. Combining Functions Functions defined over a common domain may be added. The cost C as a function of distance may be determined as follows: C1d2 = 1. multiplied and even divided so long as we do not divide by zero. Example 10 Let f1x2 = 2x + 3 and g1x2 = x2 + 4x + 4 (b) f1x2 . The next example illustrates one which you may have some familiarity with.1 = $11. each of these graphs represents a function. we see in Figure 7 that it intersects the graph it two different points. Published by Pearson Learning Solutions. (a) f1x2 + g1x2 = 12x + 32 + 1x2 + 4x + 42 = x 2 + 6x + 7. Inc.50 for each additional mile or part traveled.1 = $21. Copyright © 2007 by Pearson Education.501d .g1x2 (c) f(x)g(x) (d) f(x)/g(x) determine (a) f1x2 + g1x2 Solution. (b) What is the fare for a ride that is 5 miles long? (c) 8. rounded up to the next integer when d is not an integer (so 1.1. so we have C1d2 = 1. Often functions arise in many different formats. Applied Calculus for Business. d .g1x2 = 12x + 32 . for x Z . Walter O.2x .x 2 . this violates the vertical line test and therefore. (c) Since we round up.50. (c) f1x2g1x2 = 12x + 321x2 + 4x + 42 = 2x 3 + 11x 2 + 20x + 12. These operations are easily performed. (since d = 5 which is greater than 1) which gives C152 = 2. (b) To determine C(5).50 for the first mile or part and $2. depending on whether the distance is less or more than one mile.2 Basic Notions of Functions Solution.4 miles is round up to 2 miles). it intersects the graph in exactly one point. by Warren B.50d . we use the description C1d2 = 2.50 if d 1 if d 7 1 the cost will be $1. (a) Determine an equation that describes this function. subtracted. (b) f1x2 .32 = C192 = 2.2. . We may combine these two pieces as follows: c1d2 = e 1. Economics.1. Note that we have two different formulas for the cost.3 miles long? Solution.50 plus $2. this cannot be the graph of a function.5192 .50 times each additional mile traveled. (d) f1x2/g1x2 = 12x + 32/1x 2 + 4x + 42.50 2. Wang. (a)Let d represent the distance traveled. In Figure 6a or Figure 6b. therefore. no matter where you draw a vertical line. C18.50.1 gives the number of additional miles traveled after the first. Figure 7: A Vertical Line Intersecting a Graph at Two Points Example 9 The fare for riding in a cab in a small town is as follows: $1.1 if if 0 6 d 1 d 7 1 A function written like this is sometimes called a piecewise function since its definition depends upon which piece you are on. These operations are defined. and Finance. Since the two functions have a common domain (what is it?). and April Allen Materowski.50d .12 = 2.1x2 + 4x + 42 = .1 if d 7 1.50 + 2. Gordon.50152 .50d . as illustrated in the next example. so there are no concerns about the domain. (b) f132 = 2132 + 3 = 9. the composite function determined by the equation y = f1g1x22 is defined. We illustrate this pictorially in Figure 8. (We note that the domain of each function . and Finance.) The next example illustrates our concern about the domain of the composite function. (c) f(g(x)) and (d) g(f(x)) Solution. and for the f machine to operate. (c) f1g1x22 = 21g1x22 + 3 = 21x2 + 12 + 3 = 2x 2 + 5. x must be in the domain of g otherwise. the conditions for y = g1f1x22 are similar and will be left as an exercise. Published by Pearson Learning Solutions. therefore f1g1322 = f1102 = 21102 + 3 = 23. . therefore g1f1322 = g192 = 92 + 1 = 82. Some care needs to be taken with regard to the domain.) (a) g132 = 32 + 1 = 10. Suppose we are given two functions defined by the equations y = f1x2 and y = g1x2. Walter O. (Verify that setting x = 3 in (c) and (d) yields the same results found in (a) and (b). Gordon. determine (a) f(g(3)). therefore the number g(x) must be in the domain of f. called the composition of functions. We wish to consider under what conditions can we define the new composite functions y = f1g1x22 and y = g1f1x22 When these functions are defined. Economics. the first is the composition of f with g and the second is the composition of g with f. For any x-value in this domain. Applied Calculus for Business. for the g machine to operate. Note. x must be in its domain. the domain of the composite function defined by the equation y = f1g1x22 is the set of all x such that x is in the domain of g and g(x) is in the domain of f. Copyright © 2007 by Pearson Education. Thus. we could not obtain g(x). (b) g(f(3)). and April Allen Materowski.q 6 x 6 q . g(x) must be in the domain of f.Section 1. Inc.2 Basic Notions of Functions * ** 69 Functions have one additional way of being combined. We examine the conditions under which we may define y = f1g1x22. (d) g1f1x22 = 1f1x222 + 1 = 12x + 322 + 1 = 4x 2 + 12x + 10. by Warren B. Wang. Composition x+ g(x)+ f (g(x))+ g f Figure 8: A Pictorial Representation of f(g(x)) Example 11 Let f1x2 = 2x + 3 and g1x2 = x 2 + 1. Second we are using the value g(x) to compute f(g(x)). First. as in the previous example. that is.3. is called decomposition. y). We prefer the latter notation. and April Allen Materowski. the cost is a function of four variables. cost of steel. Thus. You substitute for each variable as it appears in the equation. Gordon. and u = g1x2 = 12x2 . capital and labor. that is. g1 . and y is the number of units of labor available. The definition extends is a natural way to any number of variables. written z = f1x. The definition of function extends in a very natural way. y2 = 100x4y4. as the next two examples illustrates. then u is to be the expression inside the parenthesis. Note that some texts write 1f g21x2 to mean f(g(x)).22 is not defined).1 (verify!). Computations with functions of two or more variables is handled the same way as with a function of a single variable. Wang. Function of Several Variables Breaking apart a composite function. Copyright © 2007 by Pearson Education.2 and . Solution. we may often need to deal with functions of more than a single variable. Applied Calculus for Business. we need only solve the equation x + 1 = 0 x + 2 whose solution is x = . We choose u to be the inner function.2 as it is the domain of g (that is. Economics. then we may write f1g1x22 = f1u2. by Warren B. thus. where x is the number of units of capital available.70 * ** Section 1. A simple calculation yields f1g1x22 = 12x2 .2 Basic Notions of Functions Example 12 Let f1x2 = 1/x and g1x2 = Solution.325. that is. Suppose we let u = g1x2. Example 14 1 3 Suppose the productivity z is given by the equation z = f1x.3210. determine the domain of f(g(x)). Published by Pearson Learning Solutions. x + 2 If we compute f(g(x)) we have that f1g1x22 = 1/1g1x22 = 1x + 22/1x + 12 however when does this make sense? First.3210. Also note that g(x) must not be 0 as 0 is not in the domain of f. we will make the choice that is most useful when such problems arise in the study of the calculus. x + 1 . productivity is a function of the two variables. fiberglass and rubber. verify this is correct. . we associate a unique value z. The cost of an automobile may depend on the labor. Decomposition Example 13 Consider the function defined by the equation y = 12x 2 . we choose u = g1x2 = 2x 2 . and Finance. the domain of the composite function f1g1x22 = 1x + 22/1x + 12 is all x except . To determine when g1x2 = 0. To each pair. we observe that x Z . There are many possible choices for the two functions. (x. find f(x) and g(x) so that f1g1x22 = 12x2 . Note that another possibility is to choose f1x2 = x 2. Inc. It follows that in realistic applications. Productivity may depend on the amount of available capital as well as the number of units of labor available. Walter O. Determine f(81.1. y2. if we view f1 2 = 1 210. Consider first the case of a function of two variables. 16).3210. 3 4 81 = 3. y. 3. z2 = xy .1.f1x. Gordon. 1Recall 814 = 2 1 3 1 4 16 B 3 = 23 = 8. as the next example illustrates. 22 = 1 . 22.12122 + 3132122 = 17.2 and that 164 = A 2 Example 15 Given the function defined by the equation w = f1x. Published by Pearson Learning Solutions.) Calculator Tips Applied Calculus for Business.xz + 3yz.1 .f1x. Example 16 Given z = f1x. However. 3. y + k2 . (To get the Ú symbol press . the vertical line key to the left of the 7 key. There is a key that allow us to insert the domain. y2 . y2 . Inc.12132 . y + k2 . by Warren B.f1x. y + k2 = x 2 + 2x1y + k2 + 1y + k22 = x 2 + 2xy + 2yk + y 2 + 2yk + k2 f1x. and April Allen Materowski. y2 1x 2 + 2xy + 2xk + y 2 + 2yk + k22 . Economics. assuming neither h nor k is zero.f1x. We define the function in Figure 9. (Period key) similarly for the . f181. (b) . Copyright © 2007 by Pearson Education. y2 = 1x + h22 + 21x + h2y + y 2 = x 2 + 2xh + h2 + 2xy + 2hy + y 2 f1x. we also want to include the domain. 162 = 1001812411624 = 100132182 = 2400.3.2 Basic Notions of Functions * ** 71 Solution. determine f1 . (a) f1x + h. Suppose you want to draw the graph of the function whose equation is y = f1x2 = 22x .1x 2 + 2xy + y 22 = k k 2 k 1 2 x + 2 y + k 2 2xk + 2yk + k = = 2x + 2y + k k k We remark that a function of two variables defines a three dimensional graph. and Finance.Section 1. y2 = x2 + 2xy + y 2 f1x + h. f1 .1. Wang. y2 f1x. y2 . y2 = x2 + 2xy + y 2. Of course. we realize the domain is x Ú 3/2. Solution. The idea of the difference quotient also can be generalized. . determine f1x + h. Walter O. y2 = h 1x2 + 2xh + h2 + 2xy + 2hy + y 22 . y2 = x2 + 2xy + y 2 f1x. (a) h k Solution. We proceed as follows: We go to the Y = window and define y1. 0.1x 2 + 2xy + y 22 = h h12x + h + 2y2 2xh + h2 + 2hy = = 2x + 2y + h h h (b) f1x. 3 We now press the graph key. . Published by Pearson Learning Solutions.2 Basic Notions of Functions Figure 9: Defining y1 = 22x .3 in the Default Window We can improve this graph by changing the window. Walter O. Copyright © 2007 by Pearson Education. Figure 10: y1 = 22x .72 * ** Section 1. and Finance. Figure 11: Changing the Window Applied Calculus for Business. Economics. see Figure 11. and April Allen Materowski. by Warren B. Gordon. Inc. Wang. and obtain Figure 10. y = f1x2 = c 2x .3. their product. Copyright © 2007 by Pearson Education. What happens if you try to compute y11y21 . we enter these as illustrated in Figure 13. Wang. Gordon. you press * Enter to obtain 1. if your are in the HOME screen (press the HOME button to get to this screen) and want to evaluate this function when x = 3. and April Allen Materowski. Consider the following example. . What happens if you try to compute y1(1)? Why? The composition of two (or more) functions is easy to do as well.73205. for example. Economics.3 -1 2 x .122? Piecewise functions can be drawn several ways. their compositions y1(y2(x)) and y2(y1(x)). Inc. Compute. and Finance. The calculator gives the answer as 23. Figure 12: Redrawn Graph of y1 = 22x . Applied Calculus for Business. Published by Pearson Learning Solutions.Section 1. Perhaps the easiest way is to define each piece separately. Consider the functions defined in Example 12. y11x2 # y21x2. (You can determine how many decimal places will be shown using the MODE button and then modify Display Digits).17 if if if x 1 -2 1 6 x 6 4 x Ú 4 The functions are input as indicated in Figure 14.3 It should be noted that calculations are easily done on the calculator with functions.2 Basic Notions of Functions * ** 73 The graph is redrawn in Figure 12. Walter O. As y1 = 22x . If you want a numerical approximaion. just enter one as y1(x) and the other as y2(x) in the Y = screen 1*F12. by Warren B. you need only enter y1(3) and press Enter. Figure 13 Returning to the HOME screen will allow us to do computations with these two functions. We remark that the calculator can easily perform most functional notation operations including the computation of the difference quotient. Inc. Published by Pearson Learning Solutions. Computations with functions of two variables can also be performed on your calculator. The Alpha key is needed to type in the and. We shall examine the concept of continuity more fully later in the next chapter. You must first press the MODE key and change FUNCTION to 3D (three dimensional) by scrolling down. and April Allen Materowski.2 Basic Notions of Functions Figure 14: Piecewise Defined Functions Note that the space bar key is located above the 1 . and Finance. see below. Figure 15: Graph of the Piecewise Function The graph looks like it has a jump at x = 4. Figure 16 Applied Calculus for Business. Economics. . by Warren B.74 * ** Section 1. Once this is done then the Y = screen becomes a Z = screen. We next press the graph key and obtain Figure 15. see Figure 16. because of the way it displays the graph by means of pixels being on or off. The calculator cannot always accurately portray the continuity of the graph. Gordon. but that is how it is displayed by the calculator.2 key on the bottom of the calculator. Copyright © 2007 by Pearson Education. it does not. Walter O. Wang. 92. and April Allen Materowski. (e) f1x + h2 x + 1 19. 14.2 Basic Notions of Functions * ** 75 Suppose z = f1x. W = 511. .2226 8. . 17. (e) f1x + h2 18. 1926 4. Exercise 8 17. see Figure 18. s = 513.12. 15. Given the function defined by the equation f1x2 = determine 2x . Figure 18 Remember to press MODE and change GRAPH back to FUNCTION when you are done. (e) f1x + h2 20.Section 1.1 determine (a) f(0). and Finance. Copyright © 2007 by Pearson Education. 926 2. we illustrate in Figure 17. 19. Given the function defined by the equation f1x2 = 3x . (b) g1 .2 In exercises 1 8 determine if the given set is a function. T = 5112. This is illustrated at the end of Section 3. 52. 12. Published by Pearson Learning Solutions. 13. 14. .1.5. 13. 1. 32.52. Exercise 3 12. 16). 17.12. 72. 32. 52. . 112. (c) f(1). f = 511. 22. 72. 52. 16. (b) f1 . Exercise 1 10. 1 . 52.52. g = 512. Walter O. and we want to evaluate f(81. 111.12. y2 = 100x 1/4y 3/4. Given the function defined by the equation g1x2 = 2x 2x + 1 determine (a) g(0).12. 14. (c) f(2) (d) f(2x). h = 513. .52. 13.126 7.3. 15.7. 15. Exercise 2 11. 22. 52. 12. 112. 13. 32. (c) f(3). Exercise 7 16. 526 5. 19.3 (a) f(0). Economics. 1126 3. (b) f1 . 1126 6. 16). Exercise 4 13. Exercise 6 15. (d) f(2x). 72.22.12. . 15. by Warren B. 119. . found in the Catalog.52. (c) g(3). V = 514. (e) g(2x) Applied Calculus for Business. Gordon. Wang. where we use z1 in place of z. (d) f(3x). 42. (d) g(8). Given the function defined by the equation f1x2 = 2x 2 . . Figure 17 We next return to the home screen and enter z1(81. Exercise 5 14. Inc. 17. 1 . 232. EXERCISE SET 1.12. The calculator computes the difference quotient using the expression avgRC. determine (a) f(0). . (b) f1 . . r = 512. 526 In exercise 9 16 determine the domain and range of the set given in 9. 9x In exercises 46 54 use the vertical line test to determine if the given graph may represent a function. f1x2 = x 41. f1x2 = 22x . f1x2 = 3x . 50 Figure 19: Ex. 33.4 40.5 34. w1x2 = 23x . and April Allen Materowski. g1x2 = 2x .2x2 + 4 37. 48 49. 51. g1x2 = 44. g1x2 = . f1x2 = 43. f1x2 = x2 . .2 Basic Notions of Functions f1x + h2 . f1x2 = x3 29.3 x + 1 2x + 5 5x + 3 6x2 .x . f1x2 = 2x2 + 3x .3 In exercises 42 45. f1x2 = 5x2 . 47 48. determine the domain of the function defined by the given equation. Figure 21: Ex. r1x2 = 22x + 1 39. h1x2 = 45. Walter O. f1x2 = x2 + 3x + 9 26. f1x2 = mx + b 24. f1x2 = 5x + 7 23. Inc. Published by Pearson Learning Solutions. h Whenever possible. 46 Figure 23: Ex.5x + 3 25. 51 Applied Calculus for Business. f1x2 = 4x + 1 35. 49 x3 . Gordon. 46. 32. In exercises 33 41 (a) determine the domain. f1x2 = 2x + 3 Hint: refer to Example 19 Section 5.1 Hint: refer to Example 19 Section 5. g1x2 = x + 5 36. by Warren B. simplify the expression so that the resulting expression is defined when h = 0.3 Figure 20: Ex.12 2x + 1 50. h Z 0.2 22. r1x2 = 3x + 2 2x . 42. f1x2 = 3x . Wang. Figure 22: Ex. Economics. 31. 2 Figure 21: Ex.76 * ** Section 1. Copyright © 2007 by Pearson Education. In Exercises 21 32 compute the difference quotient .4x + 11 28. f1x2 = x + 1 x .f1x2 47.7 27. f1x2 = 2/x 30. and Finance. (b) sketch the graph and (c) determine the range of the function defined by the given equation. h1x2 = 1x 38. 21. Sketch the graph of y = flr1x2 + x for . Wang. and Finance. 53 y = x = e 54. f1g1x22 = 2x2 . Sketch the graph of y = x + 2 . f1x2 = 22x + 3 g1x2 = x2 . 74. The Floor and CEILING functions are defined as follows: -x x x 6 0 x Ú 0 Figure 26: Ex. 3.5. the following function f1x2 = 4x + 3x + 2x + 1x. The fare for riding a cab is as follows: $1.Section 1. otherwise x if x. f1g1x22 = 12x8 .3 In exercises 83 86 determine two simple functions who composition is f(g(x)).g1x2. if x is an integer integer to x s left.3 60. $0.52.2 52. is an integer integer to x s right otherwise. 3. (a) flr(5).5. f1x2 = 2x + 1 69. f1x2 = x . (b) ceil1 . by composition. Copyright © 2007 by Pearson Education. assume there is no growth after age 19.52. determine (a) f1x2 + g1x2. Determine the domain of the composite function defined by y = g1f1x22. Economics.7x2 + 3x . (b) flr1 . f1x2 = 2x . The charge for the first hour or part is $80. f1g1x22 = 2 3 x 4 . What is the height at age (b) 9? (c) 15? In exercises 58 62.2 . (d) flr1 . Applied Calculus for Business.3 82. 63.3 79. (b) How much does it cost for a 10 mile ride? 56.3 x x 3.2 87. Ex.2x + 23218 84. Inc. Basic Notions of Functions * ** 77 In Exercises 63 69 determine (a) f(g(x)) and the domain of the composite function. (c) f(x)g(x). Sketch the graph of y = ceil1x2 + x for . f1x2 = 3 x + 2 g1x2 = 4x + 7 g1x2 = x 2 g1x2 = 2x . (d) ceil1 . (a) Determine C(d) an equation representing the cost C of the ride in terms of the distance d traveled. Gordon. From age 12 to age 19 growth is 2 inches per year. (a) ceil(5). 54 55.4 2 y = flr1x2 = e y = ceil1x2 = e x. Walter O. What is the charge for (b) 4 hours. Height is a function of age. Sketch the graph of y = flr1x2 for . 71. (a) Determine an equation which give the height h as a function of age a. 57. b. Sketch the graph of y = 2x .3 2 Figure 24. Let f1x2 = ax + b and g1x2 = cx + d determine the conditions on a. 83. (c) 7 hours. (c) ceil(5.5 67. f1x2 = 2x + 3 64. c and d if f1g1x22 = g1f1x22.2x + 19254 85. 81. and for each hour or part above 6 hours the hourly charge is $48. Published by Pearson Learning Solutions.3 86. 76.12 77.11x5 .75 for the first mile or part.3 . The absolute value function is defined as follows: Figure 25.2 59. f1x2 = 3x + 7 61. 52 53. f1x2 = 3x . Use these functions to solve exercises 76 81. f1g1x22 = 13x2 . 80. f1x2 = x2 + 5 62. 66. The charge for each additional hour or part up to 6 hours is $55.50 for each additional mile or part. (c) flr(5.3 x + 1 g1x2 = 2 4x + 2 g1x2 = 2 x . Sketch the graph of y = ceil1x2 for . Sketch the graph of y = x . f1x2 = x + 3 x + 2 65. x x 3.12 78.1). Show how to build.1 g1x2 = 70. (b) f1x2 . The Elite Limo company charges by the hour for use of its cars. by Warren B. Ex. f1x2 = 2x . 73.3 2 g1x2 = x 2 + 3 g1x2 = 2x . (d) f(x)/g(x) when defined. . Sketch the graph of y = x . Determine a relationship between flr(x) and ceil(x). and April Allen Materowski. Suppose at birth the average male is 16 inches tall and grows 3 inches per year until age 12. 58. 75. (b) g(f(x)) and the domain of the composite function.1).7 3 x2 . 72.1 g1x2 = 12/x g1x2 = g1x2 = x + 5 2 x . f1x2 = 3x + 7 68. (a) Determine C(h) an equation represent the cost C of the ride in terms of the hours h. f12. z + l2 . discuss what effect (if any) the overland and under water costs have on the determination of x.f1x. Given f1x.3 Applications of Linear Functions neither h nor k is zero. y2 .f12. by Warren B. Walter O. D.12. y. y2 = 8x3y2 determine (a) f (3. y2 . y + k. A power station is on one side of a straight river which is five miles wide. in this section we consider linear functions in some detail. 22. y2 = x2 . Applied Calculus for Business. then the total monthly cost. determine Fig. 5). y. if the total cost of the power line is to be as small as possible (a sketch of the grap or a table may be useful) h f12 + h. Suppose further that the cost of producing each of a particular item is m dollars. y. Thus. determine (c) f1x + h.2. (g) In (c) and h k (e) what happens if you allow h to equal 0 at the end of your calculations? (h) Same question for (d) and (f) is you allow k to equal 0 at the end of your calculations. z2 h f1x. determine (a) f11. 1. 20 miles up-river. and the y-intercept is the overhead. y2 h .) 89. y + k2 .f1x. y. If it costs $80 per mile to run the line under water and $50 per mile to run it overland. zero.22 if neither h. (d) f1x. 90. y2 . k. 12 . (f) . To start with.f1x. Suppose that a manufacturer has fixed overhead costs of D dollars per month.f1x. . Gordon. determine (a) f12. h k f12 + h. 27). (b) f1 . 1 + k2 . y2 k f1x. Referring to the previous two problems. Given f1x.f1x.3 Applications of Linear Functions » » » » Break-Even Analysis Depreciation Piecewise Linear Graphs Calculator Tips In the previous section we defined and examined the notion of a function. Economics. l is zero. (d) . z2 .3y 3z2 + z2. 88 (This problem and the next will be re-examined when we learn more aboutcalculus. is given by the rule Cost = Overhead + 1Cost per item2 # 1Number of items2.f1x.2xy2 . 1. (e) f1x + h. y. Inc.f1x. The slope of the line y = C1x2 is the cost per item m (also known as the marginal cost). Many real processes are modeled by linear functions. Given f1x. Wang. and a factory is on the other side. y + k2 . A power line is to be run from the power station. (b) f(2. . Suppose the power station in the previous exercise is moved one mile inland find the point A at which the power line enters the river.f1x. In symbols. z2 k . P 5 miles x A 20 .78 * ** Section 1. . y2 . We will look at some examples from business. . z2 = x 2 . (g) In (c) and (e) what happens if you allow h to k equal 0 at the end of your calculations? (h) Same question for (d) and (f) if you allow k to equal 0 at the end of your calculations.f1x. z2 l . 32 (e) . 27: Ex. Published by Pearson Learning Solutions.x F |--------------20 miles ------------------| 92. and Finance. . which is x miles upriver from P. the point on the other side of the river where the power line comes out of the river. z2 . 91. under the river to some point A. Copyright © 2007 by Pearson Education. and April Allen Materowski. if (c) (e) f1x + h.1. and then over land to the factory (see Fig. .3y3. y. 3 + k2 . y2 93. estimate x. C(x) is a linear function of x. If the assembly line produces x items per month. this is C1x2 = D + mx. If neither h nor k if f1x. 88. C(x). 32 (b) f10. the simplest description of production costs is a linear one.2xy 2 . determine (c) (d) (f) f1x. y2 . y2 f1x. y2 . 2). and the y-intercept is zero. is the break-even point. is given by Revenue = 1Price per item2 # 1Number of items2. D/1p . by Warren B.. the slope is p . the revenue line would have a higher slope than the cost line but starts at the origin for zero sales. the x-intercept of the line. Walter O.m2. revenue function. R(x). BBZ Wineries has a fixed monthly overhead of $2000 and a cost of $20 per bottle to make and bottle wine. which is revenue minus cost. P(x) is negative (i. p . Wang. To the left of this value. The profit function is also linear.m2x . you get nothing.m and the y-intercept is . Alternatively. the manufacturer is selling things for less than it costs to make them and will surely never make a profit! The x-value for which P1x2 = 0. otherwise. This time. you could simply consider the profit.m)x . See Figure 1(a).m should be positive.1D + mx2 = px . and write P1x2 = R1x2 . Break-Even Analysis y = R or C R(x) = px y=P P(x) =(p . the slope is the selling price per item p. Published by Pearson Learning Solutions. suppose that the manufacturer sells each item produced for p dollars per item. R1x2 = px Thus. (b) Find the profit function. that is. That is.D C(x) = mx + D break even point x break even point x Figure 1(a): Linear Revenue and Cost Here is a simple numerical example. (a) Find the cost function.C1x2 = px . and Finance. ( You sell nothing. and break-even point. In this case. See Figure 1(b). We shall use P(x) to denote profit (note upper case P represents profit. Copyright © 2007 by Pearson Education. and April Allen Materowski. the revenue.D . The wine sells for $30 per bottle. Figure 1(b): Linear Profit Example 1. lower case p represents price).D. Economics. Its x-coordinate is the number of sales for which revenue would equal cost. is also a linear function of x. Inc. R(x).e. (c) How large must sales be in order to achieve a profit of $1500 per month? Applied Calculus for Business. In such a case. Notice that the slope. and to the right. a loss). The total revenue to the producer.D. . ) The selling price per item p should be higher than the cost of producing it. The point at which revenue equals cost is called the break-even point. Gordon.e..mx = 1p . a profit).Section 1.3 Applications of Linear Functions * ** 79 Now. P(x) is positive (i. revenue will eventually match and finally exceed cost. As x increases. Depreciation Another common business situation where a linear function arises is in modeling the process of depreciation of assets. Suppose that a company or individual invests in a piece of equipment. only to have it drop to value zero in the year in Applied Calculus for Business. Walter O. Published by Pearson Learning Solutions. building. . the winery shows a monthly profit of $1500.12000 + 20x2 = 30x . and p = 30. The graph of this situation is shown in Figure 2(b). on sales of 350 bottles. R12002 = C12002 = $6000. (a)Let x = the number of bottles produced per month. Wang. and April Allen Materowski.3 Applications of Linear Functions Solution.80 * ** Section 1. that is.2000 (c) (The break-even point can be determined (again) by setting P(x) equal to zero. C(x) = 2000 + 20x R(x) = 30x P(x) = 10x . when 30x = 2000 + 20x 10x = 2000 x = 200. This gives x = 2000/10 = 200. when sales are 200 bottles per month. the winery breaks even (See Figure 2(a)). The cost equation is C1x2 = 2000 + 20x The revenue equation is R1x2 = 30x (In the general notation. D = 2000. or other tangible asset that does not have an unlimited life expectancy. as above.) We want P1x2 = 1500.2000 . Thus.) The two lines intersect when R1x2 = C1x2. and Finance.20x = 10x . Copyright © 2007 by Pearson Education. Gordon.2000 Figure 2(a): Linear Revenue and Cost (b) The profit function is Figure 2(b) Linear Profit P1x2 = 30x . Inc. Economics. It is not reasonable to carry that asset on your books at value V year after year. Let the value of that item when new be V. so we solve 1500 = 10x .2000 3500 = 10x x = 350 That is. That is. m = 20. by Warren B. 2000). For example. Walter O. Notice also that the slope is negative since the value is decreasing. Letting t be time measured in years.10. called linear or straight-line depreciation. The slope of the line must be m 2000 . suppose that a taxicab driver values his cab at $40. of which only a few are acceptable to the business community and the Internal Revenue Service. It is perfectly natural for the cab driver to assume that there will be some salvage value to the car at the time it is disposed of.000. the rate at which the taxicab is depreciating to zero is the total change in value divided by the length of time. Table 1: Depreciation of a $40. the value of the item plotted versus time will have a constant negative slope .000 2 20.000 when new.000 3 10. In other words. we know that the line passes through the two points (0.V/T and y-intercept V: A1t2 = 1 .000.40.V/T2t + V Of course.Section 1. Therefore. or .0 5 and since we know the y-intercept is 13. Instead. you normally write off the value of the item by some systematic plan over the item s expected lifetime. Applied Calculus for Business.2200 5 . Suppose also that he anticipates a usable life of 4 years. Inc. the value of the cab will drop to zero in T = 4 years. it is not necessary that an asset be depreciated to zero. it must lose 1/4 of its original value each year.000 (that is 1 4 of $40. That is. The simplest way is to assume that the asset depreciates by a fixed amount each year. Gordon. T. the salvage value of the car will be 2. A(t)) lie along the straight line with slope . by Warren B. The equation is A1t2 = . Economics.000/4. and Finance. we are using linear depreciation. Example 2. . There are several ways this can be done. Published by Pearson Learning Solutions. Ms. Thus.000 4 0 It is not difficult to see that these points (t. Solution. the car is worth 13.000 Taxi over Four Years t A(t) 0 40. we have Table 1. at which time he will junk the cab. Winslow buys a new car for her business for $13.13000 .000t + 40. because the asset (car) loses value over time. Consider the following example.000) each year.000.2200t + 13000. Notice.000 and at time t = 5. and A(t) be the value of the cab at time t. Copyright © 2007 by Pearson Education. A1t2 = .000. In general. Since at time 0. it must depreciate $10. She expects to use the car for five years and then sell it to a used car dealer for $2.11000 = = . Find the function A(t) that describes the value of the car after t years 10 t 52 if she uses straight line depreciation. for this method. Notice again that the slope is the rate of change of value and it is negative. and April Allen Materowski.3 Applications of Linear Functions * ** 81 which it finally wears out.10000 and y-intercept 40. we may interpret the slope as the rate of depreciation.13000) and (5.000 1 30.000.000. Wang. If the cab is assumed to depreciate a constant fixed amount each year and drop to zero after 4 years. the graph is a straight line passing through (3. What about the transition points x = 1 and x = 3? At x = 1? In this case. In addition.1. Applied Calculus for Business. by Warren B. It is easy to calculate that the slope is . for x 6 1. The graph of a function can be distinguished by the fact that every vertical line intersects the graph in at most one point. the straight-line depreciation relationships are true only for integer values of x and the graphs really consist of isolated dots rather than continuous straight lines. Wang. We see that to the left of x = 1. f1x2 = 2x. So the equation of this line is just y = 2x. Finally.82 * ** Section 1. if x is to the right of 3. If our answer for a break-even point had not turned out to be a whole number. we can figure out an algebraic equation represented by the graph and express f(x) that way. Graphs such as this are called piecewise linear . We express the function in algebraic form. this graph can be described as: f1x2 = c 2x 2 -x + 5 1 if x if 1 6 x if x 7 3 3 A good example of a piecewise linear graph is offered by an Income Tax Form. we would have been forced to decide precisely which was intended. since both give the same y-value.3 Applications of Linear Functions Piecewise Linear Graphs Actually. the next larger integer value would give the first profit. y x Figure 3: A Piecewise Linear Graph The transition point x = 3 is handled similarly. however. Given such a graph. since every x must define exactly one y. that is easy. and using the point-slope formula. Thus. So we can pick either. we shall continue to let x be any real number. the graph is a straight line that passes through (0. if we are lucky. Copyright © 2007 by Pearson Education.0). Next. then we could make the following natural interpretation: that is. its slope is 2 and its y-intercept is 0.0) and (1.) At any rate. For given straight lines. . and April Allen Materowski. For instance when the x-values are not expressed in units but in hundreds or thousands of units as is common in most real business problems. (If the two rules gave different values at such a point. it is certainly the graph of a function (by the vertical line criterion). For simplicity. we see that between x = 1 and x = 3. Economics. Published by Pearson Learning Solutions. Walter O.x + 5 as its equation. the graph is a horizontal line with height y = 2. Inc. and Finance. consider the graph in Figure 3. we get y = . 2. for 1 6 x 6 3. f1x2 = 2. So.2). By inspection.2) and (5. Therefore. However. there are cases in which fractions make perfect sense. Gordon. Remember the vertical line criterion for functions. it does not matter whether you use the rule y = 2x or y = 2. 10x.800 $143.Section 1.688002 + 14010 if 0 6 x 7000 if 7000 6 x 28400 if 28400 6 x 68800 if 68800 6 x 143500 We show how to use your calculator to sketch the graph of this function below.400 $68.000 $28.000.000 $28.284002 + 3910 0.500. then the tax is 700 plus 15% of 1x .7. the rule gives the same value for t. The graph shown for T(x) in the previous example may be drawn with your calculator by entering each of the pieces of the function separately on the Y = screen. Wang.4002 + 3. Copyright © 2007 by Pearson Education. to get the Ú symbol you press and then the # (period) key.800. Gordon. and for x between 68.010 10% 15% 25% 28% $0 $7. Calculator Tips Figure 4(a): Entering the Function From Example 3 Applied Calculus for Business.500 This Amount Plus This % Of the Excess Over $0 $700 $3. find t = T1x2 for the domain 0 6 x 143.400 and 68.010. Proceeding similarly. .151x .2 key and you must press Alpha and then this key.3 Applications of Linear Functions * ** 83 Example 3. Inc. If x 7.281x .400 $68. the space bar is located above the 1 .910. then t = 0.251x . Thus. Note that at the splitting point. In short The graph of T(x) is shown in Figure 4. and April Allen Materowski.000 $28.8002 + 14.251x . The condition (in this case the interval of definition) is inserted using the vertical bar key to the left of the number 7 key. Solution.910 $14.0002 + 700. Walter O. and Finance. we have t = 0.800 and 143.68. Table 2 is a portion of this schedule.500.10x 0. for x between 28. 0.800 But Not Over $7.800 The Tax is Letting x be your taxable income and t be your tax.400 $68.000 and 28. Economics. t = 0.28. Published by Pearson Learning Solutions. If x is between 7. by Warren B. The 2003 Schedule X is used by single taxpayers to determine their tax. To get the symbol you press and then the 0 key. similarly.281x .400. Some remarks: first.0002.151x .7. we have t = 0.70002 + 700 T1x2 = d 0. We illustrate in Figures 4(a) and 4(b). Table 2: Portion of Schedule X Schedule X If Taxable Income Is THEN Is Over Single $0 $7. 500. that means there is not enough room to show the rest of the line on the screen and you need to scroll (use the arrow keys) to see it. The calculator shows a slight bend at these x-values. by Warren B. so we need to set an appropriate window. 2. in this window we can press F2 (Zoom) and we obtain different zoom options. and selling price of $45. Now we know we are working with larger numbers here. what is the new break-even point? Applied Calculus for Business. to see them more clearly would require you to zoom in near each transition point. y3 and y4. zooming in at these points will better show the transition (bend) from one piece of the graph to the next. One possible window is given in Figure 4(b). Walter O. Published by Pearson Learning Solutions. Wang. (Note that the first switch in pieces occurs when x = 7000. the first bend is difficult to see in the present window as it has a relatively smaller y-value. Rework Example 1 for fixed overhead of $2. the shifts from one piece of the graph to another is not easily seen. Gordon. If his overhead remains $2. How does this affect the break-even point? 3.400 and the third at x = 68800. and Finance. Figure 4(b): A Window to View T(x) Using this window. We leave it to you to experiment with these options. Figure 4(c): Sketch of T(x) EXERCISE SET 1. Since this graph has such a large range of values. Copyright © 2007 by Pearson Education.3 Applications of Linear Functions Notice the black arrow to the right of y2. . the second at x = 28. cost per bottle of $30.200.500 and cost per bottle remains $30. Suppose that the manufacturer in Exercise 1 is forced to cut the selling price to $35 in order to clear inventories. Suppose that the manufacturer in Exercise 1 can manage to reduce the overhead to $1. In most cases. The default window will not be accurate. Again. Economics. Alternately. and April Allen Materowski. we have the sketch in Figure 4(c). you choose the portion of the graph you want to zoom in on by moving the cursor to the left and pressing enter and then move to the right and press enter.84 * ** Section 1. Inc.3 1. A house and lot valued at $100. How many points of intersection do you see? What are their coordinates? 20. f112. (a) Sketch the graph of the line y = 2x . Since only the value of the house depreciates. Schedule Y-1. (b) Plot cost and revenue on the same axes and locate the break-even point. Wang.Section 1.22. Consider the function given by . 1) (3.2y + 2 = 0. f102. 22. Economics. Consider the function given by f1x2 = e x + 3 -x + 3 if x 0 if 0 6 x Find f1 . f(1). If each six-pack sells for $1. 9.2y + 1 = 0 on the same axes as the graph of y = f1x2 from Exercise 18.20.52. Applications of Linear Functions * ** 85 (-1. 8. to be used by married tax payers in 2003 is given in Table 2.000 sixpacks of cola. 2) (-1.22.000 and it is assumed to have a scrap value of $100. f(2).) 13. After 10 years the book value of the asset is $64.000. f142. Gordon. At a selling price of $50. f(0).. -1) Figure 5 (c) 15. f11*22. Copyright © 2007 by Pearson Education. How many points of intersection do you see? What are their coordinates? (b) Sketch the graph of the line x .3 4. and f(5). and f(4) Applied Calculus for Business. a lamp company breaks even on total sales of $3000. Plot the profit function.2) + + + (3. find and graph the relation. Plot the graph of y 17. find the marginal production cost per lamp and find the profit on sales of 130 lamps. and expects to sell at least 100 items.12.000 to prepare 15.000. Plot the cost graph. and April Allen Materowski. Plot the graph of y = f1x2.500 when the production is 10. Find f1 . Find f1 . by Warren B. Assuming that cost is a linear function of number of items. Walter O. what should he set the selling price to be? 6.000. and f(3). f(1. A manufacturer of vases discovers that it sells 1800 vases at $12 per vase but only 1300 at $15 per vase. and $8. Published by Pearson Learning Solutions.000 flashlights.000. If the manufacturer of Exercise 4 wishes to guarantee profits of at least $600. and f(7).000 profit? 10. 1) (2. Sketch the graph of the line x . Plot the graph of y = f1x2 18. How must the price be set to make possible a $90. If a manufacturer has fixed costs of $700. 2) (1.22. Consider the function given by 2 f1x2 = c x + 2 6 = f1x2. (3. Figure 5(a) shows the graph of a piecewise linear function f. A small bottling company finds that it costs $6.5). Find the algebraic formulation of f(x). At what time will the machine be worth $5000 according to this model? 12. A manufacturer of small glass figurines discovers that it costs $1. if x 0 if 0 6 x if 4 6 x 4 Find f1 . Figure 5(b) shows the graph of a piecewise linear function f. Assuming that demand is linearly related to price. 2) ° + (4. (Ignore inflation. A machine worth $10. If the company s overhead is $1200. find the assumed value of the land. 1) (1. Wong Industries is depreciating the value of its machinery over a thirty-year life. How many points of intersection do you see? What are their coordinates? 21. Find the algebraic formulation of f(x). Figure 5(c) shows the graph of a piecewise linear function f. 11. .800 for a production run of 1000 and $2200 for a production run of 1500. on the same axes as the graph of y = f1x2 from Exercise 17. f(2). find the overhead and the marginal cost of a figurine.000 new and having a scrap value of $500 is to be depreciated over a ten-year life. Find the algebraic formulation of f(x). how should he set the selling price to guarantee breaking even? 5.500 on a production of 6000 flashlights and a profit of $10. and Finance.000 six-packs.000 to prepare 10.1) + (2. Find a piecewise linear function that describes the broken-straight-line depreciation that the company is using. Inc. and f(5). f102. Find f1 .3 on the same axes as the graph of y = f1x2 from Exercise 16.2x + 1 2x .4) + (1.000 is being depreciated over 25 years by the straight line method. it is decided that the machinery still has another twenty years of life left but it will then be worth nothing.25 apiece. After 20 years. (a) If flashlights sell for $2. The Amalgamated Flashlight Company shows a profit of $4. f(3) and f(5). 19. (c) Suppose that the capacity of the flashlight assembly line is 30. -1) ° Figure 5 (a) Figure 5 (b) 14. 2) (1.-5) (0. find and plot the graphs of the total cost function and the revenue function. find the cost equation for flashlight production.3 -x + 6 if x 1 if 1 6 x if 3 6 x f1x2 = c 3 Find f(0). 16. Its original value was $1. 7. Find the function that describes straight line depreciation for this situation. a cost per item for production of $20. Find the break-even point. 001 1x = 20. It turns out that once we understand how two special cases appear. Your tax is 125216. on 26. We charge 8¢ per copy for 100 copies or less. and 4¢ per copy for all over 1000. Economics.400. we examine the special case y = x2. find t = T1x2 for the domain 0 6 x 311. that is.000 earnings.000. equal and opposite x-values have the same y-value. Express the tax.999 1x = 19. therefore this graph is symmetric with respect to the y-axis.0012? 1.800 $114. then we almost know how any other parabola will appear.000 $56. you may subtract 1000.000. So.389. .700 $311. the portion of the graph on each side of the axis is a mirror image of the other. the y-value is 0. Wang.000 $56. Similarly. and is positive for any other x-value. The graph of this function is called a parabola. you may subtract 2000.9992 or $20.000. for example. given the quadratic equation ax2 + bx + c = 0.700 $311.096. Inc. Published by Pearson Learning Solutions.820.50 per thousand dollars of earnings. In particular.000. if you earn under $10. Write C(x) algebraically and plot its graph.650 $174.00 $22. Gordon.000 and 20.00 Plus This % 10% 15% 25% 28% 33% 35% Of the Excess Over $0 $14.650 $174. Would you rather earn $19. 24.4 Quadratic Functions Parabolas » » » » » » » » Scaling Vertical Translation Axis of a Parabola Horizontal Translation Locating the Vertex Graphing a Parabola in the form y * ax 2 + bx + c Applications to Optimization Calculator Tips After studying the linear equation ax + b = 0. Copyright © 2007 by Pearson Education. Walter O. and April Allen Materowski. it is natural to next consider the quadratic function defined by y = f1x2 = ax2 + bx + c.282. and Finance. Plot the graph. Let x be the number of copies ordered and C(x) be the cost of the job. A local photocopying store advertises as follows. if between 10. and then its applications. Applied Calculus for Business.50.800 $114.00 $7.52 = $162. Metro City s Non-Resident Earnings tax is $6.800 $114. However.000 $56.4 Quadratic Functions Parabolas Table 2: Schedule Y1 Schedule Y1 If Taxable Income Is THEN Is Over Married Filing Jointly or Qualifying Widow(er) $0 $14. Moreover. and then we plot these points to obtain the graph of y = f1x2 = x 2.950 The Tax is Letting x be your taxable income and t be your tax. by Warren B. as a function of x in thousands.950 This Amount $0 $1. you may subtract $3000 from your earnings. the y-axis behaves like a mirror.650 $174. where a Z 0.50 $39. 23. and if between 20. T(x).950 and plot its graph. the next step was to examine the linear function defined by y = f1x2 = ax + b.86 * ** Section 1.700 $311. We observe that when x = 0.50 $84. you may subtract 1000 and pay tax on 25. In Table 1 we choose x-values and compute the corresponding y-value.000 and 30. 6¢ per copy for each copy over 100 but not over 1000.950 But Not Over $14. and April Allen Materowski.4 .1022 = 0 . Observe that the parabola is a U shaped graph. Inc. What happens to the shape of the graph if we change the sign of the coefficient of the x2 term? In the previous example its coefficient was positive. As our next example. 92 1 .92 Applied Calculus for Business. that is.x2 y y y y y y y = = = = = = = . Changing the sign will.1 .3. 0). as we shall see change the way the graph opens. Gordon. . .12 (0. y = x2 Figure 1: The Graph of y = f1x2 = x2 Table 2: Points used to Plot y = .1 . . The lowest point on the graph.122 = . Economics.122 = 1 1022 = 0 1122 = 1 1222 = 4 1322 = 9 2 point on graph 1 . 0) (1.1 .9 2 point on graph 1 .222 = 4 1 .42 13. 0) 11. . 42 1 . The graph is symmetric with respect to the y-axis.1.2.92 . Table 2 gives the required points.x 2 x-value -3 -2 -1 0 1 2 3 y = . Published by Pearson Learning Solutions. (0.4 Quadratic Functions Parabolas * ** 87 Table 1: Points used to Plot the Graph of y = f1x2 = x 2 x-value -3 y = f1x2 = x 2 y y y y y y y = = = = = = = 1 .12 12.1222 = .1122 = .4 . .3. resulting in the graph opening upward. 1) (2.32 = . . 12 (0. Wang. 4) (3. Copyright © 2007 by Pearson Education.2.x 2.222 = .1 . the line x = 0. 9) -2 -1 0 1 2 3 The graph is plotted using these point in Figure 1.9 . and Finance. 42 1 .Section 1.1322 = . Walter O.1 . by Warren B. we consider the parabola y = .1. its turning point or vertex is the origin.32 = 9 1 . in this case opening upward.1 . x 2 and y = x2 are identical. In fact. Its turning point. symmetric with respect to the y-axis . Table 3. Note that this graph looks very much like the graph of y = x 2 except it rises faster.2. 22 (0. for the same x-value its has twice the y-value. Gordon. We have seen the effect of multiplying x 2 term by a constant. Notice that the graph of the equation y = . 8) (3. 182 1 . Table 3: Points used to Plot y = 2x 2 x-value -3 -2 y = 2x 2 y y y y y y y = = = = = = = 21 . 0) (1. Notice that the graph of y = 2x 2 is narrower.1. for the same x-value. . The reason is that for the same x-value. Thus.x2 is also a U shaped graph opening downward.122 = 2 21022 = 0 21122 = 2 21222 = 8 21322 = 18 2 point on graph 1 . its y-value is doubled. Economics. to better see that. is the highest point on the graph. by Warren B. y = f1x2 = ax2 + bx + c will have an upward shaped appearance U when a 7 0. Wang. and April Allen Materowski.32 = 18 21 .222 = 8 21 . In fact the graphs of y = . we see multiplying the coefficient of x2 by a constant does nothing more than scale the y-value. gives the points needed to plot this curve. Inc. 82 1 .4 Quadratic Functions Parabolas The graph is given in Figure 2. Published by Pearson Learning Solutions. a vertical translation. Scaling Applied Calculus for Business. Walter O. Copyright © 2007 by Pearson Education.88 * ** Section 1. Consider the graph y = 2x2.the line x = 0. if we turn one of them upside down we obtain the other. 18) -1 0 1 2 3 The graph is given in Figure 3. How does this graph differ from y = x 2? The answer is simple. and Finance. that is. 2) (2. y = -x2 Figure 2: The Graph y = . That is.x2 We shall see soon that all parabolas will have one of the above shapes. the y-value on one curve is the negative of the other. and a downward shaped one when a 6 0. The effect of adding a constant to y = x 2 has the effect as moving the curve vertically. or vertex. we plot them both on the same coordinate system in Figure 4.3. 1. Economics. . and Finance. Walter O. 22 (0. 102 1 .4 Quadratic Functions Parabolas * ** 89 y = 2x2 Figure 3: The Graph y = 2x2 y = 2 x2 y = x2 Figure 4: y = x2 and y = 2x2 Consider the graph of y = x2 + 1. Gordon.32 + 1 = 10 1 .2. and April Allen Materowski.122 + 1 = 5 1022 + 1 = 1 1122 + 1 = 2 1222 + 1 = 5 1322 + 1 = 10 2 point on graph 1 .222 + 1 = 5 1 . We indicate in Table 5 the points used to sketch its graph which is given in Figure 5. 1) (1. Copyright © 2007 by Pearson Education. 10) Applied Calculus for Business.Section 1. Published by Pearson Learning Solutions. by Warren B. Inc.3. Table 4: Points used to Plot y = x 2 + 1 x-value -3 -2 -1 0 1 2 3 y = x2 + 1 y y y y y y y = = = = = = = 1 . 52 1 . 2) (2. 5) (3. Wang. Walter O. by Warren B. . It is convenient to examine y = 1x . in general. where the graphs are plotted together.90 * ** Section 1. and Finance. Copyright © 2007 by Pearson Education.4 Quadratic Functions Parabolas y = x2 + 1 Figure 5: y = x2 + 1 Note that in comparing the graphs of y = x2 + 1 and y = x2. Consider the quadratic function defined by y = x 2 . Economics. this graph of this function only differs from the graph of y = x 2 . they are identical. without any change in the parabola s shape. we see that for the same x-values. The only remaining term we need consider is the x-term in the quadratic function. and April Allen Materowski. that is.2x + 1 = 1x .122. we can see what effect the x-term has on the quadratic. by studying either. but the first one is one unit above the other. As indicated y = x2 + 1 y = x2 Vertical Translation 2 2 Figure 6: The Graphs of y = x2 + 1 and y = x2 above. we see. Published by Pearson Learning Solutions. Wang.2x by one unit. Inc. that the c term in the quadratic function y = ax2 + bx + c moves the graph vertically upward or downward depending on c being positive or negative. Thus. This is most easily seen in Figure 6. Thus. Gordon. the y-value of the first graph is one unit above the y-value of the second graph.122 since that graph s y values are most easily obtained by the substitution of its Applied Calculus for Business. that is y = ax 2 + bx and y = ax2 + bx + c are identically shaped but only differ in their vertical position by c units. 3 . Walter O. 9) We plot this graph in Figure 7. This is most easily observed if both graphs are plotted together as in Figure 8.3. 42 (0.1.122 = 9 1 . 0) (2. Economics.122 = 0 12 . 1) (3.h22 + k.12 = 16 1 .122 = 1 11 . y = (x .122 y y y y y y y y = = = = = = = = 1 . we shall consider other means.122 = 9 2 point on graph 1 . Wang.4 Quadratic Functions Parabolas * ** 91 x-values.1 .2 .122 = 4 14 . and Finance. as considered below. Axis of a Parabola Horizontal Translation Applied Calculus for Business. by Warren B. Published by Pearson Learning Solutions. . If we move the graph of y = x 2 one unit to the right. that is. this form is not always the one presented.122 Observe that this graph is still symmetric with respect to the vertical line drawn through the vertex. Table 5 contains the points we need to draw its graph. 92 1 . The vertical line passing through the vertex is often called the axis of the parabola or its axis of symmetry. Thus. therefore.122 = 4 10 . and April Allen Materowski. the most convenient form of the parabola for plotting purposes is the form y = a1x . In fact. 1) (1. 162 1 . Summarizing. except now the vertex is at the point (1. Notice that this graph is identical to the graph y = x 2 except it is shifted one unit to the right. 4) (4. we have that quadratic function defined by y = ax 2 + bx + c. the x term in the quadratic function is the term that moves the graph horizontally.2. Gordon.Section 1. However. The b term effects the horizontal position of the vertex and the c term its vertical position.122 = 1 13 . causes a horizontal translation. Copyright © 2007 by Pearson Education.122. Table 5: Points used to Plot y = 1x . 0) so the graph is symmetric with respect to the line x = 1. with a Z 0 is always a U shaped graph. it coincides with the graph of y = 1x .122 x-value -3 -2 -1 0 1 2 3 4 y = 1x .1)2 x=1 Figure 7: The graph of y = 1x . Inc. it opens upward if a 7 0 and downward if a 6 0. 21x . The vertex occurs when x = h. The x-coordinate of the vertex is x = 3.42 16. and then choose a few points to its left and right to obtain enough points to reasonably plot the graph. Table 6: Points used to Plot y = . in either case the point at which x = h is the vertex. Walter O.214 . 2) 15. all other y-values will be beneath it.216 32 322 322 322 322 322 322 2 point on graph = = = = = = = . . This is done in Table 6. plot the points. and Finance.14 + + + + + + + 4 4 4 4 4 4 4 10. Solution. We begin with this x-value and choose several values to its left and right.h22 + k. or the highest point on a parabola opening downward. .4 Quadratic Functions Parabolas y = x2 y = (x . which means that we should first locate the vertex of the parabola.213 .42 (2. compute their corresponding y-values.322 + 4 x-value 0 1 2 3 4 5 6 y = . we get the graph as given in Figure 9. Thus. by Warren B. . We observed that the vertex is always the lowest point on a parabola opening upward.142 Plotting these points. if a 7 0 then all other yvalues will be above this one or if a 6 0. Inc. Copyright © 2007 by Pearson Education. The question that remains is. since at this x-value the parenthesis term is zero.1)2 Figure 8: The Graphs of y = 1x . as was done above.212 .122 and y = x2 We observed that when the parabola is given in the form y = a1x . 4) (4. .92 * ** Section 1.142 11.210 .211 .21x . Gordon.21x . Published by Pearson Learning Solutions. We choose several x-values to its left and right and compute their y-values. how can we most easily draw the graph of any given parabola? A good representation of the graph should indicate its U shape. and April Allen Materowski. and then draw the graph. Consider the parabola Applied Calculus for Business.322 + 4. Economics. .215 .322 + 4 y y y y y y y = = = = = = = . it is a simple matter to choose the appropriate points needed to plot the graph. Wang.14 -4 2 4 2 -4 . Example 1 Sketch the graph of the parabola y = . 2) (3. We know even before plotting its graph that it opens upward (why?).3)2 + 4 Figure 9: The Graph of y = . Inc. The procedure by which we locate the vertex is nothing more than a variation of the method of completion of the square. Copyright © 2007 by Pearson Education. and Finance.4 Quadratic Functions Parabolas * ** 93 x=3 y = -2(x . We remarked above that it is an easy matter to plot the parabola when it is written in the form y = 1x + 222 + 1.5 + 4 = x 2 + 4x + 4 or y . remembering that whatever we do to one side of an equation. We first isolate the x-terms and write y . we could construct a table of values including the x-value . so we have y . and choose a few values to the left and right.322 + 4 y = x2 + 4x + 5. if we were to plot this parabola we could begin our table as indicated in Table 7. we must do to the other.2. Thus.5 = x2 + 4x We next complete the square on the right hand side. Walter O. that if we knew the x-coordinate of the vertex. the vertex of the parabola is at the point V1 . our objective is to plot the graph using the original form of the equation. by Warren B. However. This means.2. and at this value. . 12. This will happen when x = . Economics. Thus. 1 2 142 = 2 and 22 = 4. Therefore. Table 7: Constructing a Table of Values x-value -5 -4 -3 x-coordinate at vertex : .2 -1 0 1 y = x 2 + 4x + 5 point on graph Applied Calculus for Business.1 = 1x + 222 or y = 1x + 222 + 1 Now observe that the lowest possible value for y will occur when the term in the parenthesis is zero. Gordon. we shall search for its vertex which is its lowest point. Published by Pearson Learning Solutions.Section 1. y = 1. Wang.21x . and April Allen Materowski.2. c/a = x2 + b/ax 1b b b 2 b2 We now complete the square. including the vertex as well as three x-values larger and smaller than this x-value. (6) Draw and label the axis of the parabola . (2) Make a table of x-values. . (4) Plot each of the seven points determined from the above steps.2b a. Wang. where a Z 0. = .94 * ** Section 1. Graphing a Parabola in the form y * ax 2 + bx + c (1) Locate the x-coordinate of the vertex which is x = . y c b2 b b2 + = x2 + x + 2 a a a 4a 4a2 or y c b2 b 2 + b = ax + 2 a a 2a 4a b2 b 2 = aax + b 4a 2a b 2 b2 b + c 4a 2a multiplying by a gives y .2b a .the vertical line through the vertex. and Finance. and a b = .2ba. This is conveniently done by choosing three successive integer values to the right and left of the x-coordinate of the vertex. Therefore. doing so may prevent errors in drawing its graph. Published by Pearson Learning Solutions. we conlude that given the parabola y = ax 2 + bx + c. we want to locate the vertex and use the original form of the equation. we divide by a so the coefficient of the x 2 term is 1.c + or solving for y gives y = aax + We can combine fractions and write this last equation as y = aax + b 2 4ac . by Warren B. three points uniquely determine it). let s see what happens. giving y/a .4 Quadratic Functions Parabolas Remember. to quickly draw a parabola we need only determine the x-coordinate of the vertex and then construct the table of values needed to get enough points to accurately obtain the graph. First.b2 b + 4a 2a (1) Locating the Vertex Graphing a Parabola We see that the term in the parenthesis will be zero when x = . as we shall see in the exercises. Gordon. Walter O. and April Allen Materowski. Applied Calculus for Business. You really do not need seven points to graph a parabola (in fact. This gives y/a = x2 + b/ax + c/a We next isolate the x-terms. The y-values are obtained by substituting the x-values into the equation of the parabola. Thus. Can we determine a formula that will easily give the x-coordinate of the vertex? If we do what we did above to the general quadratic function defined by y = ax 2 + bx + c. Economics. Copyright © 2007 by Pearson Education. Inc. the x-coordinate of its vertex is x = . (5) Draw a smooth graph through these seven points. (3) Complete the table by computing the y-values corresponding to each of the above x-values. 2a 2a 2a 4a 2 Therefore. however. 81 .2.2 + 210. The x-intercepts are usually not convenient points to plot. Notice. are x1 = . Table 8: Points used to Plot y = . as given in Table 8.8x + 12 y y y y y y y = = = = = = = . and April Allen Materowski.8x + 12 = 0 However.1. 202 1 .21 . 122 1 . 182 V1 . Gordon.81 .2.222 = .8x + 12 Note that the above example.) V x = -2 y = -2x2 .122 . Economics.22 + .21022 . by Warren B. the y-intercept is indicated on the graph.2x2 .2 and x2 L 1.2 -1 0 1 y = .210. as to find them we need to solve a quadratic equation.21122 .32 + . and Finance. We will choose three x-coordinates above and below this value and compute their y-coordinates. Of course. 2) We use the points from Table 8 to plot the graph given in Figure 10.5.81 .81 . we can read off the approximate solutions to this equation from our graph. Copyright © 2007 by Pearson Education.3. found by the quadratic formula.2 . Published by Pearson Learning Solutions. because the coefficient of the x 2 term is negative.82/121 .52 + . we know that the graph opens down.8x + 12 x-value -5 -4 -3 x-coordinate at vertex : .2x2 . that even before we begin the graph.12 + .1 .52 .322 . since the parabola is always symmetric about its Applied Calculus for Business.8112 + 12 2 point on graph 12 = 12 = 12 = 12 = 12 = = 12 = 2 2 12 18 20 18 1 .Section 1.2x 2 .422 . 182 (0.4.4 Quadratic Functions Parabolas * ** 95 Example 2 Sketch the graph of the parabola y = .5.8x + 12. corresponding points on either side of the vertex had the same y-coordinates.2. 12) (1.21 .2x 2 .8x + 12 Figure 10: The Graph of y = . x2 = .21 .222 . Wang.42 + . Inc.81 . this is always an easy point to plot since we need only set x = 0 and solve for y to find it. In fact. 22 1 . in the tables used above. Walter O. (Note.21 . . The x-coordinate of the vertex is x = . In the above case we have x1 L .21 . Solution.2x 2 . The exact answers. the equation .8102 + 12 . In the above case. However. Economics. Gordon.12 31022 .142 13.15 10 = . In such a case.81 . we will choose integers values to the left and right of this x-value to minimize the arithmetic in computing the y-coordinates.8x .22 31 . We shall choose integer values to the right and left of this x-value.102 and is easily found. and April Allen Materowski.122 .10 31 .1. 6) x = 4/3 y = 3x2 .8x .10 x-value -2 -1 0 1 x-coordinate at vertex : 4/3 2 3 4 y = 3x 2 . Solution. this will always be the case for symmetrically chosen x-values. .1 .152 V14/3. Table 9: Points used to Plot y = 3x 2 . .46/32 12.2. .10. Example 3 Sketch the graph of the parabola y = 3x2 . This is true when the x-coordinate at the vertex is not an integer. Applied Calculus for Business. 182 1 . Walter O.81 .8122 .8x . .22 31222 .10 31122 .81 . Copyright © 2007 by Pearson Education.8112 . Note that the only fractional arithmetic needed is to compute the y-value at the vertex. the y-intercept is 10. The next example illustrates this remark. . The graph is given in Figure 11.4 Quadratic Functions Parabolas axis.8142 . it may not always be convenient to plot the symmetric points.10 y y y y y y y y = = = = = = = = 31 . and Finance. since we chose integer values for x at the other points.10 Note that in the above example.10 31322 . Inc.96 * ** Section 1. The details are indicated in Table 9. 12 10.22 . . The x-coordinate at the vertex is x = .8102 . Note the integer to the right of 4/3 is 2 and the one to its left is 1.8132 . Wang.102 11. Published by Pearson Learning Solutions.8x .72 (4.10 V(4/3. -46/3) Figure 11: The Graph of y = 3x2 .10 31422 .8x . .82/121322 = 4/3.10 = .222 . by Warren B.14 = -7 = 6 1 .46/3 = .10 2 point on graph 10 = 18 10 = 1 = . so it too must be non-negative.2 or 5. the ball is at its maximum height. Since t represents time.5 seconds to reach the maximum height which is 196 feet above the ground. A typical problem that arises in calculus is to determine the maximum or minimum value assumed by a given function. it makes no sense to use negative t-values. Using these points.Section 1.1 246 4 + 246 Doing so. Since we must have a positive value for t. Published by Pearson Learning Solutions. we now have h as a function of t.8x . 160) (1. by Warren B. Since the height.16t 2 + 48t + 160 h = . Gordon. 96) Once we recognize the function to be optimized is parabolic.161322 + 48132 + 160 = 160 . x2 = . we find them to be x1 = . and April Allen Materowski. and what is this height? (b) how long does it take for the ball to hit the ground? Solution. Applied Calculus for Business. Economics. 192) (3/2. Consider the following examples. it is no longer necessary to plot its graph as illustrated in the next two examples. (a) We see that at the vertex. It can be shown that its height h measured in feet above ground is given by the equation h = .10 = 0. therefore it takes 3/2 = 1. (b) The ball hits the ground when its height h = 0.9 and 3. Example 5 A ball is thrown vertically upward from the ledge of a building 160 feet above the ground with an initial velocity of 48 feet per second.161222 + 48122 + 160 = 192 .6 respectively. From our graph.48/121 . 4 .1622 = 3/2.16102 + 48102 + 160 = 160 h = . and Finance. The integer t-values to the left and right of this value are 1 and 2. .1613/222 + 4813/22 + 160 = 196 . Solution. Wang. we 3 3 can see that their approximate values are .4 Quadratic Functions Parabolas * ** 97 Example 4 Determine the x-intercepts in the graph given above. We construct Table 10. Copyright © 2007 by Pearson Education. 192) (3. where t is the time measured in seconds. Note that instead of y being a function of x. That means its t-intercept is the point at which this occurs.0. h represents the height of the ball above the ground level. Determine (a) how long it takes the ball to reach its maximum height. The exact intercepts are found by solving the equation 3x2 .16t 2 + 48t + 160 t-value 0 1 t-coordinate at vertex : 3/2 2 3 4 h = . Inc. we plot its graph.161422 + 48142 + 160 = 160 2 point on graph (0. Applications to Optimization Table 10: Points used to Plot h = . we plot the graph given in Figure 12. as a function of time is a parabola.16t2 + 48t + 160 = 0 and find that t = . Walter O.16t2 + 48t + 160. 160) (4. we have that it takes 5 seconds for the ball to hit the ground. 196) (2.161122 + 48112 + 160 = 192 h h h h = = = = . then it is clear from its graph that its optimal (maximum or minimum) value will occur at the vertex. If the function is quadratic. We solve the quadratic . The t coordinate of the vertex is given by t = . This is somewhat unusual. Copyright © 2007 by Pearson Education. The problem is to maximize the total area A = xy. Inc.2x + 600 therefore.2x2 + 600x Once again. and April Allen Materowski.2x + 6002 or A = . . since the total amount of fencing used in constructing the enclosure is 4x + 2y = 1200.16t2 + 48t + 160 Example 6 A farmer will construct a rectangular enclosure from 1200 feet of fencing. and Finance. as A is a function of two variables.4 Quadratic Functions Parabolas V(3/2. Published by Pearson Learning Solutions. Solving for y we have y = . Applied Calculus for Business. 196) = 3/2 h = -16t2 + 48t + 160 Figure 12: h = . by Warren B. A = xy = x1 . Walter O. x and y. y x x y x x Figure 13: A Rectangular Enclosure of Width x and Length y Solution. Since x represents a dimension. Gordon.98 * ** Section 1. We let x and y represent the dimensions of the enclosure as indicated in Figure 13. Determine the dimensions of the enclosure that maximize the enclosed area. Economics. as illustrated in Figure 13. Wang. we can eliminate one of these variables. He plans to subdivide the enclosure into three parts. we have an equation of a quadratic function whose graph is a parabola. However. we must have that x 7 0. In the event that these functions are quadratic.20002/121222 = 500 The producer should produce 500 bicycles to minimize his cost. The cost function is a parabola opening upward. Consider.60x + 475 Applied Calculus for Business. The next example illustrates.Section 1. Calculator Tips Figure 14: y1 = 2x2 . From the equation y = . (The cost in producing the 500 bicycles is $100. the width of the enclosure is to be 150 feet and its length 300 feet. Economics. initial labor costs. Solution. we have. Gordon. etc.000 and the average cost per bicycle is 100000/500 = $200. y = 2x 2 . Example 7 The cost in dollars of producing x bicycles is given by the equation C = 2x2 . Thus. we find that the corresponding length is y = . Published by Pearson Learning Solutions. Since this is a parabola opening down. see Figure 14.2x + 600. Therefore. To make sure you obtain the U-shaped portion of the graph in your window. (why?) the vertex is the highest point on the graph. Inc. Copyright © 2007 by Pearson Education. for example.211502 + 600 = 300. you need only first calculate the x-coordinate of the vertex and make sure this is included in the window selected. This procedure may be easily used to plot the graph of any parabola of the form y = ax2 + bx + c. We go to the Y = window and define y1. minimize costs or maximize revenue.4 Quadratic Functions Parabolas * ** 99 The x-coordinate of the vertex is x = . for example.000.2000x + 600000. The maximum area is 1150213002 = 45. the lowest point on the graph is the vertex. Typical business problems arising in applications are to maximize profit.600/121 .000 square feet. by Warren B. Wang.1 . the width x = 150 feet. Thus.60x + 475. machinery. Walter O. so at the vertex the area is maximized. Determine the number of bicycles that should be produced to minimize the producer s cost. This number is sometimes called the startup or overhead cost needed to begin production. In the previous section we indicated how to plot the graph of a function using the calculator. and April Allen Materowski.) In the previous example. Thus. the cost of rent. x = . when the area is maximized. .222 = 150. the methods of this section are applicable. The x-coordinate of the vertex is therefore the number of bicycles that should be produced to minimize the cost (which is the y-coordinate of the vertex). and Finance. for the area to be maximized. you might have asked yourself why the cost of producing zero bicycles is $600. Gordon. from the home screen we calculate y11152 = 25. the difference in succeeding x-values listed in the table. . Figure 17: Choosing tblStart = 15 and ¢ tbl = 1 Applied Calculus for Business. we obtain Figure 15. Wang. we compute the x-coordinate at the vertex. Thus. Figure 16: Setting Up the Table tblStart is the x-value at which we want the table to begin and ¢ tbl is the incremental value. and April Allen Materowski. Walter O. Therefore. We can generate a table of values using the TABLE key 1*F52. x = . See Figure 17.100 * ** Section 1. Published by Pearson Learning Solutions.602/121222 = 15. so we use F2 (Setup). that is. Inc. Note. We will choose tblStart = 15 and (the x-value at the vertex) and ¢ tbl = 1.1 . by Warren B. and Finance. 25). we see that the vertex is (15.4 Quadratic Functions Parabolas Before we press the graph button. Economics. we want to include the vertex in our window. Figure 15: Using the TABLE Key The table generated does not include the vertex. Copyright © 2007 by Pearson Education. we obtain Figure 16. Gordon. Walter O. Figure 20: y1 = 2x2 . by Warren B. from the table. Figure 19: Choosing an Appropriate Window The graph is now obtained by pressing the Graph key 1* F32. Copyright © 2007 by Pearson Education.). Published by Pearson Learning Solutions.Section 1. it is clear that we should could use a window containing x-values between 12 and 18 and the corresponding y-values would be between 25 and 43. .4 Quadratic Functions Parabolas * ** 101 Next press Enter twice to get the table. see Figure 18. Wang. see Figure 20. and April Allen Materowski. If we also want to include the origin (we don t have to) then we could choose the window given in Figure 19 (there are many other reasonable choices for a window. Inc. Figure 18: The Table Starting at x = 15 Scrolling up and down will give different points on the graph. and Finance. Thus. Economics.60x + 475 Applied Calculus for Business. y = x2 + 4x .4 (a) In Exercises 1 21 draw the graph of the parabola. A barnyard is to be fenced is as indicated in Figure 23.102 * ** Section 1.8 21. 4 x x . measured in seconds by the equation h = . y = 31x + 122 .2x2 .12 23.4 Quadratic Functions Parabolas Once again. If 6000 feet of fencing is to be used in its construction determine the dimension x and y that will maximize its total area? 2y x y 2x y y y x 2x 2x Figure 22 29. determine (a) how high the bullet will go and how long it takes to reach that height. A ball is thrown vertically upward from the ground.2x2 + 4x + 8 18. y = 1x .21x . If 3600 feet of fencing is to be used in its construction determine the dimension x and y that will maximize its total area? 2 2 y x x x y x x Figure 21 28. y = 31x + 12 + 4 10.31x + 122 7.3 9. y = 41x . y = . and April Allen Materowski. estimate the x-intercepts. (c) Check your estimate by finding the x-intercepts exactly. 1. y = 1x + 122 2.3x + 5x .*+ x .2x2 + 4x . Published by Pearson Learning Solutions. .2x2 + 7x + 12 20. y = 2x2 + 4x + 5 15. y = 1x + 322 5. (b) how long does it take for the ball to come back to the ground? 26. measured in seconds by the equation h = . y = 1x . Copyright © 2007 by Pearson Education. y = . Gordon. Economics.322 4. (b) From your graph.5x + 12 22.21x .4x2 . y = . y = 3x2 + 12x .5 13.4x .322 + 5 11. y = .12 25. y = . EXERCISE SET 1. A gun is fired upward from the ledge of a 176 foot cliff. y = . The bullet leaves the gun with a velocity of 80 feet per second.122 6. by Warren B. determine (a) how high the ball will go and how long does it takes to reach that height. and Finance.122 8. Inc. If 4200 feet of fencing is to be used in its construction determine the dimension x and y that will maximize its total area? s2 Hint: The area of an equilateral triangle with side s is 23. (b) how long does it take for the bullet to come back to the ground? 27.222 3. y = .*+ x Figure 23 Applied Calculus for Business. Wang. y = 2x2 + 5x + 9 24.16t2 + 96t. You should experiment with various choices.4x + 9 16. y = .*+ x y .6 14.21x + 422 + 7 12.15 17. y = . y = 2x2 + 5x .*+ x y y . we stress that there are many other window choices that could have been made that would illustrate the U-portion of the parabola. A barnyard is to be fenced is as indicated in Figure 22. y = .16t2 + 80t + 176. y = x2 .4x2 . If its height h measured in feet is given in terms of time t. Walter O.5x . A barnyard is to be fenced is as indicated in Figure 21. If its height h measured in feet is given in terms of time t. y = .9 19. Student Government at a University is chartering a plane for Spring Break. k + p2 and directrix y = k . Wang. y) be any point on the circle. (a) Suppose the The Circle * ** 103 focus is located at (0. The given point is called the center of the circle and the distance each point is from the center is called the radius. . Let P (x. therefore. This definition may be used to determine the equation of circle if we are given the coordinates of the center and the length of the radius. Determine the equation of the parabola passing through the points (0. you are drawing an arc of a circle. Find the equation of the parabola with focus on the line y = x at (h. Keeping the string taut move the pencil. The revenue R. Determine the equation of the parabola passing through the points (0. (The axis of the parabola is the line y = x.2h.p.3x2 + 600x + 1000. 35. 1. our objective is to determine an equation relating x to y. The plane can seat 150 passengers. 02. and April Allen Materowski. The above method for sketching a circle may be used to define it. Economics. 10) (1. obtaining. y) using the distance formula is Definition of a Circle Applied Calculus for Business.x . Walter O. Continuing once around. we have the distance from the center C (1.1.x21120 + 15x2. (b) How many seats should be unused to maximize the airline s revenue? (c) What price would each passenger pay if the airline maximized its revenue? (d) Is this a good deal for the Student Government? 33. and Finance. so we may factor to obtain. a circle on a piece of cardboard. (a) Show that the airline s revenue R = 1150 . 36. in dollars. what is the average cost per set? 31. determine the equation of this parabola. 4ay = 4a 2x2 + 4abx + 4ac Now add b2 both sides of the equation yielding b2 + 4ay = 4a 2x2 + 4abx + b2 + 4ac Note the first three terms on the right hand side are a prefect square. in producing x large screen televison sets is given by the equation C = 4x2 . (2. 6) and 1 . h) and directrix with equation y = . determine the equation of this parabola. 8). How would you go about doing it? One approach is to attach a pencil to one end of a taut string and hold the other end fixed.b/2a. the circle is drawn. Given the parabola y = ax 2 + bx + c (a) determine the relationship between a. Gordon.p2 and directrix y = k + p. (a) How many items should be sold to maximize the revenue? (b) What is the revenue? 32. p) and the directrix has the equation y = .) 39. Solution. The cost C. by Warren B. Let x represent the number of unsold seats. 30). Given y = ax2 + bx + c multiply both sides of the equation by 4a. (b) Suppose the focus is located at 10. b2 + 4ay = 12ax + b22 + 4ac Solve for y and then show that it follows that the vertex occurs when x = . This example illustrates another way of finding the vertex of a parabola (and deriving the quadratic formula). b and c if the parabola is to have no real x-intercepts. The circle is defined to be the set of all points equidistant from a given point. Example 1 Find the equation of the circle with center at the point C (1.5 The Circle » » » » » » Definition of a Circle Equation of a Circle Graphing a Circle Tangent Line The Ellipse Calculator Tips Suppose someone asks you to draw. Copyright © 2007 by Pearson Education. The airline will charge $120 per passenger and added to this a surcharge of $15 per passenger for each unsold seat. (b) When will its graph lie above the x-axis? (When will it lie below the x-axis? 34.5 30. 2) and radius 3.2000x + 600000. 37. We illustrate in the following example.Section 1. in dollars. Inc. We know that every point on the circle is 3 units from the center.p.p2 and the directrix has the equation y = p. 38. Find the equation of the parabola with (a) focus at 1h. 9) and (4. As you are doing so. Also show that setting y = 0 results in the quadratic formula. as accurately as you can. Published by Pearson Learning Solutions. derived from selling x computers is given by R = . . k . (a) How many television sets should be produced to minimize the cost? (b) What is the minimum cost? (c) At this level of production. A parabola may be defined as the set of all points equidistant from a fixed point called the focus and a fixed line called the directrix. 2) to the point P(x. (b) focus at 1h. k22 = r or squaring both sides we have 1x . k = 4. equation (1) reduces to x2 + y2 = r2 (3) The only difference between the circle whose equation is given by (1) and the one given by (3) is their location.2x + 4y . . we could multiply out and combine like terms and write the equation in the form x2 + y 2 . and if it is. as it is not immediately obvious that it is indeed a circle (as we shall see).122 + 1y . and Finance.2ky + h2 + k2 . and is not the most useful form for the equation of the circle.h22 + 1y . then it is a distance r from the center.r 2. Thus.5 The Circle 21x . Each point on the circle given by equation (1) is translated horizontally by h units and vertically by k units. we may do in general to obtain the equation of the circle centered at the point C (h. 4) and the radius is r = 216 = 4.122 + 1y .2k and c = h2 + k2 .104 * ** Section 1. Applied Calculus for Business.r 2 = 0 If we let a = .2h. Wang.222 = 9 This is one way we can leave the equation. by Warren B. Comparing with (1) we have immediately h = 2. it is not immediately apparent where its center is located nor what is its radius. Inc. Solution. We illustrate this remark in Example 5 below. determine is center and its radius. When the circle is centered at the origin. h = k = 0.222 = 3 If we square both sides of the equation we obtain 1x .222 + 1y . y) is any point on the circle. we have 21x . and April Allen Materowski. Example 2 Given the equation of the circle 1x . or if we like. Economics. If the point P(x. Gordon. and r 2 = 16.k22 = r2 (1) Equation (1) is called the standard form and the most convenient form of the equation of a circle. Therefore. Copyright © 2007 by Pearson Education. by the distance formula. that is. Walter O. the center of the circle is C (2. k) with radius r. then we may write the equation in the form x2 + y2 + ax + by + c = 0 (2) Equation (2) is called the general form. we obtain x2 + y 2 .2hx . if equation (1) is expanded. Published by Pearson Learning Solutions. b = .422 = 16. However. Inspection immediately yields the center and radius.4 = 0 Equation of a Circle What we did in Example 1.h22 + 1y . 5. 3. Move horizontally to the left from the center a distance equal to the radius and plot this point. We illustrate this approach with the following example. but we suggest the following approach which uses four points in addition to the center.52. The radius is r 2 = 25 or r = 5. Connect these four points with a smooth curve. Graphing a Circle When the general form is given for the equation of the circle we must first transform it into the standard form so we may easily locate its center and radius. Example 4 Given the circle whose equation is x2 + y 2 . The points are plotted and the circle is drawn in Figure 1. we must complete the square.9 Applied Calculus for Business.3 = 0 and y + 5 = 0. 4) yields (6. we see that h is the value of x that makes the first parenthesis zero and k is the value of y that makes the second parenthesis vanish. Solution. The circle is centered at C (2.5 The Circle * ** 105 Looking at (1).k = 0. once for the x terms and once for the y terms. We must also remember that whatever we do to one side of an equation we must do to the other. and April Allen Materowski. but in this case. Using the observation made above. determine is center and its radius. We choose the four points suggested above. and Finance. Given the equation of a circle we can sketch its graph fairly easily. Example 3 Sketch the circle whose equation is given in Example 2. yielding x = 3 and y = . Example 3 Given the equation of the circle 1x . Solution.2. Plot the center 2. Move vertically below the center a distance equal to the radius and plot this point. Graphing a Circle 1. Gordon. 8) and moving below 4 units yields (2. the center of the circle is C13. (a) Determine its center and radius. 6. by Warren B. Economics. Solution. (b) Sketch its graph. Wang. In order to do this. Moving up 4 units yields (2. 0).6x + y 2 + 10y = . 42. Move horizontally to the right from the center a distance equal to the radius and plot this point.322 + 1y + 522 = 25. 4) and has radius 4. 4. Moving to the left 4 units yields 1 .6x + 10y + 9 = 0.5.h = 0 and to find the y-coordinate we may set y . Move vertically above the center a distance equal to the radius and plot this point.Section 1. (a) We rewrite the equation as x 2 . 4). We illustrate the procedure in the next example. Thus. Moving to the right 4 units from (2. Walter O. thus. three points uniquely determine a circle. Actually. set x . Copyright © 2007 by Pearson Education. . . twice. Published by Pearson Learning Solutions. to find the x-coordinate of the center we may set x . Inc. Wang. . Gordon.9 + 9 or 1x .52 and has radius 5. Published by Pearson Learning Solutions. and 1 . and April Allen Materowski. Figure 2: 1x .3. Walter O. .422 = 16 We first complete the square for the x terms: 1 2 1 . We see that the circle is centered at 13. so we have x2 .322 + 1y + 522 = 25 This is the circle given in Example 3.106 * ** Section 1.322 + y 2 + 10y = 0 We next complete the square for the y terms: 1*21102 = 5 and 1522 = 25. and Finance. so we have 1x .322 + y 2 + 10y + 25 = 0 + 25 or 1x .5 The Circle Figure 1: 1x . Inc.6x + 9 + y 2 + 10y = .322 + 1y + 522 = 25 Applied Calculus for Business. Copyright © 2007 by Pearson Education.222 + 1y .322 = 9. by Warren B. Economics.62 = . by Warren B.52. Consider the next example. Gordon.2)2 = 9 x2 + y2 = 9 Figure 3: The graphs of 1x . moving to the left from the center we have 1 .6y + 15 = 0 Solution.52.102. We plot these points and the graph in Figure 2. The sketch of the two circles is given in Figure 3.6y + 13 = 0 is rewritten as Applied Calculus for Business. 2) then the two graphs would be coincident. We complete the square to understand what is happening here.Section 1.5 The Circle * ** 107 (b) we plot its graph by using the four points described above.222 = 9 and x2 + y2 = 9 If we moved the circle centered at the origin so that its center would be at (1. 2). and the points used to plot the graphs are indicated. (x . . 0) and moving 5 units down from the center we have 13. (b) x2 + y 2 + 4x . and April Allen Materowski.6y + 13 = 0. Inc. That is. 0). . take the point (3. Wang. Not every equation of the form x2 + y 2 + ax + by + c = 0 is a circle. (a) x2 + y 2 + 4x . . Solution. Economics. . Published by Pearson Learning Solutions. For example.1)2 + (y .0) is moved one unit to the right and 2 units up then we get the corresponding point on the other circle.122 + 1y . Moving up 5 units from the center we have (3. 13 + 1. Moving to the right from the center 5 units we have the point 18. Example 5 Sketch the circles 1x . Example 6 Classify each of the following: (a) x 2 + y 2 + 4x . if any point on the circle centered at (0.122 + 1y .2. and Finance. Copyright © 2007 by Pearson Education.222 = 9 and x2 + y 2 = 9 on the same coordinate system. 0 + 22 yields the corresponding point (4. Walter O. so we have 1x + 222 + y 2 .11 + 9 or 1x + 222 + 1y .x 2 (Note that to each x-value there corresponds two y-values.11 1 *21 .322 = 9.15 1 *2142 = 2 and 22 = 4.6y = .322 = . Walter O. it results in a contradiction. That is.6y = . Drawing a vertical line through the circle intersects the circle at two points with different y-values. x2 + y 2 = 9 solving for y we have y2 = 9 . by Warren B. .6y + 15 = 0 is rewritten as x 2 + 4x + y 2 .6y + 9 = .322 = 9. Economics.2. so we have x2 + 4x + 4 + y 2 . Therefore the equation given defines neither a curve nor a point.) Applied Calculus for Business. (or the circle of radius zero centered at 1 . namely the point 1 .2. therefore we have x = . the equation reduces to a single point. This may be seen at once from the vertical line test. Consider the circle centered at the origin with radius 3. 29 .6y + 9 = . and Finance.322 = 0 The only way a sum of squares can be zero is if each square term is zero.3. Copyright © 2007 by Pearson Education. a point or a contradiction.5 The Circle x 2 + 4x + y 2 .108 * ** Section 1.6y = .6y = . 322. Published by Pearson Learning Solutions.9 1 *21 .3. so we have x2 + 4x + 4 + y 2 .9 + 9 or 1x + 222 + 1y . Inc. so we have 1x + 222 + y 2 .2 It is impossible for the sum of squares to total a negative number.13 + 4 or 1x + 222 + y 2 . The sum must be either zero or a positive number.6y = .x2 or y = . Wang.6y = . and April Allen Materowski. and 1 .13 1 *2142 = 2 and 22 = 4.62 = . 32. The last example indicates that the equation x 2 + y 2 + ax + by + c = 0 can either be a circle. (b) x2 + y 2 + 4x .2 and y = 3. and 1 . Note that a circle does not describe a function. Gordon.15 + 4 or 1x + 222 + y 2 .62 = . this is the upper half of the circle and to each x-value their corresponds one y-value. Copyright © 2007 by Pearson Education. and may be used as its defining property. suppose we consider y = 29 . by Warren B. Economics.Section 1. Wang. Gordon. In Figure 1.29 . Thus. and Finance. . and April Allen Materowski.x2. Figure 5: The Graph of y = . We shall make an observation about the tangent line to the circle which is carried over to other curves. Inc. Figure 4: The Graph of y = 29 . Walter O.x2 The tangent line to a circle at a point P on its circumference is the line perpendicular to the radius of the circle at P. This half of the circle does indeed define a function. While the tangent line to a circle has the property that it is perpendicular to the radius at the point of tangency.x 2 also defines a function. Let us look at a specific example. it is not this property which generalizes to other curves.29 .x2 The graph of y = . Its graph is given in Figure 5. a circle does not define a function. Its graph is given in Figure 4. We may easily determine the equation of the tangent line to this circle at Tangent Line Applied Calculus for Business. Published by Pearson Learning Solutions. Let the equation of the circle in Figure 6 be x2 + y 2 = 25. The line T is the tangent line which is perpendicular to the radius of the circle at the point P. but its upper or lower halves taken separately do.5 The Circle * ** 109 However. and Finance.97 2. Note that near P.9925 3.04 3. Copyright © 2007 by Pearson Education.90 2.98 2.065944 4. its equation is found to be y = -3 4x + 25 4 Let us compute y-values on both the tangent line and the circle for x-values near the point P(3. In Table 1. Notice the definition says nothing about the tangent line being perpendicular to other lines at the point of tangency.3/4 and passes through P(3. Economics.06 4.969685 3.0122 L 3. we can solve for the y-value on the upper half of the circle which is found to be y = 225 .95 2.92 2.044305 4.4).985 3. by Warren B.073082 4.4).029690 4.938731 3.06 3. as in the case of the circle where it is perpendicular to the radius.022325 4.923009 Applied Calculus for Business. using the pointslope formula.3 4x + x 2. Published by Pearson Learning Solutions. and its equation is y = 4 3 x (why?). 4) on y = .96 2.0675 4. (The term smooth used in the definition will be clarified later in this text when you study calculus. DEFINITION The tangent line to a smooth curve at a point P is the best linear approximation to the curve at that point.94 3.3/413.984922 3.0525 4.x 2 When x = 3.110 * ** Section 1. among all different lines that touch the curve at the point P.051555 4. while the corresponding value on the circle is y = 225 .955 3.99248.9775 3.10 y = -3 4x + 4.07 3.015 4. and near the point of tangency. 4). The notion of perpendicularity is unique to the circle. Note the closer we get to the point P.962007 3.930890 3. Inc.13. the better the y-values agree. . Wang.9325 3.01 3.97 3.08 3. That is.0075 4 3.007481 4 3.99 3 3.954289 3. the y-values on the circle and line are nearly the same.02 3.045 4. Gordon.05 3.03 3. and April Allen Materowski. we observe that the radius is a segment of the line passing through the origin (0. 0) and P(3.x 2 4.992480 3.9625 3. This observation motivates an alternative definition of a tangent line to a smooth curve at a point P. its slope is .91 2.977323 3.037016 4. we indicate other corresponding values as we vary x near P. where you will discover that some curves do not have tangent lines at points which are sharp.03 4.9475 3. we find the y-value on the tangent line is y = .01. ) Table 1: Comparing the y-values near P(3.075 4. First.058768 4. Walter O.925 25 4 25 4 and x 2 + y 2 = 25 y = 225 . Since the tangent line is perpendicular to this line.946530 3.012 + 25/4 = Figure 6: A Tangent Line to a Circle 3.09 3.5 The Circle T the point P(3.014922 4. 4). the tangent line is the one whose y-values best approximate the y-values on the curve near the point of tangency.0225 4.0375 4.94 2.93 2.9925. 4) The observation that a tangent line approximates the curve near the point of tangency is an extremely useful tool. Gordon. Tangent Line y = x3 Figure 7: The Tangent Line to y = x3 at the Point P(2. the tangent line to the graph of y = x 3 at the point (2. The Ellipse 4x2 + 9y2 = 36 Figure 8: The Graph of 4x2 + 9y2 = 36 Applied Calculus for Business. A circle centered at the origin has the equation x2 + y 2 = r2. . Such a graph is called an ellipse.3). the y-values are almost identical. Its graph is given in Figure 8. and Finance. Copyright © 2007 by Pearson Education. Inc. one that is used over and over again in mathematics and is sometimes called linearization or linear approximation. we indicate in Figure 7. Published by Pearson Learning Solutions. that is. What happens to the graph if we change the coefficients of the squared terms so that they are not the same? For example. The determination of the equation of the tangent line is studied later in this text. and April Allen Materowski. Economics. let us consider the graph of the equation 4x 2 + 9y 2 = 36. Wang. Walter O. Observe how well the tangent line approximates the curve near the point.Section 1.5 The Circle * ** 111 As an illustration. by Warren B. The equation 4x2 + 9y 2 = 36 may be rewritten in the standard form by dividing by 36 to obtain x 9 + Example 7 Sketch the graph of the ellipse 2 2 y2 y2 4 = 1. 0). 2). any equation of the form x2 + 2 = 1 is the equation of an ellipse a b centered at the origin.3 or 3. Thus. . To find the x-intercepts. More generally. and Finance. Wang. yielding the points 10. we could rewrite this equation as 9x2 + 25y 2 = 225.5 The Circle The graph is obtained by finding the four intercepts. we saw that they were identical except the first one is centered at the origin and2 the second is centered at (h. thus the points 1 . 3). To find the y-intercepts. .3 or 3. 0). 02 and (3. they are also identical except that the second 25 9 25 9 one has its center 1x . 02 and (5. Economics. or y 2 = 9. and April Allen Materowski.h22 + 1y .22 and (0.222 + = 1 and + = 1. and solving we have x = . We shall leave the analysis of these translated ellipses to the exercises.5. Walter O. k). we set x = 0 and obtain 25y2 = 225. 2). thus the y-intercepts are 10.2 or 2. this equation has the form Ax2 + By2 = C. If the fractions are cleared. Now consider the graphs of x2 y2 1x .3.112 * ** Section 1. Published by Pearson Learning Solutions. Figure 9: The Graph of y2 x2 + = 1 25 9 When we examined the graphs of the two circles x 2 + y 2 = r 2 and 1x . Copyright © 2007 by Pearson Education.32 and (0. we obtain 4x2 = 36. Gordon. solving. 25 9 Solution.12 1y . see Figure 10. Setting x = 0 we obtain 9y2 = 36 or y = . + = 1 is identical to 2 2 x2 2 a + y2 b 2 a b = 1 except its center (and all its other points) are moved h units in the x-direction and k units is the y-direction. By setting y = 0. we set y = 0 and obtain 9x 2 = 225. Note that multiplying by the LCD which is 225. thus the x-intercepts are 1 .k22 = r2. . y2 x2 + = 1. Applied Calculus for Business. The graph is plotted in Figure 9. solving. y = . x = .5 or 5. in general.k22 at the point (1. or x2 = 25. Inc. by Warren B. It is most easily drawn by finding the x and y intercepts and plotting them.h22 1y . . To sketch the graph and the tangent line together. We do this by pressing F2 Setup and then entering 2.1. One way of automating this calculation is to Figure 10: The Graphs of 1x . Calculator Tips Figure 11 Figure 12: Comparing the y-values Near x = 3 Applied Calculus for Business. a more extensive table could be generated using a spreadsheet like Excel. That is how the larger table given in Table 1 above was generated.222 y2 x2 + = 1 and + = 1 25 9 25 9 use the Table feature of the calculator.1 for ¢ tbl and then press Enter.x .3 4 x + 4 and y2 = 225 . but we need to determine the proper x-values. Walter O. A table is displayed. See Figure 12. Economics. by Warren B. Published by Pearson Learning Solutions.Section 1. while the : .122 1y . See Figure 11.8 with an increment 0. What we do is define y1 to be the equation of the tangent line and y2 to be the equation of the circle. Alternately. Since we want to examine these two functions near x = 3. For the example considered above. Choose an appropriate window and they are both drawn. to see their value appear on the screen. Gordon. You can trace either. and Finance. 25 2 y1 = . the q p cursor arrows move you from one curve to the other. Copyright © 2007 by Pearson Education. and April Allen Materowski.5 The Circle * ** 113 Consider the problem of comparing the y-values on the tangent line near the point of tangency with the y-values on the circle.8 in tblstart and 0. Wang. To trace. We next set up the table (*F5 or Apps then 5). is a simple matter on the calculator as they are already entered in the Y = screen. we will choose out starting point to 2. We must first go to the Y = Window 1*F12 and define these functions. Inc. move the cursor along the selected curve. press F3. (c) f(1).222 + 1y + 322 = 4.12 + 1y + 22 + 4 = 0 18.122 4 1x + 422 16 43.12. (c) g1 . 1x . 29.62. (b) g(2). 16x2 + 16y 2 . 1x + 322 + 1y + 222 = 9 7. Sketch the graph of + + 1y + 222 9 1y .5 In exercises 1 15 determine the center and radius of the given circle and sketch its graph. x2 + y 2 + 6x . b. In exercises 36 41 sketch the graph of the given ellipse.42. Let y = g1x2 describe the upper half of the circle x2 + y2 + 8x . On the same coordinate axes. 32.62. Economics.2x + 4y + 12 = 0 23.48x + 60y . (b) f(0).12. y2 x2 + = 1 9 25 y2 x2 + = 1 25 9 y2 x2 + = 1 4 16 y2 x2 + = 1 16 4 40. (b) g1 . (c) f(2). . Let y = f1x2 describe the upper half of the circle x2 + y 2 + 6x . x2 + y 2 + 8x . x2 + y 2 + 8x + 6y + 16 = 0 26. (c) f1 .9 = 0 20. 12). 16. 9x2 + 4y 2 = 36 41.222 = 4 2.322 + 1y + 222 = 25 5. and April Allen Materowski.27 = 0 15. 39.42. 36x2 + 36y 2 . x + y + 1 = 0 17.283 = 0 In exercises 16 25 determine whether the given equations is a circle.42. 1x + 12 + 1y + 42 = 4 6. Determine (a) f1 . (c) f1 . Determine (a) f(2). Walter O. . (d) a contradiction? 27. 2x2 + 2y 2 .322 + 1y . Determine (a) g1 . labeling all intercepts. 1x + 42 + 1y . 38.222 = 9 3. Let y = g1x2 describe the lower half of the circle x2 + y2 = 16. by Warren B. x2 + y 2 .5 The Circle EXERCISE SET 1. Let y = f1x2 describe the upper half of the circle x2 + y 2 + 8x .6y + 9 = 0.12.10y + 9 = 0 13. 9x2 + 9y 2 .32.422 = 4 How are the graphs related. Determine (a) f1 . 36.322 + 1y + 222 = 0 22.422 + 1y . (b) Compare the y-values on the tangent line with those on the circle near x = 3. x2 + y 2 . and 25 9 y = g1x2 its lower half. 1x + 422 + 1y . sketch the graph of: (a) x2 + y2 = 4 (b) 1x .6y + 9 = 0 12.22. a point. (d) g1 .32 . 1x . . 34. Determine (a) g1 .10y + 9 = 0 10. Sketch the graph of 1x 44. x2 + y 2 = 25 9. and Finance.12x + 24y .10y + 9 = 0. 31.32. Under what conditions on a.2x + 4y + 1 = 0 11.6y + 9 = 0. Inc. (a) Determine the equation of the tangent line to the circle x2 + y2 = 169 at the point 112.6y + 13 = 0 24. (a) Determine the equation of the tangent line to the circle x2 + y2 = 25 at the point 13.114 * ** Section 1. (d) g(1). . (b) f1 . 35. (b) a point. (b) g(0). 4x2 + 9y 2 = 36 42. Wang. 1. Copyright © 2007 by Pearson Education. (c) 1x + 122 + 1y . Published by Pearson Learning Solutions. 33. (b) Compare the y-values on the tangent line with those on the circle near x = 12. 1x . x2 + y 2 + 4x . x2 + y 2 . c. Gordon. (c) a line. 1x + 322 + 1y .322 4 = 1. and d is ax2 + ay 2 + bx + cy + d = 0 (a) a circle.12.48x + 8y .10x + 4y + 29 = 0 25.52. (b) g1 . 28.6x .222 = 0 21. and y = g1x2 the lower half of the circle. 1x . 37. (a) Determine the equation of the tangent line to the circle x2 + y2 = 169 at the point (5.101 = 0 14. 1x . or a contradiction (no real graph).22. Determine (a) f1 .322 = 16 4. (c) g(2). = 1 Applied Calculus for Business. (b) Compare the y-values on the tangent line with those on the circle near x = 5. Let y = f1x2 describe the upper half of the circle x2 + y 2 = 16. 1x + 422 + 1y + 322 + 9 = 0 19. x2 + y 2 = 4 8. 2 2 2 2 2 2 2 2 30. Let y = f1x2 denote the upper half of the ellipse y2 x2 + = 1. Published by Pearson Learning Solutions.6 Economic Functions » » » » » » Supply Function Demand Function Market Equilibrium Revenue. and April Allen Materowski. if p2 7 p1. suppose that the relation between price and quantity is p . it is traditional to plot this graph with x on the horizontal axis and p on the vertical axis. the producers will be willing to provide more of the item and hence the supply will rise. so should x. . Now as p increases. Graphically. this would mean that plotting x (horizontally) versus p (vertically). the supply of a given commodity is a function of the price of the commodity. Gordon. Inc. In particular. the relationship is p . by Warren B. So. p Ú 0 Applied Calculus for Business. it is assumed that as the price increases. we could think of x as an increasing function of p or we could think of p as an increasing function of x. That is. For example. That is.6 Economic Functions * ** 115 1. However. Walter O. in our example. Economics. There are applications in which one form might be preferred to the other. Copyright © 2007 by Pearson Education. In mathematical notation. We would say that supply is an increasing function of price.3x = 42 p = 3x + 42 Which is a straight line with positive slope 3. all other things being equal. then S1p22 7 S1p12. p Supply Function p2 p1 x x1 x2 Figure 1: A Supply Function In the simplest case. then x = S1p2 says that the amount supplied is a function of price. S(p) is a linear function with positive slope. let x be the number of units of the commodity supplied and let p be the price per unit. and Finance. Cost and Profit Functions Marginal Functions Calculator Tips In elementary microeconomic theory there is the fundamental assumption that. That is. x Ú 0. as we move from left to right the graph goes upward (see Figure 1).Section 1. Wang.3x = 42. the graph of the supply function is restricted to the first quadrant. We must also require that the values of both p and x remain non-negative (Why?). 3x = 42 Figure 2: Linear Supply We see that when x = 0. that is.11/22p Or. you should read this the other way: when the price is 42 (or less) there will be no supply. Thus. Again. in the form x = P1p2. Solving for x. not only is x a decreasing function of p. x = 36 . Copyright © 2007 by Pearson Education. In this rudimentary theory. by Warren B. and April Allen Materowski. This is very typical of a linear supply function. p = 42.6 Economic Functions The graph is shown in Figure. no supplier will bother to produce the commodity in question.2 p . as p increases.11/22p x x1 x2 Demand Function p p1 p2 x Ú 0. Interestingly. Inc. it is called the demand function.14 This is a linear supply function with positive slope 1/3. p Ú 0 that is. When solved for x in terms of p. normally one plots p as a function of x. x should decrease. Published by Pearson Learning Solutions. In symbols. Wang. solving for p. In other words. A typical example of a linear demand equation might be 2x + p = 72. Walter O. which must be of the form p = mx + p0 where p0 is the lowest price for which there will be any supply. we note that you could just as easily have solved for p in terms of x. it is called the Price function. but if viewed from the other perspective. there is also a relationship between the amount that can be sold and the unit price. we would write it p1 7 p2 then P1p12 6 P1p22. Gordon. That is. There is a basic assumption that for this relationship. Of course. p is a decreasing function of x.116 * ** Section 1.2x Applied Calculus for Business. Economics. Solving for x in terms of p yields 3x = p . Figure 3: A Demand Function p = 72 . When solved for p in terms of x and written in the form p = D1x2. x is a decreasing function of p (see Figure 3). This relationship is called the demand equation. and Finance.42 x = 11/32p . at price less than 42. P1p2 = 36 . As above. . would seem to be the maximum demand. even if free. in theory. Wang. when p = 0. if the price is below the equilibrium price. the supply will increase and the demand decrease. the market is always trying to reach its equilibrium position.Section 1. Most economists feel that such upper and lower bounds on price. p = 72. If the price goes above this value. x = 36. In order to find this position. In either form. Gordon. However. this can be quite difficult. and April Allen Materowski. Inc. We shall see that for nonlinear supply and demand equations this identification may be generalized. supply and demand will be equal. at a price greater than or equal to 72. When supply and demand equations are plotted on the same set of axes. supply. When x = 0. there appear to be two significant values.6 Economic Functions * ** 117 that is. they should not be given much credence near their end points at the axes. D1x2 = 72 . Example 1. it is necessary to solve the two equations simultaneously. it is easy to see that the graph is a straight line with negative slope (see Figure 4) p = -2x + 72 Figure 4: Linear Demand For this simple case of linear demand. Similarly. The price. . no one would be interested in buying this commodity. their intersection is called the point of market equilibrium. real markets are always changing so things are rarely stable at an equilibrium point. by Warren B. Suppose the market for humidifiers is governed by the two equations Market Equilibrium Applied Calculus for Business. Published by Pearson Learning Solutions. It is easy to identify the supply or demand equation when each is linear. The supply equation must have positive slope since it is an increasing function. and demand do not provide a realistic model. Walter O. Economics. although linear models are used to approximate more complex ones or to simplify computations. This amount. as will now be illustrated. p = 72.2x. the most you could give away of this commodity is 36 units. Therefore. it is easy. It is characterized by the fact that at the equilibrium price. Of course. If the equations are not linear. x = 36. and Finance. would appear to be the highest price possible. That is. while the demand equation has negative slope because it is a decreasing function. causing an oversupply and forcing the producers to lower their price. Copyright © 2007 by Pearson Education. there will be a shortage followed by an upward adjustment in the market. For linear equations such as the ones given above. That is. Similarly. 3x = 42 Demand: p + 2x = 72 where p is in dollars and demand is in thousands of humidifiers. as the next example illustrates. at which time we will sell 6.3162 = 42 p . yielding p = 12 . Example 2 Suppose the demand and supply functions are as follows: Demand: p + x2 = 12 Supply: p .18 = 42 p = 60 (Substitution into the demand equation would also yield p = 60. Economics.) Thus. Solution. . The graph is shown in Figure 5. p 3x = 42 p + 2x = 72 Figure 5: Finding Market Equilibrium It might happen that neither the supply nor the demand function is linear. we get . Walter O.5x = . Wang. Published by Pearson Learning Solutions. and April Allen Materowski. as can be verified.118 * ** Section 1. substituting into the supply (first) equation we get p .6 Economic Functions Supply: p . We solve simultaneously.30 so x = 6. Copyright © 2007 by Pearson Education. Gordon.000 humidifiers 1x = 62. Solution. and We can solve each equation for p.x2 = 4 Determine Market Equilibrium. Inc. Subtracting the demand equation from the supply equation. Find the point of market equilibrium.x 2 p = 4 + x2 Applied Calculus for Business. by Warren B. and Finance. the market is in equilibrium when the price is $60 per humidifier. Then. 2x.Section 1. In the above example. p = 4 + 1222 = 8 thus. did we need to specify which was the demand and supply curve? p = 4 + x2 p = 12 . The revenue is the producer s income derived from the sale of his product. Inc. you multiply the number of items sold by the price per item. Cost and Profit Functions R = xp or (1a) R = xD1x2 (1b) For example. p Ú 0 and x Ú 02 substituting gives. the total revenue function R is Revenue.x2 = 4 + x 2 or 2x2 = 8 x2 = 4 or x = 2 1remember. suppose the demand equation is linear. Market Equilibrium occurs at (2. The price p paid for the product by the consumer is given from the demand equation p = D1x2. Walter O. as in Example 1. Wang. Gordon. Published by Pearson Learning Solutions. we have 12 . by Warren B. and Finance. Copyright © 2007 by Pearson Education. 8). Thus. p = 72 . Economics. namely. then the Revenue function Applied Calculus for Business. and April Allen Materowski.6 Economic Functions * ** 119 by substitution. cost and profit.x2 Figure 6: Market Equilibrium for Non-Linear Demand and Supply We next examine the other common functions arising in Economics: revenue. . see Figure 6. This is an example of quadratic cost.000.x This is a nonlinear function which we plot by choosing x values between 0 and 8 to calculate the corresponding R-values and then plot the points.000 when x Ú 0. (If x represented the actual number of units and p was the price in dollars. sketch its graph and estimate the point at which the revenue is maximized. we will learn how to use the calculus to determine this point exactly. the best we can do now is estimate that the revenue is maximized when x is between 5 and 6. In Section 1. Figure 7: R1x2 = 72 . or x 8. when x = 10. Since this function is not parabolic. Inc. This makes perfectly good sense as there Applied Calculus for Business. Walter O.4. by Warren B. Gordon. . that is C102 = 100.120 * ** Section 1. determine the revenue function. then the revenue is $520.x. Copyright © 2007 by Pearson Education. suppose the cost function in dollars is given by the equation C1x2 = 2x 2 . and April Allen Materowski. Economics.2x2 = 72x .22 = 18 at this value for x we have p = 72 . Let us assume that we are either given data by which we may determine the cost function or it is determined for us. We have already examined this particular problem in Section 1.x Ú 0.000.1722/21 .3 we examined a linear cost function.) A sketch of the revenue function is given in Figure 7. Wang. We observe the domain is determined by the condition that 8 . For example.200x + 100.2x 2 So. it may take many different forms. Solution. Looking at the graph. p = 52 and the revenue is 520. We recognize the graph of the revenue function to be a parabola. The graph is given in Figure 8.2x2 A useful question to ask is when is the revenue maximized and what is the price and revenue when it is maximized. and Finance.21182 = 36 and R = 181362 = 648. Later on in the text. we know its maximum occurs at the vertex which is x = . Cost need not be linear.6 Economic Functions R = xp = x172 . Once again observe that even if the producer produces no items his cost is $100. Example 3 Suppose the demand equation is given as p = 28 . for example. Consider the problem when demand is non-linear as in the next example. Published by Pearson Learning Solutions. we have no simple formula to determine its turning point. The revenue function is R = xp = x 28 . Inc. By selection a reasonable number of points (or by using the graphing capabilities of your calculator) we sketch the graph given in Figure 9. it appears that the minimum occurs at an x-value between 2 and 2. Economics. denoted by C. Example 4 Suppose the cost function is given by the equation C1x2 = 80x + 400/x + 500. by Warren B. However.Section 1.5. The definition is Applied Calculus for Business.6 Economic Functions * ** 121 Figure 8: R = xp = x 28 .x are expenses arising even before production begins. is to estimate the minimum cost. This particular example is quadratic so we could easily determine the minimum cost. and April Allen Materowski. Gordon. when the cost function is not quadratic. Sketch its graph and estimate the point at which the cost is minimized. A typical problem is to minimize the producer s cost. x 7 0. Walter O. Economists define a related function which averages the overhead cost among all the items produced. This cost is sometimes called the overhead or fixed cost. it is called the average cost. We saw that the overhead figures substantially into the cost function. Published by Pearson Learning Solutions. at this time. Solution. for example purchase of machinery and labor. Wang. as we illustrate in the next example. the best we could do. . Figure 9: C1x2 = 80x + 400/x + 500 From the sketch. and Finance. Copyright © 2007 by Pearson Education. Wang. except for very simple profit functions. we shall be content with determining the profit function and estimating its maximum as illustrated in the next example. the x-value.000 The price charged p = 28000 . P18002 = 800 28000 . the producer s objective is to maximize the profit.218002 = 26400 = $80 A sketch of the profit function is given in Figure 10. .C1x2 (3) Usually. Copyright © 2007 by Pearson Education. that is.2x.200x + 100.000 = 95.000 while C1502 = 95000/50 = 1900. when x = 800.1000x + 20000. Published by Pearson Learning Solutions. Once again. For now.20000 = $204. Economics. Applied Calculus for Business. the average cost is the total cost divided by the number of items produced. Solution. and Finance. We shall leave that determination to later on in this text. Example 5 Determine the total and average cost of producing 50 items if the total cost is given by the equation C1x2 = 2x2 . Example 6 The demand for some commodity is given by the equation p = D1x2 = 28000 .6 Economic Functions C1x2 = C1x2 x (2) Thus.2001502 + 100. at this time.000. (Assume the price and cost are given in dollars) Solution. we are unable.1000x + 200002 or P1x2 = x 28000 .20000 Therefore.180022 + 100018002 .1x2 . Determine the Profit function and determine the profit and price charged when the level of production is 800 items. we have P1x2 = R1x2 . The profit function is P1x2 = x 28000 .2x . Which of the two costs do you think is more useful to use? Why? Profit is the difference between the money you take in (revenue) and the money you expend (cost) therefore. Walter O. We have.x2 + 1000x . while the production cost is given by the equation C1x2 = x2 . by Warren B. C1502 = 215022 . that results in the maximization of profit. Gordon.218002 .122 * ** Section 1. and April Allen Materowski. if we represent profit by P.2x . to determine the exact level of production. Inc. owns a large apartment complex containing 400 identical apartments.6 Economic Functions * ** 123 P(x)= . Find an expression which may be used to represent their monthly income and determine the monthly rent to be charged to maximize their income.1000 Figure 10: P1x2 = x 28000 .1802 = 60 Applied Calculus for Business.21600/21 . Example 7 BBC Rentals Inc.3n Thus. by Warren B. . so we know the maximum for R occurs at its vertex which is at n = .1000 Note that our graph suggests the profit is maximized when the level of production. Table 1 Monthly Rent 800 No. Solution. and April Allen Materowski. Table 1 indicates what is happening with regard to rent and the number of rented apartments if there are 1. is about 550. we see the monthly revenue (income) is the product of the monthly rent and number of rented apartments. Observe their monthly income is the revenue derived from the rent.180n2 + 21600n + 320000 Fortunately for us. Inc. Published by Pearson Learning Solutions. the R = 1800 + 60n21400 . therefore.x2 + 1000x .Section 1. If the monthly rent for each apartment is $800 all the apartments are rented. and Finance. We give two different approaches to the solution to this problem. We are looking for a pattern for n monthly increases in the rent. x. of Rented Apartments 400 800 800 800 800 + + + + 11602 21602 31602 n1602 = = = = 860 920 980 800 + 60n 400 400 400 400 - 1132 2132 3132 n132 = = = = 397 394 391 400 . The exact value of x may be found using calculus. as we shall see later on in this text. Copyright © 2007 by Pearson Education. Wang.3n2 = . METHOD 1. 3 increases of $60. Gordon. Economics. opening downward. 2.2x . Walter O. the graph of R is a parabola.x2 + 1000x . However. for each $60 monthly increase in rent 3 apartments become vacant. the monthly income is the product of the monthly rent and the number of rented apartments. then the best we could presently do would be to find an approximate solution using the methods described below. the number of rented apartments is then 400 . 800) the number of rented apartments when there are no increases. by Warren B. Gordon. Entering. The calculator can be used for determining the intersection of two or more graphs (equations).3x = 42 and y + 2x = 72 (note we replaced p by y since y requires fewer keystrokes on the calculator).E1x2 (4b) Calculator Tips Later in this text. and (397.800 = . More generally. Economics. we have p . cost and profit are usually considered. Thus. Published by Pearson Learning Solutions.20x + 88002 = . so the marginal cost. there are 180 vacant apartments when the monthly income is maximized. In the above example.4002 or p = . we shall give another definition of a marginal function that is virtually identical with (4). and Finance. the slope is 1860 .201x . one is to use the solve command. when we study applications of the calculus.20x + 8800 The monthly revenue is R = xp = x1 . that is. Consider Example 1 where we needed to find the simultaneous solution to y . C(x) is the cost of producing x items.400. C1x + 12 is the cost of producing x + 1 items. The marginal revenue. This is a linear relations between the rent p and the number of rented apartments x. and April Allen Materowski.4002 = 60/1 .20x 2 + 8800x This is a parabola opening downward so the maximum value for R occurs at its vertex which is at 8800 x = = 220 21202 thus 220 apartments are rented when the revenue is maximized and the monthly rent is p = .31602 = 220. Wang. . In Economics.20. how is x related to n? Note that we were able to solve this problem exactly because it was a quadratic function. Inc. 860) when there is one $60 increase. If it were not. METHOD 2 For every $60 increase in rent. let E(x) represent the equation of any realistic economic function then its marginal value is E1x + 12 . the marginal cost is defined by MC = C1x + 12 . their difference is the cost of producing the x + 1 item. on the Home screen Applied Calculus for Business.124 * ** Section 1. Copyright © 2007 by Pearson Education.32 = .6 Economic Functions Thus the monthly rent to be charged is 800 + 601602 = $4. Similarly the marginal revenue would be the revenue derived by producing the last item.C1x2 (4a) Marginal Functions Its interpretation is clear. so in particular when we have the following two points on the line (400. 3 apartments become vacant. Note. Using the point slope formula. There are two approaches.2012202 + 8800 = $4400.8002/1397 . For example. Walter O. a function based on another function is also defined. namely the marginal function. NOTES: (1) The reason the calculator asks you to confirm the curves. Next we check our Window 1* F22 to make sure it produces a useful sketch in the first quadrant. To indicate that x Ú 0 and y Ú 0. often. this produces the desired one. Walter O. it will give a numerical approximation. the calculator produced the exact point of intersection. we enter solve 1y + x2 = 12 and y . In this example. Similarly. and y21x2 = 4 + x2 x Ú 0. We are interested in the fifth one. We then have the calculator sketch the graph. if the window is not a good one. ZOOM (press F2) and select ZOOMFIT. We illustrate by considering the problem of estimating the minimum cost for the function defined by the equation C1x2 = 80x + 400/x + 500 with x Ú 0. the key is to the left of the 7 key. if we want to solve the pair of equations in Example 2. Now it asks for a Lower Bound. We inserted the condition x Ú 0. Later on.x2 x Ú 0. see Figure 15. y62 You will observe the calculator also gives the negative solution for x. it may not be a very good one. since we recognize these graphs as parabolas and they are also defined for negative values of x. You could then refine your window and select values that give a better window and improve upon your estimate. as you study the calculus. press Enter again.3x = 42 and y + 2x = 72. move the cursor to any point on the graph to the right of the intersection point and press Enter. Economics. Scroll down and press Enter. Copyright © 2007 by Pearson Education. It then asks for an Upper Bound. (2) It asks for a Lower Bound. you could enter instead solve 1y + x2 = 12 and y . The calculator indicates a point on the first curve. is that you may have more than two curves on the screen and the calculator needs to confirm which curve s intersection(s) you want to determine. the Intersection. An alternate approach is to have the calculator graph the functions and then deter. Wang.6 Economic Functions * ** 125 solve 1y .Section 1. then it indicates a point on the second curve. To do Example 2. press Enter. xc = 2 and yc = 8. Note that braces enclose x and y. We can now use the calculator to determine the point of intersection.2).x2 = 4. 5x. either by creating a table of values for the given function or by sketching the graph of the function. see Figure 14. We can now press Graph 1* F32 to get the sketch of the two graphs. . and an Upper Bound.) The calculator can be very helpful in estimating the points at which functions achieve their maximum or minimum values. a point to the left of the intersection. a point to the right of the intersection so as to sandwich in the correct intersection. y62 (Recall the space bar is Alpha 1 . Pressing the F5 key gives a screen of options. It might be that there are several points of intersection. see Figure 12. We enter the equation in the Y = screen. press Enter when you are to the left of the intersection. Gordon. you will learn the exact method for locating such points. but by tracing the graph (F3) you should see the lowest point on the graph is between x = 2 and 3. and April Allen Materowski. in this case.x 2 = 4.Figure 11: Defining the Functions mine the point of intersection. The calculator produces the intersection point on the bottom of the screen. move the cursor to any point on the curve to the left of the intersection point (as you move it indicates the point on the curve on the bottom of the screen as Figure 12: Setting the Window xc and yc. we solve for y and in the Y = screen we let y11x2 = 12 . see Figure 13. and Finance. See Figure 11. Try this and see how well you estimate the x-value at he minimum point. 5x. (This is also true of the solve command. There are several ways we may obtain the approximate values. by Warren B. 5x. y6 x Ú 0 and y Ú 02 Recall Ú is obtained by pressing * # . Published by Pearson Learning Solutions. You should now have a picture of the graph. Figure 13: A Sketch of the Two Graphs Figure 14 Applied Calculus for Business. Inc. Suppose. p . 20. and Finance. use the lower and upper bound to sandwich the minimum point on the graph and then press Enter to obtain the coordinates of the minimum. starting at 2 and take the change ¢ tbl to 0.5p + 2x + 70 = 0.8x2 = 20 7. Express x in millions of bushels. When the cord price is $50. 2p . there will be no supply. p + x = 100 2p .x p = 236 . . (a) Determine the price per radio when 500 radios were demanded. when the cord price is $100.11p = 0 2 13. Economics. and April Allen Materowski. Wang.3x = 16 3. (b) the cost 8p . . Suppose the cost of producing x items is given by the equation C1x2 = x2 . When the price was increased to 50 cents.3x = 17 2p + x = 30 21. As before. the demand will be for 3 million bushels.x . 400p = 1000x + 16000 6. find the demand function and the price function. The supply of wood to a mill is found to vary linearly with the price per cord. Find the linear supply function satisfying the following conditions: when the price is $4.6 In Exercises 1 -7 (a) Determine if the given equation is a supply or demand equation. How does this change effect the point of market equilibrium? 27. p = 2x + 14. 12. (c) determine the coordinates of market equilibrium. Plot both on the same set of axes and find the point of market equilibrium. Walter O.11000 = 0 5. If the demand equation is linear. When the price is $11.1. If you want to improve upon this estimate go back and redefine your table.7p + 2x + 7 = 0. In each exercise (a) sketch their graphs on the same set of axes. x + 4p = 16 2. Gordon.300 = 0. The commodity suddenly increases in popularity which is reflected in an upward adjustment in the demand curve. the number of paying riders per month was about 50 million.12 = 0 4.3x . The supply equation for a certain commodity is x = 4p .5x2 = 3 p + 3x2 = 75 26. 2p . (b) identify which is the supply and demand curve and the appropriate domain. Scrolling down the table the y1 value is minimized around x = 2. This adjustment creates a new demand curve. 16. where x represents the number of items in hundreds. in Example 7. When the price per radio was $15. Applied Calculus for Business.126 * ** Section 1. find the demand function and the price function. it costs the BBC realty company. parallel to the old one and such that at every price the demand is 3 units higher than previously.5 units. the supply will go up 2.2x2 = 1 p + 4x2 = 25 23. x + 6x .1. Find the new demand equation.000 tons. 2p + x . find the maximum possible demand and the maximum price that can be charged. Does a linear model seem appropriate near p = 0? In Exercises 20 25 you are given a pair of supply and demand equations. 8. 1. on the average. For every dollar increase in price.x x + 8x + 220 .6 Economic Functions Alternately press APPS and select TABLE. by Warren B. (b) Plot the graph of the given equation. Determine (a) the overhead cost.950 = 0 24. find the minimum price for which there will be any supply. What monthly rent should they charge if their profit is to be maximized? 28. The demand equation is 2x + 4p = 15. For a demand equation. 2p + x .490 = 0 p = 216 .3. Copyright © 2007 by Pearson Education. 9. . ridership dropped 20%. Published by Pearson Learning Solutions. (a) How many cords are supplied when the price per cord is $175. (b) above what price will the demand be zero? (c) Why isn t the demand infinite when the price is zero? 14. 500p + 700x . When the price of a New York City bus ride was 35 cents.3x + 2300. 18. and press Enter twice. press F5 and then scroll down and select Minimum. The highest price for which there is any demand is $40 per ton. 3p + 6x2 = 12 In exercises 8 12. (d) determine the revenue equation and (e) determine the revenue at market equilibrium.01 or smaller.384 + 12p = 0 2 17. Figure 15 EXERCISE SET 1. one representing a supply curve and the other a demand curve. 19. The third method: with the graph displayed on the Y = screen. 2p . 1000 radios were demanded. Plot the graph and determine the maximum monthly ridership if bus rides were free.120 = 0 p = x + 6 25. $800 each month to maintain each vacant apartment. Find the linear supply function satisfying the following conditions: the minimum price for which there is any supply is $3. there will be 600 units supplied. 3p + 15x . 11. For a supply equation. If the demand equation is linear. The maximum possible demand for a certain commodity is 20. When the price of a certain farm product is 40 cents per bushel. and C the cost in dollars.7x = 60 22. 10. p . (b) What is the lowest price above which cords will be supplied? 15. If the price increases to 60 cents per bushel. p . where p is the per unit price for x items.200 = 0 2p + 5x . 700 radios were demanded.200 = 0 5p + 14x . the demand will be for 6 million bushels. you are given a pair of equations. Press SETUP (F2) set tblStart = 2 and ¢ tbl = . identify which is supply and which demand. Find the demand equation assuming that it is linear. In a small town it has been discovered that the relationship between the number of radios demanded varies linearly with its price. thirty cords are supplied. ten cords are supplied. Inc. when the price was $30. When the price for a color television is $240. Let us. (c) the revenue derived from the sale of the 20th item the marginal revenue when x = 20. that this particular function may be easily factored. Published by Pearson Learning Solutions. These techniques will usually not yield all the information we require but we will resolve these deficiencies later. (f) the level of production which minimizes the cost and the minimum cost. Wang. (d) the average cost of producing 50 items. (c) the average cost function. Walter O. In Example 7.5x . in dollars. At this point in the text we have not yet developed the machinery to allow us to use this technique. Suppose the demand equation is p = 22000 . find the relationship between n and x in the two methods used to solve the problem. of producing x items is C1x2 = 100x + 2000/x. namely. and April Allen Materowski. x = 3/2 and x = . at which profit will be maximized. Consider the quadratic function defined by the equation f1x2 = 6x 2 . we can use some simple techniques like factoring and sign analysis of the function (reviewed in Section 0. Copyright © 2007 by Pearson Education. Determine (a) the cost of producing 50 items. Gordon. How many color televisions should be sold to maximize monthly sales and what price should be changed to maximize sales? 33. Using the sign analysis of Section 0. by Warren B. 29. Of course we know this is a parabola. 30. where x is the number of items sold when the per unit price is p dollars. 32. (b) the cost of producing the 50th item. Inc. (d) the level of production that maximizes the revenue.7 of producing 100 items. the marginal cost when x = 50. (e) the average cost of producing 100 items.5) to get some information about the function. Suppose the cost. and the maximum revenue. (c) the cost of producing the 100th item (the marginal cost at x = 100). However. Suppose the demand equation is 200x + 10p = 10000.3213x + 22 This factored form immediately provides the x-intercepts (also called the zeros) of the function.5x. Determine (a) the revenue function. everywhere else it must be either positive or negative. . 31.6. (b) the Profit function. (e) estimate the level of production which minimizes the cost and estimate the minimum cost. (b) the domain revenue function. Determine (a) the revenue function. (d) estimate the level of production that maximizes the revenue. Suppose it has been determined that the demand (in thousands of dollars) for 20 a certain item is given by the equation p = . and we have f1x2 = 12x .7 More on Functions » » » » » » » » Using the Zeros Even Functions Symmetry about the y-axis Odd Functions Symmetry about the Origin Rational Functions Vertical Asymptotes Horizontal Asymptotes Translations Calculator Tips One of the most effective techniques for sketching the graph of a function is the calculus. for the moment. Using the Zeros Applied Calculus for Business. (b) the domain of the revenue function. and can sketch its graph very well using the methods of Section 1. Observe. For each $5 increase in price. (c) the revenue derived from the More on Functions * ** 127 sale of the 100th item the marginal revenue when x = 100. where x is the number of items sold when the per unit price is p dollars. (d) from your estimate in (c).5. (c) estimate the level of production x. the monthly sales for this item at a department store is 450. what price should be charged to maximize profit? 34. Determine (a) the revenue function.4. pretend not to recognize this function. (d) the average cost function. Economics. That means that the only time the value of this function is zero is at these two x-values. and Finance. and the maximum revenue. and the cost of producing x 1x hundred items is given by C1x2 = 5x + 10. 1.2/3.Section 1. the monthly sales fall by 5 units. we have Figure 1. There is still a lot we do not know about the function.92 = x1x . Published by Pearson Learning Solutions.9x Applied Calculus for Business. Using sign analysis we have Figure 3. we analyze the sign of the function and draw a possible graphical representation. Example 1 Sketch the graph of f1x2 = x3 . a graph illustrating these properties is given in Figure 2. However. . Therefore.3213x + 22 This tell us the y-values are positive when x is less than .3. with the understanding that.9x. more information about functions will become available.2/3 or greater that 3/2 and negative elsewhere. We have f1x2 = x1x 2 . Economics. We consider a cubic equation in the next example. for example its turning points or how it increases or decreases. as your mathematical development increases. if we did not recognize this graph that it did not Figure 2: A Representation of f1x2 = 12x . Once the zeros are located (meaning we can factor the polynomial). Inc. The best we can do at this point is to graph a possible representation of a function. the zeros of the function are at x = . Solution.128 * ** Section 1. Gordon. this information does not yield the coordinates of the turning point which of course. Wang. does it do it quickly or slowly? These questions are most effectively answered using the calculus. 0 -3 + 0 0 0 3 + x Figure 3: The Sign of f1x2 = x3 . Nor would we know. This idea may be extended to higher order polynomial functions provided we can determine all the zeros of the function. Walter O. This cubic may be easily factored. 0 and 3. is the vertex. note the use of the word possible. and Finance.7 More on Functions 0 + 2/3 0 + 3/2 x Figure 1: The sign of 12x .321x + 32 Therefore. Once again. by Warren B.3213x + 22 Using Sign Analysis have additional turning points. and April Allen Materowski. Copyright © 2007 by Pearson Education. as we shall see later on this section. Wang. it cannot cross the x-axis. 4). Therefore its graph must either be above the x-axis or below it.3 or when 0 6 x 6 3 and positive when . This function is always positive it has no real zeros therefore. What do we do when we cannot factor the polynomial? At this point. it is clear that we cannot yet determine the exact coordinates of the turning points of this graph which we shall see occur when x = . 23. Figure 5: f1x2 = x4 + 4x2 + 4 While this graph may look like a parabola. and Finance. Applied Calculus for Business. Consider the function defined by f1x2 = x 4 + 4x 2 + 4. it may happen that a function has no zeros. Copyright © 2007 by Pearson Education. however. Economics.9x Once again. the y-values are negative when x 6 . Figure 4: A Representation of the Graph of f1x2 = x3 . what does this mean? If it has no zeros then the sign of the function must either always be positive or always negative (why?). Gordon. and April Allen Materowski. by Warren B. which may be factored and written as f1x2 = 1x2 + 222. A sketch is given in Figure 5. it is not. Published by Pearson Learning Solutions. It is flatter near the origin and rises more quickly than a parabola.7 More on Functions * ** 129 Therefore. A sketch of a graph satisfying these conditions is given in Figure 4. Sometimes. not very much. Inc. Walter O.Section 1..3 6 x 6 0 or x 7 3. the calculator can be very helpful in such cases. Because of the symmetry of this graph (an idea we examine next) it follows that the lowest point on this graph is at the y-intercept which is (0. . complete its sketch for x 6 0.23 = 0 which becomes y3 . The first symmetry we examine is symmetry with respect to the y-axis. by Warren B.3x 2y . Gordon. the same equation we started with. This translates into if f1 + x2 * f1x2 then f is an even function (1) Sometimes. as we shall see. Figure 6 Applied Calculus for Business. the y-values must be the same. the graph of the equation is symmetric with respect to the y-axis.31 .x leaves the equation unchanged. Copyright © 2007 by Pearson Education.x24 . We can rephrase the condition as follows: The graph of an equation is symmetric with respect to the y-axis if replacing x by . we considered the axis of symmetry. That is.3x2y . for equally spaced x-values to its left and right. . the portion of the graph to its left was a reflection of the graph to its right. Wang. Economics.23 = 0 is symmetric about the y-axis. as we illustrate below. that means for equal and opposite x-values on its graph.130 * ** Section 1.7 More on Functions Even Functions Symmetry about the y-axis When we plotted parabolic functions.x22 + 12 = x 4 . Knowing a function has a symmetry is often useful in sketching its graph.3x2 + 12 = f1x2 satisfying (1).x we have y 3 . Walter O. that means we need only worry about sketching its graph for x Ú 0. there are functions which are not polynomials that also exhibit this symmetry. and April Allen Materowski. (b) If we replace x by . the equation representing the function is not given in this form. the vertical line about which.x2 = 1 .x22y . (b) Show the graph of the equation y3 . We are interested in determining under what conditions functions may have two other kinds of symmetries. (a) f1 .3x 2 + 12 is even. If a function whose equation is given by y = f1x2 is to be symmetric with respect to the y-axis.23 = 0. functions which have this symmetry are called even functions. and Finance. it had the same y-values on the graph. However. the axis of symmetry was like a mirror. Published by Pearson Learning Solutions. x must appear only to even powers. Example 2 (a) Show the function whose equation is y = f1x2 = x 4 . Therefore. If we know a function has even symmetry. Do you see why we call functions which have symmetry with respect to the y-axis even functions? For polynomial functions to have this type of symmetry.31 . Solution. the rest of its graph is just its reflection about the y-axis. Example 3 Assume the function whose graph for x Ú 0 is given in Figure 6 is even. Inc. Published by Pearson Learning Solutions. it is symmetric with respect to the y-axis.2x5y 3 = 0. More generally. Walter O.x2 = .y2 . in this case 1 . If a graph is symmetric with respect to the origin. (b) Show the graph of the equation xy .f1x2 then f is an odd function (2) Odd Functions Symmetry about the Origin Sometimes.x21 . by Warren B.x2 or . Applied Calculus for Business. Therefore.x251 .y23 = 0 or xy .2x5y 3 = 0 is symmetric with respect to the origin.Section 1. .7 More on Functions * ** 131 Solution. Another example is 2 the function defined by the equation f1x2 = x3. 8) is on the graph.21 .82.14x 3 .2x2 = .x2. it follows that f1 .f1x2. it has the required symmetry. Observe that a continuous even function (by continuous.4x 3 + 2x = . You should verify this conclusion by sketching the graph of this function. if (x. and April Allen Materowski. y) is a point on the graph so too is 1 .x2. we mean the graph has no holes or jumps ) must have a turning point at its y-intercept (why?) The other symmetry we wish to consider is symmetry with respect to the origin.x.y = f1 .x2 = 41 . that is. that means for example. Economics. Consider the absolute value function defined by the equation y = f1x2 = x . Copyright © 2007 by Pearson Education. functions which have this symmetry are called odd functions. Thus . we write the condition as if f1 . Example 3 (a) Show the function defined by the equation y = f1x2 = 4x3 . (b) replacing x by . The complete graph is given in Figure 7. Gordon. . Figure 7: Completion of the Graph by Symmetry Non-polynomial functions can also exhibit even symmetry. then its mirror image with respect to the origin is also on the graph.x23 .2x is odd.2. . Solution. Wang.21 . if the point (2. Now if we define y by y = f1x2.1.x and y by . and Finance. (a) f1 . therefore this function is even. multiplying by .x and y by .y we have 1 . Note the portion drawn for x 6 0 is just the mirror image about the y-axis.y leaves the equation unchanged. then the y-coordinate at . We also leave as an exercise for you to verify that the product or quotient of even functions is also an even function.x2 = . the equation representing the function is not given in this format. Inc.y2.f1x2 = f1 . We can rephrase the condition as follows: The graph of an equation is symmetric with respect to the origin if replacing both x by .x is f1 .x2 = f1x2 (why?). the same equation we began with. (1. 1 .y leaves the equation unchanged. However. by Warren B.1.7 More on Functions Example 4 The portion of the graph for some odd function is given for 0 8.5. Inc. Walter O. and conversely the graph of any equation which satisfies this property is symmetric with respect to the x-axis. Functions cannot exhibit this property (why?). 11. p1x2 A function defined by an equation of the form r1x2 = where p(x) and q(x) are q1x2 polynomials is called a rational function. There are two questions that we shall examine: (1) Division by zero is undefined. You might note we did not consider symmetry about the x-axis.5. 1).2. y x 2 is given in Figure Figure 8 Solution. . and Finance. Copyright © 2007 by Pearson Education. Also note that the shape of the curve changes at this point as well. 12 and 1 . We shall examine this more fully in the exercises. 1 .5.0. the graph of an equation which has this kind of symmetry has the property that replacing y by . 0).1. Wang. 0) are respectively 1 .5. and April Allen Materowski.12 and (2. . .132 * ** Section 1. 02. 02. Figure 9: Completion of the Graph by Symmetry Note that continuous odd functions pass through the origin (why?). Note that the reflection of (0. Sketches of such functions illustrate their interesting features. We shall talk about this point more when we consider the calculus. Complete the sketch giving the graph for .12. See Figure 9. The shape is sometimes called concavity. Gordon.2 x 6 0. Economics. Suppose there are x-values at which the denominator vanishes (while the Rational Functions Applied Calculus for Business. We need only reflect this graph about the origin to obtain the complete sketch. Published by Pearson Learning Solutions. What about negative values.x2 = 1 . by Warren B. in the above example we examine what happens if we select x-values near 2 and compute the corresponding y-values (we round the y-value to the nearest integer).2 or 2. .000001 2.000.000 50. Walter O.99999 1.000.000 or . and Finance.999999 2 2. Suppose.001 r1x2 = x for x near 2 x . suppose.x2 tion is symmetric with respect to the origin. however. x suppose r1x2 = 2 . Such a quotient is said to be either positively or negatively infinite. x . we substituted either of these values. Inc.500. its numerator is an odd function and its denominator is an even function. it is clear that this function is defined for all real numbers except x . Table 1: The behavior of r1x2 = x 1. say .000. the result would have the form N 0 Where N is a non-zero number. See Table 1. x2 . to see why. Economics. and the ratio of an odd and even function is always odd.10.9999 1.000 Undefined 500.50.2 and 2. Copyright © 2007 by Pearson Education. In the previous example.4 1 . that is.00001 2. the denominator vanishes. Gordon.5000 . What happens to the behavior of the functional values (the y-values) at x-values near where the denominator approaches zero? (2) What happens to the functional values as x becomes a very large positive number or a very small negative number (note.000? We shall soon consider these questions. We ask you to examine this and other symmetry combinations in the exercises.100)? For example.000 or even larger.x2 2 -x = 2 = .4 Note that r(x) is the ratio of two polynomials. it is an odd function.1. Published by Pearson Learning Solutions.000 .1. How does the function behave near these two points? What happens to the functional values as x becomes very large. when x is either . but first observe for this particular example that first examining its symmetry reveals useful information.4 2 x x2 . Let us now examine the first question. say 1.000 5000 500 Applied Calculus for Business.500 .999 1.50 .4 x = .0001 2. and April Allen Materowski.99 1. Wang.000.000 or 10.000. The domain of this function consists of all real numbers except these two values. Example 5 Discuss the symmetry of the function defined by the equation r1x2 = Solution. Suppose there are x-values at which the denominator of a rational function is zero while the numerator is not zero.4 .000 is a very small negative number compared to .Section 1. .r1x2 therefore the func4 x .7 More on Functions * ** 133 numerator does not). We have that r1 . Sometimes we say that y-values become infinite at these points.00001 . consider the graph of this function near x = 2. Figure 11 is a sketch of the graph of the function. and Finance. see Table 2.500 Thus. The function behaves similarly near x = .2 and x = 2 to show the functional values become infinite near these lines.134 * ** Section 1. the larger their y-values are in absolute value.4 50 500 5000 50.7 More on Functions Observe that if we take x-values to the left of 2 and approach 2.2.50. and April Allen Materowski.5000 . Economics. Published by Pearson Learning Solutions. . by Warren B. Wang. that is the closer we get to these x-values. We have a similar behavior of the function near x = .2.1. ultimately becoming infinite as we get closer to x = 2.000 .1. Copyright © 2007 by Pearson Education.000 Undefined .000 . Gordon. Applied Calculus for Business.1. As we approach 2 from the right the y-values become larger and larger.999999 -2 . the y-values become very large (in absolute value). its graph is symmetric with respect to the origin).000 500. Inc. To illustrate graphically.99 . x=2 Figure 10: The Graph of r1x2 = near x = 2 x x .2.999 .4 2 x x2 .500.2. Table 2: The behavior of r1x2 = x .1.001 r1x2 = x for x near . Walter O.0001 . see Figure 10.2 x . as the x-values get close to this vertical line. the y-values become more and more negative and their absolute value is very large.2 (which does not surprise us because the function is odd. note we include the vertical lines x = . near x = .000001 .1.99999 .2 and x = 2. These lines are called vertical asymptotes.9999 .4 2 Note that we included the vertical line x = 2 in the sketch to indicate what happens as the curve approaches it.2. In the previous example. when x = 3. then the line x = a is called a vertical asymptote of the function.3211x + 12 Applied Calculus for Business.Section 1. The domain of this function is all real numbers except x = . given any function. and Finance. the zeros of the denominator (which are not zeros of the numerator) are the vertical asymptotes of the function.x . going from one piece to another on either side of the asymptote results in an infinite jump. If we factor. Walter O. therefore the two vertical asymptotes of the given function are x = . . Thus.1 is both a zero of the numerator and denominator. suppose when x = a the denominator vanishes and the numerator does not. 2x 2 . the y-value is 0.1 and 3/2. Note at either of these values the numerator is not zero.1 and x = 3/2.321x + 12 = 0. The zeros of the numerator provide the x-intercepts or zeros of the function. Economics.4 2 Note that the above rational function is one kind of an example of a discontinuous function. Inc. we have. Gordon.3 This point is not in the domain of the function since the denominator is undefined there. Example 6 Determine the vertical asymptotes of the function defined by f1x2 = 2x .3 2 Vertical Asymptotes Solution. by Warren B.7 More on Functions * ** 135 x= 2 x=2 Figure 11: r1x2 = x x .3 = 0. f1x2 = 21x + 12 12x . The denominator vanishes when 2x2 . (Alternately. that is. the numerator is zero when x = 3.x . Copyright © 2007 by Pearson Education. observe x = . the line x = a is a vertical asymptote if the functional values become infinite near x = a. 2x . this function has separate pieces. Wang.x . and April Allen Materowski.1 and x = 3/2. More generally.) The domain of a rational function consists of all x-values with the exception of those at which the denominator is zero. or 12x . Published by Pearson Learning Solutions.6 . Nor does the function have a vertical asymptote at this point. This yields x = . Consider the rational function defined by the equation 2x + 2 f1x2 = . 2/5. Copyright © 2007 by Pearson Education. We shall see there are two possibilities. Note that the function defined in the previous example is not a rational function as the numerator is not a polynomial. in addition to rational functions. the next example illustrates one such case. x . later on in this text. Consider k/x p. If they do.1 2x . no matter what the constant value of k Applied Calculus for Business.1 (because this value is not in the original function s domain). A sketch of the function is given in Figure 12. Economics. The line y = b is said to be a horizontal asymptote of a function if the functional values are near b as the x values get very large. and Finance. thus you could write f1x2 = 2 for all x = . Wang. therefore x = 3 is a vertical asymptote. by Warren B. It will be helpful to first make some observations. and April Allen Materowski. .3 Solution.3 Note that if we did not exclude this x-value you might be tempted to write f1 .3 Horizontal Asymptotes We now examine the question of how a rational function behaves when the x-values are large. (We discuss this notion more fully. In a sense. then the horizontal line they approach is called a horizontal asymptote. in absolute value. have vertical asymptotes. Many different types of functions. Gordon. in absolute value. The domain of this function is all x Ú 0 and x = 3. The denominator is zero when x = 3 and the numerator is not. sort of a steady state value of the function.) We indicated that near a vertical asymptote the y-values become infinite. x=3 Figure 12: f1x2 = 1x x . Walter O. Published by Pearson Learning Solutions.1. the y-values on the curve level off and approach a horizontal line or they do not.136 * ** Section 1. where k is a constant and p is any positive number. you may do so providing you indicate that after the cancellation x may not equal .2/52. In fact the graph of this function has a hole or discontinuity at the point 1 .7 More on Functions While you might be tempted to cancel the common factor 1x + 12 from the numerator and denominator. we may think of the y-values stabilizing as the x-values get large. . Inc.12 = . Example 7 Determine the vertical asymptotes for the function defined by f1x2 = 1x . As x gets very large in absolute value what happens to this quotient? As the denominator gets very large. The above example illustrates that a cancelling zero of the numerator and denominator cancels the infinite behavior of the function near this x-value. written q .7 More on Functions * ** 137 is.3x + 11 6x4 + 3x 3 + 2x .q (The arrow is to be read approaches. instead of stating as x gets very large. 10. Try large values of x like 1000.10. . It seems clear from Table 3 that as x : q . as x gets larger and larger.2x 3 . similarly if x is . Published by Pearson Learning Solutions. We illustrate two related methods. .4994169 0. Copyright © 2007 by Pearson Education. For example x 10 will be a large positive number for a negative x-value. Thus we have the following: If x is a large number in absolute value.3 + 4b x x x x x x = 3 2 3 2 11 11 x4 a 6 + + 3 .000 (of course these are finite values. There is a simple algebraic procedure by which we may come to this same conclusion. Sometimes we say it approaches positive infinity. on the other hand x11 will be negative for negative values of x. 1.1.000 Á the quotient also approaches 0. Table 3 indicates the behavior of the function as x : q . 10. 3x4 . When x is a negative number whose absolute value is very large. then k/xp : 0 and xp may either : + q or . p 7 0 and k are constants. ) From now on. the quotient becomes. you can see what is happening.4999417 0.3 + 4b a3 . we shall write x : q . With your calculator.1000.2x 3 .11 3x 4 .3x + 11 6x 4 + 3x 3 + 2x . What happens to x p as x becomes a very large positive number? It should be clear p that x is a very large positive number as well.3x + 11 f1x2 = = 6x4 + 3x 3 + 2x .4b a6 + + 3 . .4b x x x x x x Applied Calculus for Business.Section 1. 0.3x + 11 as x gets large 6x 4 + 3x 3 + 2x . Gordon.000.11 x4 a 3 2 3 11 2 3 11 . and Finance. but its sign depends on whether or not p is an odd or even integer. 3x 4 . for example 100/x.000.2x 3 .00001 and so on.000 f1x2 = 0. Consider the function defined by the equation f1x2 = 3x4 . similarly.4999994 METHOD I Factor out the dominant (highest) power from both the numerator and denominator and perform any cancellation that may occur. Inc. 0. Wang.01. .11 Table 3: The behavior of f1x2 = x 1000 10. and April Allen Materowski.000. Walter O. 0. the y-values get arbitrarily close to 1*2. .000. but they give a picture of what is happening to the y-values as x gets large). instead of writing as x is a negative number which gets very large in absolute value we shall write x : . by Warren B.2x 3 . Consider. say 1000. Economics.11 Let us examine the behavior of this function as x : q (a very large positive number sometimes we will say as x becomes infinite).000.q . 1. the quotient gets very close to zero. then the result is infinite as well.000.000 1. 3x 3 + 17x + 21 12x4 : = 4 5x2 . METHOD II Keeping the dominant terms in the numerator and denominator.4 + 3 x4 a 2 . This second method indicates that we need only keep the dominant terms in the numerator and denominator to determine if there are horizontal asymptotes.3x + 11 1 = : 4 3 4 2 6x + 3x + 2x .q . that means the only terms left in the above expression are the 3 in the numerator and the 6 in the denominator. the dominant term in the denominator. Inc. and April Allen Materowski.q . the dominant term in the numerator is the one with the highest power. all other terms are insignificant with respect to it for very large values of x. that is y = 1*2.q . k/xp : 0 as x : q or as x : . so we have.138 * ** Section 1.2 + 3x 4 and (b) x : . is the term with the highest power. we already observed above that every term of the form k/xp approaches zero. Published by Pearson Learning Solutions. Applied Calculus for Business. METHOD 2.2 + 3x 4 x4 a 12 = 3 17 21 3 17 21 + 3 + 4b 12 + 3 + 4 x x x x x x = 5 2 5 2 .q . Gordon. Example 8 12x 4 . Economics. That is as x : q . Similarly. for very large x-values. Copyright © 2007 by Pearson Education. is also a horizontal asymptote when x : . as x : q or as x : . METHOD I f1x2 = 12x4 .q ) or equivalently. thus the horizontal asymptote is y = 1*2. we see that f1x2 = 12x 4 . Note that x is a negative number. the above analysis is identical and the curve approaches y = 1*2. The next example illustrates that it is possible for a function to behave differently as x : q and as x : . y = 4 is the horizontal asymptote as x : q or as x : .2 + 3x 4 3x4 Yielding the same conclusion as in (a).q 12 - (Remember. by Warren B. very large in absolute value. As x : q .4 + 3 x2 x In either (a) or (b).3x 3 + 17x + 21 5x2 . Walter O.3x 3 + 17x + 21 Examine the behavior of the function f1x2 = as (a) x : q 5x2 . .11 6x giving y = 1*2 as the horizontal asymptote. and Finance. Solution. all other terms are insignificant with respect to it. f1x2 = 3x4 3x 4 . Wang.4 + 3 b x2 x x x 1 17 21 + 3 + 4 x x x : 12/3 = 4 5 2 .7 More on Functions as x : q . f1x2 : 3/6 = 1*2.2x 3 . 4 is a horizontal asymptote. Economics. 4x 2x + 1 2 : 4x 2x 2 x if x Ú 0 . Using Method II. y = 4 is a horizontal asymptote and as x : .q f1x2 = recall that 2x 2 = e thus. we have as x : q or as x : . Inc.x if x 6 0 : 4x 2x 2 = 4x = 4 x and when x : . as the next example illustrates. 0) and since the denominator is always positive. It may happen that a rational function may get infinite as x : q or as x : . It is straight-forward to sketch this function (which is odd). Wang. Walter O. A sketch of this function is given in Figure 13.q . the sign of the function will be determined by the sign of the numerator. and Finance. Note the two different horizontal asymptotes should not be surprising as the function in the previous example is odd.7 More on Functions * ** 139 Example 9 Determine the horizontal asymptotes for the function defined by f1x2 = Solution. this function has two different asymptotes. Note it passes through (0. which is positive for x 7 0 and negative for x 6 0. Published by Pearson Learning Solutions. when x : q (x is positive) f1x2 = 4x 2x + 1 2 4x 2x 2 + 1 .q (x is negative) f1x2 = 4x 2x + 1 2 : 4x 2x 2 = 4x = -4 -x Therefore. Gordon. y = . Copyright © 2007 by Pearson Education. y=4 y= 4 Figure 13: f1x2 = 4x 2 x2 + 1 .Section 1. by Warren B. as x : q .q . . and April Allen Materowski. Applied Calculus for Business. Inc. in each case.2x 2 + 3x + 11 Investigate the behavior of the function defined by f1x2 = as 2x2 + 5x + 11 x : q and as x : .3x 5 -3 3 = x : 2 2 2 2x + 5x + 11 2x as x : q . f1x2 = 3x2 .q .3x5 . and April Allen Materowski. Wang. k/xp : 0.2x2 + 3x + 11 f1x2 = 2x2 + 5x + 11 Example 11 Determine the behavior of the function defined by f1x2 = and as x : . Solution. .2x2 + 3x + 11 . Walter O. f1x2 : .3x 5 . See Figure 15. and Finance.q . as x : . Published by Pearson Learning Solutions.2x + 9 3 3/5 3x2 = = 5 : 7 7 4 5 5x 5x . We have.3x 5 . no horizontal asymptotes.11x 4 + 21 as x : q or as x : . therefore. . that is.q we have f1x2 : q .2x + 9 f1x2 = 5x7 . as x : q or as x : .q . A sketch illustrating the behavior of the function is given in Figure 14. Copyright © 2007 by Pearson Education. Figure 14: . when there is no horizontal asymptote or when the x-axis is a horizontal asymptote. There is no leveling off of the y-values. that is. therefore f1x2 : 0.7 More on Functions Example 10.140 * ** Section 1. and large in absolute value. we see that a large number to a power is large and multiplying it by a negative number makes it negative. by Warren B.axis. y = 0 is a horizontal asymptote.) Applied Calculus for Business.11x4 + 21 Did you observe in the last two examples. Gordon. what happens with regard to the relative sizes of the dominant terms in the numerator and denominator? (See Exercise 65. On the other hand. Solution.11x + 21 5x x 3x 2 . x : q .q . We have as x : q or as x : . Figure 15: 3x2 .q f1x2 = . and decreases as you move to the right. Economics.2x + 9 as x : q 5x7 .q . x : q or as x : . Note how the function increases without bound (becomes infinite) as you move to the left.q the graph approaches the x. that is. 0) (0. The sign of the function is determined by the sign of the numerator and denominator. x + 3 We see to the left of the vertical asymptote x = . Published by Pearson Learning Solutions.3 (why?). x + 3 Solution. The y-intercept is 10. 2x . by Warren B. Note that the graph does not cross the horizontal asymptote. to its right they are approaching .4 = 0. or x = 2. . and as x : q or as x : . x + 3 We now connect these segments. Inc. we can verify this by trying to find those points at which f1x2 = 2 (see Exercise 71). Copyright © 2007 by Pearson Education. following the procedure described in Section 0.) ND + -3 0 + 2 x Figure 16: Sign of f1x2 = 2x . and Finance. the y-values are approaching q (since the sign of the function is positive). Gordon.q since the sign is negative. We have a vertical asymptote at x = .5. Walter O.7 More on Functions * ** 141 Example 12 Sketch the graph of the function defined by f1x2 = 2x . we have Figure 16 where ND means not defined (a vertical asymptote. The graph is given in Figure 18.4 Graph of f1x2 = . connecting the various segments of the graph.4 .4 . Applied Calculus for Business.4/32. there are no solutions to this equation. y = 2 is the horizontal asymptote (verify this!). The zero of the function occurs when the numerator is zero.3. Economics. and complete the sketch.Section 1. and April Allen Materowski. remembering that y = 2 is a horizontal asymptote. y=2 x = -3 (2. We begin our sketch in Figure 17 showing these basic facts. Wang.-4/3) Figure 17: Segments of the 2x .q . meaning the graph does not intersect the asymptote. . verify these statements. The next example illustrates the procedure. y = 0 is the horizontal asymptote. Figure 19: Beginning a 3x Sketch of f1x2 = 2 x + 9 Notice the piece of the graph on the right illustrating the fact that y is positive for x 7 0 and as x gets large it approaches the asymptote y = 0. the sketch will in general.0) (0. Copyright © 2007 by Pearson Education. not provide the location of the points where the graph turns. The calculus will provide those details. observe that f1x2 = 3x 3x 3 : 2 = : 0 as x : q or as x : .q . Wang. however. Example 13 Sketch the graph of the function defined by the equation f1x2 = 3x . We also know the y-intercept is (0. and also has y = 0 as x : . x + 3 Examining the sign of the function. Walter O. Published by Pearson Learning Solutions. We begin our sketch in Figure 19. Since the denominator is always positive. .*4/3) Figure 18: The Graph of 2x . Economics. There are no vertical asymptotes because the denominator is never zero (in fact it is always positive). the piece on the left illustrates that y is negative when x 6 0. Similarly. Gordon. We first observe that this is an odd function. by Warren B.7 More on Functions y=2 x = *3 (2.142 * ** Section 1. 0). x2 + 9 Solution. The graph Applied Calculus for Business. so its graph is symmetric with respect to the origin. we can sketch rational functions. the sign of the function will be determined by the numerator which is positive for x 7 0 and negative when x 6 0.4 f1x2 = . Inc. To find the horizontal asymptote.q x x + 9 x 2 Thus. and April Allen Materowski. and Finance. by Warren B. Economics. Walter O. we do not know the coordinates of the turning points. but a graph can never cross its vertical asymptote (why?). products. The calculator is also a useful tool in obtaining sketches of portions of their graphs. (At this point. or rational powers of polynomial functions is called an algebraic function. See Figure 21. and April Allen Materowski. Wang. . which results in the two turning points on the graph. We will have to wait until we study the calculus to find them. We shall see that the calculus is an essential tool in examining the behavior of such functions and sketching their graphs. there are functions for which the graph may cross its horizontal asymptotes one or more times. Inc. any function that can be expressed in terms of sums differences. Published by Pearson Learning Solutions. giving the middle piece. logarithmic and trigonometric functions. Copyright © 2007 by Pearson Education. to the left it is negative. and Finance. Gordon. in fact. A function which is not algebraic is called transcendental.Section 1.7 More on Functions * ** 143 passes through the origin.) Figure 20: The Graph of 3x f1x2 = 2 x + 9 Observe from Figure 21 that the graph crosses its horizontal asymptote at the origin. See Figure 20. examples of which are exponential. to the right it is positive. quotients. Applied Calculus for Business. Figure 21: A Graph Intersecting Its Horizontal Asymptote In general. We need only connect these pieces. 21x . Wang.21x . the graph representing f1x .21x .122 . Published by Pearson Learning Solutions.2x + 3 and f1x . by Warren B.144 * ** Section 1. Gordon.h2? We examined this question previously for the parabola and the circle. Inc. Copyright © 2007 by Pearson Education.12 + 3 x -1 0 1 2 3 4 5 f1x .12 = 1x .12 + 3. . f(x) = x2*2 x + 3 f (x * 1) = (x * 1)2 *2(x * 1) + 3 Figure 22: f1x2 = x2 .2x + 3 x -2 -1 0 1 2 3 4 f1x2 = x 2 .122 . consider f1x2 = x 2 . Walter O.12 = 1x . If we construct a table of values for each what do we observe (Table 4a and Table 4b)? Table 4a: f1x2 = x 2 .2x + 3 and f1x .12 + 3 11 6 3 2 3 6 11 Translations We observe the y-value for each x-value in Table 4b corresponds to an x value which is one less in Table 4a.12 = 1x .122 .122 . and April Allen Materowski. and Finance.12 = 1x . Economics.7 More on Functions Consider the graph of the function defined by the equation y = f1x2.12 = 1x .21x . how is it related to the function defined by the equation y = f1x .122 .21x . See Figure 22.12 + 3 Applied Calculus for Business. In other words. For example.2x + 3 except it is moved one unit to the right.2x + 3 11 6 3 2 3 6 11 Table 4b: f1x .12 + 3 is identical to the graph of f1x2 = x 2 . Gordon. we ask the question how do the graphs y = f1x2 and y = f1x2 + k differ? It is apparent that once again the graphs are identical with the y-value on the second graph k units above the first if k 7 0 and k units below the first if k 6 0. y = f1x2 and y = f1x .2x + 3 and g1x2 = x 2 . More generally. b) +* (a -h.b) +* (a. Inc. the middle curve at x = a and the curve on the right at x = a .7 More on Functions * ** 145 Observe the graphs are identical. y = f1x2 and y = f1x2 + k for k 7 0 For example.b) y = f(x+h) y =f(x) y = f(x . and Finance.h2 + k differ from it? It should now be ap- Applied Calculus for Business. except the graph on the right is positioned one unit to the right of the graph on the left. Observe that the curve on the left at x = a + h. Published by Pearson Learning Solutions. that is.h each have the same y-value. Note how the graph of f is translated 2 units upward to obtain the graph of g.h ) Figure 23: Horizontal Translation y = f1x + h2. b +k) +* y = f(x) + k (a. see Figure 24.h2. y = f1x . Wang.h2 is identical to the graph of the equation y = f1x2 except each point on the first graph is translated h units to the right if h is positive and h units to the left if h is negative. if we are given the graph of y = f1x2 how does the graph of y = f1x .2x + 5 (observe.Section 1. g1x2 = f1x2 + 2). Walter O. the graph given by the equation y = f1x . b) +* y = f(x) y = f(x) k +* (a. that is. their graphs are illustrated in Figure 25. each point on the graph on the left is translated one unit to the right. consider f1x2 = x2 . Copyright © 2007 by Pearson Education. and April Allen Materowski. Economics. b k) Figure 24: Vertical Translation y = f1x2 . We can now combine the two translations.h2 and y = f1x + h2 for h 7 0. . See Figure 23 where we have y = f1x2. (a.k. h 7 0 Next. by Warren B. +* (a+h. . Applied Calculus for Business. and April Allen Materowski. by Warren B. (*2.146 * ** Section 1. take every point on the original graph and subtract 2 from its x-coordinate and add 3 to its y-coordinate. That is. 0) (*1. to the left it is negative) and translated k units vertically (up if k 7 0. Inc.2x + 3 and g1x2 = f1x2 + 2 = x2 . sketch the graph of y = f1x + 22 + 3. Walter O. Gordon. See Figure 27. f(x) = x2 * 2x + 3 g(x) = x2 * 2x + 5 Figure 25: f1x2 = x2 . Wang.7 More on Functions parent that each point on the second curve is translated h units horizontally (to the right if h is positive. Published by Pearson Learning Solutions. for some function. 0) (2. To sketch the required graph the graph in Figure 27 needs to be move 2 units to the left and 3 units upward. *2) Figure 26: y = f1x2 Solution. Example 13 Given Figure 26 which gives the graph y = f1x2. 0) (1. and Finance. Copyright © 2007 by Pearson Education. 2) (*3.2x + 5 We illustrate with an example. down if k 6 0). Economics. even with a calculator.5. this theorem provides an initial window to use with our calculator. we have z 6 1 + max5 .5. 02 : 1 . 1 . Published by Pearson Learning Solutions.2. a2 . 1 and 3/2.) Clearly. 1) (0. we see that use of this theorem gives a window for the zeros.1xn . For example.5 and xmax = 2.1. Economics. and Finance. Walter O. Polynomials whose zeros are not easily found are hard to sketch without the calculus.3. 22 : 1 . (Note the zeros are actually x = . . there is a theorem (whose proof requires the more advanced mathematics studied in Complex Variable Theory) that is useful in determining a reasonable window in studying graphs with a calculator.4. by Warren B. 3) (-3. 3) (-1.5 means that . and April Allen Materowski.22 + 3 Thus.1 + a n . We state the following theorem.2 . See Figure 27. compute the absolute value of each coefficient and call the maximum value A) then z 6 1 + A The requirement that the coefficient of the highest term is one presents no problem as dividing a function by a constant does not change the position of its zeros. a n .5 6 z 6 2. Thus. Gordon.5 This means all the zeros lie within this interval.2x n . 3) Figure 27: y = f1x . Given the polynomial function defined by the equation f1x2 = xn + a n .2 + Á + a 2x 2 + a1x + a 0 Let z be any real zero of this function and A the maximum of 5 a n . Thus. 52 and so on for the other indicated points.3 2 x . when we cannot easily determine the exact zeros. in the above example. 5) (-5. Thus.2. Inc.3x 2 . g was obtained from f by dividing by 2. if we wanted the calculator to sketch f1x2 = 2x 3 .5 z 6 2.2. Á . for exam3 2 ple f1x2 = 2x 3 .2x + 3. we choose a window such that xmin = . Calculator Tips Applied Calculus for Business. Wang.2x + 3 and g1x2 = x 3 .1 . the window for the zeros of f are obtained from g.3/2 . .1 .7 More on Functions * ** 147 (-4. a 1 . However. 3/2 6 = 1 + 3/2 = 2. 1 . Copyright © 2007 by Pearson Education.x + 2 have the same zeros. Sometimes this can be difficult. a 0 6 (that is.3x2 . We first need to determine a reasonable window. 32.Section 1. Inc. Economics. if we need to change that part of the window. we can do so afterwards. but that is not always the case.5/3x 2 . so we have z 6 1 + 4 = 5. see Figure 30 Figure 29 (We note that the usual default window would have produced a reasonable graph.2x 4 + 12x 3 .) We can also sketch rational functions with the calculator. Wang.148 * ** Section 1. Solution.5x 2 . and April Allen Materowski. .6.2/3x 4 + 4x 3 . Copyright © 2007 by Pearson Education. Gordon. Walter O. Applied Calculus for Business. we divide by 3 to obtain g1x2 = x 5 .3x .7 More on Functions Example 14 Illustrate the above theorem for determining a window containing all the zeros fo the function defined by f1x2 = 3x5 . Published by Pearson Learning Solutions. by Warren B. Thus we choose as our window as indicated in Figure 28.2 The coefficient whose absolute value is largest is 4. Note that we kept the default values for y. Use your calculator to sketch the graph. Figure 28 We next have the calculator graph the function. Since the theorem requires the coefficient of the highest term be one.9x . this can be dealt with by choosing a reasonable window in which to view the graph. Therefore all zeros of g and therefore f are in the interval . but you need to be careful near the vertical asymptotes. and Finance.5 6 z 6 5. 9 Applied Calculus for Business. f1x2 = 2x + 3 10. origin or neither. if they exists. f1x2 = 34. f1x2 = 2x2 + 1 14.4 x2 .5 36. f1x2 = 40.5 2x2 + 5 4x2 + 9 19. Published by Pearson Learning Solutions. f1x2 = . f1x2 = 33. f1x2 = 2x2 + 7 3x + 4 21x . xy = 1 25. 12.x In Exercises 12 23 determine if the function defined by the given equation is odd. x3 . f1x2 = 37. f1x2 = 18.5 x + 5 x .x2 11.x + 3 4. . An equation is symmetric with respect to the x-axis if replacing y by . x 2 + 4xy + xy 3 = 8 30.92 8. (a) Why can t such an equation represent a function? (b) In Exercise 24 29. f1x2 = 2x2 . f1x2 = 49. f1x2 = 2x . f1x2 = 1x x . f1x2 = 45. f1x2 = 9x3 .x Z 0 x In Exercises 24 29. even or neither.x + 3 15. f1x2 = 44. f1x2 = 1x3 .2 x2 . vertical and horizontal asymptotes along with the sign of the function to sketch its graph.y 3 = 0 29.3x2214x2 . x2 + y 2 = 5 28. f1x2 = 41. f1x2 = 17.2 3x2 35. Wang. x2 + x2y 2 + y 2 = 4 26. f1x2 = x4 .1221x . 2x x . which of the equations has this symmetry? 3x3 + 7 3x + 2 42.x3 7. and Finance. f1x2 = x2 . f1x2 = x5 21. f1x2 = 3 13. f1x2 = 38 determine the vertical asymptote(s) if one exists. f1x2 = x314x6 + 11210 20. 1 x . f1x2 = 2x + 3x .9 3. f1x2 = x2 x x 23.x5 6. determine the zeros and then. 1. determine if the graph of the given equation is symmetric with respect to the y-axis.3 In Exercises 39 39. f1x2 = 2x . f1x2 = 12x .52 7x4 + 2 2x 9x2 + 1 In Exercises 47 55 use the zeros. f1x2 = 46. f1x2 = 32. f1x2 = 2x2 . Walter O.12 2. x2 + 4 + 3y = 0 27.x 212x .x + 32 9.7 In Exercises 1 11. 47. Economics. f1x2 = 46 determine the horizontal asymptotes. Gordon.y leaves the equation unchanged. and April Allen Materowski.3 43.2x . f1x2 = 29 .5 2 x . using the sign of the function. draw its sketch. f1x2 = 2x3 .2 x . f1x2 = 1 x . f1x2 = x . Inc.Section 1. f1x2 = x x + 1 x x + 1 3x 2 4 5 2 2 In Exercises 31 31.3x 16.7 More on Functions * ** 149 EXERCISE SET 1. by Warren B. 24. f1x2 = 4x .2221x + 12 x3 + 3x + 2 12x .2 2x x + 1 x + 1 4x2 .3 x x2 + 5x + 6 2 x2 . f1x2 = x 21 . f1x2 = x x 22.4x2 5. Copyright © 2007 by Pearson Education. f1x2 = 48.x .3 38. 1/42.9 3x 2 x2 . draw the graph with the corresponding points clearly labeled of (a) f1x .7 x x . Walter O. f1x2 = 225x2 . (c) f1x2 + 3.1.1 57. f1x2 = 52. locate the coordinates of the turning point of the function. (d) f1x2 . 63. an odd with an even is odd and an even with an even is even.22.22 + 1 and (f) f1x + 12 .2 50. by Warren B. (e) f1x .2 Figure 33: Ex. (b) f1x + 12. and Finance. and April Allen Materowski. f1x2 = 54.x . Figure 31: Ex.2 Figure 30: Ex. Show the product (or quotient) of an odd function with and odd function is even. . (d) f1x2 .7x + 12 6x2 . draw the graph with the corresponding points clearly labeled of (a) f1x . f1x2 = 56. Given the graph of the function y = f1x2 in Figure 33. (b) f1x + 12.1. (c) Find its range by solving for x as a function of y.122 1x .9241x .25231x + 725 graph cross its horizontal asymptote? 5x 225x2 + 1 5x (b) How many times does the 55. Show the graph of the function defined by f1x2 = horizontal asymptote.150 * ** Section 1.16 4x2 x2 . f1x2 = 51. and (b) Sketch its graph clearly showing all asymptotes. 57 58.3 (a) Determine its domain. (c) f1x2 + 3. Copyright © 2007 by Pearson Education. Does there exist a function which is both odd and even? Applied Calculus for Business. 60 61.16 x31x2 .22 + 1 and (f) f1x + 12 .4221x2 . 58 64.2 Figure 32: Ex. Given the graph of the function y = f1x2 in Figure 30. Economics.22.6x . (c) f1x2 + 3. 2x . (c) f1x2 + 3. (d) Using (c).1. Gordon. 59 60. (d) f1x2 . draw the graph with the corresponding points clearly labeled of (a) f1x .1/42 + 1 and (f) f A x + 1*2 B . draw the graph with the corresponding points clearly labeled of (a) f1x . Given the graph of the function y = f1x2 in Figure 31. Inc. (e) f1x . (e) f1x . (b) f A x + 1*2 B . (d) f1x2 . (a) f1x2 = x2 . f1x2 = 53.36 2 More on Functions 59. Published by Pearson Learning Solutions. Given the graph of the function y = f1x2 in Figure 32.1. Wang. Consider the function defined by the equation f1x2 = 1 2x2 .22 + 1 and (f) f1x + 12 . (b) f1x + 12.22. (e) f1x .4 does not cross its x + 3 62. Given the function defined by the equation Regression * ** 151 y = 2x + 4 f1x2 = a nx n + a n . Published by Pearson Learning Solutions. Using your calculator.12x + 5 2x + 5 4x3 .q . g1x2 = 4x2 . and Finance. f1x2 = 70. In the previous exercise. In the real world. other times quadratic or perhaps some other shape. Suppose f is not a constant function with f1x + y2 = f1x2f1y2. When presented with a real set of data. .f1x2 f1h2 . Gordon. 1x2. For example. 69. Note how the graph approaches the line y = 2x + 4 as x gets large. do they exhibit any other kinds of symmetry? 66. (c) n 7 m. must f be a rational function? Suppose the rational function r1x2 = q1x2 + l1x2 where l(x) approaches 0 as x approaches + q or . r1x2 = 4 2x + 6x + 8 = 2x + 4 + x + 1 x + 1 2 as x approaches + q or .5 Determine the horizontal asymptotes (if they exist) if (a) n 6 m.1 + Á + a 1x + a 0 .8 Regression » » » » » Scatter Plot Line of Best Fit Linear Regression Correlation Coefficient Non-Linear Regression Calculator Tips In applications. When l(x) is linear it is called a slant or oblique asymptote. Show that the graph of the function defined in Example 12 does not cross its horizontal asymptote. find all the asymptotes and plot the graph of the given function. Á . (a) f102 = 1. show f1x + h2 .1 + Á + b1x + b0 2x2 + 6x + 8 x + 1 and its Slanted Asymptote Figure 34: r1x2 = y = 2x + 4 In Exercises 69 70. (b) h h 67. almost look linear. However.9 71. one is faced with the problem of finding a line that best fits the data.q . do m and n have to be integers. 1xn. and if there are.8 65. y32. Sometimes the points. Walter O.1x m . due to experimental or measurement errors. y22.2x2 + 3x . called a scatter plot. We first consider the case when the data appears linear. (b) n = m. determine the vertical asymptote of the function defined in Example 11. Wang. y12. 4/1x + 12 : 0. so r1x2 : 2x + 4. by Warren B. bmx m + bm . when the model is linear or when the data appears to lie close to Scatter Plot Applied Calculus for Business. then r1x2 : q1x2 as x approaches + q or . Inc. yn2 and seeks to determine a relationship between their x and y coordinates. it would be almost miraculous for a set of many pairs of points to lie exactly on the same line even when the underlying model is exactly linear. Do there exist graphs of equations which are symmetric with respect to the y-axis and origin. Copyright © 2007 by Pearson Education.q .Section 1. or we say r(x) is asymptotic to q(x). other cases will be considered at the end of this section. 68. Economics. 72.1xn . one is given a number of data points 1x1. Therefore y = 2x + 4 is an asymptote for r(x).1 = f1x2 a b. that is. and April Allen Materowski. 1x3. 1. if these points were taken from measurements. x2 . when plotted. A sketch is given is Figure 34. yi) (x2. . That is. Suppose that we have n data points 1x1. and Finance. Although this may sound almost impossible. the squared error at a given xi is 1yi . The proof of the technique is considered later in the study of calculus. 1x3. (xn. one looks at the sum of the squares of the errors. it is actually very simple. and the value of y predicted by the equation. We define the terms needed in this section. y3) (xi.152 * ** Section 1. Economics. Gordon. Now for each x-value. Let us write y = ax + b for the equation of the line. Inc.1axi + b222. xi. but let us look at the idea behind it. y1) yi . Published by Pearson Learning Solutions. Wang. It does require us to use some new notation. y32. Copyright © 2007 by Pearson Education. by Warren B. y12. y2) (x1. yn2.8 Regression some straight line. y22. However.(axi + b) = Error at xi Figure 1: A Scatter Plot and Regression Line What we would like to do is to pick the line so as to minimize the total error over all points. Á . there is the observed y-value. there is a standard technique to find the so-called line of best fit or the regression equation. yi. x1 + x2 + Á + xn 1average of the x values2 n y1 + y2 + Á + yn 1average of the y-values2 y = n sx = x1 2 + x2 2 + Á + xn 2 1sum of the squares of the x-values2 sx 1average of the sum of the squares of the x-values2 sx = n sy = y1 2 + y2 2 + Á + yn 2 1sum of the squares of the y-values2 sy 1average of the sum of the squares of the y-values2 sy = n = x1y1 + x2y2 + x3y3 + Á + xnyn 1sum of the products of the x and y values2 x = sxy Applied Calculus for Business. 1x2. We plot these points and see that they lie roughly along a non-vertical straight line.1axi + b2 is known as the error at x = xi. This difference yi . and April Allen Materowski. positive and negative errors would then cancel out and the fit might be very bad. 1xn. that is y = axi + b (see Figure 1). One looks at the sum of these numbers for all the points and tries to determine the values of a and b that minimizes this sum. yn) (x3. Walter O. In order to avoid this. Section 1.121/621159085/62]/[91/6 . Therefore.112121/62 = 22142. sx = 91/6 = 15.11249.ax (1) (2) Line of Best Fit Linear Regression It is clear that the calculations can be tedious and a calculator is useful. Gordon. the letter s is to remind you that you are summing squares of the variable in the subscript.) Using calculus. we set x = 10 (remember we are counting years from 1982 as 1) and we forecast that the median family income at that time will be y = 1124921102 + 22142 = 34629. Table 1: Median Annual Income 1982 1987 Year Income 1982 23433 1983 24580 1984 25948 1985 27144 1986 28236 1987 29744 Solution. we have y = 1249x + 22142 as the line of best fit.e. i.17. by Warren B. Wang. and sxy sums the products of the x and y.5. for 1991. is given by x = 21/6. Next we need the sum of the squares of the x-values. We shall show how the calculator finds the regression line very quickly at the end of this section Example 1 Median family income per year. so n = 6. Economics. let 1982 be year 1. Use the line to predict the median family income for 1991. Copyright © 2007 by Pearson Education.4. Plot this data and find and plot the line of best fit.6. and Finance. For simplicity.629 per year. The sum of the products sxy = 1121234332 + 1221245802 + 1321259482 + 1421271442 + 1521282362 + 1621297442 = 578657 and the average is sxy = 578657/6.121/622] = 1249. The points and the best fit line (regression line) are shown in Figure 2. the notation is easy to remember as follows: a bar above any symbol means it is an average. the sum of the x values is 1 + 2 + 3 + 4 + 5 + 6 = 21 and the average x. it can be shown that the line of best fit. The years are now 1. in dollars.3 Rounding these numbers to the nearest integer. divide by n. .x 2 and b = y .11 b = 159085/6 . $34. Inc. Walter O.3. sx = 12 + 22 + 32 + 42 + 2 5 + 62 = 91 and the average.2. in the United States for the six year period from 1982 to 1987 is given in Table 1. measure x in years and y in dollars. Substituting into the formula we have a = [578657/6 .8 Regression * ** 153 and sxy = sxy n 1average of the sum of the products2 (Note.. Therefore. The sum of the y-values is 23433 + 24580 + 25948 + 27144 + 28236 + 29744 = 159085 and their average is y = 159085/6. Applied Calculus for Business. Published by Pearson Learning Solutions. and April Allen Materowski. that is. the regression line has the equation y = ax + b where a = sxy x#y sx . some trends are linear but it is difficult to find a simple explanation for the observed phenomena. Notice that this result predicts that the number of deaths by falls seems to be declining at a rate of about 381 per year. We let y be the number of deaths and find that x = 3. and substituting. y = 12981.2. Wang. What might be surprising is that the increase was so close to being perfectly linear during the mid-1980 s. of Deaths 1978 13690 1979 13216 1980 13294 1981 12628 1982 12077 Find the line of best fit and plot it. One would expect mortality numbers to increase with increasing population. Gordon. and Finance. we have y = . Walter O.8 Regression y = 1249x + 22142 Figure 2: The Best Fit Line for Example 1 In the preceding example. and April Allen Materowski. The results are graphed in Figure 3. by Warren B. Solution. Economics.154 * ** Section 1. Published by Pearson Learning Solutions. Table 2 shows the number of deaths by falls in the United States over a five year period from 1978 through 1982. Table 2: Deaths by Falls over a Five Year Period Year No.2 Therefore. The median family income going up is a reflection of the general increase in the costs of goods and services. we saw the process of inflation in the American economy. sx = 11.381.4 and b = 14125. of course. Example 2. sxy = 38180. Why do you think falls should be an exception? Applied Calculus for Business. On the other hand. . Copyright © 2007 by Pearson Education. Again.381x + 14125. we find a = . even though the population of the United States is growing. which. rounding off. must be balanced by an increase in salaries and other forms of earnings. Consider the following example. Inc. we measure x in years and let the first year (in this case 1978) be x = 1. the closer the points are to a straight line. Wang. To free the measure from units we divide the sum of products by another product of the same units. It is clear that both involve some elementary but tedious computations. we can show that) r = sxy . assume the coordinate axes are translated so the origin corresponds to the point 1x. if r is near .ny22 (3b) Either of these expressions may be used to compute the correlation coefficient.Section 1. by Warren B. and their magnitudes will depend on these units. y2 in the original coordinate system. x and y have units in which they are measured. if the points do appear to be linear then they will be mostly in quadrants I and III or in quadrants II and IV (why?). Inc. if we now return to variables in the original coordinate system (where X = x .1 or 1 then the Applied Calculus for Business. Copyright © 2007 by Pearson Education. and Y = y . so without loss of generality. In this new X-Y coordinate system. especially if the data set is large.8 Regression * ** 155 y = *381x +14125 Figure 3: The Best Fit Line for Example 2 Given a scatter plot we need a measure to determine if the correlation between the variables is indeed linear. Linearity is unaffected by the position of the coordinate axis. or as we shall see at the end of the section with our calculator. We divide s XY by 2sXsY. and negative if they are in quadrants II and IV.nx y 21sx . . and April Allen Materowski. Note that in this translated coordinate system.y. and Finance. Fortunately. the points lie on a straight line. the sum of the products will be positive if most of the points are in quadrants I and III. It can be shown that if r = . this product has the same units as sXY. it is useful in making comparisons to have a measure which is independent of units. Thus. Published by Pearson Learning Solutions. we can automate the calculations using a spreadsheet like Excel. so the quotient is dimensionless. therefore. Gordon. Economics. If r is near zero then there is no linear relationship between x and y. It is therefore reasonable to conclude that the larger sXY is in absolute value. However.nx221sy . We.1 or r = 1. Walter O.x. r. define the correlation coefficient with respect to these translated variables as r = sXY 2sXsY (3a) Correlation Coefficient however. 5 2.6 A 6 B Note that r is close to one.) Example 3 Determine the correlation coefficient in Example 1. Suppose the data now appears parabolic. we have sXY = 21859. Similarly. a and b had to be determined. suggesting a strong linear correlation between x and y.694444446 396690. we Applied Calculus for Business.3081.2.58333333333 21859. Gordon.999207 Table 3: Computing r in Example 1 x 1 2 3 4 5 6 21 3.5 y 23433 24580 25948 27144 28236 29744 159085 26514. sX = 17.083333333334 314. Published by Pearson Learning Solutions.2.8333 3229. Walter O.5 L 0.027777776 2964710. Economics.3611111 27348356.5 Sum Mean Alternately. and Finance.6 A 21 6 B B 14245354561 .167 . and we need to find a.5 Y2 9493588. Inc.5.0.02777779 3741000. in the quadratic case we need to find a.5 1. r = 159085 578657 . The first thing we must do when determining a regression equation with the calculator is input the data. If we use (3a) we need to use the translated variables X and Y. b.25 6.5. and c so that the parabola y = ax2 + bx + c best fits the data points. x and y. Instead of writing down the formulas for these coefficients.91666666667 2901. we shall show below.8333333. the computations are straight-forward.1. . b and c. (We actually generated the Table 3 using Excel.5 Y = y . we may use (3b) which uses the original variables. and d.6 A 21 6 6 B 2A 6 91 Non-Linear Regression Calculator Tips We shall see at the end of the section that these calculations are done almost automatically for us with the calculator.5. Either way. c. We shall use the data from Example 1 to illustrate. the slope of the regression line. giving r = sXY 2sXsY = 217.02777777 10431823. The formulas are messier than the linear case (that is to be expected). (Note that from our discussion above r has the same sign as a. sx = 91 and sy = 234332 + 245802 + 259482 + 271442 + 282362 + 297442 = 424534561.5 . and = sY 27348356.156 * ** Section 1.25 283.83333 218595.5 2. Our goal should be to find the parabola that best fits the data. we found sxy = 578657. suppose we want to find a. In the linear case.167 629. In Example 1. Solution. We may use either (3a) or (3b) to compute the correlation coefficient. n = 6.5 .25 0.566.x .69444445 320544.25 2. and leave the determination of the explicit formulas to you until after you study the calculus. That is.5 0. What if the scatter plot looks like a cubic equation? Then we would expect a regression curve of the form y = ax 3 + bx 2 + cx + d. Basically. and April Allen Materowski.25 17.916666666666 2582.8333333 XY 7702. y = 1599085/6 = 26514. by Warren B.y .) Using these values. Wang. how to obtain the required equation and curve using the calculator.1934.5 # 27348356.25 0.1666667. Copyright © 2007 by Pearson Education.75 8074.8 Regression correlation between the points is linear.8333 X2 6.8333 1721. b.167 . and if it near zero the relationship is not linear.16667 X = x .999207 159085 2 2 . x = 21/6 = 3. the computations are summarized in Table 3. and L 0. We proceed as follows: 1. enter y1 for RegEq. Copyright © 2007 by Pearson Education. and in C2 the y-values. We will name it xx . Also. see Figure 6. Press F5 (Calculate). Press the MODE button on your calculator and check that the first line indicates FUNCTION as the Graph. name x C1 and y C2. see Figure 5. the second column y. and select Data/Matrix Editor. Enter in C1 (column 1) the x-values. change Calculation Type to LinReg (scroll down). Press the APPS (applications) button.8 Regression * ** 157 input a table for the given data sets and then will have the calculator compute the equation of the regression line and the correlation coefficient. and April Allen Materowski. Figure 6: Naming the Columns Applied Calculus for Business. Walter O. Economics. (Note that Figure 5 shows only four of the entries. Make sure Type is set to Data. Gordon. Wang. this tells the calculator to name the first column in the data set x. Press Enter twice. by Warren B. and Finance. Scroll down to Variable and give it a name and press Enter twice. all six have been entered.Section 1. select New. see Figure 4. . Inc. Published by Pearson Learning Solutions.) Figure 5: Entering the Data After you enter all the data. and it stores the regression equation in memory as y1. Figure 4: Naming the Data You now see a new screen (the Data/Matrix Screen) that looks like a table. Set Plot Type to Scatter. See Figure 7. Walter O. There is an alternative way of obtaining the same results with your calculator. its y-intercept. Close the screen by pressing Enter. Figure 8: Defining the Scatter Plot Display the Y = Editor for y1(x).and then we obtain Figure 9. Mark to Box. Press F5 (Calculate) and make sure all the settings are as before including Store RegEq which should be set to y1(x). Economics. the only difference between what is given here and the manual is that we did not select MedMed as the Calculation Type we selected LinReg and we stored the equation as y1(x) not y2(x). Gordon. Inc. press Delete and then Enter. 1press * F12 and then select Style (2nd F1) set the Display Style to DOT and press Enter. press F1 to define Plot 1. The information we entered is still stored in the calculator s memory. . and April Allen Materowski. and Finance. Published by Pearson Learning Solutions. You may proceed as follows: Figure 9: The Scatter Plot and Regression Line Applied Calculus for Business..158 * ** Section 1. by Warren B. the calculator can also provide the scatter plot along with the regression line. b = 22142. Scroll up to highlight Plot 1. Wang. you may delete it by selecting VarLink (2nd-) scroll down to xx press F1(Manage). the scatter plot and regression line. Save by pressing Enter twice. press F2(Zoom) and select ZoomData. Note that these directions can be found in the TI 89/92 Plus Manual beginning on page 83. a = 1249. as do the values for a and b as well as the correlation coefficient and its square. Next.266667 and the correlation coefficient. Figure 7: The Regression Equation and Correlation Coefficient Note we have the slope of the regression line. See Figure 8. Copyright © 2007 by Pearson Education.114286. Press F2 to display the Plot Set up Screen.8 Regression The information required for the regression line now appears. r = 0. Moreover. x = C1 and y = C2.999207 (the last value is the square of the correlation coefficient). However. Nevertheless. Consider the following: Table 4 indicates the number of students who worked full-time while attending classes at a small urban university. as follows: 51. and Finance. The calculator steps will be almost identical to the linear example considered above with one exception. namely. In principle. of Students Year 1023 1991 475 1992 430 1993 304 1994 411 1995 531 1996 982 1997 Find the parabola (quadratic regression curve) that best fits this data and estimate the number of students who will work full-time in 2004. As before. 5. 3. Since the curve is non-linear. Wang. Inc. see Figure 11. t2 and press ENTER Choose WINDOW 1*F22. Regeq(x) STO y1(x) and press Enter (This stores the regression equation as y1(x)) then enter on the entry line NewPlot 1. that is.Section 1. 27144. where we name the variable zz1 Figure 10: Naming the data As before. t1. Copyright © 2007 by Pearson Education. 25948. the mathematics used to find the regression curve is the same as the linear case. . 4. Published by Pearson Learning Solutions. data may resemble non-linear curves. 297446 STO t2 press ENTER Make sure you are using curly braces to enclose these two lists (2) Next enter on the entry line of the HOME screen (use the Catalog or type) LinReg t1. then press Zoom (F2). see Figure 10. and scroll down and select ZOOMDATA (or press 9). 1991 1997 No. (The regression line is now drawn. the calculator automates the determination of the regression curve. 24580. 28236. Walter O. 66 STO t1 press ENTER 523433.1. Economics. 2. there will be more equations to solve to determine the coefficients (which will involve more complicated algebraic expressions). Gordon. C1 and C2 for x and y respectively and store RegEQ to y1(x). and April Allen Materowski. Table 4: Students Working Full Time. enter on the entry line in the HOME screen. We enter this data into the calculator. Applied Calculus for Business. In some cases. to minimize the square of the error between the data points and the best fit curve.8 Regression * ** 159 (1) Store the x-coordinates and y-coordinates as lists named t1 and t2. you may need to press *F3) Often. we will choose QuadReg instead of LinReg. by Warren B. let 1991 correspond to year 1. we enter the data and then when we Calculate: we choose QuadReg (for quadratic regression). t2 and press ENTER then ShowStat and press ENTER (This produces Figure 7) (3) To draw the regression line. Economics. with QuadReg replacing LinReg. . Walter O. our estimate for the number of students who work in 2004 (year 14) is y11142 = 7.160 * ** Section 1. Gordon. and April Allen Materowski. by Warren B.767. we have the commands CubicReg and QuartReg for cubic and quartic regression curves. the alternative method may be used to produce the above. Copyright © 2007 by Pearson Education. Figure 13: Quadratic Regression Curve and Scatter Plot As with linear regression. Published by Pearson Learning Solutions. see Figure 13. Inc. As before. see Figure 12. and Finance. In particular. Wang. Figure 12: The Quadratic Regression Curve Rounding to the nearest integers.8 Regression Figure 11: Defining the Columns Pressing Enter twice gives the required information. we can have the calculator draw the scatter plot and the regression curve. the parabola of best fit is y = 75x2 . Likewise.600x + 1496. Applied Calculus for Business. Section 1.8 (Note. and 872 in 1989. of men and women in the United States for the years 1974 through 1980.1 4.8. . (c) From the equation of the line. (a) Find a line of best fit for this data and (b) use it to project the popcorn consumption in 1999. If you know something else about the data.232 = 9024). Louis No. (a)Find the line of best fit for this data. You may be asking yourself the question that if there are several possible regression curves for the same scatter plot. in the answers.1.16. What is your conclusion? 2.16. which one do we use? The answer is not always obvious. (c) In what year would you expect consumption to first exceed one billion pounds? Applied Calculus for Business. Table 8 contains the data for the sixteen National League teams in the 2002 season. (c) Does this indicate anything about the attractiveness of American investments to citizens of other countries? 5. (d) Is the gap widening or shrinking? (e) What would this data predict about the earnings gap in 1986? (The actual gap in 1986 was 25.2). 1521. (a) Find and (b) plot the line of best fit and (c) determine the regression coefficient.7). Table 5 shows the gross national product (GNP) of the United States in billions of dollars for the years 1978 through 1983. 7. 1218.7). then it is clear that this is the regression curve to use. in dollars. 8. Inc.3. and it suggests that quadratic regression best fits the model.2 5. (7. (a) Find the line of best fit and the regression coefficient.0 5.8. (1. The amount of popcorn consumed in the United States (in millions of pounds) was: 353 in 1970. Economics. (b) Plot the line. Table 6 shows the median incomes.0 5.6.17.8. 741 in 1987. Data: (10.4 4. (The actual figure was 4488. and (b) the regression coefficient. 1981. 807 in 1988. 393 in 1975. 6.4 Table 5 Year GNP 1978 2128 1979 2414 1980 2626 1981 2926 1082 3073 1983 3311 (d) What would this line predict about the divorce rate in 1987? See Exercise 11 for a continuation. we shall assume that the first year for which data is given is labeled as year 1. (2. by Warren B. other considerations come into play.8 Regression * ** 161 Other types of regression equations may be calculated the same way and will leave their examination to the exercises. 700 in 1986. 670 in 1985. (a) Find the line of best fit. and 1983 were 1833.3. Copyright © 2007 by Pearson Education. Baseball experts believe that there is a strong linear correlation between the number of games a team wins in a seasons and the teams batting average. and April Allen Materowski. Published by Pearson Learning Solutions.) 1. Table 8 Team Arizona Atlanta Chicago Cincinnati Colorado Florida Houston Los Angeles Milwaukee Montreal New York Philadelphia Pittsburgh San Diego San Francisco St. (a) Find the line of best fit for women s incomes versus time and (b) the line of best fit for men s income versus time.8).5 3. The numbers of foreign investors in American enterprises for the years 1980. Walter O.7 4.9. EXERCISE SET 1. The following (x.7). predict the gross national product in 1987. y) pairs give the advertising dollars (in millions) spent on direct mail and newspaper advertisements respectively for the same years. (b) Roughly speaking. 1982.) Table 7 Year 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 Rate 3.5). (8.256 . how do the two expenditures relate to one another? (c) Do you think that advertising people believe that one of these forms of advertising can substitute for the other.5.6 4.14.5. (a) Use the data from Exercise 2 to find the line of best fit for men s income as a function of women s income. Wang. (c) Plot both lines on the same set of axes. of Wins 98 101 67 78 73 79 64 92 56 83 75 80 72 66 95 97 Team Batting Average 267 260 246 253 274 261 262 264 253 261 258 259 244 253 267 268 Table 6 Median Income Year 1974 1975 1976 1977 1978 1979 1980 Women 6970 7504 8099 8618 9350 10169 11197 Men 11889 12758 13455 14626 15730 17045 18612 3. (b) What is the slope of this line? (c) Does it predict that the earnings gap is widening or shrinking? 4. The divorce rate (number of divorces per thousand couples) in the United States for the decade of the 1970 s is given Table 7. and 632 respectively. Gordon. (4. and Finance. Sometimes. 568 in 1980.9 5. 14. Table 11 Year Rate 1980 1981 1982 1983 1984 1985 1986 1987 5. Find the equation of the best fit quartic for this data.05. 10. (6. 15.8 300 5. and April Allen Materowski.1 700 1. Wang. (b) quadratic. (7.3 5.95. Given the data set (1. Given the data set 1 . 12. 6. .3).9. (a) Is there any obvious place where the data breaks into two straight lines? (b) Fit a piecewise linear function to the data.9).0 13.10. 13.1.0.8 134. (5. (3.01.1. Since 1979 the divorce rate statistics (see Exercise 5) are given in Table 11. . The revenue (in thousands of dollars) resulting from the demand for a given item is indicated in Table 13.8 200 3. 4.0 4. Published by Pearson Learning Solutions.9 155. 146). (b) Can you conclude anything? Sometimes when one plots a set of data.1 5.16. find the best fit cubic. (d) quartic. it appears to be piecewise linear. Table 9 Year 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 91 115 131 159 147 424 429 470 523 Number 94 Table 12 Year 1977 1978 1979 1980 1981 1982 1983 1984 1985 Number 141.82.92.1 11. 10. Cost and Profit Functions Marginal Functions Using the Zeros Even Functions Symmetry about the y-axis Odd Functions Symmetry about the origin Rational Functions Vertical Asymptotes Horizontal Asymptotes Translations Regression Applied Calculus for Business.9 5.9).1. . .5).12. Gordon. Walter O. 2. Consider Exercises 9 12. (5. 9. causing one to ponder an explanation for the change in pattern at the corner(s). (1. Economics. 30. 16. (c) Does this information say anything about the safety of riding a bicycle? Table 10 Year 1960 1965 1970 1975 1980 1982 1984 1985 680 780 1000 1200 1100 1100 1100 Number 460 Table 13 Demand Revenue 100 0.022.8). (2.8 CHAPTER REVIEW Key Ideas Two Dimensional Coordinate System Horizontal and Vertical Lines The Slope Intercept Form Graphing The Point-Slope Equation The Slope Formula Economic Applications The General Linear Equation Definition of a Function Functional Notation Difference Quotient Domain and Range Dependent and Independent Variables Vertical Line Test Combining Functions Composition Functions of Several Variables Break-Even Analysis Depreciation Piecewise Linear Functions Scaling Vertical Translation Axis of a Parabola Horizontal Translation Locating the Vertex Graphing a Parabola in the form y = ax 2 + bx + c Applications to Optimization Definition of a Circle Equation of a Circle Graphing a Circle Tangent Line The Ellipse Supply Function Demand Function Market Equilibrium Revenue.01. 5.8). (4.0 4. . Over the 9 year period. (2. 1977 through 1985. 3.2 500 5.8 4. 11. (a) Fit a piecewise linear graph to this data.2 131. Find the lines of best fit for the data as so divided. Inc.9 152.2 5. (c) cubic. the other from 1983 through 1987. find the quartic of best fit.1). . the number of legal abortions (in thousands) performed in Bulgaria are given in the Table 12.1 132.4 147. Find the best fit regression (a) line.9. Given the data set 10. For the United States. Copyright © 2007 by Pearson Education.7 147. (c) Does this indicate any significant change in people s behavior? 12.7 147. (3.0.92.162 * ** Chapter Review (a) Find the line of best fit for this data (label 1980 year 11 so as to continue with Exercise 5). Which best fits the data? 15. The graph is easily visualized as two straight lines: one from 1977 to 1982.05.9. 0.7 400 7. The numbers of federal officials convicted of corruption in the period 1977 to 1986 is given in Table 9.1). 14. the numbers of cyclists (not motorcyclists) killed in accidents with automobiles is given in Table 10.932.12. by Warren B. (b) Combine this with the result of Exercise 5 to obtain a piecewise linear function that describes the American divorce rate over the eighteen year period.92. and Finance.6 600 4.2). . 9.2). Determine the domain and sketch the graph of the function defined by 3x . 81). Gordon. (a) The cost of producing x-items is given by the equation C1x2 = 2x2 .2 f1x2 = d 7 15 .4 x2 . (b) quadratic of best fit. 28. Plot the points 11.12 = 0. 42 with slope 2/3. h f12 + h2 .32 5. determine (a) . 12). y. 3x . Sketch the graph of f1x2 = 2x2 + 5x . . (b) g(1). Wang. determine the approximate solution of . identify which of these could represent a demand function and which a supply function. Determine the equation of the ellipse centered at the origin with one x-intercept at (4.52. for which x-values is it not defined? 14.Chapter Review 1.4x + 6y + 13 = 0. For the functions defined in the previous exercise. (b) y = f1x + 22. 10. 3 .12y + 4 = 0. 30. Determine the equation of the line with x-intercept 3and y-intercept .200x + 6000. . Determine the domain of the function defined by f1x2 = 17. . 35. (a) Give a qualitative sketch of the graph of the polynomial whose equation is f1x2 = x1x . 6.01. (c) y = 4. Give a better description of this circle. by Warren B.13x + 6 25 + 2x 15. 18.32 and radius 4. z2 = 2xyz + 3x2y3 + 9y2z2. 7. (a) Given p = 2x . and then determine the average cost of producing the 99th item. (b) four years? 21. Determine the equation of the line passing through 13. indicate all asymptotes and zeros. at which time it is has a scrap value of $50.3.200x + 6000. determine (a) f(x)g(x). 34.5. determine f(2.12 . . 20. Determine the equation of the line passing through 1 . and g1x2 = 4 + 3x. (c) y = f1x2 + 2.2x 1 * ** 163 22. Published by Pearson Learning Solutions.f122 f1x + h2 . (e) f1x2 = x2 x 3 . (4. x + 9 36. (b) From your sketch.x. What is its value at the end of (a) two years.x x3 .2x x-values is it not defined. (b) 3x = 12.2 and p = 28 . Copyright © 2007 by Pearson Education. 9. Inc. Sketch the line (a) x = . (c) . (a) Determine the x and y. 33. determine the (a) line of best fit.7x.4x + 6y .f132 12. 8. 19. Determine the symmetry of the function defined by (a) f1x2 = 5x3 . 27. 2). 4 3 32. (b) Sketch its graph and (c) determine its range. Given f1x2 = . 12). determine the cost of producing the 99th item. 0) and one y-intercept at 10. determine f(16. (d) Substitute x = 2 and then h h x = 3 into (c) any observations? 13. Given a function defined by Table 1. 1 . and compare your approximate solutions with the exact solution. Determine the coordinates of the center of the circle and determine its radius if its equation is x2 + y2 . (f) y = f1x + 22 + 4. and g1x2 = 3 . (d) f1x2 = .3x + 9. (e) y = f1x .f1x2 (b) . The cost of producing x-items is given by the equation C1x2 = 2x2 . Give its sketch.2x if if if if x 1 1 6 x 6 3 3 6 x 4 x 7 4 2 + x .2.4x + 8 = 6. Given f1x.32. 5). 4. Sketch the graph of the ellipse whose equation is 4x2 + 3y 2 + 8x . Economics.3y = 7 and (b) sketch its graph.*2 and passing through the point 10. 3. Given f1x2 = 2x2 . (b) Determine the average cost function. Determine the equation of the line passing through the points 12. Walter O. determine the number of items to be produced to minimize the cost as well as giving the minimum cost. 11. Does the equation 4x . (b) f(x)/g(x).y 2 = 6 define a function? Explain.122. (a) Determine the coordinate of the vertex of the parabola y = .32 and 1 . Determine the equation of the circle with center at 12. (a) Determine the equation of the tangent line to the circle of radius 13 centered at the point (*5. . Applied Calculus for Business. 72. for which 3 .22 + 3. .5. (b) Compare the y-values on the circle with the yvalues on the tangent line when x = . . (b) Over what domain are they defined? (c) Determine the coordinates of market equilibrium. (a) Determine the domain of the function defined by f1x2 = 23 .2. 8) and (6. Given the data set (1. Determine the domain of the function defined by f1x2 = 2 .22.22. x . determine. determine (a) f(g(x)).intercepts of the line 2x . 3). 24. Given f1x2 = (c) f1g1 .7x + 12 2 + x 16. find them exactly and compare. 25.12 x2 . y2 = x3/4y 1/4. and Finance. Determine the coordinates of the center of the circle and determine its radius if its equation is x2 + y2 .2x. 26.4. 2. determine the graph s approximate intercepts. (b) g(f(x)). in tabular form. (c) From your sketch.6.2x2 .4y = 15.1 . (b) perpendicular to the line 3x . (c) cubic of best fit. and April Allen Materowski. (b) estimate where the turning points are on the graph. (c) f1x2 = x3 . Table 1 x f(x) -2 3 -1 4 0 -2 1 -3 2 1 3 4 4 7 5 9 23. Sketch this circle.4x.2x2 . Sketch the graph of the ellipse with equation + = 1.222 1y + 422 31. x2 (b) f1x2 = 2 . 29.4x + 8 and sketch its graph. (d) y = f1x2 .2. A $1200 computer depreciates linearly of five years. Given f1x. 6x2 . 5). each of the following: (a) y = f1x .2221x + 32312x + 52. (c) for what x values is (b) not defined? (d) for which x-values in g(x)/f(x) not defined? f13 + h2 . 1x . (2. (3. . determine (a) f(3). 42 and 1 . (d) 3y + 9 = 6.22 and (a) parallel. 1. Determine the equation of the line with slope . by Warren B.Applied Calculus for Business. Economics. Published by Pearson Learning Solutions. Inc. Walter O. Gordon. . and April Allen Materowski. and Finance. Wang. Copyright © 2007 by Pearson Education. Economics. and then learn how to find the derivative of functions defined implicitly. . Walter O. We describe the fundamental method of finding this so-called derivative function from the definition and then develop the simplified rules for rapid calculation of derivatives that have been discovered by earlier generations of mathematicians. we begin the study of the differential calculus of one variable. Starting with the notion of the slope of a straight line. we show that the derivative can also be interpreted as a rate of change and look at some applications of that interpretation. Wang. The last section shows how the tangent line can be used to approximate the zeros of a function Newton s Method. Copyright © 2007 by Pearson Education. and April Allen Materowski. Applied Calculus for Business. Published by Pearson Learning Solutions. Next. Inc.2 An Introduction to Calculus In this chapter. by Warren B. Gordon. and Finance. we shall define the slope of a general curve as a function of x. the tangent line at any point is the line that approximates the direction of the curve at that point. for a linear function. that is as the line among all possible lines. Figure 2 shows the curve with several lines drawn at the point P. the line is vertical. Copyright © 2007 by Pearson Education. By the slope of the curve at the point P(x. In that case. f(x)) so that it approximates the shape of the curve at this point. It is possible for a line to meet the intuitive sense of tangency. These include tax rate. Published by Pearson Learning Solutions. whose y-values are closest to the y-values along the curve near the point P.166 * ** Section 2. Gordon. That is. it is the slope of the line which gives us most of the information about the behavior of the function. No other line touching the curve at this point. A zero slope tells us that the line is horizontal and y is constant. and marginal cost. We shall define the slope of an arbitrary curve and see how this concept can be used. we resort to a physical notion. that is easy. When the slope is undefined. slope has different interpretations which depend upon the context. y is not a function of x. Even though the road may be curving. cross it again elsewhere. Therefore. touching the curve in only one point. Walter O. It is that straight line that we would call the tangent to the curve. Think of the curve in Figure 1 as a road and imagine yourself driving along the road. The slope itself can be thought of as the rise of a line divided by its run. It is easy to pick out the tangent line. In classical geometry a tangent line to a circle is a line that touches the circle in only one point. and April Allen Materowski. Our object here is to generalize the notion of slope. Applied Calculus for Business. In fact. locally but. Economics. we could define the tangent line as the best linear approximation to the curve near the point of tangency. and x is constant. will parallel the curve as well as the tangent line. this definition must produce the slope of the line. at any instant you are looking in some direction that you think of as straight ahead. rate of depreciation. What do we mean by a tangent line? For a circle. It is even possible. we mean the slope of the tangent line to the curve at this point. and Finance. for a tangent line to cross the curve at the point of tangency. f(x)). Wang. of course. Moreover.1 Slope of a Curve 2. Of course. When the slope is negative the y-values decrease as x increases. Let us see how we should define the slope of a curve. Take a ruler and try to place it at the point P(x. for arbitrary curves this definition will not suffice. This line is the tangent line. or the change in y with respect to a change in x. as you will see. .1 » » » » » » » Slope of a Curve Slope of a Tangent Line The Slope as a Limit Slope of a Curve Equation of a Tangent Line A Place Where No Tangent Exists The Derivative Calculator Tips We have seen that when x and y are linearly related. Consider the function whose graph is sketched in Figure 1. However. A positive slope indicates that as x increases so does y. by Warren B. among others. Inc. Section 2. So. Inc. you would not expect the slope of a curve to be constant. We would like to find the slope of this curve (the slope of the tangent line to Applied Calculus for Business. f(x)) Figure 1: Tangent Line at P y = f(x) P(x. how do we find the slope of a tangent line? Before getting to the general definition. Gordon. R. by Warren B. To indicate that the slope of the curve depends upon the point P(x. Q. Copyright © 2007 by Pearson Education. Consider the function defined by the equation f1x2 = x 2 + 1. Walter O. You should recall that the graph of this function is a parabola. Economics. Notice that the slope of the curve is different at each of these points. and S.1 Slope of a Curve * ** 167 Tangent line at P y = f(x) +* P(x. . let us first look at a specific example. Wang. f(x)) we write it as mtan1x2. we have drawn tangent lines to the curve at each of the points P. Published by Pearson Learning Solutions. In Figure 3. and Finance. we said that we would define the slope of a curve at a point to be the slope of its tangent line. However. and April Allen Materowski. f(x)) +* Figure 2: The Tangent Line and Other Lines through P For a linear function the slope is constant. Now remember. Applied Calculus for Business. Using a ruler. Copyright © 2007 by Pearson Education. let the x-coordinate of Q be 2 + 0.5) In Figure 4.004001. we have a sketch of the curve and its tangent line at (2.5). f(x) = x2+1 Tangent Line Figure 4: f1x2 = x2 + 1 and the Tangent Line at P (2.5). Inc. and call it Q. Gordon. you should try this approach and compare your answer to the one we obtain below (see Exercises 1 3).001 = 2.001. . Nevertheless. To find the slope of the line exactly. and April Allen Materowski. our measurement of the slope will be at best an approximation. Imagine that we have magnified our sketch and that Figure 5 reflects this magnification. Q is so close to P that you really could not tell them apart unless the graph was greatly magnified. Since f1x2 = x 2 + 1. unless the graphs of the curve and tangent line are drawn exactly.5.00122 + 1 = 5.1 Slope of a Curve Q P R S Figure 3: The Slope of the Tangent Line at Different Points along the Curve this curve) at the point P(2. Unfortunately. Published by Pearson Learning Solutions.168 * ** Section 2. as an exercise.0012 = 12. Walter O. by Warren B. In fact. we could try to draw the tangent line at this point. Wang. Let us choose Q to be just to the right of P.001. the y-coordinate of Q is f12. find the coordinates of two points on the line and then compute the slope. and the point Q may be labeled Q(2. Make sure that this point is very close to the point P. we shall proceed as follows. Choose a point on the graph either to the left or to the right of the point P. Economics. and Finance.004001). Actually. In fact. with x-coordinate 2 + 1 . and April Allen Materowski. Call the x-coordinate of Q.0001. and the slope of the secant line connecting this point to P turns out to be 3. 2 + h. msec = f12 + h2 .f122 4h + h2 5 + 4h + h2 . It is this so-called limiting value that we shall define to be the slope of the tangent line. Thus.004001) msec This slope is nearly equal to the slope of the tangent line.00040001. we probably could not tell them apart. Letting h approach zero 14 + h2 becomes 4. Summarizing. Compute the slope of the secant line through the two points P and Q. thus. Published by Pearson Learning Solutions. Wang. Now we can simplify by dividing by h. They write mtan122 = lim msec = lim 14 + h2 h :0 h :0 The symbol lim is read the limit as h approaches zero of . combine the instructions into one by writing.001 2. we obtained secant lines with slopes just a little more than 4. say with an x-coordinate 2 + 0. in the unmagnified state.1 Slope of a Curve * ** 169 We next draw a line connecting Q to P. mtan122 = lim msec = lim h :0 f12 + h2 . Walter O. In fact. we have the slope of the tangent line at P. This is not the tangent line but. which is very close to P(2. the slope is clearly approaching the value 4. Economics. It is due to the magnification that we are able to see any difference at all. and Finance. . msec = 5. Choose a point Q.000122 + 1 = 5.5). 5.9999. That is. h is a small negative number. Points to the left produced slopes just a little less than 4. . Inc. h is a small positive number. you have probably guessed the slope of the tangent line at the point P(2. . msec = 4h + h2 = 4 + h h 3.Section 2. Can we do better? Why not take a point even closer to P.5 = = 12 + h2 . we should get better and better approximations to the slope of the tangent line.9999. it is expected to be close to the tangent line. When Q is to the left of P. 5) Figure 5: The Points P and Q on the Magnified Curve Slope of a Tangent Line Note that the difference in the x-coordinates is h.5 = 4.001.2 Q(2. (When Q is just to the right of P. It is just the difference of the y-values divided by the difference of their x-values.004001 . Thus. We could. Now let Q get even closer to P by letting h approach 0. We can compute the slope of this secant line which we shall denote by msec.2 h h P(2. Copyright © 2007 by Pearson Education. Using points just to the right of P. As we let the points get closer and closer to P.001 . Similarly. we do the following: 1.0. and the slope of the secant line connecting this point to P will turn out to be 4. Letting h approach zero. The basic idea is that as we take the points closer and closer to P. denoted by mtan122 Mathematicians have a special way of writing the instruction given in Step 3.99960001.0001? We find the corresponding y-coordinate to be 12. Its y-coordinate computes to 4. Let us call the line connecting P to Q a secant line. the slope of the h :0 tangent line is the limit as h approaches zero of the slope of the secant line. 5). using this notation.00012 = 1.) The y-coordinate of Q is f12 + h2 = 12 + h22 + 1 = 4 + 4h + h2 + 1 = 5 + 4h + h2 2.f122 h :0 h Applied Calculus for Business. Gordon. suppose we took a point just to the left of P. the limit as h approaches zero of 14 + h2. by Warren B. . Gordon. we did not let h approach zero until after there was a cancellation of the common factor h in the numerator and the denominator. we must be very careful when we let h approach zero.170 Section 2. and Finance.62 = 3x . Example 1.6xh + 3h + 22 = x2 + 2. by Warren B. Published by Pearson Learning Solutions. In general. (a) As h approaches zero.6 changes as h approaches zero and the limit is 3x . if we let h approach zero. .6xh + 3h + 22 (d) lim x 2h + 4xh + 3h + h2 h :0 h Solution. lim 13x + 4h . lim 110 + h2 = 10. it does not change. Therefore. neither 3x nor . Compute (a) lim 110 + 4h2 h :0 h :0 h :0 (b) lim 13x + 4h . (c) Both 6xh and 3h are multiples of h. Economics. Since 10 is a constant. Copyright © 2007 by Pearson Education. h :0 h :0 h :0 lim 1x 2 . both these terms go to zero and the limit is just x2 + 2.62 (c) lim 1x 2 . (d) If we try to let h approach zero as this expression now stands. and the limit is just 10. Inc. (b) As in part (a). let us generalize this with the following definition: The Slope as a Limit DEFINITION 1 The slope of the tangent line (when it exists) to the function defined by the equation y = f1x2 at the point P(x. so we can treat it as a constant. we will obtain an expression of the form 0/0. also must go to zero? If h is a number like 0. That is. f(x)). so that also goes to zero and we have the form 0/0. only after the possibility of getting 0/0 eliminated do we let h approach zero. Wang. what is h2?) The denominator is h. 4h (which is simply 4 times h) must also approach zero. Now we can cancel the common factor h in numerator and denominator.6. We give a more detailed examination of limits in Section 3. all the terms in the numerator are multiples of h and will go to zero. and April Allen Materowski. in this exercise. When the calculations were done in our first example. Therefore. 6x does not change as h changes. Thus. 3x is a variable but it does not depend upon h. If we do it immediately.3. which is an indeterminate form and always requires more investigation. However. Walter O. h2. Here are some examples of limits taken as h approaches zero. that is.f1x2 h:0 h (1) In trying to apply this definition. (Do you see that the last term.001.6. Thus.1 Slope of a Curve In fact. 4h must approach zero. is given by mtan1x2 = lim f1x + h2 . Although x is a variable. we can factor an h from each term in the numerator to get h1x2 + 4x + 3 + h2. which is indeterminate. any constant multiple of h will approach zero. In fact. h1x 2 + 4x + 3 + h2 x 2h + 4xh + 3h + h2 = lim h:0 h:0 h h = lim 1x 2 + 4x + 3 + h2 = x 2 + 4x + 3 lim h:0 Applied Calculus for Business. at the points where the pieces join) may have other peculiarities that will make it impossible to define a tangent line at one or more points in their domain.1 Slope of a Curve 171 We remark that this method may not always yield a real value for the slope. (d) mtan112 = 2112 .2 = . and cancel it with the one in the denominator. In addition. Wang. some functions (such as piecewise linear functions. when we let h approach zero. and April Allen Materowski. We know that the slope of a vertical line is undefined. factor out h in the numerator. (c) x = .f1x2 lim h:0 h h12x + h . f1x + h2 . Just as we saw above.2x + 1 when (a) x = 3. We compute the two terms in the numerator. Published by Pearson Learning Solutions.22 = 2x . Equation of a Tangent Line Applied Calculus for Business. (c) mtan1 .) To avoid any algebraic errors. but since the computation for each one is the same.f1x2 = 2xh + h2 . Gordon. so at such points our method is doomed to failure. (d) x = 1 Solution.2 = 4.2. Now form the quotient.1.Section 2. mtan1x2 = lim h:0 f1x + h2 . Economics. (We could compute each problem separately.2x + 1 at each of the points indicated in Example 2.2h = h:0 h h = lim 12x + h . a function may have points at which the tangent line is vertical.2x + 1 Slope of a Curve Notice that we line up similar terms.2 = .22 = lim 2xh + h2 .2h 3.12 = 21 .2 h h 4. Take the difference to complete the numerator.22 2xh + h2 . Inc. (b) mtan102 = 2102 .21x + h2 + 1 = x 2 + 2xh + h2 . by Warren B. and Finance. We use Definition 1 to compute mtan1x2 as a function of x and then substitute the various values for x. Walter O. We can now answer the questions. it makes sense to compute the slope of the curve at an arbitrary point x and then substitute for x at the end.2h + 1 f1x2 = x2 . Example 2 Determine the slope of the curve defined by the equation f1x2 = x2 .2 became simply 2x .12 . (a) mtan132 = 2132 . For example.2h = = 2x + h . we compute the slope in four steps.4. anticipating the next step. Copyright © 2007 by Pearson Education. f1x + h2 = 1x + h22 . Take the limit as h : 0. 2x + h . 2. .2 = 0 Example 3 Determine the equation of the tangent line to the curve y = x 2 . f1x + h2 .2. 1.2x .2 h:0 Check that last step. We see that there are two terms in the numerator of (1) and we must compute their difference. which is to simplify algebraically. (b) x = 0.f1x2 h = h12x + h . Thus the equation of the tangent line is y . before we allowed h to approach zero.0. we had to factor and reduce the fraction in order to insure that we did not get the form 0/0.4 = .25158.4x.2 = mtan1321x . It is y . and April Allen Materowski. . and 0. f13 + h2 = 213 + h2 + 1 = 24 + h f132 = 2 2. If we try h = . 0. 0.122 or y = . and Finance. we shall replace x by 3 in equation (1). f13 + h2 . y = f132 = 1322 .1 . Economics.1.4. We found in Example 2 that mtan122 = 4. So we try calculating the value of the fraction when h is 0. If we allowed h to approach zero. and substitute into the above equation.2.21 .02 or y = .8. the equation of the tangent line is y .1 = .2112 + 1 = 0.172 Section 2. let us see if a calculator will help. Therefore. However. y = f102 = 1022 . Gordon. Example 4 Determine the equation of the tangent line to the curve defined by the equation f1x2 = 2x + 1 when x = 3.001 we obtain 0. But what happens if we cannot factor and reduce the fraction? Let us take a look at such a situation.4 = 41x . Inc. f13 + h2 . we would have the form 0/0. y = f1 . Wang. Therefore. . We first note that f132 = 23 + 1 = 2.02 or y = 0. the equation of the tangent line is y . The equation of any line may be found using the point slope formula which was derived in Chapter 1. Therefore. We found in Example 2 that mtan1 .41x .001. Again the same limit as h approaches zero from the left (through negative values). (What kind of line is this?) We emphasize that whenever we try to find mtan1x2. Walter O.32 or y = 4x .1 Slope of a Curve Solution.12 + 1 = 4.2132 + 1 = 4.25 = 1/4.2 = h h Note that now there is no obvious cancellation.21x .0. (d) When x = 1.2 3. (a) When x = 3. We found in Example 2 that mtan112 = 0. the equation of the tangent line is y .2102 + 1 = 1. Therefore. Published by Pearson Learning Solutions. by Warren B. Solution. See Table 1.f132 24 + h .25001 respectively. Since we want the slope at a particular point.y1 = m1x .01.2x + 1.1. the equation of the tangent line is y .1. (b) When x = 0. We found in Example 2 that mtan102 = . We proceed as follows: 1.f132 = 24 + h . and 0. We cannot let h = 0. We feel safe in asserting that mtan132 = 1/4 and the equation of the tangent line is Applied Calculus for Business. but we can let h be close to zero. It certainly appears that the limit as h approaches zero will be 0. y = f112 = 1122 .122 . (c) When x = .25016.01 and .12 = 1 .0 = 01x .12 = .32.x12.0. What remains is to determine the slope of the curve at x = 3. Copyright © 2007 by Pearson Education. let us see that we could actually have found the limit by algebra without resorting to the calculator.25158 0. The algebraic device used in such problems is rationalization.1 . mtan122 = 1*4.c 2d B = a2b . and Finance. yielding y . we can get the appropriate cancellation by means of multiplying the numerator and denominator of the fraction by the conjugate of the numerator.2 h 0. we would have the form 0/0.0. Wang.2 = 1*41x . which is equivalent to the equation found above. Inc. If we allow h to approach zero. we have A a 2b + c 2d B A A a 2b .f132 1 1 1 = lim = = h :0 h : 0 24 + h + 2 h 4 24 + 2 lim Thus. In Example 4 we had f13 + h2 .001 0.2 = 1*41x . we have.x + 4y = 5 Although this is correct.24984 0. A 24 + h . producing a rational expression.2 B A 24 + h + 2 B h # A 24 + h + 2 B = 14 + h2 .32. Published by Pearson Learning Solutions.01 0.25016 . by Warren B. Gordon. Step 4. f13 + h2 . that is. rationalizing the numerator.c2d Note that the binomial expressions are identical with the exception of the connecting sign and are called conjugates of each other. and April Allen Materowski.Section 2. Copyright © 2007 by Pearson Education. . Solution. Example 5 Use rationalization to find the equation of the tangent line to the curve defined by the equation f1x2 = 2x + 1 when x = 3.0.0.24998 0.1 f13 + h2 . Recall that when the two following binomial expressions are multiplied. However.4 h A 24 + h + 2 B = h h A 24 + h + 2 B = 1 24 + h + 2 Therefore.2 = h h There is no obvious cancellation of h.1 Slope of a Curve 173 Table 1: Calculating the Difference Quotient for Small h h . Applied Calculus for Business.250016 0.24845 y . Thus. Economics.f132 24 + h .01 . It is precisely this observation which allows us to perform the cancellation needed in the previous example.32.f132 h = 24 + h . Walter O. which can be simplified to . The multiplication of these conjugates results in the clearing of all radicals.001 0. . That is. At such points the graph will not have a tangent line. it must not matter which side of P it is on. y = mx + b.x if x 6 0 A Place Where No Tangent Exists Applied Calculus for Business. But what does smooth mean? In words. The points where the slope changes are sharp points. As another example. Copyright © 2007 by Pearson Education. Remember. a graph is smooth if it has no corners or sharp points . it may be possible to get around this difficulty by using a calculator. Thus. as Q approaches P. Economics. In Figure 5. it might be that the tangent line to the curve is vertical. Consider any of the piecewise linear functions discussed in Chapter 1. We indicated above that the slope of a curve might not exist at every point on the curve. that is. and Finance. Inc. However. the slope of the tangent line is the same at each point. the slope of the tangent line at this point is not well-defined. and draw its tangent line. if a curve is smooth then the tangent line will exist at each point. x if x Ú 0 . by Warren B. However. The linear function is the only function which has this property. As we have seen. we choose a point Q either to the left or right of P and draw the secant line connecting P to Q. the limiting value for the slope of the secant line is the slope of the tangent line. consider the absolute value function. Some curves with corners and sharp points are sketched in Figure 5. Generally speaking. if there is a tangent line at a point on the curve it must be unique. Example 6 Show that the absolute value function f1x2 = x = e has no tangent line at x = 0. It is important to observe that the equation of the tangent line to each point along the graph of the linear function f1x2 = mx + b is. and equals the slope of the line. the algebraic steps needed to determine the slope of the curve using (1) are not immediately obvious.174 * ** Section 2. This observation gives us a whole class of examples of curves with sharp points. For example. the result may only be as accurate as the accuracy of the calculator used. given any linear function in the form f1x2 = mx + b. Published by Pearson Learning Solutions. Walter O. Nevertheless. As we let h approach zero. there are other possibilities. we shall usually choose it over a numerical one. If there is a simple algebraic technique.1 Slope of a Curve Sometimes. Wang. When we pick Q. Gordon. Why is this so? Choose any point on the line. the limiting value for the slope will not be the same as when we approach it from the right. sometimes guessing the answer this way will help you to find the right trick. mtan1x2 = m. The tangent line is precisely the line itself. in fact. and April Allen Materowski. At a sharp point this is not the case. y y P P x x Figure 5: Curves Which Are Not Smooth at P What makes a point a corner? Remember how we proceed when trying to find the slope of the tangent line at a point P. you can see that if we approach P from the left. Thus. We shall use them interchangeably. by Warren B. The interpretation depends upon what the function is modeling. but if we approach P from the right.f1x2 represents the slope of the tangent line at x. then the expression will represent the marginal cost of producing one more item. whether you are asked to find the slope of a curve. the slope of the tangent line does not exist at (0. f ¿ 1 x2 dy dx y ¿ 1 x2 d [f1x2] dx 1read fprime of x2 1read dydx not dy over dx2 1read y prime of x2 1read the derivative of f1x22 Applied Calculus for Business. .f1x2 the same thing that is. Thus. Below.1 Slope of a Curve 175 Solution. Each of the different interpretations of the derivative may be thought of as a particular brand name for the generic item. Therefore. Wang. Referring to Figure 6. or the marginal cost. a generic name is assigned to this expression. Walter O. We h shall soon see that this expression may also have other meanings which depend upon the interpretation of the function. The expression lim h :0 The Derivative NOTATIONS: If y = f1x2 then each of the following may be used to represent the derivative at x (the full name is the derivative of f with respect to x.1. you would do f1x + h2 . and April Allen Materowski. Economics. Inc. The derivative will be defined precisely in Definition 2.0). It is called the derivative. Or. the slope coming from the right is not the same as the slope coming from the left. if f represents position and x time. Published by Pearson Learning Solutions. we shall see that the expression is to be interpreted as a velocity. 0) from the left.0). Gordon.). y = *x y=x P(0. For example.Section 2. the tangent line has slope . Since at (0. compute lim . we list the most commonly used ones. we see that if we approach P(0. h :0 h There are various symbols that are used to represent the derivative. the slope of the tangent line is + 1. 0) Figure 6: f1x2 = x f1x + h2 . and Finance. Copyright © 2007 by Pearson Education. the velocity of a particle. if f represents total cost of producing x items. f ¿ 1x2 = 3x2. Sometimes. (c) This is just another way of phrasing (a). Walter O. Similarly. Next press the math key (F5) and scroll down to the Tangent option and press Enter. (b) The slope of the tangent line at x = 2 is f ¿ 122 = 31222 = 12. Gordon. Thus. and Finance.176 Section 2. f ¿ 1a2. (a) We must find the derivative of the given function. we could write f ¿ 132. Wang. If we want to indicate the derivative at x = a. So without additional work.f1x2 h13x 2 + 3xh + h22 3x2h + 3xh2 + h3 = = 3x 2 + 3xh + h2 = h h h f1x + h2 . f1x + h2 = 1x + h23 = x 3 + 3x 2h + 3xh2 + h3 f1x2 = x3 2. . we say that it is differentiable at that point. Next Applied Calculus for Business. We use (2). [f ¿ 1x2] x = a.f1x2 h:0 h (2) Notice that although we used the f ¿ 1x2 notation in (2). f ¿ 1x2 = lim Thus. we have as the derivative. we may write any of the following: dy d ` . y ¿ 1a2. Economics.1 Slope of a Curve Note that the various alternatives allow f(x) and y to be used interchangeably. by Warren B. We may now rewrite Definition1 as follows: DEFINITION 2 The derivative (when it exists) of the function defined by the equation y = f1x2 at the point P(x. Copyright © 2007 by Pearson Education.f1x2 = lim 13x 2 + 3xh + h22 = 3x2 h:0 h:0 h 4. Inc. instead of saying find the derivative of the function . d 3 1x 2 = 3x 2. Example 7 (a) Differentiate the function f1x2 = x 3. We proceed as follows: let the calculator draw the graph in the standard window. f1x + h2 . 1. to indicate the slope of the tangent line at dx x = a dx x = 3. if a function has a derivative at a point. Suppose we have entered on the Y = screen. d 3 1x 2 (c) Find dx Solution. y11x2 = x 2 + 1 and want to draw the tangent line at x = 2. and April Allen Materowski. (b) Determine the slope of the curve defined by f1x2 = x 3 at x = 2. Published by Pearson Learning Solutions. any of the alternatives would have been equally acceptable. f1x + h2 . we say differentiate the function.f1x2 = 3x2h + 3xh2 + h3 3. dx Calculator Tips The calculator can be used to visually illustrate the concept of a tangent line. f(x)) is given by f ¿ 1x2 = lim f1x + h2 . Section 2. h). y = 4x + 3 It is also a simple matter to compute the derivative. 6) P(8. Inc. see Figure 7 If you want to show additional tangent lines at other points. Gordon. scroll down to the Tangent option press Enter. h. P(4. x. via its definition. Figure 7: The tangent line to y11x2 = x2 + 1 at x = 2. then the derivative is nothing more than the limit of this quotient as h approaches zero. x.1 Slope of a Curve * ** 177 enter 2 (for x = 2) and press Enter. In the Calculator Tips Section 2. 9. . enter the x-value of the point where you want the tangent line. and press Enter. Economics. choose another point on the tangent line to determine the slope of the curve at P. 1. Walter O. that is. using the calculator. and 10. that is limit(avgRC(f(x). and the tangent line is drawn and its equation is given. press F5. by Warren B. 2. Published by Pearson Learning Solutions. h) gives the difference quotient. just repeat the process. EXERCISE SET 2. Wang. Copyright © 2007 by Pearson Education. In each case. 5) Figure 8: Ex 1 Figure 9: Ex 2 Applied Calculus for Business. and April Allen Materowski. and Finance.1 Exercises 1 3 refer to Figures 8.7 we illustrate that avgRC(f(x). 0). Given the curve whose equation is f1x2 = x 0.178 3. 26 f1x2 = . (b) Determine f ¿ 1x2 if x 6 1. Given the curve whose equation is f1x2 = x2 + 3. 25. f(x) as defined in Exercise 5. y = x2 at the point (3. f1x2 = e Applied Calculus for Business.001 (ii) 1. Use this information to sketch the curve.12x + 5. 6.x 2 at the point 1 . 4. if Q has as its x-coordinate: (i) 1. 02 27. by Warren B. 19. 16.25.999 (vi) 4. 42 P12. f1x2 = 1x 26. f(x) as defined in Exercise 6. (a) f1x2 = 53 (b) Give a geometric explanation for your result. 27 28. Given f1x2 = x3 . 21. 18.2 if x 1 x + 1 if x 7 1 (a) Sketch the graph of this function. f1x2 = x2 + 3 f1x2 = 2x + 1 f1x2 = x 3 P11. f1x2 = 1x 20. Published by Pearson Learning Solutions. 8. (b) have slope 15. Economics. (b) Using your sketch. if Q has as its x-coordinate: (i) 5. f1x2 = . Given the curve whose equation is f1x2 = 2x + 4. Let P be the point (5. 15.9999. Inc. Let P be the point (1. f1x2 = 3/x 24.00001 (iii) 0.0001 (iv) 0.3.2x2 + 3x + 3 Figure 11: Ex. 22 P12. if Q has as its x-coordinate: (i) 1.x2 + 2x . f1x2 = 3 . 12.9999. 42.99 (v) 0.7x + 9. 22. Walter O.1 9.1 Slope of a Curve 23.01 (ii) 5.0001 (iv) 4. (a) Determine the slope of the secant line joining P to Q. At which point will the curve have slope (a) 0.1. (c) have slope 36? 30.001 (iii) 5. 13. 12 P13. See Figure 11.999 (vi) 0. (b) What limiting value does the slope of the secant line appear to be approaching as Q approaches P? 10. 14.2/x 2 (compare with exercises 19 and 24). (a) Determine the slope of the secant line joining P to Q. (a) Determine the slope of the secant line joining P to Q.3). and then draw the tangent line at the point P. At which points will its tangent line (a) be horizontal.999 (iv) 0.1).9). Wang. See Figure 12.4). (b) What limiting value does the slope of the secant line appear to be approaching as Q approaches P? 11. 7. (c) Use (1) to determine the exact value of the slope at P. (a) sketch the graph of the given function. f1x2 = . f1x2 = . Copyright © 2007 by Pearson Education. Given f1x2 = 1x (a) At which point will the tangent line be vertical? (b) What can you say about the derivative at this point? 31. 82 P11.2x . short segments of the tangent lines are given at various points along a curve. 7) Figure 10: Ex 3 In Exercises 4 8. Let P be the point (1.3x2 + 7x .001 (iii) 1. f1x2 = mx + b Figure 12: Ex.12x.11.01 (ii) 1. and April Allen Materowski. (c) . approximate the slope of the curve at P. 17. Given f1x2 = 3x2 .9999.99 (v) 4. (b) 6. (b) What limiting value does the slope of the secant line appear to be approaching as Q approaches P? In Exercises 12 17.6? 29. P(8. Gordon. f1x2 = e 4x . f(x) as defined in Exercise 4. f(x) as defined in Exercise 7. * ** Section 2. (c) Determine f ¿ 1x2 if x 7 1 (d) What can you conclude about f ¿ 112? x2 if x Ú 0 x if x 6 0 32. . x In Exercises 26 and 27. 5. Find f ¿ 1x2 in Exercises 18 . and Finance. f1x2 = 2x2 . determine the derivative at the given point on the curve using equation (2). Find the slope of the secant lines passing through P(0. something better has been done. without resorting to the definition. .0.8) crosses the curve. Of course. f1x22 for Q instead of 1x + h.1. Walter O. f1x2 = e 9x + 5 if x Ú 1 (a) What is f ¿ 1x2 if x 7 1? (b) What is x2 + 7x + 6 if x 6 1 f ¿ 1x2 if x 6 1? (c) What is the slope of the curve just to the left of x = 1? Derivative Rules 1 * ** 179 37. and Finance. Find the point on the curve y = x at which the tangent line at (2.1.0. it may be necessary to go through a great deal of algebraic manipulation.0001 (as h approaches 0 from the right).2 (a) Sketch the graph of this function. this may be also be proved from equation (2) of the previous section. you would assume that tables of derivatives would have been put together by working mathematicians.0001. .0) and Q(h. In the previous section we indicated that the derivative of a linear function is its slope.5 . call the equation y3(x) and enter it the Y = screen (c) x = 2.2 » » » » » Derivative Rules 1 Derivative of a Linear Function The Simple Power Rule The Constant Multiplier Rule The Sum Rule Calculator Tips At this point. 39.1. Published by Pearson Learning Solutions. Let y11x2 = x2 + 1.01.001 and 0. Suppose that. In fact. call the equation y2(x) and enter it the Y = screen (b) x = 2. have been developed. Show that the definition of the derivative will then have the following alternate form: f ¿ 1x2 = lim f1x22 . the only way that we can compute the derivative is by applying the definition. f(h)).Section 2.05. Given f1x2 = 2x . 0)? Why not? (b) Now repeat the process for g(x). as you were asked to do in the preceding exercise set.x x2 : x (d) What is f ¿ 112? 35. However. Use the alternate form of the derivative given in Exercise 37. we wrote (x2. (a) At what point is the function not differentiable? (b) What is the derivative to the left of this point? (c) What is the derivative to the right of this point? 34. Inc. Let us state this as a rule. We call this special case Rule 2. the justification for these rules will rest upon the definition. (b) Exercise 22. This followed directly from the interpretation of the derivative as the slope of a tangent line. Copyright © 2007 by Pearson Education.025. (e) Have the calculator add the tangent line at x = 2. Using h = . in the development of the definition of the derivative. Economics. and h = 0.. (as h approaches 0 from the left). call the equation y4(x) and enter it the Y = screen (d) choose an appropriate window so the curve and all these secant lines can be seen.0. f1x + h2). What is happening to the secant lines as x approaches 2? 2. Derivative of a Linear Function Applied Calculus for Business. What is the difference in the behavior at P(0. If nothing else.f1x2 x2 . Does mtan1x2 exist at (0. Gordon. (b) Determine f ¿ 1x2 if x 6 0 (c) Determine f ¿ 1x2 if x 7 0 (d) What can you say about f ¿ 102? 33. (c) Exercise 23. RULE 1 THE DERIVATIVE OF A LINEAR FUNCTION d 1mx + b2 = m dx Note that in the special case in which m = 0. (d) Exercise 24.) 36. Wang. the line is horizontal (it has zero slope) and the derivative is zero. by Warren B. Of course.001 and . (You may want to use the result of Exercise 17.0) for the two functions? 3 38. . when you realize that the calculus has been used for over three centuries. determine the equation of the secant line through each of the following x-values and x * 2: (a) x = 2. We have seen that in order to apply the definition. you must expect that some shortcuts would have been discovered. equation (2) of the previous section. A collection of simple rules for producing derivatives.01.0.0. to compute f ¿ 1x2 for the function defined in: (a) Exercise 21. 0. Our objective in this section is to begin to determine those rules that will let us calculate derivatives without using the definition. and April Allen Materowski. Consider the two functions: f1x2 = x1/3 and g1x2 = x4/3 near x = 0. . x 100 h :0 h We must expand the first expression in the numerator and cancel the h from the denominator. how will we be able to let h approach zero without encountering the form 0/0 ? Of course. dx We prove this rule in the special case in which N is a non-negative integer. You may recognize (1) as another way of writing the binomial expansion of 1x + h2N. with f1x2 = xN. we have not really expanded our knowledge. Walter O. we need only consider N 7 1. (Actually. Consider expressions of the form 1x + h2N. where N is a positive integer. In which case.2 Derivative Rules 1 RULE 2 THE DERIVATIVE OF A CONSTANT d 1b2 = 0 dx In words. or N = 1 (why?). by Warren B. we have.) If N = 2. Thus.1 where N is any real number. and April Allen Materowski. Wang. where T4 = 16x2 + 4xh + h22 Do you see the developing pattern? In general. since we have Rule 1. 1x + h22 = x 2 + 2xh + h2 1x + h23 = x 3 + 3x 2h + 3xh2 + h3 = x3 + 3x 2h + h213x + h2 = x3 + 3x2h + h2T3 where T3 = 13x + h2 1x + h24 = x 4 + 4x 3h + 6x 2h2 + 4xh3 + h4 = x 4 + 4x 3h + h2T4. Inc. so good but. . Applied Calculus for Business. Otherwise.x N h :0 h We now use (1) from above to replace the first term in the numerator and obtain. We would like to be able to determine the derivative of any function whose equation is of the form f1x2 = xN. and develop a derivative formula for these functions. we want more than the solution to this specific problem. Consider the problem of finding the derivative of the function whose equation is f1x2 = x100. we need only prove it for any integer N Ú 2. Published by Pearson Learning Solutions. Using the definition of the derivative. (2) of the previous section.1h + h2TN (1) The Simple Power Rule where TN is a polynomial in x and h. f ¿ 1x2 = lim 1x + h2N . the derivative of a constant is zero. you know the exact form of TN. we have that 1x + h2N = x N + Nx N . So far. Gordon. Let us now look at some simple non-linear functions. Economics. Rules 1 and 2 just summarized our observation that the slope of a curve reduces to the slope of the line if the function is linear. We already know that Rule 3 is correct in the case N = 0. If N = 3. RULE 3 THE SIMPLE POWER RULE d N 1x 2 = Nx N . and Finance.180 * ** Section 2. Copyright © 2007 by Pearson Education. If N = 4. We apply the definition and obtain f ¿ 1x2 = lim 1x + h2100 . Walter O. Also note the form of our answers.2 = x .we usually leave the answer in radical form. by Warren B. 5 5 dx 5x 2/5 d d 1/2 1 1 2 1 1 1 1 (c) 1 1x2 = 1x 2 = x 2 . (a) f1x2 = x43 (b) f1x2 = x 3/5 1 (c) f1x2 = 1x (d) f1x2 = 2 .2x -3 = . The Constant Multiplier Rule RULE 4 THE CONSTANT MULTIPLIER RULE d 1Cf1x22 = Cf ¿ 1x2 dx Applied Calculus for Business.n bn a n an a b = n b b Given the differentiable function defined by y * f(x). We illustrate its use in the following example. and Finance. Example 1 Determine f ¿ 1x2 for each of the following. assume its truth for any constant N. Copyright © 2007 by Pearson Education. Inc. is multiplied by h. d 43 1x 2 = 43x 43 . Published by Pearson Learning Solutions. Gordon.x N Nx N .1 = 43x 42 dx d 3/5 3 3 5 3 2 3 (b) 1x 2 = x 5 .1 + hTN2 lim = lim 1Nx N . Table 1: Review of the Laws of Exponents bn = b # b Á b 1bm2n = bm n 1ab2n = a nbn b -n = 1 bn b m # b n = bm + n b0 = 1 bm/n = 2bm = A 2b B m n n # bm = bm . It might be a good idea for you to review these rules which are summarized in Table 1. Economics.1 h:0 h:0 h Notice that because TN. similarly.1 + hTN2 = Nx N .2x -2 . For now. which is a polynomial in x and h. .1h + h2TN = lim = h:0 h:0 h h h1NxN .Section 2. In later sections we shall show why this rule is valid for values of N which are not nonnegative integers.2 Derivative Rules 1 181 f ¿ 1x2 = lim x N + Nx N .3 dx x 2 dx x (a) Notice that in (c) and (d) we used various rules involving exponents and rewriting radicals in terms of fractional exponents. we rarely leave answers with negative exponents as in (d). and April Allen Materowski. that entire term goes to zero as h goes to zero.1 = .2 = = 1/2 dx dx 2 2 2 1x 2x d 1 d -2 2 (d) a b = 1x 2 = . suppose we multiply this function by a constant C. if we start with a radical. Wang. x Solution.5 = x -5 = .1h + h2TN . as in (c). How will the derivative of Cf(x) be related to the derivative of f(x)? Rule 4 answers this question. Certainly s132 = 5 + 7 = 12. The denominator remains simply h. s1x2 = f1x2 + g1x2. which in the limit becomes the sum of the individual derivatives. s1x2 = f1x2 + g1x2. we may handle f1x2 .g1x2 by simply noting that f1x2 . dx dx d 15x 72. for a given x-value. Economics. Published by Pearson Learning Solutions.g1x22 = f ¿ 1x2 .[f1x2 + g1x2] = [f1x + h2 . We can form a new function. Inc. Applying Rule 4. Thus. RULE 5 THE SUM RULE d 1f1x2 + g1x22 = f ¿ 1x2 + g ¿ 1x2 dx d 1f1x2 . you should find yourself doing the intermediate steps mentally and writing d 15x 72 = 5 # 7x 6 = 35x 6.182 Section 2.f1x2] + [g1x + h2 . It is not hard to see why this rule is true.f1x2] Thus. and Finance. we have d d 15x 72 = 5 1x 72 = 517x 62 = 35x 6. When we sum two functions. called the sum. and April Allen Materowski. the y-value of the sum is the sum of the two individual y-values. dx The Sum Rule Suppose we have two differentiable functions f(x) and g(x).g1x2 = f1x2 + 1 .12g1x2 and applying the constant multiple rule. Gordon. Copyright © 2007 by Pearson Education. Applied Calculus for Business. Multiplying a function by a constant means that each y-value is multiplied by the constant. Example 2 Determine Solution. dx After doing a few examples using this rule. Wang. this rule states that the derivative of a sum is the sum of the derivatives and the derivative of the difference is the difference of the derivatives. if g1x2 = Cf1x2.2 Derivative Rules 1 In words.g1x2 = Cf1x + h2 . Thus the quotient becomes a sum of the individual components. . Of course. by Warren B.s1x2 = f1x + h2 + g1x + h2 . the derivative of a constant times a function is the constant times the derivative of the function. This rule generalizes to more than two functions and we may restate it in words by stating that the derivative of a sum of functions is the sum of their derivatives. Is that clear? Suppose f132 = 5 and g132 = 7. Walter O. Therefore. the numerator of equation (2) of Section 2. The next rule shows how the derivative of s(x) is related to the derivative of the two component functions.1 becomes s1x + h2 . the entire numerator of the difference quotient in equation (2) of the previous section is multiplied by C (without any change in the denominator) resulting in the derivative being multiplied by C.Cf1x2 = C[f1x + h2 .g1x2].g ¿ 1x2 dx In words. The details of the formal proof are given at the end of Section 3.3. A little thought should indicate why this rule is true. then g1x + h2 . Substituting this information into the point-slope equation for the line.31322 + 5132 . As you can see. 24) on the curve. Solution.3 # 2x -4 = 20x3 . mtan182 = 4182-1/3 = 4/2 = 2. we need the slope and one point. 24). at the point where x = 3.6. rather than writing it out in detail. Find the exact value of y at x = 9 on the curve and on the tangent line. we have y . y ¿ = 3x 2 .2! Let us look at some applications of these rules. by Warren B. then on the curve.1 that the tangent line is the straight line that best approximates the curve near the point of tangency. 4. At the point with x = 3. Find the equation of the tangent line to y = 6x 2/3 at the point (8. y = 14x . an error of less than 2 parts per hundred was a small price to pay for the greater simplicity of computing the value on the tangent line function rather than the value of f(x). The derivative is. Economics.24 = 21x . Since we are given the point (8. Applied Calculus for Business. The second term was handled similarly.3x 2 + 5x . and Finance. At y = 9. Therefore. which represents a percent error of 10. For this we need the derivative. we have y = 2192 + 8 = 26. Solution.33 Example 5 We remarked in Section 3.960521100%2 = 1. Gordon. so. Copyright © 2007 by Pearson Education. we have mtan132 = 31322 . f ¿ 1 x2 = d d d 15x4 + 2x -32 = 15x 42 + 12x -32 = 4 # 5x3 . Hence.52%. we need the derivative. and decreased the power of x by 1. To find the slope.9605 (to four decimal places on a calculator). in the days before calculators. Inc.32 = 14x . y = 2x + 8. Find the error and the percent error that you have by using the straight line approximation.6x + 5. . and April Allen Materowski.Section 2. 5. y ¿ = 612/32x2/3 . we multiplied the coefficient. yields . reducing the number . At x = 8.9 = 141x . The error in the approximation is 0. Remember. Example 4 Find the equation of the line tangent to y = x3 . 9). find f ¿ 1x2. If x = 3. the point of tangency is (3. we have y = 61922/3 = 25.3 by 1.6 = 9.6132 + 5 = 14. Published by Pearson Learning Solutions.2 Derivative Rules 1 183 Example 3 If f1x2 = 5x4 + 2x -3. Solution. we need only find the slope of the tangent line in order to get its equation.6x -4 dx dx dx Note that in applying Rule 3 to the first term. Using our new rules. on the tangent line. y = 1323 . The point is easy to find.1 = 4x -1/3. On the curve. the equation of the tangent line is y .4 not .0395. Wang. by the power of x. In order to find the equation of the line. Walter O.42.0395/25.82. . Published by Pearson Learning Solutions. On the keypad. . the alphabetic d will not work! Applied Calculus for Business. by Warren B. Note the with key is the symbol located to the dx left of 7 on the keypad. Gordon. and the closing parenthesis. Inc.2 Derivative Rules 1 Calculator Tips The TI 89 can find derivatives very easily. x2 as in Figure 1. Our notation for derivative is d/dx (f(x)) which tells us to take the derivative of f with respect to x. Economics. To access this d. is different from the alphabetical d which requires you first press the alpha key and then the comma key.184 * ** Section 2. Only the d above the number 8 key is used for differentiation. We must add the function we want the derivative of. 15x 82 would dx be entered as d15x ¿8. Walter O. . in orange is the letter d. The TI 89 does it a little differently. Figure 1: Using the TI 89 to Find Derivatives Pressing Enter gives the result as seen in Figure 2. and April Allen Materowski. For example. Wang. Figure 2: Finding the Derivative on the TI 89 Figure 3: Computing the Value of a Derivative with the TI 89 Be careful. you press the 2nd key followed by the number 8. and Finance. Using the with key we can have the calculator compute the value of the derivative at a given value for x. . For example d 15x 82 x = 2 is illustrated in Figure 3. the variable we are d differentiating with respect to. Copyright © 2007 by Pearson Education. the d used for differentiation is above the number 8 key (and first requires you press the 2nd key). above the number 8. Doing so presents on the screen d( which means the derivative of . lim h:0 2 38 + h . Gordon. - 1 x2 .17) to the curve y = 2x . + 2 3 x + 2.1x .x2. 3 u 4 du h:0 11.2 30. 25. f1x2 = 1x . use the appropriate rules to determine the derivative. (c) Compare the slopes of the tangent and the chord. lim 29. Compute 3 7 + 3/4 + 5. (c) Find the equation of the tangent line to the demand curve at p = 2. (b) Find the drop in ridership if the price is raised to $2.x h 1 28. Suppose that M = 4. f1x2 = 3x2 64 x = 2 h:0 .12 and (3. where p is the price of a commodity and x is the quantity of the commodity that can be sold.2x + 1.221x + 22. f1x2 = 2 3x . 6. (a) Find the ridership when the price is two dollars. (b) Find the point on the curve at which the slope of the tangent line is the same as the slope of the chord. Show that the area of the triangle in the first quadrant formed by the tangent line to any point on the curve y = k/x. Find the x-coordinate on the parabola where the tangent line to the curve has the same slope as the chord.3x4 + 2/x2 . y = 3x2 .2x 2 + 7.20. 1. y = 3x4 . (e) Find the ratio of the value found in (d) to the change in price.1.x + 1 at x = 2.x2.13) to the curve y = 4 . Walter O.7. 4 2 2x + h . y = 3x2/3 . Find the points on the curve y = 2x3 . y = x2. 17. lim 1x + h257 .2 Derivative Rules 1 * ** 185 EXERCISE SET 2.7x + 9. and Finance.5 + 293 b dx x dy 3x5 .10) (b) (4. Determine the equations of the tangent lines from the point (0. 13) on the curve. f(31. by Warren B. Use your knowledge of the derivative to compute the limit given in Exercises 26 31. f1x2 = 2x7 8. lim 27. Hint: Divide first. 36. 9) on the curve. find f ¿ 1x2. w = 32v1/4 16 v2 + 7v2 + 2. suppose you do not have a calculator and want to approximate f(7. This is called the marginal demand.7x 2 + 2 5.2x3/2 . 24. Find the slope of the chord (see Exercises 34 and 35) connecting the points with x-coordinates r and s. find y ¿ . x = 2 16. y = x2 + 3x . . (a) For what values of x is f ¿ 1x2 7 0? (b) For what values is f ¿ 1x2 6 0? (c) At which point(s) will the tangent line be horizontal? 21. 33.2 . d 7 a 3x5 .12. find ds .9).1. Wang. Inc.x 57 . f1x2 = 3x . This line is called a chord. Economics. dv 9. x4 x 7. 13. Published by Pearson Learning Solutions. (a) Find the slope of the line connecting the points (1.1 4 x + 7x + 32x4 + 71. k 7 0 and the coordinate axes is a constant. 34.Section 2. 4). Copyright © 2007 by Pearson Education. In Exercises 12 15. (c) Compare your result with the answer given by your calculator. k = 1/2. lim 1x + h22 h h:0 2. Consider the function defined by f1x2 = x4 + 8x3. find f ¿ 1 . s = . Given the parabola. Approximate. Find the point(s) on the curve y = 6x1/3 at which (a) the slope is 1*2. Find the equation of the line perpendicular to the tangent line to the curve y = x3 . 1) and (3. Let y = ax2 + bx + c. . h:0 h h:0 31. h 32. (a) Find the equation of the tangent line at x = 8. Determine the equations of the tangent lines from the point (a) (1. h 2/3 2/3 1x + h2 . determine the equation of the tangent line at the indicated x-value.2 In Exercises 1 11. (Note that this exercise gives an alternative method for locating the vertex of a parabola. Recall that lines are perpendicular if the product of their slopes is . Find the points on the curve y = x . y = 4x + 2/x . 3.25. p is the price per ride on a New York City subway. dt dw . find f ¿ 1x2. and x is the number of riders per day (in millions). dx 3x3 4. 35. r = + .9. lim . find 3 dy dx ` x=2 23.) 19. One of the most commonly used mathematical models for a demand function in micro-economics is x = Mp -k. 26. find f ¿ 1x2.7 dr 10. y = . and April Allen Materowski. 18.8x + 3 at which the tangent line is horizontal. find u 5 3u . using the method of the previous exercise. 20. Find the point on the parabola y = ax2 + bx + c where the tangent line is horizontal. (b) Find the slope of the tangent line to the curve at (2. if f1x2 = x2/5. y = 2 3 x + 2x . (b) the tangent line is vertical. Given the parabola.99). find .5. find .16t2 + 256t + 128.54x + 1 at which the tangent line is horizontal. 12. (b) Compute the y-value on the tangent line at x = 7.25. Applied Calculus for Business. x = 8 x2 14. h 17 17 12 + h2 . (d) Use the value of x on the straight line to find the approximate drop in ridership if the price is raised to $2. Consider f1x2 = 4x2/3. 22. x = 8 15. (a) Find the slope of the line connecting the points 11. x :3 The Limit Solution. it will only be necessary for us to understand it intuitively. However. there are times when the function is not defined at the point at which x = a. In the following definition.3 » » » » » » » » Limits and Continuity The Limit Limits by Substitution One Sided Limits Jumps and Holes Continuity Removable Discontinuities Differentiability and Continuity Calculator Tips The limit is one of the most commonly used tools of the calculus. Remember. and so on.0001. Copyright © 2007 by Pearson Education. when the x-values are very close to 3. However. At this level.99992 = 9. suppose x is 2. we assume y is a function of x and given by the equation y = f1x2.9992 + 3 = 9.00002.9992 = 212. We leave its formal definition to more advanced courses. as well as the meaning of the words very near.00001. Published by Pearson Learning Solutions. For example. . division by zero is always undefined. These terms will all be illustrated in the examples. by Warren B. either just to its left or just to its right. f13. Therefore. and Finance. The closer the x-values are to 3 (either just to its left. Inc.000012 = 10. You might be asking yourself why not just substitute x = 3 into the expression and obtain the value 10 for the limit? In many cases. very close. Of particular concern are fractions in which the denominator is zero. Walter O. and the rule changes at the point x = a. and even know how to determine limits in many cases. A few examples should illustrate the meaning of the definition. So we pick some values of x that we would think of as very near 3. In the preceding two sections. Now you are probably wondering what this means. Thus. DEFINITION 1 x :a lim f1x2 = L. these terms are more carefully defined. just to the right or just to the left. is read the limit of f(x) as x approaches a is L. Applied Calculus for Business. If x = 3. or just to its right). and April Allen Materowski. f12. you have already seen one of its applications.186 Section 2.9998. Let us consider examples that illustrate these cases.00012 = 10. Other times the function is defined in pieces. In these or other situations. Example 1 Find lim 12x + 42. In more advanced courses. Wang.999. special care must be taken. the closer the y-values get to 10. We want to find the y value that f1x2 = 2x + 4 is very near when x is very near 3. f13. that is all you have to do to get the correct answer. Gordon. once you understand what they mean intuitively.9999. L (when it exists) is the unique number that the y-values are very near when the x-values are very close to a. the transition to a more rigorous definition is not very difficult. Economics. 10 is that unique number that the y-values are very close to. Now f12.0002.998. If x is 2. x :3 lim 12x + 42 = 10. If x = 3.3 Limits and Continuity 2. the problem reduces to the kind of situation considered in Example 1.9 = lim 1x + 32 x :3 x . you see that substitution (in this case. When x is near 3.3 Solution.9 = = x + 3 x .) Table 1: Examining the Value of the Function Near x = 3 Observe that as x approaches 3 from either direction. choosing values of x just to the left and right of 3. Copyright © 2007 by Pearson Education. refer to the Calculator Tips in Section 1. let us proceed as above. x + 3 is near 6. Inc. you would obtain the form 0/0 . However. Economics. Published by Pearson Learning Solutions.Section 2. x . We show our results in Table 1. the values of f(x) get very close to 6. If the function is nice . We will discover that nice usually means that the function has no Applied Calculus for Business.3 Limits and Continuity 187 Example 2 x2 . Again. (The entries in the y1 column are the y-values corresponding to the x-values in the first column.3 x :3 lim Thus. Therefore. Walter O. it would not change the limit. since. The form 0/0 is called an indeterminate form and requires closer examination. and Finance. when we consider the limit. x2 . after the quotient is simplified) gives us the correct answer. the limit would remain 6.3 1x .9 Find lim . x is near but never equal to 3.32. the cancellation is valid. all that matters in determining the limit is what happens to the y-values near x = 3. It did not matter that the function was not defined at x = 3. the limit is 6.32 If we let x = 3. Therefore. by Warren B. x :3 x . the function is undefined. Gordon. But observe that 1x . the function could have been defined to have any y-value at all at x = 3. it will usually turn out that all you have to do is substitute to determine the limit.5) You might be convinced that substituting x = 3 directly into the given expression in the previous example is not useful and you would be correct. In fact. Wang. if we were to define f132 = 21. . (Note: For details in setting up a Table.321x + 32 x2 . For example. (Why? Because the denominator of the fraction is zero!) If you tried to substitute x = 3 directly into this expression.3 = 0 and we could not cancel the common factor 1x . Therefore. At x = 3. and April Allen Materowski. we see that since the function is not defined at x = 3. Except at the point x = 3. THEOREM 2 x :a lim x = a (To see why these two limits are true. . there is a hole in its graph at the point (3. the limit of a constant is the constant itself. However. To see what we mean by nice . Economics. and Finance.3 Limits and Continuity jumps or holes . Suppose that we try to sketch the graph of y = f1x2 from Example 2. and April Allen Materowski. Gordon. x :a lim kf1x2 = k lim f1x2 = kL x :a In words. Nonetheless.) THEOREM 3 If k is a constant. Applied Calculus for Business. We again stress that most functions that you shall encounter in this text will be nice except at an occasional isolated point. THEOREM 1 x :a lim c = c In words. Walter O. Copyright © 2007 by Pearson Education.3 Limits by Substitution In general. Wang. and that is the limit as x approaches 3. the function is defined by the equation y = x + 3. Published by Pearson Learning Solutions. and use the definition of the limit. All of these theorems should seem obvious. by Warren B. when x is close to x = 3. The graph of this function is simply a straight line of slope 1 with y-intercept 3. 6 x=3 x Figure 1: f1x2 = x2 . then lim f1x2 = f1a2. Inc. In all of these theorems. although the proofs of some of them require several steps.9 = x + 3 if x Z 3 x . draw the graph of y = c and y = x.188 * ** Section 2. and lim g1x2 = M x :a each exist. the following is true: If the function defined by the equation y = f1x2 is a nice function.6) (see Figure 1). let us state x :a some theorems (without proof) that will make the notion more precise. we assume that x :a lim f1x2 = L. the limit of a constant times a function is the constant times the limit of the function. the y-values are all close to 6. Section 2.3 Limits and Continuity 189 THEOREM 4 x :a lim 1f1x2 ; g1x22 = lim f1x2 ; lim g1x2 = L ; M x :a x :a In words, the limit of a sum (difference) is the sum (difference) of the limits. THEOREM 5 x :a lim f1x2g1x2 = A lim f1x2 B A lim g1x2 B = LM x :a x :a In words, the limit of a product is the product of the limits. THEOREM 6 If M Z 0, lim lim f1x2 f1x2 L x :a = = x : a g1x2 lim g1x2 M x :a In words, the limit of a quotient is the quotient of the limits, as long as M Z 0. THEOREM 7 If N is any real number, lim 1f1x22N = A lim f1x2 B N = LN x :a x :a (if L 6 0, and if N = p/q where q is even, then L is not a real number. Therefore, we exclude this case from the theorem.) In words, the limit of a function (raised) to a power is the power of the limit (subject to the restrictions indicated above). Let us look at some examples of how these theorems could be used. Example 3 Apply Theorems 1 through 7 to evaluate the following limits. (a) lim 6x x :2 N (b) lim 1x 2 - 32 x :2 (c) lim 16x1x 2 - 322 x :2 (d) lim 1x + 42 x :2 (e) lim 6x1x 2 - 32 x :2 x + 4 Solution. (a) lim 6x = 6 # lim x (Theorem 3) x :2 x :2 6122 = 12 (Theorem 2) (b) lim 1x2 - 32 + lim x 2 - lim 3 (Theorem 4) x :2 x :2 x :2 x :2 2 lim x = 4 (Theorem 7), and lim 3 = 3 (Theorem 1) x :2 2 x :2 thus, lim 1x - 32 = 4 - 3 = 1 (Theorem 4) (c) lim 16x1x 2 - 322 = lim 6x # lim 1x 2 - 32 1Theorem 52 = 12 # 1 = 12 x :2 x :2 x :2 (d) lim 1x + 42 = 6 (Theorems 1, 2 and 4) x :2 (e) In part (c) we found the limit of the numerator is 12. In part (d) we found the limit of the denominator is 6. Therefore, by Theorem 6, the limit of the quotient is 2. The point of the last example is to evaluate the limit in part (e). It seems like a lot of steps to find the limit of one function. However, every algebraic function is defined by a similar Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 190 Section 2.3 Limits and Continuity series of component steps. All of these steps are made up of the operations of addition, subtraction, multiplication, division, powers, and roots. Therefore, by application of these seven theorems, any algebraic function is nice. The limit as x approaches a of f(x) will be just f(a) as long as nothing special (such as division by zero) occurs at x = a. Thus, we could have simply evaluated the limit in part (e) above by just substituting 2 directly for x wherever it appeared in the fraction. Example 2 above is a bit more complicated because of the appearance of a zero denominator. However, once we cancelled the 0/0 term, substitution was appropriate. This is a good trick to look for. Any time you are faced with an expression in which substitution yields something of the form 0/0 , try some algebraic manipulations first, and then substitute. Example 4 Compute lim h :0 29 + h - 3 . h Solution. If we try to substitute 0 for h in the expression, we obtain the form 0/0 . Instead, we first perform some algebraic manipulations rationalizing the numerator by multiplying by the conjugate. Observe that a 19 + h2 - 9 29 + h - 3 29 + h + 3 ba b = h 29 + h + 3 h A 29 + h + 3 B = Therefore, we have, h :0 h h A 29 + h + 3 B = 1 29 + h + 3 lim 29 + h - 3 1 1 = lim = . h : 0 29 + h + 3 h 6 In finding limits we often use this technique. (We could have also taken values for h just to the left and right of 0 and constructed a table to estimate the limit.) One Sided Limits When considering a limit, we need to consider what happens just to the left and right of the point in questions. It proves useful to introduce symbols which represent the behavior of the function to the left and right of the point. Again, assuming y = f1x2, we define lim f1x2 read the limit of f(x) as x approaches a from its left the left-handed limit, x : awhich, when it exists, is the unique number that the y-values are very near when the x-values are just to the left of a, and lim+ f1x2, read the limit of f(x) as x approaches a from its x :a right the right-handed limit, which, when it exists, is the unique number that the y-values are very near when the x-values are just to the right of a. It then follows from the definition of the limit, that the limit exits at a if and only if these two one sided limits are equal to each other, that is lim f1x2 exits if and only if lim- f1x2 = lim+ f1x2. x :a x :a x :a Our next example illustrates one-sided limit through the use of the notion of piecewise linear functions. Example 5 Given f1x2 = e - 2x + 6 3x + 1 Determine lim f1x2. x :1 if x 6 1 if x Ú 1 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.3 Limits and Continuity 191 Solution. If x is just to the left of 1, then x 6 1, and we are on the piece of the function defined by f1x2 = - 2x + 6. Just to the left of x = 1, we have lim- f1x2 = lim-1 - 2x + 62 = 4. Similarly, if x is just to the right of 1, then x 7 1, x:1 x:1 and we are on the piece of the function defined by f1x2 = 3x + 1, so we have lim+ f1x2 = lim+ 13x + 12 = 4, Since the left and right handed limits are the same, x:1 x:1 number, we have lim f1x2 = 4. Figure 2 illustrates graphically why the limit is 4. x :1 y=4 x=2 Figure 2: f1x2 = e - 2x + 6 3x + 1 if x 6 1 if x Ú 1 The next example illustrates the notion of a jump in a function. You will see that when a function has a jump, it is impossible for the y-values just to the left of the jump to be near the y-values just to its right. Therefore, the limit does not exist. Example 6 Given, f1x2 = e - 2x + 6 if x 6 2 3x + 4 if x Ú 2 Determine lim f1x2, if it exists. x :2 Jumps and Holes Solution. Just to the left of x = 2, we are on the piece of the function defined by f1x2 = - 2x + 6. Therefore the left-hand limit, lim- f1x2 = lim- 1 - 2x + 62 = 2. Just x :2 x :2 to the right of x = 2, we are on the piece of the function defined by f1x2 = 3x + 4, and the right hand limit, lim+ f1x2 = lim+ 13x + 42 = 10. Since we are not close to the x :2 x :2 same y-value on either side of x = 2, that is, the left and right handed limits are not equal, the limit does not exist. From Figure 3 it is evident why the limit does not exist at x = 2. There is a jump in the y-value at x = 2. In general, at any point in the domain of the function at which it has a jump, the limit does not exist. However, there is a difference between a jump and a hole. Example 2 has a hole, but the limit exists because the limiting y-values on either side of the hole are the same. In general, a function will have a limit at any point at which its graph has a hole. In fact, the limit is precisely the y-value needed to plug the hole. Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 192 Section 2.3 Limits and Continuity y = 10 x=2 y=2 Figure 3: f1x2 = e - 2x + 6 3x + 4 if x 6 2 if x Ú 2 The next example illustrates how to determine a limit when there are no obvious ways of cancelling the 0/0 term. It requires the use of the calculator. Example 7 3x - 1 . x x :0 Determine lim Solution. If you try to substitute x = 0 into this expression, you will find that we obtain the indeterminate form 0/0. There are no obvious algebraic tricks to try, so we resort to the definition, and use the calculator. We shall choose x-values just to the right and just to the left of x = 0, to see what y-value, if any, we are near. We summarize our calculations in Table 2. We see from Table 2, that as x gets very close to 0, (on either side), the y-value gets very close to 1.099 (to three decimal places). In fact, we shall see, the exact answer is ln 3, the natural logarithm of 3, which is 1.098612289 to 9 decimal places. 3x - 1 Table 2: lim . x:0 x Since a numerical approach is not a rigorous proof, whenever possible we shall show how to determine a limit algebraically and use the calculator as a check. However, when we are not able to handle the problem algebraically, a numerical approach may be very useful Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.3 Limits and Continuity * ** 193 and will usually suffice for our needs. In more advanced courses, techniques are developed so that one could actually prove the validity of such numerical results. Essentially, we have demonstrated that to compute a limit, when there are no jumps, substitution is an effective method. But, you must be careful when to substitute if you obtain the form 0/0. There are functions which do not have limits for reasons other than jumps but they will not be considered here. As we indicated earlier, if f(x) is a nice function, for example, a polynomial, or a rational function (the ratio of two polynomials), or a radical function, then lim f1x2 = f1a2. In the case of a rational function, we assume that x :a the denominator is non-zero at x = a. In the case of a function that involves a radical with an even index, we assume that any expression under the radical is non-negative. (In Exercise 27 of Section 3.5, we consider another method for determining the limit when you obtain the indeterminate form 0/0 it is called L Hôpital s rule.) The time has come to attach the correct name to a nice function. The kind of function that we have in mind is what we call a continuous function. Roughly speaking, any function whose graph is without holes or jumps is a continuous function. Any point in the domain at which there is a hole or jump in the function is called a discontinuity. There are other types of discontinuities, a discussion of them is left to more advanced courses. Continuity Figure 4: Illustrating Discontinuities Since polynomials are functions without holes or jumps, they are functions which are continuous everywhere. Similarly, at any point at which its denominator is non-zero, a rational function is continuous. To illustrate a discontinuous function, we need only construct a function which has either a hole or a jump at some point in its domain. In Figure 4, we have a function which is undefined at x = 1, has a hole at x = 2, and has a jump at x = 4. At each of these x-values, the function is discontinuous. Thus, if we have a function without undefined points, holes, or jumps, we have a continuous function. Now that we have a feel for the physical definition of continuity, we give a formal definition. DEFINITION 2 A function defined by the equation y = f1x2 is continuous at x = a, if the following three conditions are all satisfied: Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 194 Section 2.3 Limits and Continuity (1) f(a) is defined (2) lim f1x2 = L exists x:a (3) f1a2 = L Condition (1) means that at x = a, we know how to compute the corresponding y-value, f(a). Condition (2) implies that there is no jump in the graph at x = a, and Condition (3) means that there is no hole at x = a. Many texts shorten the definition of continuity at x = a by just writing that the function is continuous at x = a, if lim f1x2 = f1a2. The x :a assumption being that by writing f(a) and its limit, we understand that both are defined. We could replace the statement of continuity at a x :a lim f1x2 = f1a2 by the equivalent statement h :0 lim f1a + h2 = f1a2 If h is near zero, (that is, a + h is near a) the y-values are near f(a); so the limit is f(a). We leave the verification of their equivalence as an exercise. Now that we have a formal definition of continuity at a point, we can say that a function is continuous on a given interval if it is continuous at every point of the interval. Let us apply the definition to a function to determine where it is or is not continuous. Example 8 Determine those points, if any, at which the function defined by y = f1x2 = e is not continuous. Solution. For x 2, the rule is y = x 2 + 1, which is a polynomial and therefore continuous. Similarly, for x 7 2, the graph is a line and therefore continuous. The only questionable point is at x = 2. Let us check the three conditions that must be satisfied in order that the function be continuous at x = 2. 1. f122 = 1222 + 1 = 5. 2. lim f1x2 = 5. (Did you check the limit from both sides?) x:2 x2 + 1 3x - 1 if x 2 if x 7 2 3. f122 = lim f1x2. x:2 Thus, the three conditions are satisfied and the function is continuous at x = 2, and therefore continuous for all x. Figure 5, is a calculator sketch of the graph, it is not always clear from such sketches if the function has a hole or not at a given point. Example 9 Consider the function defined by the equation |x| if x Z 0 x f1x2 = d 3 if x = 0 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.3 Limits and Continuity 195 Figure 5: The Graph of x2 + 1 if x 2 y = f1x2 = e 3x - 1 if x 7 2 At what values, if any, is it discontinuous. Solution. We have that Recalling that |x| = e - x if x 6 0 x if x Ú 0 |x| - 1 if x 6 0 = e x 1 if x 7 0 We now may write f1x2 = |x| = c x - 1 if x 6 0 3 if x = 0 1 if x 7 0 The only point at which continuity is questionable is at x = 0 (why?). Since f102 = 3, the function is defined at x = 3. However, just to the left of x = 0, the y-values are - 1, and just to its right the y-values are + 1. Thus, there is no common y-value that both are near and so the limit does not exist. Condition 2 is violated, so the function is not continuous at x = 0. Note that the geometrical evidence for the discontinuity is the jump in the graph at x = 0 (see Figure 6). (0,3) +* Figure 6: f1x2 = c -1 3 1 if x 6 0 if x = 0 if x 7 0 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 196 Section 2.3 Limits and Continuity x 2 - 16 Determine what f(4) should be if for x Z 4 if the function defined by f1x2 = is x - 4 to be continuous. Solution. If x Z 4 f1x2 may be rewritten as f1x2 = x + 4, (Why?). It then follows that for continuity, we must satisfy condition(3), lim f1x2 = lim 1x + 42 = 8, therex :4 x :4 fore, for continuity, we need f142 = 8. Example 10 Removable Discontinuities Differentiability and Continuity Hole discontinuities are often called removable discontinuities, because defining or redefining the function at the hole so that its y-value equals the limit at the point eliminates the hole. All other types of discontinuities are called non-removable. For example, in the last example by defining y to be equal to 8 at x = 4, that is, f142 = 8. The function becomes continuous because the hole that would have appeared without this definition is removed. The following theorem is an immediate consequence of the alternate formulation and the definition of differentiability. THEOREM 9 At any point in its domain at which a function is differentiable, it is continuous. PROOF Because f(x) is differentiable at x = a, we know that f ¿1a2 = lim f(a + h) -f(a) exists. This implies that f(a) exists. We need only show that h:0 h lim f1a + h2 = f1a2 to prove that the function is continuous at x = a. We proceed with h:0 the following observation: f1a + h2 = a f1a + h2 - f1a2 h b h + f1a2 We take the limit of each side as h approaches zero to obtain, h:0 lim f1a + h2 = lim a c h:0 f1a + h2 - f1a2 d h + f1a2b h = lim c h:0 f1a + h2 - f1a2 d lim h + lim f1a2 h:0 h:0 h (Note the application of the product and sum rule for limits.) The first limit on the right is f ¿ 1a2 the second is 0, and the limit of a constant is the constant. Thus, we have lim f1a + h2 = f1a2 # 0 + f1a2 or, lim f1a + h2 = f1a2, h :0 h :0 and by the alternate formulation of continuity, we have that at any point at which the function is differentiable, the function is continuous. Theorem 9 gives a quick method of determining if a function is continuous. If you can determine its derivative, then it is continuous. The rules developed in the previous section of this chapter may prove useful in this connection, as the next example illustrates. Example 11 Prove the quartic function defined by f1x2 = 3x 4 + 2x 3 - 5x 2 + 8x + 12 is everywhere continuous. Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.3 Limits and Continuity 197 Solution. We have f ¿ 1x2 = 12x 3 + 6x 2 - 10x + 8. That is, the derivative exists for every value of x, therefore the function is everywhere continuous. However, you must be warned that the converse of the theorem is not true. That is, a function may be continuous but not differentiable. We have already considered examples of this in Section 3.1, where we observed that a sharp point on a curve (a function is continuous at a sharp point), the derivative does not exist, consider the next example. Example 12 x - 3 (a) At what points is f discontinuous? (b) One point is a removable x2 - 3x discontinuity. How would you define y at that point so as to remove the discontinuity? (c) At what points does f not have a derivative? (d) If you redefine f so as to remove the removable discontinuity, is the function differentiable there? Let f1x2 = Solution. (a) The denominator is x 2 - 3x = x1x - 32. This is equal to zero at x = 0 and x = 3. So the function is discontinuous at x = 0 and x = 3 because it is undefined at these two places. (b) No matter what you do as x approaches 0, f(x) increases without bound. For example at x = 0.0001, f10.0012 = 1000, as x gets closer to 0, the y-values grow even larger. Therefore, there is no way to define f(0) that will make the function continuous. However, the discontinuity at x = 3 is removable. If we simply divide out the factor 1x - 32 from numerator and denominator, we have f1x2 = 1/x, for x Z 3. Hence, lim f1x2 = 1/3. If we define f132 = 1/3, the function as redefined is continuous at x = 3. x :3 (c) The function is not differentiable at x = 0 or x = 3 because it is not continuous at these two points. (d) Once the value has been properly defined at x = 3, we have simply f1x2 = 1/x = x -1, which has derivative f ¿ 1x2 = - x -2 = - 1/x 2, except at x = 0 where f remains undefined and, therefore, neither continuous nor differentiable. Before you jump to any conclusions, let us say that it is not automatic that when a discontinuity is removed the function becomes differentiable at that point. In fact, examples to the contrary may be found in the exercises. As one further indication of how the limit definitions may be used, we give a formal proof of Rule 5, Section 2.2, the Sum Rule. We are given that f(x) and g(x) have f ¿ 1x2 and g ¿ 1x2 as their respective derivatives. We must show that s1x2 = f1x2 + g1x2 has derivative s ¿ 1x2 = f ¿ 1x2 + g ¿ 1x2. We use the definition of the derivative. It is useful to note that s1 2 = f1 2 + g1 2, therefore, 1. s1x + h2 = f1x + h2 + g1x + h2 g1x2 s1x2 = f1x2 + 2. s1x + h2 - s1x2 = f1x + h2 - f1x2 + g1x + h2 - g1x2 Therefore, s1x + h2 - s1x2 f1x + h2 - f1x2 + g1x + h2 - g1x2 = h h f1x + h2 - f1x2 g1x + h2 - g1x2 = + h h s1x + h2 - s1x2 4. s ¿ 1x2 = lim h:0 h 3. Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 198 Section 2.3 Limits and Continuity f1x + h2 - f1x2 g1x + h2 - g1x2 + lim = f ¿ 1x2 + g ¿ 1x2. h:0 h:0 h h Note that the proof really follows directly from the fact that the limit of a sum is equal to the sum of the limits. = lim We can compute limits with functions of several variables, consider the next example. Example 13 Given f1x, y2 = 2x 2y 3 determine f1x + h, y2 - f1x, y2 (a) lim , h :0 h f1x, y + k2 - f1x, y2 (b) lim . k :0 k Solution. f1x + h, y2 - f1x, y2 21x + h22y 3 - 2x 2y 3 (a) lim = lim = h:0 h:0 h h 2 3 3 2 2 3 2x y + 4xhy + 2h y - 2x y lim = h:0 h h14xy 3 + 2hy 32 4xhy 3 + 2h2y 3 = lim 14xy 3 + 2hy 32 = 4xy 3 lim = lim h:0 h:0 h:0 h h f1x, y + k2 - f1x, y2 2x 21y + k23 - 2x 2y 3 = lim (b) lim = k:0 k:0 k k 2x21y 3 + 3y 2k + 3yk2 + k32 - 22y 3 lim = k:0 k 2 3 2 2 2 2 2 3 2x y + 6x y k + 6x yk + 2x k - 2x2y 3 lim = k:0 k 6x2y 2k + 6x2yk2 + 2x 2k3 lim = k:0 k 2 2 2 2 2 k16x y + 6x yk + 2x k 2 = lim 16x 2y 2 + 6x 2yk + 2x 2k22 = 6x 2y 2 lim k :0 k :0 k Calculator Tips Notice in the above example that the limit requested in (a) looks almost like a derivative with respect to x. In fact, observe that if you look at the expression, y does not change, so if you treated y as a constant it looks like the derivative with respect to the variable x. Similarly, the limit in (b) looks almost like a derivative with respect to y. We will have more to say about this when we study differentiation of functions of two or more variables. The TI 89 calculator can compute limits directly, but some care needs to be taken. The syntax for having the calculator compute a limit is limit (expression, variable, point [, direction]). The [,direction] portion of the command is optional and is only used for one-sided limit. For example, to compute 3x - 1 x x :0 lim we enter limit 113¿x - 12/x, x, 02, Figure 7 indicates the input and result. The calculator gives the answer as ln(3), which we shall see is the natural logarithm of 3, which is approximately 1.09861 (press * Enter). Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.3 Limits and Continuity * ** 199 Figure 7: Computing Limits with the Calculator To determine a one-sided limit we need to add the direction option, - 1 for a leftx handed limit and + 1 for a right-handed limit. For example, to compute lim- , we prox:0 x ceed as in Figure 8 (note abs(x) is the way we enter x on the calculator). Figure 8: A One Sided Limit Using the Calculator EXERCISE SET 2.3 In Exercises 1 26, find the indicated limit. 1. lim 5 2. lim 9 x:7 x:0 8. lim 13x2 - 72 9. lim 14x3 + 9x2 + 52 x:1 x:2 3. x : -1 x:2 x:0 x:2 lim a 5x + 1 b 2 10. Let f1x2 = x:3 x2 - 4 , determine each of the following limits: (a) lim f1x2, x:2 x + 2 x : -2 4. lim 17.1 - 3.4x2 5. lim 1x + 5x + 32 6. lim 12x 2 - 3x + 52 7. x : -2 2 (b) lim f1x2, (c) lim f1x2. x2 - 9 11. Let f1x2 = , determine each of the following limits: (a) lim f1x2, x : -3 x - 3 (b) lim f1x2. x:3 lim 12x2 + 6x - 112 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 200 * ** Section 2.3 x Limits and Continuity 6x - 5 x2 36. If f1x2 = d 5x + 14 70 x:5 12. lim + 5x2 a2 + 5 13. lim a:2 a x:1 3 14. f1x2 = 7x2 + 2h - 3, determine (a) lim f1x2 (b) lim f1x2. 15. lim 16. lim x2 - 7x + 12 x:3 x - 3 2t2 + 1 t : 2 3t2 h:0 x:0 if if if if x:5 x 6 5 5 6 x 6 7 x 9 7 x 7 9 x:7 x : 77 Determine (a) lim- f1x2 (b) lim+ f1x2 (c) lim- f1x2 (d) lim + f1x2 (e) lim- f1x2 (f) lim+ f1x2 (g) at which x-values is f discontinuous, and classify x:9 x:9 the discontinuities. (h) graph the function. In Exercises 37 39, decide whether the function whose graph is shown is continuous. If it is not continuous, identify the x-values at which it is discontinuous and classify the discontinuity. + 1 1x - 3 17. lim x:9 x - 9 18. lim a:9 2a - 5 a - 5 x2 + 5x + 2 19. lim + 2x + 1 23 + h - 23 20. lim h:0 h 2 21. lim A 1 2 t + 1t B t:0 x : 0 2x2 22. lim 1t + 1221t - 227 1t + 222 1x + 2h22 - x 2 2h t:3 23. lim h:0 21 + x + h - 21 + x 24. lim h:0 h 25. f1x2 = x , determine (a) lim f1x2 (b) lim f1x2 (c) lim f1x2 26. f1x2 = |x| , determine (a) lim f1x2 (b) lim f1x2 (c) lim f1x2 x:4 x : -4 x : -a x (d) lim f1x2 where a 7 0. x:a x:1 x:0 x : -1 Figure 9: Ex. 37 Figure 10: Ex. 38 27. (a) Draw the graph of a function which is continuous at each point in its domain. (b) Draw the graph of a function which is continuous at every point in its domain but is not differentiable at x = 0. 28. Determine any function which is discontinuous at x = 1 and x = 5 but which has a derivative at every other point. 29. Where is the function f1x2 = 1/1x - 52 discontinuous. Why? 1 if x 0 30. (a) Graph the function defined by f1x2 = e 2 if x 7 0 (b) Where is the function discontinuous? Why? 31. Is the function defined by f1x2 = e x 1 (b) x = 0? 1 Figure 11: Ex. 39 37. Figure 9 38. Figure 10 39. Figure 11 In Exercises 40 46 decide whether or not the function is continuous. If it is not continuous, identify the points at which it is discontinuous. 40. f1x2 = x2 + 2x + 5 41. f1x2 = x2 - 3 x + 2 1 x if x Z 0 continuous at (a) x = 2? if x = 0 2 if x 6 1 4 if x = 1 3 - x if x 7 1 32. (a) Graph the function defined by f1x2 = c (b) Where is it discontinuous? 42. f1x2 = x2 + 43. f1x2 = e 44. f1x2 = e (c) What kind of discontinuities does it have? 33. Discuss the continuity of the function defined by f1x2 = 34. Determine (a) lim+ x:3 x if x 6 5 x2 if x Ú 5 0 if x 6 0 x if x Ú 0 x2 - 9 x - 3 x - 3 x - 3 x - a x - a (b) lim(c) lim(d) lim+ x:3 x - 3 x:a x - a x:a x - a x - 3 x3 x (b) limx:0 45. f1x2 = x 46. f1x2 = x - 5 47. Given f1x2 = c 1 if x 6 1 5 if x = 1 2 - x if x 7 1 35. Determine (a) lim+ x:0 x3 x Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.3 (a) Sketch the graph of the function. (b) Where is it discontinuous? (c) Redefine the function so as to remove the discontinuity. (d) Is the function as redefined differentiable at this point? In Exercises 48 52, each function is discontinuous at x = a. If the discontinuity may be removed, redefine the function so as to make it continuous at x = a and decide if the function as redefined is differentiable at x = a. 48. f1x2 = x2 - 4 x - 2 a = 2 a = 2 a = 2 Limits and Continuity * ** 201 64. Suppose f1x, y2 = 3x2 + 2y2 + 5x - 11y + 7, determine f1x + h, y2 - f1x, y2 f1x, y + k2 - f1x, y2 . (a) lim (b) lim h:0 k:0 h k 65. (a) Give an example of a function which is defined for all values of x except x = 1 and x = 2; and discontinuous at x = 1 and x = 2. (b) Give an example if both discontinuities are removable. 66. Can a jump discontinuity be removed? Explain! 67. Suppose f is continuous at x = a and that f1a2 7 0. Show, by means of a sketch, that f1x2 7 0 in some vicinity of x = a. If f is not continuous at x = a, show the conclusion need not be true. 68. Discuss the discontinuities of D1x2 = e 0 1 if x is irrational if x is rational 49. f1x2 = 1/1x - 22 50. f1x2 = c x2 if x 6 2 3 if x = 2 6 - x if x 7 2 if x 6 3 x3 51. f1x2 = c 15 if x = 3 5x + 12 if x 6 3 a = 3 x2 - 5x + 6 52. f1x2 = a = 2 x - 2 In Exercises 53 62, discuss (i) the continuity and; (ii) the differentiability of the given function. (iii) If a discontinuity is removable, redefine the function so as to make it continuous there. (iv) Is it now differentiable? x2 - 9 53. (a) f1x2 = x - 3 x + 3 (b) g1x2 = e 4 if x 2 if 2 6 x if x 7 4 if x Z 3 if x = 3 (c) h1x2 = x + 3. 69. (a) Suppose f is continuous on 1 x 3, with f112 = 7, and f132 = 12. Show that every y-value between 7 and 12 is assumed at least once. (Hint: draw a sketch) (b) More generally, if f is continuous on a x b, with f1a2 = M, and f1b2 = N, then each y-value between M and N is assumed at least once. (This is known as the Intermediate Value Theorem.) 70. (a) Using the previous exercise, prove that there is some x-value between 2 and 3 at which f1x2 = x2 - 1 equals 6. (b) Find this value of x. 71. Using Exercise 69, prove that x 5 + x - 5 = 0 has a root between x = 1 and x = 2. 72. On July 2, Sam weighed 120 lbs. On August 2, he weighed 130 lbs. (a) Prove that at some time between these two dates his weight was 125 lbs. (b) Could his weight ever have been 135 lbs between those dates? 73. At 7 AM the outdoor temperature was 68*F and at 11 AM the temperature was 76*F, is there a time between 7 and 11 AM when the temperature was (a) 73*F? (b) 71*F? (c) 65*F? -1 54. w1x2 = c 3 -2 55. r1x2 - 1/x2 4 56. (a) L1x2 = e x - 1 2 x - 1 2 if x 6 3 if x 7 3 if x 3 if x 7 3 (b) M1x2 = c x - 1 1 2 if x 6 3 if x = 3 if x 7 3 if x 6 3 if x Ú 3 (c) N1x2 = e (d) P1x2 = e x - 1 2 74. Mary begins her climb at the bottom of the mountain at 8 AM Monday morning and reaches the top of the mountain at 1 PM that afternoon. She camps there overnight and begins her descent on Tuesday at 8 AM and reaches the bottom of the mountain at 1 PM that afternoon. Is there a common time on each day when she is at the same elevation on the mountain? Explain. x b. Show that the function attains both a 75. Suppose f is continuous for a maximum and minimum value. That is, there are at least two numbers, c1 and c2, between a and b, such that, f1c12 = M and f1c22 = N, where f1x2 M, and f1x2 Ú N, for all x such that a x b. This is called the Extreme Value Theorem x 3. (a) What is the maximum value 76. Consider f1x2 = x2 - 1 on - 2 that this function attains? (b) Its minimum? (c) For which x-values does it attain the maximum and minimum? 77. A manufacturer estimates that his daily cost, in dollars, of producing x television sets is C1x2 = 2000 + 500x + 8000/x. He must manufacture at least 2 sets, and due to storage limitations, he cannot produce more than 250 sets. Prove, by using Exercise 75 that there is some number of sets which will minimize his cost. x b. 78. Suppose y = f1x2 is a non-constant continuous function on a Also suppose that f1a2 = f1b2. Why must the graph of this function have at least one turning point between a and b? 57. f1x2 = x2 - 2 x - 5 x x2 if x 6 1 x3 (b) g1x2 = e 2 if x Ú 1 x if x 6 2 if x Ú 2 58. (a) f1x2 = e 59. f1x2 = 1 + 1x x2 60. t1s2 - 21 - s2 61. d1x2 = e 24 - x 2 x + 2 if x 6 0 if x 7 0 if x 3 if 3 6 x 6 4 if x Ú 4 f1x + h, y2 - f1x, y2 h 2x + 1 62. g1x2 = c 3x - 7 4 63. Suppose f1x, y2 = 3x2y 2 determine (a) lim (b) lim f1x, y + k2 - f1x, y2 k k:0 h:0 . Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 202 * ** Section 2.4 Limits at Infinity, Infinite Limits and Asymptotes 82. (a) Show lim x2 = 4. How close to 2 should the x-values be chosen to guarantee that the y-values are (b) between 4 - 0.1 and 4 + 0.1? (c) between 4 - .01 and 4 + .01? (d) between 4 - H and 4 + P? where P is any positive number less than 4. 83. Show lim f1x2 = f1a2 is equivalent to lim f1a + h2 = f1a2. x:a h:0 x:2 79. Show that an alternate definition for the derivative at a is given by f ¿ (x) = lim f1x2 - f1a2 x - a x:a . Hint: let a = x + h in the definition given for the derivative. 80. Using the alternate definition of the derivative from the preceding exercise, compute determine the derivative of each of the following: (a) f1x2 = 2x2 (b) f1x2 = 4/x. 81. Suppose in Example 2 we want to select x-values so the y-values lie (a) between 6 - .01 and 6 + .01, (b) between 6 - .001 and 6 + .001, (c) between 6 - P and 6 + P, where P is any positive number less than 6, determine how close to 3 the x-values need to be? (Hint: let x = a + h.) 2.4 » » » » » » Limits at Infinity, Infinite Limits and Asymptotes Limits at Infinity Dominant Terms Horizontal Asymptotes Infinite Limits Vertical Asymptotes Calculator Tips Limits at Infinity In this section, we shall re-examine the notion of a rational function which was discussed in Section 1.7, using the limit as a means of defining a horizontal and vertical asymptote. In Section 1.7, we saw that much of the information about such graphs can be determined by examining two facets of its behavior; the behavior of the function as the x-values get very large (in absolute value), and considering its values near the zeros of both its numerator and denominator. We saw that this investigation coupled with the sign analysis of the function itself will reveal the essential shape of the graph. Recall that a rational function has the form r1x2 = p1x2/q1x2, where both p and q 5x4 - 2x 3 - 7 are polynomials. For example, r1x2 = is a rational function. 3x7 + 8x 2 - 11 We begin with the investigation of the behavior of a function for large values of x. For example, what happens to the values of the function whose equation is f1x2 = 3x 2 - 7 2x2 + 1 as x assumes arbitrarily large values? This question is reduced in mathematical shorthand to, What is lim f1x2? Here, to say that lim f1x2 = L, means that L is the unique x:q x:q number (if one exists) that the y-values get very near, when the x-values get arbitrarily large. If the values of f(x) become unbounded, we say that the limit is infinite. Let us examine 3x 2 - 7 x : q 2x 2 + 1 lim for large values of x. Consider Table 1 which examines what happens to the y-values as the x-values get large. As x becomes arbitrarily large, the y-values approach 1.5, therefore, Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.4 Limits at Infinity, Infinite Limits and Asymptotes 203 L = 1.5. We remark that for larger values of x, say 100,000 or higher, most calculators will give 1.5 as the answer! In fact, as far as your calculator is concerned, when it does the calculations for large values of x, the 7 in the numerator is Table 1: Examination of the Behavior as x Becomes Large x f(x) 100 1.4996 1000 1.499996 10000 1.49999996 insignificant compared to the term 3x 2. Similarly, in the denominator the 1 is negligible compared to the 2x 2 term. Hence the calculator is eventually just evaluating 3x2/2x 2 = 3/2 = 1.5. The following observation gives us a very simple method for computing limits, often called limits at infinity of rational functions (when they exist). When x is very large only the terms having the largest exponents in the numerator and denominator have any effect upon the limit. In Section 1.7, we referred to these terms as the dominant or leading terms. (Sometimes we shall say for x being infinite when we mean for x very large or arbitrarily large.) We illustrate this observation with several examples. Example 1 8x 3 - 2x + 3 . x : q 4x3 - 5x 2 + 1 Dominant Terms Evaluate lim Solution. The dominant term in the numerator is 8x3, and the dominant term in the 8x 3 = 2. denominator is 4x3. Thus, the problem reduces to computing lim x : q 4x 3 It is important to realize that only the dominant term matters when x becomes arbitrarily large. All other terms, no matter how large their coefficients may be, are insignificant in comparison to the dominant term when x is large. Example 2 Evaluate lim Solution. 5x2 - 300 - 3x3 + 2 x : q 2x5 . Ignoring all but the dominant terms, we have lim 5x2 - 300 5 3 x : q 2x - 3x + 2 = lim 5x 2 5 x : q 2x = lim 5 . x : q 2x 3 As x becomes arbitrarily large, the denominator gets very large, the entire fraction gets arbitrarily close to zero, and the given limit is zero. The observation made in the preceding example generalizes, and we have the following: THEOREM 1 lim k p x:q x = 0 where k is a constant and p is a positive constant. The above theorem, which is nothing more than a limit formulation of an observation we made in Section 1.7, provides an alternative method for computing limits as x becomes Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 204 Section 2.4 Limits at Infinity, Infinite Limits and Asymptotes large. To compute such limits, factor the dominant term from both the numerator and denominator and then apply Theorem 1. In Example 1, we may write 2 3 2 + 3b 8 - 2 + 2 8x - 2x + 3 x x x lim = lim = lim x : q 4x3 - 5x 2 + 1 x:q x:q 5 1 5 4 + + 3b x3 a 4 x x x 3 x3 a 8 - 3 x3 1 x3 Using Theorem 1, we see that the second and third terms in both the numerator and denominator approach zero and the limit is 8/4 = 2. Example 3 Evaluate lim Solution. 2x 3 - 5x + 3 . x:q 3x2 + 5 Looking only at the dominant terms, we have 2x 3 - 5x + 3 2x . = lim x:q x:q 3 3x2 + 5 lim As x becomes arbitrarily large, 2x/3 is itself unbounded. That is, it too becomes arbitrarily large number. Furthermore, it is always positive. We indicate this behavior by saying that the limit is plus infinity 1 + q 2. Thus, the required limit is + q . When we write + q for a limit, it means that the function is increasing without bound. In other words, the limit does not exist. When a limit does exist, it means that when x becomes arbitrarily large, the y-values tend to stabilize, or approach a finite equilibrium value. This equilibrium value is the limit. We shall call this the limit at infinity. Conversely, when the function has no limit as x takes on arbitrarily large values, it means that the function does not stabilize to an equilibrium value. Sometimes we find different behavior for a function as x decreases to - q . That is, when x takes on negative values which are, in absolute value, arbitrarily large. If f is such a function, and we want to examine such a limit, we write, lim f1x2. However, the procedure for finding such limits is x: -q exactly the same as when x is a large positive number. Example 4 Evaluate lim Solution. 3x 2 - x . x : - q 2x2 + 1 Keeping the dominant terms, we have, 3x2 - x = x : - q 2x 2 + 1 lim 3x 2 = 3/2. x : - q 2x 2 lim We summarize the results found in the above examples with the following theorem which generalizes Theorem 1 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.4 Limits at Infinity, Infinite Limits and Asymptotes 205 THEOREM 2 a nx n + a n - 1 x n - 1 bkx k + bk - 1x k - 1 an bn + Á + a0 = d0 + Á + b0 ;q if k = n if k 7 n if k 6 n x: ; q lim where n and k are positive integers. Note that the theorem tells us that the limit at infinity of the ratio of two polynomials is determined by the degrees of the polynomials. Remember, the degree of the polynomial is just the exponent of the dominant term. Thus, that if the numerator and denominator are of the same degree, then for (absolutely) large x the limit is the ratio of the coefficients of the dominant terms. If the denominator is of higher degree than the numerator, the limit is 0. Finally, if the degree of the numerator exceeds the degree of the denominator, then the limit is infinite. The sign depends upon the sign of the coefficients of the dominant terms, whether k - n is an odd or even integer, and whether x is positively large or is an absolutely large negative number. The eight possibilities are examined in Exercise 37. It can be shown that, under appropriate conditions, the theorem generalizes, and is true even when n and k are not integers. Example 5 Let P be the population of a certain species of fish. P is a function of time, t. When t is measured in months and P in hundreds of thousands, P is given by the equation P = f1t2 8t2 + 3 = 2 . Determine the equilibrium population of the fish, that is, the limiting popula4t + 9 tion of the fish. By the equilibrium or limiting population, we mean the population after 8t2 + 3 many months. That is, when t is arbitrarily large. We have, lim 2 = 2. Thus, the t : q 4t + 9 limiting or equilibrium population is two hundred thousand fish. Solution. For simplicity, in what follows we shall use the expression x gets large to mean x gets large in absolute value . Where there is a difference in behavior for positive and negative values of x, we shall point it out carefully. With that understanding, let us examine the behavior of a function as its x-values get large and see how its graph looks. In Example 1 we found that lim 8x 3 - 2x + 3 8x 3 - 2x + 3 Similarly, lim = 2. = 2. This means x : q 4x3 - 5x 2 + 1 x : - q 4x3 - 5x 2 + 1 that in a sketch of the function, when x is any very large value, the associated y-value must be essentially equal to the equilibrium value of 2. That is, the graph should begin to look like the line y = 2. We recall, from Section 1.7 that the horizontal line y = 2 is called a horizontal asymptote for the curve. In Figure 1 the horizontal asymptote y = 2 is drawn as a dashed line. Note that the graph is sketched only for large values of x. (The connection of the various component pieces is accomplished using the method of Section 1.7.) We can now redefine a horizontal asymptote using the limit. Horizontal Asymptotes Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 206 * ** Section 2.4 Limits at Infinity, Infinite Limits and Asymptotes y=2 Figure 1: 8x3 - 2x + 3 f1x2 = for Large x 4x3 - 5x2 + 1 DEFINITION 1 Suppose L and M are finite numbers. If lim f1x2 = L, then y = L is x:q a horizontal asymptote of f as x approaches q . If lim f1x2 = N, then y = N is a horix:-q zontal asymptote of f as x approaches - q . = 0. As x approaches - q we ob- 3x + 2 tain the same limit. (See Theorem 2.) This means that as x gets arbitrarily large the x-values approach 0. That is, y = 0 is the horizontal asymptote. Figure 2 gives a sketch of the graph of the function for absolutely large values of x, Figure 2a gives a sketch as x approaches - q , and Figure 2b gives a sketch as x approaches q ; note that in each case, the curve approaches the x-axis - y = 0, as the horizontal asymptote. To join these two components into one graph, you will need to determine the vertical asymptote and zeros, and proceed as in Section 1.6. This will be left to the exercises, where you will need a calculator to help you determine the approximate location of the vertical asymptote. A reasonable question at this time is, how do we know whether the curve is approaching the horizontal asymptote from above or below? Choosing a large positive (negative) number and then calculating the y-value at this number will indicate the direction. For example, in Figure 2, choose x = 100 and use your calculator to show that the corresponding y-value is about 2.0253. In Example 2 we found that lim x : q 2x 5 5x 2 - 3 Figure 2a: As x Approaches - q Figure 2b: As x Approaches q 5x2 - 3 2x5 - 3x + 2 Figure 2: The Behavior of f1x2 = Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.4 Limits at Infinity, Infinite Limits and Asymptotes * ** 207 This means that the curve is a little above the asymptote. Similarly, if we let x = - 100 in Figure 3 we find that x is about - 0.0000025, just below the asymptote y = 0. The next question we examine in this section is, What happens if we seek a limit for a function which, after substitution, has the form N/0 and N is not zero? To understand what happens, we need to consider several examples. We shall see that the limit is undefined (sometimes said to be infinite). By infinite or undefined, we shall mean that the value of the function is (in absolute value) arbitrarily large. 1 Consider, for example, lim . Clearly the function is undefined at x = 2. Let x :2 x - 2 us examine the behavior of the function just to the right of x = 2. Suppose we choose x to be 2 + .01. Substituting, we obtain y = 1/12 + .01 - 22 = 100. Similarly, if we choose a number even closer to 2, say 2 + .00001, we get y = 10,000 as the result. Thus, if we choose any number just to the right of 2, the result is an arbitrarily large positive number, which we write as + q . Thus, we write, lim 1 = + q. - 2 Infinite Limits x : 2+ x That is, if we approach 2 from the right, (indicated by 2+), then the y-values obtained are arbitrarily large positive numbers, indicated by + q . Actually, our reasoning goes something like this. The numerator is always 1. If x is slightly larger than 2, then x - 2 is a very small positive number. Now 1 divided by a very small number gives a very large result. Since both the numerator and denominator are positive, we say that the limit is + q . Similarly, if we choose a number just to the left of 2, say 1.99999, we obtain - 100,000 as our result. Hence, as we approach 2 just from the left, (indicated by 2-) we 1 = - q . Since the problem asks us what happens as x approaches 2, have, limx :2 x - 2 without regard to direction, we say the limit does not exist. If the direction was indicated, say as x approached 2+, then we would indicate the limit by + q . Visualizing what is happening is easy. Since x = 2 is not in the domain of the function, its graph may not cross the vertical line x = 2. (If it did, at the crossing, there would be a y-value defined for x = 2.) To the left of this line, the y-values are arbitrarily large and negative, while to the right of x = 2 the y-values are arbitrarily large and positive. The line x = 2, which is drawn as a dotted line in Figure 3 is recognized to be a vertical asymptote. Note that to the left of x = 2 the y-values are negative and (absolutely) large, while to its right the y-values are positively large. More formally, we may redefine a vertical asymptote as follows: Vertical Asymptotes DEFINITION 2 If lim+ f1x2 = ; q or if lim- f1x2 = ; q , then the function f has a x :a x :a vertical asymptote at x = a. Note that those x-values at which the denominator is zero (and the numerator is not zero) usually produce the vertical asymptotes. Therefore, particular attention must be paid to the zeros of the denominator. Example 6 Determine lim+ x :3 x . x - 3 Solution. Choose a value just to the right of 3, say 3.00001. We have, upon substitution, 3.00001/13.00001 - 32 = 300,001. This is a large positive number. Thus, we con- Figure 3: The Vertical Asymptote of 1 f1x2 = x - 2 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 208 Section 2.4 Limits at Infinity, Infinite Limits and Asymptotes lude that, lim+ x :3 x = + q . It follows from Definition 2 that x = 3 is a vertical asx - 3 mptote. We may generalize the above example to obtain the following theorem, in which we assume n is a positive integer. THEOREM 3 1 q n = + x : a 1x - a2 1 + q if n is an even integer limn = e q x : a 1x - a2 if n is an odd integer lim+ With appropriate modifications, the various methods used for evaluation of limits are valid for infinite limits and limits at infinity. However, it should be clear that for continuous functions, such as polynomials, one can calculate the limits intuitively as follows: Suppose f1x2 = P1x2/Q1x2, and P1a2 = N Z 0 while Q1a2 = 0. If we try to evaluate lim f1x2 by substitution we will have something of the form N/0. This really means that x :a in the limit, we are dividing a non-zero number very close to zero. This will give a result of very large magnitude. Hence we need only determine the sign of the quotient to know if the limit is + q or - q . Example 7 Determine: 0.5x (a) lim+ x :2 1 ; 1x - 223 (b) lim+ x :2 -x ; 1x - 223 (c) lim x : 4/3 2x - 7 1 - 3x + 425 . Figure 4: Illustrating an Incorrect Result Provided by the Calculator Solution. (a) This follows immediately from Theorem 3 and we conclude that the limit is + q . (b) As x approaches 2+, the numerator approaches - 2. The denominator is 1x - 223, which must be positive since x 7 2. Therefore, we have - 2 divided by a very small positive number. The quotient is negative and, hence, the limit is - q . (c) As x approaches 4/3, the numerator approaches - 13/3. The denominator is 1 - 3x + 425, which is positive since x 6 4/3. Therefore, we have a negative number divided by a very small positive number, and the quotient is negative. The limit is thus - q. Calculator Tips Using horizontal an vertical asymptotes, along with the zeros and sign of the function we were able to produce a rough sketch of the graph of the function. However, missing from this sketch are the determination of any turning points of the functions, its peaks and valleys. We shall examine how to include such points, when they exist, in the next chapter. The Calculator can be used to determine limits, even as x becomes infinite (note that the q symbol is inserted by pressing *CATALOG). However, you need to be careful, as sometimes the calculator may give the wrong answer. Consider the following limit, lim 11 + .07/x20.5x. Using the calculator, we enter the limit and press Enter, see Figure 4. Note the calculator gives zero as the incorrect answer, the correct answer is approximately, 1.4191. How could you check to see if the calculator gives the correct result? x: q Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.4 Limits at Infinity, Infinite Limits and Asymptotes * ** 209 EXERCISE SET 2.4 In Exercises 1 19 compute the indicated limit. 1. lim 2. lim 3x - 7x 2x - 3 5x2 + x + 3 6x2 + 7 3x4 + 7x - 11 2 2 x: q 26. f1x2 = x/1x2 - 12 27. f1x2 = x/11 + x22 28. f1x2 = 4x2/1x2 + 42 29. f1x2 = 4x2/1x2 - 42 30. f1x2 = 3x2/1x + 122 In Exercises 31-32 compute the indicated limit. 31. (a) lim 2x 2x 2 + 8 2x 2 - 1 x + 3 x: q x: q 3. lim 5x5 - 3x + 2 2x3 - 7 4. lim x: q x + 1 x: q (b) lim (b) lim 2x 2x 2 + 8 x: -q . (Hint: 2x2 = - x if x 6 0.) 5. (a) lim x2 x: q (b) lim x 2 x: -q 3 (c) lim x3 x: q (d) lim x 3 x: -q 3 6. (a) lim 7. lim 8. - 5x + x + 3 x: q 6x2 + 7 3 5x - 2x + 1 2x - 4 3x5 - 7x + 2 x - 7 2x8 + 3x5 + 7x 5x 14 4 3 (b) lim - 5x + x + 3 6x2 + 7 32. (a) lim x: q x: -q 2x 2 - 1 . x + 3 x: -q x: q In Exercises 33-36: (a) Find the x-intercept(s); (b) Find the vertical asymptotes; (c) Find the horizontal asymptotes. (d) Sketch the graph 33. f1x2 = 13x - 2212x + 322 x: -q lim 9. (a) lim x: q + 7x + 2 9 (b) lim 2x8 + 3x 5 + 7x 5x14 + 7x9 + 2 x: -q 10. lim 11. lim 12. lim 14x - 3221x - 12 - 121x + 122 x . + x2 12x - 123 - 1221x + 52 1 x3 (b) limx:0 x : q 12x x215x - 32 2x 34. f1x2 = . 2x 2 + 1 2x 35. f1x2 = . 2x 2 - 1 1x - 121x + 2221x - 322 . 36. f1x2 = 1x + 12212x - 323 37. Normally if we say a limit L exists, we mean that L is a finite number. In what follows, we allow L to stand for either + q or - q . Let anxn + a n - 1xn - 1 + Á + a 0 lim = L If the integer n - k 7 0, show q x : bkx k + bk - 1x k - 1 + Á + b0 that L is as given in Table 2 for each of the eight cases indicated: x: q 1 x : q 13x 13. (a) lim+ x:0 1 x3 . 1 1x - 322 x 1x - 22 2 14. (a) lim+ x:3 1 1x - 322 x 1x - 22 - 32 2 (b) limx:3 . Table 2 . 15. (a) lim+ x:2 (b) limx:2 n - k even even even even odd odd odd odd Sign of a nbk positive positive negative negative positive positive negative negative x approaches +q -q +q -q +q -q +q -q L +q +q -q -q +q -q -q +q 16. x : 3 - 1x lim 1x + 22 x2 - 11x + 30 x2 - 11x + 30 17. (a) lim+ (b) lim + x:5 x : -5 x + 5 x + 5 1 1 . 18. (a) lim+ (b) lim + x : 5 1x + 523 x : - 5 1x + 523 19. x:2 lim- x - 2 x2 - 4 In Exercises 20 30: (a) Find the x-intercept(s); (b) Find the vertical asymptotes; (c) Find the horizontal asymptotes. (d) Sketch the graph. 20. f1x2 = 2x/1x - 12 21. f1x2 = 12 - x2/1x + 12 22. f1x2 = 12x - 32/1x - 12 23. f1x2 = 11 + 2x2/1x - 32 24. f1x2 = 12x + 42/14 - x2 25. f1x2 = x/12x - 32 38. (a) Determine the x-intercepts, (b) the vertical asymptotes, (c) the horizontal asymptotes and (d) Sketch the graph of the function whose equation is 8x3 - 2x + 3 f1x2 = . 4x3 - 5x2 + 1 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 210 * ** Section 2.5 Derivative Rules 2 yields f1x2 = 2x - 4 + 14x2 - 44x + 5 2x3 - 7x + 9 39. (a) Determine the x-intercepts, (b) the vertical asymptotes, (c) the horizontal asymptotes and (d) Sketch the graph of the function whose equation is 5x2 - 3 f1x2 = . You ll need to use your calculator to determine the 2x5 - 3x + 2 vertical asymptotes. Exercises 40 44 deal with oblique asymptotes. The line y = mx + b is called a slant or oblique linear asymptote of the function whose equation is y = f1x2 if either x: q This implies that the oblique asymptote is y = 2x - 4. Using long division, find the oblique asymptotes in Exercises 41 42. 41. f1x2 = 12x2 + x - 62/1x - 12 42. f1x2 = 13x4 - 2x3 + 2x2 - 3x + 22/1x 3 - 12 43. Sketch the graph of the function in Exercise 41. 44. Sketch the graph of the function in Exercise 42. 45. The calculator gave the wrong limit for lim 11 + .07/x25x, how could you obtain an approximately accurate result? x: q lim 5f1x2 - 1mx + b26 = 0 or x: -q lim 5f1x2 - 1mx + b26 = 0 40. Show that the given linear equation is a slant asymptote of y = f1x2 for (a) f1x2 = x2/1x + 12, y = x - 1; (b) f1x2 = 1x3 + 2x + 32/ 1x2 + 5x - 32, y = x - 5 Long division of polynomials may be used to determine the oblique asymptotes of a rational function in which the degree of the numerator exceeds the degree of the denominator by one. For example, performing long division on f1x2 = 4x4 - 8x 3 + 2x + 5 2x3 - 7x + 9 46. Find the x-values at which the graph in Exercise 33 crosses its horizontal asymptote. 47. Find the x-values at which the graph in Exercise 36 crosses its horizontal asymptote. 2.5 » » » Derivative Rules 2 The Product Rule The Quotient Rule Calculator Tips The Product Rule In Section 2.2, we began to develop the rules for finding the derivatives of simple functions. We can find the derivatives of powers of x and constant multiples of these powers. We can also handle sums and differences of functions whose derivatives we know. We shall now consider finding the derivatives of products and quotients of functions. Unlike the derivative of a sum, the derivative of a product is not the product of the derivatives nor is the derivative of a quotient the quotient of the derivatives. For example, consider y = x2. We know that y ¿ = 2x. However, we could write y = x2 = xx. If the derivative of the product were the product of the derivatives, then y ¿ = 1 # 1 = 1, which is clearly wrong. Therefore, let us start with the product rule. Assume that we have two differentiable functions, F(x) and S(x). (F for first , S for second .) RULE 6 THE PRODUCT RULE d 1F1x2S1x22 = F1x2S ¿ 1x2 + S1x2F ¿ 1x2 dx In words, the derivative of a product is the first (function) times the derivative of the second (function) plus the second times the derivative of the first. (The proof of the product rule can be found at the end of this section.) Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.5 Derivative Rules 2 211 Example 1 Find f ¿ 1x2 if f1x2 = x512x 3 + 12 Solution. We do not have to use the product rule to find f ¿ 1x2, all we need do is distribute the x5 term and use the rules developed in Section 2.2. As an exercise, you should do it that way and compare your solution to the one obtained using the product rule. f ¿ 1 x2 = x 5 d d 12x 3 + 12 + 12x 3 + 12 1x 52 = x5[6x 2] + 12x3 + 12[5x 4] dx dx (We enclose the factor that was differentiated in square brackets [...], for easy detection.) Now simplifying, we have f ¿ 1x2 = 6x 7 + 10x 7 + 5x 4 = 16x7 + 5x 4. As we already mentioned, Example 1 could have been done without using the product rule. However, there are products which can not be done any other way, as we shall soon see. Beginners often confuse the product rule with the constant multiplier rule. While it is true that you may apply the product rule in finding the derivative of an expression like 3x5, it should not be done. By the constant multiplier rule and power rule, its derivative is simply 5 # 3x 4 = 15x 4. The product rule is meant to be applied when both factors involve the variable. If one of the factors is a constant, use the constant multiplier rule! Let us now consider the quotient rule. Again, assume that we have two differentiable functions, N(x) and D(x) The Quotient Rule RULE 7 THE QUOTIENT RULE D1x2N ¿ 1x2 - N1x2D ¿ 1x2 d N1x2 a b = dx D1x2 [D1x2]2 In words, the quotient rule says the derivative of the quotient is the bottom (denominator) times the derivative of the top (numerator) minus the top times the derivative of the bottom divided by the bottom squared. (The proof of the quotient rule is left for you as an exercise.) Example 2 Find f ¿ 1x2 if f1x2 = Solution. 3x - 2 4x - 5 Applying the quotient rule, we have, 14x - 52[3] - 13x - 22[4] 14x - 52 2 f ¿ 1x2 = = 12x - 15 - 12x + 8 -7 = 2 14x - 52 14x - 522 (Again note that the square brackets [ ] indicate the expression that was differentiated.) Note that the sign in front of the 8 is + . To avoid any distribution errors, observe that - 13x - 22142 = - 413x - 22. It is absolutely essential that each term be enclosed within its own parenthesis. Not doing so is incorrect, and will often result in algebraic errors. Example 3 Find f ¿ 1x2 if f1x2 = x2 . 4x 2 + 1 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 212 Section 2.5 Derivative Rules 2 Solution. Applying the quotient rule, we have, 14x2 + 12[2x] - 1x 22[8x] 14x2 + 122 = 8x 3 + 2x - 8x 3 2x = 2 2 2 14x + 12 14x + 122 f ¿ 1x2 = (Again note that the square brackets [ ] indicate the expression that was differentiated.) Example 4 1x2 - 621x - 32 Find the equation of the line tangent to y = at the point when x = 4. 2 1x + 1 Solution. First, substitution of x = 4, yields y = 2. Now take the derivative to find the slope of the tangent line. Notice that the numerator of the fraction is itself a product. First use the quotient rule, 52 1x + 16 y¿ = d d 51x 2 - 621x - 326 - 1x 2 - 621x - 32 52 1x + 16 dx dx 12 1x + 122 Now, the first term in the numerator has a factor that is the derivative of a product. For this we need the product rule. In addition, in order to take the derivative of +x, in the second -1/2 . numerator term, remember that 1x = x1/2. Thus, 1 1x2 ¿ = 1 2x Thus, y ¿ 1x2 = 12 1x + 1251x2 - 62[1] + 1x - 32[2x]6 - 51x 2 - 621x - 326[x -1/2] 12 1x + 122 We would ordinarily simplify this expression algebraically, but here we only need the value of the derivative at x = 4. Since it is easier to do arithmetic than algebra, let us sub-1 stitute x = 4 directly. Note that 24 = 2 and 4 *2 = 1/ 24 = 1*2. mtan142 = 5152[1102 + 112182] - A 1*2 B 11021126/1522 = 85/25 = 17/5 Hence, the equation of the tangent line is y - 2 = 117/521x - 42 or 5y = 17x - 58. d -2 -2 1x 2 = - 2x -3 = 3 however, dx x we have not yet proven the power rule when the power is negative. We can use the quotient rule along with the power rule for positive powers to establish its validity. In this particular case, We know, by direct application of the power rule, that d -2 x2[0] - 1[2x] d 1 - 2x -2 1x 2 = a 2b = = = - 2x -3 = 3 4 4 dx dx x x x x More generally, let N be any positive integer, then we shall show, that d -N 1x 2 = - Nx -N - 1 dx That is, we bring down the power and decrease the power by one. Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.5 Derivative Rules 2 213 x N[0] - 1[Nx N - 1] d -N d 1 1x 2 = a Nb = dx dx x 1xN22 = - NxN - 1 = - Nx N - 1 - 2N = - Nx -N - 1 x 2N When the power rule was applied, it was on the denominator term, xN, which has a posibM tive exponent. In the next to the last step we used the law of exponents N = bM - N. b Thus, we have proven that the power rule is valid for any integer. We have yet to show that it is true for all rational and irrational powers. Example 5 x4 - 2x 5 + 7 , find f ¿ 1x2. If f1x2 = 3x2 Solution. We could apply the quotient rule, but it is not necessary. x4 2x 5 7 2 3 7 -2 2 + = 1 3x - 3x + 3x 2 2 3x 3x 3x 2 14 -3 3 x Observe that we may write f1x2 = so 2 f ¿ 1 x2 = 2 3 x - 2x - Alternately, by combining fractions, we may write f ¿ 1x2 = 2x 4 - 6x 5 - 14 . 3x3 The quotient rule is usually applied when the denominator is a sum (or difference) of two or more terms. If the denominator consists of one term, as in the above example, it is often more efficient to rewrite the expression and apply the other rules. We conclude this section with a derivation of the product rule. The proof uses the definition along with a very clever trick of adding and subtracting a function that gives us recognizable difference quotients. The same kind of trick may be used to prove the quotient rule, which we leave to the exercises. Consider P1x2 = F1x2S1x2, that is, P12 = F12S12. Then: Step 1. P1x + h2 = F1x + h2S1x + h2 P1x2 = F1x2S1x2 Step 2. P1x + h2 - P1x2 = F1x + h2S1x + h2 - F1x2S1x2 Step 3. P1x + h2 - P1x2 F1x + h2S1x + h2 - F1x2S1x2 = h h We do not yet recognize the difference quotient. However, if we add zero written as the terms - F1x + h2S1x2 + F1x + h2S1x2 then we may rewrite the numerator of the difference quotient as Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 214 * ** Section 2.5 Derivative Rules 2 F1x + h2S1x + h2 - F1x + h2S1x2 + F1x + h2S1x2 - F1x2S1x2. Factoring we have the numerator as, F1x + h25S1x + h2 - S1x26 + S1x25F1x + h2 - F1x26 Dividing the numerator by h, we have, F1x + h2 a S1x + h2 - S1x2 h b + S1x2a F1x + h2 - F1x2 b h If we now allow h to approach zero, then the difference quotients enclosed in the large parentheses become S ¿ 1x2 and F ¿ 1x2 respectively. Since F is continuous, F1x + h2 approaches F(x) as h approaches 0, thus, Step 4. P ¿ 1x2 = F1x2S ¿ 1x2 + S1x2F ¿ 1x2. * It certainly was a clever trick to add on and subtract off the term F1x + h2S1x2. It is surprising how often mathematical proofs hinge upon just such tricks. However, we shall see another way of deriving this result when we study logarithmic differentiation. It may happen that your answer for a derivative appears different from the answer given in the text. Sometimes it may take some clever algebraic manipulations to show the answers are equivalent. There is a way around this using your calculator, using the solve command. Suppose text answer is the answer given by the text, and your answer is the one you obtained. We assume the independent variable is x. Enter solve1text answer = your answer, x2 If, after pressing Enter, the calculator responds with TRUE then you know the answers are equivalent, if it responds FALSE then they are not. For example, suppose your an1x + 12-2 1 s answer is swer is and the text , then we need only enter x - 1 1x + 123/21x - 121/2 Ax + 1 solve11x + 12¿1-22/*11x - 12/1x + 122 = 1/1x + 12¿13/22*1x - 12¿11/22, x2 If this is entered properly the TI 89 responds with TRUE. Calculator Tips EXERCISE SET 2.5 1. f1x2 = 17x3213x4 - 92. Find f ¿ 1x2 in two different ways. 2. f1x2 = 15x + 2212x + 92. Find f ¿ 1x2 in two different ways. 3. f1x2 = 1x2 + 221x4 - 72. Find f ¿ 1x2 in two different ways. 4. f1x2 = x612x4 - 3x + 42. Find f ¿ 1x2 in two different ways. 5. Find f ¿ 1x2 if f1x2 = 6. Find y ¿ if y = dy x2 2 6 4 8. Find y ¿ if y = 9. Find 10. Find 4x3 + 5 2x2 + 7 . d 3 a b by (a) using the power rule and; (b) using the quotient rule. dx x2 2x + 3 3x - 2 d x5 - 2x 2 + 3 a b by: (a) writing it as a sum of powers and; 4x dx (b) using the quotient rule. d 4x6 + 3x 3 - 8x a b by: (a) writing it as a sum of powers and; dx 6x5 (b) using the quotient rule. x + 1 3 - 5x 7. Find if y = . dx x + 3 11. Find Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 28. 27.8 x2 + 3 The Chain Rule * ** 215 line crosses the x-axis.x32 (why?). f1x2 = 13. . and T are each differentiable functions. what do you think y ¿ is. Repeat Exercise 16. Wang. x3 + 12 15.3x + 22 x4 + 1 4 x + 3x2 . This two stage process is typical of a composite function. x = 2. it is an expression raised to a power. Now differentiate both sides and solve for a b. Hint: Use the alternate definition of the derivative as given in Exercise 37 in Section 3. Economics. and 5 and f102 = 30. and then raise the result to the 34th power. (b) f1x2 = 1x + 12/x 1 *2. determine the equation of the tangent line to the given curve at the indicated x-value. Show that in words it says: to multiply two factors by the derivative of the third factor. we are severely limited in the types of functions that we can differentiate. and x = 3. Do you see any relationship between these roots and the intercept of the tangent line? [See Exercise 24 below.x221x . D1x2 dx D1x2 22. and April Allen Materowski. where F. Hint: let g1x2 = 1/f1x2 and find g ¿ 1x2 using the quotient rule.4x = 0. Find all points at which the derivative of y = 1x2 . has the third root as its x-intercept. (a) What happens to the product rule if F1x2 = S1x2? (b) Derive a product rule for y = F1x2S1x2T1x2.x12 1x . 17. 12. and do this in all three possible ways and add the results. y02. Consider lim . Observe that N1x2 = N1x2 d N1x2 D1x2a b . Copyright © 2007 by Pearson Education. Walter O.4x in factored form and find the roots of x3 . We can only handle functions like x N or sums. you first find 2x 3 + 5.8 x4 . to compute F(x). use them to get an idea of the shape of the graph of y = f1x2 between x = 0 and x = 4. x3. Use the results of the previous exercise to compute each of the following limx3 . Find those points at which the tangent line to y = f1x2 is horizontal for: (a) f1x2 = x/1x2 + 12. 20. (b) Verify the result obtained in Exercise 24 for the tangent lines drawn to the curve at x = 3 and x = .3. for: (a) f1x2 = x 1 *2/1x + 12. 16. (a) Show that lim f1x2 = f ¿ 1a2 g ¿ 1a2 18. f1x2 = x2 a b. 21.] 19.2x2 + 321x3 + 12. (c) Write the expression x3 . That is. x = 2 x = 2 x = 1 14. as yet. Here is another way to derive the quotient rule.4x. how do we find the derivative of F1x2 = 12x3 + 5234? It is important that you understand why we cannot. that is. x2. the point where the x : a g1x2 provided g ¿ (a) Z 0.3 1x + 121x2 .1 29. Prove the quotient rule using the same kind of trick that was used for the product rule. where N is a positive integer? 23. (e) Using (c). 26. (d) Rewrite (b) if F = S = T. . (c) Sketch short segments of the tangent lines found in (b).3x . (b) Show that if the derivatives of both f and g are continuous functions then the above result may be written as x : a g1x2 lim f1x2 = lim f ¿ 1x2 x : a g ¿ 1x2 This result is a special case of a theorem known as L Hôpital s rule. its: (a) lim (b) lim x:2 x . and Finance. . may be written as y = a1x . Show that the tangent line to the curve at 1x0. remember that A2 = A # A. [Hint: To find the derivative. x2 .1 . y02 is midway between two successive roots. where f and g are differx : a g1x2 entiable functions with f1a2 = g1a2 = 0. S. x to a power.] 24. Published by Pearson Learning Solutions. x = 1. 25. Thus. Applied Calculus for Business. find the derivative of this expression. For the function y = x/1x2 + 12: (a) Find the derivative.Section 2. A cubic equation of the form y = ax3 + bx 2 + cx + d having three distinct roots (x-intercepts) x1.1. where F. if y = [F1x2]N. This exercise illustrates how derivatives may be used to compute limits in one f1x2 very special and important case.6 The Chain Rule » » » The Chain Rule The General Power Rule Calculator Tips Even with the product and quotient rules. Gordon. (a) Find the equation of the line tangent to this curve at x = 1. 2. by Warren B. S. T and W are each differentiable functions. where 1x0. (b) Find the equations of the tangent lines at x = 0. f1x2 = x = 1.422 is equal to zero. It is not simply of the form.2 x:1 x . where g1x2 = 2x3 + 5. However. products and quotients of such functions.] (c) Guess the product rule for y = F1x2S1x2T1x2W1x2.1. Explain where the fact that a differentiable function must be continuous was assumed in the proof of the product rule for derivatives. Let y = x3 . (b)Find the x-intercept of the tangent line. f1x2 = 1x4 . 1. [Hint: Consider 5F1x2S1x26T1x2 and apply the product rule twice. Inc. F(x) can be thought of as [g1x2]34. Explain the relationship between the answers in parts (a) and (b) of Exercises 16 and 17.6 In Exercises 12 15. (a) Find the cubic y = f1x2 with roots . (b) f1x2 = 1x and g1x2 = 3x2 + 5. f1g1x22 = 1x3 + 11223.) x+ g (x) + f(g(x)) + g f Figure 2: Composite Function Box Diagram for f(g(x)) We note that in order to obtain the final output. The boxlike figure indicates the rule or correspondence used in determining y. Thus.216 * ** Section 2. Wang. Economics. and April Allen Materowski. (a) f12 = 1223 and g12 = 123 + 11. We note that if the boxes were switched. Thus f1g1x22 = 1g1x2223. Copyright © 2007 by Pearson Education. if x is in the domain. and Finance. Solution. g1x2 = 2x3 + 5. the corresponding range value is y = f1x2. we fill the parentheses in g() with f(x). But. y = g1x2 and we represent the process by box-like figures which are attached in series as in Figure 2. which is called a composite function. Published by Pearson Learning Solutions. Example 1 Find the composite functions f(g(x)) and g(f(x)) if: (a) f1x2 = x23 and g1x2 = x 3 + 11. To find f(g(x)). In our example. Walter O. we know that g1x2 = x 3 + 11. All such x will give us the domain of the combined function y = f1g1x22. we fill the parentheses in f() with g(x). x+ + f(x) = y Figure 1: Function Diagram y = f1x2 Thus. Suppose now that we have a second function. It is easy to determine the composite function once we are given the component functions. Inc. . g1f1x22 = 1f1x223 + 11 = 1x2323 + 11 = x69 + 11. namely y = g1f1x22.6 The Chain Rule Before we give the rule which will enable us to differentiate such functions. Notice the answers are not the same. we review the notion of a composite function as discussed in Section 1. we would obtain a different composite function. to find g(f(x)). (For example. and g(x) is in the domain of f. Applied Calculus for Business. Similarly. Gordon. x goes into the g-box and 2x 3 + 5 comes out. the f box first followed by the g box. For example. and so. x must be such that it is in the domain of g. This then goes into the f-box and comes out 12x3 + 5234.2 where we saw that it is sometimes convenient to represent a function by a diagram as given in Figure 1. we might have f1x2 = x34. by Warren B. as the next example illustrates. Inc. but this will not be a problem for us. Even when using the power function. (a) D(x) looks like something to the 54th power so we call the expression within the parentheses. In fact.6 The Chain Rule 217 (b) In this case. this kind of decomposition is not a natural one. For example. that is. Wang. We assume that all component functions are differentiable over common domains. that we want to be able to differentiate are composite functions. . then H1x2 = 1g1x2234. can we decompose it into its component functions? The answer is yes. it is more useful to think of this one as T1x2 = f1g1x22 + 5h1g1x22. but it could be done. Economics. Copyright © 2007 by Pearson Education.2. We will indicate why this is so shortly. There are other decompositions that one could choose. and y = H1x2 = f1u2.226 Solution. Walter O. The rule is called the chain rule. If we have a function H(x) that we want to decompose as f(g(x)). As a guideline. Now we can return to the original question posed at the opening of this section. you should look for the natural decomposition. but the decomposition may not be unique. that is g1x2 = 34x3 + 9x + 11 and f1x2 = x54. by Warren B. There is almost always a natural or suggested decomposition that will be obvious. Each term involves within the parentheses. consider the function H1x2 = 1x2 + 228 we could write this as 21x 2 + 2216. the natural decomposition is f1x2 = x 8 and g1x2 = x2 + 2. f(x) is the power function. f1x2 = x 2 and h1x2 = x 6 The choice of decomposition soon becomes very natural. you will soon see that for purposes of finding derivatives. Think of H1x2 = f1g1x22. we can write H(x) as the composition of two very simple functions. the expression x 3 . then we could choose f1x2 = *x and g1x2 = 1x 2 + 2216. (b) Inspection of T(x) shows that it is the sum of two terms. That is. Similarly. f12 = 212 and g12 = 3122 + 5.2. will be the power function x N. the only good choice for f(x). For example. Now. and April Allen Materowski. Hence. the outer function. and Finance. as in (b) above but they are not very helpful. We shall give a rule that will enable us to determine the derivative of the composite function by thinking of it in terms of its components. f1g1x22 = 21g1x2) = 213x2 + 112. where g1x2 = x2 + 9. choose f(x) to be a simple function and g(x) to be the messy or inner part. We shall sometimes refer to g(x) as the core or inner function. where g1x2 = x3 . Since we can handle the derivatives of sums of functions independently. for now. given the function H1x2 = 1x2 + 9234. Functions defined by expressions such as 12x 3 + 5234. so let f12 = 1234. and T1x2 = f1g1x22. since the function H(x) was some expression to a power. Now. Of course.222 + 51x 3 . g(x). This is not the natural thing to do. D1x2 = f1g1x22. then H ¿ 1x2 = f ¿ 1g1x22g ¿ (x) Sometimes the chain rule is written a little differently. the expression being g(x) and the power function being f(x). f1x2 = x 34. RULE 8 THE CHAIN RULE Given H1x2 = f1g1x22. Then we may write dy dy du dH = = dx dx du dx The Chain Rule Applied Calculus for Business.2. g1f1x22 = 31f1x222 + 5 = 31*x22 + 5 = 3x + 5 Now we can pose the reverse question.Section 2. If we let g1x2 = x3 . Let u = g1x2. try the following. we let f1x2 = x 2 + 5x6. Example 2 Rewrite each function as the composition of two functions. (a) D1x2 = 134x 3 + 9x + 11254 (b) T1x2 = 1x3 . Gordon. then T1x2 = [g1x2]2 + 5[g1x2]6. Published by Pearson Learning Solutions. It was very natural to choose this as the decomposition. given a composite function. Thus. 3x . At this point.12-2132. Example 3 Find H ¿ 1x2 if H1x2 = 13x2 + 7223 Solution. . let us rewrite it as y = 413x . when x = 1.122 Applied Calculus for Business.1. that is bring down the power and decrease the power by one and then multiply this expression by the derivative of the inner part. This is known as the General Power Rule.12. Note that we never really need to do the decomposition in this detail. Inc. We must multiply by a correction factor. Gordon. g ¿ 1x2 = 6x. we have: The General Power Rule RULE 8A THE GENERAL POWER RULE If H1x2 = 1g1x22N then H ¿ 1x2 = N1g1x22N .12-2 = .1g ¿ 1x2 or more simply as du d N u = NuN . Walter O. the derivative of the core function. since f is a composite function. you first apply the usual rule on f. f ¿ 1x2 = 23x22.12-1 and use the chain rule. Since f ¿ 1x2 = Nx N . and Finance. Thus.1. Published by Pearson Learning Solutions.6 The Chain Rule The chain rule states the following: To find the derivative of a composite function. Thus. this is not all there is to it. Therefore. We may write H1x2 = f1g1x22 where f1x2 = x23 and g1x2 = 3x2 + 7. which is g ¿ 1x2. We illustrate the chain rule (general power rule) with several examples. the slope of the tangent line.218 Section 2. To find the slope we must find the derivative. we find y = 2 (verify!).2 = m1x . the most important application of the chain rule is to functions of the form [g1x2]N. Now. thus we immediately have. we could think of the function as a quotient and use the quotient rule. It is derived by simply letting f1x2 = xN. Economics. However. we have H ¿ 1x2 = f ¿ 1g1x22 g ¿ 1x2 = 2313x 2 + 7222 6x = 138x13x 2 + 7222. y = . Wang. the general power rule states that we use the simple rule on the power.12 13x . f ¿ 1g1x22 = 231g1x2222 = 2313x 2 + 7222. we need to find m.1213x .1 dx dx where u is a differentiable function of x. In addition. Of course. Instead. Copyright © 2007 by Pearson Education. # # # Example 4 Find the equation of the line tangent to y = 4 at the point where x = 1. H ¿ 1x2 = 2313x 2 + 7222 6x = 138x13x 2 + 7222. The required equation is then y . and April Allen Materowski. Basically. where the factor 3 is the derivative of the inner or core function 3x .1 Solution. By the general power rule y ¿ = 1 . by Warren B.4213x . Economics. With practice. We have concluded the calculus portion of the problem. Wang. that is.2 = . mtan112 = . combining like terms within the braces. # # # The algebra probably takes as least as long as the differentiation. we may use it along with some of the other rules to differentiate more complicated functions. by Warren B. Published by Pearson Learning Solutions. Inc. yields f ¿ 1x2 = x31x 2 + 3x + 52455x12x + 32 + 41x 2 + 3x + 526. or to determine the equation of the tangent line at a particular point. B x2 + 8 We write f1x2 = a x 2 + 3 1/2 b . The more complicated part could well be performing the algebraic simplifications. x2 + 8 Applied Calculus for Business. reduces the expression to f ¿ 1x2 = 2x31x2 + 3x + 524114x2 + 27x + 202. For example. you will be doing the differentiation part of the problem very quickly. In such cases. Factoring the common factor x31x2 + 3x + 524. Walter O. as you will see in the following chapters. The application of the various differentiation rules is usually the easier part. Let us do that now. you might need to work with the derivative function. by the chain rule. Example 5 If f1x2 = x41x 2 + 3x + 525. and factoring out a 2. x2 + 3 . Distributing the 5x. therefore.12/22 = . Of course. we get f ¿ 1x2 = x4 51x 2 + 3x + 524 12x + 32 + 1x 2 + 3x + 525 4x 3. However. and April Allen Materowski.12 or y = . we have. To show you how easy it is to find the derivative of a complicated expression if it is to be evaluated at a particular point. find f ¿ 1x2.Section 2. All that has to be done is to use the appropriate rules and then substitute a number for the independent variable immediately after the differentiation. then the algebra is not necessary. asking us to compute the derivative.3x + 5 Now that we have the chain rule.6 The Chain Rule 219 At x = 1.3. Thus. we have Applying the product rule with F1x2 = x4 and S1x2 = 1x2 + 3x + 525. you would have to simplify f ¿ 1x2. and Finance. You might even need to know the derivative of the derivative function. find f ¿ 112.31x . We illustrate both approaches in the next few examples. Example 6 If f1x2 = Solution. Gordon. you might want to know where the derivative is equal to zero. Solution. Copyright © 2007 by Pearson Education. the equation of the tangent line is y . d 2 d 1x + 3x + 525 + 1x 2 + 3x + 525 1x 42 dx dx f ¿ 1x2 = x4 Applying the chain rule (general power rule) to the derivative of S(x). in many problems. consider the next example. . if the problem is numerical. 5y 4 . Wang. f ¿ 112 + 1 4 -1/2 9122 . any time we take the derivative of a variable that is different from x. but the derivative of y5 is not simply 5y 4. then the chain rule must be applied whenever g1x2 Z x. if the differentiation were with respect to any other variable. if we want to determine 1y 2. as usual.1x + 32[2x] a 2 a b b 2 x + 8 1x2 + 822 We may now substitute x = 1 to find. Gordon. if we want to differentiate f(g(x)) with respect to x.6 The Chain Rule f ¿ 1x2 = Using the quotient rule. there is no question that the chain rule is not applied because we are differentiatdx ing xN with respect to x. f ¿ 1x2 = 1 x 2 + 3 -1/2 # d x 2 + 3 a a b b dx x2 + 8 2 x2 + 8 2 2 1 x2 + 3 -1/2 1x + 82[2x] . However. 2 2 A 4 81 2 2 81 54 2 9 9 We could have simplified before we did the evaluation.4122 1 9 10 1 3 10 5 a b a b = a b = = . Copyright © 2007 by Pearson Education. Observe that we are differentiating with respect to the variable x. More precisely. To differentiate y 5 properly. More dx generally. It is understood that x is a function of t. by Warren B. therefore. We have. that is.1 dx dx we recognize this derivative as a restatement of the extended power rule. Consider d N 1x 2. There is another way of describing when the chain rule must be applied. then we would need to use the chain rule. (a) The derivative of 6 will be zero and the derivative of 2x 2 is 4x. remember that y is some function of x and use the chain rule. as the derivative. whenever the inner expression is different from the independent variable. Economics. Example 7 Determine: (a) d 16 + 2x2 + y 52 dx (b) d 2 3 1x y 2 dx Solution. and Finance. the chain rule must be applied. Walter O. but that would have involved steps that are not needed in determining the derivative at the given x-value. Inc. and April Allen Materowski. the chain rule must be dt dx dy applied since the variables are not in agreement. dy d 16 + 2x 2 + y 52 = 4x + 5y 4 dx dx Applied Calculus for Business. d Consider 1x22. . Thus. with respect to the variable t. of dt x2.220 Section 2. we have dy d N 1y 2 = Ny N . Published by Pearson Learning Solutions. the chain rule must be applied. We want to find the derivative. the chain rule must be applied and dx d 5 the derivative is 2x . Similarly. then the chain rule must be applied. Walter O. and variable is the differentiation variable. Find y ¿ if y = 2 .6t + 127.729. 24. The syntax is d(expression.32 1x + 2x + 828.121/2. 4. Find dy/dx if y = x61x2 . f1x2 = 1/x. see Figure 3. The answer it gives may not always be in the form we want it. y = 1x2 . 8. g1x2 = 2x3 . 26. 1x + 124 28. .7x + 223/4. f1x2 = 3x . g1x2 = x 1 . g1x2 = 1/x. Ax .123. y = .1 3. find dy/dt. 9. h1x2 = 13x2 . y = 12x2 . .Section 2. 23. 2x . f1x2 = x + 5.123 + 512x . dx B x2 + 8 Figure 3: Differentiation Using the TI 89 EXERCISE SET 2. Gordon. variable) where d is the derivative symbol above the number 8. Consider x2 + 3 d a b .x In Exercises 8 12. Copyright © 2007 by Pearson Education.1 25. 2x2 + 4 13. Published by Pearson Learning Solutions.7x + 32 . Find dy/dx if y = x + 1 .928. 4 . find the equation of the tangent line to the curve at the given x-value. As we mentioned before. find two functions f and g. 12. but that can be easily resolved. 25u2 + 9 18. 2. 16. it does all the work for us. h1x2 = a 2x + 3 3 b . 1x + 123 2 2 2 20. the calculator can differentiate functions very easily. 17. Find dv/dt if v = t # 2 3 t2 . y = 1x + 822/315x3 . find dy/dx.1 20 . h1x2 = 15x . dy (b) d 7 1y 2.6x + 223. f1x2 = 1x2 + 3x + 1217. Given any differentiable function. Find f ¿ 1x2 if f1x2 = 22x3 + 3x + 2. dy dy d d d 2 3 1x y 2 = x 2 1y 32 + y 3 1x22 = x 23y 2 + y 32x = 3x2y 2 + 2xy 3 dx dx dx dx dx Calculator Tips We shall have a lot more to say about this in Section 2. Find ds/dt if s = 13t2 + 1221t3 . f1x2 = g1x2 = 5.2 .222/31r2 . Applied Calculus for Business. by Warren B. 22. 1 . f1x2 = x3. find f ¿ 1x2. The key idea is the following: if you are differentiating a function of one variable with respect to a different variable.122 + 5. f1x2 = .6 In Exercises 1 7. 10. Find dr/du if r = 21. (b) g(f(x)). thinking of x 2 as F(x) and y3 as S(x). Economics. Compute: (a) d 7 1y 2. g1x2 = 1x. dx (c) x = 2 x = 1.4 3 3/5 11. y = 15x2 . Inc.6 The Chain Rule 221 (b) In this case. we must start with the product rule. 15. h1x2 = 12x . and Finance.7x + 1621/2 30. Find y ¿ if y = 1x . find: (a) f(g(x)) and. x .123 19. where we will want to find the slope of a curve whose equation is not in the explicit form y = f1x2.8. 29.3x + 7. x = 0 x = 0 d A 1x5 + 12y7 B dx 1 x . f1x2 = x2 + 2x + 1. we can find its derivative with respect to x without any thought to the process using the calculator. whose composition f(g(x)) will result in the given function h(x). 1. x .2. Find dw/dr if w = 15r2 . Find dv/dt if v = t2 # 2 3 12t + 822 In Exercises 26 29. h1x2 = 14. and April Allen Materowski. expression is the function to be differentiated.1 7. 6.7x + 1241 . 1x2 + 123 27. x + 4 g1x2 = x 2. 1x2 . y = 13t2 + 2t2-1/2. Wang. g1x2 = x. (a) Rewrite H(x) in the form f(g(h(x))). that for a manufacturer. try to find f(x) by guessing. (b) Repeat for f1x2 = x 2 . The slope is known as the marginal cost.1 2 32. economic situation. determine A T1x22 B .42106 *2.92 + * when x = 1. Given f ¿ 1x2 = 18x13x2 + 922. so-called. determine A L1x2 + 12 B . and y = f1x32. triple composition. If f ¿ 1u2 = .7 (a) Marginal Functions and Rates of Change (b) d 3 1y + 3y 2 + 42. and Finance. by Warren B. dx dx d d 1 A L1x2 B = . (a) Given f1x2 = x2 . we shall consider other kinds of functions which will utilize the chain rule. Find B 1x2 + 227 2. We will first examine how the derivative is used in microeconomics. find dy/dx. but possible. In Exercises 34 36 we anticipate other applications of the chain rule. For the linear function. dx 43. you will find the derivative being used as a quantitative tool. for more complex situations. He buys the pretzels from a wholesaler for 50¢ or 1*2 dollar apiece. 34. (c) What can you deduce about the relationship of the zeroes of f(x) and g(x) and the zeroes of their derivatives? (d) In general. Suppose. In many disciplines. Wang. His rent for the stall.4210. 1*2. find dy/dx. and y = f1x22. 39. d d 36. Hint: let y = 12x + x422. is the y-intercept and the cost per item.222 * ** Section 2. determine A E1x22 B . However. u = x3 dy dy du and use = . Walter O. This expression is called a linear total cost function. Thus.7 » » » » » Marginal Functions and Rates of Change Marginal Functions Average Cost Velocity Average and Instantaneous Rates of Change Calculator Tips Marginal Functions The slope of a curve is only one interpretation of the derivative. Gordon. of course. Find d A 32x + 23x + 14x B . try to find f(x) by guessing. It is the cost of acquiring (or in the case of a manufacturer. what can you say about the points of horizontal tangency of f(x) and [f1x2]N? 41. Suppose y = E1x2 where 35.3x + 2215. Published by Pearson Learning Solutions. if f(x) is any function. and April Allen Materowski. You can see that his overhead. 45. Find the points of horizontal tangency for f(x) and g(x). he or she is doing nothing more than examining the derivative of a function.1. dx 40. dx dx 37. such as in the natural or social sciences.423 + 1x2 . $800. . You will see that when an economist discusses marginal functions. where x = the number of pretels that he buys. 5 dx 1x + 12 1 31. in dollars. as we have been dealing only with algebraic type functions. for the most part we examined the special case of the generalized power rule. insurance. Given f ¿ 1x2 = 6x 51x 6 + 5210. in dollars. producing) one additional item. Notice that the overhead cost is C102 = $2000. dx du dx 44. is the slope. (c) Find a general rule for the derivative of this. dx x dx 33. and g1x2 = 1x2 . Compute (c) d 3 1y + 3y2 + 42. Given H1x2 = 551x2 .4. u + 1 While we have stated the chain rule. the total cost will not be a linear function. of producing x items is given by the equation C1x2 = x 2 + 70x + 2000. Given f ¿ 1x2 = 48x31x 4 + 5212. That is. Differentiate 12x + x422 with respect to x3. Let us begin by looking at a very simple. dx 3 2 5 3 d 13x + 122 12x . dy d 3 1y + 3y2 + 4210 dx u2 . If f ¿ 1u2 = u2 . it costs the producer money even if he Applied Calculus for Business. of doing business each month is given by the function C1x2 = 1*2 x + 800. for example. Economics. (b) Find the derivative. his total cost. Later on in the text. Suppose y = T1x2 where A T1x2 B = S21x2. the slope and hence the marginal cost. Suppose you are given a function defined by the equation y = T1x2 whose d derivative T ¿ 1x2 = S21x2. Suppose y = L1x2 where d d 1E1x22 = E1x2.3x + 2 and g1x2 = 1x2 . A man sells pretzels at a small refreshment stand in a shopping mall. find A T12x32 B . 38. Copyright © 2007 by Pearson Education. is a constant. the total cost. and other fixed costs total $800 per month. try to find f(x) by guessing. 42. utility bills. Inc. of producing x short wave radios is C1x2 = x2 + 80x + 3500. then we have. by Warren B. C11002 = 1002 + 8011002 + 3500 = 21. the marginal cost is approximately equal to the derivative of the cost function.1 (remember.000 . Economics.C1992. then E ¿ 1x2 L In most real cases it happens that h = 1 makes this approximation valid. which is $19. is approximately equal to the marginal function at level x or level x . the revenue derived by the producer when x + 1 items are sold.E1x .C1x2. Economists call this difference the marginal cost when the production level is 99. If we denote the total cost function by C(x). Example 1 Given that the cost. E1x + 12 . the derivative. Wang.Section 2.$18. and Finance. If we let E(x) represent any of these economic functions. R1x + 12 . and the marginal profit is approximately equal to the derivative of the profit function. (b) Find C ¿ 11002 and interpret this derivative. we have E ¿ 1x2 L E1x + 12 . we can approach the point on the curve from either the left or right).P1x2. Thus. then we shall show that the marginal economic function. the marginal revenue is approximately equal to the derivative of the revenue function.C1992. in dollars. Copyright © 2007 by Pearson Education. Applied Calculus for Business.1. The cost of producing 100 items is C11002 = 1002 + 7011002 + 2000 = $19. and April Allen Materowski. and the total profit function by P(x). Assuming that each of these economic functions is differentiable. is called the marginal revenue at x. Walter O. P1x + 12 . (ii) the 101st radio. the derivative of the economic function is approximately equal to the marginal function.E1x2 That is.E1x2 h:0 h E1x + h2 . We should also observe that if we allow h = . DEFINITION 1 C1x + 12 . is called the marginal profit at x.E1x2 = E1x2 .R1x2. Solution. Gordon.E1x2 h Now suppose that h is sufficiently small. then we would obtain the approximation. Consider the expression C11002 . the producers profit due to the x + 1st item.7 Marginal Functions and Rates of Change * ** 223 does not produce anything. while the cost of producing 99 items is C1992 = 992 + 701992 + 2000 = $18.500. (a)(i) We need to compute C11002 . E ¿ 1x2 L E1x . from the definition of the derivative that E ¿ 1x2 = lim E1x + h2 . that is. is called the marginal cost at production level x.731 = $269.E1x2 L E ¿ 1x2.12 -1 Thus. E ¿ 1x2. then we have the following definition.000.731.12 . It represents the cost of producing the 100th item. (a) Find the cost of producing: (i) the 100th radio. the total revenue function by R(x). Published by Pearson Learning Solutions. . the cost of producing the x + 1st item. Inc. while there are others that you seek to maximize (revenue and profit). Wang. C ¿ 11002 = 280.C11002 = $281.00. Thus. R ¿ 1x2 the marginal revenue at x. the marginal cost at production level 100 is $281. Inc.7 Marginal Functions and Rates of Change C1992 = 992 + 801992 + 3500 = 21. by Warren B.) In the above example. Therefore. (Notice.E1x . C11012 = 21. C ¿ 11002 may be used to approximate the marginal cost at level 99. Copyright © 2007 by Pearson Education. we find p = . (e) What is the price for each color television when the marginal profit is zero? (f) Sketch the profit function. (or for E1x2 . Thus.224 Section 2. Applied Calculus for Business.781. denoted by C1x2 is defined as. (b) Now. we found that the cost of producing 100 radios was $21. that the error in either case is $1. and April Allen Materowski. Published by Pearson Learning Solutions. (a) If we solve the demand equation for price. we shall call C ¿ 1x2 the marginal cost at x. the marginal cost at level 99 is C11002 . (b) Determine the marginal revenue function. The percent error in the approximation is less than 0.500.4%. from Chapter 1 that the total revenue.C1992 = $279. What is the average cost per radio? All we have to do is to divide the total cost by 100 to obtain the average cost per radio as $215. or level 100. Economics. where p is the per unit price in dollars. P ¿ 1x2 the marginal profit at x. R1x2 = 1price per item2 # 1number of items2 = p # x Example 2 Suppose that the relationship between price of and demand for a certain type of large color television set is given by the demand equation 10p + x = 10. Gordon. That is.221.000.E1x2. Example 2 below examines one such question when the demand and cost functions are linear. If you were a producer. Recall. (ii) We need to compute C11012 . and x is the number of sets demanded.0. Average Cost C1x2 = C1x2 x There are natural questions that can be posed at this time. We shall answer these questions in the next chapter. drawing the tangent line when the marginal profit is zero. Thus. C11012 . If the producer s cost is C1x2 = 600x + 3000: (a) Determine the revenue function. C ¿ 1x2 = 2x + 80.C11002. we have the following definition. We shall assume that the derivative E ¿ 1x2 is such a good approximation for the marginal function E1x + 12 . (d) Determine the marginal profit function. Walter O.1x + 1000.12) that we shall not distinguish between them. and C ¿ 1x2 the marginal average cost at x. DEFINITION 2 The average cost function. and Finance. (c) Determine the profit function. what should the level of production be to minimize your total cost? How do you maximize your revenue? How do you maximize your profit? How do you minimize your average cost? Notice that there are certain functions that you want to minimize (cost and average cost). What does this profit represent? (g) What can you conclude about the profit at the price obtained in (e)? Solution. More generally. . 0.0. R ¿ 1x2 = C ¿ 1x2. it is time to stop. (This is true. P1x2 = R1x2 .C1x2 = .0. In words. Copyright © 2007 by Pearson Education. or x = 400/0. That is. 397. Economics.7 Marginal Functions and Rates of Change 225 Therefore. . In other words.000) Figure 1: P1x2 = . the profit is maximized.1x + 10002x = .1x 2 + 400x . Published by Pearson Learning Solutions.1x 2 + 400x .2x + 400 = 0. That means that the cost of producing one more item exactly equals the revenue obtained by producing it.2 = 2000. Let us now consider an application of the derivative that occurs in the natural sciences.1600x + 30002 = . We sketch the parabola in Figure 1.3000.2x + 1000.0. R1x2 = p # x = 1 . the point at which the marginal profit 1P ¿ 1x22 is zero. by Warren B. When R ¿ 1x2 = 0. R ¿ 1x2 . R ¿ 1x2 = . (2000. when x = 2000 and p = 800. Gordon. we recognize it to be a parabola opening downwards. (e) When P ¿ 1x2 = 0.4. Wang.0. Thus the production level when the marginal profit is zero is 2000.000.C1x2.1x2 + 1000x . it pays to keep producing more items. in general for all parabolas.1120002 + 1000 = 800. (b) The marginal revenue function.0. That is.1x 2 + 1000x. We shall see that the derivative may be interpreted as an instantaneous rate of Velocity Applied Calculus for Business. (g) Observe that the vertex (turning point) of the parabola occurs at the point at which the tangent line is horizontal.3000 Do you think that is an accident that the profit is maximized where the marginal profit is zero? Since the profit is P1x2 = R1x2 . Its maximum value is P120002 = $397.C ¿ 1x2.1x2 + 400x . and Finance.0. (f) Since the profit function is P1x2 = . When the marginal cost becomes equal to the marginal revenue. see Section 1. We are interested in those values of x in its domain for which x Ú 0 (why?). and the corresponding price per color television is p = .0.Section 2. (d) P ¿ 1x2 = .0. As long as marginal cost is less than marginal revenue.) At the vertex.0. the marginal profit is P ¿ 1x2 = R ¿ 1x2 . Walter O.3000. (c) The profit function. we have .2x + 400.C ¿ 1x2 = 0. the price per color television is $800. Inc. the marginal revenue equals the marginal cost. and April Allen Materowski. there is no gain to be made by making the next item. where s is position and t is time.1 Suppose we reduce the time interval from [1.1 0. the average velocity over the one-tenth of a second time interval [1. that is.1 You are probably anticipating our next step.f112 564.f112 2 .7 Marginal Functions and Rates of Change change. 1.01 of a second.12 = .01 Suppose now. Inc.f112 0. Applied Calculus for Business. and April Allen Materowski. At the end of one second. then the average velocity over this time interval is f11 + 0.4 ft/sec. The distance of the particle measured from some fixed point. That is. The average velocity over this h second time interval is f11 + h2 . say 0. that the time interval is [1. from the ledge of a building 512 feet high. and Finance.560 = = 31. 0. the instantaneous velocity when t = 1.04.1 Does this difference quotient look familiar? By writing f11 + 0. Wang.560 = 16 ft/sec. Let us determine the average velocity during this one-tenth of a second interval. from t = 1 to t = 2. That is.226 * ** Section 2.161122 + 64112 + 512 = 560 feet.01]. Copyright © 2007 by Pearson Education. Published by Pearson Learning Solutions. 2 . v112 = lim f11 + h2 .1 = 576 . is given by the equation. [1.122 + 6411. from a time interval of length one second to one of length one-tenth of a second.1) we have the average velocity over this one-tenth of a second time interval as f11 + 0. f11. the time interval is h seconds. In this application. That is. Its average velocity during this one second time interval is the total distance traveled divided by the total time elapsed. At the end of two seconds.2]. .84 ft/sec 0. That is. it should be clear how we proceed. That is.12 .04 . 1. s = f1t2.560 = 16 feet.12 .f112 h:0 h This is recognized to be the derivative at t = 1. f122 . s = f1t2 = . f11.01 0.1]. We shall later show that its height s at time t is governed by the equation. is f122 . Walter O. the height is f122 = .12 in place of f(1.f112 = 576 . To begin. The distance traveled during the time interval [1. More specifically.f112 h If we ask what the velocity of the ball is at the instant when t = 1. 1 + h]. We let h approach zero in the above difference quotient.012 . 1. Therefore. Gordon. Economics. assume a ball is thrown vertically upwards with an initial velocity of 64 feet per second. the height of the ball is f112 = .560 = = 30. 2] to [1.16t2 + 64t + 512.f112 560.1611. by Warren B.3184 . That is.161222 + 64122 + 512 = 576 feet. Suppose we make the time interval even smaller.12 + 512 = 563.1] is. we assume that a particle is moving along a straight line. 161222 + 64122 + 512 = 576 ft. we mean the instantaneous rate of change (the derivative).32132 + 64 = .32t + 64. Example 3 A ball is thrown vertically upwards with an initial velocity of 64 ft/sec from the top of a building 512 feet high. we mean the instantaneous velocity.16t2 + 64t + 512. we may interpret a derivative as an instantaneous rate of change. if s = f1t2 then the average velocity over the time interval [t. The negative sign indicates that the ball is moving downward. t + h] is vavg = and the instantaneous velocity at time t is v1t2 = s ¿ 1t2 = lim Observe that in the above example. f ¿ 1x2. . the average rate of change of the function on the interval [x. what is its velocity the instant before it strikes?) (g) Sketch the graph of s. DEFINITION 3 Given the differentiable function defined by the equation y = f1x2.7 Marginal Functions and Rates of Change * ** 227 v112 = ds = f ¿ 112 ` dt t = 1 f1t + h2 . the maximum height. Gordon.16t2 + 64t + 512. the ball will reach its maximum height at the instant when its velocity is zero. and April Allen Materowski. Assume the altitude of the ball (measured in feet) at time t (in seconds) is governed by the equation s1t2 = . which occurs when v = 0.32 ft/sec. Walter O. Similarly. Inc. (b) Determine its velocity at the end of 3 seconds. (a) Determine the height at the end of 3 seconds. the average velocity over the h second time interval corresponds to the slope of the secant line. If the velocity were positive. (e) To determine when the height is zero.f1t2 h:0 h Average and Instantaneous Rates of Change More generally.f1t2 h f1t + h2 . the ball would continue to rise.32t + 64. and the instantaneous velocity corresponds to the slope of the tangent line. we have the following. Thus. Wang. A negative velocity means that it is moving downward. Copyright © 2007 by Pearson Education. it is at its maximum height when t = 2. x + h]. its velocity is v132 = . we may interpret the marginal cost as the instantaneous rate of change of cost with respect to production level. we set s = 0 and solve the quadratic equation Applied Calculus for Business.161322 + 64132 + 512 = 560 ft.Section 2. Therefore. (c) If the velocity is zero we have 0 = . (b)The velocity is v1t2 = s ¿ 1t2 = . (That is. when we ask for a rate of change. That is. Therefore. Published by Pearson Learning Solutions. when we ask for velocity. The height is governed by the equation s1t2 = . Solution. we find that t = 2 sec. (a) The height at the end of 3 seconds is s132 = . (d) A positive velocity means it is moving upward. This is such a useful interpretation that we often delete the adjective instantaneous. (c) At what time will its velocity be zero? (d) What is the maximum altitude reached by the ball? (e) At what time will its height be zero? (f) What is the impact velocity of the ball when it hits the ground. Similarly. Thus. is s122 = . or t = 2. and Finance. Summarizing.f1x2 h the (instantaneous) rate of change at x is the derivative. At the end of 3 seconds. In general. Solving for t. Economics. we may interpret the slope of the secant line as the average rate of change of y with respect to x on the interval [x. x + h] is yavg = f1x + h2 . by Warren B. and the slope of the tangent line as the instantaneous rate of change of y with respect to x. Why?) Thus. s ¿ 1t2. s(t). Inc.192 ft/sec. the velocity when s = 0. subject only to the force of gravity. particles may move in many other ways. if t is between 0 and 2. Dividing each side of the equation by . of any object that moves vertically. and April Allen Materowski.5).3t2 for t 7 0.228 Section 2.32182 + 64 = . Economics. thus s is moving in the positive direction when t 7 2. while v0 is negative if the object is projected downward.821t + 42 = 0. Figure 3 shows the path of the particle along the s-axis. Figure 2: s1t2 = . You may assume equation (1) in the exercises. we obtain. and Finance.22 7 0 The inequality is solved easily using sign analysis (see Section 0. is positive. . Therefore. Of course. (f) The velocity at any time is v1t2 = . that is. For positive values of t. Copyright © 2007 by Pearson Education.8 meters/sec 2. (g) The graph of s versus t is a parabola and is given in Figure 2. The impact velocity. Since s ¿ 1t2 = 3t2 .6t 7 0 3t1t .32t + 64. The particle is moving in the positive direction when the velocity. Gordon. (We reject the root t = . If the object is projected upwards.7 Marginal Functions and Rates of Change . it is possible for s(t) to take different forms. x + h] was defined as Figure 3: s1t2 = t3 .3t2 for t 7 0 yavg = f1x + h2 . is v182 = . and for t 7 2 the product will be positive. Factoring. we shall show that the height. Wang. the negative sign indicates that the ball is moving downward. we need only determine where 3t2 . Published by Pearson Learning Solutions. (1) The constant g is called the acceleration due to gravity and is approximately equal to 32 ft/sec 2. and an initial height s0.2 will be negative.4.16t2 + 64t + 512 = 0. when t = 8. v0 is positive. the product of 3t and t . The average rate of change on the interval [x.16t2 + 64t + 512 When we study integral calculus. by Warren B. we have 1t . Walter O. with an initial velocity v0. That is.32 = 0. One additional remark. Example 4 The location s of a certain particle from its starting position is given by the equation s1t2 = t3 .4t . t2 . Again.1*2 gt2 + v0t + s0. Consider the following example. is given by the equation s1t2 = . it takes the ball 8 seconds to reach the ground. the solution is t = 8.f1x2 h Applied Calculus for Business.6t. 9.16. For what values of t is it moving in the positive direction? Solution. or using metric units. Section 2.7 Marginal Functions and Rates of Change * ** 229 suppose we rename the interval [a, b], that is, we set x = a and x + h = b then h = b - a and we have yavg f1b2 - f1a2 = b - a Calculator Tips This is called the average rate of change of the function over the interval [a, b]. We have seen that the difference quotient has various interpretations. Geometrically, it represents the slope of the secant line, and in this section, it was interpreted as an average rate of change. We can use the calculator to compute the difference quotient by pressing the Catalog key and scrolling to avgRC (which stands for average rate of change). The syntax is avgRC(expression, var, h). Suppose y11x2 = 3x 2 - 4x + 7, has been entered in the Y = screen, then avgRC(y1(x), x, h) returns 6x + 3h - 4, as it should, see Figure 3. (Note that h is optional, if it is omitted, the calculator assumes you are using h = 0.001.) Figure 4: avgRC to Compute the Difference Quotient EXERCISE SET 2.7 1. The cost, in dollars, of producing x bicycles is given by C1x2 = 60 + 10x + 1000/x 1x Ú 102. (a) What does it cost to produce: (i) 99; (ii) 100 bicycles? (b) What is the cost of producing the 100th bicycle? (i) Find it exactly; (ii) Use the derivative to find it approximately. (c) What is the error in using the derivative to approximate the marginal cost of the 100th bicycle? (d) What is the average cost per bicycle when producing the 100 bicycles? 2. Let C1x2 = 2x2 + 300x + 50 be the cost of producing x items. What is the cost of producing the 89th item? Give both the exact and approximate answers. 3. Let 10p + x = 100 be the demand equation, where p is the price per item when x items are demanded. (a) Find the total revenue when the level of production is: (i) 40; (ii) 41; (b) Find the exact revenue derived from the 41st item. (c) Find the approximate revenue derived from the 41st item. (d) What is the error if the derivative is used to approximate the marginal revenue? 4. Suppose that the number of riders per day on the New York City subway system is 4 (million) when the fare is $2.00. Suppose that the ridership will drop to 3.8 (million) when the fare is raised to $2.25. Assuming that the relationship between demand and price is linear: (a) Find the demand equation. (b) Find the revenue and marginal revenue functions. (c) Graph the revenue function and show that the maximum revenue is at the point where the marginal revenue is zero. (d) Find the price and number of riders that will maximize the total revenue. In Exercises 5 and 6, the cost and demand equations are given. (a) Determine the profit function. (b) What is the profit when the level of production is 100? (c) Find the marginal profit function. (d) What is the price per item when the marginal profit is zero? (e) Sketch the profit function. (f) What is the level of production when the marginal profit is zero? What does it represent? 5. C1x2 = 20x + 500, 20p + x = 1000. 6. C1x2 = 30x + 1000, 10p + x = 700. In Exercises 7 and 8, find the average velocity over the given time interval, if s = f1t2 is the equation of height as a function of time. 7. s = - 16t2 + 128t: (a) [0, 3]; (b) [3, 4]; (c) [0, 4]. 8. s = - 16t2 + 160t: (a) [0, 3]; (b) [3, 6]; (c) [3, 10]. 9. A ball is thrown vertically upward from the ground with an initial velocity of 176 ft/sec. (a) Determine its height s, as a function of time (use equation (1)). (b) Find the average velocity of the ball during the 2 second time interval [1, 3]. (c) Find the distance traveled over the time interval [t, t + h]. (d) Find the average velocity during the time interval [t, t + h]. (e) What limiting average velocity do you obtain if h approaches zero in (e)? (f) What is the velocity when: (i) t = 1? (ii) t = 3? (g) When will the velocity be zero? What does this mean? (h) How high will the ball go? 10. Suppose in Exercise 9, the ball was thrown vertically upward from a ledge 192 feet high. (a) How high will the ball go? (b) How long does it take for the ball to hit the ground? (c) What is its impact velocity with the ground? 11. A rocket is launched vertically upward with an initial velocity of 6400 ft/sec. (a) When will the velocity of the rocket be zero? (b) What is the maximum altitude the rocket will attain? 12. A ball is thrown vertically upward from the ledge of a building 256 feet high with an initial velocity of 16 ft/sec. What is the ball s impact velocity with the ground? 13. A helicopter is stationary at an altitude of 512 feet. A package falls vertically from it. Determine the time it takes the package to hit the ground and its impact velocity with the ground. In Exercises 14 17, find the average rate of change of y with respect to x on the given interval. 14. y = 3x + 2 (a) [0, 1]; (b) [3, 5]; (c) [x1, x2] interpret this answer 15. y = mx + b; 2 [x1, x2], interpret this answer. 16. y = 2x - 7x + 2; [2, 3]. 17. y = 3x2 - 7x - 5; [3, 6]. In Exercises 18 22, find the instantaneous rate of change of y with respect to x at the given x-value 18. y = 3x + 2; 19. y = mx + b; x = 5 x = a x = 2 20. y = 3x2 - 7x + 2; Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 230 * ** Section 2.8 Implicit Differentiation 31. If y = 13x2 + 923/2, find the rate of change of y with respect to (a) x; 3x (b) 2 . x - 1 32. If y = 1x3 - 122, find the rate of change of y with respect to (a) 2x; 2x (b) mx + b; (c) 2 . x - 1 33. Find the rate of change of the volume of a cube with side x with respect to (a) a side; (b) its (surface) area. 34. Find the rate of change of the area of an equilateral triangle of side x with rex2 spect to its: (a) side; (b) perimeter. (Hint: A = 23 ) 4 The relative rate of change of y = f1x2 with respect to x at x = a is defined by f ¿ 1a2 . Find the relative rate of change of f(x) with respect to x, (or f(t) with respect f1a2 to t) in Exercises 35 38. 35. f1t2 = 3t3 - 4t2 - 9; 36. f1t2 = - 16t2 - 128t; x + 3 ; Ax - 2 a = 2. a = 8. a = 0. 21. y = x + 3 ; Ax - 2 x = 6 x = 0 22. y = 1x2 - 1221x 2 - 3x + 123 In Exercises 23 25, (a) Find the instantaneous rate of change of s as a function of t. (b) Find the value(s) of t for which the instantaneous rate of change is positive. 23. s1t2 = 3t2 - 12t 24. s1t2 = 4t2 - 8t 25. s1t2 = t3 - 3t2 - 9t 26. Let C1x2 = ax2 + bx + d represent the cost of producing x items. Find the error if C ¿ 1x2 is used for the marginal cost of the xth item. 27. Given the demand equation p + mx = b 1m 7 0, b 7 02. (a) On the same set of axes, graph the demand and revenue equations. (b) Show that the x-intercept of the marginal revenue function occurs at that x-value for which the revenue is maximum and is 1*2 the x-intercept of the demand equation. 28. An object is thrown vertically upward with an initial velocity v0 from a height s0. (a) Show that it takes the object the same time to go from s0 to its maximum height as to fall from its maximum height to s0. (b) Show that the velocity of the object when it returns to height s0 is - v0. (c) Why must (a) and (b) be true physically? (d) Give an expression for the maximum height assuming that v0 7 0. 29. A ball is to thrown vertically upward from the ground so that it may be caught by an individual on a ledge 128 feet high. What is the minimum initial velocity with which it may be thrown? 30. With what initial velocity must an object be projected vertically upward from the ground to reach a height of 600 feet? 37. f1x2 = 1x2 - 1221x2 - 3x + 123; 38. f1x2 = a = 6. 39. Suppose the horizontal position x a particle has traveled in t seconds, 0 t 5 is given by the equation x = 5 - 25 - t. (a) How far will the particle have traveled when t = 1? (b) What is the velocity of the particle after t seconds? (c) What happens to the velocity as t gets close to 5? 2.8 » » » Implicit Differentiation Finding a Tangent Line Finding the Derivative Calculator Tips With the exception of some of the examples on related rates, we have only determined the derivative of functions of the form y = f1x2. That is, we have y explicitly as a function of x. What if x and y are related by some equation for which it is not possible or desirable to write y as a function of x. For example, suppose we have x = y3 + y + 1 In this case, x is a function of y. You may draw its graph by substituting values for y and finding the corresponding x-values, see Figure 1. From the graph you can see that y is also a function of x. But could you solve this equation for y in terms of x? Probably not.* However, for each x, it is possible to determine y graphically or by various numerical schemes. The important question is, can we easily determine the derivative dy/dx? The answer is yes. How? By the chain rule! Recall, from Section 3.6, when the dependent and independent variables are the same, there is no need for the chain rule, but when they are different, *There is a formula by which one can solve a cubic, but it is very cumbersome to use in practice. Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.8 Implicit Differentiation * ** 231 d 5 1x 2 = 5x 4 (note the independent and dependent variwe must use it. For example, dx ables agree), Figure 1: x = y3 + y + 1 dy d 5 1y 2 = 5y 4 . So, how do we find dy/dx dx dx when we do not have y explicitly in terms of x? The idea behind the method is simple. Even though we may not be able to solve for y in terms of x, we assume that y is a function of x. Then we differentiate the equation that gives the relationship between x and y, and solve for the derivative. This method is called implicit differentiation. One main difference is that when we find the derivative, it will usually be a function of both x and y, unlike the explicit case, where the derivative is obtained as a function of x alone. Remember, the purpose of the derivative is to determine the slope of a curve, the velocity, or a particular marginal value. To compute the numerical value of the derivative we only need to substitute. It should not bother us that we are substituting for both x and y. The next examples illustrates the idea. but if y is a function of x then by the chain rule, Example 1 Find dy /dx if x = y3 + y + 1 Solution. Since the two sides of the equation are equal to one another, their derivatives must be equal. We take the derivative of both sides with respect to x. On the left side we have just x, and the derivative of x with respect to x is 1. However, on the right side, we have an expression in y, not x. Remember that y is a function of x even though we do not know its exact form. Thus, we must use the chain rule on the right hand side, and get 1 = 3y 2 dy dy + dx dx (Do you see what happened on the right? The derivative of y3 is 3y 2 dy/dx by the chain rule. The derivative of y is just dy/dx and the derivative of 1 is zero.) Now we solve for dy/dx. Factoring the dy/dx, gives 1 = 13y 2 + 12 and dividing by 13y 2 + 12 yields dy dx Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 232 Section 2.8 Implicit Differentiation dy 1 = 2 dx 3y + 1 In this case, it happens that the derivative is entirely in terms of y. Finding a Tangent Line Example 2 Given the equation x3 + 3y 5 + 2x = 3y 7 + 2x 2 + 4. Determine the equation of the tangent line to the curve defined by this equation at the point (2, 1). Solution. We know that the answer to the problem is y - 1 = m1x - 22, where m, the slope of the tangent line is the derivative at the point (2, 1). Since we have the equation relating x and y, all we have to do is differentiate each side of the equation to find an equation involving x, y, and dy/dx from which we can determine the derivative. Differentiating each term of the equation we have, 3x2 + 15y 4 dy dy + 2 = 21y 6 + 4x dx dx (Do you understand why the derivative factor is present after differentiating each term involving y? y is a function of x, so the chain rule had to be applied.) To complete the solution, we need only substitute x = 2 and y = 1. We obtain, 31222 + 151124 dy dy + 2 = 211126 + 4122 dx dx Simplifying, combining like terms, and transposing, we have, - 6dy/dx = - 6, or dy/dx = 1 Thus, the equation of the tangent line is y - 1 = 11x - 22, or y = x - 1 Finding the Derivative Example 2 was a numerical problem. In any numerical problem, where you are given a point at which to compute the derivative (slope, velocity, or marginal value) substitute immediately after the differentiation step, that will make the computations the simplest. Suppose you choose instead, to find the derivative in terms of x and y, and then substitute afterwards. Then you must first group together all terms involving the derivative on one side of the equation, and everything else on the other side of the equation. Example 3 Find dy/dx for the function in Example 2. Solution. Let us save some writing by using y ¿ for dy/dx. In the above example, we would obtain, (writing y ¿ instead of dy/dx). 15y4y ¿ - 21y 6y ¿ = 4x - 3x 2 - 2 Factoring the left hand side of the equation, we obtain 115y 4 - 21y 62y ¿ = 4x - 3x 2 - 2 dividing by the term in parentheses, yields, y¿ = dy 4x - 3x2 - 2 = dx 15y4 - 21y 6 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.8 Implicit Differentiation 233 If you now substitute x = 2 and y = 1, you will find, as before, that y ¿ = 1. As you can see, it is far easier to do the substitution immediately after the differentiation, rather than after the algebra. However, if you need to find the derivative at several points, or to find a general expression for the derivative, then perform the algebraic manipulations. The next example illustrates the use of the product rule in finding the derivative implicitly. We will use y ¿ to indicate dy/dx whenever we find it convenient to do so. Example 4 Find y ¿ if 3x2y 4 - 2x 2 + 3y 5 = 17. Solution. We differentiate each side of the equation. Note that the first term involves a product, so the product (and chain rule) must be applied. We have, 3x2[4y 3y ¿ ] + y 4[6x] - 4x + 15y 4y ¿ = 0. or, y ¿ 112x 2y 3 + 15y 42 = 4x - 6xy 4. Dividing, we find that y¿ = 4x - 6xy4 12x2y3 + 15y4 Even when we can solve for y in terms of x, we may prefer to find the derivative by implicit differentiation. The next example illustrates this. Example 5 The equation x2 + p2 = 25 represents a demand equation when x and p are each in the first quadrant, that is, x Ú 0, and p Ú 0. Find: (a) dp/dx; (b) dx/dp; (c) dR/dx (d) dR/dp. Solution. We could solve for p as a function of x and find p = 225 - x 2, and then differentiate with respect to x. It is simpler to use implicit differentiation. (a) Differentiating with respect to x, we have, 2x + 2p dp/dx = 0, or dp/dx = - x/p (b) Differentiating with respect to p, we have, 2x dx/dp + 2p = 0, or dx/dp = - p/x (c) The revenue function is R = xp; therefore, by the product rule, dR/dx = x dp/dx + p. Substituting from (a) for dp/dx, we have, dR/dx = x1 - x/p2 + p = 1p2 - x22/p. (d) To find dR/dp. we again use the product rule. R = xp, dR/dp = x + p dx/dp. Substituting from (b) for dx/dp, we have dR/dp = x + p1 - p/x2 = 1x 2 - p22/x Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 234 * ** Section 2.8 Implicit Differentiation Note that dx/dp = 1/1dp/dx2. This will always be true as long as neither of these derivatives is zero. The reason is very simple. If we think of p as a function of x, say p = P1x2, then dp/dx represents the slope of the curve at the point (x, p). If we take the same curve and think of x as a function of p, say x = D1p2, then the slope is the same. But now suppose we think of x as the vertical axis, then the slope at the point (p, x) is dx/dp. Since slope is change in y divided by change in x, interchanging the axes makes the slope into the change in x divided by change in y. Thus, interchanging axes has the effect of taking the reciprocal of the slope. Hence, the derivatives are reciprocals. Now go back and look at the solution to Example 1. Compare the result there to what you would get by just taking dx/dy. We close this section with a proof of the power rule for rational exponents. To this point, we have proved the rule only for positive and negative integers. POWER RULE FOR RATIONAL EXPONENTS d m Axn B = dx where m and n are integers. m m n -1 nx PROOF Let y = x m/n, then, yn = 1x m/n2n or yn = x m. Differentiate each side of this equation with respect to x, giving, nyn - 1y ¿ = mxm - 1. Solving for y ¿ , we have, y¿ = Substitute for y, using, y = xm/n, we have, y¿ = * bM # = bM - N and 1bM2N = bM N.) N b Thus, once again, we have shown that the power rule states that to find the derivative of x to a power, multiply by the power and decrease the power by one, that is for any rational (Note the two rules of exponents that were used, d r 1x 2 = rx r - 1. We shall examine the case when the exponent is irrational dx when we consider logarithmic and exponential functions. The TI 89 can find the derivative, implicitly, but we have to do it two stages. First, when the equation is entered, we need to let the calculator know that y is a function of x so we must always write y(x) in place of y, wherever it appears. Thus, if we wanted to find dy/dx when the equation is x2 + y 2 = 9, we would enter d1x ¿2 + y1x2¿2 = 9, x2 and then press enter, what the screen displays is the equation differentiated, see Figure 2. We can now have the calculator solve this equation for the derivative, but we need to give the derivative a new name otherwise an error message is produced. Suppose we call the derivative D, then we enter number r, solve12y*D + 2x = 0, D2 This is indicated in Figure 3, along with the solution. m m xm - 1 m xm - 1 m m m = = x m - 1 - Am - n B = x n - 1 m n A x n B 1n - 12 n xm - m n n n m xm - 1 . n yn - 1 Calculator Tips Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.8 Implicit Differentiation * ** 235 Figure 2: Find the Derivative Implicitly Using the TI 89 Figure 3: Finding the Derivative Thus, we have dy/dx = - x/y. (Be careful, the D entered as uppercase and exhibited on the screen as lowercase is the letter D located above the comma key and requires the pressing the alpha key first. This is different from the d used for differentiation which is above the number 8 key and requires pressing the 2nd key first.) EXERCISE SET 2.8 In Exercises 1 4, find dy/dx at the indicated point in two different ways: (a) Solve for y as a function of x and differentiate; (b) Differentiate implicitly. 1. y2 = 3x + 1; 3. x2y + 1 = 2x; 15, - 42 12, 12 11, 12 2. 3x2 - 4xy - 4 = 0; 10. x2/3 + y 2/3 = 1 11. 3x2y - 4y2x + 7 = 2x3y3 12. 2/x - 3/y = x 2y2 = 2x + y3 x2 + y2 14. 3x 3/4 - 2y 2/3 + 7y2 - 3x = 9x2 - 5y 4 13. 15. 2 1x + y - 1x - y = 2. Find dy dx x2 - y2 4. 2x2 - 3xy + 4y2 - 5y = 2; 11, 22 (Hint: Use the quadratic formula to solve for y.) In Exercises 5 14 find dy/dx using any method. 5. x3 + y 3 = 10 6. x2 9 y2 4 ` 14,02 16. 5s21v3 - 12 = 7. Find (a) ds/dv; (b) dv/ds. 17. 1/t + 1/s = 1. Find (a) ds/dt; (b) dt/ds, (c) Show that 1ds/dt21dt/ds2 = 1. 18. 1x + y + 1x - y = to the curve at (5/2, 3/2). x2 y3 + 31 . Determine the equation of the tangent line 27 - = 1 7. xy = 7 8. x - xy + 7 = 0 9. 5x 2 + 6x2y 2 = y 2 + 15 2 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 236 * ** Section 2.9 Elements of Geometry 23. (a) Determine the equation of the tangent line at any point (h, k) which lies on the curve x2/3 + y2/3 = a 2/3 1a 7 02. (b) Let the tangent line to this curve intersect the x and y axes at the points 1x1, 02 and 10, y12 respectively. Show that the length of the line segment joining these two points is a constant. 24. Show that the tangent line to any point on the circle x2 + y 2 = a 2 is always perpendicular to the radius drawn to the point of tangency. 19. x2 + y 2 = 25. (a) Determine the equation of the tangent line at each of the following points: (i) (3, 4); (ii) 13, - 42; (b) Find the equation of the tangent line at (3, 4) by finding the equation of the line passing through (3, 4) which is perpendicular to the (radius) line segment whose end points are (0, 0) and (3, 4). 20. 2x2 + 3y 2 = 14. Determine the equation of the tangent line at each of the following points: (a) (1, 2); (b) 11, - 22. 21. A demand equation is given by p = 29 - x, where p is the price per item when x items are demanded. Find dR/dx when x = 1. 22. (a) Find the equation of the tangent line to the curve x3 + 3y2 - 12x - 1 = 0 at (1, 2). (b) At what point(s) on the curve will the tangent line be horizontal? (c) Show that the tangent line is vertical for some x0 where 3 6 x0 6 4. 2.9 » » » » » Elements of Geometry Vertical Angles Parallel Lines Similarity Congruence Midpoint Formula This section examines some of the elementary geometric concepts that may arise while solving applied problems in the calculus. Consider the angle illustrated in Figure 1, As OA rotates counterclockwise, the angle increases. When AOB becomes a line, the angle is said to be a straight angle whose measure is 180*, the symbol for degree is *. A A O B O B A __________________ B O Figure 1 Consider Figure 2, where two straight lines intersect at E. A C B E D Figure 2: The Intersection of Two Lines Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.9 Elements of Geometry * ** 237 Since CD is a straight line, in degrees, the measure of 1 AED + 1 AEC = 180° ( 1 is the symbol for the measure of an angle, also note that when three letters are used to describe the angle, the middle letter will always represent the vertex of the angle.) similarly, since AB is a straight line 1 AEC + 1 CEB = 180° so it follows that 1 AED = 1 CEB and similarly, 1 AEC = 1 BED These equal pairs of angles are sometimes called vertical angles. Thus, when two lines intersect, their vertical angles are equal. We also note that when two angles sum to 180*, the angles are said to be supplementary angles. Thus, the equality of vertical angles is a consequence of the fact that supplements of equal angles are equal to each other. In Section 1.1, we discussed equations of parallel lines from an intuitive point of view. More precisely, two non-vertical lines are said to be parallel if the angle they make with the positive x-axis is the same. This angle is sometimes called the inclination of the line. See Figure 3, where lines L1 and L2 intersect the x-axis at the same angle a, and are therefore parallel. (Note that a 6 180°) 0 L1 L2 Vertical Angles Parallel Lines + + x-axis Figure 3: Parallel Lines Let us now examine Figure 4, where the parallel lines L1 and L2 are intersected by the line T L2 L1 T , . 0 / + - * 2 Figure 4: Parallel Lines Intersected by a Transversal Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 238 * ** Section 2.9 Elements of Geometry Since the lines are parallel the angles measured by a and b are the same. To convince yourself, rotate the figure so the line T, sometimes called a transversal, is horizontal (i.e., is the x-axis), then these angles are the inclination angles and since the lines L1 and L2 are parallel, they must be the same. Since T is a straight line, it then follows that angles g and d are also equal. Thus, when two parallel lines are cut by a transversal, the corresponding angles are equal, that is, in Figure 2, 0a 0g 0m 0P = = = = 0b 0d 0s 0h The angles m and d are called alternate interior angles as are a and h. It is easy to see that as a consequence of vertical angles being equal as well as the supplements of two equal angles are equal, that when two parallel lines are cut by a transversal, the alternate interior angles are equal, that is 0m = 0d and 0a = 0h Example 1 Given the parallel lines in Figure 5, and the given angle, determine remaining angles. Solution. L2 L1 140o T , / . + - * 1 Figure 5: Determining the Angles We have, 140 + 0 b = 180, therefore, 0 b = 40° a is a corresponding angle to b , h, P are vertical angles, therefore 0 a = 0 P = 0 h = 40° g is an angle corresponding to the 140° angle, m is its vertical angle, and s its corresponding angle, thus, Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.9 Elements of Geometry 239 * g = * m = * s = 140° The next notion we discuss is similarity of triangles. Two triangles are said to be similar if two angles of one triangle are equal to two of the other. For example, in Figure 6, given the triangles joining the vertices A, B and C, and A ¿ , B ¿ and C ¿ written ¢ ABC and ¢ A ¿ B ¿ C ¿ , suppose the angle at vertex A equals the angle at vertex A ¿ and the angle at vertex B equals the angle at vertex B ¿ then the triangles are similar, sometimes written as ¢ ABC ' ¢ A ¿ B ¿ C ¿ . It follows from the fact that the sum of the three angles in any triangles is 180* that the remaining two angles are also equal to each other. B B' Similarity A' A C C' Figure 6: Similar Triangles Essentially, two triangles are similar to each other if one is a reduced version of the other, or one can be obtained from the other by a uniform compression without distorting any of its angles. Consider Figure 7, A C' B' C B A' Figure 7: Similar Triangles Assume the angles at vertices A and A ¿ equal each other and the angles at vertices at C and C ¿ are also equal. Then these two triangles are similar to each other. Note that one is a uniform compression of the other, the fact that one is also rotated in no way changes the fact that they are similar. Thus, physically, we can describe similarity of triangles (or of any geometric figure) as a uniform compression, possibly along with a rotation, or reflection of one figure. One would expect similar figures to have other related geometric properties as well. The fact that one is a uniform compression of the other, implies that there is a simple relationship between their sides. Consider the two triangles in Figure 8 Assuming the angles at vertices A, B (and C) equal the angles at vertices A ¿ , B ¿ , (and C ¿ ), then the lengths of their corresponding sides, a, b, c, and a ¿ , b ¿ , and c ¿ are in proportion, that is, a b c = = a¿ b¿ c¿ This follows intuitively from the visualization of one triangle being a uniform compression of the other. Since the triangle is being compressed uniformly, it follows that the Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 240 * ** Section 2.9 Elements of Geometry B c' c A b a A' C B' a' C' b' Figure 8: Similar Triangles and Corresponding Sides sides are as well, resulting in the ratio of corresponding sides being in proportion. (This can be proven formally using the notions of parallelism. We omit the details.) The fact that corresponding sides of similar triangles are in proportion is a very useful property that may be used in a variety of applications. Jules Verne, in his novel The Mysterious Island, has the engineer, Cyrus Harding, use this property to compute the height of a cliff. Example 2 Cyrus Harding had provided himself with a straight stick, twelve feet long, which he had measured as exactly as possible by comparing it with his own height, which he knew to a hair. Herbert carried a plumb-line which Harding had given him, that is to say, a simple stone fastened to the end of a flexible fiber. Having reached a spot about twenty feet from the edge of the beach, and nearly five hundred feet from the cliff, which rose perpendicularly, Harding thrust the pole two feet into the sand, and wedging it up carefully, he managed, by means of the plumb-line, to erect it perpendicularly with the plane of the horizon. That done, he retired the necessary distance, when, lying on the sand, his eye glanced at the same time at the top of the pole and the crest of the cliff. He carefully marked the place with a little stick. Then addressing Herbert Do you know the first principles of geometry? he asked. Slightly, captain, replied Herbert, who did not wish to put himself forward. You remember what are the properties of two similar triangles? Yes, replied Herbert; their homologous sides are proportional. Well, my boy, I have just constructed two similar right-angled triangles; the first, the smallest, has for its sides the perpendicular pole, the distance which separates the little stick from the foot of the pole and my visual ray for hypothenuse; the second has for its sides the perpendicular cliff, the height of which we wish to measure, the distance which separates the little stick from the bottom of the cliff, and my visual ray also forms its hypothenuse, which proves to be prolongation of that of the first triangle. Ah, captain, I understand! cried Herbert. As the distance from the stick to the pole is to the distance from the stick to the base of the cliff, so is the height of the pole to the height of the cliff. Just so, Herbert, replied the engineer; and when we have measured the two first distances, knowing the height of the pole, we shall only have a sum in proportion to do, which will give us the height of the cliff, and will save us the trouble of measuring it directly. The two horizontal distances were found out by means of the pole, whose length above the sand was exactly ten feet. The first distance was fifteen feet between the stick and the place where the pole was thrust into the sand. The second distance between the stick and the bottom of the cliff was five hundred feet. These measurements finished, Cyrus Harding and the lad returned to the Chimneys. The engineer then took a flat stone which he had brought back from one of his previous excursions, a sort of slate, on which it was easy to trace figures with a sharp shell. He then proved the following proportions: 15:500 * 10:x From which it was proved that the granite cliff measured 333 feet in height. Verify Harding s calculation. Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. Section 2.9 Elements of Geometry 241 Solution. The engineer made some obvious approximations, but the result certainly produced an approximate determination of the cliff s height. Pictorially, the situation is illustrated in Figure 9. B x (Cliff s Height) D 10 A 15 E |---------- 500 -----------| C Figure 9: Harding s Determination of the Cliff s Height Note that angle at vertex A in ¢ ABC is the same angle in ¢ ADE, and each triangle has a right angle, so the two triangles are similar. Therefore, their corresponding sides are similar, so we have 15 10 = x 500 Giving, 15x = 5000 or to the nearest foot, x = 333 feet. Example 3 A water tank in the shape of a right circular cone is 20 feet high, with base radius 4 feet (see Figure 10). Water is entering the tank at some given rate. When the water depth is 12 feet, determine the radius at the top of the water. Solution. 8 x 12 4 20 Figure 10: Determining the Radius of the Water Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 242 Section 2.9 Elements of Geometry Figure 10, which is not drawn to scale, illustrates the water tank and all given information. By similarity (Verify!), we have 8 20 = x 4 20x = 32 x = 1.6 feet Congruence If two triangles are similar, and if in addition the connecting side in one triangle is equal to the corresponding connecting side in the other, then the triangles are said to be congruent. Therefore, it follows that if two triangles are congruent, their corresponding sides (and angles) are equal to each other (the similarity constant of proportionality is one). It turns out that we can prove two triangles to be congruent with less information. For example, if three sides of one triangle are equal to three sides of another triangle they are congruent, or if two angles and their connecting side in one triangle are equal two angles and their connecting side in another triangle, then the triangles are congruent. We can use the geometric concepts developed above to determine the midpoint of any line segment. Suppose we wish to determine the midpoint of the line AB joining the points 1x1, y12 to 1x2, y22. We proceed as in Figure 11. B(x2, y2) M D A(x1, y1) - x1 E C(x2, y1) Figure 11: Determining the Midpoint Coordinates Midpoint Formula Let the midpoint of the line be denoted by 1x, y2. We draw the dotted lines as indicated and label the points as shown (verify!). Since M is the midpoint of AB, it follows that AM = MB. Moreover, ¢ AME is similar to ¢ MBD (verify this hint: use parallel lines.) Thus, we have two similar triangles with corresponding connecting sides equal to each other 1AM = MB2, therefore these triangles are congruent. It then follows that corresponding sides are equal, giving x2 - x = x - x1 solving for x, we have x = similarly y2 - y = y - y1 x1 + x2 2 Applied Calculus for Business, Economics, and Finance, by Warren B. Gordon, Walter O. Wang, and April Allen Materowski. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education, Inc. 0). forms a traingle similar to the given triangle. Determine the equation of the perpendicular bisector of the line joining the points 11.3). you obtain midpoint formula. The sides of a triangle measure 5. (b) 1 . 8). For the triangle given in the previous example. These lines are called the medians of the triangle. 13. 150 yards and 200 yards. and Finance. . Find the sides of a similar triangle if the side corresponding to the 15inch side of the given triangle is 5 inches. Hint: Choose the vertices of the triangle to be (0. 65o / * + . 16. 70o + * . Consider Figure 13. and April Allen Materowski. * EXERCISE SET 2. 14. Inc. 6. Determine the coordinates of the point P which divides the line segment AB where A is the point 1x1. Published by Pearson Learning Solutions. determine the scaled measurements of the drawing. Given the triangle with vertices A(1. 13). see Figure 14. c) and (a. Determine the coordinates of the midpoint of the line segment joining the points (a) (9. which shows the intersection of two lines. 0).282. . (c) show these medians intersect in a point which is 2/3 of the way from each vertex to the opposite side. * + 11.12. Wang. .192 10. Determine the coordinates of the point P which divides the line segment AB where A is the point (2.32 and (7. The shadow of a tree is 40 feet long at the same time that the shadow of a 4foot flag pole is 2 feet long. Find the sides of a similar triangle if the side corresponding to the 6inch side of the given triangle is 9 inches. The two parallel lines L and M are intersected by T. 9. 12 and 15 inches. 4. which shows the intersection of two lines. Figure 13 3.9 Elements of Geometry * ** 243 and we have y = y1 + y2 2 Thus. show that the line segment joining the midpoints of any two sides is equal in length to one-half the third side. by Warren B. Consider Figure 12. (b. Copyright © 2007 by Pearson Education. B(3. what is its area? 8. The sides of a triangle measure 9. - L M 18. Determine the indicated angles. Figure 12 2. 132 and 12. If the corresponding side of the other triangle is 12. (b) Determine the length of each median. What is the height of the tree. The sides of a triangular parking lot are 120 yards. y22 so that AP/PB = m/n.4) and B is the point (5.Section 2. Walter O.2. (b) 13. the medians intersect in a point which is 2/3 of the way from each vertex to the opposite side. Economics. . Show that for any triangle. 152. A 6 foot man walks away from a light sitting atop a pole 16 feet above ground. Gordon. Suppose two triangles are similar to each other and the first triangle has a side of length 8 inches and its area is 24 square inches. An architect is drawing a scale drawing of the lot with 1 inch representing 50 yards.9 1. 17. (a) How long is his shadow when he is 8 feet from the pole? (b) What is the distance from the base of the pole to the tip of his shadow? 12. 52 and (6. the coordinates of the midpoint of a line segment is the average of the coordinates of the endpoints. Determine the indicated angles. . Find all the indicated angles.12) so that AP/PB = 2/3. Show that a line parallel to the base of a triangle intersecting the other two sides. 6 and 8 inches.72 and 1 . 5.5. 7. 11) and 116. 7). 15. y12 and B is the point 1x2. Verify that in the case m = n. 1). Figure 14 Applied Calculus for Business. Determine the coordinates of the midpoint of the line segment joining the points (a) 1 . and C(9. 120o . (a) Determine the equation of the three lines drawn from each vertex to the midpoint on the opposite side. and the independent variable represents time. W and S. y12 to the line Ax + By + C = 0. 23.10 Related Rates » » » » A Geometric Example An Ecological Example An Economic Example Using Similarity In preceding sections we saw that a derivative may be interpreted as a rate of change. if the dependent variable represents position (distance). Copyright © 2007 by Pearson Education. we shall encounter cases in which several variables are related to each other and are also changing with time. 3). To determine the distance d from the point P1x1. For the triangle with vertices at A(1. (If either was zero. Gordon. B(5.) . Determine this point. drawn to the hypothenuse of a right triangle (the line drawn from the right angle and perpendicular to the hypothenuse) is the mean proportional between the two segments along the hypothenuse. 22. Wang. (b) Show that * RPS = * SWT. In many real situations. . Since the variables are related their rates of change will Applied Calculus for Business. by Warren B. why?) Draw the vertical line from P to T. Published by Pearson Learning Solutions. Show that the slope of a line is independent of the points on the line used to compute it. proceed as follows: Consider Figure 15. Economics. Walter O. the problem is trivial. and April Allen Materowski. (d) Find the coordinates of Q. that is c1/h = h/c2. the absolute value was included as distance must always be positive. 19. Show the statement of the preceding exercise is true for any triangle. Show that the altitude h. see Figure 15 c1 h c2 21. (e) By similarity. then the rate of change may be viewed as velocity. (a) Show that * PSR = * TSW. (c) Show that ¢ PRS is similar to triangle ¢ WOQ. Inc. y1) d Q(0.10 Related Rates Assume neither A nor B is zero. and Finance. In particular. 5) (a) find the equation of the medians and (b) show they intersect in a single point.244 * ** Section 2. 9) and C(9. conclude that d PR = OW QW (e) Substitute for these distances and simplify. to show that d = Ax1 + By + C 2A2 + B2 (Note. -C/B) R S O T W Ax + By + C = 0 Figure 16 2. The medians of a triangle are the lines drawn from a vertex to the opposite sides which it bisects. P(x1. Figure 15 20. Gordon. x. and Finance. that x. by Warren B. requiring the application of the chain rule. 2. y. each variable is now a function of t. 2. t. we see that at any time t. x. and dx/dt = 3 mi/sec. y. after the launch. 3. S z (distance from Launch) y (altitude) L (Launch Site) x (Range) Figure 1: Shuttle Location at time t It should be obvious from the geometry of the problem. we let x be the range. Suppose that a space shuttle S is launched from Cape Canaveral. Remember. and z are not independent of one another. and April Allen Materowski. We may differentiate the equation x2 + y 2 = z2 to find the relationship between the velocities. We need a relationship among the three variables. Thus. . dx/dt. The following specific example will illustrate the way in which the variables and their velocities are related. their rates of change. dy/dt. Its altitude. In order to fix the central ideas. Copyright © 2007 by Pearson Education. A Geometric Example Find: dy/dt when z = 1500 miles and y = 900 miles. At what rate is the altitude changing when the shuttle is 1500 miles from the launch site and at an altitude of 900 miles? Solution. y. let us consider a typical case. we see a diagram showing the location of the shuttle at some time. y the altitude. In other words. z. We have the relationship x 2 + y 2 = z2. are all changing with time. Consequently. we have a right triangle. 1. Differentiating this equation with respect to t. It is important for you to observe that we attack the problem in three distinct steps: 1. Wang. In Figure 1. Inc. Using Figure 1. and straight line distance from the launch site.10 Related Rates * ** 245 be related and hence problems of this type are usually referred to as related rates problems. Walter O. Economics. its range (ground distance from the launch site). Published by Pearson Learning Solutions. we find that 2x dy dx dz + 2y = 2z dt dt dt (1) Applied Calculus for Business. It is a good idea to reformulate the problem as follows: Given: dz/dt = 8 mi/sec. and z the distance from launch at time t. Looking at Figure 1. Differentiate each side of the equation to obtain an equation relating their rates. and dz/dt are not all independent either. Example 1 A space shuttle is launched and is moving away from the launch site with a constant velocity of 8 miles per second. Substitute the specific numerical data to solve for any unknown quantity. Suppose that the range is increasing at the rate of 3 miles per second. Determine an equation that relates the variables in general. and z are all functions of t.Section 2. it contains. The next example illustrates such a case. dN/dt when a = 5 lb. . we must use the physical nature of a problem to determine a necessary equation relating the variables. Differentiating the equation N = 13a2 .02 per week. Published by Pearson Learning Solutions. the altitude is increasing at a rate of 9 1 3 mi/sec. as the next ecological example illustrates. and Finance. Find. revenue. Copyright © 2007 by Pearson Education. If the quantity of algae is increasing at a rate of 2 lb/week. Gordon. by Warren B. N. Example 3 The retail price per gallon of gasoline is increasing at $0. are each functions of time. and April Allen Materowski. Sometimes the equation is given.20 # 22 = . this is the case. What does this suggest to you about the amount of algae in the stream and about its rate of increase? An Economic Example Economic problems often fall into this category. An Ecological Example da/dt = 2 lb/week. The equation is given by N = 13a2 . at what rate is the fish population changing when the pond contains 5 lb of algae? Solution. we have that dN/dt = 413[5]2 . in a small pond depends upon the number of pounds of algae. using the chain rule.20a + 2624.246 Section 2. Substitution into (1) yields 2112002132 + 219002 dy/dt = 2115002182 or dy/dt = 9 1 3 mi/sec. dN/dt = 413a2 . Example 2 The fish population.20[5] + 262316 # 5 # 2 . 2. and profits are related to one another and are always changing with time.x2 = 0 where p is the price per gallon.2356 . Economics.20a + 2624. x = 215002 . 3. Given. That is. we have. We may now complete the problem. When the equation was differentiated. cost. Inc. The demand equation is given by 10p . Wang. We note that when z = 1500 and y = 900. Often.20a + 262316a # da/dt .20 # da/dt2 3. In the above example. Substituting a = 5 and da/dt = 2. N and a. Walter O. That is. the fish population is increasing by 80 per week at the instant when the stream contains 5 lb of algae. since we have been given the relationship between N and a. For example.9002 = 1200. Step 1 has already been completed. we needed to use the geometry of the problem to find an equation which related the variables. it gave their related rates.10 Related Rates Notice that each term is multiplied by a derivative term which resulted from the application of the chain rule. when Applied Calculus for Business. a.80 per week Thus. in dollars. dR/dt = x dp/dt + p dx/dt Now. we have that dR/dt = 1010. dp 1 1 dx = 1356 . finding p = when x = 10 p = 1/10125621/2 = 1. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education.02 dollars per week Find: dR/dt when x = 10 million gallons.x 22-1/2 a . substitute into (2) and have R in terms of x. Figure 2 shows the essential features of the problem. Note that Using Similarity There is a large class of geometrically-based related rates problems that use the properties of similar triangles as examined in Section 2.0.6. Gordon. 3.022 + 1. Substituting. we find that dx/dt = . x + y x = 5 12 Cross-multiplying.02 and x = 10. 12 5 C y D x A E Figure 2: Girl Walking away from a Lamppost Applied Calculus for Business. and Finance. the revenue is decreasing by about 0. their sides are in proportion.x21/2.0. B (Top of Lamppost) Example 4 There is a light atop a 12 foot high lamppost. 1. let us take a different tack. Let us differentiate using the product rule.61 . by Warren B.32 million gallons per week. thus. How is the length of her shadow changing? Solution. We find dx/dt by differentiating the demand equation. However. Economics. 1 10 1356 . 12x = 5x + 5y. Inc.31 million dollars per week.0.Section 2. We may now complete the solution. Our last example in this section is typical of this type. Using the chain rule. Wang. when dp/dt = 0. 2. the ratio of x to 5 is the same as the ratio of 1x + y2 to 12. That is.2x b dt 10 2 dt Thus. we solve the demand equation for p. . and April Allen Materowski. we have. Therefore.322 = .312 That is. (The negative sign indicates that the demand is decreasing). At what rate is the revenue changing when 10 million gallons are demanded? Solution. The triangles ABC and AED are similar. The total revenue in millions of dollars is given by the equation R = xp (2) We could solve the demand equation for p in terms of x. In symbols. Here y is the girl s distance from the base of the lamppost and x is the length of her shadow. Given: dp/dt = 0. A five-foot-tall girl is walking in a straight line away from the light at a rate of 6 feet/sec.8.10 Related Rates 247 x million gallons are demanded. Walter O. 1994.) Suppose that the average price of a VCR during the same time period was governed by the rule p = 820 . and 2000. 20. at what rate is the: (a) demand changing. 7 dx/dt = 5 dy/dt. EXERCISE SET 2.25 per week. where t is the time in years. (b) diagonal changing. The demand for apples is given by the equation xp = 150. If the price is increasing at a rate of $0. its volume is 4 cubic inches. The radius of a circle is increasing at the rate of 1 inch per hour. (a) At what rate is the area of the circle changing when the radius is 4 inches? (b) At what rate is the circumference changing when the radius is: (i) 4 inches. when the demand is 10 pounds? (Can you solve (b) by inspection?) 17. and Finance. Published by Pearson Learning Solutions.) 21. 6) is increasing at a rate of 7 units per second. find dy/dt when x = 2 and y = 1. find dy/dt when x = 5 and dx/dt = . 2). (ii) 5 inches? 10. At what rate is a side changing when it is 1 inch long? 11. miles per hour. Walter O. 1. 8.50t. cubic feet per minute.) 23. If y = x3 . 7x = 5y 2. with 1990 being t = 0. If dQ/dt = 6 and dx/dt = 2. 1994. at what rate is the string being paid out when its length is 130 feet? 14. Suppose that the oil spill from the damaged hull of a ship forms a circular slick whose thickness is uniformly 1 millimeter and whose radius is increasing at a rate of 20 kilometers per day. If y = 2x . where r is the radius and h its height. Inc. At what rate is the area of the circle changing when the radius is 12 feet? 2 -3 3 2 16.10 1. At the instant when the radius of the oil slick is 15 kilometers: (a) Determine the rate at which the spill is flowing out of the ship. (a) Find the number sold in 1990. we have the relationship between x and y. If y = x3 . Find dx/dt at (1.6 6. The area of a square is increasing at the rate of 1 square inch per minute. Gordon. 7 dx/dt = 152162. A point moves along the circle x2 + y2 = 5 in such a way that the distance from the point (4. find the rate at which the pressure is changing when the pressure is 20. Finding the derivative. For a particular gas it is known that when the pressure is 10. (b) Find the rate of change of sales in 1990. Suppose that Q = x y . determine dx/dt when x = 2 and dy/dt = 4. and April Allen Materowski. find dx/dt when x = 8 and dy/dt = 64.2. If the volume is increasing at 8 cubic inches per second. where P is the pressure in pounds per square inch (psi) and V is the volume of the gas. (The 3 volume V of a sphere of radius r is given by the formula V = 4 3 pr . Two cars leave an intersection at the same time. For example. If x2 + y2 = 169. Water is flowing into a conical reservoir 50 feet high with a top radius of 10 feet. by Warren B. At what rate is the bottom of the ladder moving away from the wall when the top of the ladder is 10 feet above the ground? 15. when the side is 6 inches long? 12.000 psi. (b) Find the rate of change of total revenue for 1990. At what rate is the tip of his shadow moving towards the lamppost when he is 5 feet from the lamppost? Applied Calculus for Business. 3. 1994. Economics. (b) At what rate is the distance between the two cars changing at the end of two hours? 9. where x is the number of pounds demanded and p is the price per pound. 19. A pebble thrown into a pond creates a circular ripple whose radius is increasing at 2 feet per second. The number. 2. Notice that a rate has units that are per unit time. dollars per month etc. If Q = x/y and R = xy.22x. and 2000. and 2000. and the other travels east at 30 miles per hour. Boyle s law for gasses states that PV = constant. At what rate is the depth of the water in the reservoir changing when the water is 10 feet deep? (The volume 2 of a cone is 1 3 pr h. find dy/dt when x = 2 and dx/dt = .10 Related Rates 7x = 5y Hence.000 psi. If z represents the distance between the cars. A kite maintains a constant altitude of 50 feet and is moving horizontally at the rate of 2 feet per second. find dx/dt and dy/dt if dQ/dt = . 1994. dx/dt = 30/7 Since x is the length of the shadow. Wang.3. One travels north at 40 miles per hour. (b) revenue changing. 22. At what rate is the radius of the balloon changing when the radius is 2 feet. at the rate of 16 cubic feet per second. and 1 millimeter = 10-6 kilometer. 7. determine dy/dt when x = 2 and dx/dt = 3. A large cube of ice is melting uniformly at the rate of 6 cubic inches per second. If y = 23x2 .3 and dR/dt = 9. Copyright © 2007 by Pearson Education.3x.248 Section 2. (a) describe how the rate at which the cars are separating is related to the rates at which they are traveling. we see that the shadow is getting longer at a rate of 30/7 feet/sec. and 2000. 4. feet per second. and substituting dy/dt = 6 3. 18. N (in million) of VCR s sold in the United States for the years 1990 to 2001 may be approximated by the formula N = 112t + 12/1t + 22. At what rate is the: (a) side of the square changing.) 13. (Continuation of previous exercise. (b) What effect does a 12 hour time delay in plugging the leak have on the spill? (Note: the volume of a circular disc of radius r and thickness d is p r2d. gallons per week. A 26 foot ladder is leaning against a vertical wall and its base is on level ground. A 6 foot tall man is walking towards a 24-foot lamppost at the rate of 8 feet per second. It is usually very easy to determine the given rates in a specific example. Assume that there is no slack in the kite-string. (a) Find the price and total revenue from sales for 1990. 5. . The top of the ladder slides down the wall at a rate of 2 feet per minute. when x = 6 and y = 3. A spherical balloon is being deflated at a rate of 10 cubic feet per second. Wang. and Finance. Its area is growing at the rate of 10 cm2/sec. Economics. An alternative way to do implicit differentiation is to introduce a mythical extra variable t. Think of x and y as both functions of t. find the relationship between the rates of change of x and y as in the previous section. Published by Pearson Learning Solutions. this tangent line has no x-intercept. (c) Use the result of parts (a) and (b) to find y ¿ . called the first iterate. Find the rate of change of each side. To solve the equation f1x2 = 0 we need only find those x-values at which y = 0. (This first choice may be obtained from a sketch of the curve y = f1x2 and choosing x0 as the approximation to the actual intercept. we obtain x1 = x0 .f1x02/f ¿ 1x02 (Note that if f ¿ 1x02 = 0. Is it getting longer or shorter? 25. (b) For the equation x2 + y 3 = 6. . Now. y = f ¿ 1x021x . find the rate at which the length of the man s shadow is changing. At a certain time one side is 4 m long and the other side is 3 m long.11 24. x1 is found by setting y = 0 (see Figure 1). why?) Newton s Method y y = f( x ) r x3 x2 x1 x0 x Figure 1: Illustrating Newton s Method Applied Calculus for Business. At a certain time one side is 50 cm long and the other side is 20 cm long. Are they growing or shrinking? Newtons Method * ** 249 27. Gordon. Walter O. 28. The perimeter of a rectangle is growing at the rate of 12 m/sec.4xy + 7y3 = 5 at 12.x02 + f1x02 The x-intercept of this line. Inc.Section 2.12. Use the method developed in the previous exercise to find the equation of the line tangent to x2y 2 . Find the rate of change of each side. Consider the smooth function defined by the equation y = f1x2.11 » » Newton s Method Newton s Method Calculator Tips This section deals with Newton s Method for determining the roots of the equation f1x2 = 0. . Its diagonal is growing at the rate of 4 m/sec. f1x022 is given by the equation. that is the x-intercepts of the curve. The perimeter of a rectangle is growing at the rate of 6 cm/sec. We begin by selecting any value that is near a root (x-intercept) and call this value x0. by Warren B. For the case described in the previous exercise. 2. and April Allen Materowski. Are they growing or shrinking? 26. use the fact that dy dt = dy dx dx dt (a) Solve for dy/dx.) The equation of the tangent line to the curve at the point 1x0. Copyright © 2007 by Pearson Education. x1.2 2xN . (called 1f1x222 the second derivative). We obtain y = f ¿ 1x121x . and April Allen Materowski.1 Suppose we choose as our first guess. We illustrate the method in the following example. Notice.1 x2 N-1 . x0 = 1. where f 1x2 is the derivative of the derivative. This is a repetitive process.5. Note that when N = 3 the fifth decimal place no longer changes.41412356. thus to four places. then we have f ¿ 1x2 = 2x and (1) becomes xN = xN .1C4¿2 . We copy this formula to the succeeding cells in its column to automate the process.11 Newtons Method Now we determine the equation of the tangent line to the curve passing through the point 1x1. Let f1x2 = x2 . and Finance. so the easiest way of performing the calculations is with a calculator or even more easily with a spreadsheet. generating a sequence of x-intercepts (iterates). Gordon.12 (1) For many well behaved functions xN is an excellent approximation to the root of f1x2 = 0. it can be shown in a more advanced course that this method will produce a root f1x2f 1x2 if ` ` 6 1 near the root. In fact. x3. where we can copy one row to the next and automate the calculations.) Applied Calculus for Business. that is to how many decimal places do we want the approximate root? Suppose we want the root to five decimal places. the next iterate (approximation). the zero is 1. the second iterate. the formula (1) in the form * C4 .22/12*C42. Walter O. x2. Wang. the calculation is identical to the one above with x0 being replaced by x1. xN where xN = xN . we need to decide when to stop. When the sequence of iterates approaches the root (converges). like Excel.f1x12/f ¿ 1x12 We repeat (iterate) this process. (Note that we just computed the square root of 2. Example 1 Determine the positive root of the equation x 2 . Solution.2. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education. Á .12/f ¿ 1xN . then we stop when the sixth decimal place no longer changes with the next iteration. by Warren B. It is clear that an excellent approximation to the zero of the equation is 1. f1x122.2 = 0. then we generate the following spreadsheet indicated in Table 1.1 .f1xN . x2 is x2 = x1 .x12 + f1x12 and its x-intercept. Economics. .4142.250 * ** Section 2. Inc. x1 is obtained by entering in the spread sheet in the cell C5. the cell C4 is x0. and April Allen Materowski.212/ 120*C4¿3 .21.14x + 11 so we have 3 2 5x4 N .1 .21 2 20x3 N . Published by Pearson Learning Solutions. We need a first guess.2*C4¿3 .7x 2 + 11x .Section 2.1 .14*C4 + 112. Observe.2 Example 2 Determine the positive root of f1x2 = 5x4 .15*C4¿4 .6x 2 . Walter O. In Example 2. Table 2: Finding a Root of f1x2 = 5x 4 . so there must be a root between 1 and 2 (why?).1 + 11 xN = xN . the sequence generated by the method moves away from the root. Newton s method doesn t always work. Copyright © 2007 by Pearson Education.1 - (1) We put this into a spreadsheet. Note that the entry which does all the calculations is cell C5 its formula is = C4 .14xN . we leave it as an exercise for you to locate it using Newton s method. Inc.2x 3 .1 . . f112 = . Solution. see Table 2.1 . we shall locate it below using the TI 89.21 Newton s method may be used directly on the calculator by realizing that each iterate is a composition of the previous. f ¿ 1x2 = 20x3 .6*C4¿2 . and Finance. Gordon. Wang. We examine these cases in the exercises. We choose x0 = 1.7x N .7*C4¿2 + 11*C4 .1 . we have.6x N .7x 2 + 11x .14 and f122 = 37. Thus.1 + 11xN . rather than towards it.2x 3 . to seven decimal places. This entry is then copied into the cells beneath it.5367405 The function defined in Example 2 also has a negative root.2x N . It can also happen that the sequence cycles. Sometimes. again not approaching a root. by Warren B. define y1(x) as Applied Calculus for Business.11 Newtons Method 251 Table 1: Using Newton s Method to Approximate the Positive Root of f1x2 = x 2 . Economics. the root is 1. 7. f1x2 = 5x4 .21 20x3 . EXERCISE SET 2. It avoids the use of the derivative. It is usually a good idea to have it sketch the graph first to get a visual location of the root you want to find and then use the solve command along with the with feature to obtain the desired root. that is.2x 3 .3x 3 . Wang. Published by Pearson Learning Solutions. and April Allen Materowski.3 to six decimal places.3.7. 8. Find all the roots of f1x2 = x3 + 3x2 .1. Find r. and Finance.7. and determine its x-intercept.2x 3 .3 to six decimal places. compare with the results in Exercise 5.1. obviously. Explain why x0 = 3 will not produce a root of f1x2 = x 3 . 14.1 (secant) line through the points 1xN. Another way of using the calculator to locate the root is to first have it sketch the graph.4x + 3. f1xN2 . Applied Calculus for Business. Try Newton s method with x0 = 1.22. We enter Figure 2: Newton s Method and the TI 89 solve15x ¿4 . that is. what happens if you try to use x0 = 2 to locate the root of this function. Use the Secant Method to find the roots of the function of f1x2 = x3 + 3x2 .10. Actually. suppose we want to find the negative root of the function given in Example 2.000 car loan is paid off with $250 monthly payments over three years. Consider f1x2 = 1x .11 + i2-36 i b. x3 = y11y11y111222 and so on. x2 = y11y111. that is. Newton s method can be used to approximate reciprocals. Gordon. 3). give an approximate factoring of the quartic polynomial. An $8.11 1. Hint: rewrite the above expression as a polynomial. and for upper bound you move the cursor to the right of the desired root and press enter. but this will not always be the case. Continuing this way will improve the accuracy of the iterates to the root. by Warren B. (b) 1/9. 12. you move the cursor to the left of the desired root and press enter. What happens to the successive iterates? 10. to six decimal places. what if x0 6 2? 9.20x2 + 4x + 1 to four decimal places. x0 is actually the root of f(x). Suppose you want to approximate 1/b.6x 2 . Let f1x2 = x3 . it produces both roots. then x1 = y111. Copyright © 2007 by Pearson Education. 13. f1xN .12 f1xN2 .21 = 0. with x = 1 + i. The required approximation to the root is then displayed. Find the fifth root of 25. (b) Determine the equation of the tangent line to this curve at x = 2. Why does this produce the root? 3.6x2 + 12x . What seems to be happening? Try to explain it. 5.122. and then press F5 to locate the required zero. Inc. A variation of Newton s Method is called the Secant Method. where r = 12i. Find the fourth root of 7 to six decimal places.2x ¿3 . x2 x 6 0 The calculator gives.7x2 + 11x . suppose you take x0 = 2.14x + 11 this is essentially (1) with the subscripts deleted.121/3. then let f1x2 = 1/x . (a) show that a root of this function lies in the interval (1. see Figure 2.xN . 11.2. x = 1 is the root. It replaces the derivative by the approximation. the derivative is replaced by the slope of the f ¿ (xN) L xN . after a few seconds be patient it can take a few seconds (or longer) for more complicated functions. Find the roots of f1x2 = x4 + x .11 Newtons Method Calculator Tips y11x2 = x - 5x 4 . (e) Continue the above and produce a root of the given function to six decimal places. they each need to be determined. Thus.xN . to six decimal places.f1xN12 . we have xN + 1 = xN f1xN21xN . Use this observation to compute (a) 1/6. Economics. if you forget to enter the with condition. Then if x0 = 1.21..b. 6. .7x ¿2 + 11x . x0 and x1. Walter O. (d) Determine the equation of the tangent line to the curve at the x-intercept found in (c) and its x-intercept. f1xN22 and 1xN . For example. (c) Determine the equation of the tangent line to the curve at the x-intercept found in (b) and determine the x-intercept of this tangent line. Given f1x2 = x2 . 2. The TI 89 has a variation of Newton s method built into its solve command. 4. what monthly interest rate r is the bank charging? It can be shown that this problem translates into the equation 8000 = 250 a 1 . (a) Find all the roots of f1x2 = 5x4 . (b) Using the roots.1. . Observe that this function has more than one root.f1xN . Hint: find a zero of x 4 .9x2 + 27x + 5.12 for N Ú 1 Note that this means we need two starting values near the root.2x . you need to be careful that it locates the root you are looking for. f1x2 = x3 . Determine the root of this function. For the lower bound. why? Find its root. what happens why? Suppose you take x0 7 2 what happens. However. Suppose you make a lucky starting choice. to six decimal places.7x 2 + 11x . (c) 1/13.252 * ** Section 2.7032256. find the coordinates of the point P. and Finance. 5. Find the equation of the tangent line to 1x2 + y221/2 . The bottom of the ladder is being pulled away from the wall at a rate of 4 feet per minute. At what rate is his shadow changing when he is 15 feet from the lamp post? 18.1).x2.9 x 20. 22. In Exercises 19 21: (a) Sketch the graph of the given function. 3).9 x:2 x .2x2.2x . ( a) g1x2 = (b) h1x2 = 1x . The line segment connecting the points A(5. .82 3 x + 12 (b) Find the equation of the tangent 8.24 at the point 12. Find the derivative of the indicated function. g1x2 = x + 3 if if if if if x 6 1 1 6 x x 0 0 6 x 6 1 1 x x2 .2x4 + 4x 3 7x 2 + 11x . f1x2 = c 2x . Calculate the indicated limits: (a) lim 2 . where x is the number of items sold at price p. (b) Find the approximate revenue obtained by producing and selling the 7th VCR. 16. Using Newton s Method.x2 6.15 to five decimal places. Applied Calculus for Business. A 15 foot ladder is leaning against a vertical wall and its base is on level ground. Inc. Published by Pearson Learning Solutions. If the position of a particle at time t is given by the equation s1t2 = 6 + 3t2 . (b) lim 2 x : -3 x . Gordon. (a) By using the definition of the derivative. find y ¿ if y = equation of the tangent line to the curve at the point (3. how high is the grain piled? 15. where p is in hundreds of dollars: (a) Find the total revenue function.1.12 13x + 42 2 5 6 x3 + 1 B 2x3 + 5 . (c) Find those values of t for which the velocity is positive. 10. by Warren B. Wang. 9) and B( 12. At what rate is the top of the ladder descending when the top of the ladder is 12 feet above the ground? 17.3x2 + 1x gent line at 11. 20) is to be divided at a point P so that the ratio of the distances AP/BP is 5/8. (a) Find f ¿ 1x2 if f1x2 = x2 2x 3 + 1 (b) Find the equation of the tangent line to the curve at the point where x = 2. . If the total cost of producing x toaster-ovens is given by the formula C1x2 = 40 + 12x + 0. has a 10 foot base radius. 4). .x .2x + 4y = 15 at the point (3. (a) Find f ¿ 1x2 if f1x2 = (b) Find the equation of the tangent 2x + 10 line to the curve at the point where x = . If a ball is thrown upward with initial velocity 28 feet/sec. Walter O. f1x2 = c 0 1 . (b) For what values of x is the function discontinuous? (c) For what values of x is the function not differentiable? (d) If any point of discontinuity is removable. (a) Find y ¿ for y = 2x4 . 11. 3 x2 9. Calculate the indicated limits: (a) lim x + 2 x . (a) Find y ¿ for y = line at (1.22.Chapter Review * ** 253 CHAPTER REVIEW Key Ideas Slope of a curve Tangent Line Derivative Constant Multiplier Rule Sum Rule Product Rule Quotient Rule Limit Continuity Discontinuity Removable Discontinuity Differentiable Composite Function Chain Rule Extended Power Rule Marginal Function Geometric Concepts Rate of Change Related Rates Implicit Differentiation Newton s Method 1. Find the equation of the tangent line to x 2y .4t3: (a) Find the velocity as a function of time. and April Allen Materowski.52. 3. find the approximate cost of producing the 30th toaster-oven. 19. (c) Find the point(s) at which the tangent line is horizontal. If the demand for VCR s is described by the equation 5p + 3x = 60. Economics. 2.1 x2 21.x 5 x2 (b) Find the equation of the tan- 5 x4 . 12. A 61*2 foot tall man is walking away a 13-foot lamp post at the rate of 5 feet per second. and it strikes the ground 10 seconds later: (a) How far above the ground was the ball when it was first thrown? (b) What was the highest point the ball ever reached? (c) How fast was the ball moving when it hit the ground? 14. Copyright © 2007 by Pearson Education. (b) Find the times and locations at which the velocity is zero. When the radius of the enclosed grain is 8 feet. (b) lim 2 x:5 x + 5 x + 1 x 2 + 3x 4 . find y ¿ if y = 6 . find the root of f1x2 = 3x5 . 23. redefine the function at that point to make it continuous and determine whether the function as so redefined is differentiable at that point.9). (a) By using the definition of the derivative. 4. x:2 2 3 4 (b) Find the x + 1 13.3xy 2 = y 3 . (b) Find the equation of the tangent line to the curve at the point (1.2 7. A 20 foot high grain silo in the shape of a cone. Wang. by Warren B. and Finance. Walter O. Copyright © 2007 by Pearson Education. Inc. Gordon. Published by Pearson Learning Solutions. and April Allen Materowski. . Economics.Applied Calculus for Business. Gordon. we introduce the notion of the differential and see how it is used for approximations. Wang. In particular. we begin a more detailed investigation of some of these applications. That is. Walter O. Applied Calculus for Business. In this chapter. These methods will be applied to geometric problems and examples from Economics and Finance.6. In Section 3. . and Finance.3 Applications of the Derivative In Chapter 2. we shall be concerned with finding extreme values. Inc. Published by Pearson Learning Solutions. we shall be looking for ways to find the largest and smallest possible values attained by a given function. and April Allen Materowski. by Warren B. Economics. we developed rules for finding the derivatives and began to see some of the uses to which the derivative may be put. We shall also see how limits and derivatives can be used to help us in curve sketching. Copyright © 2007 by Pearson Education. Since polynomials are functions without holes or jumps. . In the business world it is usually desirable to maximize profit and/or minimize cost. Similarly. one would want to use the least amount of material. Before we begin it is useful for us to recall what is meant by a continuous function.1 Extrema of a Function 3. It is sufficient for this chapter to think of a continuous function as one whose graph has no holes or jumps. Walter O. Wang.256 * ** Section 3. To illustrate a discontinuous function. a rational function (the quotient of two polynomials) is continuous at every point at which its denominator is not zero. In Section 2. There are numerous applications of optimization.1 Extrema of a Function » » » » » » Continuity Maximum and Minimum Values Extreme Value Theorem Relative Maxima and Minima Critical Numbers and Critical Points Calculator Tips Continuity The dictionary defines optimum to be the best or most favorable degree or amount. Copyright © 2007 by Pearson Education. in designing containers of fixed volume. we shall consider the problem of optimizing a non-linear function of a single variable. by Warren B. For example. Gordon. In Figure 1. it is often a straight-forward matter to determine the optimal solution. Later. We must first develop the necessary tools with which to perform the analysis. they are everywhere continuous functions. Inc. we have a function which has a hole at x = 2. We shall discover that once we model an application by a mathematical equation. Each point in the domain at which there is a hole or jump in the function s graph is called a discontinuity.3 a precise definition was given. and a jump at x = 4. and April Allen Materowski. At 2 4 Figure 1: A Discontinuous Function with a Hole at x = 2 and a Jump at x = 4 Applied Calculus for Business. In this section. Published by Pearson Learning Solutions. we need only construct a function which has either a hole or a jump at some point in its domain. To optimize is to find an optimum or optimal value. and Finance. Economics. we shall generalize the problem of optimization to cases in which a function may depend upon several variables. To optimize a function means to find its largest or smallest possible value. ) We note that some texts use the terminology absolute extrema. Similarly. Thus. but it never attains the y-value 2. B(b. For such cases. the function is discontinuous. by Warren B. DEFINITION 2 A function f has f(c) as its maximum value on the interval a 6 x 6 b.00000001 and 2. Consider f1x2 = 2x2 on the open interval 1 6 x 6 2. A function f has f (g) as its minimum value if f1g2 f1x2 for all x in its domain. the minimum value is the smallest possible y-value attained by the function over its domain. f (g)) Figure 2: Illustrating Maximum and Minimum Values DEFINITION 1 A function f has f (a) as its maximum value if f1a2 Ú f1x2 for all x in its domain. Consider the graph of the function given in Figure 2. or half-open. f (e)) D(d. f(e)) and G(g. Sometimes. f (c)) E(e. This may at first surprise you. However. This means that the maximum value is the largest possible y-value attained by the function over its domain. Notice that we have labeled the points A(a. We say that this function has its maximum value occurring at x = a and its minimum value at x = g. to distinguish an extremum from a relative extremum (which we consider below). you see why this happens. Note that this interval does not contain x = 1 or 2. You can see that A is the highest point on the graph and G is the lowest. When can we be certain that a function will actually attain both its extrema? The above example Applied Calculus for Business. we would have to have x be 1 but then x would not be in the interval. A function f has f(g) as its minimum value on the interval a 6 x 6 b.Section 3. . but never there. but it is not defined at either end point. Copyright © 2007 by Pearson Education. Its maximum value is f(a) and its minimum is f(g). Gordon. Economics. The function is defined and continuous on the open interval. but it is possible that a function may never achieve its extrema. The interval may be open or closed. f(a)). we shall not do so here. It gets arbitrarily close. Walter O.000000000001. Maximum and Minimum Values A(a. f(a) is the highest y-value attained by the function. f (b)) C(c. it attains any y-value with as many zeros as you like after the decimal point followed by a 1. Wang. E(e. if f1g2 f1x2 for all x in this interval. (The plural of extremum is extrema. D(d. between 1 and 2. Similarly. intuitively. C(c. when we want to refer to either a maximum or minimum we use the term extremum. Its precise mathematical definition may seem confusing at first. f(d)). and f(g) the lowest y-value. It gets very close. In order for y to reach 2. and in fact. Of course. continuity is a simple notion. Not every function has a maximum and/or minimum value. this function never attains a y-value of two. Therefore. We sketch this function in Figure 3. f(c)). this function never attains the y-value 8. we shall make a small change in Definition 1. for example it attains the y-values 2. and Finance. f(g)). and April Allen Materowski. we have the following definition. Published by Pearson Learning Solutions. More generally. f (a)) B(b. that is. f (d )) G(g. f(b)).1 Extrema of a Function * ** 257 each of these x-values. if f1c2 Ú f1x2 for all x in this interval. Inc. There are cases in which we shall be concerned with the extrema of a function on some interval not necessarily the entire domain of the function. For simplicity we shall state the revised definition for an open interval. x2 has a maximum value. Published by Pearson Learning Solutions. then it attains both a maximum and minimum value on this interval. . f has its maximum value. Wang. or possibly a or b themselves.1 Extrema of a Function o (2.1). THEOREM 1: THE EXTREME VALUE THEOREM If the function defined by y = f1x2 is continuous on the closed. Applied Calculus for Business. We need to keep the function continuous and include the end points of the interval. that at the vertex of the parabola V(0. Solution We sketch the function. which is the parabola in Figure 4. Economics. bounded interval a x b.x2 It is clear from our sketch. only that they exist. Inc. that is. at which the function has a maximum and a minimum. That is. 2) o Figure 3: f1x2 = 2x2 on 1 6 x 6 2 indicates the answer. and April Allen Materowski. This suggests the following theorem. Walter O.258 * ** Section 3. 8) (1. It does not tell us how to find the extreme values. Copyright © 2007 by Pearson Education. by Warren B. known as the Extreme Value Theorem. its maximum value is 1. This theorem says there are x-values between a and b. Gordon. close the interval. V(0. We shall soon see that they are easily found. and Finance. Note that this function does not have a minimum. Example 1 Show that the function f1x2 = 1 .1) 1 -2 -1 0 -1 -2 -3 0 x 1 2 Extreme Value Theorem Figure 4: f1x2 = 1 . there may be someone at a higher elevation. nearby (locally). A B Figure 5: A Discontinuous Function with a Maximum and Minimum We would like to find those points at which the function attains its extrema. get larger). and Finance. its y-values increase (that is. A function may even be discontinuous on an open interval and yet attain a maximum and minimum. While the function is everywhere continuous. Thus. and April Allen Materowski. Theorem 1 does not exclude such situations. The precise definition of relative extrema for a function defined on a 6 x 6 b follows. and at any point at which it has a valley it has a relative minimum. by Warren B. because each is the lowest point in some vicinity of itself. it just tells us what conditions are needed to guarantee attainment of the extreme values. However. then the function may or may not attain extreme values. its domain is not a closed interval. a function is said to decrease on an interval. f(c) is locally a maximum. It has a relative minimum value at f(c) if there is an open interval a 6 x 6 b which contains c. If the conditions of the theorem are not met. the function also has a relative maximum at E.1 Extrema of a Function * ** 259 Why not? Note that the Extreme Value Theorem is not applicable here. Economics. The function whose graph is given in Figure 5 attains its maximum value at its turning point A and its minimum at its endpoint B. no one is higher than you. its y-values Applied Calculus for Business. . each of the points B and D is a relative minimum. Similarly. Published by Pearson Learning Solutions. it is the highest point.Section 3. if as we move from left to right on that interval. if as we move from left to right on that interval. and f1c2 f1x2 for all x in this interval. if you were standing on top of a mountain. Copyright © 2007 by Pearson Education. since locally or relative to its neighboring points. In other words. we must first examine Figure 2 a little more closely and introduce a few more definitions. Near C. As above. if we wish to refer to either a relative maximum or relative minimum we shall use the term relative extremum. Points C and E resemble mountain peaks. We would really like to have a simpler way to pin down relative extrema. Inc. as asserted by Theorem 1. Similarly. Relative Maximum and Minimum Values DEFINITION 3 A function defined by the equation y = f1x2 has f(c) as a relative maximum value if there is an open interval a 6 x 6 b which contains c. Gordon. Following the same reasoning. Wang. and f1c2 Ú f1x2 for all x in this interval. Before we can. Recall that a function is said to increase on an interval. The function in Figure 2 is said to have a relative maximum at C. further away. it has a relative maximum. For example. at any point at which the function has a peak. Walter O. Perhaps the easiest way to do this is by noticing where a function is increasing or decreasing. sufficient conditions to guarantee extrema. the function is decreasing between A and B. and April Allen Materowski. Wang. to find the maximum and minimum values of a continuous function over a closed interval. get smaller). it is the way most of us think of them. (Why must we exclude functions with plateaus?) A B Figure 6: A Function with a Plateau Now that we have defined a relative extremum we may re-examine Theorem 1. However. we need only examine the y-values at the endpoints and at the relative extrema. or c reveals that they may also occur at interior points of the interval. between C and D. and between E and G. 4 5 4 3 2 2 1 1 0 y 2 1 0 5 4 3 3 0 0 1 2 x 3 4 0 0. Inc. RULE OF THUMB A function defined by the equation y = f1x2 has a relative maximum at (M. It is increasing between B and C. Where may the extrema of a continuous function occur? From Figure 2 we see that they may occur at the end points of its interval of definition. on its left it is decreasing and increasing on its right. Copyright © 2007 by Pearson Education. b. and between D and E. this Rule of Thumb could serve as an equivalent definition of a relative maximum and a relative minimum. Economics. At a valley . Gordon.5 2 0 1 2 x 3 4 Figure 7a Figure 7b Figure 7c Applied Calculus for Business.260 * ** Section 3. such as in Figure 6 . f(M)) if the function is increasing just to the left of M and decreasing just to the right of M. In Figure 2. Except for extraordinary cases where the graph actually has a plateau. Published by Pearson Learning Solutions. f(m)) if the function is decreasing just to the left of m and increasing just to the right of m. Notice that just to the left of a peak the function is increasing. Walter O. by Warren B. If an extremum occurs at an interior point.5 1 x 1. as asserted by the Extreme Value Theorem. Indeed. Thus. . The function has a relative minimum at (m. a glance at Figure 7a. and Finance. and just to its right it is decreasing. the reverse occurs. then it will also be a relative extremum.1 Extrema of a Function decrease (that is. Inc. and April Allen Materowski. or. To find the y-coordinates. from Figure 8 we see that not every critical point is a relative extremum.6x . Gordon.5 P 2 1. by Warren B. The function in Figure 8a has a horizontal tangent line at the point P. It follows from this definition that every relative extremum is a critical point. The tangent line is horizontal at any point where the derivative is zero.5 1 1. However. as at points D and E (Figure 2). f(c)) is called a critical point. and Finance. If c is a critical number for the function defined by y = f1x2 then the point (c. At the point P in Figure 8b the function has a vertical tangent line. At the point P in Figure 8c the derivative fails to exist. Then. This will be zero if x = .5 3 2. First we shall locate all points where the derivative is zero or fails to exist. Since the derivative always exists we need only find those points at which it is zero.Section 3. the curve has a sharp corner where a tangent line does not exist. Figure 8 illustrates the contrary.9.5 2 2. we need only examine those points to determine if they are relative extrema.5 1 P 0 0. as we shall see. However.1 or 3. all is not lost. and P is not a relative extremum. They are the x-coordinates at the critical points. Factoring.5 8 4 0 -4 -8 0 0. Copyright © 2007 by Pearson Education. as at points B and C (Figure 2). and P is a relative extremum. Economics.5 1 1.6x . Example 2 Find the critical points of f1x2 = x3 . However.9 = 31x2 . These are the critical numbers.5 3 Figure 8a -6 -4 -2 P Figure 8b 0 2 4 6 Figure 8c DEFINITION 4 Any number in the domain of the function at which the derivative is either zero or fails to exist is called a critical number. Published by Pearson Learning Solutions. it is a simple matter to check the end points. we must substitute each x-value into 2 Critical Numbers and Critical Points Applied Calculus for Business. we must not conclude that all such points are relative extrema. But how do we find the relative extrema? Observe that at a relative extremum one of two things occurs: The tangent line is horizontal. The tangent line is vertical or does not exist at any point where the derivative fails to exist.9x + 15 Solution f ¿ 1x2 = 3x . and P is not a relative extremum. Walter O.2x .5 2 2. we have 3x2 .3x2 . Wang. In fact. .32 = 31x .321x + 12.1 Extrema of a Function * ** 261 If we know the interval. 12 10 8 6 4 2 0 4 3. Compute the y-values at each of these critical numbers obtaining. Example 4 Find the critical points of f1x2 = x + 2 .2 or x = 2.2.1 Extrema of a Function the original equation. Á . or x = . In this example.12 = 20.12. Procedure for Determining the Extreme Values for a Continuous Function on [a. The corresponding y-values are f1 .1 Solution The domain of this function is all x except x = 1.4 Here the derivative is a fraction. Wang. we have two critical numbers: x = . cn 3. c1. whenever f ¿ 1x2 is a quotient. when x2 . The denominator is positive for all x in its domain.122. Using the quotient rule. c3. f1c32.1. 202 and 13. Thus. 02 and (2.values at each of these points and compare to determine which is the maximum and which is the minimum. f1c12. Solution Writing f1x2 = 1x2 . we take its derivative using the chain rule. Applied Calculus for Business. c2. f1c22.2.122 Since the numerator is a constant. A fraction can only be zero when the numerator is zero and its denominator is not zero. 1x . the critical points of the given function are 1 . Example 3 Find the critical points of f1x2 = 2x2 . Observe that the maximum and minimum values of a continuous function on a closed interval can occur either at an endpoint of the interval or at critical point.4.4 = 0. we determine those values at which it is zero by setting the numerator equal to zero. In this case.262 * ** Section 3. and 2. x = 0 is not in the domain of the given function. Inc. and f132 = . and April Allen Materowski. However. Published by Pearson Learning Solutions. (why?).42-1/2 # 2x = 2 2 2x . this function can never be zero. Á . Gordon. 2. . by Warren B. Thus the critical points are 1 . f1cn2 4. that occurs when x = 0. The largest y-value in Steps 1 and 3 is the maximum value. and Finance. we need only compute the y. b] 1. Economics. x . the smallest is the minimum value. Thus. Compute f(a) and f(b). Walter O. we -3 find f ¿ 1x2 = .22 = 0 and f122 = 0. 0). The derivative will fail to exist at those x-values at which the denominator is zero.421/2. it is rejected from consideration. We have that f1x2 = 1 2 x 1x . therefore. Therefore. . Therefore. Now that we can locate critical points we may finally give a procedure with which to obtain the extrema of a continuous function on the closed interval a x b. Differentiate and determine the critical numbers. Copyright © 2007 by Pearson Education. the function has no critical points. We find f1 . Therefore. it cannot jump up or down to a new extreme value.12. Thus the maximum is 15 (the extreme point is (0. and the minimum is . 3. 15)). the curve cannot turn around and go lower than it is at P.5 0 0 0. 4. Similarly. since the function is assumed differentiable it is also continuous.1 is rejected since it lies outside the given domain. Inc.5. Similarly. Theorem 2 says that if the only critical point is a relative maximum. (This theorem can be 5 4 3 8 Q 6 P 4 2 1 0 2 -0. in Figure 9a. and April Allen Materowski. We find f102 = 15 and f142 = . Walter O. then it is the minimum for the function. . then it is the maximum for the function.5 2 2. The largest of the y-values in steps 1 and 2 is 15 and the smallest is . To understand why the theorem is true. Published by Pearson Learning Solutions. Thus. then f(c) is the extremum of the function.9x + 15 on 0 x 4.1 or x = 3. However. and Finance. Since it is the only critical point.5 0 0.5 Figure 9a Figure 9b Figure 9: Illustrating The Only Critical Point Test Applied Calculus for Business. What if the interval is not closed or even unbounded? What should we do then? In such situations there is no guarantee that the function has extrema.5 1 1. Gordon.5 1 1. THEOREM 2: THE ONLY CRITICAL POINT TEST Suppose the differentiable function defined on an open or unbounded interval by the equation y = f1x2 and it has a relative extremum at x = c. If it did.12 (the extreme point is 13.12. 2. by Warren B. x = .5 -0. x = 3 is the only critical number. we need only examine Figure 9b which shows a relative minimum at P. Q is a relative maximum. Solution 1.5 2 2. They are x = . We shall refer to the key notion as The Only Critical Point Test. Thus. f132 = . Thus. Economics. and x = c is the only critical number for this function.122). the function has a minimum at P. if the only critical point is a relative minimum. In addition. There can be no higher point on the curve since it is the only critical point. We see that if we have a continuous function defined on a closed interval we can easily determine the extrema. there is one case that often arises in applications for which the question is easily resolved. . it would have to have another critical point.1 Extrema of a Function * ** 263 Example 5 Find the extrema of f1x2 = x3 . Copyright © 2007 by Pearson Education. This theorem applies to any case in which end points are not present. Wang. It says that any time a function has exactly one critical point which is a relative extremum it is the (only) extremum.Section 3.3x2 . The critical numbers were found in Example 2. (which occurs when x = 0). therefore. it will be zero only when x = 0. q 2 has a single x + 1 critical point which is a maximum. Walter O. we can easily answer the second question. Thus. Economics. Inc. the maximum of this function is 1*2. x We also note that lim 2 = 0. Published by Pearson Learning Solutions. or x = . From the discussion above. This notation will be used in the exercises. Wang. it is common to write [a. We illustrate with the function of Example 6. This function is entered as y1(x) and the with command is used to include its domain. The TI 89 calculator can locate relative extrema. 1*2 B the graph has a relative maximum.1 Extrema of a Function generalized to the case of a continuous function where the derivative fails to exist at the single critical point. we have a single critical number at x = 1. Instead of writing a x b. Gordon.x2 = 0. y = 0 is a horizontal asymptote.) Example 6 x on [0. b). and April Allen Materowski. as x gets large. Copyright © 2007 by Pearson Education. 0) rises and must turn at least once at some critical point (which is necessarily a relative maximum) and eventually approach the x-axis as its horizontal asymptote. . Moreover. Observe that the bracket indicates inclusion of the end point. b]. and by the Only One Critical Point Test. Does it have a minimum? Show that the function defined by the equation f1x2 = 2 Solution Note that since the domain of the function is x Ú 0. this derivative is zero only when the numerator is The derivative is f ¿ 1x2 = 2 1x + 122 zero.5).264 * ** Section 3. it is a maximum. Thus.x2 . the minimum of this function is 0. the graph begins at (0. while a parenthesis indicates exclusion of the end point. it is clear that at the critical point A 1. 1 . thus.1 is not in our domain. b] would stand for a 6 x b. and Finance. We close this section with a reminder of some notation used to represent intervals (see Section 0. x: q x + 1 Therefore. Exercise 38. by Warren B. Calculator Tips Figure 10: Defining y11x2 = x x2 + 1 Applied Calculus for Business. (Recall the Ú is obtained by pressing *) See Figure 10. and the graph of this function lies entirely in the first quadrant. both the numerator of this function (and the denominator) are never negative.1 or 1 Since . (a. Suppose the function is stored as y1(x). namely 1 . In place of a 6 x 6 b we often use (a. f1x2 = x2 . . Economics. Yc: . Figure 11: The Graph of y11x2 = x x2 + 1 We now press the MATH (F5) key which gives us various options. 3. it then asks for an Upper Bound. and April Allen Materowski.9t2 . Inc.5 at the point (1. s1t2 = 1t .x . . that is the maximum is 0.1241t + 323 21. f1x2 = 4x5 13. .5 x 3. In each of Exercises 11 27 locate all critical points.1 1. Gordon. see Figure 11. Sketch the graph of a continuous function that has a relative maximum at A 1. Sketch the graph of a continuous function increasing on . 10. We choose maximum. Draw the graph of a function such that the maximum is also a relative maximum. Sketch the graph of a continuous function defined on . w1x2 = 23. by Warren B. and Finance.1 Extrema of a Function * ** 265 We next have the TI 89 plot the graph for us.52 and a maximum at 1 . 12. (we choose a window containing our domain). Copyright © 2007 by Pearson Education. h1x2 = 4x3 . 4x2 + 9 1 x2 . that has relative maxima at 1x = . r1x2 = 4x3/4 + 2 19. Which of the points in Figure 12 are: (a) maxima? (b) minima? (c) relative maxima? (d) relative minima? 8.3 3x . (b) What must be true about the y-values at the maxima? 9. number 3 is MINIMUM and number 4 is MAXIMUM. f1x2 = 22. Published by Pearson Learning Solutions. Sketch the graph of a continuous function decreasing on . 7.1 6 x 6 2 and decreasing on 2 6 x 6 4. Sketch the graph of a function which is increasing to the left of x = 1 and decreasing to its right. 11.5.3. Wang. Indicate the point M on your graph which is a relative maximum. 6.5). v1t2 = t6 . 0. h1x2 = 112 . 1*2 B a minimum at 1 .2 1 Figure 12 . Sketch the graph of a function which is decreasing to the left of x = 1.7 17.32. 5.x2 . Indicate the point m on your graph which is a relative minimum. g1x2 = ax2 + bx + c 14. s1t2 = 2t3 . EXERCISE SET 3. Draw the graph of a function such that the minimum is also a relative minimum.x22 *2 18. f1x2 = 2x4 + 2x3 . move the cursor to any point to the right of the maximum and press enter.2. (a) Draw the graph of a function which has two maxima and two relative maxima. increasing to its right and passes through 10. move the cursor to any point to the left of the maximum and press enter.5 6 x 6 3 and increasing on 3 6 x 6 7.3t2 + 5 D B E A C G F 20.13x2 + 12x + 9 16. The TI asks you for a Lower Bound.Section 3.12 and 1x = 32 and a relative minimum at (x = 0). Walter O. 4. h1x2 = x + 3 .60t + 5 15. x . 2. Applied Calculus for Business. It then gives xc: 1.2x + 3 12. f1x2 = 4x3/4 + 2 on: (a) [0. (c) (2. (c) (0. Gordon. (b) [ . f1x2 = x 22 . f1x2 = 1x2 .2 The First Derivative Test » » » » Increasing and Decreasing Functions The First Derivative Test Sign Diagrams Calculator Tips Increasing and Decreasing Functions In the preceding section we defined a critical point as any point in the domain of the function at which the derivative of the function is zero or does not exist. We have indicated the tangent lines at a few points on the graph. Copyright © 2007 by Pearson Education.12x . f(x) = 2 x + 1 40. 3].*2 on the interval (0.4 x2 26. f1x2 = x/1x2 + 12 on: (a) [0.q . w1x2 = x 2x . 35. 1]. Let f(x) be defined on the closed interval [0. that if a continuous function with a single critical point that is a relative extremum. 28. Let f1x2 = x-2/3 on the interval [ .9 27. 8]. wherever the derivative is pos*It is possible for a function to be increasing (or decreasing) on an interval and yet have its derivative equal to zero at one or more points on the interval. 2].6x3 + 12x2 + 2 on: (a) [1. f1x2 = x2 . f1x2 = 4 . Notice that each tangent line has positive slope. 39. (a) Does f have extrema on [0.1. 2]. The function is decreasing as we approach the minimum from the left and increasing to its right. 1]? (b) Is this a violation of Theorem 1? 3. (a) Does f satisfy the conditions of Theorem 1? (b) Does it have a maximum value? (c) Does it have a minimum value? 37. Walter O. x . 3] 30. In Figure 1 we show the graph of an increasing function. (c) [0. (b) [2. Let us now consider the question of how to classify the critical points.2 The First Derivative Test 36. (a) Determine the extrema of the function defined by the equation x2 f(x) = 3 on [0. In Exercises 28 34 determine the extrema on the given interval. The function is increasing on any interval containing x = 0. .2x + 3 on: (a) [0. at every point where a function is increasing and the derivative exists it is non-negative*. (b) [ . f1x2 = 2x3 + 3x2 . Wang.5.2.1. (b) [ .6 on: (a) [ .q . 2]. x + 1 (b) Does this function have extrema on 1 . there are three possibilities: The point is a relative maximum. 2]. as we shall see. 2]. 3]. 25. 32 29. 0]. and Finance. Recall that it is generally the case that just to the left of a relative maximum the function is increasing and just to the right it is decreasing. which has f ¿ 102 = 0. We shall see that we need only examine the sign of the derivative.x2/3. the Only Critical Point Test requires that we classify the critical point as either a relative maximum or relative minimum before it can be applied. by Warren B. or neither.266 * ** Section 3.x on [0. (b) 1 .1. 2] 34. Fortunately. by means of a sketch. 3] 31. 16). (b) (0. More importantly for our purposes.3. At a relative minimum the reverse occurs.1. Consider f1x2 = x3. Justify your conclusions.922/3. there is a simple procedure by which we may make the classification. f1x2 = 3x + 9 on: (a) [ . a relative minimum. Let f1x2 = x. q 2. How do we classify a critical point? For example. g1x2 = 2 . 0]? 1 24. Of course. f1x2 = x4 . Applied Calculus for Business. (d) 1 . 3/2]. Justify your conclusions. Determine the extrema of the function defined by the equation x on 1 . 22 33. That is.2. 16) 32. Published by Pearson Learning Solutions. Such critical points pose no problem. 16]. Does f have extrema on this interval? 38. 2). 1] by the rule: f1x2 = b 2x2 1 if 0 6 x 6 1 if x = 0 or x = 1 . and April Allen Materowski. Economics. Show. Inc. There is one problem that we have not resolved. then this critical point is also an extremum. and so the function is decreasing. then it is a constant. where we assume f is continuous at the critical point. the derivative of the function is positive just to its left and negative just to its right (Figure 3a). if m is a relative minimum. Economics. in any interval on which f ¿ 1x2 6 0 the function is decreasing. . Similarly. Gordon. and in any interval on which f ¿ 1x2 = 0 the function is constant. Walter O. Applied Calculus for Business. Similarly. We summarize this in the following theorem known as The First Derivative Test. But. Wang. and Finance. the derivative is negative just to its left and positive just to its right (Figure 3b). The First Derivative Test M m Figure 3a and b: Derivatives near a Relative Maximum **With the exception of certain singular functions which are analyzed in more advanced courses. Figure 1: An Increasing Function Figure 2: A Decreasing Function These observations give rise to the following theorem. We know that the derivative of a constant is zero everywhere and it can be shown that the converse is true**. Figure 2 is the graph of a function whose derivative is always negative.) THEOREM 1 In any interval on which f ¿ 1x2 7 0 the function is increasing. by Warren B. If a function is neither increasing nor decreasing. Copyright © 2007 by Pearson Education.2 The First Derivative Test * * * 267 itive the function is increasing. Inc. (We assume f is a differentiable function on a given domain. and April Allen Materowski. Published by Pearson Learning Solutions.Section 3. what does this tell us about relative maxima and minima? It means that if M is a relative maximum. Observe that its tangent line has negative slope at each point at which it exists. ) Thus.6x. This theorem is graphically illustrated by the sign diagrams of Figure 4.32 is positive if x 7 3. and April Allen Materowski. we now know that the minimum value for f1x2 = x2 . An abridged discussion follows. f is increasing + M f is decreasing sign of f '(x) Figure 4a: M is a relative maximum f is decreasing sign of f '(x) m f is increasing + Figure 4b: m is a relative minimum The First Derivative Test when used in conjunction with The Only Critical Point Test can be an extremely powerful tool for analyzing the behavior of a function.268 * ** Section 3. we immediately conclude that f(3) is a relative minimum. In Section 0.9. yields x = 3 as the only critical number.6 Since this exists for all x.6 = 0. (Of course.6x is . Applied Calculus for Business. . Inc. Putting 2x . and f ¿ 1x2 is negative just to the left of m and positive just to the right of m. the graph of y = f1x2 is a parabola opening upward. By The Only Critical Point Test. . Solution We find the derivative f ¿ 1x2 = 2x . and f ¿ 1x2 is positive just to the left of M and negative just to the right of M.32 is negative if x 6 3 and 21x . Published by Pearson Learning Solutions. Hence.6132 = . Example 1 Find and classify the critical point(s) of f1x2 = x2 . (2) Suppose m is a critical number for f. Since f132 = 32 . then f(m) is a relative minimum. and Finance. then f(M) is a relative maximum.9. Consider Example 1 below. by Warren B. we have. Wang. the critical point is 13.2 The First Derivative Test THEOREM 2: THE FIRST DERIVATIVE TEST (1) Suppose M is a critical number for f. we have reduced the classification of relative extrema to examination of the sign of the derivative.32. the only critical points are where f ¿ 1x2 = 0. Copyright © 2007 by Pearson Education. Gordon. Walter O.5 we give a detailed discussion of this procedure and recommend you review it before proceeding. Economics. It is not hard to see that 21x .92 and this is a relative minimum. Now factoring f ¿ 1x2. We shall examine the sign of the derivative by using a sign diagram. (3) If c is a critical number of f and f ¿ 1x2 does not change its sign around c then f(c) is neither a relative maximum nor a relative minimum. f ¿ 1x2 = 21x . ) Applied Calculus for Business.3. The sign of the quotient in each interval is indicated in Figure 4. If x is any number between . If an odd number of these are negative then the sign of the quotient is negative. they occur at x = .521x . Wang.9x + 15 is increasing and decreasing. and Finance. This method is quite general and we use it to draw the sign diagram for any function whose key numbers have been determined. those values of x at which the quotient becomes zero) are x = 5/3 and x = 2. The quotient can change sign only where one of the three is zero. we have determined the sign of the function throughout the interval.2x . Published by Pearson Learning Solutions. otherwise the sign of the quotient is positive. and April Allen Materowski.3.3 (choose any number you wish as long as it is less than .22.) each factor is negative. Draw them on the number line and test any point between each pair of successive key numbers to determine the sign of the function at the test point. 1x . Now that we are able to determine the sign of a function it is a simple matter to classify critical points. (The critical numbers are the zeros of the derivative. by Warren B. Locate the key numbers of the function 2. note that the possible sign changes of the function can occur only at these key numbers. Example 2 Determine where f1x2 = x3 . Use this information to classify its critical points and sketch its graph.32 = 31x . usually. once we test any point in an interval between the key numbers. . 1x . Inc. The zeros of the numerator (that is. and 1x + 32. f ¿ 1x2 = 31x2 . Thus.22/1x + 32. Walter O. all factors are positive and the sign of the quotient is positive.22/1x + 32 Observe that for any x 6 . Thus. Factoring. we have. Again. Record its sign on the number line. Sign Diagrams DETERMINING A SIGN DIAGRAM FOR A FUNCTION 1.6x . the quotient is negative. The classification follows immediately from the first derivative test and analysis of the sign diagram of the derivative.Section 3. + 5/3 2 + Figure 4c: sign of 13x .521x .9. the critical numbers correspond to the key numbers for f ¿ 1x2. Let us look at some examples that involve curve sketching.22 is negative the other factors are positive.3x2 . resulting in a positive quotient. If x is any number between 5/3 and 2 (say 1. Copyright © 2007 by Pearson Education.1 and x = 3. If x is any number larger than 2 (say 5). The zero of the denominator (that is. Economics.2 The First Derivative Test * ** 269 Consider the quotient 13x . Solution We must first determine the critical numbers f ¿ 1x2 = 3x2 . Gordon.321x + 12 There are two critical numbers.3 and 5/3 (say 0) both factors in the numerator are negative and the denominator is positive. We shall refer to these as key numbers. the value at which the quotient is undefined) is x = .9). and the sign of the quotient is negative. Suppose we indicate these key numbers on the number line (Figure 4). As you can see.5. Observe that the sign of the quotient is determined by the signs of the three expressions: 13x . say .52. Similarly.122 is the relative minimum.) M(*1. the sign of the derivative is negative. (You may substitute into either the factored or non-factored form of the derivative. To the left of x = 3 the function is decreasing and to its right it is increasing. which occurs when . Copyright © 2007 by Pearson Education. by Warren B. . We record the signs giving Figure 5b. that is f102 = 15. and April Allen Materowski. 202 is the relative maximum and m13. In Figure 6b. Note that we have labeled an additional point. *12) Figure 6a: A Partial Sketch Figure 6b: Completing the Sketch Figure 6: A Sketch of the Graph f1x2 = x3 . the function has a relative minimum at x = 3. Gordon.321x + 12 Choose any number less than .2x .12 we conclude that M1 . 0 *1 0 3 Figure 5a: sign of f ¿ 1x2 = 31x2 .32 = 31x .) The sign is positive. See Figure 5a. Walter O. it is decreasing when f ¿ 1x2 6 0. Draw the number line indicating the critical numbers.3x2 .1 and decreasing to its right. the tangent line is horizontal at these values of x). it does not matter which is used.321x + 12 The function is increasing over any interval on which its derivative is positive.1. from Figure 5b we see that the function is increasing if x 6 . + 0 *1 0 3 + Figure 5b: sign of f ¿ 1x2 = 31x 2 . *12) m(3.1 6 x 6 3. at x = .2 The First Derivative Test We proceed as follows: 1. Wang.1 the function has a relative maximum.32 = 31x . Place a 0 above each critical number to remind us that the derivative is zero at that value of x (that is. To sketch the curve we begin by drawing Figure 6a which illustrates that the tangent line is horizontal both at the peak (M) and the valley (m). Figure 5b indicates that the function is increasing to the left of x = . and Finance. Inc. (We remind you that the y-coordinate of any point on a graph is obtained by substituting the appropriate x-coordinate into the equation of the function. 20) m(3. Thus. Since f1 .270 * ** Section 3.1 (say . (0. Economics. Choose any number between .9x + 15 Applied Calculus for Business.2x . .12 = 20 and f132 = . Choosing any number larger than 3 (say 4) we find the sign is now positive.2) and evaluate the sign of the derivative at this value of x. Thus. Therefore. 20) M(*1. 15) which is the y-intercept of the function. It is found by setting x = 0. Published by Pearson Learning Solutions.1 and 3 (say 0). the sketch is completed by joining the relative extrema with a continuous curve.1 or x 7 3. . Copyright © 2007 by Pearson Education. 0 0 0 3/2 + Figure 7b: sign of f ¿ 1x2 = 8x 3 . Choosing any number greater than 3/2.12x 2 = 4x 212x . Inc. 0 0 0 3/2 Figure 7a: sign of f ¿ 1x2 = 8x3 . Economics. Since f ¿ 1x2 does not change sign at x = 0. and April Allen Materowski.12x2 = 4x212x . Note that f102 = 7 and f13/22 = 29/8 = 3.4x 3 + 7 Applied Calculus for Business. Walter O. 7) (0. The function is increasing if x 7 3/2.2 The First Derivative Test * ** 271 Example 3 Sketch the graph of f1x2 = 2x4 . that is.32 Choosing any number less than 0. Wang. and Finance.32 The critical numbers occur where the derivative is zero. it is neither a relative maximum nor a relative minimum. 29/8) Figure 8a: A Partial Sketch Figure 8b: Completing the Sketch Figure 8: A Sketch of the Graph of f 1x2 = 2x 4 . indicating its relative extrema. At x = 3/2 the function has a relative minimum since at this point it changes from a decreasing to increasing function. We label both these points in Figure 8a. by Warren B.4x3 + 7. f ¿ 1x2 = 8x3 . Label the intervals on which it is increasing and decreasing.32 The function is decreasing if x 6 0 or if 0 6 x 6 3/2.625. (0. We label these critical numbers on the number line as indicated in Figure 7a.Section 3. We record this information and obtain Figure 7b. 29/8) m(3/2. 7) m(3/2. also indicating that we have a valley at x = 3/2. Gordon.12x2 = 4x212x . Solution We start with the derivative. Published by Pearson Learning Solutions. we find that f ¿ 1x2 6 0. the tangent line is horizontal at this point. Choosing any number between 0 and 3/2 we find that f ¿ 1x2 6 0 on this interval as well. we find that f ¿ 1x2 7 0. We complete the sketch in Figure 8b. when x = 0 or x = 3/2. Nevertheless. Inc. x = 23 = 8 Since f102 = 0 and f182 = 8. . Published by Pearson Learning Solutions. Solution f ¿ 1x2 = 4x-1/3 . Economics.2 = 212 . x = 0. Figure 9. Note that f102 = 0 and f182 = 8. If any number less than 0 is tested. This information is summarized in Figure 9. the derivative does not exist at x = 0. 8) M(8.x1/32/x From Figure 9 we see that the function has a relative maximum at x = 8 and a relative minimum at x = 0. 8). because the tangent line is vertical there. Gordon. Since the derivative does not exist at x = 0. Second.2x Applied Calculus for Business.x1/32 4 4 . Wang. If any number between 0 and 8 is tested. 0) m(0. If any number greater than 8 is tested the sign is negative. and Finance. we see that the derivative fails to exist at the zero of the denominator. the derivative is zero at the zero of the numerator. and April Allen Materowski. we place DNE (Does Not Exist) above it in the sign diagram. In this case. by Warren B. First. We first sketch Figure 10a and then complete the sketch in Figure 10b. Walter O. 0) Figure 10a: A Partial Sketch Figure 10b: Completing the Sketch Figure 10: A Sketch of the Graph of f1x2 = 6x2/3 . Example 4 Sketch the graph of f1x2 = 6x2/3 . 8) m(0. M (8. the critical points are (0. when x1/3 2 . Copyright © 2007 by Pearson Education. the sign of the derivative is positive. 29/8 is the minimum of this function.x1/3 = 0 = 2.2x1/3 2 = = x1/3 x1/3 x1/3 This function has two critical numbers. DNE 0 + 0 8 Figure 9: sign of f ¿ 1x2 = 212 . the sign of the derivative is negative (remember the cube root of a negative number is negative).2 The First Derivative Test Note in the previous example.272 * ** Section 3. 0) and (8.2x. the maximum revenue is at the highest point on the parabola. that is the demand is zero and there is no revenue. by Warren B. Therefore. Economics. Therefore. How could you determine that this second root is x = 27? We next consider some examples from Economics. (Figure 11) + 0 200 Figure 11: Sign of R ¿ 1x2 = . where p is the price of a set in dollars and x is the number of sets demanded.6x + 1200 = 0. Copyright © 2007 by Pearson Education. Walter O. Example 5 Suppose that the relationship between price and demand for a certain brand of color television set is given by the equation p = . it is not always an easy algebraic matter to determine the highest or lowest point on a curve. x = 0. Thus. R1x2 = xp = 9x1300 . we must have 0 x 400. Now. Since x stands for the number of bicycles. or x = 200. it is the value that maximizes the revenue. Its graph is a parabola opening downward. The maximum revenue is $120. What price should be charged per set if the total revenue is to be maximized? Solution The total revenue is given by R = xp = x1 . the total revenue equation is R1x2 = . otherwise the radicand would be negative. Inc. Here the price is zero so there is no revenue. After all. The revenue is maximized when the cost for each set is $600. The critical number occurs when .6x + 1200. what price should be charged for each bicycle? Solution Observe that x 30. Wang. Example 6 The price of a bicycle is given by the equation p = 91300 . . This is a continuous function defined on a closed interval. which is similar to the previous one except that the demand equation is no longer linear. Gordon. At one end point.312002 + 1200 = $ 600. Using sign analysis on R ¿ 1x2 we obtain the following sign diagram.10x21/2. For x = 200.2 The First Derivative Test * ** 273 Note that the above graph has two roots.Section 3. at its vertex. in order that neither x nor p be negative (verify this!).10x21/2 where 0 x 30.3x + 12002 or R = .3x2 + 1200x. If the total revenue is to be maximized. the revenue has a relative maximum when x = 200 (200 television sets are demanded).3x + 1200. and April Allen Materowski. However. we must have x Ú 0. Published by Pearson Learning Solutions. Consider the following example. Now you might be thinking that the calculus was not needed to solve this example. where p is the price per bicycle in dollars when x bicycles are demanded.3x2 + 1200x.000. p = . we find R ¿ 1x2 = .6x + 1200 Thus.00. Differentiating. By the Applied Calculus for Business. At the other end point x = 400 gives p = 0. This is the only critical number. the first at the origin and the second not shown on the sketch. and Finance. 10x21/2 # 9 2 . The problem with this approach is that the window used to display the graph may not clearly show all the relative extrema. when 9 2300 . we can solve for the zeros of the denominator of the derivative. then we enter solve1y21x2 = 0. r1t2 = 4t3 .45x R ¿ 1x2 = + 9 2300 . To determine the x-coordinates where the derivative is zero. f1x2 = 3x . As mentioned before. Wang.10x = 0 Clearing fractions and dividing by 9.a21x2 + ax + a22. The price per bicycle is then p = 9 2300 . or 15x = 300. 0). x2. Suppose the equation of the function is already stored as y1(x) in the calculator. Economics. we first determine d(y1(x). press F5. we have that 1 R ¿ 1x2 = 9x # 1300 .18 6.24x2 + 36x + 96 8.10x That is. h1x2 = 2x2 . that a maximum occurs when x = 20. Walter O. Both the x and y-coordinates are found this way. Published by Pearson Learning Solutions.4.21 7.10x = 5x. Thus. that is. the maximum must occur at a critical point. by Warren B.) Applied Calculus for Business. enter an appropriate lower and upper bound. we find 300 .2 In Exercises 1 14 use the first derivative to determine where the given function is increasing and decreasing. x = 20. suppose we then save its denominator as y2(x).10x = 45x 2300 . the TI 89 can differentiate functions using the d key (located above the number 8).101202 = $ 90. g1x2 = 4x3 . the relative maxima and minima may be found by having the calculator graph y1(x). EXERCISE SET 3. (The tangent line to the curve is vertical at (30. g1x2 = . Inc. the maximum must occur at either a critical point or an end point.) The other critical point occurs when R ¿ 1x2 = 0. x).1 5. but it means you have to have the graph displayed. Copyright © 2007 by Pearson Education.10x2-1/21 .10x The derivative fails to exist at x = 30. x2. It follows from the Extreme Value Theorem.32x (Hint: x3 . f1x2 = mx + b.274 * ** Section 3. 1.45x 2300 . R1202 = 200. if: (a) m 7 0.a3 = 1x . Alternately. and April Allen Materowski. The y-coordinates of the critical points are obtained by substituting the critical numbers just found into y1(x). we need only enter solve1d1y11x2.102 + 1300 . x2 = 0. s1t2 = t2 + 4t . Calculator Tips We shall examine additional applications of the derivative to optimization theory in Section 3. (b) m 6 0 3. This means the TI 89 can determine the critical points of a function.15t2 + 18t + 2 9. . To find the points at which the derivative does not exist. q1x2 = x4 . v1x2 = x4 + 5 10. Now. Using the chain rule. g1t2 = t2 .10x 2300 . choose maximum (or minimum) and for each relative extremum. .10x + 9 2300 .5x + 3 2. Since R102 = R1302 = 0. and Finance.2 The First Derivative Test Extreme Value Theorem.2x + 1 2 4. Gordon. f1x2 = x2 . f (a)) Figure 12 45. 42. Show that if two functions f and g have the same derivative on the same interval. h1x2 = 4x3 . f1x2 = x/1x2 + 12 36.a Theorem. is given by the equation p = 101147 . and April Allen Materowski. Published by Pearson Learning Solutions.1241t + 323 In Exercises 25 35 sketch the graph of the function defined in the given exercise. Copyright © 2007 by Pearson Education. Inc. r1x2 = 4x3/4 + 2 23. Wang. where x is the number of magazines demanded. 33. What should the price be to maximize the publisher s total revenue? Justify your conclusion! 39. a 6 c 6 b. Hint: let x1 6 x2 be any two points in the interval in question. 26.Section 3. suppose that f1a2 = f1b2. f1b2 . This result is known as Rolle s Theorem.02x21/2. Exercise 22. Exercise 7. b).60t + 5 19. f1x2 = 4x5 17. Exercise 20. Exercise 23. Show that ax2 + bx + c always has the same sign as the constant term c if b2 .3t2 + 5 24. Additionally. decreasing or remains the same when it is evaluated along the line B(b. and Finance. Justify your conclusion! 38. If the cost of the top and bottom is twice the cost of the sides. f(b)) is horizontal. Use the Mean Value Theorem to prove the First Derivative Test.x221/2 22. Sketch various possibilities for the graph of f. or The Law of the Mean. Justify your conclusion! 37. f1x2 = 2x4 + 2x3 . such that the tangent line at x = c is parallel to the x-axis 1f ¿ 1c2 = 02. such that the tangent line at x = c is parallel to the line joining A to B. Find two numbers whose sum is 100 such that their product is as large as possible. g1x2 = ax2 + bx + c 18. f1x2 = x 2x + 1 The First Derivative Test * ** 275 y = mx + b. in cents. Show from your graphs that there must be some point c.13x2 + 12x + 9 20. Gordon. Use all the information obtained from the first derivative. p1x2 = 1x . with a 7 0 and b 7 0.922/3 35.x1/3 12. . then their difference.) 41. Thus.7 21. Economics.g1x2. f (b)) A (a.4ac 6 0.2 11. f1x2 = 3x . 28.2x + 3 16. by Warren B. Exercise 24. j1x2 = 28 . show that the price at which the total revenue is maximized is independent of b. 31. For the demand equation p = 1a . then the rotated figure satisfies all the conditions of Rolle s Theorem. (Hint: Substitute for y in the objective function and apply the first derivative test. where a 6 c 6 b. s1t2 = 2t3 . h1x2 = 112 . Applied Calculus for Business.x2 .x12f ¿ 1c2.2x 14. 30. determine the frame s dimensions if the total cost is to be minimized.9t2 . 27. s1t2 = 1t . b] and differentiable on (a. Exercise 21. Show that if the figure is rotated so that the dotted line joining A(a. Hint: Define D1x2 = f1x2 . 15. 29. f1x2 = 1x2 .bx21/2. 43. Walter O. 40. 25. Exercise 11. v1t2 = t6 . except that f1a2 Z f1b2. Suppose that a function satisfies the same conditions as in the previous exercise. 44.f1a2 show that f ¿ 1c2 = . Show that the objective function P = Ax + By + C is either increasing. A rectangular picture frame is to enclose an area of 72 in2. Suppose a function is continuous on [a. f1x2 .f1x12 = 1x2 . In Exercises 15 24 classify all critical points. The price of a magazine. 34. show D ¿ 1x2 = 0 on the interval and then deduce the result as a consequence of the first derivative test. Exercise 15. f(a)) to B(b.g1x2 = constant on the interval. Exercise 8. Therefore.12/1x + 12 13.0. Apply the Mean Value Theorem using these points to deduce that f1x22 . This result is known as the Mean Value b . One such function is illustrated in Figure 12. there must be at least one point c. 32. dx We see that f ¿ 1x2 is itself a function of x. Copyright © 2007 by Pearson Education. Walter O. It tells us where the curve is increasing and decreasing and allows us to classify the critical points.7x + 9. it is a differentiable function. Wang. its rate of decrease must slow until it stops decreasing altogether at the minimum. After passing the minimum.7. and April Allen Materowski. as it nears the low point. we need to discuss the notion of higher order derivatives. 2 . Economics. think about the way a curve looks as the function nears its minimum (for example.72 = 24x . it may be decreasing very quickly. first slowly and then more rapidly. Published by Pearson Learning Solutions. P Figure 1 But. Thus. What it does not do is tell us how the curve increases or decreases. For example.276 * ** Section 3.2x2 .3 Concavity and the Second Derivative 3. by Warren B. what is the derivative of the derivative? The answer is very simple. y . To analyze this behavior. The following symd2y bols are commonly used to denote the second derivative: f 1x2. . and Finance. We now ask. Gordon. at point P in Figure 1). the function must begin to increase. dy Suppose y = f1x2 = 4x3 . Furthermore. Inc. It is d 112x2 . (You might be dx The Second Derivative Applied Calculus for Business.4x .4. At first.3 Concavity and the Second Derivative » » » » » » » » The Second Derivative Higher Order Derivatives Velocity and Acceleration Concavity The Second Derivative Test for Concavity The Second Derivative Test for Relative Extrema Implicit Differentiation and Curve Sketching Calculator Tips We have learned how examination of the first derivative can help us to sketch a curve. then f ¿1x2 = = 12x2 . dx The derivative of the derivative is called the second derivative. its rate of change is itself changing.4x . Economics. and April Allen Materowski. Instead we would indicate it by f12721x2 or y1272.3 Concavity and the Second Derivative * ** 277 wondering about the appearance of the 2 in non-symmetrical positions in 2 d2y dx2 . y1n2. for instance.2 Higher Order Derivatives Example 2 1 Find the second derivative of f1x2 = 1x2 + 12 *2 Solution For the first derivative. the following notation is used. Inc.x2 . if we wanted to indicate. Copyright © 2007 by Pearson Education. In the above example suppose we want the derivative of the second derivative. and Finance. We could go on in this manner defining fourth. Wang. It is denoted by f 1x2 = y = . dx dx dx dx 1dx22 dx2 We may extend the above notion and ask for third.*2 Factoring out 1x2 + 12-3/2.100x-6 1Second derivative. The reason is that the second derivative is d1dy2 dy d2y d dy a b .5x-4 Solution The derivatives are taken successively by the usual rules. y ¿ = 20x4 + 20x-5 1First derivative. fifth.x12x2 1x2 + 122 = 1 . we need the product rule and the chain rule. The nth order derivative of y = f1x2 is denoted dny by any of the following: f1n21x2. Walter O.Section 3. 1x2 + 122 Applied Calculus for Business. dx Example 1 Find the first three derivatives for y = 4x5 . 1x + 123/2 2 1 1 1 1 Example 3 Find f 1x2 if f1x2 = x/1x2 + 12. and even higher order derivatives. f 1x2 = x C A . the 27th derivative it would not be wise to write f with 27 primes. Gordon. dx3 f 1x2 = 24. (For our example the fourth and higher order derivatives are all zero. we have f 1x2 = 1x2 + 12-3/2[x2 + 1x2 + 12] = 1 .*2[1] f 1x2 = x21x2 + 12-3/2 + 1x2 + 12. In our example. d3y which is called the third derivative. or n . . fourth or even higher order derivatives. in general. f ¿ 1x2 = 1x2 + 12 . we need the chain rule.*2 For the second derivative. More generally.2 y = 80x3 .1*2 B 1x2 + 12-3/212x2 D + 1x2 + 12. by Warren B. Published by Pearson Learning Solutions. It has the form or . we have. f ¿ 1x2 = 1*21x2 + 12. Solution Applying the quotient rule once again. written as .*212x2 = x1x2 + 12.2 y = 240x2 + 600x-7 1Third derivative.) However. and Finance. . the velocity is v122 = 31222 .60 (a) At t = 2. which means the velocity is decreasing.60 = .278 * ** Section 3. The acceleration is a122 = 6122 . which is the absolute value of velocity. (b) Find the time at which the acceleration is zero. You should notice that when a1t2 = 0.x22[21x2 + 12[2x]] 1x2 + 122 = . The acceleration measures how velocity changes with time. (a) Find the velocity and acceleration of the particle at time t = 2. The first deals with particle motion and the second with curve sketching. and we write dv d2s a = = 2 . The velocity of the particle is becoming more negative.2x1x2 + 12[1x2 + 12 + 211 . Gordon. the velocity. The acceleration is also negative. So we say that the velocity is decreasing but. (c) At t = 10. You should verify that this point 110. Now the derivative of velocity.x22 1x2 + 123 Velocity and Acceleration The more complex the original function.2852 gives the minimum velocity. Wang. The particle is moving in the negative direction. the more involved it is to find the various derivatives.285 cm/sec. along with the chain rule. we consider s = f1t2. To find the second derivative in Example 3 required a double application of the quotient rule.601102 + 15 = . is the instantaneous rate of change of position with respect to time and dv ds v = s ¿ 1t2 = .2x] . where s is position and t is time. (b) The acceleration is zero when 6t . by Warren B. Applied Calculus for Business. its speed. the velocity is negative.60122 + 15 = .3 Concavity and the Second Derivative f 1x2 = 1x2 + 122[ . We have several interpretations for the (first) derivative. We know v. the velocity is v1102 = 311022 . actually the particle is going faster. v = v1t2 = s ¿ = 3t2 . is getting larger.11 . Published by Pearson Learning Solutions.30t2 + 15t + 23.93 cm/sec.2 Thus. dt dt Example 4 The position (in centimeters) of a particle as a function of t (in seconds).60 = 0.2x13 . .x22] 1x2 + 124 = . Copyright © 2007 by Pearson Education. . What about higher order derivatives? We shall only be concerned with two interpretations of the second derivative. That is. at t = 10 sec. is given by the equation s = t3 .48 cm/sec. Walter O. Inc.60t + 15 a = a1t2 = v ¿ = s = 6t . Economics. we have a critical point for v(t). That is. it is becoming more negative. Notice this interpretation. First. This derivative is called the acceleration. (c) What is the velocity at that time? Solution We find the first two derivatives. and April Allen Materowski. is the instantaneous rate of change of dt dt velocity with respect to time. Applied Calculus for Business. Similarly. and Finance. You might be wondering what those points are called at which the curve changes its concavity. We now have a similar situation to the one we had in Section 3. However. Wang. More precisely. Copyright © 2007 by Pearson Education. and from D to F. From A to B the graph is U-shaped. It is concave downward from B to D and from F to G. . while from B to C the graph is upside down U-shaped. Inc. We know what an inflection point is. and April Allen Materowski. we have the following definition.Section 3. Between C and E the function is decreasing. where the graph is concave upward at P and concave downward at Q. from C to D it is upside down U-shaped. Concavity C G B D F E A Figure 2: Illustrating Concavity Consider the function whose graph is sketched in Figure 2. Similarly between E and F it is U-shaped and between F and G it is upside down U-shaped. Economics. the function in Figure 2 is concave upward from A to B. the way it increases is different. P Q Figure 3: Concavity and Tangent Lines DEFINITION 1 Any point on the graph of a function at which the concavity changes is called an inflection point. Walter O. See Figure 3. by Warren B.3 Concavity and the Second Derivative * ** 279 What about the geometric interpretation of the second derivative? What shall see that it determines the shape or concavity of a function.2. Mathematicians use the term concavity to indicate how the curve is shaped. Gordon. Thus. Published by Pearson Learning Solutions. how do we determine the concavity of the function? We shall see that the answer to this question is analogous to the determination of where a function increases or decreases. the tangent line lies beneath the graph. but how do we locate it? That is. They are called inflection points. It should be pointed out that at any point in an interval on which the graph is concave upward. Observe that the graph is increasing from A to C as well as from E to G. in any interval on which the function is concave downward the tangent line lies above the graph. and from D to E it is U-shaped. dx Similarly. which is called the Second Derivative Test for Concavity. THEOREM 2: THE SECOND DERIVATIVE TEST FOR CONCAVITY In any interval on which f 1x2 7 0. Let us now look at the relationship between the second derivative and concavity. along this curve. Q and R gets larger. by Warren B. while the sign of the second derivative reveals where the function is concave upward or downward. is also true. . If the function is concave downward at x = c.3 Concavity and the Second Derivative You might be thinking that concavity. THEOREM 1(PRELIMINARY VERSION) Suppose f is a differentiable function with f ¿ 1c2 = 0. Thus. Observe that at C the function is concave downward. In any interval on which f 1x2 6 0. then the function has a relative maximum at x = c. the function is concave downward. After all. If the function is concave upward at x = c. In each. Similarly. Notice that as we follow the graph from left to right the slope of the tangent lines drawn at P. how can concavity aid us? Another look at Figure 2 should convince you otherwise. Walter O. or f 1x2 7 0. Inc. Note how similar the uses of the first and second derivatives are. is not very useful. Gordon. Therefore. Consider the concave upward function whose graph is given in Figure 4a. R P Q Q P Figure 4a: Slope of the Tangent Line Increases as x Increases Figure 4b: Slope of Tangent Line Decreases as x Increases d 1f ¿ 1x22 6 0 or f 1x2 6 0 along this curve. the slope is an increasing function (the slope increases). the sign of the first derivative indicates where the function is increasing and decreasing. which we will revise shortly. that is. the major focus of the previous two sections was the location of extrema. and April Allen Materowski. Applied Calculus for Business. Copyright © 2007 by Pearson Education. the function is concave upward. by the first derivative test its derivative is positive. d 1f ¿ 1x22 7 0. Economics. as we move from left to right on the concave downward function in Figure 4b. a relative minimum. We have the following preliminary theorem. and if we can determine the concavity at that point. Thus we see that if we have a point at which f ¿ 1x2 = 0. But the slope is the derivative. then the function has a relative minimum at x = c. It can be shown that the condx verse. Published by Pearson Learning Solutions. Thus f ¿ 1x2 is an increasing function. Point C is a relative maximum. and Finance.280 * ** Section 3. Thus. we see that the slope is decreasing (the tangent lines become less steep). the function is concave upward. at E. However. Wang. if we follow the slope of the tangent lines. while being a nice geometrical property. then we can classify the critical point by its concavity. we need only examine the sign of the appropriate derivative. 2. that means finding these key numbers for the second derivative function. Notice that determining the inflection points is analogous to classifying relative extrema.2 x (3 .x22 Figure 5 Choosing any number to the left of x = . Usually. 23. Walter O. The Second Derivative Test for Relative Extrema We are now ready to revise Theorem 1. Substituting into the equation of the function to find the y-values. . The second derivative 1x2 + 123 always exists (Why?). To analyze the sign of f 1x2. say . Identify the points of inflection for the graph of y = f1x2. sign of f" ( x) = . 0 and 23. Choose any number between .23/4 B . We may then apply sign analysis on the second derivative.2x13 .x 2)/( x 2+ 1) 3 0 0 0 0 The Second Derivative Test for Concavity Solution In Example 3.1. .23 and 0. and Finance. Therefore. since it is at precisely those points where the sign of f 1x2 may change. thus in this interval. Example 5 Determine where f1x2 = x/1x2 + 12 is concave upward and where it is concave downward. Continuing in this manner.23 6 x 6 0 or x 7 23.2 x (3 . we found that f 1x2 = . Note that the concavity changes when x = . Economics.23. 02 and I3 A 23.23. Wang. . 0. and April Allen Materowski.x2 = 0. called the Second Derivative Test for Relative Extrema. CD sign of f " ( x ) = . say . we have the inflection points. Published by Pearson Learning Solutions. so we need only consider values at which it is zero. Copyright © 2007 by Pearson Education.23. Inc. we begin with Figure 5. we find f 1x2 7 0.23. that is. The second derivative is zero when x = 0 or 3 . To find the inflection points we locate those points at which the second derivative is zero or fails to exists. we find that the sign of f 1x2 is negative.Section 3. Since we can determine the concavity of a function in terms of the second derivative. Gordon. it is concave upward (CU) in this interval. It is concave downward if x 6 . we have the completed sign diagram for f 1x2. x = .x 2)/( x 2 + 1) 3 0 CU + CD 0 0 0 CU + Figure 6: The Concavity of f 1x 2 = x /1x 2 + 12 The given function is concave upward if . we restate Theorem 1 as a corollary to the Second Derivative Test. the function is concave downward (CD). These are the x-values at the inflection points. I1 A .3 Concavity and the Second Derivative * ** 281 To determine the sign of the second derivative we need only find those points in the domain of the function at which the second derivative is zero or fails to exist. by Warren B. 23/4 B . given in Figure 6. Applied Calculus for Business. I210.23 or 0 6 x 6 23. We see that f 1x2 = 6x .12. Consider f1x2 = x3. Wang.6 = 61x .9x + 15. and h has a relative maximum (verify these statements).x4. 4) is an inflection point. Sign analysis on f 1x2 yields Figure 7. Example 7 Sketch the graph of f1x2 = x3 . Each of these functions has a critical point at x = 0 and the second derivative at x = 0 is also zero (verify!). 202 and a relative minimum at m13. the second derivative is zero at x = 1. Similarly. Therefore 13. The critical points are at x = .6x .6.3x2 . f 1 . at x = 0. Copyright © 2007 by Pearson Education. Since f 1x2 = 6x . by Warren B. (Note that at f 1 . Therefore.) Example 6 Use the Second Derivative Test to classify the critical points of f1x2 = x3 . If f 1c2 7 0. However.1. the test is inconclusive. Solution In Example 6 we found that the function has a relative maximum at M1 .122 is a relative minimum. then (c. CD 0 sign of f " ( x ) 6( x . Published by Pearson Learning Solutions. and April Allen Materowski. 1 . The completed graph is given in Figure 8b. A slight modification of this test is sometimes used to classify the critical point when f (c) = 0.1.1) 1 CU + Figure 7 Since the concavity changes at x = 1.12 6 0.122.9 = 31x . we first plot the relative extrema and inflection points as in Figure 8a. In addition. where it changes.3 Concavity and the Second Derivative THEOREM 1: (REVISED) SECOND DERIVATIVE TEST FOR RELATIVE EXTREMA Let f be a differentiable function with f ¿ 1c2 = 0. Solution f ¿ 1x2 = 3x2 . Thus. Economics.) To sketch the graph of the function. then (c. indicating all relative extrema and inflection points. and Finance. g has a relative minimum. remembering that the concavity is downward until we reach the inflection point. .321x + 12. and f 132 7 0 illustrating the second derivative test for classifying the two critical points. f has an inflection point.9x + 15. f(c)) is a relative minimum. . If this occurs. Gordon.3x2 . (See Exercises 71 and 72. f 132 = 12 7 0. We then join the points.12 = . Inc. we have that I(1. Walter O. and the first derivative test (or other methods) could be used. f(c)) is a relative maximum Note that the case f 1c2 = 0 is omitted. 202 is a relative maximum.1 and 3. g1x2 = x4 and h1x2 = . .12 6 0. Applied Calculus for Business. If f 1c2 6 0. we can use the second derivative to determine the concavity and the points of inflection for the curve.282 * ** Section 3. the function has a relative minimum at m A .3x2 . 23/4 B . 1 .2x13 . Wang.23.20) (0. Inc. 4 M(-1. Applied Calculus for Business. Gordon. 15) I (1. The sign analysis of 1x2 + 123 the second derivative was done in Example 5 and is given again in Figure 9. . 15) I (1. 1*2 B . 02 and I3 A 23.23/4 B . f 112 6 0.9x + 15 Example 8 Sketch the graph of f1x2 = x/1x2 + 12 indicating all relative extrema and inflection points. -12) Figure 8a: Starting the Sketch Figure 8b: Completing the Sketch Figure 8: Sketching the Graph of f 1x2 = x3 . -12) m(3.1 and x = 1. The critical numbers are x = .1.3 M(-1. therefore M A 1. and Finance.Section 3.x2 Solution In Example 1 we found that f ¿1x2 = 2 and that 1x + 122 .x22 f (x) = . Published by Pearson Learning Solutions. We begin by plotting the points in Figure 10. 1*2 B is a relative maximum. I210. Copyright © 2007 by Pearson Education. 4) Concavity and the Second Derivative * ** 283 m(3.12 7 0. The inflection points are at I1 A . Economics. . by Warren B. and April Allen Materowski. 20) (0. CD 0 sign of f " ( x ) = -2 x (3 -x 2)/( x 2+1) 3 - CU + CD 0 0 0 CU + Figure 9 From the sign diagram we see that f 1 . thus. Walter O. +*) Figure 12: Completing the Sketch of f 1x2 = x /1x 2 + 1) Applied Calculus for Business. 3 /4) I2 (0. . Gordon. I3 I2 I1 Figure 11 To finish the graph. and Finance. 3/4) I2 (0. To the left of I1 it is concave downward. Inc. -.q (verify this!).. and April Allen Materowski.3 /4) Figure 10: Plotting the Relative Extrema and Inflection Points In Figure 11 we join the points.3.+*) I3 ( 3. Copyright © 2007 by Pearson Education. . by Warren B.284 * ** Section 3. M(1.3 Concavity and the Second Derivative I3( 3 . Walter O. to the right of I3 it is concave upward. Wang. and. Between I2 and I3 it is concave downward. Economics.3/4) 1 m(-1.0) I1(. remembering that: Between I1 and I2 the function is concave upward.3. . we should observe that the x-axis is the horizontal asymptote for x approaching both + q and . 0) I (. The completed graph is given in Figure 12. Published by Pearson Learning Solutions. 4x3 CU + sign of f " ( x ) 12 x ( x .27. Example 9 Sketch the graph of f1x2 = x4 .41323 = . indicating all relative extrema and inflection points. I2 (2.3 Concavity and the Second Derivative * ** 285 Note that the function in the previous example is odd. At the Inflection point I1 its tangent line is horizontal. Copyright © 2007 by Pearson Education. Therefore. Wang. so the graph is symmetric with respect to the origin. Remember. 0) We see that the function has an inflection point at I110. and Finance.24x = 12x1x .32 I1 (0. 0 sign of f '( x ) 4 x 2( x . However. so the two x-intercepts (zeros) are x = 0 and x = 4. The y-value at this point is f132 = 1324 .3) 0 0 3 + Figure 13: sign of f ¿ 1x2 = 4x21x . -27 Figure 15: Beginning the Sketch of f1x2 = x4 . We next calculate the first two derivatives. We see that f 102 = 0 and f 132 = 36 7 0. the test fails. The critical numbers are found where f ¿ 1x2 = 0. Let us try to apply the second derivative test in order to classify these points. The sign diagram is given in Figure 14. by Warren B.-16) m(3. Economics. . The second derivative is zero when x = 0 or 2. we can analyze the sign of f ¿ 1x2 in order to classify the point (0. 0) The sign diagram confirms our conclusion that m13. we only really needed to determine how the graph looked for x 7 0. Let us analyze the concavity of the curve. However. The sign analysis of f ¿ 1x2 is given in Figure 13. it is easy to see that the critical numbers are x = 0 and x = 3. Since f ¿ 1x2 has been factored above. 0) is neither a relative maximum nor a relative minimum.4x3 Applied Calculus for Business. From I1 to I2 the graph is concave downward. .32 I1 (0. the curve is decreasing until it reaches m.Section 3. Therefore. On the other hand. it also reveals that (0. at x = 0. We begin the sketch in Figure 15. f ¿ 1x2 = 4x3 . The completed sketch is given in Figure 16.272 is a relative minimum. otherwise it is concave upward. and the rest of the graph could be drawn by symmetry. the tangent line there is horizontal and this will be indicated in our sketch. Published by Pearson Learning Solutions.3) CD 0 0 0 2 CU + Figure 14: sign of f 1x2 = 12x1x .162. It has a second inflection point at I212. at x = 3.32 f 1x2 = 12x2 . -27) Figure 16: The Completed Sketch of f1x2 = x4 . Gordon. Nevertheless.22.4x3 = x31x . Inc. we have a relative minimum. I2 (2. Walter O. and April Allen Materowski. where f102 = 0.4x3. 02. where the tangent line is horizontal.-16) m(3.42.12x2 = 4x21x . 0) (4. . 0). Solution Note that f1x2 = x4 . Referring to the sign diagram. + sign of f '( x ) ( x 1/3 .3 Concavity and the Second Derivative For smooth functions the concavity around a relative maximum is downward. we see that we have a relative maximum at M(0.3x2/3. Gordon. 0).42.2 = . while around a relative minimum it is upward.2x-1/3 = 1 2 x1/3 . Published by Pearson Learning Solutions. where the derivative does not exist the tangent line is vertical. This happens because of the graph not being smooth at this point. in the graph on the left. and Finance. The numerator is zero when x1/3 . 1/3 x x1/3 The critical points occur when the derivative is zero or when the derivative does not exist. if the point is not smooth. the concavity is downward both to its left and right. Economics.4 and f102 = 0. Sign analysis of the first derivative yields the sign diagram given in Figure 18. Solution We start with the first derivative. When this is not the case.2 = 0 whose solution is x = 8. and April Allen Materowski. That is. 0) and a relative minimum at m18. . Remember. it is possible to have any kind of concavity near a relative extremum. Applied Calculus for Business.22/x1/3 We find the corresponding y-values: f182 = . We first begin the sketch in Figure 19 and give the completed graph in Figure 20. Note that even though M(0. The second derivative is given by f 1x2 = 2 -4/3 2 x = . Example 10 Use the first and second derivatives to sketch the graph of f1x2 = x . 3 3x4/3 Note that f 1x2 is positive except at x = 0. while it is concave downward to the left and upward to the right of the relative minimum on the curve on the right. a point cannot simultaneously be a relative extremum and an inflection point. Figure 17: Concavity Around a Non Smooth Critical Point It also follows that for smooth functions. Inc. . See Figure 17 where we illustrate a relative minimum. The denominator is zero when x = 0.286 * ** Section 3. see the graph on the right in Figure 17.2)/ x 1/3 DNE 0 0 8 + Figure 18: sign of f ¿ 1x2 = 1x1/3 .0) is a relative maximum. f ¿ 1x2 = 1 . Wang. However. this could occur. the graph is concave upward both on its left and right. Therefore. by smooth we mean that a graph has no sharp points. by Warren B. when the numerator or the denominator is zero. That is. its derivative exists everywhere. Copyright © 2007 by Pearson Education. At M(0. Walter O. the graph is always concave upward. where it fails to exist. x. Copyright © 2007 by Pearson Education.32 1x2 . and April Allen Materowski. (Note. Hence. To prove it is the minimum. Finding the derivative. It may often be complemented with the Only Critical Point Test. 0) m(8.52 7 0.7 and 2. and Finance. Walter O. We can now include examination of the first and second derivative to improve the sketch.25 = 0.32. this gives x1/3 = 3. the domain is the unbounded open interval.5 gives a relative minimum. There we sketched the graph by examining the zeros. x = 5/2. -4) Figure 19: Beginning the Sketch of f1x2 = x . as the next example illustrates. so the x-intercepts (zeros) occur when x = 0 or x1/3 . Example 12 Sketch the graph of the function defined by f1x2 = 2x3 . Setting C ¿ 1x2 = 0. 1.3x2/3 M(0. Published by Pearson Learning Solutions. . 2 M(0. and by the Only Critical Point Test. 23. we find x = 27 as the second x-intercept.3 = 0. the negative square root is not considered since the domain is x 7 0. we obtain C ¿ 1x2 = 4 . even though the function has vertical asymptotes at these x-values. the second derivative test is often useful when finding the extrema of a function.3 Concavity and the Second Derivative * ** 287 Note that in the previous example. as they determine the sign of the second derivative. We illustrate with an Example. We find C 1x2 = 50/x3. we have that 4x2 . we considered rational functions. is given in thousands.Section 3. since x Z 0. Example 11 The average cost of producing a doll by a toy manufacturing company is given by the equation C1x2 = 2000 + 4x + 25/x. we need only show it is a relative minimum.4. We see that we need not worry about the denominator vanishing.25/x2.) Applied Calculus for Business. Inc. the minimum. giving C 12. 1 the function has vertical asymptotes. f ¿ 1x2 = 2x2 1x2 . x2 . 4x2 = 25 x = 25/4. Figure 20: The Sketch of f1x2 = x . we can write f1x2 = x . in theory. Sign analysis of the second derivative is given in Figure 21. we must have x 7 0. so the only critical numbers are x = 0. -4) That is.3x 2/3 In Sections 1.) The only critical point of the function is at x = 2. Wang. The second derivative is f 1x2 = 4x1x2 + 32 1x2 .3x2/3 = x2/31x1/3 .1 Solution You should verify that this function has no horizontal asymptote and has vertical asymptotes at x = . Economics. x 7 0.122 We know that at x = . Of course. 0) m(8. How many dolls should be manufactured if the average cost is to be minimized? Solution The reality of the problem restricts the domain of the cost function. horizontal and vertical asymptotes of the function. Gordon. We take the derivative to obtain. (Note that we need to consider the zeros of the denominator. x = 2. Thus. x = 2. Obviously. and x = . Therefore. The average cost C1x2 is given in dollars per thousand dolls and the number of dolls.123 It will be zero when x = 0.5. by Warren B. the average cost of making no dolls makes no sense. It should be noted that in applications.5 thousand dolls. and one cannot manufacture a negative number of dolls. there is no upper bound to the number of dolls we can build. cubing both sides. 8x + 18yy ¿ = 0. At x = . x . sketch the curve whose equation is 4x2 + 9y2 = 36. why?). It fails to exist when y = 0. x = -1 x=1 m I (0. 22. To find the second derivative. Gordon.1 2 Implicit Differentiation and Curve Sketching The final example of this section uses implicit differentiation as developed in Section 2.3 23 B and at x = 23 the graph is concave upward so there is a relative minimum at m A 23. it follows that at x = 0. 3. 3 23 B .3 x 3. at the x-intercepts 1 . they are x = . and Finance. . the function has an inflection point (the tangent line is horizontal at this point as well.23. and April Allen Materowski. Economics. using the quotient rule to find y . Thus. Similarly.4x2.3 Concavity and the Second Derivative CD VA sign of f "( x ) 4 x ( x 2 + 3)/( x 2 -1) 3 -1 CU + CD 0 0 VA 1 CU + Figure 21: Sign of f 1x2 = 4x1x 2 + 32/1x 2 .288 * ** Section 3. . Therefore. The graph is sketched in Figure 22.23 the second derivative is negative so the graph is concave downward so we have a relative maximum at.123 From the second derivative test. The x-intercepts occur at the endpoints of the curve. Copyright © 2007 by Pearson Education. or y ¿ = . (Why?) We find the derivatives implicitly. 3. the y-intercepts occur when y = 0. they are not relative extrema.8x/18y = . Inc.4x/9y The derivative is zero when x = 0. That is. They are y = . Applied Calculus for Business. we again differentiate implicitly. at the y-intercepts 10. M A . . Walter O. by Warren B. 2. Published by Pearson Learning Solutions. 02.8. 236 . Example 13 Using the first and second derivatives. Wang. 0) M Figure 22: f1x2 = 2x3 . Solution The permissible x-values may be found by solving for y which gives 1 y = . This curve is not the graph of a function. . 3 We note that the y-intercepts of the curve occur when x = 0. derivative and second derivative. 4C 9 y .5.c 9 y2 Substituting for y ¿ . Published by Pearson Learning Solutions. This curve is.4x b 9y S y = - Simplifying the complex fraction. by Warren B.16 . an ellipse. or zeros are outside or not visible in the usual window. this only works if you know where to look for the inflection points. first by plotting the critical points in Figure 23. We sketch the curve. . It is clear that for higher order polynomials it can be difficult to determine the zeros of the function. the 2 indicating the second derivative. the syntax is as follows: d(y1(x). 0) m(0.22 is a relative minimum. Copyright © 2007 by Pearson Education. Economics. 0) m(0. selecting inflection and then choosing lower and upper bounds. M(0. -2) (3. and April Allen Materowski. and completing the curve in Figure 24. To find the zeros of the second derivative (and possible inflection points). Inc. we obtain.xa y2 . M(0. the differentiations may be messy as well.3 Concavity and the Second Derivative * ** 289 4 y[1] . Calculator Tips Applied Calculus for Business. we need only enter solve 1d1y11x2. we have. and m10. Wang. Inflection points may also be obtained from the GRAPH window by using F5. we have that 4x2 + 9y2 = 36. However.c 9 9y3 However. -2) Figure 23: Beginning the Sketch of 4x2 + 9y2 = 36 Figure 24: Completing the Sketch of 4x2 + 9y2 = 36 The second derivative may easily be found using the TI 89. and Finance. or for other kinds of functions. from the equation of the curve. 2). assuming the function was named y1. It is precisely in these cases that the calculator is a useful tool in providing a qualitative sketch of a graph. Gordon. Consider the following example.41362 919y 2 3 = . . 4 4x2 + 9y2 d y = . inflection. x. as we saw in Section 1. and it is negative (concave downward) when y 7 0. x2. Walter O. 2) (-3. Thus. we see that y = . x. A qualitative sketch is needed when given a function whose relative extrema.Section 3. 9y3 Note that the second derivative is positive (concave upward) when y 6 0. 2) M(0. Substituting. 22 = 0. 2) is a relative maximum.x[y ¿ ] d y¿ = . (b) Find s 1t2 if s1t2 = 21 . Find the first five derivatives of f. d2y (c) Find 2 if y = 3x2 . Published by Pearson Learning Solutions.825x4 + 6000x32 100000 EXERCISE SET 3. Wang. x2 = 0. so we cannot expect to get anything reasonable on our calculator. x2 respectively.123. find f ¿ 1x2. However. What happens for the sixth and higher order derivatives? 8.123 5 2 4 2 5. f1x2 = x4. (a) Find f1321x2 if f1x2 = 1/x2. 7.3 1. and which zeros of the second derivative are inflection points.t. 3. by Warren B. After doing so. If f1x2 = a4x4 + a3x3 + a2x2 + a1x1 + a0. rela100000 tive extrema and inflection points. (b) f1x2 = 2x + 3x + 1. f1x2 = 1x3 . dx Applied Calculus for Business. Copyright © 2007 by Pearson Education. all zeros. and April Allen Materowski. It is an easy matter. Economics. 6. We will leave these to you as an exercise. For the xvalues found. Gordon.290 * ** Section 3. Find f 1x2 for f1x2 = 1x . x2 and solve 1d1y11x2. solve 1d1y11x2. we can find the zeros of the function. Find 2 . d2y 9. If f1x2 = a5x5 + a4x4 + a3x 3 + a2x 2 + a 1x + a0. What do you observe for f1n21x2 when n Ú 5? 2. Solution The entire graph of this function is not viewable in any reasonable window. and Finance. f1321x2 and f1421x2. on your sketch. Walter O. Figure 25: A Qualitative Sketch of 1 y11x2 = 12x6 .825x4 + 6000x32. The breaks in the axes indicate the graph is not drawn to scale. f 1x2. Find f 1x2 for: (a) f1x2 = (b) f1x2 = 1x2 + 1210 1x2 . find (a) f ¿ 102 (b) f 102 (c) f132102 (d) f142102 (e) f1n2102 for n 7 9. we can provide a qualitative sketch of the function as given in Figure 25. if x1/2 + y1/2 = 6. find f1n21x2.12/13x + 12. Show. we use the calculator to compute the corresponding y-values.24x5 . .3 Concavity and the Second Derivative Example 14 Sketching the graph of the function defined by 1 y11x2 = 12x6 . x2.22 = 0. x. where n is any positive integer. to verify which zeros of the first derivative are indeed relative maxima and minima. Inc. Find f 1x2 if: (a) f1x2 = x + 3x + 5.24x5 .7x + 9 dx 1x2 + 122 4. first and second derivative using solve 1y11x2 = 0. Draw the smooth graph through these points. I3. if x2 + y2 = 1. The position of an object at any time is given by the equation s1t2 = 3t3 .3t5.4t + 11. Figure 26: Exercise 19c 20. the relative minimum m. I3. and list all inflection points. (b) Does this function have an inflection point? 19.1 x 6 2 and decreasing and concave upward on 2 6 x 5. Assume that the derivative fails to exist at M. f1x2 = . Walter O.y1/2 = 6. (See Figure 26) m Figure 27: Exercise 20 21.12/1x + 12 30. 34. (a) Draw the graph of a function which is increasing on (0. 36. and April Allen Materowski. M. Find 12.2y2 = 6. m and the inflection points I1. f1x2 = 1x . f1x2 = . by Warren B. if 3x2 . 16. f1x2 = 2x + 1. Economics.20x3 + 10x + 30 29. Suppose that the position of a particle as a function of time is given by the formula s1t2 = 4t2 .6x2 + 2x + 1 26. Find d2y dx d2y 2 Concavity and the Second Derivative M * ** 291 .3 10. Copyright © 2007 by Pearson Education. f1x2 = 1 . f1x2 = 4x3 .Section 3. f1x2 = x4/5 . In Exercises 22 39 determine where the function is concave upward and downward. and I5. Repeat Exercise 16 for s1t2 = 5t3 . .2x3 + 5x2 + 7 27. (c) Find the time at which the velocity is maximum. f1x2 = . (a) Find the velocity and acceleration as functions of time. . . t 7 0. f1x2 = 3x/14x2 + 92 Applied Calculus for Business. f1x2 = x 23 . f1x2 = 29 . . sketch the graph of the function passing through the relative maximum. 1) and concave downward on (1.2x3/5 38.3x2 + 2 24.13x2 + 12x + 9 42. f1x2 = 4x3 . 35. 31. M.x.x2. 18. f1x2 = 3x4 + 16x3 + 6x2 . the relative minimum. and Finance. Find 11. dx2 d2y dx d2y 2 I3 I4 I2 I1 dx2 14. I4. 39. 4x2 + 9 33. Find the equations of its velocity and acceleration. As in the preceding exercise.8t + 21.1221x + 322 44. 17. but f ¿ 1x2 and f 1x2 exist everywhere else. Sketch the graph of a function having a critical point that is simultaneously a relative extremum and an inflection point. I2. f1x2 = 3x5 + 5x4 . if x1/2 . (b) Can such a function have any relative extrema or inflection points? (c) Suppose you have located the relative maximum.2x + 3 41. f1x2 = ax2 + bx + c: (a) a 7 0. (b) a 6 0 25. 22. Wang. if 3x2 + 2y2 = 6.x3/4. f1x2 = x2 . (b) Find the time at which the velocity is zero and the time at which the acceleration is zero. Find 13. f1x2 = 1x . (d) Find the time at which the acceleration is maximum. f1x2 = x . f1x2 = x 2x . Find the equations of its velocity and acceleration. f1x2 = 29 + x2. Gordon. Published by Pearson Learning Solutions.6x2 + 9x .t4 for t 7 0. 37. 2) and such that it is concave upward on (0.4 3x 32. The position of an object at any time is given by the equation s1t2 = 5t3 . I2. 15. and the inflection points I1.x2/3 In Exercises 40 44 use the second derivative test to classify the critical points. f1x2 = x4 . and I4.72x + 20 43. 2). Inc. as given in Figure 27. f1x2 = 4x2 23. f1x2 = 4x2 x .1 2 M I4 I3 I5 I2 I1 m .2 28. (a) Draw the graph of a continuous function increasing and concave downward on . 40. x 0 sign of f " (x ) -4 + 0 -1 0 0 + 0 2 0 5 + Figure 28b: Ex 67 Figure 31: Ex 70. f1x2 = . Suppose that f ¿ 1c2 = f 1c2 = 0. Show that the function has a relative maximum (minimum) at x = c. Suppose the graph of f ¿ 1x2 versus x is given in Figure 30. 67. 4y2 . Published by Pearson Learning Solutions.7). Also include the intercepts. Include concavity in your sketch. 72.6x + 8x 50. The Graph of f ¿ 1x2 versus x 70.42 4 5 3 3 2 2 + sign of f '( x ) DNE 1 0 3 + Figure 29a: Ex 68 + sign of f " x) DNE 1 0 2 + Figure 29b: Ex 68 69.1.4x3 + 12x2 . Draw the graph of a continuous function which is concave downward on 1 .15 51. 52 and increasing on (5. f1x2 = x . Copyright © 2007 by Pearson Education. f102 = 5. sketch a possible graph of f(x) versus x.22 = . (b) f1x2 = 1x . The Graph of f ¿ 1x2 versus x 71. f1x2 = 3x . f1x2 = x 2x + 4 60. f1 . f1x2 = 3x/14x2 + 92 59. Sketch the graph of a function satisfying the following sign diagrams (Figure 28a and 28b. Gordon. 55. 5 6 f ' (x) x Figure 30: Ex 69.5x + 6 46.6x + 6 47. Suppose the graph of f ¿ 1x2 versus x is given in Figure 31. by Warren B. f1x2 = x3 + 3x2 . f1x2 = x . f1x2 = x3 . In Exercises 45 65 use the first and second derivatives to sketch the graph of the given equation.x4 + 6x3 .8x + 112 57. 72. Economics.12 = 1.2x3/5 63. and f152 = 2. Sketch a function satisfying the sign diagrams given in Figures 29a and 29b.x2 = 16 66.x2/3 64.).5x 61. sketch a possible graph of f(x) versus x. Include concavity in your sketch. f1x2 = 5x .24x2 52. (c) f1x2 = . Inc.1231x + 324 In each case.292 * ** Section 3. f1x2 = 1*41x4 . f1 .3x 56.12x 48.224. f '(x) 0 sign of f '( x ) -2 + 0 4 Figure 28a: Ex 67 additionally. whenever they are easily determined. but f 1x2 is negative (positive) just to the left and right of x = c. f1 . and Finance.12x2 + 10 54.1x .2x2 58. f1x2 = 1x . Applied Calculus for Business.3 Concavity and the Second Derivative 68. .24x2 + 25 53. 45.223. x2 + 4y2 = 16 65. f1x2 = 1*4x4 .224.9x . Show that the second derivative test fails for: (a) f1x2 = 1x . Walter O. f142 = 9. classify the critical point(s). f1x2 = x3 + x + 2 49. f1x2 = 5 + x2/3 62.2. decreasing on 1 .1.42 = 3. Wang. (d) f1x2 = 1x . and April Allen Materowski. f1x2 = x4 . f122 = 7. f1x2 = x2 . f1x2 = x4/5 . Use this result to classify the critical points in Exercise 71. f1x2 = x4 . f1x2 = 6x . that relate the variables so that the objective function is of the form y = f1x2. and Finance. Show that a smooth odd function passing through the origin has an inflection point there. The object of this section is to look at different types of problems which may be solved using those procedures developed in the last few sections. f ¿ 1x2 = 6x .4 73. Show that y = . the function to be optimized involves more than one variable.) 75. Example 1 is a case in which there are no constraints. show that ak = where the symbol k! # Á k! = k1k .12x + 17. Therefore. Thus. the coordinates of the inflection point are the averages of the corresponding coordinates of the relative extrema. Since this is a positive number.12x2 + 10 0 (Hint: Exercise 52. b. 78. Wang.1)x m . Sometimes. c and d such that the graph of f1x2 = ax3 + bx2 + cx + d will have both a relative maximum and relative minimum? (b) Show that. if they exist.h2 (b) lim h:0 h 74. Example 1 Find the minimum value of f1x2 = 3x2 . we find the derivative. Assuming each limit exists. Gordon. Sometimes there will not be. the methods of this chapter can only be used if you can determine equations. We begin with two simple examples which illustrate the basic procedure. Solution As we have learned previously. Referring to Exercise 7. use the preceding exercise to find the values for the ak s. we wish to determine its maximum or minimum value. We shall see that the solution to such optimization problems requires an examination of the derivatives of the function. The symbol 0! is defined to be equal to 1. compute them.x4 + 6x3 . Economics.1 if ax m + bym = c.f ¿ 1x . Sometimes there will be an additional constraint on the possible values of the variables. Applications I Geometric Optimization Problems * ** 293 77. Calculating the second derivative. f ¿ 1x + h2 . We have found the solutions to several such problems in the previous two sections. called constraint equations.4 Applications I Geometric Optimization Problems » » » » » Area and Perimeter Optimization Procedure Volume Distance and Velocity Calculator Tips Given any function defined by the equation of the form y = f1x2.12. Copyright © 2007 by Pearson Education. 4! = 4132122112 = 24. The only critical point is at x = 2. one good place to look for optimum values is at critical points. and April Allen Materowski. We shall see that many optimization problems may be reduced to problems of the form given in Example 1 or 2. If f1x2 = 11 + x25 = a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5. . Show that any smooth even function has a relative extremum at its y-intercept. we get f 1x2 = 6. Example 2 illustrates a situation that involves two variables and one constraint. Show that . The symbol ! is read factorial. the function has a relative minimum which Applied Calculus for Business. by the second derivative test.121k . 76. Inc. Walter O. In such cases. We shall call the function to be optimized the objective function. 81. (a) What condition must be imposed upon a.Section 3. by Warren B. when this condition is satisfied. Published by Pearson Learning Solutions. 80. f1k2102 79.9x . Determine the number of real roots of the equation x3 + 3x2 .ac(m . 3. a function of a single variable.2 b 2 y 2m .15 = 0.f ¿ 1x2 (a) lim h:0 h f ¿ 1x + h2 .22 2 1. Walter O.162 + 2 A x2 . One other remark. Since the objective function contains two variables. Gordon.25 L 14. From the constraint equation. Second.4 Applications I Geometric Optimization Problems is its minimum (by the Only Critical Point Test) at x = 2. you should usually do two things: First. namely. we obtain.1*2 B 2x = 4x3 . dx2 Thus. Published by Pearson Learning Solutions. Example 2 Minimize D = 1x .1622 + A y . at (2. Determine the dimensions of the corral if its total area is to be maximized.8 = 0. However the minimum value of S = 2208. D = 1x . The equation y = x2 is a constraint. we have.43. Copyright © 2007 by Pearson Education. D has a relative minimum at x = 2 and by the Only Critical Point Test. .1*2 B 2 = 208. and the minimum is D = 12 . or x = 2. Notice that S = D. the minimum occurs when x = 2.82.32 = 41x3 . Economics.25. 4).1 2 B . Our objective is to determine x and y so that the total area is maximized.1622 + A y . The corral is to be subdivided by two fences parallel to the sides. Substituting for y from the constraint equations. Suppose. Area and Perimeter Example 3 A rancher has available 1600 feet of fencing to construct a rectangular corral. y = x2. and. which we can use to eliminate one of the variables. we have y = 4. When the objective function involves two variables. we try to find a constraint equation. and Finance.1622 + A 4 . d2D = 12x2 7 0 if x = 2. x and y. dx The only critical point occurs when x3 .1*2 B 2 subject to the condition that y = x2.294 * ** Section 3.1*2 B 2 We now differentiate dD = 21x . in the preceding example you were asked to minimize 2 2 s = 31x . Wang. Applying the second derivative test to classify the critical point. it tells us what the relationship between x and y is. Example 2 illustrates a very important principle. Solution Our objective is to minimize D. Applied Calculus for Business. identify all the variables. and April Allen Materowski. Let x be the width of the corral and y its length (see Figure 1). Note that we did not need calculus to solve the last example. we use the constraint equation to eliminate one of them. Clearly a parabola opening upward has its minimum at its vertex.1622 + A x2 . Let us now look at some geometric problems. and that the minimum of s will occur at the same place that D has its minimum. by Warren B. Inc. Solution When looking at any geometric problem. The minimum value is f122 = 5. sketch a figure illustrating the problem. and April Allen Materowski. we have that 4x + 2y = 1600. . The previous examples suggests a general procedure that may be applied to many optimization problems. Optimization Procedure Volumes Applied Calculus for Business. Find the critical points of the objective function. 6.4 6 0. Gordon. the rancher should construct a corral with length 400 feet and width 200 feet. Therefore. When appropriate. Substituting for x in the constraint equation. Define the variable or variables to be determined. The rancher has 1600 feet of fencing available for the corral. and Finance. Justify your conclusion Example 4 A manufacturer wishes to produce a (rectangular) box with an open top with maximum volume from rectangular sheets 8 inches by 15 inches. A = xy. 4. Using the second derivative test. implying that A achieves its maximum value at x = 200. Solving this constraint equation for y. 2.212002 = 400.2x Thus. Published by Pearson Learning Solutions. The relationship is precisely the constraint equation. we must find a relationship between them. and then turning up the sides to form the box.4 y Applications I Geometric Optimization Problems * ** 295 x x x x y Figure 1: Illustrating the Corral To achieve this objective we must maximize the total area. Copyright © 2007 by Pearson Education. Therefore. Use the constraint equation and substitute into the objective function reducing it to a function of one variable. The objective function has two variables.2x2 A ¿ = 800 . by Warren B. The box will be constructed by cutting out an equal square from each corner. Economics. 5.2x2 = 800x .Section 3. we have A = . To eliminate one of them. OPTIMIZATION PROCEDURE 1. we obtain y = 800 . Inc. A = xy = x1800 . Wang. Walter O. when appropriate. to maximize the total area. we have.4x and we have the only critical point at x = 200. Determine the objective function to be maximized or minimized 3. y = 800 . Determine the dimensions of the box. determine the constraint equations. sketch a figure illustrating the problem and the variables. by Warren B. Rather than apply the product rule we multiply the factors obtaining V = 4x3 . since the smaller of the two dimensions is 8.2x2.2x218 . Wang.2x. Note that the height of the box is x. Setting V ¿ 1x2 = 0. Also observe that a constraint is imposed upon x. Figure 2a shows the rectangular sheet with the square corner to be removed drawn dashed. and April Allen Materowski. The dotted lines in Figure 2b indicate where the sheet is to be folded to construct the box. Economics.2 x x Figure 3: The Box 2. and Finance.46x2 + 120x Differentiating. 5. V ¿ 1x2 = 413x2 . the width is now 8 .23x + 302 = 0 413x . 8 8 . x must be less than 4. Copyright © 2007 by Pearson Education. V¿1x2 = 12x2 . Walter O.4 Applications I Geometric Optimization Problems Solution 1.2x x x x x x x x x 15 15 .62 = 0 x = 5/3 or x = 6 Applied Calculus for Business. The volume of a (rectangular) box is the product of the height. That is. 3.2x. This step is not necessary as the objective function contains only one variable. The problem reduces to maximizing V = LWH = x115 . .2x Figure 2b 15 .521x .2 x x x x x x x x x Figure 2a 8 . Published by Pearson Learning Solutions. Folding along the dotted lines gives the completed box as illustrated in Figure 3. Figure 2b shows the sheet with the corners cut out.92x + 120. This constraint dictates the domain of the function to be optimized.296 * ** Section 3. Gordon. 1. Let x be the width of each square corner. otherwise there is no material left to construct the sides of the box. The constraint is 0 6 x 6 4. length and width. we look for critical points. 4. and the length is 15 . Note that since x inches are cut from each corner. Inc. That is. and 14 35 2450 3 length 15 . as shown in Figure 4. Where on the river should the pumping station be located in order to minimize the length of piping used in connecting the towns? Solution 1. Copyright © 2007 by Pearson Education. located (see Figure 5). 6. . L ¿ 1x2 = 1*21x2 + 362-1/22x + 1*21x2 . as we shall see.18x + 90 9 .92. but its solution requires some algebraic manipulation. Note that 0 C 3 River 9 D C 3 9-x P x D 6 A A B 6 B Figure 4 Figure 5 2. and the volume = A 5 3 B A 3 B A 3 B = 27 in . P. and V 15/32 = . we must cut out a square of side 5/3 inches from each corner. is x 9. respectively. Walter O. and Finance. we write x 2x + 36 2 x 2x + 36 2 + x .18x + 90 3. The next example is easy to set up.x 2 = 0 = 2x . and April Allen Materowski.52. We have that 0 x (Step 4 is not necessary. Differentiating. Gordon.18x + 902-1/212x .Section 3. Example 5 Towns A and B are located 3 and 6 miles respectively. Wang.2x = 14/3 inches.x22 = 2x2 . the distance from P to B is 2x2 + 62 = 2x2 + 36. Let x be the distance along the river from town B to where the pumping station. by Warren B.9 2x . Economics. The distance between points C and D is 9 miles. width 8 . and setting L ¿ 1x2 to zero.) 5.182. L ¿ 1x2 = To solve for x. and the distance from A to P is 232 + 19 . we must minimize L = 2x2 + 36 + 2x2 . Inc. from a straight river. Simplifying. we have.2x = 35/3 inches. Let C and D be the points on the river bank closest to A and B. Thus. 9.18x + 90 2 Applied Calculus for Business. V 1x2 = 24x . By the Pythagorean Theorem. The two towns are to be supplied water from a pumping station situated on the river between them.4 Applications I Geometric Optimization Problems * ** 297 Since x must be less than 4. Published by Pearson Learning Solutions.18x + 90. The resulting box will have height x = 5/3 inches. the only critical point in the domain is at x = 5/3. Therefore. the volume is maximized when x = 5/3 inches. 71860214 6 0 and L ¿ 16. the physical and mathematical explanations are in complete agreement. Distance and Velocity In Section 2. Copyright © 2007 by Pearson Education. and by the Only Critical Point Test. we find that the critical point is at t = 3. Wang. To classify the critical point it is easiest to apply the first derivative test. Inc. A computation gives. and then over land to the Applied Calculus for Business. The maximum height of the ball is s132 = . How high will the ball go? Assume that the height of the ball as a function of time is s1t2 = .18x + 90 2 Clearing fractions (cross multiplying) we find. Solution Since we have s as a function of t already. x = 6 is the only critical point in the given domain. or 1x .18x + 81 = 2 x + 36 x . A power line is to be run from the power station.648x + 36x2.161322 + 96132 + 192 = 336 feet above the ground. . by Warren B.0. Economics. which is x miles up-river from P. a maximum.621x .12 = 0. (Note that we could have also determined the minimum using the Extreme Value Theorem. Gordon.648x + 2916 = 0 Dividing by 27 x2 . The derivative is s ¿ 1t2 = . Thus. Exercise 29 indicates a geometric method. and Finance.4 Applications I Geometric Optimization Problems Squaring both sides gives x2 x2 . From a physical point of view. we begin at step 5. implying that at the only critical point L is a minimum. Walter O. x4 . 27x2 . and April Allen Materowski.712926292 + 0.92 = 0. under the river to some point A. Example 7 A power station is on one side of a straight river which is five miles wide. 20 miles up-river.7 we solved similar problems by setting the velocity v = 0. Since x 9. 6. s 132 6 0. Thus.182 = 0. 6. L ¿ 15. the ball will reach its maximum height at the instant when its velocity is zero. how?) It is interesting to observe that this problem may be solved without the use of calculus. Setting this to zero.18x3 + x4 + 2916 . Therefore. Example 6 A ball is thrown vertically upward from the ledge of a building 192 feet high with an initial velocity of 96 feet per second.24x + 108 = 0.32t + 96. thus the critical point is a relative maximum.695022096 7 0 Thus.16t2 + 96t + 192. the pumping station should be 6 miles from D and 3 miles from C. Published by Pearson Learning Solutions. and a factory is on the other side.298 * ** Section 3.18x3 + 90x2 = 81x2 . the sign of L ¿ 1x2 changes from negative to positive at x = 6.701139952 . and April Allen Materowski. and Finance.x2 x Ú 0 and x 20 This can be done fairly easily with pencil and paper. if the total cost of the power line is to be as small as possible. Walter O.25. Solution P Calculator Tips 5 miles x A 20 . the solution is straight-forward and left as an exercise for you. by Warren B. now determine the point on the other side of the river where the power line is to come out of the river. . the point on the other side of the river where the power line comes out of the river. the calculations are considerably reduced using the calculator. Applied Calculus for Business.24. Wang.4 Applications I Geometric Optimization Problems * ** 299 factory (Figure 6). and the cost at this value is approximately $1312. The critical point turns out to occur at x = (Verify!). then the problem is to minimize y11x2 = 80 2x2 + 25 + 50120 .Section 3. However. Copyright © 2007 by Pearson Education. P 1 x A 20 . Economics. Solution Figure 7 illustrates the problem to be solved.x F |------------20 miles --------------| Figure 6 If we let y1 represent the cost. y1102 = $ 1400. however. this function is continuous on a closed interval. find x. 25 239 39 L 4. Published by Pearson Learning Solutions. y11202 = $ 1649.x22 x Ú 0 and x 20 This problem is very difficult to do without a calculator.0032 miles Example 8 Suppose the power station in the previous example is moved one mile inland. Gordon. therefore we need only compute the y-values at the two endpoint and critical points to determine the minimum cost. By the Extreme Value Theorem. To determine when the derivative is zero requires solving a fourth degree equation. Inc.x 5 F Figure 7: The Power Station One Mile Inland The problem is then to minimize the cost function y21x2 = 50 2x2 + 1 + 80 225 + 120 . with a calculator. If it costs $80 per mile to run the line under water and $50 per mile to run it overland. Wang. Consider the farmer in Exercise 7 whose field borders the river. Determine what the side of the square cut from the corners should be if the box is to have as large a volume as possible.4 1. If 400 feet of fencing will be used on the three other sides. Inc. what dimensions will minimize the cost of construction? 15. Suppose the tin for the top and bottom of the can cost 8 cents per square inch. (a) A farmer has a field bordering a straight river. A rancher has divided a plot of ground into an L shaped region. Gordon. A farmer wishes to build a rectangular pig pen using as little fencing as possible. Copyright © 2007 by Pearson Education. An aquarium is to be made in the shape of a rectangular solid with a square base and an open top. 10 17. Figure 11: Ex. The volume of the aquarium is to be 108 cubic inches. determine the dimensions x and y which maximizes the total area. Show that the answer to Exercise 3 follows from Exercise 2. (a) A wholesale manufacturer of canned corn wants a tin can (a right circular cylinder) that will have a volume of 54 cubic inches. The sum of two positive numbers is 100. Economics. What should the diameter of the can be if the cost of the can is to be minimized. Show that the rectangle with fixed area and minimum perimeter is a square. what should the dimensions of the plot be if its area is to be maximized? (b) What are the dimensions if the river side is also to be fenced? 8. find the dimensions that maximizes its area. 180. and April Allen Materowski. The larger section to serve as a corral and the smaller as a training area (see Figure 9). what should the numbers be if their product is to be as large as possible? 4. Determine its height and diameter. Show that the rectangle of maximum area with fixed perimeter is a square. Minimize the amount of fencing needed to construct the region shown in Figure 10. 3. 7. 5. What is the least amount of fencing needed? (see Figure 8. If the area of the pen is to be 225 square feet. A 12 centimeter square sheet of cardboard is to be made into a box by cutting equal squares from each corner and then folding up its sides. by Warren B.300 * ** Section 3. y Figure 9: Ex. A Norman window is one in which the window is constructed by capping a rectangular region with a semicircular region (see Figure 11). while the tin for the remainder of the can costs 1 cent per square inch. (b) Measure a can of corn from your local supermarket. Does its dimensions yield a minimum? (c) Why do you think manufacturers choose not to use this type of design? x 2y x 16. 2. Find the dimensions of the largest rectangle whose perimeter is 144 feet. what are its dimensions? 6. If the volume of the tank must be 50 cubic feet. 12 13. 11 12. 9 10. Walter O. If 480 feet of fencing is available. y /2 x /2 x y Figure 10: Ex. . Assume he wants the area of the plot to be 800 square feet. What should its dimensions be if he wants to use the least amount of fencing? 9. If the total perimeter of the window is to be 30 feet. A holding pen for fish is to be made in the form of a rectangular solid with a square base and open top. He wants to design a rectangular plot for an organic crop with the river acting as a natural border. Consider the can in the previous exercise. The base will be slate that costs $4 per square foot and the sides will be glass that costs $5 per square foot.) 11. What dimensions will minimize the amount of material needed to build it? 14. and Finance.000 square feet of land is to be enclosed in a rectangular plot which will then be subdivided into three plots by a pair of fences both parallel to the sides. As little tin as possible as possible is to be used in the construction of the can. if the total area is 480 square meters.4 Applications I Geometric Optimization Problems EXERCISE SET 3. Applied Calculus for Business. Published by Pearson Learning Solutions. x y Figure 8: Ex. Section 3. 3). Copyright © 2007 by Pearson Education. 29 31.4 18.a122 + 1x . where along the beach should she land so that she may get to John in the least amount of time? Solve the problem if John is (b) 100 miles. the location of the pumping station. by Warren B. Find the coordinates of its vertices if its area is to be minimized. Find the dimensions of the largest rectangle with lower base on the x-axis and upper vertices on the curve whose equation is (a) x2 + y2 = 4 (b) 9x2 + 4y2 = 36. Find the point on the curve y = 1x closest to (3/2. A right triangle is formed in the first quadrant by a line passing through the point (3.a222 has its minimum value when x = 1a1 + a22/2. 19. Economics. Deduce a contradiction by using the fact that the sum of the lengths of two sides of a triangle exceeds the length of the third side. 21 22. Gordon. 25. Use the theorem of Pythagoras along with similar triangles to find the total length. (c) 1 mile from A. and April Allen Materowski. 27. How should the wire be cut if the sum of the areas is to be maximized? 28. 4 miles from the beach (see Figure 12). Make sure to justify your conclusions. The top and bottom margins are each 1 inch wide. Where this line intersects the horizontal line is the required point P.a322 has its minimum value when x = 1a1 + a2 + a32/3. 5) and the coordinate axes. A 12 inch piece of wire is to be cut into two pieces. Walter O. A low stone wall 27 inches high is 64 inches away from a building. (b) Show that f1x2 = 1x . . Hint: The computation may be easier if you try to minimize d2 rather than d.22. on a straight beach. 0). Mary is in a boat in the sea at C. Find the points on the circle x + y = 9 (a) closest to and (b) farthest from the point (8. A rectangular page is to contain 96 square inches of print. Applied Calculus for Business.) 4 C Figure 12: Ex. Assuming Mary can row at 3 miles per hour and jog at 5 miles per hour. 2) (c) (2. 26. What should the dimensions of the page be if the least amount of material is to be used? 20. Inc. Wang. Find the dimensions of such a package with square ends whose volume is to be a maximum. Suppose postal requirements are that the maximum of the length plus girth (cross sectional perimeter) of a rectangular package that may be sent is 144 inches. See Figure 13. 2 2 2 2 A' P A B Figure 14: Ex.5 inches. 21.a122 + 1x . and Finance. 12). and the margins on each side are 1. 0) (b) (6. 29 A 10 B 30.) Pole Building 27 64 Figure 13: Ex. Draw a line connecting A ¿ to B. 10 miles from A. and from the fence to the building. Show that the solution to Example 5 may be found as follows: reflect A to the other side of the river to A ¿ as shown in Figure 14. Applications I Geometric Optimization Problems * ** 301 29. Complete the solution to Exercise 8. (a) John is at B. .a222 + 1x . the portion from the ground to the fence. What is the length of the shortest pole that passes over the fence and reaches the building? (Hint: The length of the pole is composed of two portions. (c) Generalize the above. Published by Pearson Learning Solutions. (a) Show that f1x2 = 1x . 24. One piece is to be used to form a square and the other to form a circle. Find the points on the curve whose equation is x + y = 16 nearest and farthest from (a) (8. Find the point on the line 3x + 4y + 7 = 0 closest to the point 11. 23. (Hint: Assume that some other point gives the minimum length. we have only a demand curve. From Figure 1 we see that 0 x 20. However. A sketch is not necessary here. Where p is the price per car in dollars and x is the number of cars that will be purchased at that price. Hence.000 and 0 p 200. 1. Instead. Thus.5 Application II Business and Economic Optimization Problems 3.000. . in business applications. we have that p = . Demand and Revenue In this section we shall apply the same methods used in the previous section. Thus. negative values for x and p are usually not allowable. That is.10x + 200000. Economics. What price should be charged per car if the total revenue is to be maximized? Solution We follow the steps in the procedure indicated in the previous section. The only difference will be that the examples to be considered come from economics and business. we consider a market situation in which there is no competition. Published by Pearson Learning Solutions.5 Applications II Business and Economic Optimization Problems » » » Price. we wish to maximize R 2 and 3. Figure 1: p = . Applied Calculus for Business. Gordon.000.) 4. That is. the quantity and sales price of a commodity is not the market equilibrium point determined by supply and demand curves. (Remember. Wang. Walter O. R = xp = . and the price is set by the monopolist producer. Solving the demand equation for p. and April Allen Materowski. R = xp. and Finance. We shall make some observations which follow directly from our knowledge of the derivative that are usually considered theorems in micro-economics. Our first example shall deal with the revenue available to a monopolist. Inc. Example 1 Suppose that the price and demand for a particular luxury automobile are related by the demand equation p + 10x = 200.10x2 + 200000x. Demand and Revenue Cost and Average Cost Elasticity of Demand Price.302 * ** Section 3. We recall that the total revenue is defined as the product of the price per item times the number of items sold. Copyright © 2007 by Pearson Education. if we sketch the demand equation we will see that there are constraints imposed on the variables x and p. by Warren B.10x + 200000. it really does not matter which variable is eliminated. a maximum. implying that the revenue has a relative maximum at x = 10. Published by Pearson Learning Solutions.20 6 0. the average cost is minimized when x = 70. R ¿ 1x2 = . Let us look at some such examples. and R = xp = 1 .2. at the critical point. we solved for p in terms of x and then found R in terms of x. Thus. We remark that in Step 4.000 apiece. x = 70.0002p = . Copyright © 2007 by Pearson Education.000. producing a revenue of one billion dollars. the optimum strategy for the manufacturer is to produce 10. C has a relative minimum.000 = $ 100. we have a class of business problems in which the object is to minimize cost (or average cost). by Warren B. in dollars. Setting 2x . it is the minimum. Applied Calculus for Business. As you can see.000 cars to sell at $100. x = .20x + 200. Since this is the only critical point.0. Inc. Thus. x2 Setting this equal to zero. and the minimum average cost is C1702 = $ 138. The minimum cost is C112 = $ 4899. What is the minimum cost? Solution We may begin immediately to solve the problem. to find the critical point. The only critical point is where R ¿ 1x2 = 0. we have seen one typical type of business or economic application. C ¿ 1x2 = 2x . . of producing x hundred bicycles is given by C1x2 = x2 .000p.2x + 4900. we obtain x = 10.10110. or x2 = 4900. The object was to maximize revenue. Therefore. Cost and Average Cost Example 3 In the preceding example find the minimum average cost.0.0002 + 200.1p2 + 20.5 Application II Business and Economic Optimization Problems * ** 303 5. therefore.000.000 = 0. R 1x2 = .1p + 20.0. To complete the problem.1p + 20. C1x2 x2 2x 4900 Solution The average cost. and April Allen Materowski. p = .000. In Example 1. we must find the price per car when the revenue is maximized. we get x = 1. C1x2 = = + = x . C 1x2 = 2 7 0. Next differentiate with respect to p to find that the revenue is maximized when p = $ 100. On the other side of the coin.20x + 200. we have 4900/x2 = 1.Section 3.2 = 0.2 + 4900x-1 x x x x Differentiating. the price per car.000. we find C ¿ 1x2 = 1 4900 . C 1x2 = 9800/x3 7 0 when x = 70. Gordon. Solving. Economics. Setting . More realistic examples would involve maximization of profits. we could solve for R as a function of p by first finding x in terms of p in the demand equation. and Finance. Wang.000. Alternately.000 and. The only critical point is where C ¿ 1x2 = 0. When x = 10. We leave the remaining details as an exercise.000. That is. 6. Example 2 Suppose that the cost. Walter O. and April Allen Materowski.0.170x + 8002 Applied Calculus for Business. That is.5 Application II Business and Economic Optimization Problems Economists often solve problems like the one considered in Example 3 by using the fact that the average cost is minimized at the level of production at which the average cost equals the marginal cost. The remainder of the solution follows as above. the only critical point comes from the zero of the numerator. Since the cost is given in terms of x. Dividing by x. The demand equation is 20p + x = 18000. when the marginal cost equals the average cost. . we have that x = 4900/x.2 = x . we have 2x .C1x2 = 0 or equivalently. PROOF The proof follows from the quotient rule. Thus. the marginal cost is C ¿ 1x2 = 2x .2. (a) What level of production maximizes profit? (b) What is the price per stereo when profit is maximized? (c) What is the maximum profit? Solution Recall that the profit. P.05x + 900. R1x2 = xp = . Wang.304 * ** Section 3. when xC ¿ 1x2 . or x2 = 4900 as above.C1x2 = . Published by Pearson Learning Solutions. THEOREM 1 The average cost is minimized at a level of production at which marginal cost equals average cost. Walter O.C1x2 x2 .2 + 4900x-1.0. C1x2 = C1x2/x. (See Theorem 1 below. setting C ¿ 1x2 = C1x2. Inc.0. Economics. when C ¿ 1x2 = C1x2. Example 4 The cost in dollars of producing x stereos is given by C1x2 = 70x + 800.2 + 4900x-1 Simplifying. rather than in terms of p. From the demand equation we have p = . Since x 7 0. is the difference between the revenue and the cost. we have C ¿ 1x2 = C1x2/x = C1x2. Gordon. differentiating. and Finance.05x2 + 900x P1x2 = R1x2 . we shall find the revenue in terms of x. and the average cost is C1x2 = x . by Warren B.) In the above example. Copyright © 2007 by Pearson Education. that is. C ¿ 1x2 = xC ¿ 1x2 . xC ¿ 1x2 = C1x2. That is.05x2 + 900x . 443. the level of production at which profit is maximized is x = 8300. . Therefore.C1x2 thus P ¿ 1x2 = R ¿ 1x2 .0. Walter O. For each ten cents that she raises the ticket price she will sell three fewer seats. the profit is maximized when the marginal revenue equals the marginal cost.1 6 0.051830022 + 830183002 . Inc.0.0. Published by Pearson Learning Solutions. (c) The maximum profit is P183002 = . the profit is maximized when the marginal revenue equals the marginal cost.C ¿ 1x2 The critical point occurs when R ¿ 1x2 . or R ¿ 1x2 = C ¿ 1x2. Copyright © 2007 by Pearson Education. Again. that is.0. At $4 per ticket.0. The remainder of the solution continues as above.5 Application II Business and Economic Optimization Problems * ** 305 or P1x2 = . That Applied Calculus for Business. C ¿ 1x2 = 70. we find that P ¿ 1x2 = . she sells 204 tickets. One ten cent increase brings the price to $4. Gordon. P1x2 = R1x2 . Differentiating.1x + 830. and Finance.800 = $ 3.05x2 + 830x . Let us see what the given information tells us. An economist will use the fact that for a realistic economic function. since this is the only critical point the relative maximum is the maximum. Solving. That is. We could have used Theorem 2 to do Example 4 as follows: find the marginal revenue and marginal cost.700. the profit has a relative maximum at x = 8300.00 per ticket. as the next example illustrates.05183002 + 900 = $ 485. THEOREM 2 For realistic economic functions.C ¿ 1x2 = 0. by Warren B. Example 5 A theater has 204 seats. There are situations where we are given data from which we must construct the function to be optimized. p = .0. and April Allen Materowski. The manager finds that she can fill all the seats if she charges $4.1x + 900. Equating the marginal revenue with the marginal cost yields .10 and leaves three seats unsold.800.1x + 900 = 70.0. P 1x2 = . (a) The level of production that maximizes profit is x = 8300. when R ¿ 1x2 = C ¿ 1x2 PROOF The proof is an immediate consequence of the sum rule for derivatives. (b) The price charged per stereo comes from the demand equation. R ¿ 1x2 = . What ticket price should she charge to maximize the ticket revenue? Solution The ticket revenue is the product of the number of tickets sold and the price per ticket. when the marginal revenue equals the marginal cost.Section 3. Economics. Wang. The critical point is at x = 8300. 40.2132 Let n = the number of ten cent increases. when the price is $4.6 6 0. 4) and (201. Gordon. our measurement should be independent of units. The price per ticket when there are x increases is 4 + n1. The resulting ratio is a dimensionless quantity.n132 = 204 .102 = 4 + 0.1132 tickets are sold. Therefore. Elasticity of Demand Given a demand function. 204 .2132 seats are sold and so on. DEFINITION 1 The elasticity of demand. The price is increased by 141.6n + 8.306 * ** Section 3. the number of tickets sold is 204 and when the price is $4. then the value of dx/dp is multiplied by a factor of 2000 making it 4000 pounds per dollar. Walter O.1132 204 . . Certainly. we divide dx/dp by the quantity x/p. Published by Pearson Learning Solutions.0.20 and lowers the number of seats sold to 198. the total ticket revenue is R1n2 = 14 + 0. 4. denoted by PD is defined by dx dp p dx = = # x dp x p PD (1) Applied Calculus for Business. If we change our units of measurement from tons to pounds. We tabulate our observations. Price per Ticket 4 Number of Seats Sold 204 4 + 11.1n.102 4 + 21. 204 .102 and so on. 5 30 x 54 + . Setting the marginal revenue to zero gives x = 162 and 15 5 substitution into the demand equation gives p = $ 5. if we want to measure the rate of change of demand with respect to price. Copyright © 2007 by Pearson Education. Inc. The derivative. The number of seats sold is 204 .10) and its equation is p = and we have R1x2 = We find R ¿ 1x2 = 1 2 54 x + x 5 30 1 54 x + 1Verify!2. by Warren B. we could have solved this problem by observing that the demand equation is linear.3n2 + 8. dx/dp measures the change in demand with respect to price. The new ticket price that maximizes the ticket revenue is $5.10 the number of tickets sold is 201.0. Economics. For example. That is. only 201 = 204 . Thus the demand equation (line) passes through the points (204. Wang. To that end. suppose x is measured in tons and p in dollars and dx/dp = 2 tons per dollar.40. which has the same units. and April Allen Materowski. R 1n2 = .102 = $ 1. Alternately. and Finance. Two ten cent increases brings the price to $4.1n21204 .40. suppose we solve for x in terms of p and write this relationship as x = D1p2.4 The only critical point occurs when n = 14.0.3n.3n2 = .4n + 816 R ¿ 1n2 = . this is not a very useful measure as it is dependent upon the units used.5 Application II Business and Economic Optimization Problems is. However. This ratio is called the elasticity of demand and we designate it PD. Thus the ticket revenue is maximized when n = 14. 3p. in this price range. by Warren B. For nonlinear demand equations the elasticity is not exactly equal to the ratio of the relative change in demand to relative change in price.5 Application II Business and Economic Optimization Problems * ** 307 Note that in economic theory. At this price range. it is inelastic. Economics.f1q12 f(q1) (2) To better understand this terminology. the demand is elastic. (b) Show that the elasticity equals the ratio of the relative change in demand to the relative change in price when p changes from 1 to 2. The demand is relatively inflexible. represents a 3/97 or roughly a 3% change in demand.3p. the ratio is . Solution (a) for x = 100 . Gordon. If PD = 1. about a 3% change in price. . Example 6 (a)If the demand equation is x = 100 . when p = 2. the terminology elastic and inelastic carry over naturally to nonlinear demand functions. the relative change in price is 12 . demand decreases as the price increases. and Finance. x changes from 97 to 94 and the relative change in demand is 3/97. a 30% change. Of course. Thus. from p = 1 to p = 2. a small change in price results in a relatively large change in demand. That is. every demand function is a decreasing function.f1q12 * 100% f(q1) (3) f1q22 . Wang. dx/dp 6 0. x = 94. Thus. Copyright © 2007 by Pearson Education.3/97. 1/1 = . find the elasticity of demand when p = 1. a one unit change in p.Section 3. the corresponding change in demand is from x = 10 to x = 7. represents 100% change in price. Applied Calculus for Business. The demand is called elastic if PD 7 1. from x = 97 to x = 94. Thus a 100% change in price causes only a 3% change in demand.12/1 = 1. In this case it is easy to show that the elasticity is exactly equal to the ratio of the relative change in demand to the relative change in price (see Exercise 22). Thus. and April Allen Materowski. Therefore. PD = dx # p dp x dx = . we have been analyzing the case in which the demand equation is linear. In fact. that is. consider the demand equation x = 100 . and the demand is inelastic if PD 6 1. DEFINITION 2 The relative change of a function whose equation is p = f1q2 as q changes from q1 to q2 is relative change = The percentage change is defined as percent change = f1q22 .3/97 (b) As p changes from 1 to 2. we may obtain equation (1) by taking the limit of the ratio as the change in price approaches zero (see Exercise 23). We first need an additional definition. we have x = 97. The 3 unit change in x.3p. Published by Pearson Learning Solutions. x = 97. and when p = 1.3/97 . When p = 1. Therefore. Inc. the elasticity is said to be unitary. as p changes from p = 30 to p = 31. Since both x and p are positive it follows that the elasticity of demand will always be negative. therefore dp PD = .3. Walter O.311/972 = . but it does give approximately this value. On the other hand. The reason for choosing PD = 1 as the transition from elastic to inelastic demand will be given shortly. 1/8. . If PD 6 .2p. it is easy to show that the elasticity of demand may p x = dp dx be written as PD (4) Exercise 24 asks you to derive equation (4). Solution The total revenue is R = xp. the revenue is maximized when PD = 1. Inc. at p = 18 the demand is inelastic. Thus. Applied Calculus for Business. dR/dp = x + p1dx/dp2 We may write p dx dR = xc1 + a b d = x dp dp dx dp T xD1 + x p or dR = x[1 + PD].1 6 PD 6 0. To the left of the maximum the demand is elastic and to its right it is inelastic. Example 8 Show that when the revenue is maximized.1/82/18/182 = . Therefore.9/32. However. dp Since x 7 0. when p = 18. and April Allen Materowski.1. Wang.2p p = 18 = .21182 = 8. Copyright © 2007 by Pearson Education. that is. The next example indicates why unitary elasticity is the value at which we say the elasticity changes from elastic to inelastic.1. PD = 1 . PD = 1. using the product rule. . by Warren B. and when PD 7 .5 Application II Business and Economic Optimization Problems Example 7 Given the demand equation x = 2100 . PD = 9/32 6 1. that is when PD 7 1.308 * ** Section 3.1. it is often the case that the price determines the demand. Gordon. Walter O. find the elasticity of demand when p = 18. It should be noted that in the above discussion we have considered demand as a function of price. when PD 6 1. After all. demand is inelastic. the only critical point occurs when PD = . If you do so. we could just as well have considered price as a function of demand. x = 2100 . demand is elastic. Is the demand elastic or inelastic at p = 18? Solution dx/dp = 1 2100 . Economics. Published by Pearson Learning Solutions. that is. and Finance. thus. p = 3. and its tangent line at the point B. (d) px = 8.x = 100. In Exercise 5. and a cost equation C1x2 = x + 1. and the cost equation is C1x2 = 3x + 1. where x is in carats and p is in thousands of dollars.x. Assume that the demand equation is linear. Wang.x2 + 4x. the demand is for 5 coats. (b) p = 1x + 22 (e) p2 . A fur dealer finds that when coats sell for $4000. Walter O. Each month an automobile dealer makes a profit of $200 on each car that she sells if not more than 50 cars are sold.Ap/x.Section 3. (c) p/x = 10 at p = 5. Inc. (c) p + x = 75. where c and A are constants. Let price change from p to p + h. If the cost equation in Exercise 1 is C1x2 = 0. For each five additional trees planted. (b) If overhead is $2500 per month and the production cost per coat is $2000. Find the minimum average cost if the total cost function is C1x2 = x2 + 5x + 9. Gordon. If the demand equation is p = 29 . 4. 20. 18 19. Published by Pearson Learning Solutions. (f) p = 29 . Given the demand equation p = 100/1x + 12. Find the maximum revenue if the demand equation is: (a) p = 100/1x + 12. 14. and April Allen Materowski. A producer finds that demand for his commodity obeys a linear demand equation p + 2x = 100 where p is in dollars and x in thousands of units. Figure 2 indicates the graph of a demand equation. Find the elasticity of demand at the indicated price for each of the following. (e) px2 = 12. For every car above 50 that she sells her profit per car is decreased by $2. A diamond dealer finds that the demand for flawless diamonds is governed by p = .c. (b) p = 29 . 9. (a) Find the level of production that will maximize revenue. (a) p + 4x = 80 at p = 40. (c) x = 100/1p + 122. Applied Calculus for Business.5 1. find the maximum profit. monthly sales are 6 coats.p. Economics. find the cost equation and profit equation. Compare the result with the exact answer found by using either equation (1) or (4). 22. what tax should the government impose on the producer if it wishes to maximize its tax revenue? 18. Let p change from p0 to p0 + h. Suppose x = D1p2 is a nonlinear demand equation. (b) 10p + x = 500.x + 3. (a) Find the demand and revenue equations. Interpret A. Suppose in Exercise 5 that the producer is subjected to a tax of $10 per thousand units. p C B x A Figure 2: Ex. by Warren B.x at p = 2. 3. 23. In Exercise 6. (c) 4p + x = 100. Minimize the average cost if the total cost function is C1x2 = 3x2 + 200x + 12. the yield per tree decreases by 10 peaches. What should his production level be in order to maximize profits? 15. (a) For what values of x is this a possible demand equation? (b) Find the number of carats to be made available to maximize revenue. (f) p2 + x2 = 50. Show that elasticity of demand is the limit as h approaches zero of the relative change in demand to the relative change in price. What is the price and revenue for this amount? 10.x42 . p = 40. Show that PD = . In Exercise 6 if the government applies a luxury tax of 20% per coat. Copyright © 2007 by Pearson Education.dAB/dBC. and Finance. 13. How many trees should be planted to maximize the total yield of the orchard? 12. 8. how are profits affected? 16. . p = 2. (b) x = 2100 . Derive equation (4). p = 46. 150 . When the price increases to $5000. what price should be charged to maximize profit? 6. (a) p + 4x = 80. 21. 5. find the maximum profit. For a linear demand equation show that PD = . p = 4. 7. How many cars should the dealer sell monthly to maximize her profit? 11. Show that the ratio of the relative change in x to the relative change in p is exactly equal to the elasticity of demand at p0. (c) Find the level of production that maximizes profit. what tax should the government impose on the producer if it wishes to maximize its tax revenue? 17.5x2 + x + 1. Use the results of the preceding exercise and a scale drawing to find the elasticity of demand for each of the following. p = 0. If a demand equation has the form xpc = A show that PD = .5 Application II Business and Economic Optimization Problems * ** 309 EXERCISE SET 3. An orchard contains 300 peach trees with each tree yielding 800 peaches. (d) p2 + x2 = 100. where dAB is the distance from the point B to the x-intercept A and dBC is the distance from B to the p-intercept C. what should his level of production be to maximize profits? 2. (a) x + 4p = 200. 24. Given a linear demand equation x = mp + b. If the producer s costs are given by C1x2 = 2 + 3x. Find the elasticity of demand for each of the following demand equations. To illustrate the notion of linearization we consider several examples. we expand upon this notion and illustrate how this is used to linearize a function. we have the following definition. f1x022 is y . Recall that given a differentiable function defined by the equation y = f1x2. near the point of tangency we have the approximation f1x2 L T1x2 (2) (1) We shall see that (2) is useful in numerical evaluations. and Finance. f1x022 is given by y = T1x2 = f ¿ 1x021x . the tangent line lies beneath the graph. DEFINITION 1 The equation of the tangent line to the differentiable function defined by y = f1x2 at the point 1x0.x02. Published by Pearson Learning Solutions. see Figure 1a. Copyright © 2007 by Pearson Education.x02 + f1x02 With this is mind.x02 + f1x02 T(x) is called the linearization of f near the point 1x0.6 Linearization and Differentials » » » » » Linearization Differentials The Differential Approximation Differentiable Functions Differential Formulas Linearization In our discussion of the derivative. Thus. by Warren B. while it lies above the graph if f is concave downward near x0. Gordon. Walter O. Solving for y.310 * ** Section 3. f1x022. Note that if the graph of f is concave upward near P1x0. and April Allen Materowski. we have y = f ¿ 1x021x .6 Linearization and Differentials 3. . Economics. see Figure 1b. In this section. the equation of the tangent line at the point 1x0. Inc. P P Figure 1a: Concave Upward near P Figure 1b: Concave Downward near P Applied Calculus for Business.f1x02 = f ¿ 1x021x . f1x022. Wang. we saw that the tangent line is the best linear approximation to the function near the point of tangency. However. T1x2 = 11x . Wang. Copyright © 2007 by Pearson Education.000833333. T1x2 = . Solution If we let f1x2 = x1/3 then we must compute f(8. Solution We have f102 = 1 and f ¿ 1x2 = 1*211 + 2x2-1/2122 = 11 + 2x2-1/2. Thus.2x + 1. f ¿ 1x2 = 4x . we will linearize the function near x0 = 8. Our next example illustrates how this may be done. f ¿ 1x2 = therefore f ¿182 = 1 -2/3 x . therefore.01.1.6. approximate 2 3 8. Since we have to compute f1x02 = 1x021/3. and Finance. . the graph is concave upward and y = T1x2 is below y = f1x2. and f ¿ 112 = . we shall find it. Therefore. we obtain.012 = 1/1218 . Published by Pearson Learning Solutions. We have f182 = 81/3 = 2.12 . Example 2 Linearize f1x2 = 21 + 2x near x0 = 0. by Warren B. To make the approximation reasonably accurate we must choose the point at which the function is linearized close to x = 8. f18. Economics. giving f ¿ 102 = 1.2. Gordon.02 + 1 or T1x2 = x + 1 Let us not lose sight of the fact that linearization is nothing more than finding the equation of the tangent line. Example 3 Using linearization.01. We may now exploit this fact and use linearization as an effective tool for approximation. the purpose of the example is to illustrate the method of linearization.01. but we must point out it is much easier and probably more accurately done by a calculator. Using a calculator we find that Applied Calculus for Business. 2/3 3 3142 12 12 3182 Since f1x2 L T1x2 for values of x near x0 = 8. Inc.82 + 2.6 Linearization and Differentials * ** 311 Example 1 Linearize f1x2 = 2x2 .21x .1. calculations are especially simple. Since the linearization of a function is much easier to deal with than the function itself.Section 3. and April Allen Materowski.012 L T18. 8 will be a good choice for x0 since it is a perfect cube and near 8. Linearization is used to replace a complicated expression by a simpler one.8. Walter O. The next example illustrates that once this is done. 3 1 -2/3 1 1 1 1 182 = = = . to nine places. Note that f 1x21x2 = 4 7 0. We have that T1x2 = 1x . Substituting into (1). or T1x2 = . Solution We have f112 = .01).6x + 3 near x0 = 1.012 + 2 = 2 + 1/1200 L 2. f1x0 + h22 to R1x0 + h. Its tangent line y = T1x2 is drawn at P1x0. and Finance. which is not equal to dy. f1x022. which is exactly what occurred. Copyright © 2007 by Pearson Education. called differentials. and equals f1x0 + h2 . In addition. the distance dSR. Walter O. dRP = h.2/19x5/32 6 0 if x 7 0. dy = f ¿ 1x02 ¢ x. Wang. and not dy divided by dx . In fact.f1x02. it is much simpler to solve the above example with a calculator. Differentials Certainly.T1x02 = [f ¿ 1x021x0 + h . However. that is. f( x 0)) y = f( x ) x0 x0 + h Figure 2: y = f1x2 and its Tangent Line at P1x0. the tangent line is above the graph near x = 8. Published by Pearson Learning Solutions.000832986. The error is in the sixth decimal place. Gordon. Since f 1x2 = . We shall find differentials useful when we discuss the reversal of the differentiation process later in this text . f1x022. We shall give meaning to each of the dx expressions dy and dx. from S to R. . f( x 0 + h )) y = T (x) *y S dy R ( x 0 + h . Let h = ¢ x then dSR = T1x0 + h2 . f1x022. We let S be the point where the tangent line crosses the vertical line QR. therefore our approximate value will be on the large side. Observe that the distance from Q to R is denoted by ¢ y or ¢ f. That is. this must be the same as the slope of PS computed by using the change in y divided by the change in x.6 Linearization and Differentials 8. the sketch shows the horizontal line drawn from P to R. If we set dQS = P ¢ x then we have Applied Calculus for Business. so that we may interpret their ratio as being the derivative. and the vertical line drawn from Q1x0 + h. f ( x 0) ) f ( x 0) P ( x0 . y = T1x2 Figure 2 is a sketch of a smooth function y = f1x2. divided by the horizontal distance from P to R. f( x 0 + h ) Q ( x 0 + h .011/3 = 2. Inc. We shall also show how the differential may be used to approximate small changes. When developing the notation for the derivative we were careful to read the expresdy sion as dydx. we have ¢ y = dy + dQS. by Warren B. but the importance of the example is to illustrate that linearization reduces a more complex calculation to a simpler (linear) one.312 * ** Section 3.x02 + f1x02] . and April Allen Materowski. Economics. By definition.f1x02 = f ¿1x02h = f ¿1x02 ¢ x We define this quantity to be the differential dy. the slope of the tangent line is f ¿ 1x2. f ( x 0) ) y = T(x) y = f( x ) x0 x0 + h Figure 3: y = f 1x2 and its Tangent Line at P1x0.1x2 . then ¢ y 6 dy. Applied Calculus for Business. f 1x022. Figure 2 shows the graph as being concave upward near P. Inc. Thus.122x ¢ x = 4x1x2 . Summarizing.6 Linearization and Differentials * ** 313 ¢ y = f1x0 + h2 . However. (b) ¢ y = f1x + h2 . we have the following definition in which we delete the subscript on x. f ( x 0) ) dy R ( x 0 + h .122 . S f( x 0 + h ) Q ( x 0 + h . the distance dQS. we have the approximation that ¢ y = dy. if the graph is concave upward near P.f( x 0 + h )) * y f( x 0) P ( x0 . compute: (a) dy. as ¢ x approaches zero. since P may represent any point on the graph DEFINITION 2 h = ¢ x = dx dy = f ¿ 1x2 ¢ x ¢ y = f1x + h2 .Section 3. Walter O. which we called P ¢ x becomes negligible. and Finance. Where f is Concave Upward Example 4 Given y = f1x2 = 1x2 .f1x2 = dy + P ¢ x (3) (4) (5) It is instructive to observe that as Q approaches P. (d) What happens to P as ¢ x approaches zero? Solution (a) dy = f ¿ 1x2 ¢ x = 21x2 . as in Figure 3. a small change in f is approximately equal to a corresponding change in T.122 = 4x3h . . Published by Pearson Learning Solutions. P 7 0 and if it is concave downward near P. (c) P. Thus. The reason for the unusual choice of notation P ¢ x will be explained shortly. (b) ¢ y. and we see that in this case that ¢ y 7 dy. by Warren B. if the graph is concave downward near P.f1x2 = 11x + h22 . Wang. That is. Copyright © 2007 by Pearson Education. Economics. P 6 0. Gordon. y = T1x2.12 ¢ x.4xh + 6x2h2 + 4xh3 + h4 .f1x02 = dy + P ¢ x. and April Allen Materowski.2h2.122. and factoring. we mean and by percentage change we mean the relative f change written as a percent. we have that P . by Warren B. When t = 16. ¢x : 0 Observe that P depends upon both ¢ x and x.6 Linearization and Differentials Replacing h by ¢ x. we see that P ¢ x = 16x2 ¢ x + 4x ¢ x2 + ¢ x3 . where the population at time t is given in millions while t is measured in seconds. we have t 60 dP = 1/4 10. therefore. as well as ¢ x the distance from Q to P. It is precisely this observation that makes differentials useful in approximating ¢ y if ¢ x is small. Economics.8011623/4 = 2. Solution We are required to find ¢ P but will instead find dP.) Sometimes. what is the corresponding percentage increase in the cost? ¢x dx ¢C dC = = 5%. The Differential Approximation Example 5 The population of a particular bacterial culture was found to be governed by the equation P1t2 = 80t 3/4.1 seconds. Published by Pearson Learning Solutions. Find the approximate change in the population between 16 and 16. Wang. where x is the number of transports produced. We must determine L . Walter O. and Finance. the relative change may be approxidf mated by . f Example 6 The cost of producing x military transports. Of course if ¢ f L df. or P = 6x2 ¢ x + 4x ¢ x2 + ¢ x3 . P depends upon the value of x at P. Copyright © 2007 by Pearson Education. the term relative change or percentage change is used. is given by C1x2 = 6x3.2 ¢ x2 ¢ x. By ¢f the relative change in f.2 ¢ x. If there is a 5% increase in the number of transports produced. We are given that 60 ¢ t = dt = 0. Solution We are given that 18x2 dx 18x2 dx dx dC = = = 3 = 315%2 = 15%. in place of change. the exact change is ¢ P = 80116. (Note.12 ¢ x + 16x2 ¢ x + 4x ¢ x2 + ¢ x3 .2 ¢ x2 ¢ x. Thus. to nine places.314 * ** Section 3. The next example illustrates this approximation.997662312. we find dP = 1/4 dt. the approximate change in the population of the culture is 16 3 million. .lim 16x2 ¢ x + 4x ¢ x2 + ¢ x3 . 3 x C C 6x Applied Calculus for Business. in millions of dollars.1.12 = 3. we have ¢ y = 4x1x2 .2 ¢ x2 = 0. Inc. That is. (d) As ¢ x approaches zero. Gordon. Differentiating. and April Allen Materowski. We x x C C 2 have that dC = 18x dx. Example 4 illustrated that it is often much easier to compute dy than ¢ y.0123/4 . (c) From ¢ y = f ¿ 1x2 ¢ x + P ¢ x. Inc. a 5% increase in the number of transports produced results in a 15% increase in cost.f1x2 . then dy can dx Differential Formulas Applied Calculus for Business. the case y = f1x2 but suppose that x is a function of some other variable t. However. Wang. there is a little more to it than that. Copyright © 2007 by Pearson Education. x = g1t2. Walter O. if we have y = xN. the ratio of dy to dx is precisely the derivative and. for example.f ¿ 1x2.6 Linearization and Differentials * ** 315 Thus. Consider. Using this new terminology. d [f1g1t22] = f ¿ 1g1t22g ¿ 1t2. this difference approaches zero as ¢ x = h approaches zero. Gordon. It is apparent that the quantity P is of some significance when we use differentials. it is now possible to rewrite all our derivative formulas as differential formulas.dy = f1x + h2 . f ¿ 1x2 = f ¿ 1g1t22 Therefore. we have. For a differentiable function. in fact. Thus.f ¿ 1x2 ¢ x. Exercise 33 illustrates how (6) may be used to prove the chain rule. Differentiable Functions DEFINITION 3 A function whose equation is y = f1x2 is differentiable if the change in y may be written in the form ¢y = D ¢x + P ¢x (6) where P approaches zero as ¢ x approaches zero.f1x2 h . All the rules for derivatives will carry through to differentials. and April Allen Materowski. Thus. The interpretation of P follows directly from (5). once again. .Section 3. Or. dy = f ¿ 1x2 ¢ x and dx = g ¿ 1t2 ¢ t By the Chain Rule. for instance. we have P = f1x + h2 . P ¢ x = ¢ y . In fact we may define differentiability of a function as follows. thus dt dy = f ¿ 1g1t22g ¿ 1t2 ¢ t However. Say. the differential of a sum is the sum of the differentials because the derivative of a sum is the sum of the derivatives. Dividing by ¢ x = h. We now see that the differential gives the approximate change in y for a small change in x. the symbol be considered a fraction. It is precisely this observation that is used to generalize the definition of differentiability to a function of several variables. D is then the derivative of the function. dy = f ¿ 1x2 dx. Thus. Economics. Published by Pearson Learning Solutions. by Warren B. P measures the difference between the slope of the (secant) line through P and Q and the slope of the tangent line at P. and Finance. Now. Gordon.12x2 dx (b) d1x4 .1 could be rewritten as dy = NxN . Example 7 Find d1x4 . and April Allen Materowski. Published by Pearson Learning Solutions.6x2 + 112 = d1x42 .1 dx or d1xN2 = nxN . Later in dx this text it will be convenient for us to use the differential form of the product rule. Solution To compute the differential you could either: (a) Find the derivative of the entire function and multiply by dx.12x dx = 14x3 . or. and Finance. (a) f ¿ 1x2 = 4x3 . written both in differential and derivative form.6x2 + 112 = 14x3 . Economics. by Warren B.12x2 dx Applied Calculus for Business.12x Therefore. Walter O.6d1x22 + 0 dx = 4x3 dx . then the product rule d1uv2 dv du + v = u dx dx dx may be written as d1uv2 = u dv + v du Table 1 summarizes all the derivative formulas up to this point.1 dx.316 * ** Section 3. (b) Find the differential of each term separately using the fact that the differential of a sum is the sum of the differentials. .6x2 + 112. Inc. Copyright © 2007 by Pearson Education.6 Linearization and Differentials dy = NxN .d16x22 + d1112 = 4x3 dx . d1x4 .6x2 + 11. Wang. (a) Letting f1x2 = x4 . Thus. For example if u and v are two functions of x. 1 dx d1uv2 = u dv + v du N D dN .01. Economics.9924. (c) P. and Finance.98. If y = f1x2 = 3x3 + 1 find (a) dy.1 dx dx dv d du u = 1uv2 . (c) P.01).v du EXERCISE SET 3. Wang.2x + 5 (a) x0 = 0. (iii) x = 1 and ¢ x = .00124. (a) 2 3 28. (a) 2 3 26. 9. x0 = 0. (b) 11. f1x2 = 3x2 + 12x . approximate f(1. 7. (b) x0 = 2 2. 2 8. (d) Evaluate each of these quantities if: (i) x = 10 and ¢ x = 1.6 Linearization and Differentials * ** 317 Table 1: Derivatives and their Differential Form Derivative Form d 1c2 = 0 dx df d 1cf1x22 = c dx dx d N N-1 1x 2 = Nx dx dv du d 1uv2 = u + v dx dx dx dN dD .N D d N dx dx a b = 2 dx D N du d 1f1u22 = f ¿ 1u2 dx dx du d N 1u 2 = NuN .Section 3. 11. (c) what about x0 = . 14. Let f1x2 = 24 + 5x. Walter O.1).9. Let f1x2 = 3x4 . (b) 2 3 26. 1. (d) Evaluate each of these quantities if: (i) x = 10 and ¢ x = 1.1. 13. 10. (b) 132. (b) ¢ y. and April Allen Materowski.27. Approximate f(1.6 In Exercises 1 4. (c) Using an appropriate linearization. In Exercises 7 11 use linearization to approximate the given quantity. If y = f1x2 = 1x3 + 522 find (a) dy.3 (a) x0 = 0.01. f1x2 = 2 3 1 + x.2. linearize f (a) near x0 = 0. 12. f1x2 = 21 + x (a) x0 = 0. . (b) ¢ y. by Warren B. (a) 13. Published by Pearson Learning Solutions. 4.01. Copyright © 2007 by Pearson Education.2 3. (b) x0 = . (b) near x0 = 1. (ii) x = 1 and ¢ x = 0. (b) x0 = 1. (a) 14.00325. linearize f near x0. In each case determine whether the result is too large or too small.1 du u dv = d1uv2 . does it present a problem? 5.01. Gordon. If y = f1x2 = 5x2.0423/5. (c) P.v dx dx dx Differential Form d1c2 = 0 d1cf2 = c df d1xN2 = NxN . 6. (b) 2 3 . (c) Explain why one of the above choices for x0 yields a better approximation then the other. Inc. f1x2 = 3x . Applied Calculus for Business. (b) near x0 = 0 and computing T(1. find (a) dy. (ii) x = 1 and ¢ x = 0. (b) 215.7x + 10.0. (b) ¢ y.01) by linearizing f (a) near x0 = 1.N dD b = D N2 da d1f1u22 = f ¿ 1u2 du d1uN2 = NuN . (a) 216. (a) Find the exact change in area. (a) Let T11x2 be the linearization of f(x). and T(x) the linearization of F1x2 = f1x2 + g1x2. find (a) dy. (a) Show that ¢ y = f ¿ 1u2 ¢ u + P1 ¢ u where P1 approaches zero as ¢ u approaches zero. 22. (b) y = x612x4 . Inc. and Finance. 31. Using differentials. (a) find ¢ y. (c) Draw a picture of the original and the expanded square.0. (b) Choose x0 = 9. show that its linearization near x0 = 0 is given by T1x2 = 1 + akx. 30. This exercise uses (6) to prove the chain rule. Published by Pearson Learning Solutions. What is the maximum possible error in the measurement of the side of the cube? 21. 33. (b) Show ¢ y = D ¢ x + P ¢ x. what is the corresponding percentage increase in the volume of the blood flowing through it? (b) Show that for a relative change in the radius of the artery. and find T(x). determine (a) the change in the revenue as x changes from 36 lbs to 36. (c) Instead of solving f1x2 = 0. (a) Suppose there is a 15% increase in the radius of the artery. 16.975. The area of a square with side length s is A = s2. (a) Find the exact change in volume. (b) Find the differential approximation to the increase in area.6. such that f1r2 = 0. (a) show that the given function is continuous for all x Ú 0. The volume of blood flowing through an artery whose radius is r is proportional to the fourth power of the radius. A rubber ball (a sphere) is expanding in such a way that its radius increases from 8cm to 8. (a) What is the approximate change in its volume? (b) Is the approximate change larger or smaller than the actual change? 4 a V = pr3. Find df. The volume of a cube with side length s is V = s3. (c) Identify D.2 2 x (b) Find dy if y = 2 .1 = 0. (a) Find dy if y = .1 lbs.72. (b) ¢ y.1 25. 29.1 lbs.01. Explain why (c) and (d) together prove the chain rule. where k is any real number. By slightly increasing the radius of an artery they are able to obtain a significant increase in blood flow. (b) Find dy if y = x51x 2 . 8 6 r 6 9.005in. (c) P. Show that T1x2 = T11x2 + T21x2. Find dy. Consider a square whose side has been extended to length s + ¢ s.04cm. (d) Show that P approaches zero as ¢ x approaches zero. and April Allen Materowski. Economics. If y = f1x2 = 1x3 + 521. find (a) dy. (b) Let f1x2 = 11 + 2x21/3 + 11 + 3x21/4. x . find: (a) dy. by Warren B. 17. Wang.0012 cubic inches in computing the volume must be allowed. 2x + 3 24. use the approximation f1x2 L T1x2 and instead solve T1x2 = 0 to approximate r. A cube is measured and each side is found to be 4 inches with a possible error of at most 0. Given f1x2 = 1x2 + 423. Consider the function f1x2 = 1x . 18.975. there is a fourfold increase in the volume of blood flowing through it (Physicians exploit this relationship between the volume of blood flowing and the radius of the artery. each near x0. (d) Give a geometric interpretation for the approximation and for the error.6 Linearization and Differentials 26. b 3 19. Walter O. Due to heat conditions a possible error of 0.318 * ** Section 3. f182 6 0 and f192 7 0. Copyright © 2007 by Pearson Education. The total revenue in cents in selling x pounds of coffee is given by R1x2 = 10x1100 . (b) the percent change in the revenue as x changes from 36 lbs to 36. 15.1 . If f1x2 = 11 + ax2k. (a) Find dr if r = 20 25u2 + 9 . (ii) x = 3 and ¢ x = 0.x21/2. x + 1 3x . The volume of a cube is measured to be 8 cubic inches.2 3 x . where P approaches zero as ¢ x approaches zero. Suppose y = f1u2 and u = g1x2 are each differentiable functions. 28. (a) Find df if f1x2 = 2 . Use (a) and the previous exercise to quickly find the linearization of f near x0 = 0. (b) ¢ y.2 3 x . Find an approximate solution of 1x . 32. (c) Substitute (b) into (a) to show that ¢ y = f ¿ 1 u2 g ¿ 1 x 2 ¢ x + P ¢ x where P = P1g ¿ 1x2 + P2f ¿ 1u2 + P1P2. (c) P. Gordon. Consider a cube whose side has been extended to length s + ¢ s.3x + 42. (a) f1x2 = 1x2 + 221x4 . (b) Find the differential approximation to the increase in volume. T21x2 the linearization of g(x).) 23. (c) P. 20.7.928 27. (b) ¢ y.3 (b) Find f ¿ 1x2 if f1x2 = 22x3 + 3x + 2. If y = f1x2 = 2x2 + 2x + 3. (d) Evaluate each of these quantities if: (i) x = 3 and ¢ x = 1. 2x . . Conclude that there is a number r. Determine the approximate error in computing the volume of the cube. Applied Calculus for Business. Let y = f1x2 = 12x + 120. (b) Show that ¢ u = g ¿ 1x2 ¢ x + P2 ¢ x where P2 approaches zero as ¢ x approaches zero. if any for: (a) y = x2 x1/3 . . Walter O. (b) G1x2 = . 2.6x2 + 9x .9x + 5. find the maximum and minimum values of f on the interval [ .x23.3x.2. (a) For g1x2 = x3 . (a) For f1x2 = 2x2 . y2 = D1 ¢ x + D2 ¢ y + P1 ¢ x + P2 ¢ y. x + 1 x + 2 8. y2 = x2y3. 3.1. use the first derivative test to determine where the given function is increasing and where it is decreasing. (a) Sketch the graph of a continuous function defined on . 3.1. x + 4 x + 4 2 Applied Calculus for Business. 4.x23. (a) compute ¢ f = f1x + ¢ x. if any for: (a) h1x2 = x1/3 . find the maximum and minimum values of g on the interval [ . 42.f1x.4 x 4. (c) For g1x2 = x1/3 . (a) f1x2 = 3x2 .4x2.4x2. (b) f1x2 = x3 + 6x2.1. (b) For f1x2 = x3 . Wang. Gordon. Find all critical point(s). (b) Sketch the graph of a continuous function decreasing on .18x + 21.12x. (a) g1x2 = x3 . Find all critical point(s). We shall see later in the text that D1 and D2 are called partial derivatives with respect to x and y respectively.Chapter Review 34. 1). find the maximum and minimum values of g on the interval [ .18x + 21. that has a relative maximum at (1. Demand and Revenue Cost and Average Cost Elasticity and Demand Linearization The Differential Approximation Differentiable Function Differential Formulas 1.3x.4x2. find the maximum and minimum values of g on the interval [ . and Finance. that has relative a maxima at 1 .2 x 2. CHAPTER REVIEW Key Ideas Continuity Maxima and Minima Relative Maxima and Minima Critical number and Critical Points The Only Critical Point Test Increasing and Decreasing Functions First Derivative Test The Second Derivative Higher Order Derivatives Velocity and Acceleration Concavity The Second Derivative Test for Concavity The Second Derivative Test for Relative Extrema Implicit Differentiation and Curve Sketching Optimization Procedure Area and Perimeter Volume Distance and Velocity Price.12x. (b) h1x2 = x211 . (c) f1x2 = x3 .2 6 x 6 7.14. (d) g1x2 = x211 . and April Allen Materowski. (b) y = 2 . y2 * ** 319 (b) Show that ¢ f = f1x + ¢ x. . if any for: (a) g1x2 = x3 .52 and a maximum at 13. Find all critical point(s).x23. x2 3x . find the maximum and minimum values of f on the interval [ . for: (a) f1x2 = 3x2 . 5]. by Warren B. (b) g1x2 = x4 .14. where P1 and P2 each approach zero as both ¢ x and ¢ y approach zero.f1x. Indicate the point M on your graph which is a relative maximum. 2]. (b) f1x2 = x3 + 6x2. Consider the function of two variables f1x.3x. Find all critical point(s). Classify the critical point(s). Inc. find the maximum and minimum values of g on the interval [ . (c) g1x2 = x1/3 . 6. 2].3 6 x 6 .1*2 6 x 6 1*2 and increasing on 1*2 6 x 6 2. 12]. Published by Pearson Learning Solutions. y + ¢ y2 .1.12x2. Copyright © 2007 by Pearson Education.2 and decreasing on . 11.6x2 + 9x . 5. a relative minimum at 10. y + ¢ y2 . (a) Sketch the graph of a continuous function increasing on . (c) f1x2 = x3 . (b) Sketch the graph of a continuous function defined on .6x2 + 9x . if any for: (a) F1x2 = 7. (c) For f1x2 = x3 + 6x. (d) For g1x2 = x211 . if any. 8].1. Economics.52 and a relative minimum at (0. (b) g1x2 = x4 . Find all critical point(s). find the maximum and minimum values of f on the interval [0.14. 10. (b) For g1x2 = x4 . 4]. Indicate the point m on your graph which is a relative minimum. In Exercises 10 through 13.6. 2). if any. 4]. 9. 30. Published by Pearson Learning Solutions. (a) Linearize f1x2 = 2 4 1 + 2x near x = 0. Find two non-negative numbers whose sum is 20 and such that: (a) The sum of their squares is: (i) a minimum.x23. (a) Find f 1x2 for: f1x2 = 1*2 x4 . (ii) maximized.12. by Warren B. 15. 2x .006.16t2 + 64t + 192. (d) Find the position when the acceleration is maximum. Sketch the graph of: (a) y = x1/31x . how many gimcracks should you make in order to minimize the average cost of a gimcrack? What is the minimum possible average cost? 26. 16.t4. Economics. If the demand for x gimcracks at price p is determined by the equation 3x + 5p = 1500: (a) What production will maximize revenue? (b) Using the total cost function in the previous exercise.3x2 + x -3 + 23.12x2 . In Exercises 16 and 17. Use the differential to find the approximate value of: (a) 224. (b) given in Exercise 6(b). (b) Find the times at which the velocity is zero and the times at which the acceleration is zero. (c) f1x2 = x3 + 6x (a) g1x2 = x3 . In Exercises 14 through 15 locate the points of inflection (if any) for the given functions. (a) Find the equations of its velocity and acceleration. (a) f1x2 = 2x2 . (b) f1x2 = x3 . If the total cost of producing x gimcracks is C1x2 = 144 + 40x + x2.9x + 5. Copyright © 2007 by Pearson Education. (b) y = 3x4 . what dimensions will maximize the area of the window? (a) The function F in Exercise 6(a). If the total length of the frame and sash is to be 20 feet. 31. 20. (b) Use the result of (a) to compute 2 4 1. A right triangle is to have area 18 square inches.4x2.x224. 1 . x4 . 23 24. 29. (b) The function G in Exercise 6(b). Suppose that the position of a particle as a function of time is given by the formula s1t2 = 4t3 . (a) (b) (c) (d) g1x2 g1x2 g1x2 g1x2 = = = = x3 . (b) y = x511 . 19.14. Walter O. 14. (c) y = 4x + 3 x2 + 6 .12x. Figure Ex. (b) The function defined in Exercise 7(b). Sketch the graph of the function (a) given in Exercise 7(a) (b) given in Exercise 7(b) 28. (c) y = x3 . 21. (b) g1x2 = x4 .12x2. Gordon.6x2 + 9x .3 Applied Calculus for Business. (b) 2 3 64. What should the lengths of the legs be. 23).5x3. * ** Chapter Review 23. (b) Find the fourth derivative for f1x2 = 2x + 1 22.8x3. (b) Find its position at the time when its velocity is zero. (a) f1x2 = 3x2 .320 12. (c) Find the position at the time when the velocity is maximum. and Finance. and April Allen Materowski. Inc. x1/3 . x211 .60x + 100.1221x + 223. (c) f1x2 = x3 .6. (b) f1x2 = x3 + 6x2. 27. A window is to be made in the shape of a rectangle with a sash across it (Figure Ex. (a) Find the velocity and acceleration as functions of time.14.1 . Wang. determine the production level that will maximize profit. (ii) a maximum.6x2 + 9x .4x2.3x. if the length of the hypotenuse is to be as small as possible? 25.18x + 21. . (b) The sum of the square of the first plus twice the square of the second is: (i) minimized. 17. (a) The function defined in Exercise 7(a). (d) y = 1x . 18. The position of an object at any time is given by the equation s1t2 = . classify the critical points by use of the second derivative test and the sketch the graph of the given function. Find dy if: (a) y = 7x2 . Sketch the graphs of the function (a) given in Exercise 6(a). 13. Gordon. by Warren B. Applied Calculus for Business. Inc. and then examines how they arise in natural applications. and April Allen Materowski. We shall see that calculus is an indispensable tool in examining these functions. Copyright © 2007 by Pearson Education. and Finance. Economics. Walter O. Published by Pearson Learning Solutions.4 Exponential and Logarithmic Functions This chapter introduces the exponential and logarithmic functions. Wang. . show the function is one-to-one. Applied Calculus for Business. This observation is called the horizontal line test. Published by Pearson Learning Solutions. contrary to the definition of a function being one-to-one. Example 2 From the sketch of the graph of f1x2 = 3x + 7. 14.12.1 Inverse Functions 4. and April Allen Materowski. (Or equivalently as: if x1 Z x2 then f1x12 Z f1x22. Since the condition states that different x-values correspond to different y-values. 12. 326. 13. 10. what about the set we obtain by interchanging the x and y values? Is it also a function? The set obtained is 513.) Example 1 Is the linear function defined by the equation f1x2 = 3x + 7 one-to-one? Solution Suppose we have f1x12 = f1x22. Observe both 13. Horizontal Line Test The notion of a function being one-to-one is easily determined from its graph.322 * ** Section 4. 22. Walter O. This translates into a very simple statement regarding the original function. by Warren B. then it follows from this equation that 3x1 = 3x2 or x 1 = x2 thus. 5) belong to this set. 02. it means for a function to be one-to-one. then the same y-value would correspond to two different x-values. 526. 17. What can we do to insure that the set obtained by interchanging the coordinates also yields a function? It should be clear that we need the second set to satisfy the uniqueness property as well.12 and (3. Solution The sketch of the graph is given in Figure 1. Copyright © 2007 by Pearson Education. Economics. Wang. since the only way two y-values are the same is if their x-values are. . the function is one-to one. no horizontal line drawn through the graph can intersect it in more than one point. 72. 42. that is 3x1 + 7 = 3x2 + 7.1 Inverse Functions » » » » » » » One-to-One Function Horizontal Line Test Increasing and Decreasing Functions Inverse Function Composition Property Derivative of the Inverse Calculator Tips One-to-One Function Recall that in the definition of a function defined by the equation y = f1x2 we required that to each x-value there correspond a unique y-value. Gordon. If it did. 32. Inc. 15. We may rephrase this by saying that the only way two y-values are the same is if their x-values are the same. A function defined by the equation y = f1x2 is said to be a one-to-one function if (in addition to each x corresponding to a unique y) each y-value corresponds to a unique x-value. Suppose the function is f = 51 . therefore it cannot be a function (Why?). we may write this condition as follows: if f1x12 = f1x22 then we must have x1 = x2.1. . . and Finance. Walter O. It should be noted in the previous example. Gordon. What kinds of continuous functions are one-to-one functions? How do they look? From the horizontal line test we see that a horizontal line will intersect a graph in more than one point if the graph has a turning point. the graph has no relative extrema). f( x ) = 2 x 2 + 1 Figure 2: The Horizontal Line Test Applied to f1x2 = 2x 2 + 1. by the horizontal line test. intersects the graph in two points or not at all. Therefore. Note that any horizontal line that is not drawn through the vertex of the parabola. Solution The graph of the function is the parabola indicated in Figure 2. This is illustrated in Figure 2.1 y f( x ) = 3 x + 7 Inverse Functions * ** 323 Figure 1: Applying the Horizontal Line Test to f1x2 = 3x + 7 Note that if a horizontal line is drawn anywhere. Example 3 Determine if the function whose equation is given by y = f1x2 = 2x2 + 1 is one-to-one. Therefore. that we could just as well have applied the definition of a function being one-to-one to show that the given function is not one-to-one. as long as a graph does not turn (that is. Economics. Published by Pearson Learning Solutions. the function is one-to-one. Inc. . by Warren B. and April Allen Materowski. Wang. this is not the graph of a one-to-one function.Section 4. it intersects the graph in exactly one point (meaning each y-value corresponds to exactly one x-value). a horizontal line will only Increasing and Decreasing Functions Applied Calculus for Business. and Finance. Copyright © 2007 by Pearson Education. Therefore. 10. 32. This is not a function. therefore. Economics. 12. 18. Published by Pearson Learning Solutions. 15. Example 4 Is the set obtained by interchanging the x and y-values in each of the given functions also a function? (a) f = 51 .32. (b) The function g is a one-to-one function since each y-value corresponds to exactly one x-value. Observe that the first derivative test can be very useful in classifying increasing and decreasing function. 52. 82. Under what conditions can we interchange the x and y-values and have the new set thus obtained also be a function? From the above discussions we see that the original function must be a one-to-one function. 11.7 is a one-toone function. Wang. . 514. one whose y-values decrease as the x-value increase. Thus. 02. Copyright © 2007 by Pearson Education. Inc. a one-to-one function.q . 15. 22. (b) g = 51 . namely . the function is increasing and on any interval on which the derivative is negative. 526. 12. Figure 3a: An Increasing Function Figure 3b: A Decreasing Function Figure 3: Two Types of One-to-One Functions Let us return to question posed earlier. (Why?) Therefore. . 15. q 2? (b) Is there any domain on which this function may be defined where this function will be one-to-one? Applied Calculus for Business.12. we obtain 512. will be a one-to one function.7. by Warren B.1. Example 6 (a) Is the function defined by y = f1x2 = 2x 2 + 1 one-to-one over the domain 1.324 * ** Section 4. and Finance. Walter O. 42.3 and 3. 32. 13. 1126. Solution We have f ¿ 1x2 = 12x2 + 9 7 0 for all x. 120. Recall that on any interval on which the derivative is positive. 111. This is illustrated in Figure 3.72. Gordon. 526 is also a function. 13.1 Inverse Functions intersect it in one point. Therefore the set obtained by interchanging x with y. 22. . 22. 1 . or the graph of a decreasing function. f is an increasing function over its entire domain and. . the function is decreasing.3. 12. the graph of an increasing function. 13. one whose y-values increase as the x-value do. the set obtained violates the uniqueness property of the definition of a function. 2026 Solution (a) If we interchange the x and y-values. Example 5 Show that the function defined by the equation y = f1x2 = 4x3 + 9x . and April Allen Materowski. 32. This occurs because f is not a one-to-one function as y = 2 corresponds to two different x-values. Solution (a) we require that x . not the reciprocal of the function! Applied Calculus for Business.4 23 L 1. if we limit the domain of the function to be only x 7 0.1 x 1 2 3 4 5 y * f1x2 * 2x + 1 0 1 22 L 1. Walter O. (b) If we limit the domain of the function to be only x 6 0.1. More formally we have the following: Given the one-to-one function defined by the equation y = f1x2. (c) determine the range of the function. therefore. and Finance. We write the equation of the inverse function as y = f -11x2*. and April Allen Materowski. 3) (3. y = f( x ) = (4.1 Ú 0. it is the symbol for the inverse function.) Example 7 Given the function defined by the equation y = f1x2 = 2x . . or equivalently. The function obtained by interchanging the x-values with the y-values is called the inverse function. Published by Pearson Learning Solutions. Inc. See Figure 2. (e) sketch the graph of the inverse function (f) determine the domain and the range of the inverse function. (b) with our domain specified. Table 1: Points on the Graph of y = f1x2 = 2x . (b) sketch the graph of the function. See Table 1 where we choose convenient x-values. it is one-to-one on that interval. the function obtained by interchanging every x and every y-value is called the inverse function. which is negative when x 6 0 and positive when x 7 0. then it is an increasing function and therefore it is one-to-one on that interval. then it is a decreasing function and. q 2.7 2 Inverse Function (c) we now plot these points to obtaining the graph. by Warren B. Gordon. Economics. (a) determine the domain of the function. 2) Figure 4: y = f1x2 = 2x . (d) show the function is one-to-one. Wang. Copyright © 2007 by Pearson Education. therefore it decreases for x 6 0 and increases for x 7 0 (that is.1 *Note that the symbol f -1 is not to be confused with a negative exponent.1 Inverse Functions * ** 325 Solution (a) f ¿ 1x2 = 4x. has a relative minimum at x = 0) so it is not a one-to-one function on the domain 1 .Section 4. (Note that the function must be one-to-one in order for the inverse function to be defined.q . we can generate a table of values with which to sketch the graph. our domain is x Ú 1. Similarly. see Figure 4. and Finance. It is also instructive to plot both graphs on the same coordinate axes. therefore it is one-to-one. observe that since the inverse function was obtained by interchanging the x and y-values in the original function that means that the domain of the original function becomes the range of the inverse (x Ú 1 is the domain of the original function. y Ú 1 is the range of the inverse) and the range of the original function becomes the domain of the inverse (y Ú 0 is the range of the original function. however. even without the equation we have considerable information about the inverse function. This is given in Figure 5. the new x and y-values obtained by this interchange are given in Table 2. we know its domain and range. Published by Pearson Learning Solutions. We next plot the graph using the points in Tables 2. This is done in Figure 6.) 2 2x . x Ú 0 is the domain of the inverse function). Walter O. 3) Figure 5: The Graph of the Inverse Function Some observations are in order. .1 B = dx 1 7 0 on its domain. Gordon. Applied Calculus for Business. 5) ( 3. Moreover. (Note that d A 2x . Thus.1 (e) We may obtain the points on the inverse function by interchanging the x and y values in Table 1. 4) ( 2.1 Inverse Functions (d) We observe from Figure 4 that this graph increases as it moves from left to right. it is an increasing function. Inc. it should be noted that we have not yet obtained the equation of the inverse function.7 2 y 1 2 3 4 5 (f) We see from the graph that the domain of the inverse function is x Ú 0 and its range is y Ú 1. (2. Table 2: Points on the Inverse Function x 0 1 22 L 1. as well as its graph. Wang. First.326 * ** Section 4. and April Allen Materowski.4 23 L 1. Economics. by Warren B. Copyright © 2007 by Pearson Education. Section 4. We now consider the determination of the equation of the inverse function. we first write y = f1x2 = 3x + 7 We next interchange y with x to obtain x = f1y2 = 3y + 7 we solve the equation x = 3y + 7 for y to obtain y = or we write.7 3 Applied Calculus for Business. Wang. Inc. Namely. given any one-to-one function. Copyright © 2007 by Pearson Education. This is true in general.1 Inverse Functions * ** 327 y=x f( x ) = Figure 6: y = f1x2 = 2x . Example 7 Determine the equation of the inverse of the function defined by the equation f1x2 = 3x + 7.1 and its Inverse Function Observe that the function and its inverse are mirror images about the line y = x. That means we need only interchange x and y in the equation of the original function to obtain the equation of the inverse function. To find the inverse. We observed that the inverse function is determined by interchanging every x and y coordinate. by Warren B. the reflection of this function about the line y = x yields the graph of its inverse function. and April Allen Materowski. and Finance. Solution In Example 1 we showed that this function is one-to one. Economics.7 3 x . Walter O. . y = f -11x2 = x . Published by Pearson Learning Solutions. Gordon. Economics.2 2x + 5 3x . Inc. and April Allen Materowski. by Warren B. Gordon. Example 9 Given the rational function defined by the equation f1x2 = tion of the inverse function. in Example 5.1 Interchanging x with y we have x = f1y2 = 2y . we clear fractions to obtain x12y + 52 = 3y .328 * ** Section 4.2 2y + 5 3y .5x or y12x . Walter O.3y = . .32 = .1 squaring both sides.2 2y + 5 3x .2 . To determine the equation of the inverse. and sketched its graph there.12 + 5x2 3y . and Finance. Wang. we have x = f1y2 = or x = To solve for y. we have 2xy .2 or 2xy + 5x = 3y . that the domain of the inverse function is x Ú 0. we obtain x2 = y . we write y = f1x2 = 2x . determine the equa2x + 5 Applied Calculus for Business.2 isolating the y terms.1 Solution We demonstrated that this function is one-to-one in Example 5. Published by Pearson Learning Solutions.2 . Copyright © 2007 by Pearson Education.1 Inverse Functions Example 8 Determine the equation of the inverse of the function defined by the equation f1x2 = 2x . Solution We write y = f1x2 = interchanging x with y.1 or y = x2 + 1 or y = f -11x2 = x2 + 1 Note that we determined. as we shall see later on in this chapter. Gordon. The graph of the inverse function is obtained from the graph of the original function by interchanging its x and y-coordinates.3 3 . 5. Wang. Inverse functions satisfy a unique property with regard to their composition with each other. it is necessary to define new functions to implement Step 5. Published by Pearson Learning Solutions. Summary Given the one-to-one function y * f1x2 1. (see Exercise 65). The inverse function. 4. 2. .2x . and therefore is one-to one. and April Allen Materowski. we have that b = f1a2.12 + 5x2 2 + 5x = 2x . y = f -1 (x) y=x Composition Property y = f(x) Figure 7: y = f1x2 and its Inverse y = f -11x2 Since the point (b. The equation of the inverse is determined by (a) interchanging y with x in the equation y * f1x2. The point (a. Therefore. a) is on the graph of y = f -11x2 we have that a = f -11b2. y * f +11x2 exists. a) is on the graph of y = f -11x2. and Finance. Inc. Economics.Section 4. its image point (b. and (b) solving the equation x * f1y2 for y.2x We leave as an exercise for you to verify that the function in the preceding exercise is always increasing. The domain of f becomes the range of f +1. Walter O. therefore.) 3. b) is a point on the graph of y = f1x2. and since (a. Applied Calculus for Business. b) is on the graph y = f1x2. The range of f becomes the domain of f +1. (The graph of y * f +11x2 will be the mirror image of the graph of y * f1x2 using the line y * x as the mirror. Consider Figure 7 where we have the graph of y = f1x2 and its inverse y = f-11x2. Copyright © 2007 by Pearson Education. by Warren B.1 Inverse Functions * ** 329 or y = or y = f -11x2 = 2 + 5x 3 . We remark that sometimes. Should we not expect to have a simple relationship between their derivatives at corresponding image points? In Figure 8. b = f1a2 and a = f 1b2. and Finance. Gordon. by Warren B. Solution We need to show that f1g1x22 = g1f1x22 = x. that is. and April Allen Materowski. it follows that b = f1a2 = f1f-11b22 -1 (1) (2) These are the special composition properties satisfied by inverse functions. It is usually given as follows: suppose f1g1x22 = g1f1x22 = x (over the appropriate domains. Since f1x2 = 3x + 7 we have that f1g1x22 = 3g1x2 + 7 = 3 a x . Walter O.330 * ** Section 4. Our definition was geometric and (3) was obtained as a consequence of the geometry.7 13x + 72 . Economics. we want to determine how g ¿ 1b2 is related to f ¿ 1a2.72 3 are inverse functions using the compo- therefore f -11x2 = Note that in Example 6. Example 10 Show that f1x2 = 3x + 7 and g1x2 = sition property.) then g and f are inverse functions of each other and we write g = f -1.7 b + 7 = x . Geometrically. y = g(x) = f -1(x) y = f(x) y=x Figure 8: Derivatives of Inverse Functions at Corresponding Image Points Applied Calculus for Business. Inc. .7 + 7 = x 3 f1x2 . of course.72 3 1x . Copyright © 2007 by Pearson Education. Wang.1 Inverse Functions a = f -11b2 = f -11f1a22 Similarly. They are often written as follows: f +11f1x22 * x and f1f +11x22 * x (3) Some texts take (3) as the defining property of inverse functions. Suppose y = f1x2 defines a differentiable function with an inverse defined by y = g1x2 = f-11x2.7 3x g1f1x22 = = = = x 3 3 3 1x . we saw that the inverse function is the image of f when reflected about the line y = x. we found the inverse directly by applying its geometric definition. Published by Pearson Learning Solutions. therefore. functions whose equations satisfy (3) are inverse functions. Fortunately. if it decreasing so. In what follows. so there is no inverse. Published by Pearson Learning Solutions. . too. g ¿ 1b2 = however. If we differentiate. by Warren B. g) are differentiable functions. one-to-one. and Finance. Thus. g1b2 = a so we have g ¿ 1b2 = or d -1 f 1x2 dx or 1f-11x22 ¿ 1b2 = 1 f ¿ 1a2 (4d) x=b Derivative of the Inverse 1 f ¿ 1g1x22 (4a) 1 f ¿ 1g1b22 1 f ¿ 1a2 (4b) = 1 f ¿ 1 a2 (4c) Thus. (b) determine f -1192. y) when y = 9. d -1 f 1x2 x = 9. We would expect the same to be true with regard to differentiability. this x-value is f -1192. a) on f -1. Walter O. Note that when a graph has a (smooth) relative extremum (a turning point). that is. if f has a turning point. (c) determine dx Solution (a) We have f ¿ 1x2 = 6x2 + 3 7 0 for all x.1 Inverse Functions * ** 331 From a geometrical point of view. from (4) we see that the slope of the tangent line at (a. However. is its inverse. Economics. b) on f is the reciprocal of the slope of the tangent line at (b. it is not one-to-one. f1g1x22 = x Suppose f (and. f -11922 is the point on the inverse which corresponds to the image point on the function (x. therefore.Section 4. it follows that the continuity of f implies the continuity of the f -1. Applied Calculus for Business. It also follows that if a function is increasing. we let g1x2 = f-11x2. Copyright © 2007 by Pearson Education. 1f-11x22 ¿ 192. f -1192 = 1. so is its inverse. we obtain f ¿ 1g1x22g ¿ 1x2 = 1 or g ¿ 1x2 = or in particular. and its solution is x = 1. so the function is increasing and is. Exercise 11 Given f1x2 = 2x3 + 3x + 4 (a) Show f is a one-to-one function. using the chain rule. (b) We know that 19. Inc. so we set y = 2x3 + 3x + 4 = 9 and solve for x. therefore. and April Allen Materowski. Wang. We have. Gordon. this is an easy problem to solve by inspection. using the composition property. its derivative is zero. Exercise 13 Given f1x2 = 2x3 + 3x + 4 (which was seen to be one-to-one in Exercise 11). Using the solve command.3 2. dx 2 1x Solution Let (a. Inc. EXERCISE SET 4. f1x2 = 2x . g1x2 = . 1) on the inverse function s graph is the reciprocal of the slope of the tangent line at the point (1.4x 2 + 1 7. f 1x2 x = 9 = 1f-11x22 ¿ 192 = 1/9.1 Inverse Functions (c) The slope of the tangent line at the point (9. Gordon. dx Exercise 12 If f1x2 = x2 is defined on the domain x Ú 0 then its inverse function is f -11x2 = 1x. Note that b = a2 or a = 2b. r1x2 = 22 . by Warren B. we have solve 12x3 + 3x + 4 = 12. and April Allen Materowski. Published by Pearson Learning Solutions. . a). Consider the following exercise. Solution We proceed as we did in Exercise 11.7 6. We know that 112. and Finance. so f ¿ 112 = 9.27732 as an approximate solution. Copyright © 2007 by Pearson Education. w1x2 = . x2 and the calculator gives us x = 1.332 * ** Section 4. Using (4) show that 1 1x2 = .7x + 9 5.277322 L 611. Economics. d 1 (Verify!). G1x2 = . What do we do if that is not the case? We can use the solve command on the calculator and then proceed in the same manner. d -1 f ¿ 1x2 = 6x2 + 3.5x Applied Calculus for Business. so we set 2x3 + 3x + 4 = 12 and solve for x. Walter O. (Why is the negative square root not an option?) Using (4) we have d 1 1x2 dx If we replace b by x we have d 1 1 1x2 = .3x + 7 3. Therefore f ¿ 11. Wang. h1x2 = 22x + 3 8.789282 L 0. f1x2 = 2x2 . 1. this x-value is f -11122. f -111222 is the point on the inverse which corresponds to the point on the function whose y-value is 12.2773222 + 3 = 12. h1x2 = 2x + 5 4. therefore. b) be a point on f and its image point on f -1 is then (b.1 In Exercises 1 10 determine whether or not the function determined by the given equation is one-to-one. determine 1f-11x22 ¿ (12).07819. 9) on the graph of f.78928 and 1f-11x22 ¿ (12) = 1/112. dx 2 1x x=b = 1 d 2 1x 2 dx = x=a 1 1 = 2a 2 2b Calculator Tips We were able to solve the cubic equation in Exercise 11 by inspection. by Warren B. x Ú 0 20.5x 19. Fig. 16:Ex.2x + 6. 0 In Exercises 11 20 determine if the equation describes an increasing or decreasing function or neither. 24 25. x Ú 0 10. s1x2 = . In Exercises 21 28 determine if the given graph represents a one-to-one function.2x2 + 6. 14:Ex. w1x2 = . f1x2 = 2x2 . x 0 2 Fig. 26 Fig. Published by Pearson Learning Solutions. 9:Ex. Copyright © 2007 by Pearson Education. Fig. Wang. Economics.3 12.4x + 1 17. 25 Fig. 21. and April Allen Materowski. 10:Ex. h1x2 = 2x + 5 14. 12: Ex. g1x2 = . 23 Applied Calculus for Business. and Finance. 28.3x + 7 13. 21 22. G1x2 = .7 16. 22 23.Section 4. 27. 15:Ex. x 2 Inverse Functions * ** 333 24. . v1x2 = 4x2 + 1.7x + 9 15. Walter O. h1x2 = 22x + 3 18. Gordon. Inc.1 9. 27 Fig. Fig. s1x2 = . 11. 11:Ex. r1x2 = 22 . f1x2 = 2x . v1x2 = 4x2 + 1. 28 Fig. 13:Ex. 26. f1x2 = 2x2 + 1. g1x2 = 1/ 2 x 2 + 3/2 x Ú 0 60. g1x2 = 3 75. If f1x2 = 22x . f1x2 = x2. Suppose f and g are inverse functions and have second derivatives. Hint: assume both g and h are inverse functions of f. Using the results of the previous exercise. dx 3x2/3 0 x 6 1 x Ú 1 69.7. f1x2 = 2x2 + 3. f1x2 = 2x + 1 47. verify that the function in the indicated exercise is always increasing or always decreasing and therefore one-to one. by Warren B. (b) Determine 1f-11x22 ¿ (7). f1x2 = 2 1 3 x 38. x2 . (a) Given f1x2 = 2x3 + 3x .4. f1x2 = 25 .5.1. f1x2 = 22x . Exercise 37 65. g1x2 = 2 x + 7 54. and Finance. x Ú 0. using the derivative. (d) show the function is one-to-one.8 32. Consider the function defined by y = f1x2 = e . 70. Wang.12 43. x Ú 0 48. (b) f -11 . Prove the inverse function is unique.3. f1x2 = . 61.3x + 9.5 53. (e) sketch the graph of its inverse function.3. x 36.5x 2 + 3.x2.9 (b) Does this function have an inverse? Explain. using (4) that d 1/3 1 1x 2 = . f1x2 = 8x3 37.4x 2x + 7 51. show this function is one-to-one. x Ú 0 35. If f1x2 = 3x + 2 2x . g1x2 = 2x .x2. f1x2 = 3x .9 5 x + 3 2 In Exercises 29 38. 68. If f1x2 = 2x . x 49. (a) Do even continuous functions have an inverse? (b) Odd functions? 74. g1x2 = x . f1x2 = . Prove. Published by Pearson Learning Solutions.1 In Exercises 61 65. . (c) determine its range. Applied Calculus for Business. x . Are there functions that are their own inverse? 73. f1x2 = 22x . Inc. f1x2 = 3x . Consider f1x2 = 4x . find (a) f-1102. Hint: there are four cases to consider.3 52. determine how the concavity of g is related to the concavity of f.12 3 57. consider (gfh)(x). (a) Given f1x2 = 3x5 + 2x3 + 2. f1x2 = 2x + 5. Show that f 1g1x22 g 1x 2 = 1f ¿ 1g1x2223 76.2 2x + 5 39. find (a) f -1112. 67.2x 4x 0 x 6 2 x Ú 2 sketch the graph of this function and determine if it is one-to-one. f1x2 = 2x .32.4x2 + 7. for the function determined by the given equation (a) determine its domain. (b) f -1132 42. find (a) f -1132. f1x2 = 2x . (b) f -1132 40. find (a) f-1112.1 Inverse Functions 55. 45. Gordon. g1x2 = 25 . f1x2 = 29 . f1x2 = x2 + 1. and April Allen Materowski.x2 58. f1x2 = 5x . Will a one-to-one function always be a decreasing or increasing function? 72. 71. (g) determine the range of the inverse function. f1x2 = 9x . find (a) f-11 . (b) f -1142 2x + 1 . f1x2 = 26 . (b) f -1152 41. 29.334 * ** Section 4. If f1x2 = 23x + 2.3 30. (f) determine the domain of the inverse function. (b) f -1132 44. g1x2 = 29 . f1x2 = 11 . If f1x2 = 3x + 4 In Exercises 45 52 determine the equation of the inverse function. f1x2 = . f1x2 = 1x . In Exercises 53 60 verify that f and g are inverse functions using the composition property. f1x2 = 5x + 9. Exercise 35 64.3x 2x + 5 0 66. find (a) f-1122.9 46. Economics. Exercise 29 0 62.2x 33. x Ú 0 34. Copyright © 2007 by Pearson Education. g1x2 = 56.25 . Consider the function defined by y = f1x2 = e x -x sketch the graph of this function and determine if it is one-to-one. 77. (h) find the equation of the inverse function and (i) verify that f -11f1x22 = x and f1f -11x22 = x. f1x2 = . (a) Show that this function is always decreasing.3x + 7 31. show this function is one-to-one (b) Determine 1f-11x22 ¿ (18). f1x2 = 22x + 3 50. Exercise 32 63. (b) sketch its graph. If f1x2 = .x2 59.3 . Walter O. 414213 = 21. Gordon.414 = 21. we would obtain 222 = 2.665137562 2. is well defined.4142135624 = 2.4142135 = 21. plotted and then connected the points to sketch the graph. It is clear. b is not allowed to be negative because we want to deal only with real numbers for example.665144. where b is any positive number other than 1 and x is any real number. Walter O.414213562 = 21. we see that to each x-value there corresponds one y-value thus. to seven decimal places.2 Exponential Functions * ** 335 4. Inc. Published by Pearson Learning Solutions. As irrational values for x lie between rational values that we plotted. We could analyze any irrational power the same way. Applied Calculus for Business.665143104 2.664749650 2.Section 4.4142 = 21. the equation y = f1x2 = 2x defines a function.665144138 2.41421 = 21.2 Exponential Functions » » » » » » » Exponential Expressions The Graph of y * f1x2 * b x Solving Special Exponential Equations Finding the Exponential Function Growth and Decay Rates Power Function Calculator Tips Consider the exponential expression 2x. 2. since 1 raised to any finite power is 1. Note: we exclude b from being 1. how could we define 222? To ten decimal places. they are. all the laws of exponents given for rational exponents will be satisfied for irrational exponents as well.41 = 21.41421356 = 21.665119099 2. Thus. We now assume that any number of the form bx.4142135624. and Finance. 1 . .4 = 21. using a calculator. Copyright © 2007 by Pearson Education. the value of 222 is. For example. we have 22 L 1. What do we mean by the expression when x is irrational? We do not intend to give a rigorous definition here. For example. These values are illustrated in Table 1. by continuity. Wang. Now that we understand what we mean by expressions of the form bx. expressions like 22/3.639015822 2. but rather be guided by our intuition. and April Allen Materowski. that as we take more and more places. Economics. When x is a rational number 2x is well-defined.421/2 = 2i. and we have that 21. As a consequence. these powers can be computed with our calculator. certainly a trivial example.6651441.665144143 Exponential Expressions and so on. or 2-3/2 make perfectly good sense.657371628 2. You will note that we selected convenient rational values for x.665144142 2.665144027 2. Fortunately. represented on the graph as well. We shall select convenient values in its domain and draw its graph. by Warren B. 5 20 = 1 21 = 2 22 = 4 Using these points. Economics. Inc. so we see that the negative x-axis is a horizontal asymptote for this function. y =f(x) = 2x Figure 1: The Graph of y = f1x2 = 2 x Some observations are in order. the function defined by the equation y = f1x2 = bx where b 7 1 has the same properties. Published by Pearson Learning Solutions. so the range is y 7 0. we plot the graph given in Figure 1. Gordon. the graph gets closer to the x-axis. but never touches it (since 2x 7 0). we have 2x 6 1 and when x 7 0. and April Allen Materowski. The graph assumes all non-negative y-values. 2x 7 1. Copyright © 2007 by Pearson Education. and no matter what value we take for x. Note that as x becomes more negative.25 2-1 = 0. and Finance. for b 7 1 Applied Calculus for Business.125 2-2 = 0. In general. Wang. 2x 7 0.1). The domain of the function is all real numbers. when x 6 0. We also see that this is a one-to-one function as well. The y-intercept is (0.336 * ** Section 4.2 Exponential Functions Table 1: Points Used for Plotting the Graph y = 2 x x -3 -2 -1 0 1 2 y = 2x 2-3 = 0. . This observation implies the existence of its inverse function which we shall examine in a subsequent section. and its graph looks like the one drawn in Figure 2. The Graph of y * f1x2 * b x y = f( x ) = b x Figure 2: The Graph y = f1x2 = bx. by Warren B. Walter O. Gordon. the graph of y = f1x2 = bx when 0 6 b 6 1 will look like the graph given in Figure 4. Inc.25 11/223 = 0.5 11/222 = 0. for example. it is a one-to-one function as well. We observe that the domain of the function is all real numbers.1). It should also be noted that the graph is a decreasing function. Table 2: Points for the graph of y = 11/22x x -3 -2 -1 0 1 2 3 y * 11/22x 11/22-3 = 23 = 11/22-2 = 22 = 11/22-1 = 21 = 11/22-0 = 20 = 11/221 = 0. but never touches it (since 2x 7 0). The positive x-axis. therefore. . when x 7 0. and no matter what 1 x 1 x x value we take for x. y = f( x ) = (1/2) x Figure 3: The Graph of y = f1x2 = 11/22x More generally. Note that as x becomes more negative. A 2 B 7 1. and April Allen Materowski. so the range is y 7 0. Published by Pearson Learning Solutions.125 8 4 2 1 We plot these points and obtain the graph given in Figure 3.Section 4. Applied Calculus for Business. A 1 2 B 7 0. Copyright © 2007 by Pearson Education. and Finance. we have A 2 B 6 1 and when x 6 0. by Warren B. The graph assumes all non-negative y-values. or y = 0 is a horizontal asymptote as x gets large. Economics. lets examine the graph of the function defined by the equation y = f1x2 = 11/22x.2 Exponential Functions * ** 337 Next consider the case when 0 6 b 6 1. We give the values used in Table 2. The y-intercept is (0. Wang. Walter O. the graph gets closer to the x-axis. 1b 2 7 0. Example 1 Sketch the graph of the function defined by the equation y = f1x2 = 21x . and April Allen Materowski. Copyright © 2007 by Pearson Education. When b 6 1. The Range is y 7 0 3.q 6 x 6 q 2. The function is always positive. For 0 6 b 6 1 the function is decreasing. We use a calculator to determine the y-values. (see Table 3). Exponential functions where the exponents are more complicated will have the same general form. Applied Calculus for Business. and round the answer to one decimal place. it follows that when b 7 1. Published by Pearson Learning Solutions. and Finance. Economics. The graph is always concave upward Note. 4. as we see in the next example. 0 < b < 1 Figure 4: The Graph of y = f 1x2 = b x. as x : q positive x-axis. 5. the function is one-to-one. Moreover. For b 7 1 the function is increasing. as x : . When b 7 1. from the above properties. . we must have 1b 2 6 0. has an inverse. The Domain is all real numbers. 0 6 b 6 1 Summary of the properties of the exponential function defined by y * 1 2 * 1. that is . Gordon. Inc. and.2 Exponential Functions y = f(x) = bx . we must have d x d x 1b 2 7 0 and when 0 6 b 6 1. In either case. by Warren B.q the negative x-axis. We shall verify these remarks when we dx2 learn how to differentiate exponential functions. y = 0 is its horizontal asymptote.32/4.338 * ** Section 4. y = 0 is its horizontal asymptote. Wang. Walter O. therefore. since each of dx dx d2 x the graphs are concave upward. Solution We take a reasonable number of x-values and compute the corresponding y-values. 25 = 0. x -2 -1 0 1 2 3 4 5 y = f1x2 = 21x . However. Dividing the exponent by 4 changes the scale and causes the graph to rise more slowly than y = 2x. Thus.3 = 34 + 5x Since the exponential function is a one-to-one function. Gordon.8 20 = 1 2112/4 = 20.7 21-12/4 = 2-0. Economics. then so are the exponents.4 We use these values to sketch the graph which is given in Figure 5. What if the bases are different? For example. in the given problem we have that 2x . by Warren B.Section 4. the term x .2 Exponential Functions * ** 339 Table 3: Points to plot the Graph of y = f 1x2 = 21x .75 = 0. but related problem.5 21-32/4 = 2-0. we have that if br = bs. Published by Pearson Learning Solutions. Therefore. there is one special case in which the bases are different.2 2122/4 = 20. note that the y-value does not equal one until x = 3.3 = 4 + 5x or solving. . We illustrate in the next two examples. we find that x = . and April Allen Materowski.3)/4 We now look at a different.32/4. y = f( x ) = 2 (x .32/4 21-52/4 = 2-1. and Finance. Observe that the graph is a typical exponential function. we observe that if the bases are the same. Consider the problem of finding x if we are given that 32x . Copyright © 2007 by Pearson Education. and we will discuss this generalized problem when we define the logarithm. then it must follow that r = s. suppose the problem was to solve 32x .6 21-22/4 = 2-0. Wang.25 = 1.3 moves the graph 3 units to the right. Inc.5 = 1.25 = 0. but using some simple properties of exponents.5 = 0. Solving Special Exponential Equations Applied Calculus for Business. Walter O.3)/4 Figure 5: The Graph of y = f 1x2 = 21x .7/3. the equation may be modified to an equivalent one in which the bases are the same.4 21-42/4 = 2-1 = 0.3 = 54 + 5x? We will need a more general procedure to solve such exponential equations. we shall shortly learn how to solve the general case.12x or x = 20/21 Example 3 Solve the equation 1 = 93 .3x 12 2 = 124218 .3x. Let us examine exponential functions a little more closely so we can develop a sense as to when data can be represented by such a function. so we have 1 = 93 . that moving the term from the denominator to the numerator changes the sign of the exponent.340 * ** Section 4.6x = 6 . but both 8 and 16 are powers of the same base.10x or x = 3/2 Finding the Exponential Function We consider these last few examples special cases. by Warren B. Copyright © 2007 by Pearson Education. Economics.10x 36x 3-6x = 36 . Inc. we have that 9x + 12 = 32 . As mentioned above. We proceed as follows: 83x + 4 = 168 . solve for x. 272x Solution Both 27 and 9 are powers of 3.. Table 1: Equally Spaced x-value with Constant Ratios in Successive y-values x f(x) 10 200 20 460 30 1058 40 2433. and April Allen Materowski. Gordon.12x We have now changed the equation into an equivalent one in which the bases are the same. because a simple example like 3x = 2x + 1 cannot be solved using these methods. Published by Pearson Learning Solutions. The first observation is that an exponential function has the property that for equally spaced x-values. Suppose y = f1x2 = Cbx.4 50 5596.5x2 3 2x 13 2 1 = 36 . therefore.3x2 3 13x + 42 29x + 12 = 232 .) We now have the same bases. Wang. Consider Table 1. therefore. Walter O. and Finance. Solution It appears that the bases are different.10x (Recall.5x for x. the ratio of successive y-values is constant. 2.2 Exponential Functions Example 2 Given 83x + 4 = 168 . .82 Applied Calculus for Business. .5x 272x 1 = 132213 . it appears that the successive ratios are nearly equal then the assumption of an exponential relationship is reasonable. Wang.956512. Assuming the form Cbx.12h2 = bh = f1a + 1k . Cb20 b20 460 = 10 = b10 = = 2. For example 460/200 = 1058/460 = 2433.12h In the previous example.956512.82/2415 = 2. measurement is never exact. Published by Pearson Learning Solutions.) Thus.12h and the next x-value is a + kh. assuming f1x2 = Cbx. we may work backwards and determine the exponential equation. by Warren B. In the real world. we have b10 = 2. we also could have represented the relationship as f1x2 = 86.31/10 L 1.3.31/102x = 86.956511. we have f1a + kh2 = Cba + kh and f1a + 1k .1a + 1k .4/1050 = 5596. h was 10 and we found the ratio to be b10. However.3 10 200 Cb b (We could choose any successive points.08686 Therefore. Walter O. We shall illustrate this case in the calculator tips.08786102 = 2. if we have any exponential function and the x-values are equally spaced then the ratio of any two successive y-values is constant. so 200 = C11. f1x2 = C11.3C and C = 200/2. and Finance. Inc.08686x2 Note. Copyright © 2007 by Pearson Education.3 From this data. if in a data set.3 or b = 2. Economics. suppose each x-value differs from the previous x-value by h units.9565 so we have f1x2 = 86.12h their ratio is f1a + kh2 Cba + kh = ba + kh . Applied Calculus for Business.12h2 = Cba + 1k . and April Allen Materowski.Section 4. Gordon. . To show this. and the ratio of any two successive y-values is 2.08686x2 We can find C by using any of the points. for example f1102 = 200.3x/102 More generally.12h2 Cba + 1k . we have.2 Exponential Functions * ** 341 Observe these x-values are all equally spaced apart by h = 10 units.3 L 86. If the starting x-value is a then the kth x-value is a + 1k . so we expect some inaccuracies. 06 (similarly 264. it represents decay or decline. Wang.1 = 1. For example P112 = 23511. 2 and 3.06-22 = 209.06. When r 7 0.1 + . (c) What would you expect the population was in 1998? Solution (a) Let t = 0 correspond to year 2000.342 * ** Section 4.062 = 235 + .4% each year. estimate the country s population in 2005. (b) Predict the population in 2005.22 = 23511. and we need to look at successive ratios.06 = 6%. Then the successive years in the Table are t = 1. C = f102 is the y-value corresponding to t = 0. So we see that the growth of the population each year was 6%. or starting value. Economics.1 and so on for the other years.06 and 279. Example 5 The population of the Ukraine is decreasing by 0. we shall write the relationship in the form f1t2 = Cbt. in this example h = 1. Published by Pearson Learning Solutions. it represents growth. With such an interpretation. That is. If the population in the year 2000 was 49. That is.062 = 249.1 2002 264. Walter O. Gordon. so in many examples. and Finance.0652 = 314. observe that when t = 0.06 # 235 P122 = 249. the independent variable is time.06t2 (b) The population in year 2005. What interpretation can we give to r? In the last example observe that we found b = 1. as in the previous example.889/264. when r 6 0.06 # 249. Copyright © 2007 by Pearson Education.046 2003 279. we see that C = P102 = 235.889 (a) Determine an exponential equation satisfying this data. Observe that the population in any year was equal to the population of the previous year + 6% of the population of the previous year. which we could write as 1 + 0.149 million Growth and Decay Rates We can write b = 1 + r. Example 4 Suppose the population of the United States is given as follows in Table 2: Table 2: U.2. Applied Calculus for Business. Let the population be P1t2 = Cbt From the remark given above. Observe 249.1/235 = 1.06). Population 2000 2003 Year Population in Millions 2000 235 2001 249. that is. Often in such cases we define t = 0 as the initial.S. so we have P1 . when t = 5 would be P152 = 23511. Inc.153 million. we see that when r is positive. f102 = Cb0 = C.046/249. r = . so we have b2 = b1 = 1. To find bh. . it represents the growth rate or the percentage of change.483 million (c) 1998 would correspond to t = .06 b1 and P1t2 = 23511.06.2 Exponential Functions Very frequently. Thus. by Warren B.111.046 = 1. and April Allen Materowski. 3 we enter 126¿1-2. As we saw earlier.4% each year. (a) Determine an exponential equation which gives the amount A of Radon remaining at the end of t (thousand) years.15310. we state that exponential functions dominate power functions.0. Note that for an exponential expression. if we begin with a 100 lb sample of Radon. by Warren B. Suppose. one of the form xk. then the year 2005 will correspond to t = 5.35 lbs. to compute 126-2.35.15310. since b = 1 + r and r = .32 and press enter. the base is the constant and the power is the variable. Walter O. It is important not to confuse an exponential expression which has the form bx with a power expression. 100 lbs. (b) How much Radon remains after 10.000 years? Solution After 1000 years. and A(t) is immediately obtained as given above.0. 35% of it. we have b = 1 .999625 = 49. Since the population is decreasing by 0. Economics. in this example.65.004 = . we have b1 = 0. they grow more quickly and decay more rapidly. Wang. we have A1102 = 10010. the calculator is especially convenient in the finding the equation of a regression curve. thus. . we have r = .Section 4. (b) In 10. 0.652t (Remember C is the initial amount.0.) We really did not need to construct the above Table to compute b. Exponential and power functions grow (and decay) differently. To compute the value of an exponential expression we enter it normally and use the caret key ¿ to denote exponentiation. We shall illustrate this comparison in the Example 7 below where we model a data set three ways. In Section 2. and Finance. the ratios of the correspon- Power Function Calculator Tips Applied Calculus for Business.35 # 100 = 35 lbs will decay and 65 lbs would remain. Copyright © 2007 by Pearson Education. that is. whereas for a power expression the base is the variable and the power is the constant. for equally spaced x-values. A power function has an equation of the form f1x2 = ax b.99962t and we have in year 2005 P152 = 49. We can construct Table 3.65210 L 1. Gordon.25 3 27.9996.2 Exponential Functions * ** 343 Solution Let t = 0 correspond to the year 2000. Published by Pearson Learning Solutions.35 = 0. that is.4 we shall examine this remark more fully.004 and we have b = 1 . Inc. but for now. As above. Table 3: Decay of Radon Years (thousands) Radon Remaining (lbs) 0 100 1 65 2 42. where k is a constant. the equation would be P1t2 = 49.0548 million Example 6 Radioactive Radon decays at a rate of about 35% every 1000 years. The calculator makes computations with exponential expression very straight-forward. From our discussion above.4625 Thus.0.65 and we have the amount of Radon remaining A(t) at the end of t years (where t is in thousands) is A1t2 = 10010.000 years. and April Allen Materowski. Walter O.346. but they are close.066. Solution (a) If we take successive ratios.98 Determine (a) an exponential function (b) a linear function and (c) a power function that may represent this data. we obtain 1.746.99177711.1. Example 7 Consider the data set illustrated in Table 4 Table 4: Finding Different Regression Equations for the Same Data Set x f(x) 1 2. 1. (The regression curve is now drawn. 1.5. 1.163 2 2.085. 4. 2. We may then suspect that the data may represent an exponential. Regeq(x) STO y1(x) and press Enter (This stores the regression equation as y1(x)) then enter on the entry line NewPlot 1. (3) To draw the regression line. which are fairly close to each other. We therefore fit this data to an exponential regression curve. 2. We proceed as follows: (1) Store the x-coordinates and y-coordinates as lists named t1 and t2.163. enter on the entry line in the HOME screen.746 5 2. in the same way that we would do linear or polynomial regression.083102x2 Applied Calculus for Business.346 3 2. 3. t1. done exactly the same way on the TI-89 as you would for linear (or quadratic) regression with the only change being we replace LinReg with ExpReg. so it is reasonable to assume that the function we need to obtain is an exponential. as follows: 51. See Figure 7) Figure 6: Exponential Regression Figure 7: The Exponential Regression Curve f 1x2 = 1.99177711. 2. t2 and press ENTER Choose WINDOW 1*F22. and Finance. then press Zoom (F2). Inc.986 STO t2 press ENTER Make sure you are using curly braces to enclose these two lists (2) Next enter on the entry line of the HOME screen (use the Catalog or type) ExpReg t1.098. that is the exponential regression curve. . t2 and press ENTER then ShowStat and press ENTER (This produces Figure 6) So we have f1x2 = 1. Copyright © 2007 by Pearson Education.2 Exponential Functions ding y-values are not the same. The procedure with the calculator is almost identical to the linear and polynomial case with the only difference that we enter ExpReg as indicated in the following example. we may determine the best fit exponential function.344 * ** Section 4.085. by Warren B. and scroll down and select ZOOMDATA (or press 9). 2.5 4 2. and April Allen Materowski. 56 STO t1 press ENTER 52. Economics.083102x2 as the exponential regression curve. In such a case. Wang. Gordon. 2. Published by Pearson Learning Solutions. Figure 9: The Power Regression Curve y = 2. the exponential yields 4. by Warren B.277. Published by Pearson Learning Solutions.192012 So which one model do we choose. as the correlation coefficient is close to one.105729x 0. This demonstrates that exponential growth dominates linear and power growth.971. and April Allen Materowski.2034x + 1. (c) To obtain the Power function we proceed as above with PowerReg replacing ExpReg. then the answer would be the one that best supports the data. we obtain (Verify!) Figure 8. suppose x in the above data set represented years and f(x) represented the population in year x. Gordon.2 Exponential Functions * ** 345 (b) For the linear equation. replacing ExpReg with LinReg. compare values for x = 100. Inc.Section 4. exponential or power curve? If we had additional data. the population was 5. Wang. . the linear.9368 This looks like a good fit as well. and Finance. Walter O. Applied Calculus for Business. Let us also suppose that we know that in year x = 10. Figure 8: The Linear Regression Curve y = 0. 1000 and so on. the exponential values grow much faster than either the linear or power values. For example. We obtain Figure 9. It appears that the exponential better models the data than either the line or the power.49 and the power model yields 3. Copyright © 2007 by Pearson Education. Economics. For example. Also note that as x gets large. Using the linear model we obtain 3. How much of a 75 lb sample remains after (a) 2000 years (b) 10. What will be your salary in (a) 5 years. (a) determine an exponential equation representing the population as a function of year.25 -2.4x 1 = 2-x. 44. sketch the graph of the function defined by the given equation. What will be the population in year (a) 2000. and Finance. Radioactive Carbon 14 decays by about 11% every 1000 years. (c) 2010? 50.4 4 33. 5 11.346 * ** Section 4. y = f1x2 = 3x 14. where t = 0 corresponds to year 1999.232.5x 42. 93x + 2 = 272 + x 41. How much would remain in a 50 lb sample at the end of (a) 5000 years (b) 20. How much of a 10lb sample remains after (a) 3 days.78213 8. 422 9.000 per year with a 3% increase each year. by Warren B. f1x2 = 36.718-x 23. (b) 25 years? 58. (b) 2005. 1 .2 + 3 29.000 years? 55. The population is growing at about 6% each year.718 x x x 47.1 x Solve for x in Exercises 37 44 37.x2. 9 .23 32x . 2222 -3. Sketch on the same graph y = f1x2 = 2x and y = g1x2 = can be concluded about the two functions? 46.75-3/4 6. What can be concluded about the functions y = f1x2 = bx and y = f1x2 = b -x? 48.6 million.54 4. and April Allen Materowski. What 2x 1 = 3-x. f1x2 = 10x 4-2x 35. Walter O. 1 . f1x2 = 2-x 3x 34. f1x2 = 3x + 2 27.3 30. 7 12. The population P at time t of a city is estimated by the equation P1t2 = 1500122-0. (b) What will be the population in 2006? (c) What was the population in 1998? 54. Gordon. The population of Hungary is decreasing by 0. Thorium 234 decays by about 3% every day. Suppose g1x2 = f1 . f1x2 = 4 3x + 1 5 63x + 2 2x . 4. f1x2 = 2. y = f1x2 = 4-x 16. Radium decays by about 35% every 1000 years. Inc. f1x2 = 3x . 0.2x = 163x + 7 = 275 .5 2.2 25. y = f1x2 = 11.56 45.21 38.3. y = f1x2 = 3-x 15. f1x2 = 23x . What will be the population of this community in year (a) 2000.3 = 161 . The population P at time t of a rural community is given by the equation P1t2 = 1500132-0. f1x2 = 2.753 4/3 40.5. (b) What will be the population in 2010? (c) What was the population in 2003? 52. (b) 1 week (c) 30 days? 56. 1. f1x2 = 3x .652-x 21. What 3x 5.2325 7.1 82 + 3x 812 + 3x 1 814 .673. 1 8x + 1 = 161 . How many points uniquely determine (a) an exponential function? (b) a power function? 22. 82x . what can be concluded about the two functions f and g? 49.2 Use your calculator to compute the expression given in Exercises 1 12. f1x2 = 3x . 25x = 83x + 1 Applied Calculus for Business.4x 3.1 91 + 3x 10.000 years? 57.5% each year.2 28. 272x + 1 = 43. (b) 15 years? 53. Sketch on the same graph y = f1x2 = 3x and y = g1x2 = can be concluded about the two functions? In Exercises 13 36. 15. (a) Determine an exponential equation representing the population as a function of year.4x 39. f1x2 = 2 3x . A company offers you a position with a starting salary of $45. Published by Pearson Learning Solutions. (b) 2005.25t.139 million.2 Exponential Functions EXERCISE SET 4. If the population in 2005 is 296. Its population in 2000 was 10. t = 1 to year 2000 and so on. y = f1x2 = 11/32x (compare with Exercise 14) 18. f1x2 = 3x + 2 26.25t Suppose t = 0 corresponds to year 1999. t = 1 to year 2000 and so on. 1. 23x 42x . 163x + 1 = 642 .6 32. f1x2 = 312x2 24. 13.32 20. f1x2 = 512-x2 31. y = f1x2 = 4x 17. (c) 2015? 51.2 . Economics. 4. f1x2 = 3x . An IRA is expected to earn a about 5% each year. y = f1x2 = 11/42 (compare with Exercise 15) 19. .7 -3. Wang. f1x2 = 10. What will an initial $1000 investment be worth in (a) 10 years. Copyright © 2007 by Pearson Education. 7 35. Which of the three functions would best approximate this new data point? 63. Given the functions you found in Example 60. Which of the three functions would best approximate this new data point? 62. Find an exponential function determined by the data in Table 6.2 127. and April Allen Materowski. Table 8: EX. Find an exponential function determined by the data in Table 8.28 20 26. suppose you are later given another data point (5. Table 5: EX. What reasons can you give to explain the different number of deaths in the late 1980s and early 1990s as compared with the later 1990s and early 2000s? Source: Centers for Disease Control and Prevention: HIV/AIDS Surveillance Report.8 114. and so on.2 98. 65 8 20 16 17 24 14.9 68. 6) and (4.) (c) (i)Find another best fit exponential and (ii) power function that represents the data from years 1993 through 1997. Gordon.S.(Data from http://www.7 20 50.6 15 20. Copyright © 2007 by Pearson Education. and Finance.84 80 37.1 164.45 32 12. Inc.3 64.9 83. (Let 1815 be year 0.458 65. 5) and (2.7 26. 68 Year 1981 1982 1983 1984 1985 1986 1987 * Estimate 66. Given the points (1.1 140.edu/malcz/ExpFit/ data. 1825 year 1 and so on.0 14. 63 5 12 10 15. 64 10 15.24 30 160.0 190. estimate the population in 2005. Walter O.768 40 514. .) Based on your solution. Published by Pearson Learning Solutions. (b) exponential function and (c) power function they determine. 66 20 100 40 72 60 51. (Let 1981 be year 1. The population of the United States from 1815 to 1975 is given in Table 9. Table 6: EX. The number of deaths in a year attributed to AIDS is indicated in Table 10.9 214. 7) determine the (a) linear function. Table 7: EX.15 Table 10: Deaths Due to AIDS Ex. Economics.mste. Find an exponential function determined by the data in Table 5.2 59. suppose you are later given another data point (8.7 19. 1982 year 2.Section 4. 67 Year 1815 1825 1835 1845 1855 1865 1875 1885 1895 1905 1915 1925 1935 1945 1955 1965 1975 Exponential Functions * ** 347 Population (in Millions) 8.3258 Deaths 130 466 1511 3526 6996 12183 16488 Year 1988 1989 1990 1991 1992 1993 1994 Deaths 21244 28054 31836 37106 41849 45733 50657 Year 1995 1996 1997 1998* 1999* 2000* 2001* Deaths 51414 38074 21846 19005 18491 17741 18524 67. Using the years 1981 through 1993 find the best fit (a) exponential (b) power function representing this data. Given the points (2. 9).364 Table 9: U. Find an exponential function determined by the data in Table 7. by Warren B. Given the functions you found in Example 59.uiuc.4 55. (b) exponential function and (c) power function they determine. 60. Wang. Determine an exponential regression function that will best fit this data. 61.html) Applied Calculus for Business.2 44.2825 68.3 11. 21). 12) determine the (a) linear function. Population Ex. Wang. p. . but it is closer to y = 3x since p is closer to the number 3 then 4. The graphs of all three functions are given in Figure 1. even if the constant is represented by a Greek letter. you always obtain the same number. We introduced the expression px just to emphasize that we can have an exponential function whose base is any positive constant. Copyright © 2007 by Pearson Education.3 The Number e 4. we first take a somewhat circuitous route and examine a basic notion of finance. and Finance. measure its circumference and then divide the circumference by the diameter of the circle.141592654. Gordon.3 The Number e » » » Continuous Compounding of Interest The Constant e Calculator Tips If you draw any circle. It is an irrational number. and y = 4x Observe that the graph of y = px is sandwiched between the graphs of y = 3x and y = 4x. We begin by reminding you of the law of compound interest. and April Allen Materowski. Thus. Published by Pearson Learning Solutions. Walter O. Inc.348 * ** Section 4. y =* x y = 4x y = 3x Figure 1: The graphs of y = 3x. approximately equal to 3.14159 for p and then use the appropriate keys on the calculator. it would make perfectly good sense to plot the graph of the function y = f1x2 = px. If this is done. However. This constant p is well known to all of us. We could generate a table of values by using 3. If P dollars is invested into an account yielding interest at a rate r compounded n times per year for t years. we would find that its graph is very similar to both the graphs of y = f1x2 = 3x and y = g1x2 = 4x. by Warren B. We will soon introduce you to another mathematical constant and then examine exponential expressions involving this number as its base. y = px. then at the end of this time the accumulation A is given by the formula Applied Calculus for Business. Economics. observe that as n : q . let us rewrite the expression using the power property law of exponents as a1 + r nt r n t b = a a1 + b b n n Continuous Compounding of Interest Suppose we let u = r/n or equivalently. Inc.12.3 The Number e * ** 349 A = Pa1 + r nt b n What happens to the accumulation as n.12 Note that as n. as n : q ? Consider the expression a 1 + b . Copyright © 2007 by Pearson Education. Economics. we see that 11 + u2u is approximately equal to 2. Wang. (The first five frequencies are commonly used and their names are indicated next to them. we need to understand what happens to the expression 11 + u2u as u ap1 proaches zero.814. . it appears that the answer should be around $1.Section 4. increases. the frequency at which interest is given.822. Walter O. and Finance. In fact.03 $1822. From both the graph given in Figure 2. increases? Consider the concrete example of investing $1. Then the expression may be rewritten. the accumulation increases.02 $1. Gordon. 1 u : 0. and taking values of u near zero you can generate a table similar to the one given in 1 Table 2. What happens to the accumulation if interest is given continuously.11 $1822. at every instant of time? From Table 1. as n : q . In fact. u = r/n gets small. in fact. using the laws of exponents as a1 + r 1 r nt r n t b = a a 1 + b b = A 11 + u2u B t = A 11 + u2u B rt n n Furthermore. Table 1 gives the accumulation for different values of n. there is almost no difference between getting interest hourly or every minute. as n gets larger the accumulation appears to increase much more slowly. almost becoming fixed.000 Invested at 6% Compounded n Times Per Year for 10 Years n 1 (annual interest) 2 (semi-annually) 4 (quarterly) 12 (monthly) 365 (daily) 365 # 24 (hourly) 365 # 24 # 60 (every minute) A * 100011 + 0. that is.7. using your calculator.) Table 1: The Accumulation of $1. One approach is to sketch 11 + u2u versus u. we are asking what happens to the accumur nt lation as n becomes infinite.06/n210n $1.790.85 $1. Note that as u approaches zero the value of the expression approaches values a little larger than 2. by Warren B. and April Allen Materowski. moreover.11 $1. n = r/u.822. Applied Calculus for Business.40 $1. the number of times per year interest is given. that is.819. Published by Pearson Learning Solutions.000 for ten years at a rate r = 6%. What n happens to it as n : q ? First. Thus.71828 as u approaches zero. Mathematically. and Table 2. We do that in Figure 2 where we plot the graph as u takes on values from 1 to near 0.806. . Copyright © 2007 by Pearson Education. Gordon. Applied Calculus for Business.71828.718268237 2. we have x : q .704813829 2.001 . and Finance.718280469 2.716923932 2. Mathematically. e = 2.350 * ** Section 4. and April Allen Materowski. Wang. then a1 + r 1 r nt r n t b = a a 1 + b b = A 11 + u2u B t = A 11 + u2u B rt approaches ert n n and the compound interest formula becomes A * Pe rt (2) when interest is given continuously (at every instant of time).000001 .01 . Inc.718281825 1 1 1 The Constant e In more advanced courses it can be shown rigorously that the expression 11 + u2u does indeed approach a fixed value as u approaches zero and this value is denoted by the 1 number e. When t is negative equation (2) gives a past value. we have that to five decimal places. (Note that when t is positive equation (2) gives the future value of P. Economics. Published by Pearson Learning Solutions.00001 . 1 If we substitute e for 11 + u2u as u approaches zero. by Warren B.0000001 .718145827 2. then as u : 0. we define e as e = lim 11 + u2u u:0 1 (1a) Equivalently. so we may give an equivalent formulation of the definition of e as e = lim a 1 + x: q 1 x b x (1b) As seen in Figure 2 or Table 2.0001 . That is.3 The Number e ( 1 + u ) /u 1 Figure 2: The Graph of 11 + u2u Versus u as u Approaches Zero Table 2: The value of 11 + u2u as u approaches zero u . if we let x = 1/u.00000001 11 + u21/u 2.718281692 2. we define e as the value the expression 11 + u2u approaches as u approaches zero. Walter O.) We illustrate the use of (2) in the next few examples. We can now examine the function defined by the equation y = f1x2 = ex.09 Plotting the points from Table 3. ex is indicated in green above the X key.3 The Number e * ** 351 Example 1 $1500 is invested for 5 years at an interest rate of 7% compounded (a) monthly.4176252312 = $ 2. we could. Walter O.44. Published by Pearson Learning Solutions. Determine its accumulation. We have now introduced another mathematical constant. Solution For the first two parts of the problem we use the discrete formula. Since e is slightly less than 3.35 = 150011.60. n (a) We have n = 12. .) Table 3: Values Used to Plot the Graph of y = f 1x2 = e x x -2 -1 0 1 2 3 y * ex e -2 L 0. On the TI-89 calculator. first requiring the pressing of * to access it. e2 ) (-2. Since e is a constant.128.419020392 = $ 2. the accumulation increased.14 e -1 L 0. nt = 513652 = 1825 and A = 150011 + . e -2 ) (-1. Gordon. and A = 150011 + .07.128.4190675492 = $ 2. t = 5 and r = 0. e1 ) Figure 3: The Graph of y = f 1x2 = e x Applied Calculus for Business.53. e. It can be shown that like p. using a calculator. e3 ) (2. (The y-values are rounded to two places. we obtain the graph in Figure 3 y = f( x ) = e x (3.37 e0 = 1 e1 L 2. We first use a calculator to obtain Table 3. We have # A = 1500e0. e is such a useful constant in mathematics that most calculators have a key for the exponential function ex. Inc. evaluate expressions of the form ex. and April Allen Materowski. that as the frequency increased.39 e3 L 20.126.07/36521825 = 150011.1) (1. (b) daily. namely e. Copyright © 2007 by Pearson Education. (c) continuously.07 5 = 1500e0.72 e2 L 7. by Warren B. its graph will look very similar to y = 3x.07/12260 = 150011. Observe in the previous examples. (c) Since interest is now being given continuously. we use equation (2). Wang. e is also an irrational number. Economics.Section 4. (b) We have n = 365. r nt A = P a 1 + b with P = 1500. nt = 51122 = 60. and Finance. However.q ). their relative positions change. and Finance. Copyright © 2007 by Pearson Education. y = 3x f( x ) = e x Figure 4: The Graphs of y = f 1x2 = e x and y = 3x Observe that since e 6 3. Wang. Applied Calculus for Business. Published by Pearson Learning Solutions. by Warren B. it is a one-to-one function and has an inverse. It has the negative x-axis as its horizontal asymptote (x : . Therefore. The function is an increasing function on its domain. Its domain consists of all x-values and its range is y 7 0. Inc. This is done in Figure 4.352 * ** Section 4. . it is difficult to see this on this scale so in Figure 5 we change scales to better see this last observations. and April Allen Materowski. for x 7 0. y = 3x y = f( x ) = e x Figure 5: The Graphs of y = f 1x2 = e x and y = 3x for x 6 0 To summarize: the graph of y = f1x2 = ex is a typical exponential function. and when x 6 0. Gordon. y = 3x is above the graph of y = f1x2 = ex. Walter O.3 The Number e It is instructive to compare the graph of y = f1x2 = ex with the graph of y = 3x. Economics. to access the ex key on the calculator. y = f1x2 = a b e e x 27. (b) 10. y = f1x2 = ex + e . How much should be deposited into an account today if it is to accumulate to $950 in 6 years if the account bears interest at 4.000121t. The determination of k requires the taking of a logarithm. It was the mathematician Euler who first represented the constant by the letter e in 1727. We can use the TI 89 to graph any exponential function. The base of any positive exponent can be written in terms of e. (c) 25. we can enter in the Y = screen. y = f1x2 = 3e-2x 15. y = f1x2 = ex + 3 16.23% compounded (a) monthly. (b) daily.87% compounded (a) monthly.25% compounded (a) monthly.000 years. y = f1x2 = ex .000 in 10 years if the account bears interest at 6. y = f1x2 = 24. Determine the accumulation if interest is 6% compounded (a) monthly. if the base is 2. Thus.693147. then 2 = ek. $1250 is deposited into an account for 10 years.1 + 3 20. e 2. Economics. Calculator Tips EXERCISE SET 4. y = f1x2 = 2ex . (b) daily.37 22 In Exercises 13 28 sketch the graph of the function defined by the given equation. Using it. 1. (c) continuously? 12. $4750 is deposited into an account for 20 years. Gordon. y = f1x2 = ex + 1 . and Finance. solve 12 = e ¿1k2. y = f1x2 = a b 3 3 x 28.25% compounded (a) monthly. (c) continuously. k2 yields (press * Enter for a numerical answer) 0. and April Allen Materowski. Inc. Published by Pearson Learning Solutions. 8. For example. y = f1x2 = ex .1 18. is given by the equation Q1t2 = 100e -0. For example. Walter O. e3 2. y = f1x2 = ex .3 21.000 in 15 years if the account bears interest at 7. e 4. we can find a number k such that b = ek For example. The quantity Q of radioactive carbon remaining in a 100 gram wood sample at time t given in years. y = f1x2 = a b e 29. (c) continuously? e x 25. y = f1x2 = ex + 1 + 3 22.Section 4. 13. (b) daily.3 The Number e * ** 353 The constant e did not appear in the early study of finance. Wang. How much should be deposited into an account today if it is to accumulate to $15.37% compounded (a) monthly.3 In Exercise 1 4 calculate the given expression. How much should be deposited into an account today if it is to accumulate to $2100 in 7 years if the account bears interest at 5% compounded (a) monthly. Base e is so important to calculus that it is called the natural base. (c) continuously. any base can be replaced by base e with an appropriate k. 6. (c) continuously. Making sure the calculator has an appropriate window. we press * and then X. Determine the accumulation if interest is 4. (c) continuously? 10.25% compounded (a) monthly. simplifies calculations in the calculus. 9. (b) daily. that is. (b) daily. (c) continuously? 11. As mentioned above. we can use the solve command to determine k. and k is approximately 0. it will then graph this equation. Determine the accumulation if interest is 7.1 14. (b) daily. Later on in this chapter. The first reference to the constant was in 1618 in the work of John Napier. y11x2 = e ¿1-2x2.3 17. by Warren B. (b) daily. Determine the accumulation if interest is 5. to be considered later on in this chapter. e -2 3. How much radioactive carbon remains in the sample after (a) 100 years. y = f1x2 = ex . $1475 is deposited into an account for 12 years.1 . y = f1x2 = ex + 1 19. (b) daily. 7.e -x 2 5. $500 is deposited into an account for 6 years. we shall see. we shall give an explicit formula which determines k. (c) continuously. How much should be deposited into an account today if it is to accumulate to $2. However. The remark made above was that any base b can be written in the form ek. y = f1x2 = a b 2 2 x 26.3 23.x 2 ex .693147. Copyright © 2007 by Pearson Education.000 years? Applied Calculus for Business. . we would obtain the form 0/0. Copyright © 2007 by Pearson Education. Consider Applied Calculus for Business. 1 + h L eh for h near zero We now prove the theorem using the definition of the derivative and (2). THEOREM 1 THE SIMPLE EXPONENTIAL RULE d x 1e 2 = e x dx 1 (1) The Simple Exponential Rule Before we prove this theorem. Proof Let f1x2 = ex then we have f ¿ 1x2 = lim f1x + h2 . raising each side to the power h. The quantity Q of radioactive radon remaining in a 500 gram sample at time t is given by the equation Q1t2 = 500e -0. 32. and Finance. The reason is simple. the derivative in base e results in a simpler expression. It then follows from the constant multiplier rule that d 1Cex2 = Cex dx There is no other function that has this property (except the trivial function f K 0). by Warren B. From our knowledge of limits we understand that this means if h is near zero. 30.12 ex + h . We purposely consider the case with base e first. Walter O.ex exeh . and April Allen Materowski. determine k.000 years? 4. How much radioactive radon remains in the sample after (a) 100 years. (c) 10.ex = lim = lim = lim h:0 h:0 h:0 h:0 h h h h (2) 1 If we try to take the limit at any stage in the previous line.000428t. determine k. we can use (2) to replace eh. Published by Pearson Learning Solutions. To show that is the case. Economics. this gives ex1eh . we have 11 + h2h L e or equivalently. Show that if h is near zero.12 ex11 + h . and examine other bases later on in this chapter. Hint: Use (1a) with h replacing u. If 0.4 The Derivative of the Exponential Function » » » » The Simple Exponential Rule The Generalized Exponential Rule Exponential Domination Calculator Tips In this section we examine the derivative of the exponential function whose equation is y = ex. specifically f ¿ 1x2 = f1x2. eh L 1 + h.354 * ** Section 4. Inc. Fortunately. Gordon.000 years.4 The Derivative of the Exponential Function 31.f1x2 ex1eh .5 = ek. If 3 = ek. (b) 1.12 hex f ¿ 1x2 = lim = lim = lim = lim ex = ex h:0 h:0 h:0 h h:0 h h Notice that the theorem states that the derivative of ex is itself. suppose there were some other function which had itself as its derivative. Wang. 33. . we recall that lim 11 + h2h = e (this is (1a) from the preh:0 vious section with h replacing x). xe.x] = .f1x2ex exf1x2 .2 x A e 2 B = e. then that function is a constant. Wang. .Section 4. In other words. THEOREM 2 THE GENERALIZED EXPONENTIAL RULE Suppose u is some differentiable function of x. thus f1x2 = C ex or f1x2 = Cex Thus. requiring the use of the chain rule. 2 (b) f1x2 = e2x Applied Calculus for Business.exf1x2 d f1x2 0 a x b = = = 2x = 0 dx e 1ex22 e2x e If the derivative of a function is zero. Gordon.1 x2 1 a . and Finance. dx Solution 1 2 d 1 2 1 2 d .1 x2 Determine Ae 2 B. Determine dx Solution d d -2x 1e 2 = e -2x 1 . there is no other (non-trivial) function other than the exponential that can have itself as its own derivative.x2 b = e. which does not agree with the differentiation variable.2e -2x dx dx Example 2 d . which agrees with the differentiation variable. by Warren B. the exponent is u. then we have du d u 1e 2 = e u dx dx (4) The Generalized Exponential Rule Note the difference between the simple rule and the generalized rule.4 The Derivative of the Exponential Function * ** 355 exf ¿ 1x2 . In the simple rule the exponent is x.2] = .2 x dx 2 dx Example 3 Determine the equation of the tangent line at the point x = 0 if (a) f1x2 = ex. Example 1 d -2x 1e 2.2x2 = e -2x[ . Walter O.2 x [ . and April Allen Materowski. Copyright © 2007 by Pearson Education. Published by Pearson Learning Solutions. Economics. Inc. for y Z 0. if dy = y then y = Cex dx (3) Now that we know the derivative of ex we can apply the chain rule and find the derivative when the exponent is a function of x. In the general rule. 12x + 2 = 0.3x + 221x[e -3x1 . therefore.3x + 221 . The criti4 -2 cal numbers are. f ¿ 1x2 = ex. 2 d 2 (b) At x = 0. .3x + 1 .3x + 122 = e -3x19x2 .3x + 222 = xe -3x 1 . which shows that the graph does indeed change concavity at these point so they are inflection points. Walter O.043 respectively.02 and 0. Published by Pearson Learning Solutions. Using the second derivative test.32] + e -3x2 = e -3x1 . therefore the equation of the tangent line is y .195 and 1.27 6 0 so the function has a relative maximum at 4 -2 -2 L 0. Wang.4 The Derivative of the Exponential Function Solution (a) At x = 0.12x + 22 The critical points occur when the first derivative is zero. x = 0 and 2/3. 9 e B . we have f102 = e0 = 1. 22 9x2 .3x + 22 + 1 .32 + 2xe -3x = xe -3x1 .3xe -3x + 1 .3x2 + e -3x2x dx = x2e -3x1 . 0) and A 2 3.3x + 22 a x 1e -3x2 + e -3x[1] b = dx . Copyright © 2007 by Pearson Education. f 102 = 2 7 0.3] + 1 . that is when 2 . determine the relative extrema and inflection points of the function and sketch its graph.02 or y = x + 1. by Warren B. the function has a relative minimum at (0. the first derivative will be zero when x = 0 or . we obtain Figure 1a. 14/9e The inflection points occur when the second derivative is zero. we have d -3x d 1e 2 + e -3x 1x 22 = dx dx d x2e -3x 1 . and Finance. therefore. f 12/32 L .1 = 11x . and April Allen Materowski. f 1x2 = d d d 1xe -3x1 . Corresponding y-values are 0. or 3 x L 0.3x + 22 f ¿ 1x2 = x2 Using the product rule several more times we have. so f ¿ 102 = 0.1 = 01x . Using sign analysis on the second derivative. the critical points are (0. A sketch is given in Figure 1b.138. Example 4 Using the first and second derivatives of f1x2 = x2e -3x. Gordon.356 * ** Section 4. Since the exponential function is always positive. The solutions to this quadratic equation are x = . f ¿ 1x2 = e2x 12x 22 = 4xe2x . 0).3x + 2 = 0. Economics.0. so f ¿ 102 = e0 = 1. 9 e B .062 A2 3 .3x + 22 1xe -3x2 = dx dx dx d xe -3x[ . Inc. we have f102 = e0 = 1. dx and the equation of the tangent line is y . Applied Calculus for Business. Solution Using the product rule.12 or y = 1. and April Allen Materowski. which is why the product indicated in the theorem approaches zero. . so much so that their product is zero. an exponential will grow much faster and decay more quickly than a polynomial function. This observation generalizes as follows: x: q THEOREM 2 EXPONENTIAL DOMINATION OVER POWER FUNCTIONS If a and b are positive numbers then x: q lim xae -bx = 0 a (5) This theorem states that while the x grows very large as x : q .12x + 22 M * I (1. there needed to be an inflection point afer M otherwise the graph would have crossed the x-axis. Exponential functions with negative exponents decay very quickly. 0) Figure 1b: The Graph of f 1x2 = x 2e-3x Note that the function in Figure 1b is decreasing after its relative maximum point and since it is always positive. Economics. Moreover. For smaller values of x the product is large). To convince yourself of this fact. Walter O. For example. Applied Calculus for Business. lim x2e -3x = 0. . In particular. by Warren B.138. Copyright © 2007 by Pearson Education. Similarly. (Make sure you choose very large values of x.02x for large values of x. that is.1 38 x Figure 1a: Sign f 1x 2 = e-3x19x 2 . Inc. the product is 0. greater than 55. these two results indicate that an exponential function rises and decays much more quickly than most functions.043) 2 * I (. exponential functions with positive exponents grow very quickly.02) 1 m(0.000. . when x = 55.195.000 to see that the product is small. which it cannot do because f1x2 Ú 0. Wang. Gordon. it is overwhelmed by the way the exponential term approaches zero as x : q .195 0 CU + x * 1. note that y = 0 is a horizontal asymptote as x : q .4 The Derivative of the Exponential Function * ** 357 Sign of f 1x2 CU 0 CD + x * 0.Section 4. Published by Pearson Learning Solutions. That is why the companion statement to the theorem is x: q Exponential Domination lim x -aebx = q (6) Basically. The calculus confirmed this. examine the product x 100e -0. and Finance.0002053. much more quickly than algebraic functions. Note that this function is even.x 1 2 d 1 2 1 2 1 2 d . The same is true.4 The Derivative of the Exponential Function Exponential functions arise in many different applications.x2 = . Example 5 1 2 Sketch the graph of the function defined by the equation f1x2 = e -2 x . Copyright © 2007 by Pearson Education.1 x2 A e 2 B + e -2 x 1 . (The purpose of the multiplicative scaling factor is that the total area under the bell curve must be one.x C . Moreover. and April Allen Materowski. so the critical point is (0. We saw that the exponential function arose naturally when examining continuous compounding./ 2 B . f 1x2 = . that is. e -0.e -2 x = e -2 x 1x2 . However. y = 0 is. most of the graph of this bell shaped curve is on .12 dx dx 1 2 Note that f 102 = . Since the exponential function is always positive.x2 = f1x2. Inc. e . by Warren B. M(0. Gordon.xe -2 x . it is symmetric with respect to the y-axis as f1 . the exponential function is extremely important. Published by Pearson Learning Solutions. In Example 2 we found that f ¿ 1x2 = . 1. the second derivative is zero when x = . It will have its maximum M at 2p x = 0.12 A sketch of the function is given in Figure 3. that is. We shall classify this critical point using the second derivative test.) This bell shaped curve is called the normal or Gaussian distribution. 1. Figure 3: f 1x2 = e-2 x 1 2 Applied Calculus for Business. That function is n1x2 = 11 e . 1. e -0. by symmetry as x : . that is. a horizontal asymptote as x : q .5) I 2(1 . therefore. It gives rise to the so called bell or normal or Gaussian curve as illustrated in the next example. .1 which means the critical point is a relative maximum (since it s the only critical point it is also a maximum). 1). and its inflection points at x = . and Finance. the only critical point occurs when x = 0 and f102 = 1. Its graph is given in Figure 4. In the study of statistics. for x 7 3 the y-values are almost zero. e -2 x : 0. CU + Sign of f '' ( x ) -1 1 1 2 CD 0 0 CU + x Figure 2: Sign of f 1x2 = e-2 x 1x 2 . 1 2 Note that as x : q the graph approaches 0.5) The actual bell shaped curve used in statistics is almost identical to this one except for a -1 x2 2 scaling constant. 1) I 1(-1 . Solution We need to determine the critical and inflection points for this function.q . Using sign analysis (see Figure 2) we see the function has inflection points at 1 A . Economics. Wang. Walter O.xe -2 x D .3 6 x 6 3.358 * ** Section 4. if any.) 2 2 ex . f1x2 = e-x 4. (Note: this function arises frequently in engineering applications and is call the hyperbolic cosine. (a) Determine on which intervals the function is (a) increasing and decreasing. to find the in2 -1 2 x . Determine on which intervals the function is (b) increasing and decreasing. f1x2 = e-x . x. the y coordinates are found by entering y11 . Sketch the graph of the function defined by the equation f1x2 = 2xe-4x. are the extrema of the function? 25. x = 0 20.1/a2.32.4 In Exercises 1 17 determine the derivative. f1x2 = ex1x . The proof of (7) is almost identical to the one given above and is left to the exercises( Exercise 33). and the constant of proportionality is the growth (decay) rate.3y 3 2 2 2 2 Applied Calculus for Business. f1x2 = x e + e -x e 2x + 1 11. a is a constant 14. defined by cosh1x2 = e x + e -x . f1x2 = ex + 1 x2 . 1.4y2 + 2x .x . 18. but make sure you choose an appropriate window. (a) Show it is 2 an even function. Using the calculator. are the extrema of the function. 27. Sketch the graph of the function defined by the equation f1x2 = x2e -2x. Also if dx dy (7) = ky then y = Cekx dx That is.1 ex . 22 = 0. f1x2 = x2e2x . x = 0 19. x = 3 21.Section 4. f1x2 = e2x 2. x2e2x y . if any. 26.2xe2x + 4e2x 15. Economics. Given the function defined by the equation c1x2 = ex + e . e *½ ) ) I2 (1.2x2 bution n1x2 = 11 2p e EXERCISE SET 4. ex . x = 1 24. f1x2 = 2 x + 1 12.2ex + 3e2y = 3x 3 . . f1x2 = ex. f1x2 = e -3x 3. which gives x = . it follows d kx that if k is a constant 1e 2 = kekx. a is a constant 5.x 10.ey = y2 . f1x2 = x2 e 2x 6. f1x2 = eax1x . f1x2 = x4e -3x 7. determine the equation of the tangent line at the given x-coordinate. ( * 1. (c) concave upward and downward. and April Allen Materowski. f1x2 = x3e3x.x 2 17. Walter O. Sketch the graph of the function defined by the equation f1x2 = x3e -2x.12 13. e *½ I1 ) Figure 4: The Normal1 Distri. and Finance. f1x2 = x2ex 8. (c) determine any relative extrema (d) Sketch the graph. Gordon. and enter solve flection points of the normal distribution.e . (d) determine any relative extrema (e) Sketch the graph. Wang.1 ex + 1 2 In Exercises 18 23. There are many important applications whose growth or decay at any instant is proportional to itself at that instant. the only function whose derivative is proportional to itself is an exponential function. ex + 2y = x2 + y 16. For example. From the chain rule. we may easily find the derivative and sketch the graph of exponential functions. The sketch of the graph is obtained from the Y = screen. (f) What. we let y11x2 = 11 e 2p 1d1y11x2. Published by Pearson Learning Solutions. f1x2 = e21x .12 and y1(1). Inc. no matter how complicated they may be. as we shall see in Section 2.4 The Derivative of the Exponential Function * ** 359 We close this section with a generalization of (3). f1x2 = 9. f1x2 = eax . 1. Given the function defined by the equation f1x2 = 4xe-3x. f1x2 = xe-2x. (e) What. Copyright © 2007 by Pearson Education.7. f1x2 = e-x. Calculator Tips M (0 . . (b) concave upward and downward. x2. by Warren B. x = 1 22. x = 0 23. 28. 1 at which x = c is . by Warren B. and April Allen Materowski. Suppose a rectangle whose base is on the x-axis is inscribed under this curve such that its upper vertices are the inflection points of the curve. (d) determine any relative extrema (e) Sketch the graph.360 * ** Section 4. (f) What. Gordon.) 2 30. Show that the x-intercept of the tangent line to the curve y = eax at the point ac .5 Logarithmic Functions and Derivatives 31. Prove that if f ¿ 1x2 = kf1x2 then f1x2 = Cekx. where a is a positive constant. Determine on which intervals the function is (b) increasing and decreasing. In particular. defined by sinh1x2 = ex + e -x . Find the area of the triangle formed by the x-axis and the tangent lines to the curves y = eax and y = e -ax at the point where x = 0. where a is a constant. Walter O. Given the function f1x2 = e-ax . show the x-intercept of the tangent a x line to the curve y = e is always one unit to the left of the point of tangency.5 Logarithmic Functions and Derivatives » » » » » » » Definition of a Logarithm Base 10 and e pH of a Solution Graphing Logarithmic Functions The Simple Logarithmic Rule The Generalized Logarithmic Rule Calculator Tips We begin with the exponential function defined by the equation y = f1x2 = 2x whose graph was plotted in an earlier section and is reproduced in Figure 1. y =f(x) = 2x Figure 1: The Graph of y = f 1x2 = 2x Applied Calculus for Business. Wang.e . Economics. . 2 29. 33. Published by Pearson Learning Solutions. Inc.x . Copyright © 2007 by Pearson Education. 4. (c) concave upward and downward. and Finance. Determine the area of the rectangle. (a) Show it is 2 an odd function. are the extrema of the function? (Note: this function arises frequently in engineering applications and is call the hyperbolic sine. 32. Given the function defined by the equation s1x2 = ex . if any. Before we discuss the equation of the inverse let us make some observations about the function and its inverse.q 6 y 6 q Graph passes through (1. The square root symbol was introduced so that we could solve this last equation for y in terms of x. However. Before we give the equation we make an observation about defining functions. this is only a representative symbol. Gordon. how do we determine the equation of its inverse function? We proceed exactly as we have learned before: we first interchange x with y and write x = f1y2 = 2y.5 Logarithmic Functions and Derivatives * ** 361 Since this is an increasing function it is one-to-one. Sometimes. 0) The graph for the inverse function may be obtained from the graph of the exponential function by interchanging the x and the y coordinates. Copyright © 2007 by Pearson Education. and. The statement x = 25 is equivalent to the statement x2 = 5 (and x 7 0). therefore. 1) y * f +11x2 Domain: x 7 0 Range: q 6 y 6 q Graph passes through (1. in fact we could just as well have written the symbol as SQRT (5).Section 4. The unanswered question at this point is the determination of the equation of the inverse function in the form y = f -11x2. Inc. You might seem surprised by this statement. . We can now go back to our question. and April Allen Materowski. Published by Pearson Learning Solutions. That is. the statement y = 1x is equivalent to the statement that y is the positive solution to the equation y 2 = x. Suppose we ask you to write the positive number whose square is 5. by Warren B. Similarly. y * f1x2 * 2x Domain: . We do this by defining a new function! The solution to the equation x = 2y is y = log 2 x. Without thinking. Given the function whose equation is y = f1x2 = bx where b Z 1 is a positive number we have: y * f1x2 * b x Domain: . 1) y * f +11x2 Domain: x 7 0 Range: . Economics. it is convenient to define new functions. Wang. This graph obtained by the interchange is given in Figure 2. Thus we have that the following two equivalent expressions.q 6 x 6 q Range: y 7 0 Graph passes through (0. the symbol y = 1x is equivalent to x = y2 (and y 7 0). The problem that now arises is how do we solve this equation for y in terms of x. Walter O. it has an inverse. Figure 2: The Inverse of y = f1x2 = 2x The analysis for an arbitrary exponential function and its inverse is almost identical. but it is something with which you are familiar. Given the equation y = f1x2 = 2x. and Finance.q 6 x 6 q Range: y 7 0 Graph passes through (0. you would write 25. 0) We may draw the graph of the inverse function by interchanging the points used to plot the original function. namely Applied Calculus for Business. Gordon. Thus.5 Logarithmic Functions and Derivatives x = 2y is equivalent to y = log 2 x The expression on the right is read y equals the logarithm of x to the base 2. the graph of the logarithm is the mirror image of the exponential about the line y = x. . Inc. More generally.362 * ** Section 4. Economics. then the inverse function is given by y = f -11x2 = log b x is equivalent to the expression x = by. then the equation of its inverse function is y = f -11x2 = log 2 x. we have. and Finance. we obtain the graphs given in Figure 4. Wang. y = f( x ) = b x y=x Definition of a Logarithm Figure 3: The graphs of y = f 1x2 = b x and y = f -11x2 = logb x where b 7 1 When 0 6 b 6 1. Thus any exponential expression may be written in logarithmic format and any logarithmic expression may be written in exponential format. Figure 3 illustrates the graphs of y = f1x2 = bx and y = f -11x2 = log b x where b 7 1. and April Allen Materowski. Published by Pearson Learning Solutions. Note that 2 is the base of the exponential of the expression on the left. Note that in both Figure 3 and 4. if y = f1x2 = 2x. y = f( x ) = b x y=x Figure 4: The graphs of y = f 1x2 = b x and y = f -11x2 = logb x where 0 6 b 6 1 Applied Calculus for Business. given the exponential function y = f1x2 = bx where b Z 1 is a positive number. We have already discovered the domain of the logarithm is x 7 0 and its range is .q 6 y 6 q . by Warren B. Walter O. Copyright © 2007 by Pearson Education. There are two special bases deserving special attention. rewriting this expression in exponential format. by Warren B. (c) log 5 25 = . Example 3 Compute log 3 27.5 Logarithmic Functions and Derivatives * ** 363 Example 1 Write each of the following exponential expressions in logarithmic format: (a) 32 = 9. Walter O.2 becomes 5-2 = (c) log 5 25 (d) log e w = r becomes er = w For special types of problems switching from the logarithmic to the exponential form of the expression is useful. namely base 10 and base e. when base 10 is used. Four basic properties of logarithms log b 1 = 0 log b b = 1 log b b = x blogb x = x x Base 10 and e (1) (2) (3) (4) Applied Calculus for Business. (d) log e w = r. thus (a) 32 = 9 becomes log 3 9 = 2 (b) 2-3 = 1/8 becomes log 2 1 8 = -3 3 (c) 10 = 1000 becomes log 10 1000 = 3 1 (d) A 1/ 2 B 3 = 1/8 becomes log 1 = 3 2 8 Example 2 Write each of the following expressions in exponential format: 1 1 (a) log 4 16 = 2. (d) 11/ 223 = 1/8. and Finance. Logarithms to the base 10 are called common logarithms and those to base e are called natural logarithms. Economics. Solution To answer each of these questions we use the fact that x = by is equivalent to y = log b x. Solution Let log 3 27 = x. (c) 103 = 1000. Often. (b) 2-3 = 1/8. as we illustrate in the next example. we write ln instead of log e. . we write log instead of log 10 and when base e is used. (b) log 1 = 2. Inc. and April Allen Materowski.Section 4. Copyright © 2007 by Pearson Education.2. Wang. Thus log x means log 10 x and ln x means log e x. 39 Solution (a) log 4 16 = 2 becomes 42 = 16 1 2 = 2 becomes A 1 (b) log 1 3B = 39 1 9 1 25 1 = . This is a special exponential equation which is easily solved by writing 3x = 33. or x = 3. It should be noted that if the base of the logarithm was different from 3 then the resulting exponential function could not be easily solved. we have 3x = 27. Gordon. We shall examine such cases shortly. Published by Pearson Learning Solutions. Since the concentration can change over several orders of magnitude. Copyright © 2007 by Pearson Education.2 * 10-11. we have pH = . therefore x = 1.12. if the pH is about 7. it is convenient to use a logarithmic scale. so by using them we can quickly compute logarithms to these bases. by Warren B. L. . For example.3617278. we have. using a calculator. Using a calculator we obtain (to the nearest tenth) pH = 10. the solution is neutral. The TI-89 has the ln key above the X in orange and the log key is in the catalog. writing this in exponential format. It is easy to show. Graphing Logarithmic Functions Calculators are useful in plotting the graphs of logarithmic functions. Let log b 1 = x. Example 4 Determine the pH of a solution whose hydronium ion concentration is 4. Wang.log[H 3O +] + (3) where [H 3O ] represents the concentration of the ion. let log b b = x. namely that f1f -11x22 = f -11f1x22 = x We note that two special cases of (2) are log 10 = 1 and ln e = 1 (2b) (2a) pH of a Solution Calculators provide log and ln keys. expressed in logarithmic notation. P. it is a simple matter to compute logarithms or to plot the graphs of logarithms to base 10 or e. This solution is basic since the pH exceeds 7. Thus. Published by Pearson Learning Solutions.5 Logarithmic Functions and Derivatives The first property follows from our discussion above. we have bx = 1. and is therefore a translation of log x.2 * 10-11. Economics. If the pH of a solution is less than 7. Sorenson when in 1909 he defined the pH of a solution. Solution Substituting into (3). Example 5 Sketch the graph of y = ln12x . Walter O. if the pH is greater than 7. as pH = . (Other bases will be examined in the next section. that any logarithm of the form log1ax + b2 may be written as log1x + b/a2 + log1a2. thus we have bx = b0 or x = 0. ln 14 = 2. This idea was used by the Swedish chemist S. We prove the second property the same way.6390573 and log 23 = 1. and April Allen Materowski. using the calculator. Gordon.4. The acidity of a solution was originally viewed as the concentration of the hydronium ion H 3O + present in it. indicated by [H 3O +]. but b0 = 1.) Logarithms arise in a variety of applications. and Finance.log 4. as illustrated in the next two examples. Properties (3) and (4) follow immediately from the properties of inverse functions. For reinforcement. The pH is usually given to the nearest tenth. using the properties of logarithms in the next section. the solution is is called basic. Applied Calculus for Business. we illustrate it again. the solution is called acidic. Inc. Consider the following example from chemistry. rewrite in exponential format as bx = b = b1. to seven decimal places.364 * ** Section 4. 40 We plot these points to obtain the graph given in Figure 5. and Finance. we cannot use a calculator directly to obtain the y-values since there is no key for base 2.6 0 1. Economics.95 2. Walter O. 2x .12 It is a straight-forward matter to plot the graph of most logarithmic functions whose base is e or ten. for the given example. .61 1. we shall determine the equation of the inverse function. This approach is illustrated in the next example. Table 1: Points used to plot y = ln12x . However. y = ln (2 x .5 Logarithmic Functions and Derivatives * ** 365 Solution The domain of the function is found by remembering that if y = log b u. and April Allen Materowski. Thus. Applied Calculus for Business. thus the domain is x 7 1/ 2. Example 6 Sketch the graph of the function whose equation is y = f1x2 = 2 log 2 x. we can also plot such logarithmic functions by recalling that any logarithmic function is the inverse of some exponential function.12 x . where all logarithms have been rounded to two decimal places.6 1 2 3 4 6 y * ln12x + 12 . We now can choose appropriate values with which to plot a graph. Instead. Inc. We shall see in the next section how we can graph logarithmic functions to other bases with a calculator. Wang.Section 4. Published by Pearson Learning Solutions. We use a calculator to generate Table 1. Copyright © 2007 by Pearson Education.1 1. by Warren B. Solution As we observed above. then the domain is determined from the inequality u 7 0. Gordon.1) Figure 5: The graph y = ln12x .1 7 0.1. We interchange x with y to obtain. We need only use the appropriate key on a calculator to obtain the y-value corresponding to any x-value in the domain. Note that is was not necesx sary to plot the graph of y = 22 .25 0. using the points from Table 2 in Figure 6. Gordon. y = 2 log 2 x. These points are indicated in Table 2. Economics. and Finance. Table 1: x Points to plot the graph of y = 2 2 x -4 -2 0 2 4 6 4 y = 22 x = 2-2 = 0. Wang. we only needed to list points on its graph. and April Allen Materowski.5 1 2 4 8 y = 2 log 2 x -4 -2 0 2 4 6 We plot the graph. All we need to do is to interchange the two columns in Table 1 and we have points on the graph of this function. we generated Table 2. by Warren B. Inc. gives as the equation of the inverse function y = 22 x We could plot this graph easily by generating a table of values as given in Table 1. Walter O. Table 2: Points to plot the graph of y = 2 log 2 x x 0. .366 * ** Section 4.5 Logarithmic Functions and Derivatives x = 2 log 2 y or x = log 2 y 2 rewriting this expression in exponential format. Copyright © 2007 by Pearson Education. Published by Pearson Learning Solutions. Applied Calculus for Business.25 2-2 -2 2 2 = 2-1 = 0. and by interchanging the x and y-values of these points.50 1 2 = 21 = 2 22 4 = 22 = 4 22 6 22 = 2 3 = 8 We wish to plot the graph of the inverse function. Section 4. Inc. The Simple Logarithmic Rule THEOREM 1: THE SIMPLE LOGARITHMIC RULE 1 d 1ln x2 = x dx Proof Let y = ln x then we have x = ey We differentiate this equation implicitly to obtain 1 = ey or dy 1 = y dx e but x = ey. Gordon. by Warren B. log b 1 = 0.5 Logarithmic Functions and Derivatives * ** 367 y = 2log 2 x Figure 6: The Graph of y = 2 log2 x No matter what the base of the logarithm functions are. let (b. Economics. Published by Pearson Learning Solutions. as given in Figure 4. The continuity and differentiability of the logarithm function follow from the exponential function. and April Allen Materowski. Note that log b x 7 0 when x 7 1 and log b x 6 0 when x 6 1. . we have dy dx (1) Applied Calculus for Business. b) where b = ea. We next consider the derivative of the natural logarithm and after we examine the properties of logarithms in the next section we will consider the derivative of any logarithmic (and exponential) function. their basic shapes are all the same. Walter O. so we have dy 1 = x dx Alternately. a) be any point on the graph of y = ln x 1b = ln a2 then the corresponding point on y = ex is (a. Wang. and Finance. From Section 1. Copyright © 2007 by Pearson Education. using the first and second derivatives.x dx -x x dx Thus.x and from the generalized rule d 1 d 1 1 1ln1 . in either case. therefore. using the generalized rule (with u = x2 + 1). that f1x2 = ln x is (a) an increasing and (b) a concave downward function. and April Allen Materowski. Applied Calculus for Business.x22 d 2x 2 . where u is a differentiable function of x. However. Copyright © 2007 by Pearson Education. . x = .x22 = 1 . determine f ¿ 1x2. we x dx have the first derivative is always positive and. Gordon. x = x when x 7 0.2x 2 a 2 = = (b) f 1x2 = b = dx x + 1 1x 2 + 122 1x2 + 122 1x2 + 122 Example 8 let f1x2 = ln x . and since the domain of this function is x 7 0. Solution The domain of this function is x Z 0. Wang.2x[2x] 211 . Inc. so in this interval. f ¿ 1x2 = 1/x. Published by Pearson Learning Solutions. determine (a) f ¿ 1x2. when x 6 0.x2 = [ . Walter O. we obtain the generalized logarithmic rule. d 2 1 1 2x f ¿ 1x2 = 2 1x + 12 = 2 [2x] = 2 dx x + 1 x + 1 x + 1 1x 2 + 12[2] . and Finance. the function is increasing. (b) f 1x2 Solution (a) We have. THEOREM 2: THE GENERALIZED LOGARITHMIC RULE d 1 du 1ln u2 = # u dx dx Example 7 Let f1x2 = ln1x2 + 12. we have d 1 1ln x 2 = x dx Example 9 Verify. by Warren B. Economics. Solution (a) We have f ¿ 1x2 = d 1 1ln x2 = .368 * ** Section 4.1] = .5 Logarithmic Functions and Derivatives 1f-11x22 ¿ (b) = or in this case 1f-11x22 ¿ (b) = 1 d x 1e 2 dx 1 f ¿ 1 a2 = x=a 1 1 = ea b The Generalized Logarithmic Rule or replacing b with x we have (1). Using the Chain rule. 2 6 0 for all x.1 and x = 1. Figure 7a: f 1x2 = ln x CD sign of f''( x ) 0 CU + 0 CD x -1 1 Figure 7b: The Sign of f 1x2 (-1. Analyzing the sign of the second derivative in Figure 7b. and ln 1 = 0. Gordon.x2 = f1x2. giving Figure 7a. 0). so the critical point is (0. the function is concave downdx x x Observe that we now are able to give a quick sketch of the graph of f1x2 = ln x knowing that it passes through (1. By the second derivative test. The first and second derivatives were found in Example 7. ln 2) Figure 8: The Graph of f 1x2 = ln1x 2 + 12 Example 11 Determine the equation of the tangent line to f1x2 = e -2x + 1 ln14x2 + 12 when x = 1/ 2. we see that 1 . Applied Calculus for Business. The critical point occurs when x = 0. therefore. Inc. Wang. The second derivative is zero when x = .1. 0). Solution First. Published by Pearson Learning Solutions. Walter O.Section 4. and April Allen Materowski. Example 10 Sketch the graph of f1x2 = ln1x 2 + 12. we observe that the function is even as f1 . Figure 7b also indicates the concavity. Moreover. ln 2) (1. Economics. by Warren B. as x : 0+. The graph is given in Figure 8. Copyright © 2007 by Pearson Education. . is always increasing and is concave upward. 0). d 1 1 a b = . f1x2 : .5 Logarithmic Functions and Derivatives * ** 369 (b) f 1x2 = ward. and Finance.q (this follows from ex : 0 as x : q ). ln 2) are each inflection points. f 102 = 2 7 0 so the function has a minimum at (0. ln 22 and (1. you can define this function on your calculator using the STO key and place it in your calculator s memory.5 Logarithmic Functions and Derivatives Solution We have. However.1/ 2 B . see Figure 9.1/ 2 B or y = 0. using the catalog.693 = 0. We shall see that these follow from the natural base e very easily using some basic properties that we will examine in the next section.7) and press Enter.1 dx dx du d u 1e 2 = eu dx dx d 1 du 1ln u2 = u dx dx Chain Rule Calculator Tips The only other functions we will discuss in this text will be exponential and logarithmic functions to bases other than e. f ¿ 1x2 = e -2x + 1 d d 1ln14x 2 + 122 + ln14x2 + 12 1e -2x + 12 = dx dx 1 [8x] + ln14x2 + 12e -2x + 1[ . For example in determining the pH.2 ln 2 = 2 . we have y . Walter O.0.0.2 ln 2 L 0. For example. This means if you want to evaluate. Published by Pearson Learning Solutions. f A 1/ 2 B = e0 ln122 = ln 2 L 0. that is base 10. you enter lg(28. . algebraic. There are. That is why LN is prominently displayed in orange on the calculator key above X. we have f ¿ A 1/ 2 B = e0 Therefore. and Finance. Economics. logarithmic and exponential. The rules are as follows: Simple Rule d N 1x 2 = Nx N .1 dx d x 1e 2 = ex dx d 1 1ln x2 = x dx Generalized Rule d N du 1u 2 = NuN . where m = f ¿ A 1/ 2 B . Applied Calculus for Business. applications where it is convenient to use log. Wang.693. Log can be found in the Catalog of the calculator. say log 28. enter log(x) then press STO and name it lg(x). The most useful base for us is base e. Copyright © 2007 by Pearson Education.2] e -2x + 1 2 4x + 1 1 4A B + 1 1 2 2 When x = 1/ 2. Gordon.693 = m A x . however.614x + 1 1 A8A1 2 B B = 2 4 .370 * ** Section 4. by Warren B.614 A x . Inc. thus y .7. and April Allen Materowski.614 At this point you should realize that we can now find the derivatives of three kinds of functions. 4/3 11. 4. log 100 = 2 14. we suspect that the data is logarithmic. 8-1/3 = 1/ 2 5. t2 and press ENTER Choose WINDOW 1*F22. 60 = 1 6. Regeq(x) STO y1(x) and press Enter (This stores the regression equation as y1(x)) then enter on the entry line NewPlot 1. (2. We proceed as follows: (1) Store the x-coordinates and y-coordinates as lists named t1 and t2.65. Walter O.13. Suppose we are given the following data set: (1.15. and April Allen Materowski. 3-2 = 1/9 3. power and exponential functions. log 3 27 = 3 15. then press Zoom (F2). Copyright © 2007 by Pearson Education. 34 = 81 2.29). 0. Inc.2 13.71) We fit this data to a logarithmic regression curve done exactly the same way on the TI 89 as you would for linear (or quadratic) with the only change being we replace LinReg with LnReg. 19/1623/2 = 27/64 2 In Exercises 9 16 write the given expression in exponential format. 1.65). . Published by Pearson Learning Solutions.16). or if for some other reason. log b 1 = 0 12. (The regression curve is now drawn. and scroll down and select ZOOMDATA (or press 9). 1. 1. 1.) Figure 9: Defining Log on the TI 89 Figure 10: Obtaining the Logarithmic Regression Equation Figure 11: The Regression Curve EXERCISE SET 4.716 STO t2 press ENTER Make sure you are using curly braces to enclose these two lists (2) Next enter on the entry line of the HOME screen (use the Catalog or type) LnReg t1. 163/4 = 8 7.1.13). by Warren B. as follows: 51.5 In Exercises 1 8 write the given expression in logarithmic format 1. Gordon. 0. and Finance. 3. t2 and press ENTER then ShowStat and press ENTER (This produces Figure 10) (3) To draw the regression curve. 11/32 = 1/9 8. 1. t1.29. 56 STO t1 press ENTER 50. log 2 8 = 3 10. 0. log 1/2 1/8 = 3 16. log 2/3 16/81 = 4 Applied Calculus for Business. log 8 1/16 = . 2. log 1/100 = . enter on the entry line in the HOME screen. See Figure 11. 1. then we may determine a best fit regression equation just as we did for linear. 9. (3. (5. (4. Wang.5 Logarithmic Functions and Derivatives * ** 371 If a data set appears to resemble a logarithmic curve. Economics. 24 = 16 4.Section 4. f1x2 = ln x 5 50.12 37. y = f1x2 = 3 log12x + 32 40. (a) 4. log 1/3 27 21. y = f1x2 = 2 log14x . 1. Inc. 58. log 4 x = . f1x2 = e -x ln x 64.x2 42. log 1/4 x = 3/2 In Exercises 31 35 determine the pH of a solution with the given hydronium concentration. (a) 6. log 4 16 18. (b) 2. by Warren B. .032.52 39.4 * 10-5 In Exercises 36 44 sketch the graph of the function whose equation is given.912. log 10000 20. determine f ¿ 1x2 45. 1. (6. (a) 3. f1x2 = ln x e 2x + x 2 + 1 In exercises 17 24 determine the value of the logarithm without the use of a calculator. log x 1/16 = . f1x2 = x . Repeat the previous exercise for the data set: 12.2 * 10-7 (b) 2. Determine all relative extrema. Gordon.4 * 10 -11 (b) 1. f1x2 = ln 3x 46. log x 8 = 3 26. f1x2 = ln x -3 51. 60. Classify the solution as an acid base or almost neutral.482. y = f1x2 = ln12x + 12 38. f1x2 = 1ln x25 55.292. y = f1x2 = 4 log 21x . (a) Find the logarithmic regression function that best fits the following data set: (2.4 * 10-4 (b) 7.2. 14. f1x2 = ln x 3 49. y = f1x2 = 3 log 2 x Applied Calculus for Business. f1x2 = ln1x 5 + 122 54.41). 16. 13. . log 25 125 In Exercises 25 30 find x 25. log 25 x = 1/ 2 30. .522. f1x2 = e-2x1ln x + 42 x = 1 In Exercises 60 63 sketch the graph of the given function.82). .1. f1x2 = ln x 63. Economics. log 2 32 23. inflection points and asymptotes. Copyright © 2007 by Pearson Education. Walter O.65).2. . f1x2 = 2x ln x x = 1 59. 1.5 Properties of Logarithmic Functions 43. (5. using the first and second derivatives. f1x2 = ln ax.4/3 29.1 * 10-12 32. log 3/4 27/64 24. 1. (a) 5. (b) Using the regression equation. f1x2 = x ln x 61. what are the y-coordinates when x = 1 and x = 8? 65. y = f1x2 = 2 log 3 x 44. and April Allen Materowski.372 * ** Section 4. a 7 0 48.0. (4. .2 ln x 62.2 27.8 * 10-8 34. Published by Pearson Learning Solutions. (3.2x2 41. 15.3 * 10 -6 33. y = f1x2 = ln12x . log 8 4 19. log 64 1/16 22. 0.2 * 10 -5 57. f1x2 = ln xn 52. 31.07). f1x2 = ln 4x 47. f1x2 = ln14 + 3x22 53. 36. f1x2 = e-2x ln x 56. y = f1x2 = 3 ln13 . log 3 x = 4 28. 17.3 * 10-9 (b) 7.2.12 In Exercises 45 57.71). f1x2 = ln1ln x2 In Exercises 58 59 determine the equation of the tangent line at the indicated x-coordinate. and Finance. y = f1x2 = ln1 . (a) 8 * 10-8 35. Wang. when logarithms were first defined.000000003 cate that this number is outside of its range. Moreover. First. However. Inc. We can use this observation to show that if two functions have the same derivative. Walter O. Economics. most calculators will indi2. Verification for any other base follows from the change of base theorem that we give later on this section and will be left as an exercise. Wang. one of their most important uses was to calculate products. Let A and B be positive numbers. with the application of some simple properties. then log b AB = log b A + log b B log b A = log b A . base e. many of the calculations using logarithms are now no longer needed. quotients and powers of large numbers. . here. They were able to be used for this purpose because of their basic properties which we shall discuss.6 Properties of Logarithmic Functions * ** 373 4. these properties are invaluable in the study of the calculus of the logarithm. this logarithm may be easily computed.g1x2 = c We are now in a position to prove the various properties of logarithms. but 1620313221000 not all of them. we take a different approach. we observed that if the derivative is zero (on some interval) then the function is a constant (on that interval).5.6 Properties of Logarithmic Functions » » » » » » » » » Multiplicative and Division Properties Exponential Property Derivatives Using the Properties Logarithmic Equations Exponential Equations Change of Base Derivatives in Different Bases Logarithmic Differentiation Calculator Tips Historically. Copyright © 2007 by Pearson Education. by Warren B. Published by Pearson Learning Solutions. Gordon. For example try to calculate ln . We prove the properties for natural logarithms. and Finance.log b B B (1) (2) Multiplicative and Division Properties In college algebra. Suppose in particular that f ¿ 1x2 = g ¿ 1x2 Let D1x2 = f1x2 . With the appearance of the calculator. Applied Calculus for Business. they differ at most by a constant. it follows that D1x2 = c or f1x2 . and April Allen Materowski.g1x2 then D ¿ 1x2 = f ¿ 1x2 . proof of these properties is usually given by rewriting the logarithmic expression in exponential format (see Exercises 103 104). as we shall see.g ¿ 1x2 = 0. that is. therefore. we recall that when we studied the first derivative test.Section 4. (c) ln . choose x = 1. using (3). Economics.ln a a (3) As a simple illustration. in particular. then we have 1 1 0 = ln 1 = ln a a # b = ln a + ln a a or ln Now. by Warren B. we have log1211 # 3292 = log 211 + log 329. so we have c = ln a or ln ax . Similarly. Gordon.ln x = ln a or ln ax = ln a + ln x choose x = b and we have ln ab = ln a + ln b as required.374 * ** Section 4. .ln 219. Wang. 357 = ln 357 . we obtain the reciprocal property of logarithms. consider ln ax. we can prove (2). Copyright © 2007 by Pearson Education. Before we prove (2). we have ln1271 # 4372 = ln 271 + ln 437. we see that d d 1ln ax2 = 1ln x2 dx dx so we have ln ax . This result is true for x 7 0. Suppose we let b = 1/a.6 Properties of Logarithmic Functions To prove (1). Walter O.ln 1 = c but ln 1 = 0. ln 219 Example 1 Write the given expression as a sum and difference of logarithms 221 # 352 29 # 37 (a) log 1211 # 3292.ln b b b b 1 = . Inc.ln x = c We next compute c. Applied Calculus for Business. From the generalized rule. (d) log a # b 17 57 69 Solution (a) using (1). Published by Pearson Learning Solutions. we have that d 1 d 1 1 1ln ax2 = 1ax2 = a = ax dx ax x dx therefore. ln a 1 1 = ln a a # b = ln a + ln = ln a . and April Allen Materowski. and Finance. (b) ln (11/17). so we have ln a . Inc. we now use (1) to obtain 17 ln (d) 221 # 352 = ln1221 # 3522 . if r is an integer. verify that we have d d 1ln xr2 = 1r ln x2 dx dx therefore.log157 # 692 = log a log 29 + log 37 . there are no simplification rules for product and quotient expressions like 1log b A21log b B2 or 3. ln xr . Economics.r ln x = c this is true for all x 7 0. C 7 0 log b A . Wang.1log 57 + log 692 = distribution of the . Walter O. then we have by (1) log b A3 = log b A # A # A = log b A + log b A + log b A = 3 log b A The general case is proven as follows. then we have log b Ar = r log b A (4) Exponential Property This property is the natural generalization of the multiplicative property.ln 17 17 29 # 37 property 122 b = 57 # 69 property 112 log129 # 372 . Using the generalized logarithmic rule. which gives c = 0 therefore ln xr . .log 57 .ln 17. log b B then log b1ABC2 = log b1A2 + log b1B2 + log b1C2 Another useful property of logarithms is in the simplification of exponents. we have that ln = ln1221 # 3522 . Gordon. The rules generalize.B2.ln 17 = ln 221 + ln 352 .Section 4. 221 # 352 (c) Using (2). We have not given any rules to simplify expressions like log b1A + B2 or log b1A .log 69 Notes: 1.r ln x = 0 or ln xr = r ln x Applied Calculus for Business. and April Allen Materowski. Copyright © 2007 by Pearson Education. so choose x = 1.sign log 29 + log 37 . giving ln 1 . for example if A.6 Properties of Logarithmic Functions * ** 375 (b) using (2) we have ln111/172 = ln 11 . For example. Similarly. Published by Pearson Learning Solutions. by Warren B. There are no rules for simplifying the logarithm of a sum or difference! We only have rules for simplifying logarithms of products and quotients. B. 2.ln 17. Suppose r is any real number and A 7 0. say r = 3. and Finance.r ln 1 = c. 3 log w . Walter O. and Finance.A ln w 10 + ln z2 B = 1 4 ln x + 1 3 ln y . and April Allen Materowski. Applied Calculus for Business. Solution We could differentiate directly. .2 log z. (c) Using (1).ln w 10z2 = ln x4 + ln y 3 .6 Properties of Logarithmic Functions Example 2 Simplify (a) log x3. (c) log 2 x2y 3.10 ln w . but note that f1x2 = 3 ln x. x. Example 4 Find f ¿ 1x2 if f1x2 = ln x3. Solution 1 3 log x + 1 2 log y . Inc.3 log w .2 log z = log x3 + log y 2 . by Warren B. (b) ln 1x.A log w 3 + log z2 B = log x3y 2 . we have log 2 x2y 3 = log 2 x2 + log 2 y 3 = 1 2 log 2 x + 3 log 2 y 1 1 1 1 x4 1 3y w10 1z (Assume w.log w 3z2 = log + x3y 1 1 2 1 1 1 1 Using property (4) Using property (2) Using property (1) w 3 z2 * = b log a x3 1y 1 3 wz2 Rewriting exponents as radicals Derivatives Using the Properties The properties of logarithms are often useful when used in conjugation with the derivative rules.376 * ** Section 4. Gordon. Copyright © 2007 by Pearson Education. Published by Pearson Learning Solutions. Economics. Note that using the properties of logarithms change powers into constant multipliers. we have that log x3 = 3 log x. we have that f ¿ 1x2 = 311/x2 = 3/x. (b) we rewrite ln 1x = ln x2 = 1 2 ln x. y. and by the constant multiplier rule. (d) ln Solution (a) using (3).2 ln z 1 Using property (4) Distributing the minus sign Example 3 1 Rewrite the following as a single logarithm: 3 log x + 1 2 log y . as illustrated in the following examples. products and quotients into sums and differences. Wang. Using the exponential property (d) ln x4 1 3y w10 1z 1 = ln x4y 3 w 10z2 1 1 1 = Rewriting radicals as exponents Using property (2) 1 ln x4y 3 . z 7 0). Inc. Checking the second root. Published by Pearson Learning Solutions. therefore. of course.6 Properties of Logarithmic Functions * ** 377 Example 5 Find f ¿ 1x2 if f1x2 = ln A x 5 22x + 1 B . (b) We rewrite log 312x + 92 . We must check our solutions. since the logarithm is only defined for positive numbers. Copyright © 2007 by Pearson Education. and April Allen Materowski. Applied Calculus for Business. Economics. by Warren B.102 = 3 (b) log 312x + 92 .Section 4. and Finance.182 or.182 Logarithmic Equations we rewrite this expression in exponential format as 12x + 92 = 31 = 3 13x . Wang. checks since log 2 2 = 1.log 313x .182 = 1 using (2) to obtain log 3 12x + 92 = 1 13x .8 The solutions to this quadratic equation are x1 = . . 1 f1x2 = ln A x5 22x + 1 B = ln x5 + ln12x + 122 = 5 ln x + 2 ln12x + 12 1 Therefore. log 2 4 + log 2 2 = 3 or log 2 22 + log 2 2 = 3 2 log 2 2 + log 2 2 = 3 3 log 2 2 = 3 This. we reject this answer. We now write this expression in exponential format as 23 = x13x . Solution We have using the properties. clearing fractions. Substituting .10x or 0 = 3x2 . Consider the following example.2/3 for x we have in the first term the logarithm of a negative number. Gordon. (a) log 2 x + log 213x .102 or 8 = 3x2 . this is nonsense.log 313x . f ¿ 1x2 = 5 # 1 + x 1 2 1 5 1 11x + 5 [2] = + = x 2x + 1 2x + 1 x12x + 12 The above properties are useful in solving various kinds of logarithmic equations. and x2 = 4 (verify this!).10x .102 = 3.2/3.182 = 1 Solutions (a) we use property (1) to rewrite log 2 x + log 213x . Example 6 Solve each of the following equations for x. Walter O.102 = 3 as log 2 x13x . Economics. Solution Since 4x = 15. Copyright © 2007 by Pearson Education.95345 ln 4 Note that 42 = 16. and Finance. We leave the checking as an exercise for you. Gordon. so we expected our result to be smaller than 2. Published by Pearson Learning Solutions. then log b A = log b B.2x ln 9 we isolate the x terms by transposing to obtain 3x ln 5 + 2x ln 9 = 4 ln 9 . Example 7 Solve the equation 4x = 15. by Warren B. we have that ln 4x = ln 15 Using (3) we have x ln 4 = ln 15 or x = ln 15 L 1. we have 12 + 3x2 ln 5 = 14 .2x Solution We take the logarithm of each side of the equation to obtain ln 512 + 3x2 = ln 914 . Exponential Equations You will recall that we solved special exponential equations where we were able to reduce each exponent to the same base. We can now illustrate how to solve exponential equations of the more general type. Inc. and April Allen Materowski. or x = 9.2 ln 5 Applied Calculus for Business. then it follows from the logarithm being a one-to-one function that A = B.6 Properties of Logarithmic Functions 2x + 9 = 9x . We now illustrate how this is used to solve exponential equations. conversely if log b A = log b B.378 * ** Section 4. Example 8 Solve for x: 52 + 3x = 94 .2x2 (Note the use of parentheses to avoid errors. Walter O. Wang.2x2 ln 9 or 2 ln 5 + 3x ln 5 = 4 ln 9 . We first observe that if A and B are positive numbers then A * B is equivalent to log b A * log b B (4) Clearly if A = B.) Using the exponential property. .54 7x = 63. 5.01. even if the decimal portion is less than 0. or t = ln 1.012 = 4.603943 1Verify this!2 x = ln 10125 ln In the next example we use the properties of logarithms to determine the time it takes an investment to reach a specified accumulation. we have.12t = 0. .6 Properties of Logarithmic Functions * ** 379 or x13 ln 5 + 2 ln 92 = 4 ln 9 . Economics. assuming 30 days per month. compounded (a) monthly.12 = 4.Section 4.65 = ln e0. (b) Using A = Pert we have 1650 = 1000e0. or 1. This gives 2 months and the remaining fraction. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education. To the nearest day. or 1.12t ln e = 0.327501194 months = 0.65 = ln11. Walter O.65/0. using the compound interest formula.12t 1remember.01212t = 12t ln 1.01212t. Inc.193958432 # 12 months = 2.12/12212t.12t Taking logarithms.12t. Alternately.327501194 months. (a) We have. t = ln 1.650 = 11.2 ln 5 or x = 4 ln 9 . Are you surprised that the difference is only seven days? Applied Calculus for Business.82503552 days. A = P11 + r/n2nt. Example 9 How long does it take $1000 to accumulate to $1650 if money is earning interest at a rate of 12% per year. Gordon. Banks always round up to the nearest day. Thus. by Warren B. 1650 = 100011 + 0. ln e = 12 or. and Finance. Wang.2 ln 5 3 ln 5 + 2 ln 9 We may evaluate this expression with a calculator to find that x L 0. we find that t = 4 years 2 months and 3 days.603943. ln 1. 0.173127399 years. t = 4 years 2 months and 10 days. the expression for x my be rewritten using the properties of logarithms as 6561 25 L 0. (b) continuously? Solution Let t be the time in years. and April Allen Materowski. We solve this exponential equation by taking the logarithm of each side of the equation Thus.193958432 years. 0. We may compute t to the nearest day as follows: The fraction.193958432 years is equal to 0.327501194 # 30 days = 9.65/112 ln 1.65 = e0. ln 1. that is log a bx = log a N We now use the exponential property and have x log a b = log a N or x = log a N log a b proving the theorem.0824 = 8. We have the following theorem: log b N * log a N log a b (5a) where a and b and positive numbers different from 1 and N 7 0. Using base e. there is a simple way to change the base of any logarithm to any other convenient base. At the end of that time. ln N ln b (5b) Applied Calculus for Business.1 9 ln 2. Copyright © 2007 by Pearson Education. When applying (5). The method used was indirect since calculators do not provide a base two key. we usually choose a to be e as this base is most accessible on most calculators. L 0. Published by Pearson Learning Solutions. the account contains $2100. we have 2x = 7. and April Allen Materowski. we have.24% Change of Base In the previous section.1 = 9r. we let log b N = x We write this in exponential format as bx = N We take the logarithm of this expression to any other base a. What rate of interest did it earn? Solution Since interest is compounded continuously. by Warren B. However.6 Properties of Logarithmic Functions Example 10 $1000 is deposited into an account for 9 years in a bank that pays interest compounded continuously. Wang. Let log 2 7 = x. Inc. r = ln 2. Walter O. we rewrite (5a) as log b N = Example 11 Use your calculator to obtain log 2 7. using base e. . Solution We will re-derive the result in the context of this example. Economics. Gordon. Taking logarithms of each side of the equation (to base e). and Finance. we sketched the graph of a logarithmic function to base two.1 = e9r Taking the natural logarithm. then in exponential form. To prove (5). we have 2100 = 1000e9r 2.380 * ** Section 4. We need only choose a reasonable number of x-values to ln 2 find the y-values. y = f1x2 = . in Example 5 of the previous section. ln x .5 1 2 3 4 y = 2 ln x ln 2 -4 -2 0 2 3. it is a simple matter to now plot the graph of functions to bases other than e or ten as we illustrate in the next example. and April Allen Materowski.Section 4. These points are plotted and the graph is given in Figure 1. and using (5) or (5b). Note that this is the identical function plotted. Economics.6 Properties of Logarithmic Functions * ** 381 ln 2x = ln 7 or x ln 2 = ln 7 or x = ln 7 L 2. Published by Pearson Learning Solutions. Example 12 Sketch the graph of y = f1x2 = 2 log 2 x. with a = e. ln 2 2 ln x therefore.80735 ln 2 With the above in mind.17 4 y = f( x ) = 2 log 2 x Figure 1: The Graph of y = f 1x2 = 2 log2 x Applied Calculus for Business.25 0. Gordon. indirectly. The y-values are rounded to the nearest hundredth. The values are given in Table 1. and Finance. Wang. Inc. by Warren B. and then plot these points to obtain the graph. Walter O. we have log 2 x = Table 1: Points used in Graphing y = f 1x2 = 2 log 2 x x 0. Solution We change the base. . Copyright © 2007 by Pearson Education. With the chain rule. using dx (5b) we may write 1 d d ln x 1 d 1 1 1log b x2 = a b = 1ln x2 = = dx ln b ln b dx ln b x x ln b dx (Remember. ln b is a constant so we used the constant multiplier rule. and April Allen Materowski. Example 13 Determine (a) Solution d 1 1log 7 x2 = x ln 7 dx d 2 d 1 1 2x 1log 31x 2 + 922 = 2 1x + 92 = 2 [2x] = 2 (b) dx 1x + 92 ln 3 dx 1x + 92 ln 3 1x + 92 ln 3 (a) We note that if you forget (6) all is not lost. we now have the following: (7) d d 1log 7 x2 (b) 1log 31x 2 + 922. dx dx ln 7 ln 7 dx ln 7 x x ln 7 We can now consider the derivative of the exponential function to any base. we have k = ln b or bx = 1ek2x = exk = ex ln b Therefore d x d d x ln b 1b 2 = 1e 2 = ex ln b 1x ln b2 = bx ln b dx dx dx Note that we used (7) in rewriting the expression ex ln b as bx. .382 * ** Section 4. and Finance. giving a simpler result. there exists a constant k such that b = ek To determine k. d 1 du 1logb u2 = u ln b dx dx (6b) (6a) It is clear why we prefer to work in base e. 1ln 7 x2 = a b = 1ln x2 = = . for any differentiable function u. this factor is 1.) Thus. In any other base. (6) has the extra constant factor ln b. Consider 1log b x2. Economics. In base e. for example. it is now a straight-forward matter to determine the ded rivative of a logarithmic function with respect to any base. Gordon. We remarked in an earlier section that any base b may be converted to base e. Inc. Copyright © 2007 by Pearson Education. Walter O. Published by Pearson Learning Solutions.6 Properties of Logarithmic Functions Derivatives in Different Bases With the change of base theorem. Wang. we have d 1 1log b x2 = x ln b dx and by the chain rule. you need only rewrite the problem by changd 1 d ln x 1 d 1 1 ing the base. ln b = ln ek = k ln e = k thus. dx dx Applied Calculus for Business. we need only take the logarithm of each side of this equation. by Warren B. That is. This method is called logarithmic differentiation. using (8).ln13x 2 + 423 = 2 ln12x + 12 . we have 18x2 . It is also useful when no other method seems applicable. Gordon. differentiate implicitly to obtain 3. Equations (5) and (7) essentially state that we really don t need to ever use any base other than e.2x + 162 12x + 122 18x2 . Given the function defined by the equation y = f1x2. this method is valid when f1x2 6 0. Walter O. and if A and B are positive then it follows that ln A = ln B. If one is given to us. d x3 d 3 3 3 A 2 B = 2x ln 2 1x32 = 2x ln 2[3x2] = x2 A 2x 3 ln 2 B . dx Solution We have. sometimes it is convenient to find the derivative by first taking the logarithm of each side of the equation. and Finance. Remember. dx 2 3 3x2 + 4 Solution We have 12x + 122 1 1 ln y = ln a b = ln12x + 122 . by Warren B.2x + 16 = 2 [2] [6x] = = 2 2 y dx 2x + 1 3 3x + 4 2x + 1 3x + 4 12x + 1213x2 + 42 or dy 8x2 .2x + 16212x + 12 dy # = = dx 12x + 1213x 2 + 42 13x 2 + 421/3 13x2 + 424/3 Applied Calculus for Business. we have 1 1 1 dy 1 4 2x 8x2 . therefore. We first assume that f1x2 7 0. Inc. we could have written 2x = ex ln 2 and proceed directly to find the derivative. Wang.Section 4. . Solve for dy dx 1 1 dy = f ¿ 1x 2 f1x2 y dx Logarithmic Differentiation Example 15 Use logarithmic differentiation to find 12x + 122 dy if y = . Equations (6) and (8) are.6 Properties of Logarithmic Functions * ** 383 d x 1b 2 = bx ln b dx du d u 1b 2 = bu ln b dx dx (8) where u is a differentiable function of x. Copyright © 2007 by Pearson Education. see Exercise 94. Economics. Logarithmic differentiation is useful when y involves products and quotients. Example 14 d x3 Determine A2 B.2x + 16 = y dx 12x + 1213x2 + 42 substituting for y. unnecessary if we use equations (5) and (7). we may easily convert to base e using these results. and April Allen Materowski. Published by Pearson Learning Solutions.ln13x 2 + 42 2 3 2 3 3x + 4 differentiating. The procedure is as follows: Given y = f1x2 1. Write ln y = ln f1x2 2. However. dx dx 3 3 Again. if A = B. Nor is this an exponential function. ln x5 2 4 y3 10 x 1 3y 11. log 3 y 3. . you need only enter d(f(x). the quotient rule could have been used to obtain the derivative. log4 1 4z x5 1 3 yz5 12. ln x3y4z5 8. Example 16 Use logarithmic differentiation to find dy if y = xx.) To use the calculator to find derivatives of exponential and logarithmic expressions. Inc. Economics.384 * ** Section 4. ln 1xy2 9. log6 1wv5r9 Applied Calculus for Business.x) and press Enter (Remember. Taking logarithms. 1 1 y ¿ = x c d + ln x[1] = 1 + ln x y x or y ¿ = 11 + ln x2y = 11 + ln x2xx Calculator Tips Note that y = xx = eln x = ex ln x. dx Solution Note that the power rule may not be used because the power is not a constant.7) and the calculator produces the correct result.. you need only enter lg(3. log1xy 22 5. as the base is not a constant. LN is in orange above X) Now if you want to compute say log 3 7.6 In Exercises 1 1. Gordon. Walter O. Wang. log 3x 2 2. We could place into the calculator s memory the logarithm to any base as follows: on the entry line in the home screen enter LN(x)/LN(b) sto lg(b. x). and April Allen Materowski. (Remember for approximate answers press * Enter. by the product rule. by Warren B. x EXERCISE SET 4. but note how much simpler it was to use the logarithmic properties and then find the derivative. log 2xy3 10. where f(x) is the expression to be differentiated. log 10x2y3 x 3y 4 6. and Finance. log 2 xy 2 4. we have ln y = ln xx or ln y = x ln x therefore. and we may use the generalized exponential rule to find the derivative. with x 7 0.6 Properties of Logarithmic Functions In the previous example. nothing new need be done. Published by Pearson Learning Solutions. ln 5 z 12 write the given expression as a sum of logarithms 7. Copyright © 2007 by Pearson Education. y = ln A 3x4 2 5 2x2 + 9 B 83. What interest rate compounded continuously did a $1000 investment earn if it accumulated to $1250 in 3 years? 59. 3 ln x + 2 ln y 23. log1x + 32 + log1x + 22 = log 20 38. Sketch the graph of the function defined by y = f1x2 = log 415x + 22.3x = 94x . 4x + 2 = 32 . How long would it take money to double itself if it is earning interest at 11% compounded (a) semiannually. y = ln x9 71. log 4 11 34.16 # 2-x = 3 2 x -x x 79. and April Allen Materowski. 2 log 5 12 . 13.6 In Exercises 13 18 use log 2 = 0. (b) quadruple. log x + log1x + 12 = log 12 37.2 3 log w 26. Published by Pearson Learning Solutions. Gordon.1 2 [ln x .000 investment earn if it accumulated to $18. 2x . y = log2 Applied Calculus for Business. and Finance. What interest rate compounded continuously did a $275 investment earn if it accumulated to $540 in 12 years? 61. 28.3010 and log 5 = 0. log 314x + 32 .000 in 5 years? 60.2x 49. How long would it take for an investment to (a) triple. 52 . 2 ln x . For how long must $2200 be left on deposit at 9.ln1x + 12] 30.log 3 2 20. eln12x . log 10 15.Section 4. (c) monthly. 64. log 400 In Exercises 19 30 combine the given expression into a single logarithm.log 213x . 1/ 2 log 6 25 + 2 log 6 3 21. 10 log13x + 42 In Exercises 67 88 use the properties of logarithms to find the derivative. y = log3 x2 76. 2 log 3 4 . y = x342x 80. y = ln 8x5 73. 3[log 31x + 12 . y = ln A x 3 23x + 1 B 82. log 212x + 12 .5 log w Properties of Logarithmic Functions * ** 385 52. 4x = 23 45. What interest rate compounded continuously did a $15. (d) daily.5 ln v 2 27.4 ln y 24.22. Wang. Walter O. Inc. = -1 4x . log 20 16.12 = 1 44. 1/ 2 log x + 2 log y . What interest rate compounded continuously did a $1500 investment earn if it accumulated to $4500 in 18 years? 62.17 log z . log 1/3 6 35.2 ln1x + 52] 29. Sketch the graph of the function defined by y = f1x2 = log 4 x. Sketch the graph of the function defined by y = f1x2 = log 213x . 5 log x + 1 3 log y .3 log z 25. y = ln 4x 68. Sketch the graph of the function defined by y = f1x2 = log 312x + 12.2 log 5 3 22. y = ln 23x2 + 5 4x5 24x2 + 3 2x3 50. B ln1x + 32 + ln1x + 12 . 3 = 32 46. 4[ln x . Hint: It might be easier if you use the properties of logarithms first. y = ln 22x + 1 74.log b B.2 log 31x .2 log y + 1/ 2 log z . y = log7 x5 77. y = 4 .12 = 1 40. y = ln 5x6 72. by Warren B.12] . 4 log x . 124x + 3 = 157 . log 16 18. 3 . Copyright © 2007 by Pearson Education. log 2 12 32.6990 and the properties of logarithms to compute the given logarithm. 1 4 ln 1 4 [3 2 x + 2 3 ln y + 3 ln z . 66. (e) continuously? 57. Prove log b A = log b A . log 3 7 33.log 312x + 12 = 2 41.4 ln w .ln1x + 12] . 65. 31.3% compounded quarterly to reach a total accumulation of $2350? 54.x 47. log 25 17. Economics. y = ln 15x 69. y = ln x6 70.65% compounded continuously to reach a total accumulation of $3200? 56.72 = 3 43. log 5 20 In Exercises 36 51 solve for x 36. log 4 14. y = 32x 78. For how long must $900 be left on deposit at 7. do not use a calculator. For how long must $1700 be left on deposit at 6% compounded continuously to reach a total accumulation of $3200? 55.2x x2 + 1 2 81. y = 23x 2 = 1 42. log1x + 12 + log1x .12 # 3 51. For how long must $700 be left on deposit at 6% compounded monthly to reach a total accumulation of $1750? 53.2[log 31x .4-x = 1 2 85. y = ln 2 3 5x + 7 75.62 = 5 39. log 213x + 22 + log 215x . 19. . y = ln1x ln x2 84.5 48. (b) quarterly. 67.12 + log 31x + 12] In Exercises 31 35 compute the value of the given logarithm. 63. earning 8% compounded continuously? 58. and therefore y = ln1ax + b2 is a translation of y = ln x. y = ln 88. Walter O. 98.7 Applications of Exponential and Logarithmic Functions » » » » » » » » Exponential Growth Population Growth Continuous Compounding Radioactive Decay Carbon Dating Logistic Growth Richter Scale Calculator Tips Let y = f1t2. show that A = bx and B = by then 86. take logarithms and differentiate. Economics. and Finance. 101. Hint: write y = f1x2 . 89.4 that the only non-trivial function that satisfies this equation is Applied Calculus for Business. x2 + y3 = 1 + ln1x 2 + 12 + ln y (0. Given P1x2 = F1x2/S1x2. (2). Given P1x2 = F1x2S1x2. Prove (2) using the same approach as in the previous exercise.1 90. Let y = bx. Use logarithmic differentiation to obtain the usual product rule.7 2x3 2x2 + 4 2 3 3x + 7 x312x + 124 2 Applications of Exponential and Logarithmic Functions 102. using logarithmic differentiation. let x = log b A then bx = A or 1bx2r = Ar bxr = Ar rewrite this exponential in logarithmic format as log b Ar = rx conclude log b Ar = r log b A In Exercises 93 95. 105.) 4 2 2 3 4. To prove (4) using the properties of the exponential. use logarithmic differentiation to find the derivative. Show that (1). Published by Pearson Learning Solutions. log b AB = x + y = log b A + log b B 104. where F and S are differentiable functions. y = x2x 92.2x ln y = . (3) and (4) are true for any base. To prove (1) directly from the laws of exponents. using the change of base theorem. dy 99. further. 1/ 22 94.8x + 4x y + y (2. dx 100. Copyright © 2007 by Pearson Education. Show that if y = f1x2 and f1x2 6 0 then logarithmic differentiation is valid. y = ln 25x + 4 4x713x2 + 8210 22x2 + 5 12x5 + 925 2 4 3x2 + 1 2 In Exercises 89 92. f ¿ 1x2 = kf1x2. y = x ex AB = bxby next show AB = bx + y and finally. that the rate of change of the function at any time t is proportional to the function at that time. Use logarithmic differentiation to obtain the usual quotient rule. Gordon. Inc. Show that ln1ax + b2 = ln1x + b/a2 + ln a. (See Example 16. x2 ln xy + x 4 . that is. by Warren B. 93. where t represents time. Write xx as ex ln x to find its derivative. Suppose. let x = log b A and y = log b B by rewriting each logarithm in exponential format. differentiate implicitly and then determine the equation of the tangent line to the curve defined by the given equation at the indicated point. 97. y . y = x2 + 1 91.228 24x . Wang.386 * ** Section 4. y = 1x2 + 524 25x + 7 13x . y = ln 87. show = bx ln b. 103. 1) 95. where F and S are differentiable functions.8xy = 0 12. 1) 96. and April Allen Materowski. . Then it was found in Section 4. 000? Solution Let P(t) represent the population of the culture at time t. Example 1 The rate of change at which a bacterial culture increases is proportional to the population at that instant. f(0) will be the initial population. if f(t) represents a population at time t. the rate of change of population is proportional to the population itself at any time.1 = e0. we have for this model.265314152 L 37. Walter O.000. Let us begin our measurements at t = 0. (b) What will be its size in 5 hours? (c) How long does it take to reach a size of 21.265314t Dividing by 10000.681 1to the nearest integer2 (c) To find the time at which P1t2 = 21.000 (a) determine its growth constant. Often.7 = e2k We solve this exponential equation using logarithms. Wang.72/2 L 0. If f(t) represents the mass of some substance at time t. and Finance. C is called the initial value. then we may rewrite (1) as P1t2 * P0e kt (2) Exponential Growth Population Growth which is simply equation (1) with P(t) replacing f(t) and P0 replacing C. we have f102 = C. Economics.000. and April Allen Materowski.Section 4. We noted that C had a simple but special interpretation. often written as m0. . P1t2 = 10000e0. since P102 = 10. f(0) will be the initial mass. often written as P0. If the initial culture has a population of 10. Then we have P1t2 = P0ekt. 2.7 = ln e2k ln 1. ln 1. Here is a specific example. Gordon.000. Thus. we can calculate directly. We saw that any application modeled by equation (1) is said to exhibit exponential growth if k 7 0. by Warren B. and so on. Copyright © 2007 by Pearson Education. For example. yielding 21000 = 10000e0. depending upon its sign.7 Applications of Exponential and Logarithmic Functions * ** 387 f1t2 = Ce kt (1) Where C and k are constants to be determined by the specific application.000 and in two hours it has increased to 17.265314 1to six decimal places2. Substituting t = 0 into (1) we have that f102 = Ce0. or decay if k 6 0. k is called the growth or decay constant. we substitute for P(t). Therefore.265314t Applied Calculus for Business. Inc. Published by Pearson Learning Solutions.265314t (b) Substituting t = 5.7 = 2k ln e = 2k k = 1ln 1. P152 = 10000e0. P1t2 = 10000ekt (a) When t = 2. Since e0 = 1. If we let P(t) represent the population at time t. P0 = P102 = 10. we have 17000 = 10000e2k or 1. However. 265314 L 2. Example 2 Suppose the rate at which an investment grows is proportional to the investment at that instant. Copyright © 2007 by Pearson Education. is just A(0). if we let m(t) be the mass present at time t. Given the half-life. and the principal. and Finance. we have. of course. we have t = 1ln 2. the basic relationship is A * Pe rt (3) You can see that the rate. and April Allen Materowski. with m(t) replacing f(t) and m0 replacing C. Scientists have been able to determine the half-life of radioactive elements. suppose that an initial investment of $1200 grows to $1500 in 2 years.265314t = 0.111572 or writing r as a percent. P. which is about 2 hours 47 minutes and 47 seconds.7 Applications of Exponential and Logarithmic Functions Taking logarithms.1 = ln e0. which is the time it takes for a given mass to decay to one-half its original size. Inc.25 = ln e2r = 2r ln e = 2r or r = 1ln 1. As you recall.12/0. Economics. Thus. we can determine the decay constant k as the next example illustrates. corresponds to the growth constant k. In particular. we have already encountered an example of exponential growth. Gordon. Applied Calculus for Business. It is also widely used by geologists and archeologists interested in dating ancient rock strata.79645 hours. Therefore. the supposition of continuous compounding of interest is equivalent to the condition that the rate of growth of an investment is proportional to the amount present at any time. Solution From (3).16% (to the nearest one hundredth of a percent). by Warren B. . The next example illustrates the application of exponential growth to finance. Wang.265314t and solving. determine the interest rate. Published by Pearson Learning Solutions. m1t2 * m 0e kt (3) It is obvious that this exponential decay property has significant ecological importance in planning for storage and disposal of radioactive wastes. ln 1. we find that ln 2. Radioactive Decay It has been established by physical observations that the rate at which a radioactive element decays is proportional to the amount present at any time. to continuous compounding of interest. Walter O. and artifacts.388 * ** Section 4. the initial amount. r. fossils. radioactive decay is also modeled by (1). Continuous Compounding Interestingly.25 = e2r Using logarithms. we have that r = 11. we have 1500 = 1200e2r or 1.252/2 L 0. Thus. We refer. Gordon. and April Allen Materowski.25 = 112e-0. A certain fraction of all the carbon in the world is Carbon 14.457 years old. Copyright © 2007 by Pearson Education. Economics. After the plant dies.7 Applications of Exponential and Logarithmic Functions * ** 389 Example 3 The half-life of Thorium 234 is approximately 24 days.0.25 = .52/5745 L .5 or 24k ln e = 24k = ln 0.0. it stops taking in new carbon.000121t or t = . The idea behind this method is very clever. Wang. The next example illustrates the method. Walter O. Published by Pearson Learning Solutions.02888 Archeologists are able to determine the age of an artifact by measuring the amount of radioactive carbon it contains.000121. Inc. Solution Assume we begin with two grams of Thorium. the amount of Carbon 14 remaining in anything made of wood will depend upon the age of the article. which becomes incorporated into the plant s cells. we have ln 0.Section 4. When plants breath in carbon dioxide. which is radioactive. Example 4 Archeologists found remains of an ancient campfire in which the amount of radioactive Carbon in the wood fragments is 25% the normal amount.000121 L 11457 That is. Then in 24 days we should have half that amount or 1 gram.0.25 Where t is to be determined.52/24 = . Thus. . That is. we are able to determine. they absorb the carbon.5 or k = 1ln 0.1ln 0. then m1t2 = 0. the remains are approximately 11. and Finance. 0. Carbon Dating Applied Calculus for Business. Substituting into (3) yields 1 = 2e24k or e24k = 0. for t = 24 days.000121t Taking logarithms. find its decay constant. Knowing that the half-life of radioactive Carbon is approximately 5745 years. as in the previous example.252/0. Thus. we have ln e24k = ln 0. we have m1242 = 1.5 Taking logarithms. that its decay constant k = 1ln 0. by Warren B. that is m0 = 2. how old are the remains? Solution Let m0 = 1. Copyright © 2007 by Pearson Education.) For example if 2/5 of the population is infected.390 * ** Section 4. Observe that p(0) represents the fraction of the population infected at time 0. Gordon. Of course. this can be explained by defining p(t) to be the fraction of the susceptible population. eventually. Also note that as t gets very large 1t : q 2 we e -kt : 0. it is left as an exercise (Exercise 41) to show that (4) will always have 1/2 at its inflection point. Then a model often used in determining the spread of the disease is given by the equation dp = kp11 . people with natural or other immunity are dropped from the pool. However it is related to an exponential type of growth. Suppose p(t) is the fraction or percentage of the population infected at time t. therefore. On the other hand. Wang.2/5 = 3/5 is not infected. and in Example 3. Inc. t has the units of time. In fact.p2 dt This says that the rate at which the infection spreads at time t is proportional to the product of the fraction infected and the fraction not infected at that time. (The fraction not infected at time t is 1 . and. You will note that in Example 4. Notice that in (1). if t is measured in days. Economics. 1 . With such a redefinition. and so on. this is called logistic growth.7 Applications of Exponential and Logarithmic Functions Logistic Growth A word about units. Consider the growth of an epidemic in an isolated community. This is not exponential growth. k is measured in 1/years. k must have the units 1/time. by Warren B. In our model we must have 0 p102 1. Note that in Figure 1. everyone must be infected. k is measured in 1/days. and April Allen Materowski. years. However. Published by Pearson Learning Solutions. Walter O. A typical logistic growth curve is given in Figure 1.p1t2. it appears that at the inflection point p is about 1/2. thus ekt is a dimensionless quantity. replacing C by f(0). Substituting into (4) gives p102 = 1/11 + C2. Therefore C Ú 0. Figure 1: A Typical Logistic Growth Curve This means that. the right hand side is f(0) ekt. therefore. . since it represents a fraction of the total population. k has units 1/hours. if t is measured in hours. Thus. and Finance. it was measured in days. we measure t in years. in real epidemics not everyone becomes infected. f(0) has the same units as f(t). Applied Calculus for Business. It is an exercise in differentiation to show that p1t2 = 1 1 + Ce -kt (4) When k 7 0. p(t) approaches one. 75 = 3/4 and solve for t. and April Allen Materowski. as follows 3 1 = 4 1 + 3e-0.339916 weeks . Thus.4. 1 9 Applied Calculus for Business. 0. Wang.34657t We need to find the time at which p1t2 = 75%. Therefore. We are given that p102 = 25% = 1/ 4.5 Taking logarithms and solving for k. Walter O. If 25% of the population is infected on March 1. 1 9 t = L 6. we substitute p1t2 = . and Finance. Published by Pearson Learning Solutions. p122 = 40% = 0. Thus.4 + 1.0. 3 + 9e -0.34657t = solving. k = . how long will it be before 75% of the population is infected? Solution We let t = 0 correspond to March 1. substituting into (4).A 1/ 2 B ln 0. p1t2 = 1 1 + 3e -kt 1 1 + 3e-2k We find k by using the information that when t = 2. Copyright © 2007 by Pearson Education. Gordon. t = 6 weeks.6 e-2k = 0.2e -2k = 1 1. and assume that t is measured in weeks. Economics.34657 ln or.5 = 0.34657t = 4 or e -0. to the nearest hour.2e -2k = 0. by Warren B. we find C = 3. and two weeks later 40% of the population is infected.7 Applications of Exponential and Logarithmic Functions * ** 391 Example 5 Assume logistic growth as given in (4). .34657t or clearing fractions. 2 days and 9 hours.4 = Clearing fractions yields 0. we have p1t2 = 1 1 + 3e -0. Inc.34657 Thus.Section 4. we have 1 1 = 4 1 + C solving for C. 3 on the Richter scale.0 7.25 Rewriting this expression in exponential format. Can cause serious damage over larger areas Great earthquake. Destructive in areas up to about 65 miles across. Applied Calculus for Business.9 Example 8 Compare the relationship in energy between two earthquakes who magnitude differs by 1 on the Richter Scale. by Warren B.9 Over 8 Earthquake Effects Recorded but generally not felt Felt. Gordon.4 Under 6.5 3.392 * ** Section 4.5 5. we have E = 1024. Can cause serious damage in areas several hundred miles across Example 7 What is the Richter magnitude of an earthquake whose energy is 1. minor damage to constructed buildings. Published by Pearson Learning Solutions.9 7. Determine the energy of the earthquake.5M (5) where E is measured in ergs. and Finance.32 = 24.8 + 1. Economics. The relationship is given by the (base 10) logarithmic equation log E * 11. Table 1: Richter Magnitude and Earthquake Effects Richter Magnitude (M) Less than 3. .1 6. and April Allen Materowski. we have log 1. Major earthquake. Copyright © 2007 by Pearson Education. Often results in major damage to poorly constructed buildings over small areas.0 6.0899 = 11.8 + 1.23 * 1019 = 11.7 Applications of Exponential and Logarithmic Functions Richter Scale In 1935 Charles Richter working with Dr.23 * 1019 ergs? Solution Substituting into (5).5M or 19. Example 6 The 1906 earthquake in San Francisco measured 8. As a result of the logarithmic basis of the scale. Usually. Wang. but rarely causes any damage. Beno Gutenberg developed a relationship between the magnitude M of an earthquake and the amount of energy E it radiates.78 * 1024 ergs.3 in (5) we have log E = 11.8 + 1.518. Inc. Walter O.25 L 1. M = 4.5M or to the nearest tenth. we shall see that each whole number step in the magnitude scale corresponds to the release of almost 32 times more energy than the amount associated with the preceding whole number value.8 + 1. Solution Substituting M = 8. Table 1 indicates the relationship between the magnitude of an earthquake and its effects. By entering this function into the Y = window 1 + 3e -0324657t with x replacing t.5 L 31. 3 1 In Example 5.7 Applications of Exponential and Logarithmic Functions * ** 393 Solution Let E1 be the energy associated with an earthquake of magnitude M. There are standard conversions from one system of units to the next. because the energy is being expressed in different units. E2 = 101. for the lower bound. We shall not deal with them here. E2 = 1. . by Warren B. Inc.34657t22 = 3/4.6228E1 Thus. joules or calories or foot-pounds. we have log E2 . Wang.5 rewriting as a single logarithm. 0.33992. a difference of magnitude of 1 on the Richter scale translates into the energy being increased almost 32 times. move the cursor well to the left and press Enter. then. The calculator gives (3.51M + 12 Subtracting the first equation from the second. and then for the right bound. t = 6. Note that we could find the inflection point of this graph with the calculator by pressing F5 (when in the graph window) then selecting Inflection.6228 E1 or E2 L 31.8 + 1. Choosing an appropriate window (remember 0 p1t2 1) we have the graph in Figure 2. we have log or in exponential form. Calculator Tips Figure 2: p1t2 = 1 1 + 3e -0324657t Applied Calculus for Business. Using the solve com. Economics. Copyright © 2007 by Pearson Education.16996.5M and log E2 = 11.5) as the inflection point. and Finance. move the cursor well to the right and press Enter. Walter O.log E1 = 1. t2 the calculator gives us immediately. we have log E1 = 11.8 + 1. p1t2 = . and let E2 be the energy associated with an earthquake of magnitude M + 1. We can also have the calculator plot 1 this logistic curve. using (5).Section 4. Published by Pearson Learning Solutions. Gordon.5 E1 Note: Sometimes (5) may be given differently. we needed to solve = for t.34657t 4 1 + 3 e mand we have solve11/11 + 3e ¿1-.0. and April Allen Materowski. 023/year. 7% of the island population is infected. determine the population in: (a) 1970.8. verify that its decay constant is . 27. Table 2: World Population in Billions Year Population 1900 1. If radioactive Carbon 14 has a half-life of approximately 5745 years. Walter O. The rate of increase of money at a bank is proportional to the amount invested. The rate of increase of a euglena culture at any instant is proportional to the number of euglena present at that instant. Determine the energy of an earthquake if its magnitude is 4. what is the accumulation at the end of 10 years? 7. Determine the energy of an earthquake if its magnitude is 3. and its initial population is 5 million. 29. how old is the fork? Applied Calculus for Business. when will it be valued at $10. An ancient wood fork-like instrument was found to have 3% of the amount of Carbon 14 that a new one would have. (f) Letting t get very large. letting 1970 be time zero. If an ancient wooden axe handle is found to have one-tenth the amount of Carbon 14 that a new one would have.000 in 1970 and 174. Find the rate of interest earned by an investment that grows exponentially.51 1970 3. (b) 1970. Determine the energy of an earthquake if its magnitude is 8. what percent of the original sample remains at the end of 10 years? 12.47 * 1020. The culture began with a bacteria population of 200. If the average half-life of materials in a nuclear waste dump is 10. 21.000? 18. After four weeks 22% have caught the disease. what will be left of a 10 gram sample of Radium after 200 years? (b) What is the halflife of Radium? (c) What does this suggest to you about the safety of the dumping of Radium in our environment? 11. Determine the energy of an earthquake if its magnitude is 6. At what point will 90% of the students have succumbed to chicken pox? (Assume logistic growth.6.000428/year. The rate of change at any time t. If the world population in 1960 was 3 billion. (d) 1900. A radioactive substance has a half-life of 47 years. that is. Determine the magnitude of an earthquake if its energy in ergs is 1. (c) 1982. Suppose a magazine has an initial circulation of 1 million and two months later its circulation has decreased 10%. how long will it take until the radiation level reaches 5% of the present level? 17.000 and six months later it is valued at $14. Assuming logistic growth.394 * ** Section 4. Using k = 0. Determine how the energies of two earthquakes differing by 2 in their magnitude are related.000 it was discovered that at the end of 1 week 20 villagers had a cold. How long does it take for the population to become 300 million? 3. (a) Given that the decay constant for Radium is . What percent of the original Potassium 40 in the rock remains? 16. (b) What will be left of a 22 kilogram sample after 15 years? 10. If an item is presently valued at $15. TX was 149. Assume in this model. The rate of increase of money at a bank is proportional to the amount invested.000 villagers have a cold? 22. Assume the growth constant for this culture is 3/hour. Predict the world population using this model in: (a) 1965. usually when advertising has been discontinued. how old is the axe handle? 14. The rate at which an item depreciates is proportional to its value at that instant. (a) Determine its decay constant. 36.000121 as the decay constant for Carbon 14.3 billion years.7. equals twice the number of bacteria in the culture at that instant.1. Determine the energy of an earthquake if its magnitude is 7.59 * 1018. and April Allen Materowski. After 5 days.029 and a = 3. 28. P0k aP0 + 1k . A rock has been estimated to be 4 billion years old.018. the growth constant k = 2/hr. Under certain conditions.000? 19.3. Economics. 35. (a) What was the population of the culture two hours later? (b) How long would it take for the bacterial population to triple? 2. In Exercises 13 and 14 use . (a) Assuming exponential growth. 30% are infected. At what time will its circulation reach 500. The population of Lubbock. The population in 1960 was approximately 3 billion. 24.5 in their magnitude are related. Determine how the energies of two earthquakes differing by 1. find the function P(t) that describes the population of Lubbock. 26. Gordon. 34. 9.5 hours? 4. Published by Pearson Learning Solutions. 30. How long will it be from time 0 before 1. and $3400 is invested initially: (a) What is the accumulation at the end of 6 1/ 4 years? (b) How long will it take until the value of the investment is $4950? 8. and $1500 is invested initially.67 * 10-12. 23. (b) 1982. Determine how the energies of two earthquakes differing by 2. .) 20. The half-life of Potassium 40 is 1.5 in their magnitude are related. Determine the magnitude of an earthquake if its energy in ergs is 3.5. If the proportionality constant is 5 1/ 4% per year. 32. if a $1200 investment increases to $2700 in 9 1/ 2 years. (b) In what year will the population of Lubbock reach 300. by Warren B. 31.23 * 1019. (in hours) of the number of bacteria in a culture. Inc. 33. 25. At the end of two weeks 180 villagers had a cold. If the disintegration rate is 0.500. Suppose that the rate of change at any time t.000 years. Determine the energy of an earthquake if its magnitude is 5.000? 5. how long does it take for 80% of the island population to get infected. Determine the magnitude of an earthquake if its energy in ergs is 4. Wang. What is the population after 2. A radioactive substance disintegrates at a rate proportional to the mass present at any instant.0. Initially.5 in their magnitude are related.aP02e -kt 6. the population is 250 million and three hours later the population is 275 million.7 Applications of Exponential and Logarithmic Functions EXERCISE SET 4.58 1982 4. initially. of the number of bacteria is proportional to the population at any instant. and Finance. In 1837 the Dutch mathematician Pierre François Verhulst proposed a different model for population growth given by the logistic equation P1t2 = where P0 = P102.6 15. 13. Copyright © 2007 by Pearson Education. Furthermore. Compare your predicted results with the actual population as given in Table 2.6 1950 2. 3% of the students have caught the disease. If the proportionality constant is 6.86 * 1021. Measles is found on an isolated island. k = 0.7 1. suppose. Suppose that the world population P(t) at time t is governed by equation (2). Suppose that an epidemic of chicken pox hits a private elementary school. Determine how the energies of two earthquakes differing by 3. Initially. it has been shown that sales of a product at any time will decrease at a rate proportional to the sales at that instant. (d) 1950.000 in 1980.0.000.3% per year. In a small isolated village with population 2.0. (e) 1900. Determine the magnitude of an earthquake if its energy in ergs is 1.000121. (c) 1950. performed another measurement. 42. show that dw = kw dt solve for w and then show that T1t2 = Tm + Ce -kt This is known as Newton s Law of Cooling. (Assume t = 0 corresponds to 1960.) 37. A forensic examiner measured the temperature of the body at noon and found it to be 94. 40. Gordon. She returned to the body at the same spot in one hour. How did she make this conclusion?.Tm2 dt Let w = T . . Copyright © 2007 by Pearson Education. Hint: differentiate dp = kp11 . Show that in base e. Show that (4) is a solution of dp dt = kp11 . suppose soup whose temperature is 190°F is poured into a cup in a room whose temperature is 70°F.45M.6°F.p2. dt CHAPTER REVIEW Key Ideas One-to-One Function Horizontal Line Test Increasing and Decreasing Functions Inverse Function Composition Property Derivative of the Inverse Exponential Expressions The Graph of y = f1x2 = bx Solving Special Exponential Equation Finding the Exponential Function Growth and Decay Rates Power Function Continuous Compounding of Interest The Constant e The Simple Exponential Rule The Generalized Exponential Rule Exponential Domination Definition of a logarithm Base 10 and e pH of a Solution Graphing Logarithmic Functions The Simple logarithmic Rule The Generalized Logarithmic Rule Multiplicative and Division Properties Exponential Property Derivatives Using the Properties Logarithmic Equations Exponential Equations Change of Base Derivatives in Different Bases Logarithmic Differentiation Exponential Growth Population Growth Continuous Compounding Radioactive Decay Carbon Dating Logistic Growth Richter Scale Applied Calculus for Business. That is dT = k1T .4°F. Show that at the inflection point of (4) p = 1/ 2. Published by Pearson Learning Solutions. It has been shown that the rate of change of the temperature of an object at time T(t) placed in a medium with constant temperature Tm is proportion to the difference between the object s temperature at time t and the temperature of the medium. equation (5) may be given as ln E = 27. Economics. (a) What will be the temperature of the soup in another 5 minutes? (b) How long did it take before the temperature of the soup was 150°F? 39. and April Allen Materowski. Inc. and found the body temperature to be 93. * ** 395 38.Chapter Review approximate the limiting value of the world s population as predicted by this model. Wang.p2. Five minutes later the temperature of the soup is 180°F. She later concluded that death occurred at exactly 8:59.Tm. 41. Assuming Newton s Law of Cooling. Walter O.17 + 3.25 AM. by Warren B. In a room kept at a constant temperature of 70°F a dead body was discovered. and Finance. 000428/year. If 23x = ekx. For those that are one-to-one (b) determine their domain and range. determine 16. Determine the equation of the tangent line to the curve f1x2 = 3x3e -2x when x = . (V) f1x2 = 2x4 . 20. How much should be deposited into an account today if it must accumulate to $1200 in three years if it earns 4. dx 4 d 13x . (a) Given that the decay constant for Radium is . determine the population in (a) 2006.03t.2.0. (b) determine f-1122.22 a b. 3. Determine the magnitude of an earthquake if its energy in ergs is 1. 4. (VI) f1x2 = 4x2 . Classify the solution as an acid base or almost neutral.8 * 10-12. Find d -2x2 A5 B. Find 17. Copyright © 2007 by Pearson Education. Using logarithmic differentiation. (c) show that f-11f1x22 = f1f -11x22 = x. Find d 3 1x ln13x2 + 122.1. For those functions that are not one-to-one. Given f1x2 = 3x + 2x + 7. Sketch the graph of f1x2 = x2 ln12x2. Walter O. (III) f1x2 = 3x2 + 4.2% annual interest compounded continuously? 7. Sketch the graph of f1x2 = 2xe -3x. 8) and (5.396 * ** Chapter Review 9.x2. (d)sketch its graph. (IV) f1x2 = 24 . determine over what intervals they are increasing and decreasing. If $1500 accumulates to $1800 in 3 1/ 2 years. Determine the pH of a solution with the given hydronium concentration. 12). Gordon. 14. (c) its population in 1998. The population P at time t (in years) of a culture is given by the equation P1t2 = 2200122-0. (e) determine the equation of the inverse. 6. (b) 2.2 million.37 * 1019.1. (c) determine 1f -11222 ¿ . 2. (II) f1x2 = 2x3 + 4. Inc. determine an exponential function passing through them. what interest is it earning if it is compounded continuously? 8. x Ú 0. and Finance. dx d -x2 1e ln12 + 5x22. 5. . (a) determine which are one-to-one. dx 1. Find 11. Compute log 3 8.9 * 10-4. 2 12. (I) f1x2 = 2x + 5. by Warren B. 13. Given the points (3. find k. (a) 3. Applied Calculus for Business. (b) its population in 2015. (a) show that f is one-to-one. determine (a) an exponential equation representing its population as a function of time. Economics. Suppose t = 0 corresponds to the year 2000. (b) 1997. Wang. For each of the following functions. 10. and April Allen Materowski. 15. If the population of a small country is growing at 2% per year and if its population in the year 2004 was 7. how long does it take a sample to decay to 10% of its present mass? 19. dx 2 4 x2 + 9 3 18. dx d A x 1x B . Published by Pearson Learning Solutions. . Wang. by Warren B. In this chapter we examine the properties of integration and its application to area and Finance. Walter O. and April Allen Materowski. Applied Calculus for Business. Economics. Copyright © 2007 by Pearson Education. Inc.5 Integration and its Applications Antidifferentiation Integration is the reversal of differentiation. and Finance. Published by Pearson Learning Solutions. Gordon. 398 * ** Section 5. Thus. they differ at most by an arbitrary constant. you will be asked to integrate f.6. Walter O. we now introduce a notation that indicates the antiderivative with respect to x. Consider the following: d 2 1x + c2 = 2x 2x dx = x2 + c Proof: dx L L L L 5 dx = 5x + c Proof: t2 dt = t3 + c Proof: 3 d 15x + c2 = 5 dx d t3 a + c b = t2 dt 3 d 1 2u A e + c B = e2u du 2 symbol 2u e2u du = 1 + c Proof: 2e Applied Calculus for Business. f1x2 = x3 . The terms antidifferentiate and integrate are used synonymously in mathematics. and Finance. Thus. This follows from the first derivative test. find the antiderivative of f. or in fact many others. Possible choices are f1x2 = x3. and April Allen Materowski. The symbol 2 L f1x2 dx represents the antiderivative of f with respect to which looks like an elongated S is called the integral sign. L the differential dx tells us that we are finding the antiderivative with respect to the variable x. that if two functions have the same derivative. Note that all anitderivatives of 3x2 differ at most by a constant. This function f is said to be an antiderivative or integral of F. Thus the variable x. Economics. as we discovered in Section 2. that is F1x2 = f ¿ 1x2. the most general antiderivative is x3 + c. f1x2 = x3 + 50. For example.1 Antidifferentiation Integration » » » » » » Antiderivative Integration Theorems Simple Power Rule Simple Logarithmic Rule Simple Exponential Rule Calculator Tips Antiderivative What we will do in this section is examine the reversal of the process of differentiation. or simply the integral of f. Wang. Similarly. where c is a constant. And in general. given some function F. Often. instead L of saying. any two antiderivatives will differ at most by a constant. Gordon. We use the symbol L dx to indicate that whatever is to the left of the differential dx is L 3x 2 dx means to find the antiderivative (or integral) of to be antidifferentiated. in the example just considered. d We introduce some notation: The symbol 1 2 is used to find the derivative with dx respect to the variable x. That is. .7. whatever goes in the parenthesis is differentiated with respect to x. Sometimes. Inc. Copyright © 2007 by Pearson Education.1 Antidifferentiation Integration 5. and the function f which is antidifferentiated is called the integrand. find another function f whose derivative is F. by Warren B. given F1x2 = 3x2 find a function f whose derivative f ¿ 1x2 = 3x2. 3x with respect to x. Published by Pearson Learning Solutions. the f1x2 dx is call the indefinite integral. we do nothing more than examine closely the rules for differentiation. 1k is a constant2 constant multiplier rule (1) Integration Theorems L 1f1x2 + g1x22 dx = L f1x2 dx + L g1x2 dx sum rule (2) L 1f1x2 . L The constant multiplier rule indicates that we may factor a constant out of the integral.g1x22 dx = L f1x2 dx - L g1x2 dx difference rule (3) The proof for each follows from the corresponding rule for derivatives. ple. that is. Copyright © 2007 by Pearson Education. we have used different letters to indicate the variable of integration (the independent variable). Economics. So how do we actually find the antiderivative for a given function? In many cases.Section 5. we should increase the power by 1 and divide by the new power.1 is any real number N + 1 (4) Applied Calculus for Business. we have the following theorem which we call the simple power rule. For example. For examd 1f1x2 + g1x22 = f ¿ 1x2 + g ¿ 1x2. Note the special case in (1) when the constant k = 1. Walter O. . As antidifferentiation is the inverse process. where N Z . by Warren B. Wang. Sometimes. We have the following integration theorems: L k dx = kx + c. you need only differentiate it.1 Antidifferentiation Integration * ** 399 Note that verification of the antiderivative is accomplished by differentiation. to prove that the antiderivative obtained is correct. namely. d N 1x 2 = NxN .1. the two operations. Simple Power Rule SIMPLE POWER RULE L Proof d d N+1 xN + 1 1 d N + 1 N a 1x 2 + 1c2 = x + 0 = x N. Inc. Gordon. and April Allen Materowski. Observe that in the above examples. + cb = dx N + 1 N + 1 dx dx N + 1 * xN dx = xN + 1 + c. and Finance. we need only invert these two processes. Therefore. (2) is true because dx * dx = x + c. differentiation of x to a power is accomplished by multiplydx ing by the power and decreasing the exponent by 1. More formally. This follows because d a f1x2 dx b = f1x2 dx L and d 1f1x22b dx = f1x2 + c L dx a Applying the two operations in succession results in the function f. Published by Pearson Learning Solutions. and obtain the original integrand. we have. differentiation and integration are called inverse operations. we bring it to the numerator by changing the sign of the exponent. Example 3 Evaluate L Solution x3 dx. the 1 by which we increase the exponent is written in the form n/n where n is the denominator of the exponent. For example. Example 2 Evaluate 1 7 dx. That is. + c = + c = 7 dx = 7 + 1 6 x 6 x6 L L Solution We first rewrite the integrand so we may apply to simple power rule. Walter O. Published by Pearson Learning Solutions. the multiplier becomes the reciprocal of the exponent. x L 1 x -7 + 1 x -6 -1 x -7 dx = + c. by the constant multiplier rule. we perform the division m/ n n m/n mentally and write x . Wang. m Example 4 Evaluate L 3x5 dx. 2 L x3 dx = 2 x3 + 1 x3 + 3 x3 3 5 + c = + c = + c = x3 + c 2 2 3 5 5 + 1 + 3 3 3 3 2 2 3 5 Note that when working with fractional exponents.400 * ** Section 5. Gordon. it is useful to rewrite the integrand before we apply the simple power rule. as indicated in the next few examples. 4 + 1 5 In some situations. instead of writing . by Warren B. Economics. . and Finance. x6 x6 3x 5 dx = 3 x5 dx = 3 + c = + C. making this observation means we can mentally place the multiplier out front after first computing the exponent. when a single power function is in the denominator. avoiding any need to xm/n work with the complex fraction. Solution We have. or if an expression is written as a radical. L x4 dx = x4 + 1 x5 + c = + c. we rewrite it in exponential form. 6 2 L L Applied Calculus for Business. and April Allen Materowski.1 Antidifferentiation Integration Example 1 Evaluate L Solution x4 dx. Inc. Copyright © 2007 by Pearson Education. 1 When N = . Copyright © 2007 by Pearson Education. we must convert it into a form we recognize. more generally. Wang. In many cases.2 + 7x + c = . we instead added a single constant of integration in the last integration step. d 1 we showed in Example 8 of Section 3.1 Antidifferentiation Integration * ** 401 Did you see us pull a fast one on you in that last solution? We should have written L 3x5 dx = 3 L x5 dx = 3 a x6 x6 + cb = + 3c 6 2 but since three times a constant is still a constant.2 4 x x -5 + 1 x -4 4x 3 4x3 1 + 7x + c . by Warren B. Example 6 Evaluate 3x2 + 5 dx. Example 5 Evaluate L Solution L a 4x2 a 4x2 2 + 7 b dx. In fact. x5 2 + 7 b dx = x5 1 dx + 7 dx = 4 x2 dx . the integral x -1 dx = dx.Section 5. we have the x dx simple logarithmic rule. we can replace 3c by C. This is illustrated in the next example.1.2 + 7x + c = + 2 + 1 -5 + 1 3 -4 3 2x4 Again. We will almost always compute the integral this way and add on a single arbitrary constant at the end. Gordon. Published by Pearson Learning Solutions. Therefore. 3x L 1 Solution We first rewrite the expression (and convert the radical to an exponent) so we may apply the simple power rule. Simple Logarithmic Rule Applied Calculus for Business.1/3 dx + 5 x -1/3 dx = 1/3 x L x L L a 3 x5/3 dx + 5 L x -1/3 dx = L x5/3 + 3/3 x -1/3 + 3/3 3 3 9 15 2/3 x + c 3 + 5 + c = 3 x 8/3 + 5 x 2/3 + c = x 8/3 + 5/3 + 3/3 .1/3 + 3/3 8 2 8 2 You noticed that the simple power rule excluded the case when the exponent N = . Walter O. Economics.5 that 1ln x 2 = .1. in order to evaluate a given integral. and Finance. Do we know a function whose derivaLx L tive is 1/x? We showed that the derivative of ln x is precisely 1/x. Inc. 3x2 + 5 L 1 3x dx = 3x2 5 + 1/3 b dx = 3 x2 .2 x -5 dx + 7 dx = 5 Lx L L L L L 2+1 4x2 dx . . and April Allen Materowski. you should note that instead of working with three separate constants after each integral and then combining them into one. . x x L L (Note that we used the exponential property of logarithms and wrote the 2 an exponent.5x 2/3 + 4x 3 + x 4ex . Evaluate 5ex dx = 5 ex dx = 5ex + c. x L N Z -1 N = -1 (6) (5) 2 1 dx = 2 dx = 2 ln x + c = ln x 2 + c. 3x4 L x8 + 7x2 . Do you see why we were able to delete the absolute value symbols?) Solution We have. Published by Pearson Learning Solutions.6x 4 dx. Example 9 Evaluate x8 + 7x 2 . Wang. ex dx = ex + c (7) L L One final example illustrates all the rules developed in this section. by Warren B. Gordon. Walter O. Economics. This rule follows from the derivative propd x erty of the exponential. Copyright © 2007 by Pearson Education. Inc. L Solution We have. and April Allen Materowski. dx SIMPLE EXPONENTIAL RULE L Example 8 5ex dx. namely 1e 2 = ex. Simple Exponential Rule We next consider the simple exponential rule. and Finance.5x 2/3 + 4x 3 + x4ex .402 * ** Section 5.1 Antidifferentiation Integration SIMPLE LOGARITHMIC RULE 1 dx = ln x + c x L We may now combine (4) and (5) and write them as one rule as follows. xN + 1 + c xN dx = c N + 1 L ln x + c Example 7 Evaluate 2 dx.6x 4 dx = 3x4 L x8 7x 2 5x2/3 4x 3 x 4ex 6x4 a 4 + + + b dx = 4 4 4 4 3x 3x 3x 3x 3x4 L 3x Solution Applied Calculus for Business. a . The integral symbol is indicated in orange above the 7 key on the keypad. variable2 where the function may be typed directly from the L keyboard. Published by Pearson Learning Solutions. . so it is not a good idea to get used to notation which uses any particular letter to represent the variable.x -10/3 + 3 3 3 x 3 3 L 1 7 5 4 1 1 x -2 dx x -10/3 dx + dx + ex dx 2 dx = x4 dx + 3L 3L 3 Lx 3L 3L L 1 x5 7 x -1 5 x -10/3 + 3/3 4 1 + + ln x + ex . indicated by t. that often the independent variable is time.2x + c 15 3x 7 3 3 Basically. It is clear that we will need to examine how to reverse the generalized rules for differentiation of these three functions to integrate more complicated expressions. The specific functions we considered were the power. N Z . antidifferentiation.10/3 + 3/3 3 3 3 5 x5 7 5 3 4 1 . To access this key.1 N + 1 L 1 du = ln u + c Lu L eu du = eu + c Calculator Tips Just as the TI 89 has a differentiation key. and the rules obtained were the simple ones. there were no composite functions. you first press the orange button (2nd) and then the number 7. x Applied Calculus for Business. and the variable is the variable of integration. Copyright © 2007 by Pearson Education. it also has an integration (antidifferentiation) key. this section did nothing more than restate the basic rules of differentiation in terms of its inverse operation. Wang. by Warren B. Gordon. Example 10 Evaluate L Solution See Figure 1.1 Antidifferentiation Integration * ** 403 1 7 5 4 1 + ex . or it may be stored in the calculator s memory. but instead of using the variable x. Inc. Walter O. and April Allen Materowski.3.b x -7/3 + ln x + ex . We restate the three simple rules. uN du = uN + 1 + c. logarithmic and exponential functions.Section 5. a x4 7 + 3x 4/5 b dx.2 b dx = a x 4 + x -2 . We shall do so in Section 5. The syntax for the key is 1function. and Finance. we use the variable u.2x + c = 15 3x 3 7 3 3 x5 7 5 4 1 + x -7/3 + ln x + ex .2x + c = 3 -1 3 . We shall see in applications. The variable that appeared in each function was x. Economics. L 2w L L L L L L 3 a 2r3 - A 5x2 + 2 1x . 1. x3 dx 13. 3 dx 12. 15. and April Allen Materowski. . 6. 8. 1 Lx L L L L 3 11. 1 dw 21. L L 2s3 ds 2 4 x5 dx 1 dx L L dx 14. 3. x2. dx 20. then the answers are equivalent. x) for checking? EXERCISE SET 5. 7.1 Antidifferentiation Integration Figure 1: Using the TI 89 to evaluate 7 ax4 + 3x 4/5 b dx x L Note that the calculator does not include the integration constant with the result. x2 = d1calculator answer. Inc. and Finance. 2 4 t dt 18.404 * ** Section 5. L2 3 x2 L 2t7 dt 1 dw 3 + 4 b dr r2 1x dx 1t dt 16.1 Evaluate the given integral and check your answer. x2 If the calculator responds with True. Published by Pearson Learning Solutions. Economics. Gordon. 5. To check your answer (assuming the variable used is x) with the calculator s answer you can enter solve1d1your answer. 10.3 B dx 13x23 dx 12x + 122 dx 1t2 + 122 dt 22x dx L 1x L x 3/4 L w 5/7 Applied Calculus for Business. 2. Why do we not use solve(your answer = calculator answer. Walter O. by Warren B. It also may not always give the result in the same algebraic form that you would obtain manually. 1 dx 19. 1 3 x dx 17. Copyright © 2007 by Pearson Education. 4. 9. Wang. F112 = 12. 29. In effect. Often. Use this result to prove that dx a 1 eax dx = eax + C. in applications. 5. Using the result of Exercise 31. Wang. 24. use this to determine dx. evaluate (a) e2x dx. (b) What function can you now antidifferentiate? dx d 1 1 a ln1a + bx2b . F1x2 = Thus. Inc. by Warren B. then we have the so called particular solution. Solution We need to find the integral (antiderivative) of F. F1x2 = 2x2 + 7x + c. for every different choice of the constant c we obtain another antiderivative.5 b dx x 35.2 22. 28. We illustrate with examples. the general solution we found gives rise to a family of solutions.3t4 + a 4x 3 . Published by Pearson Learning Solutions. We have that 1 d 1 dx = ln x + c. x x dx L 1 that we have dx = ln ax + C. Walter O. but it also follows from 1ln ax2 = .2ex + 7 b dx L x L L a 2et . so 14x + 72 dx = 4 x2 + 7x + c = 2x2 + 7x + c 2 Particular Solutions L Applied Calculus for Business. 4x3 . Copyright © 2007 by Pearson Education. 25. where the constant a Z 0. Example 1 Find a function F such that F ¿ 1x2 = 4x + 7 and F112 = 12. Is there a problem with these two differLx ent results for the same integrand? Explain. dx b L a + bx 7ex dx 2et dt 5 a .2 Applications of Antidifferentiation » » » » » Particular Solutions Equations of Motion Marginal Functions Separable Differentiable Equations Calculator Tips We saw in the previous section that finding an integral of a function (antidifferentiation) gives an infinite number of solutions.x2. 23.2t4 + 3 dt 2t3 L 4 dx Lx 2 Lx L L dx d 1 ax a e b . and April Allen Materowski. Gordon. 27. and Finance.7x + 2 L x5 dx 31. we are given additional data that results in our being able to find a particular solution which is both the antiderivative and satisfies this additional information. 26. L L 1 (c) dx.12 b dt t 8 . 30. Compute Applications of Antidifferentiation * ** 405 5t7 . that is.Section 5.9ex + 7 . . 4x Le 33. a L 32. (a) Compute 34. Economics. (b) e -x dx. (a) Compute d 1x ln x . or c = 1 thus y = 3x + ln x + 1 Note that in the previous example. or c = 3 therefore. we have a family of parallel parabolas. see Figure 1.2 5x 2 . F1x2 = 2x 2 + 7x + 3 In the previous example.2 Applications of Antidifferentiation 12 = 21122 + 7112 + c 12 = 9 + c. by Warren B. 1x Solution We will need to integrate twice to find f. Walter O. Inc. Gordon. but involves the second derivative. x dx Solution dy 1 1 1 dx = 3x + ln x + c = 3 + means y1x2 = a 3 + b dx = 3 dx + x x dx L L Lx Figure 1: F 1x2 = 2x 2 + 7x + c for different values of c (Note.2 dx = dx = 5x3/2 dx 2x -1/2 dx = 1/2 L 1x L x L L x5/2 x 1/2 5 x 3/2 dx . For each choice of c. namely F1x2 = 2x2 + 7x + c. we were given an equation involving the derivative of y and solved for y. since we are given the second derivative. and April Allen Materowski. Example 2 dy 1 Find y if = 3 + if x 7 0. we have 4 = 3112 + ln 1 + c. We have 5x2 . and Finance. Wang. we do not need to include the absolute value in the logarithm. . An equation involving derivatives is call a differential equation.2 . the particular parabola which satisfies the condition F112 = 12 is the particular solution required. The next example is also a differential equation. the integral had a general solution which represents an infinite family of solutions.406 * ** Section 5. and y112 = 4. each integration will yield a constant. that since x 7 0. which means that in order to solve it we will have to integrate twice. we get a different solution. Copyright © 2007 by Pearson Education.4x 1/2 + c1 5/2 1/2 L L f ¿ 1x2 = Applied Calculus for Business.2a b + c1 = 2x 5/2 . Published by Pearson Learning Solutions. Economics. Geometrically.2 x -1/2 dx = 5 a b . Example 3 Find f(x) if f 1x2 = 5x2 . if f ¿ 142 = 60 and f112 = 6.) thus y1x2 = 3x + ln x + c since y112 = 4. 4x 1/2 + 42 dx = 2 # 2 . Walter O. we could differentiate and determine both the equations for velocity and acceleration. Economics. . we have. and the second derivative represents the acceleration a at time t. The first example illustrates vertical motion where the only force on the object is gravity. The acceleration due to gravity g. f1x2 = L 7/2 3/2 12x 5/2 . Determine (a) how high the ball goes and (b) its velocity when it hits the ground. Wang.8/3 + 4 + c2. suppose we are given the acceleration of some object.8 m/sec 2.4#2 + 4x + c2 = 7x 3x 1 4 7/2 8 x .x 3/2 + 4x + 3 21 7 We saw that if s = f1t2 represents position s. (The minus sign occurs because we assume the usual position of our axes positive vertical position upward.32 ft/sec 2 or in metric units. Published by Pearson Learning Solutions.8 + c1 or c1 = 4 so we have.x 3/2 + 4x + c2 7 3 We now find c2 using f112 = 6. Can we determine its position and velocity at any other time? We illustrate the procedure by which we can do so with a few examples. suppose we know both the object s initial position and velocity. f ¿ 1x2 = 2x5/2 . and Finance. moreover.4x1/2 + c1 We can find c1 as follows: f ¿ 142 = 60 = 21425/2 . the so-called equations of motion.2 Applications of Antidifferentiation * ** 407 So we have. Thus. f ¿ 1x2 = 2x 5/2 . at time t.Section 5.4x 1/2 + 4 integrating again.9. and the force of gravity is directed downward. so we have 6 = 4/7 . by Warren B. as a function of time t.) Example 4 A ball is thrown upward from a 100 foot ledge with an initial velocity of 120 feet per second. then the first derivative represents the velocity v. We now reverse the question. Equations of Motion Applied Calculus for Business. Copyright © 2007 by Pearson Education. is approximately . . Inc. or c2 = 86/21 and we have f1x2 = 4 7/2 8 86 x . and April Allen Materowski. given the equation representing the position as a function of time.4142 / 2 + c1 or 60 = 64 . Gordon. Applied Calculus for Business.30t . (a) At the maximum height. so we have 100 = . the only force acting on the ball is gravity. Walter O.75 sec.16t2 + 120t + 100 = 0 or equivalently. . and s13.2 Applications of Antidifferentiation Solution We are being told that s102 = 100 ft and v102 = 120 ft/sec.7522 + 12013. Copyright © 2007 by Pearson Education. (b) When it hits the ground.32 # t2 + 120t + c2 = . we have v = 0. Inc.16t2 + 120t + 100 Now that we determined the equations of motion. 4t2 . Moreover. we can answer the questions. so we have dv a = = .16t2 + 120t + c2 2 We are given that s102 = 100.32t + 1202 dt = . Wang. Economics.2569 sec. and Finance.32t + 120 = 0 or t = 3.32 dt therefore v1t2 = L .161022 + 120102 + c2 or c2 = 100 and s1t2 = . and April Allen Materowski.25 = 0 we find t L 8. . Published by Pearson Learning Solutions.408 * ** Section 5. by Warren B.32t + 120 v = dt therefore. so we must solve the quadratic .32t + 120 ds = . Gordon.32 dt = .32102 + c1 or c1 = 120 and v1t2 = . s = 0.752 + 100 = 325 ft.1613. s1t2 = L 1 .32t + c1 We are given v102 = 120 so we have 120 = .752 = . Why will this impact velocity (ignoring its sign) solve the problem? See Exercise 32.) Example 5 With what minimum velocity must an object be thrown upward from ground level if it is to reach an altitude of 225 ft? Solution We are given that s102 = 0 (the object begins from ground level) and we wish to find v(0) if it is to reach 225 ft.144. Inc. v = 0.400 or v0 = 120 ft/sec Therefore an initial velocity of at least 120 ft/sec will guarantee the object will reach an altitude of 225 ft. and. Applied Calculus for Business.25692 = . Another way of solving this problem is to compute the impact velocity an object that is dropped from a vertical elevation of 225 feet has with the ground. as the next example illustrates.161v0/3222 + v01v0/322 now if we set this equal to 225 (the height the object is to attain). by Warren B.Section 5. we find v1t2 = .32t + v0 and s1t2 = .161v0/3222 + v01v0/322 = 225 or v02 = 14. this need not be the case.16t2 + v0t When it reaches the maximum height. Let us turn around the problem and for the moment assume we know the initial velocity.3218. Economics. Let v102 = v0. so we have . Walter O. Proceeding as in the previous problem. Wang.22 ft/sec (The negative sign indicates the ball is traveling downward.32t + v0 = 0 or t = v0/32 now we substitute to find the maximum height. so we have smax = s1v0/322 = . and instead find the maximum height of the object. the acceleration was a constant. its height is decreasing. therefore. and April Allen Materowski. Published by Pearson Learning Solutions. v18. In the last two examples.2 Applications of Antidifferentiation * ** 409 We now substitute to find v. . Gordon. and Finance. the object was subject only to the force of gravity.25692 + 120 = . Copyright © 2007 by Pearson Education. Of course. Of course we want this to be (at least) 225 feet. we have . 4t + 62 dt = 1023 . Gordon.2 Applications of Antidifferentiation Example 6 The acceleration of some object is given by the equation a1t2 = t2 .18 + 27 + 6 + 4 = 25.4t + 6 m/sec 2. Published by Pearson Learning Solutions. v = Since v102 = 2.4t + 6 dt t3 . find the velocity and position after 3 seconds. we should be able to work backwards.2t2 + 6t + 2 b dt = 3 L a 1 4 12 t 1 4 12 t 3 2 .2 3 t + 3t + 2t + c2 dv = t2 . Walter O. via integration.000. and construct the original function.21022 + 6102 + c1 3 t3 . 3 2 .410 * ** Section 5. Inc. . Solution We have. we find c2 = 4. C102 = 10000 and Applied Calculus for Business. Solution We are given. Copyright © 2007 by Pearson Education. and Finance. a = therefore.2 3 t + 3t + 2t + 4 Marginal Functions Recall that any marginal function is nothing more than the derivative of a given economic function.2t2 + 6t + c1 3 L 1t2 . Economics. we have 2 = or c1 = 2 and v = therefore. determine the cost equation. v132 = We have. Wang. and s = and s132 = 81/12 . v = therefore. and April Allen Materowski. as the next example illustrates. s = t3 .21322 + 6132 + 2 = 11 m/sec 3 ds t3 = . given the marginal function. Therefore.75 meters.2t2 + 6t + 2 dt 3 since s102 = 4. by Warren B.2t2 + 6t + 2 3 33 . and if the fixed costs are $10. Example 7 The marginal cost for x items is given by the equation C ¿ 1x2 = 2x + 1000 (in dollars). If the initial position is 4 meters and its initial velocity is 2 m/sec. . by Warren B. 1 dy = f1x2 dx g1y2 we now integrate each side to obtain 1 dy = f1x2 dx g 1 L L y2 Assuming we can do the integration. and C1x2 = x2 + 1000x + 10000 Certain types of differential equations may be solved very easily by integration. By separable. we can determine an equation relating x and y. and letting C = 3c1. These types of problems fall into the class of problems called separable differential equations. Inc. Economics. Essentially. dx y y2 dy = x 3 dx or integrating. and April Allen Materowski. Wang. we may rewrite the differential equation in differential form with the y-terms on one side of the equal sign and the x-terms on the other. as we saw above. we mean. Example 8 Solve the differential equation dy x3 = 2. We illustrate with an example. any differential equation of the form dy = f1x2g1y2 dx is separable.Section 5. that is. and Finance. Published by Pearson Learning Solutions. we have 4 y3 = 3 4x + C Applied Calculus for Business.2 Applications of Antidifferentiation * ** 411 C1x2 = L 12x + 10002 dx = 2 x2 + 1000x + c1 = x 2 + 1000x + c1 2 10000 = C102 = 1022 + 1000102 + c1 c1 = 10000 Thus. y2 dy = x 3 dx Separable Differential Equations Solution We rewrite the differential equation in differential form (assuming y Z 0) as L We next integrate and obtain L y3 x4 = + c1 3 4 (Note that each integral produces a constant of integration.) Multiplying by 3. Walter O. Copyright © 2007 by Pearson Education. which we combine into one constant. Gordon. 4x 2y 4. Gordon. You can verify that the answers are equivalent using the calculator method suggested the previous section. as illustrated in the next example. independent variable. . Figure 2: Solving using deSolve Applied Calculus for Business. If there are several constants of integration it may give them as @1 and @2. Published by Pearson Learning Solutions. and April Allen Materowski. Pressing Enter gives Figure 2. The syntax is as follows: deSolve(differential equation. Economics. dependent variable). Sometimes we give the solution in implicit form.2 Applications of Antidifferentiation or 4 y = 2 3 3 4x + C We remark that we cannot always solve for y explicitly as a function of x.12y5 dy = 3 2 . This is done using the deSolve key which is accessed by pressing F3 (Calc key) and then scrolling down to option C and pressing Enter. we multiply by . we have L y or 1y -3 + y -52 dy = 1x -2 . and Finance.2y 4 + Cx 2y 4 Figure 1: Using deSolve on the TI 89 (Note that we renamed the constant.412 * ** Section 5. dx x 1y + 12 Solution We rewrite in differential form (assuming x Z 0 and y Z 0) as y2 + 1 y5 integrating. Walter O. Note that the symbol @1 is the way the calculator represents the constant of integration. we have L -1 1 -1 1 = + + c1 x 4y 4 2x 2 2y2 to clear fractions. and we have 2x2y 2 + x 2 = 4xy 4 .) Calculator Tips As we mentioned in the last section.x -32 dx a y2 5 dy = x . The prime symbol (located in orange above the = key) is used to enter the derivative. Example 9 Solve the differential equation 1x . integration with the calculator is especially simple using the integral key.1 dx x3 + 1 x 1 b dy = a 3 . by Warren B. Consider Example 8 which is illustrated in Figure 1. Wang. However. Be aware that the form of the answer given by the calculator may look different from the form obtained manually. the calculator can solve many differential equations for us as well.3 b dx 5 x y L x L integrating. Inc. Copyright © 2007 by Pearson Education. assume s(0) = 3m and v(0) = 9 m/s.2t2 + 12 m/s2. f112 = 5 7. Determine its velocity and position in 3 seconds if s(0) = 4 and v(0) = 2. 30. The acceleration of an object is given by the equation a1t2 = 3et + 2t . f ¿ 1x2 = x2 . . 27. This exercise generalizes Exercise 31. f ¿ 1x2 = 2x . sketch the family of solutions. f112 = 3 6. f102 = 4 4. f ¿ 1x2 = 2 + 5/x. The equation of the tangent line to a curve at the point (2.4x + 8. Wang. their tangent lines at this point are perpendicular to each other. 23. (a) First.2x + 9.y x + y . find f(x). If f ¿ 1x2 = -2 x 2 dy = axNyM. f ¿ 1x2 = 2e . Why does it then follow that this terminal velocity may be determined by setting a = 0? dy y 38. If f ¿ 1x2 = 3/x. (Use Exercise 34 of the previous Section. sketch the family of solutions. Given 13. and R ¿ 1x2 = . Economics. 14.kv where k is a constant. f 1x2 = x . f122 = 1 10. The marginal cost and marginal revenue for x items in dollars is given by the equations C ¿ 1x2 = 70. f112 = . f122 = 7 x2 26.1x + 900. f ¿ 1x2 = 3ex 2 + 1. 29. Hint: Let y = vx. (a) The Empire State Building is 1250 feet high. f ¿ 112 = 2. Why does this method work? 33. solve the given differential equation. dy dx dy dx dy dx dy dx dy dx dy dx = 2x 3 y4 = 3x 4y4 = = = 4ex y 1x + 22y3 x 41y4 + 32 1x3 + 32y2 x 41y3 + 82 5.Section 5. dy dx = 3x 2y2 32. 1. f112 = 5 2. find the equation of the curve. (b) If an object is dropped from a height of 1250 feet. f ¿ 112 = 3. and Finance. Show that Exercise 31 may be written in the form dy dx = fAy xB. (Show that the slope of the tangent lines at any point not on the y-axis to the original family of parabolas is 2y/x. f 1x2 = 2x2 .1 9. 28. 15. y) on a curve is given by 2 1x. determine the cost function. dx 35. = f A x B .3.6 1x + 2. The marginal cost for x items in dollars is given by C ¿ 1x2 = 2x + 50. Suppose an object is dropped from a plane flying at a constant altitude. If f ¿ 1x2 = 2x + 3. y = cx2 represents a one-parameter family of parabolas. suppose you somehow know beforehand that the object will have a limiting velocity. and April Allen Materowski.2. . sketch the family of solutions.2/x + x 2.000. with what speed does it hit the ground? 16.0. Applied Calculus for Business. 25. Redo Example 5 using the method suggested after its solution.1 x4 x = kyn. by Warren B. If f ¿ 1x2 = 2ex. Gordon. find its equation. In Exercises 23 31. In the previous example. f ¿ 102 = . Determine the velocity of this object and show that the velocity will approach g/k as t gets very large. The acceleration of an object is given by the equation a1t2 = t3 . there is a frictional force proportional to the velocity opposing the object s motion. f102 = 1 11. determine y. If the overhead cost is $800. 4) is y = 3x .7 at any point on the curve.1. Inc. k is a constant. f112 = 3 8.2 In Exercises 1 10. 12. f102 = 2 3. its acceleration is a = g . f ¿ 1x2 = 3 . The marginal revenue for x items in dollars is given by R ¿ 1x2 = . d 2y If 2 = 6x . Determine its velocity and position after 4 seconds. determine its impact velocity. f 1x2 = 2. x 24. . With what minimal velocity must a ball be thrown vertically upward to reach its top.) 36.2. and use the product rule. The slope of the tangent line at any point (x. 10) is on the curve. An object is thrown upward from the ledge of a building 128 feet high with an initial velocity of 32 ft/sec.) dy dx = x . This is called the terminal velocity. If the overhead cost is $5. sketch the family of solutions. f ¿ 122 = 1. with what speed does it hit the ground? (c) What is the relationship between the answers to the questions and why is it so? 17. (Note: there are four cases that need to be dx considered. and (b) the Earth s surface? 18. Published by Pearson Learning Solutions. How long does it take an object dropped from a 50 foot height to hit (a) the Moon s surface. (b) Now suppose that in addition to gravity. 21. Determine the (a) revenue function and (b) the demand function. Copyright © 2007 by Pearson Education. Find the equation of another one-parameter family of curves with the property that at any point (x. 20. determine the profit function. show that the dx change of variable v = y/x results in a separable differentiable equation. a is a constant. f 1x2 = 3e + 2. 31. y) where the curves from both families pass through the point. Hint: write y = vx. 22. if the point (4. 19. Walter O.2 Applications of Antidifferentiation * ** 413 EXERCISE SET 5. suppose that the only force on the object is gravity. Suppose. The acceleration due to gravity on the Moon is about 1/5 that of the Earth s. that is.) 37.3. 34. 39. so (1) becomes. N + 1 N Z -1 L 1 du = ln u + c u L L Reversing the Chain Rule eu du = eu + c What do we do if we have a function that does not fall into one of these three rules? In general. we multiply the derivative by dx). then we can evaluate it. we developed the rules for integrating the simple expressions. let u = g1x2 then du = g ¿ 1x2 dx or in differential form (remember to find the differential. Walter O.1. When we studied the Chain Rule. That is. that is. For example. the second part of the integrand. . Wang. Gordon. the answer to that question is not simple. uN du = uN + 1 + c. by Warren B. namely.3 The Substitution Method 5. L f1g1x22g ¿ 1x2 dx = L f1u2 du (2) Now if this simpler looking integral on the right-hand side of (2) is one we recognize. and Finance. Consider integrals of the form f1g1x22g ¿ 1x2 dx (1) L Notice that the first term in the integral is f(g(x)) and the derivative of the inner part of this function. and April Allen Materowski. but there is an entire class a functions that can be made to look like one of these simple rules. Published by Pearson Learning Solutions. du = g ¿ 1x2 dx Notice that du is precisely g ¿ 1x2 dx. Observe that L Applied Calculus for Business.3 The Substitution Method » » » » » Reversing the Chain Rule Generalized Power Rule Generalized Logarithmic Rule Generalized Exponential Rule Calculator Tips In Section 5.414 * ** Section 5. under this substitution. Let s try this substitution again and see where it takes us. Inc. we let the inner function be represented by u. g*(x) is multiplying this function. consider the integral 13x2 + 12106x dx. It looks like the integrand may have been obtained by using the chain rule on some function having g(x) as its inner part. Copyright © 2007 by Pearson Education. Economics. logarithmic and exponential rules of integration. by almost. we have the following restatement of the constant multiplier property. the constant may be inserted. Walter O. Inc. we have 13x2 + 1210x dx = 1 6 L L 13x2 + 12106x dx We now would proceed as above. Copyright © 2007 by Pearson Education. if k is a non-zero constant. For the generalized power rule.3 The Substitution Method * ** 415 d 13x 2 + 12 = 6x dx So we try the substitution u = 3x2 + 1 then du = 6x dx Again. That means if instead of the above example we had 13x2 + 1210x dx L we observe that du = 6x dx.Section 5. we considered three special cases. Such situations are easily rectified by the constant multiplier property of the antiderivative. by Warren B. . and April Allen Materowski. and Finance. Gordon. we mean within a multiplicative constant. That is. namely the power. that is f1g1x22 = ur. then f1g1x22g ¿ 1x2 dx = 1 k L L f1g1x22kg ¿ 1x2 dx (3) When we studied the Chain Rule. where r Z . Economics. Published by Pearson Learning Solutions. 1 6 L 13x2 + 12106x dx = 1 6 L u10 du = 1 6 u11 1 13x 2 + 1211 + C + C = 11 66 The point of this remark is that if you almost have the derivative of the inner function in the integrand. It may happen that it is off by a multiplicative constant. to find the differential du we find the derivative and multiply by dx. exponential and logarithmic cases.1. the form of (1) becomes ur du ur + 1 dx = ur du = + c dx r + 1 L (2a) L Generalized Power Rule Applied Calculus for Business. by the constant multiplier property. the term g ¿ 1x2 is not quite the derivative of the inner function. However. Similarly we examine three specific cases of (1) the generalized power. Wang. We have 13x2 + 12106x dx = u10 du = u11 + c 11 L L we now substitute for u and obtain 13x2 + 12106x dx = u10 du = 13x2 + 1211 u11 + c = + c 11 11 L L Sometimes. as long as it is compensated for by multiplying the integral by its reciprocal. namely. 3 is 2.3 dx = 12x . Inc. Copyright © 2007 by Pearson Education. you must get the original integrand. Let Evaluate u = 4x3 + 7 then du = 12x2 dx therefore. L Solution Observe the derivative of 2x .3 The Substitution Method The easiest way to understand this rule is by considering several examples. Economics. by Warren B. L Solution You should immediately observe that the derivative of the inner function. So once again the constant multiplier rule will be useful. we can use (2). we have. Wang. 14x3 + 7212x 2 dx = 1 12 L L 14x3 + 721212x2 dx = 1 12 L u12 du = 1 # u13 1 13 1 u + c = 14x 3 + 7213 + c + c = 156 156 12 13 Notice that after we finish the integration step. . We have a x2 term in the integral. Of course.3 then du = 2 dx we have 22x . 4x3 + 7 is 12x 2.321/22 dx = 1 2 L u1/2 du = 1 # 2 3/2 1 1 u + c = u3/2 + c = 12x . to check your answer take its derivative.323/2 + c 3 3 2 3 Example 3 Evaluate L 1x + 1213x2 + 6x + 128 dx. we need to replace u by 4x3 + 7. that is. Applied Calculus for Business.416 * ** Section 5. Remember. and Finance.321/2 dx = 1 2 L L L 12x . Gordon. if it is off by a constant multiplier. so the method will work fine. Walter O. Published by Pearson Learning Solutions. and April Allen Materowski. we are off by a factor of 12.3 dx. Example 2 Evaluate 22x . We let u = 2x . Example 1 14x3 + 7212x2 dx. if the integrand almost looks right. a constant. its derivative in the integrand as well. and April Allen Materowski. What is wrong with the following approach? u = 5x3 + 1 and du = 15x2 dx then 15x3 + 124x4 dx = L 1 15 15x3 + 124x 215x 2 dx = L x2 15 15x 3 + 12415x2 dx = L x2 15 u4 du L and now proceed using the power rule. Is it x + 1. See Exercise 33. Example 4 15x3 + 124x4 dx. Explain why the above substitution method does not work on How could we evaluate the integral in the above example? One way is to multiply out the integrand and then integrate term by term. within a multiplicative constant. that L results. . Gordon. Published by Pearson Learning Solutions.Section 5. or is it 3x2 + 6x + 1? A little reflection should make it clear it is the latter. We can only factor out constants! Observe. then du = 15x 2 dx. Walter O. Copyright © 2007 by Pearson Education. Work out each integral and see that you would obtain different Applied Calculus for Business. We let u = 3x2 + 6x + 1 then du = 16x + 62 dx = 61x + 12 dx we have L 1x + 1213x2 + 6x + 128 dx = 1 6 L 13x 2 + 6x + 1281x + 12 dx = L 1 6 13x 2 + 6x + 12861x + 12 dx = u8 du = 13x 2 + 6x + 129 1 # u9 u9 + c = + c = + c 6 9 54 54 L Let us make sure we understand why this method is working. and Finance. The substitution u = inner function reduces the problem to one we recognize. by Warren B. We do not have an x2 term in the integrand. Wang.3 The Substitution Method * ** 417 Solution It looks like there is a choice for the inner function here. Inc. This is an illegal operation. The next example illustrates when this method is not applicable. x2 dx Z x L x dx. Essentially. we must have. Economics. L Solution If we let u = 5x3 + 1. Did you see where we performed the illegal operation? We factored out the x2 from the integral. so the above substitution method is not applicable. once we identify the inner function. Published by Pearson Learning Solutions. use (polynomial) long division to reduce the degree of the numerator. Economics. using the power property of logarithms. One basic strategy when integrating a rational expression is to make sure the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator. so we use long division 4 x2 + 4 * 4x2 + 6x + 16 4x2 + 16 6x Giving. When this is not the case. Copyright © 2007 by Pearson Education. Gordon. 3 Lx + 1 u = x3 + 1 then du = 3x2 dx so we have 1 x2 1 1 dx = x2 dx = 1 3x2 dx = 1 du = 3 3 3 3 3 u Lx + 1 Lx + 1 L Lx + 1 1 3 ln 3 u + c = 1 3 ln x + 1 + c (2b) Solution If we let The answer may also be written as ln 2 3 x3 + 1 + c. 4x2 + 6x + 16 6x = 4 + 2 x2 + 4 x + 4 so we have. 4x2 + 6x + 16 6x dx = a4 + 2 b dx = 2 x + 4 x + 4 L L Applied Calculus for Business. by Warren B. Example 6 Evaluate 4x2 + 6x + 16 dx. . Walter O. and Finance. Consider the following example. Wang. and April Allen Materowski. (2) may result in the generalized logarithmic rule which may be written as Generalized Logarithmic Rule 1 du 1 dx = du = ln u + c Lu L u dx We illustrate with a few examples.3 The Substitution Method In the same way some substitutions in some problems result in the generalized power rule. Example 5 Evaluate x2 dx.418 * ** Section 5. Inc. x2 + 4 L Solution The degree of the numerator is the same as the degree of the denominator. Published by Pearson Learning Solutions. the integral transforms into one we recognize. ObL u 1 serve that this is just an alternate way of writing du. logarithmic integrals are written in the form . Copyright © 2007 by Pearson Education. and Finance. except that now the integral we obtain is an exponential. we do not need the absolute value signs (why?).) So the problem reduces to the evaluation of the second integral. Inc. That is exactly the point of (2). Lu You should realize that both the power and logarithmic cases are really the same. (2) becomes L eu du dx = eu du = eu + c dx L (2c) Generalized Exponential Rule We illustrate with several examples. Let u = x2 + 4 du = 2x dx then 6x 1 1 dx = 3 2x dx = 3 du = 3 ln u + c = 3 ln x2 + 4 + c 2 2 u + 4 x + 4 x L L L We do not need the absolute values since x2 + 4 7 0. du One final remark. by Warren B. Sometimes. 2x Le + 8 Solution The derivative of the denominator is (except for a constant) in the integrand. and April Allen Materowski. The case of the generalized exponential really offers nothing new.Section 5. e2x 1 1 dx = 1 2e2x dx = 1 du = 1 2 2 2 ln u + c = 2x u e + 8 + 8 e L L L 2x 1 2 ln 2x e2x + 8 + c = 1 + 82 + c 2 ln1e Once again. so the evaluation of the original integral is 4x + 3 ln1x2 + 42 + c or 4x + ln1x 2 + 423 + c Example 7 Evaluate e2x dx. Walter O. Economics. that is. so we let u = e2x + 8 and du = 2e2x dx and we have. once the substitution is made.3 The Substitution Method * ** 419 L 4 dx + 6x 6x dx = 4x + dx 2 2 + 4 x + 4 x L L (The constant of integration will be added at the end. Wang. . In the case of the exponential. Applied Calculus for Business. Gordon. L Solution Let u = . 3 Solution Notice the derivative of the exponent 2x3 is 6x2. you may show that for any nonzero constant a. let u = 2x3 then du = 6x2 dx and e2x 4x 2 dx = 4 3 L L e2x x 2 dx = 3 4 6 L e2x 6x2 dx = 3 2 3 L 2 2x u + c eu du = 2 3e + c = 3e 3 With the substitution method we can now determine integrals of exponentials in any base. Published by Pearson Learning Solutions. . and we have Evaluate e -5x dx = *1 5 L L e -5x1 * 52 dx = -1 5 L eu du = -1 u 5 e + c = . Recall that bx = ex ln b with this identity we solve the next example. Applied Calculus for Business. Copyright © 2007 by Pearson Education.420 * ** Section 5. Gordon. then du = . and April Allen Materowski. Walter O. Economics. ax + c eax dx = 1 ae L (4) You are asked to show this in Exercise 29. L Solution We have Evaluate bx dx = ex ln b dx L L This is exactly the integral given in (4) with a = ln b. except for a constant multiplier. So we have. by Warren B.5x 5 e + c In exactly the same way. Wang. Inc. we have this term in the integrand so the substitution method will work. and Finance.5x.3 The Substitution Method Example 7 e -5x dx.5 dx.1 . Example 9 Evaluate L e2x 4x 2 dx. Example 10 bx dx. 6. EXERCISE SET 5. L L L L x13x2 + 225 dx x513x6 . 5. 4. Walter O. None of the subL stitutions we discussed will work on this example. then we can evaluate its integral. and April Allen Materowski. by Warren B. 9. 13. 1. Published by Pearson Learning Solutions. 8.3 Evaluate the given integral. Wang.) Calculator Tips Figure 1: Evaluating L ln x dx with the TI 89 ln x dx = x ln x . and Finance. To verify that this result is correct. however is able to compute many forms of integrals in addition to the three we just considered. Economics. The TI 89. (Recall that the calculator omits the constant of integration. 12. Inc.3 The Substitution Method * ** 421 L bx dx = L ex ln b dx = 1 x ln b 1 x bx e + c = b + c = + c ln b ln b ln b (5) This section showed that if a more complicated integrand can. 11. 12x3 + 326x2 dx 24x + 14 dx 13x5 + 12815x4 dx 12x + 1292 dx e2x 4x dx 2 7. Copyright © 2007 by Pearson Education. by a simple substitution. Gordon. 3. 2. consider ln x dx.x 4 ex dx 2 Lx -3 1 2 dx L 2x + 1 3x + 2 L 13x2 + 4x + 125 dx Applied Calculus for Business. . What do we do at this stage in our mathematical development if we come across an integral which does not reduce to one of these cases? For example. 10. be reduced to one of the three forms given above. you need L only differentiate the result and obtain the integrand. See Figure 1. Thus.x + c.229 dx 22x + 1 dx 3x2 24x 3 + 3 dx 2x3 dx L L L L L L 21 .Section 5. See Figure 1. 23. n and r are conL stants. Gordon. dx x1/3 L 17. Economics. The first. 26. (c) Explain the different results. then divide. + 2x2 + 1 13x2 + x2e4x dx 32.4 Approximation of Areas » » » » » Areas by Rectangles Left Endpoints Right Endpoints Midpoints Calculator Tips Simplistically speaking.) 5. evaluate 37. 18. is how to determine the equation of the tangent line to a curve. Wang. (c) Explain the different results. . which we have already studied in some detail. and (b) by substitution. (a) evaluate this integral by first multiplying out the L integrand. 38. b. dx 28. * ** Section 5. Applied Calculus for Business. x2 dx Lx + 1 1 dx L ax + b L L L L eax dx 1ln x23 x 42x dx 2x154x 2 dx 2 3x + 2 dx L 3x + 4x + 1 2 10x4 . Most of differential calculus is just a detailed response to this question.x + 4 xn1axm + b2r dx.4 Approximation of Areas 27. Consider the integral dx dx 4x 2 . Given 12x + 322 dx. calculus answers two questions. Inc. where. e2x 21 + 3e2x dx x dx 2 Lx + 1 53x L5 3x 1 dx L x ln x 1ln x2N 34. L 15x 3 + 124x4 dx. Walter O. 19. the x-axis and the lines x = a and x = b. Published by Pearson Learning Solutions. 30. L L x2 x2 + 1 2 b x 3 36. 29. dx L x 35. m. 15. by Warren B. The second question that calculus answers is the following: find the area of the region bounded by the graph of y = f1x2.3x2 + 3x + 4 L 4x5 . and April Allen Materowski.2x 3 + 3x2 + 8x + 5 12 + 3x2/32 16. x Le + 1 (b) Write the numerator as 1ex + 12 . + 2 dx e 2x dx 2x 1 e + 123 L e 2x dx 24. a. and Finance. What relationship should there be among these constants if the method of substitution will work to evaluate the integral? 1 dx (Hints:(a) multiply the numerator and denominator by e -x.422 14. L L L L x2e4x dx 10x4e3x 5 3 dx +7 dx 3 31.ex. 22. 20. 33. 21. Copyright © 2007 by Pearson Education. By multiplying out the integrand. 2x Le + 1 a3 + 25. Thinking like a mathematician. What we do now is to approximate the area of the region we are trying to determine by rectangles. Published by Pearson Learning Solutions. Economics. Walter O. Gordon. you might then suggest we let the number of covering rectangles become infinite (this will involve a limit). and Finance. and we should then obtain the required area.Section 5. and between the lines x = 1 and x = 2. then by summing the areas of these rectangles we should obtain a good approximation to the area of the given region. We shall see that by increasing the number of rectangles covering the region the approximation will improve. we shall discover that indeed they do. see Figure 2. . and April Allen Materowski. The area of a rectangle is defined to be the product of its length and width. the x-axis and the Lines x = a and x = b What is interesting is that while these two questions seem to have no connection. If we can cover or exhaust the area with enough contiguous rectangles. the lines x = 1 and x = 2 Applied Calculus for Business. the x-axis. the rectangle.4 y y = f( x ) Approximation of Areas * ** 423 x=a x=b x Figure 1: Area of Region Bounded by y = f 1x2. Wang. Copyright © 2007 by Pearson Education. the x-axis. Areas by Rectangles A Figure 2: Area bounded by f 1x2 = x2 + 1. Inc. In order to answer the second question we begin with one simple idea. Example 1 Let us consider the problem of approximating the area A under the parabola defined by f1x2 = x2 + 1. by Warren B. x = 1. x = 1. we illustrate this in Figure 4. Published by Pearson Learning Solutions.12/4 = 1/4. and [1. 1. (For clarity.424 * ** Section 5. Figure 3: Drawing the Vertical Sides of the Rectangles How do we complete the rectangle? There are lots of choices. 1. 2]. we draw a vertical line from the x-axis to the curve. Inc. Economics. since the length of the original interval is 2 . some too big. . Copyright © 2007 by Pearson Education. the four subintervals will be [1. 1. [1. let us make the choice and use the leftmost point of each interval to complete the rectangle. Walter O.) f( x ) = x 2 +1 Left Endpoints y 1. So we draw the lines x = 1. At each x-value on the subintervals. by Warren B. each will have length ¢ x = 12 .4 Approximation of Areas Solution How can we approximate the area of the region? Suppose we partition the interval between x = 1 and x = 2 into n = 4 equal subdivisions (we chose four arbitrarily). the figure just shows the x-axis from 1 to 2.75 and x = 2. Gordon.25]. and April Allen Materowski. Wang. [1. For this example. Thus.25 1.75 Figure 4: Choosing the Left Endpoint of Each Subinterval to Construct the Rectangle Applied Calculus for Business.25. x = 1.1 = 1 unit long and we want four equal subintervals. and Finance.5]. it draws the yaxis through x = 1.75].5 x 1. Some choices will result in rectangles whose areas are too small.5.5.25.75. see Figure 3. Published by Pearson Learning Solutions. instead of choosing the left endpoint to draw the rectangles.2522 + 1 = 2.75 Figure 5: Choosing the Right Endpoint of Each Subinterval to Construct the Rectangles f11.96875 Note that this sum under-approximates the area of the required region because each rectangle does not fill up the entire region under the curve.522 + 1 = 3.52 = 11.252 + 1/414. the height of the first rectangle is f112 = 12 + 1 = 2. Wang. Copyright © 2007 by Pearson Education.4 Approximation of Areas * ** 425 We now have four inscribed rectangles. each of width ¢ x = 1/4 = 0.5625 + 3.5625 + 3.0624 + 521/4 = 3. the heights of the other three rectangles are f11.25.7522 + 1 = 4.5 x 1.06252 = 12 + 2. and April Allen Materowski. Gordon. The sum of the areas of these four rectangles is therefore s4 = 1/4122 + 1/412. Inc.752 = 11.25 1.25 f11.2522 + 1 = 2.56252 + 1/413.5625 f11.062421/4 = 2. by Warren B. see Figure 5.25 + 4.25 + 4. Walter O. Economics.7522 + 1 = 4.252 = 11. .56252 + 1/413. Applied Calculus for Business. When x = 1 on the parabola.06252 + 1/4152 = 12.252 + 1/414.522 + 1 = 3.252 = 11. but now.0625 The area of each rectangle is then the height times 1/4. The height of the first rectangle is at x = 1 on the curve.Section 5. Example 2 Redo the previous exercise. and Finance.71875 It is clear that this sum is too large as each circumscribed rectangle over-approximates the area under the curve because a portion of the rectangle is above the curve.0625 f122 = 5 The sum of the areas of these four rectangles is now S4 = 1/412.752 = 11.52 = 11.5625 f11.25 f11. Solution The y-values at each right endpoint producing the top of the rectangles are now Right Endpoints y 1. choose the right endpoint of each subinterval. and draw the top of the rectangle through that point. and Finance.890632 + 1/ 413.26563 2.89063 + 3.5] [1. T4 = s4 + S4 2. 1.375 1.51563 f(x) = x2 + 1 y 1. using the above.4 Approximation of Areas An improved approximation is obtained by taking the average of the two results.64063 + 4.5156321/4 = 3. Inc.875 y-coordinate on Curve 2. clearly the midpoints gave the best approximation.265632 + 1/ 412. Wang.625 1. Economics.125 1.375 1.640632 + 1/ 414.5 1.75 1. The rectangles using this midpoint are drawn in Figure 6 Table 1: Finding the y -Coordinate at the Midpoint of the Subinterval Interval [1. and April Allen Materowski. This will be true in most cases.96875 + 3. 1. its midpoint and corresponding y-value. Example 3 Redo the previous example by choosing the midpoint of each subinterval.71875 = L 3.875 2 x Figure 6: Choosing the Midpoint of each Subinterval to Construct the Rectangles We have. Applied Calculus for Business. 2] Midpoint 1. Solution In Table 1. M4 = 1/ 412.125 1. 1.34775 2 2 Midpoints We could construct the rectangles by choosing any point in each subinterval: compute the y-value at that point on the curve. Copyright © 2007 by Pearson Education. we indicate the interval.64063 4.75] [1.25] [1.25 1.75.25. Published by Pearson Learning Solutions. Gordon.3333.515632 = 12. Walter O. by Warren B.89063 3.328125 Notice that the errors produced by using the left or right end points almost cancel out when the midpoint is used.625 1. .426 * ** Section 5. We shall discover that the exact area of the region is 10/3 L 3.26563 + 2.5. 125.75 1. Gordon. Walter O. c1 = 1) the next x-value in this column is c2 = c1 + ¢ x. This remark is used for the spreadsheets using left.148438 Applied Calculus for Business. this time.Section 5. and April Allen Materowski. so the width of each interval will now be ¢ x = 12 . f(x ) = x 2 + 1 y 1. and Finance. we ll generate all the information on a spreadsheet so we don t have to actually do the arithmetic. we now will partition the interval into 8 equal subintervals. Economics. right and midpoints given below.4 Approximation of Areas * ** 427 Example 4 Improve upon the approximation in the previous three examples by partitioning the interval into n = 8 rectangles. This is precisely how we generated the first column of the spreadsheet by copying the previous x-value and adding ¢ x. if we call the first x-value c1 (in the above example. (In the above example 1 + 0. . Figure 7 shows the inscribed rectangles formed under this partition using the left endpoint.375 1. We first use the left endpoint of each subinterval to draw the rectangles.875 x Figure 7: Using the Left Endpoint and n = 8 Rectangles We shall first give the spreadsheet using the left endpoint. 5 1. Wang. Thus we find that the approximate area using n = 8 rectangles and the left endpoint of each subinterval is s8 = 3. Table 2: Using the Left Endpoint and n = 8 Rectangles Note that when constructing the x-value column.25) the next is c3 = c2 + ¢ x and so on.25 = 1.125 1.12/8 = 1/8 = 0. Solution Let us repeat the calculations. 25 1. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education. Inc.625 1. As before. by Warren B. and April Allen Materowski. y 1. Inc. no other changes need to be made.148438 + 3. If we choose the right endpoint. which is the spreadsheet containing all the calculations.335938 which is very close to the exact area. we will do even better if we take the midpoint of each interval.96875. Published by Pearson Learning Solutions. Applied Calculus for Business.625 1. Wang. The same observation will be true for the right endpoint and the midpoint. the number of rectangles increase.523438 Still too large but the over-approximation has decreased. we have T8 = 1s8 + S82/2 = 13. it is apparent that as we let n. See Figure 9.125 1.875 Figure 8: Using the Right Endpoint and n = 8 Rectangles We generate Table 3. Gordon. Copyright © 2007 by Pearson Education. 5 x 1. Thus.71875.5234382/2 = 3. When we used n = 4 rectangles. we found S4 = 3. If we average the two approximations for n = 8 rectangles.25 1. we have Figure 8. where we found that s4 = 2. Walter O. Table 3 may be obtained from Table 2 by changing the entry in Cell B7 from 1 to 1.428 * ** Section 5.4 Approximation of Areas Note that this is an improvement upon the case where we used the left endpoint and four rectangles. 75 1.375 1. by Warren B. and in fact is true no matter which point is used to construct the height of the rectangles. Economics. Table 3: Using the Right Endpoint and n = 8 Rectangles (Note.) Observe that the approximation to the area is now S8 = 3. and Finance. the rectangles cover more of the area under the curve and the approximation improves. As before.125. . sometimes.Section 5. by Warren B. and April Allen Materowski. . Published by Pearson Learning Solutions. Figure 10: The Quarter Circle y = f 1x2 = 21 . The equation of the circle of radius 1 centered at the origin is given by the equation x2 + y 2 = 1. and the domain is . the rectangles better filled (exhausted) the area under the curve.3333. see Figure 10.0625 1875 1. we say these sums converge quickly to the exact answer.x 2.x2. with 0 x 1. We know from elementary geometry that the area of a circle of radius 1 is p. and Finance. that is the quarter circle in the first quadrant. we find M8 = 3. To make it even simpler. Let us consider another example. we indicated that the exact answer is 10/3 L 3. Therefore. We have agreement to two decimal places with only n = 8 rectangles. Inc.6875 1.4375 1. Copyright © 2007 by Pearson Education. our approximations are getting better very quickly. Table 4: Using the Midpoint and n = 8 Rectangles. Gordon. In each case by using more rectangles. Economics.x2 Applied Calculus for Business.3125 1. remember.8125 1. let us take one-half of this semi-circle.4 Approximation of Areas * ** 429 y 1. It appears that in the above example.1 x 1. n has to be very large before the approximation is any good. Wang. Walter O. In the language of mathematics. 1. This does not always happen. the upper half of this circle is given by the equation y = f1x2 = 21 .5625 1. Thus. So we consider the area of the region bounded by the curve whose equation is y = f1x2 = 21 .332031 Notice the improvement in the approximation of the area by doubling the number of rectangles.9375 x Figure 9: Using the midpoint of each subinterval and n = 8 Rectangles Table 4 is the spreadsheet generated using the midpoint of each subinterval. 14159 to five decimal places. We generate that spreadsheet in Table 7 where we obtain 3. Published by Pearson Learning Solutions. Inc.904518. Suppose for the moment that we do not know the value of p. Perhaps we will do better using the midpoint of each subinterval (see Figure 13). We find that we obtain.152411 as our approximation. one decimal point accuracy. we obtain 2.304518 as an approximation to p. . Solution We will use spreadsheets and form a partition using n = 10 rectangles. Copyright © 2007 by Pearson Education. Example 5 Following the discussion above. we have 13. by Warren B. Walter O. 3.430 * ** Section 5. Using the average of these two values. see Table 5 (the rectangles over-approximate the area. using the left endpoint. This is not very good: recall p L 3.1.304518 + 2. and Finance. at best. So with n = 10 rectangles we get. Still not very good. this would then be an approximation of p.104518 as an approximation to p. and April Allen Materowski.02/ 10 = 0. Gordon. see Figure 12). Wang. use n = 10 rectangles to approximate p. We next generate the spreadsheet using the right endpoint in Table 6 (these inscribed rectangles under-approximate the area.4 Approximation of Areas The area of this quarter circle is p/4. see Figure 11). We could find the approximate area under this curve and then multiply that number by 4. y Figure 11: Left Endpoints with n = 10 Table 5: Using Left Endpoint to Approximate p Applied Calculus for Business. The width of each rectangle (that is the length of each subinterval) will be ¢ x = 11 .9045182/2 = 3. We first generate the spreadsheet using the left endpoint of each subinterval (circumscribed rectangles). Economics. Economics.25 . .4 Approximation of Areas * ** 431 y x Figure 12: Using the Right Endpoint for n = 10 Table 6: Using Right Endpoint to Approximate p y .75 .65 .Section 5.85 . and Finance. and April Allen Materowski.35 .45 . Wang. Walter O.05 .55 .15 .95 Figure 13: Using the Midpoint for n = 10 Applied Calculus for Business. Inc. Gordon. Copyright © 2007 by Pearson Education. by Warren B. Published by Pearson Learning Solutions. 7854842 = 3.120417 and 3. in this particular example. Á .122 ¢ x Using the right endpoint the sum Rn is Rn = 1f1x12 + f1x22 + Á + f1xn .12 B b ¢ x (3) We saw that it is a simple matter to compute these sums using a spreadsheet. We shall show. by Warren B. xn . Wang. xn = b then the sum obtained using the left endpoint Ln is Ln = 1f1a2 + f1x12 + f1x22 + Á + f1xn . increases. the number of rectangles.141937 The left and the right sums yield 3. but for a much larger value of n.160417 as approximations to p. Copyright © 2007 by Pearson Education.4 Approximation of Areas Table 7: Using midpoints to Approximate p In the first few examples. Published by Pearson Learning Solutions.02/100 = 0. then area is measured in square feet 1ft 22. That is. However proceeding as above. the convergence is slow. Suppose the interval [a. We see that while there is some improvement as n.12 + f1b22 ¢ x and the sum using the midpoint. with n = 8 (a relatively small value).785484 and p L 41. that is. and Finance. Economics. in the next section. then the width of each rectangle is ¢ x = 11 . that these sums may be written more succinctly using some simple notation. the sum converged quickly. use 100 rectangles. n will need to be very large before the approximation improves significantly. Walter O. Inc. . x2. because of its size (but is easily done on your computer). Notes (1) We could actually write out a formula for the sums that we considered. then Applied Calculus for Business. whose endpoints are a = x0. Using the midpoint of each subinterval. (2) Area is the product of two dimensions. and April Allen Materowski. we obtained a very good approximation. It will ultimately get better.1. x3. b] is partitioned into n equal subintervals. Clearly. How large should n be to get better accuracy in approximating p? If we take n = 100. still considerably inaccurate. Thus. Gordon.432 * ** Section 5. in this example convergence is considerably slower. x1. we can easily generate the spreadsheet which we do not show here. for the area under the parabola. if they are measures in meters. Mn is Mn = a f A a + x1 2 (1) (2) B + fA x1 + x2 2 x + b B + Á + f A n . if these dimensions are each measured in feet. we find M100 = 0.01. Published by Pearson Learning Solutions. Therefore. determine the x-values that form the partition of the interval. approximate the area of the region bounded by f1x2 = 4 . Using the left endpoint with n = 4.x 2. 9 Fig. and the x-axis for x between 0 and 2. as we did throughout this section. 15: Ex.4 Approximation of Areas * ** 433 the area is measured in square meters 1m22. Calculator Tips EXERCISE SET 5.x2 x Fig. Suppose you are estimating the area bounded by some curve and the x-axis. Copyright © 2007 by Pearson Education. by Warren B.2 and 2. see Figure 16. Instead of doing this. Wang. While the calculator may be used to generate the areas of the rectangles. we shall show. Suppose you are estimating the area bounded by some curve and the x-axis.x 2. Economics. the first column would be the x-values used in each subinterval. determine the x-values that form the partition of the interval. Much of what we did in this section could be done using the Table feature on the calculator. If you are using n = 4 rectangles. 6 Applied Calculus for Business. 5 6. Inc. the x-values needed in determining the heights of the rectangles. see Figure 17 f(x) = 4 . and Finance. Walter O. in the next section. If you are using n = 4 rectangles. approximate the area of the region bounded by f1x2 = 4 . . (c) 100. 14 Ex. 5. between the lines x = 8 and x = 11. approximate the area of the region bounded by f1x2 = x2 + 2. 4. and so on. 16: Ex. see Figure 18 f(x) = 4 . While we have omitted the units of area (and will continue to do so). and the x-axis for x between 0 and 2. 17: Ex. 2. 3. f(x) = x2 + 2 7.18: Ex. and the x-axis for x between . the Table will not easily compute the sums of its columns. 8 x Fig. Using the middle point with n = 4. that is. Determine the width of each rectangle if the number of rectangles to be used in the approximation is (a) 4. f(x) = x2 + 2 9. f(x) = x2 + 2 x Fig.2 and 2. (b) 10. a much better way of obtaining the areas with the calculator using Sigma notation. and the x-axis for x between . it is considerably easier to use a spreadsheet. the x-values needed in determining the heights of the rectangles. approximate the area of the region bounded by f1x2 = x2 + 2. Determine the width of each rectangle if the number of rectangles to be used in the approximation is (a) 4.4 1. see Figure 15.x2 Fig. (c) 100. and April Allen Materowski. Suppose you are estimating the area bounded by some curve and the x-axis. (b) 10. Using the right endpoint with n = 4.Section 5. it is important to keep them in mind. see Figure 14. that is. between the lines x = 3 and x = 5. Using the left endpoint with n = 4. and the x-axis for x between 0 and 2. 7 8. Suppose you are estimating the area bounded by some curve and the x-axis. Gordon. However. between the lines x = 3 and x = 5. between the lines x = 0 and x = 2. the second column the y-values. whatever units are used for measurement of length. area is measured in the square of these units. approximate the area of the region bounded by f1x2 = x2 + 2. Using the right endpoint with n = 4. and the third column the area of the rectangle. f1x2 = 3x + 2. b1 h b2 Figure 20: Area of a Trapezoid is 1/2h1b1 + b22 In Exercises 28 32. (b) the right endpoint in each subinterval and (c) the midpoint of each subinterval. and April Allen Materowski. f x) = 4 . and the x-axis for 0 x 2. it is assumed you will be using a spreadsheet. (b) right endpoint. Copyright © 2007 by Pearson Education. . 12. and the x-axis for 1 x 4. Wang. (b) the right endpoint in each subinterval and (c) the midpoint of each subinterval. (b) the right endpoint in each subinterval and (c) the midpoint of each subinterval. between x = 0 and x = 1. and the x-axis for 0 x 1. using n = 100 rectangles and (a) the left endpoint in each subinterval. f1x2 = x2. 30. Using the middle point with n = 4. (b) right endpoint. 10 In Exercises 11 14. 20. 28. approximate the indicated area by connecting succesive yvalues on the curve by a straight line. 17. (c) midpoint. Fig.4 Approximation of Areas 23. (b) right endpoint. f1x2 = x3. In the remaining exercises. the x-axis and between the given lines. 22. Estimate the area bounded by f1x2 = ln x. Do Exercise 13 with n = 8. Published by Pearson Learning Solutions. using n = 100 rectangles and (a) the left endpoint in each subinterval. Use n = 4 trapezoids to estimate the area of the region in Exercise 10. (c) midpoint.2 and 2. (c) midpoint. n = 4 using the (a) left endpoint. Do Exercise 12 with n = 16. by Warren B. 16. Applied Calculus for Business. without using a spreadsheet. 21. Do Exercise 13 with n = 16. between x = 4 and x = 5.434 * ** Section 5. (b) the right endpoint in each subinterval and (c) the midpoint of each subinterval. 15. 26.19: Ex. Do Exercise 11 with n = 16. Estimate the area bounded by f1x2 = ex. Gordon. 19. Show that the approximation of the area obtained by using trapezoids is equal to the average of the approximations using the left and right endpoints. 27.x2 24. and the x-axis for 10. 18. Use n = 4 trapezoids to estimate the area of the region in Exercise 12.2x + 3. 32. 11. using n = 50 rectangles and (a) the left endpoint in each subinterval. 1 + x2 2. Do Exercise 11 with n = 8. determine the approximate area of the region bounded by the given curve. Redo Exercise 13 with n = 100 rectangles. resulting in trapezoids. Walter O. f1x2 = . Inc. 25.x 2. Recall that the area of a trapezoid with bases b1 and b2 and height h is 1/2h1b1 + b22. and the x-axis for x between . see Figure 19. see Figure 20. Economics. Do Exercise 10 with n = 16. 14. 29. Do Exercise 12 with n = 8. n = 4 using the (a) left endpoint. Use n = 4 trapezoids to estimate the area of the region in Exercise 13. Estimate the area bounded by f1x2 = x211 . (c) midpoint. using n = 25 rectangles and (a) the left endpoint in each subinterval. between x = 0 and x = 1. Estimate the area bounded by f1x2 = 1 x 1 . (b) right endpoint. and Finance. n = 4 using the (a) left endpoint. 31. using the indicated number n of rectangles. 13. n = 4 using the (a) left endpoint. between x = 2 and x = 3. approximate the area of the region bounded by f1x2 = 4 . Do Exercise 10 with n = 8.x2. Use n = 4 trapezoids to estimate the area of the region in Exercise 11. then the average or mean of this set is the sum of these numbers divided by n. replacing the index i with 3 to begin with. if you have a set of n data points. they delete them and write x = a xj n Applied Calculus for Business. you have a variable that is being looped (in this case summed). consider 6 i=3 2 a 2i 2 We are to sum 2i . Economics. Sometimes. Published by Pearson Learning Solutions. that is. by Warren B. We write 30 k=1 Sigma Notation a xk The Greek letter sigma 1 © 2 tells us to sum. usually. Gordon. Inc. It is the index letter that is being replaced by a number when creating the terms to be summed. until i is 6. when the starting and ending values are understood. we now write this average. we begin with the value of k = 1 and we stop with the value of k = 30.Section 5. perhaps you have seen its use in statistics. . Using sigma notation. the sample mean is often written this way. the expression we are summing is xk. Copyright © 2007 by Pearson Education. x3. x2. and April Allen Materowski.5 Sigma Notation and Areas » » » » » » Sigma Notation Linearity Property Summation Formulas Reimann Sums Areas by Riemann Sums Calculator Tips Suppose you have a set of 30 data points. its index. For example. we could have used any letter we desired for the index. Walter O. j. 5x1. and continuing with the next integer value. or k. That is. as n j=1 a xj n x = In statistics. Á . 2 2 2 2 2 a 2i = 2132 + 2142 + 2152 + 2162 6 i=3 Sigma notation is used extensively in mathematics. In the above example. Á + x30 a xk = x1 + x2 + x3 + 30 k=1 This is similar to the loop command in most programming languages. xn6. x2. Á . then you would write x1 + x2 + x3 + Á + x30 Mathematicians have developed a shorthand method for indicating such sums. indicated by x. For example. x3. say 5x1. we use i.5 Sigma Notation and Areas * ** 435 5. and starting and ending values are specified. Wang. and Finance. k is called the summation index. x306 and you wanted to indicate their sum. it means we first evaluate this function F at 1 (the index s starting point). or do the repetitive summation on a spreadsheet. we have the following: n Summation Formulas k=1 ak = n1n + 12 2 (2) Applied Calculus for Business. we shall either use the summation command on our calculator. One simple property of sigma notation. For example. We could now write sx = a x2 k k=1 n When we write n k=1 a F1k2 where F is some function of the index k. Gordon. by Warren B. . For example. Walter O. We arbitrarily chose 1 as the starting value of the index. When studying regression we had the expression sx which was the sum of the squares of the n data points. and April Allen Materowski. and Finance. Statisticians also use sigma notation to indicate the sample variance. Published by Pearson Learning Solutions. but it could be any number. then at 2.436 * ** Section 5. At this point it does not matter what the terms in the sum mean. then sum these n values. Copyright © 2007 by Pearson Education. sometimes called the linearity property. We will be studying areas of regions. 2 2 a 2i = 2 a i i=3 6 6 i=3 and 3 2 3 2 a 13k . we are just interested in understanding the summation process.5 Sigma Notation and Areas We will always indicate the index with its starting and ending values.4 a k k = -2 k = -2 n n n k = -2 It also turns out that sum of powers of integers can be easily computed by simple formulas. We shall see that when n is a large number we will rarely do the computations manually. Wang. Economics. It will turn out that we will be frequently dealing with sums of the form n i=1 a f1ci2 ¢ x which translates into Á + f1cn2 ¢ x a f1ci2 ¢ x = f1c12 ¢ x + f1c22 ¢ x + f1c32 ¢ x + n i=1 Notice the index only affects the function f (the ¢ x has no index attached to it). until F is evaluated at n (the final value of the index).4k 2 = 3 a k . is Linearity Property n i=1 n i=1 n a 1aF1i2 + bG1i22 = a a F1i2 + b a G1i2 i=1 (1) where a and b are any constants. Inc. and continuing at consecutive integer values. Walter O. Inc.235 30 Even a simple problem like this last example could be done quickly on a spreadsheet. and April Allen Materowski.12 + n S = n + 1n .122 + 11 + n2 notice each sum on the right is n + 1. Applied Calculus for Business.Section 5. 2S = n1n + 12 or S = n1n + 12/2 We indicate in the exercises how (3) and (4) may be obtained. that is.) We illustrate at the end of this section how to do such sums on your calculator. Economics. Gordon.5 2 ak = n n Sigma Notation and Areas * ** 437 k=1 n1n + 1212n + 12 6 n21n + 122 4 (3) k=1 3 ak = (4) Note that (2) says that the sum of the first n integers is the product of the last integer and. Suppose we let S be the sum. S = 1 + 2 + Á + 1n . (See Table 1. so we have. divided by 2. we write it out and then write it again. and Finance. For example. 2S = 1n + 12 + 1n + 12 + 1n + 12 + Á + 1n + 12 1n + 12 occurs exactly n times on the right hand side. by Warren B. .12 + 22 + Á .12 + Á + 2 + 1 If we add these two identical sums. Copyright © 2007 by Pearson Education. 99 k=1 ak = 99 # 100 = 4950 2 It s not hard to prove (1). beginning with the last term continuing until the first. we obtain 2S = 1n + 12 + 11n . Published by Pearson Learning Solutions. one more than the last integer. backwards. k=1 30 Solution Using (2) and (3). Wang. Example 1 Compute a 12k2 + 5k2.12 + 1n . we have 2 2 a 12k + 5k2 = 2 a k + 5 a k = 2 k=1 30 30 k=1 30130 + 1212 # 30 + 12 30130 + 12 + 5 = 6 2 k=1 10 # 31 # 61 + 751312 = 21. which is why we have to subtract the second sum on the right from the first. Inc. the sum must begin at 1. by Warren B. = SUM1C4 : C332 In order to use these formulas (2). So we have. 10 i=5 a i. Wang.ai Do you see why we subtract? The first sum on the right included first four extra terms missing from the original sum.438 * ** Section 5. . we enter in this cell. Copyright © 2007 by Pearson Education. we enter in this cell. we can easily rectify the situation. then we enter in the cell under it 1B52 = B4 + 1. Consider the following example. = 101112/2 . and April Allen Materowski. We copy the content of B5 into all the cells beneath it until we reach 30. = 2*B4¿2 + 5*B4 and then copy its contents to all the cells in its column. Published by Pearson Learning Solutions. i=5 Solution We write 10 i=5 10 i=1 4 i=1 ai = ai . Gordon. Economics. (3) and (4). our stopping value in cell B33. To obtain the sum in Cell C34. and Finance. When it does not. Example 2 10 Compute a i.4152/2 = 45 Applied Calculus for Business. To get the result in C4. so we need to exclude them. Walter O.5 Sigma Notation and Areas Table 1: Computing a 12k 2 + 5k2 on a Spreadsheet k=1 30 Once the 1 is entered in cell B4. we obtain an approximation A n to the area of the region. How might we go about doing that? The answer appears obvious: do exactly what we did on the examples in the previous section and let n. Á .Section 5. This is where the calculus comes into play. The corresponding y-values will be f1c12. become infinite. you could rewrite the sum as a 17 + 0i2. Published by Pearson Learning Solutions. We saw in the last section that while it is an easy matter to use larger and larger spreadsheets to improve upon the approximation in computing the area of a region. Before we do that. Copyright © 2007 by Pearson Education. if we sum the areas of these n rectangles. n : q n3 k =1 Evaluate lim Solution We first evaluate the sum and then the limit becomes a secondary problem. we will need to take a limit. so when i=1 i = 1. the number of rectangles. respectively. Example 3 1 n 2 ak . 1 n 2 1 n1n + 1212n + 12 2n3 + 3n2 + n 2n3 1 k = lim 3 = lim = lim = a 3 3 n: q n k=1 n: q n n: q n : q 6n3 6 3 6n lim Note that the procedure for finding the limit as n : q is done exactly as when the variable is x. f1c22 ¢ x. We shall now use it to help us represent the area of a region. Let us call c1 the x-value used in the first subinterval. we need to take a limit. While we considered examples using the left. Economics. we have n k=1 a C = Cn (5) We shall see that to compute the area of a region using sums. when i = 2. and so on. we would like to obtain the exact area. Á . and Finance. 4 Alternately. We could proceed without it. using sigma notation. and the area of the n rectangles are f1c12 ¢ x. Gordon. We shall assume that f is a non-negative continuous function (we ll explain why later). by Warren B. Walter O. Wang. thus the sum is 7142 = 28. and April Allen Materowski. which produces the same result. That is. n: q lim f1n2 = lim f1x2 x: q Sigma notation is nothing more than a shorthand. Riemann Sums Applied Calculus for Business. right or midpoint of each subinterval. Inc. namely. f1cn2. we saw that it really does not matter which point we used in each subinterval to construct the top of the rectangles: we could use any point. c2 the x-value in the second interval. but it makes the writing of expressions involving sums much easier. . That is. We have n k=1 ak 2 n1n + 1212n + 12 6 therefore. f1cn2 ¢ x. we have 7. Therefore.5 4 Sigma Notation and Areas * ** 439 What does the sum a 7 mean? There is no index dependent term in the sum. giving cn the x-value in the last (nth) interval. and so on for 3 and 4. f1c22. we have 7. i=1 More generally. We let the width of each rectangle be ¢ x. we first need to write the sums of the rectangles slightly differently. The next example illustrates how this is done. of the region. A x=a x=b Figure 1: The Area A of the Region Bounded by y = f1x2. 1]. That is. and Finance.5 Sigma Notation and Areas A n = f1c12 ¢ x + f1c22 ¢ x + Á + f1cn2 ¢ x We can express this sum using sigma notation as n A n = a f1ck2 ¢ x k=1 The sum is called a Riemann sum. A. ¢ x is equal to the length of the interval 1b . called a regular partition. the lines x = 0 and x = 1. c n . Economics. the number of rectangles we use. Á .1 n . [1/n.440 * ** Section 5. Copyright © 2007 by Pearson Education. In using (6).a n (7) Example 4 Find the area of the region bounded by the curve f1x2 = x2. Published by Pearson Learning Solutions. and April Allen Materowski. [2/n. We saw that this could be accomplished by letting the number of rectangles become infinite. Inc. ¢x = b . . d n n Applied Calculus for Business. 2/n]. Wang. by Warren B.02/n = 1/n. then the n subintervals are: [0. it is usually easiest to use only the left or only the right endpoint in each subinterval. Now we want to obtain the exact area. 1/n]. the x-axis. the x-axis and the Lines x = a and x = b Note that if we partition the interval from a to b into n subintervals of the same length. 3/n].a2 divided by n. We illustrate with a few examples. Walter O. Gordon. Mathematically we write this as Areas by Riemann Sums n A = lim A n = lim a f1ck2 ¢ x n: q n: q k=1 (6) We shall illustrate how (6) is used in computing the exact area of a region like the one illustrated in Figure 1. Solution We partition the interval [0. named after the mathematician Georg Friedrich Bernhard Riemann (1826 1866). using ¢ x = 11 . Solution Suppose we choose the left endpoint of each subinterval. f11/n2. as n k 21 An = a a b n k=1 n 2 To obtain the area. f13/n2. f11n . f12/n2. by Warren B. Gordon. Then the corresponding y-values are f11/n2. Example 5 Redo the previous example using the left endpoint of each subinterval.12/n2 f10/n2 = 0 f11/n2 = 11/n22 f12/n2 = 12/n22 Applied Calculus for Business. this limit may be computed fairly easily as follows: n n k2 1 n k 21 = lim a 3 = lim 3 a k2 = A = lim a a b n: q k=1 n n n: q k=1 n n: q n k=1 3 2 1 n1n + 1212n + 12 2n + 3n + 1 2n3 1 = lim = lim = lim 3 3 3 q q n: q n n : n : 6 3 6n 6n (Note that (3) was used in evaluating the sum. n k 21 A = lim a a b n: q k=1 n n As we saw above. f1n/n2 Now note that f1x2 = x .5 Sigma Notation and Areas * ** 441 (Note that n/n = 1 but let s keep the pattern) Suppose we choose the right endpoint of each subinterval. we let n become infinite. Inc. . so in this particular example. Economics. Published by Pearson Learning Solutions. and Finance. and April Allen Materowski. then the corresponding y-values are f10/n2. f12/n2. A. Copyright © 2007 by Pearson Education. ¢ x = 1/n) f11/n2 ¢ x = 11/n221/n f12/n2 ¢ x = 12/n221/n f13/n2 ¢ x = 13/n221/n o f1n/n2 ¢ x = 1n/n221/n Therefore. A n = 11/n221/n + 12/n221/n + 13/n221/n + Á + 1n/n221/n We may write this sum using sigma notation. Á . Walter O.) Therefore. Wang. f13/n2. the area of the region is 1/3. Á . that is.Section 5. the y-values are f11/n2 = 11/n22 f12/n2 = 12/n22 f13/n2 = 13/n22 o f1n/n2 = 1n/n22 So the areas of the rectangles are (remember. ) You will note that we have assumed that the function whose area we are computing is nonnegative. Gordon.42 and the x-axis with x between 0 and 4. If we choose the right end points. 4/n. . The subinterval endpoints would then be 0.12/n22 So the areas of the rectangles are (omitting the first rectangle whose area is 0) f11/n2 ¢ x = 11/n221/n f12/n2 ¢ x = 12/n221/n o f11n . . Inc. see Figure 2. by Warren B. Copyright © 2007 by Pearson Education. and April Allen Materowski. .12/n2 ¢ x = 11n .12n12n . . Published by Pearson Learning Solutions. we obtain for the y-values. What would happen if we consider a Riemann sum on a function that lies below the x-axis? For example.02/n = 4/n. Economics.4x = x1x .12/n221/n Therefore. Figure 2: Area of Region Bounded by f 1x2 = x 2 . A n = 11/n221/n + 12/n221/n + 13/n221/n + n-1 k 21 An = a a b n k=1 n n-1 2 n-1 2 k k 1 = lim a 3 = A = lim a a b n: q k=1 n n n: q k=1 n 1 1n . with n . and Finance. Wang.12 2n3 . Applied Calculus for Business.3n2 = lim lim n: q n: q 6n3 6n3 Á + 11n .5 Sigma Notation and Areas o f11n . 3(4/n).12 + 12 = lim + n = lim 2n3 1 = 3 n : q 6n 3 (Note that (3) was again used in evaluating the sum. 4((4/n)). n(4/n).442 * ** Section 5.1211n .1 as the last term in the sum.12 + 12121n lim 3 n: q n 6 1n . 2(4/n). 4] as above and obtain ¢ x = 14 . what if we used the region bounded by f1x2 = x2 .4x and the x-axis for x between 0 and 4 We can partition the interval [0. . Walter O.12/n2 = 11n .12/n221/n 1 n-1 2 ak = n : q n3 k =1 . 4 b n n Therefore. Gordon. we have n 64 n 64 n lim a f1ck2 ¢ x = lim 3 a k2 . with the y-values differing by a factor of . by Warren B. Figure 3: Area of Region Bounded by f 1x2 = .4b = . f1ck2 ¢ x = k # 4 4 4 64k2 64k ak # .5 Sigma Notation and Areas * ** 443 f1c12 = f114/n2 = 14/n2114/n .lim 2 = q q n: n n: n 6 2 64n3 + 96n2 + 32n 32n2 + n lim = lim n: q n: q 3n3 n2 64n3 32n2 64 32 lim = . When the function is non-negative.1x 2 . we have. the Riemann sums determine the area of the region. Wang. and the x-axis for x between 0 and 4 is 32/3. The area under this curve f1x2 = . Copyright © 2007 by Pearson Education. and April Allen Materowski. Inc. If we reflect the given curve about the x-axis. so we should have anticipated this.1x2 .42 o 4 4 f1ck2 = f1k14/n22 = k # a k # .lim 2 a k = n: q k=1 n: q n k=1 n: q n k=1 64 n1n + 1212n + 12 64 n1n + 12 lim 3 .42 f1c32 = f1314/n22 = 213/n21314/n2 . .1.2 3 n n n n n and we have as the Riemann sum n 64k2 64k 64 n 2 64 n f 1 c 2 ¢ x = = k k a ak a 3 a n2 n3 k n2 k k=1 k=1 n =1 =1 n If we now let n become infinite. we obtain the congruent region given in Figure 3.4x2. and Finance. The computations are identical to the above.42 f1c22 = f1214/n22 = 214/n21214/n2 . Published by Pearson Learning Solutions.4x2 and the x-axis for x between 0 and 4 Applied Calculus for Business. Walter O. Economics.32 = lim n : q 3n3 n : q n2 3 3 We obtain a negative result! This is not surprising as all the y-values are beneath the x-axis. We can give a simple interpretation of this result.Section 5. 5 Sigma Notation and Areas When the function is negative. See Figure 5. For example. We can compute sums very easily with the TI 89. Another approach that is sometimes useful is to choose an irregular partition. the Riemann sum will be the sum of the signed areas. press F3 and scroll to 4 and press Enter. index variable. how we may solve such problems using integration. the only change is in the sign of the y-values. but the length of each subinterval approaches zero as n becomes infinite. the Riemann sums give the negative of the area. you can choose any letter. Consider the following problem: Given f1x2 = x3. 1] Calculator Tips zero. What will be the value of n f1ck2 ¢ x? A little thought should convince you that the limit of this sum will be lim n: q a k=1 Figure 5: f 1x2 = x3 on [ . Economics. Alternately. we were able to compute the sums using the formulas given in this section. A1 A2 A3 n Figure 4: lim a f 1ck2 ¢ x When the Function Changes Sign n: q k=1 Let A 1. the y-values to the left of the origin are negative. Thus. one in which the subintervals have different lengths. j or k for the index variable. ending value2 While we usually use i. 1] into n equal subintervals. and Finance. . Fortunately. as inn dicated. Therefore. see Figure 6. while the y-value to the right of the origin are positive. Selecting g (sum produces g ( on the input line.1. starting value. for a function that may assume both positive and negative values. A 1 . 1] and compute lim a f1ck2 ¢ x. For example. The areas of each region are the same. by Warren B. Suppose we partition the interval [ . Walter O. Inc. suppose we form a regular partition on the interval [ . they cancel.A 3 + A 2. Gordon. t=5 20 Applied Calculus for Business. to compute a 3t4. A 2. j or k. consider Figure 4. What do we do if we have a function that results in a sum we do not recognize? We shall discover in the next section. Published by Pearson Learning Solutions. Then n: q k=1 we will obtain as our result. Do you see why? This function is symmetric with respect to the origin. and April Allen Materowski. We leave the consideration of irregular partitions to the exercises. Copyright © 2007 by Pearson Education. but since the limit of the Riemann sums gives the sum of the signed areas.1. they are now negative nothing else has changed.2. Its syntax is as follows: a 1expression.444 * ** Section 5. and A 3 be the areas of the regions between the curve and x-axis. you need to Figure 6: Sums on the TI 89 first use the Alpha key). Make it easier for yourself when using your calculator and choose t (to use i. Wang. The calculator has in its Catalog g (sum (found directly under sum which we do not use). (c) the first n integers. by Warren B. (b) the first 999 integers. (a) If we let a = x0 and b = xn. Find the area of the region bounded by f1x2 = x 3 and the x-axis.2i2 + 11i . a 1ak + 1 . 1 + 2 + 4 + 8 + Á + 2048 7. between (a) x = 1 and 2.1 x3 xn x2 8. 41.3j52 j=1 In Exercises 26 29.a k . 40. Gordon. 36. (c) x = .2k2. between x = 1 and x = 2. lim 28. compute the given limit. (c) x = .2k6 + 7k32 k = 12 18 25.2k2.1) Applied Calculus for Business. Find the area of the region bounded by f1x2 = 3x + 1 and the x-axis. a 12 j=1 20 k 29. Determine the region whose area is given by Exercise 27. Determine a Riemann sum that would determine the area of the region bounded by f1x2 = ex and the x-axis.1 and 0. Determine the region whose area is given by Exercise 26. and Finance. evaluate the given sum 5 k= 1 4 22.1 and 1. k= 1 5 39. . 33. between x = 1 and 2 (do not evaluate this limit). a 12 + 5j 2 . show that formulas (1). (b) x = . Determine the region whose area is given by Exercise 29.5 In Exercises 1 8 rewrite the given sum using sigma notation. 31.12 10. a 14i3 . Economics. a 13k2 . Inc.2k2.5 Sigma Notation and Areas * ** 445 EXERCISE SET 5. Compute the following sum. # + # + # + Á + # 2 2 2 3 2 19 2 1 4. (b) a 12k + 1 . 1 + 1/2 + 1/3 + Á + 1/20 2.3k2 k= 1 27 18. between x = 0 and 1 (do not evaluate this limit). a 1j1j . Walter O.2i42 i=1 53 23 24. While we gave only three simple formulas for computing sums. Determine the region whose area is given by Exercise 28. 34. a i=3 i + 1 8 k= 1 20 2 n: q 27. (d) x = . a 1k1k . (c) a 12k + 1 . lim n: q 30. Compute (a) a 12k + 1 .a k . 23. In Exercises 34 39 use Riemann sums and the limit to find the area.a k + a k . Find the area of the region bounded by f1x2 = x3 and the x-axis. lim 4 n i3 a n3 n: q n i =1 n: q 2 n i 2 a3 + a b b n ia n =1 5 n i 3 a1 + b n ia n =1 12. Find the area of the region bounded by f1x2 = 3x + 1 and the x-axis. + + + + 5 7 9 11 13 6.1 and 1. a 11 + 1 . between (a) x = 1 and x = 2.a k . 15. between x = 0 and x = 1.) n 20. a + ar + ar2 + Á + ar n . Why do you think it is called a k=1 telescoping sum? n 21. 32. Published by Pearson Learning Solutions.1 = a k + 1 . use the formulas provided in this section to compute the given sum. between x = 0 and 2. (2) and (3) of the previous section for left. Find the area of the region bounded by f1x2 = 2x 2 and the x-axis.1 and 0.Section 5. 35. a 13j2 . and April Allen Materowski. 42. (b) x = . 20 n k=1 k= 1 14.12 (Hint: write k=1 a k + 1 . 1 + 6 + 11 + 16 + 21 2 3 4 5 1 5. there are many more in your calculator s memory.2 and 1. Use your calculator to compute each of the following sums. lim 3 n i2 2 n ia =1 n 9. Determine a Riemann sum that would determine the area of the region bounded by f1x2 = ln x and the x-axis. a 1ak + 1 .a k2.122 i=1 35 17. 5 + 7 + 9 + 11 + Á + 47 1 1 1 1 3. a 12k . a 12k3 . Find the area of the region bounded by f1x2 = 2x 2 and the x-axis. 1. between x = 0 and x = 1.12 + 32 j=1 7 i 11. right and middle sums may be written as (Write out the sums and observe what happens to the terms. 26.2j + 42 j=1 30 20 16. 38. a 12 j=2 In Exercises 15 18. Compute the following sum. 37. x + + + Á + 2 3 n In Exercises 9 14. Wang. Find the average of (a) the first 99 integers.12 2 13. Copyright © 2007 by Pearson Education. a 13i5 .12 + 2k32 k= 1 19. (d) Compute this limit by recalling that near x x:0 e .12 ¢ x k= 1 n that is. then there is no way the Riemann sum could represent the area of the region (why?).446 * ** Section 5. if it approaches 0. Complete the details and find S2. by Warren B. and so on.12/n2 (b) Show that for this partition. Inc. the width of each subinterval.12 lim x e1/n 1/n n : q n1e . Wang. Suppose we let Sp = a kp. Copyright © 2007 by Pearson Education. [4/n2. using the above three sums. what sum do you 1 n 1 2 obtain? (c) comparing (a) and (b) show that lim 3 a k2 = n: q 2 k= 1 3 n 49. the length of every subinterval goes to 0 as n goes to infinity. 43. and by Exercise 20. (c) If we let x = 1/n show that this limit is equivalent to xe . 1/n2]..k 2 = n n n n n where ci could be any point in the ith subinterval. 44. for 0 x 1.. When p = 1. e L 1 + x. Applied Calculus for Business.122/n2. and April Allen Materowski. A = 1e . and Finance.12 . i 2/n2]. the x-axis. 1n .k3 = 1 + 3k + 3k2 k= 1 n 3 3 a 111 + k2 . Walter O. 45. Given f 1x2 = 1 3 x. 46. between x = 0 and 1 n x = 1. Gordon. We have already established the formula in the case p = 1. Use the partition of exercise 46 and the right endpoint of each subinterval. Economics. Hint: Find an irregular partition that will enable you to deter0 mine this area: note.12 lim n We defined a Riemann sum as a f1ci2 ¢ x where we partitioned the interval i=1 [a.1 x x = 0. so we may now solve for S2. determine (a) S3. 47. This exercise shows how to obtain formulas for Sp for positive integer values of p. k=1 we may proceed as follows: 11 + k23 . b] into n equal subintervals. (c) S5. and we have n k=1 n 2 a 1 + a 3k + a 3k = k=1 k=1 n n + 3 a k + n + 3 a k2 = n + 3S1 + 3S2 k= 1 k= 1 f1ci2 ¢ x lim 7 ¢7 : 0 a i =1 (8) as the generalization to (7). we require the width of each rectangle to approach 0. Note that as n. the area bounded by f1x2 = ex and the x-axis. (b) Supposed you used a regular partition. ¢ x.. the number of subintervals. Let ¢ x1 be the width of the first subinterval. Published by Pearson Learning Solutions. n2/n2] (a) Show that ¢ xi = 12i . . we obtain (2).1 + xk 2 B ¢x (b) Write three different expressions for the area bounded by y = f1x2. (b) S4. becomes infinite. (a) Find the area bounded by the curve f1x2 = 1x and the x-axis. that is 7 ¢ 7 : 0 is equivalent to n : q . we k=1 obtain (3). 4/n2]. and ¢ xn be the width of the nth subinterval. why? (b) Suppose you used a regular partition. then the length of every subinterval will approach zero (why?).5 Sigma Notation and Areas n Ln = a f1xk . Describe a partition where n : q but 7 ¢ 7 does not approach 0. 1/n3. Show that using the right endpoints. The definition of a Riemann sum is now generalized to n i=1 a f1ci2 ¢ xi The next step is to take the limit. approaches zero. [1/n2. and the lines x = a. [1i . Á . is given by (a) A = lim a ek/n. that the endpoints should be: 0. Á . 48. ¢ x2 be the width of the second subinterval.122/n2. what sum do you obtain? (c) Comparing (a) and 1 n 1 3 (b) show that lim 4 a k3 = . There is an easier way of saying this: let 7 ¢ 7 be the length of the largest of all the subintervals. We know the value of S1. . (b) Observe that the given sum is n: q n k=1 a geometric sum and it follows that A = 1e . Using the method of the previous exercise. Consider the following irregular partition of the interval [0. So we could write n i=1 n Rn = a f1xk2 ¢ x k= 1 lim f1ci2 ¢ x = lim a f1ci2 ¢ x n: q a ¢x : 0 i=1 (7) and n Mn = a f A k=1 xk . In order to exhaust the region under the curve. n: q 3 k= 1 4 n The first sum on the left is a telescoping sum with ak = k3. and now wish to find S2 = a k2. We could divide the interval into an irregular partition where the subintervals have different lengths. we used a regular partition. (a) find the area bounded by this curve and the x-axis for x 1. when p = 2. 9/n2]. If there were some subintervals whose width did not approach zero. and so on. it is easily determined. and x = b. 8n3. 1]: [0. Gordon. Then we define b a 3 f1x2 dx.) Now we sum the infinite number of the areas of such rectangles beginning at a and ending at b. we have the symbol tinuous on the interval [a. each of whose width is approaching zero. what we do is to find the sum of the area of an infinite number of such thin rectangles. Walter O. In motivating this notation. and we indicate the a and b to show where we start b Definite Integral and end the sums. Suppose we call the width of such a representative rectangle the differential dx and its height f(x). and Finance. but it does not have to be. by Warren B. Inc. a is called the lower limit of in- tegration and b is called the upper limit of integration. and April Allen Materowski. Copyright © 2007 by Pearson Education. Economics. b]. which is an elongated S to indicate we are summing an infinite number of rectangles.6 The Definite Integral » » » » Definite Integral Fundamental Theorem of Integral Calculus Basic Properties Calculator Tips In the previous section we saw that if f is a non-negative continuous function. we saw that this limit then gives the sum of the signed areas of the component regions. It is no coincidence that we use the Applied Calculus for Business.Section 5. We draw an infinitesimally thin representative rectangle. More formally. Then the (differential) area of this representative rectangle is f(x)*dx (see Figure 1. . Thus. th f(x) x=a dx x=b Figure 1: A Representative Rectangle with Area f(x)dx In essence. Wang. Let us define a more convenient notation to represent this limit.6 The Definite Integral * ** 447 5. When the function is negative on part or all of the interval. then the area of the region bound by the curve and the x-axis between x = a and x = b is given by n lim f1ck2 ¢ x n: q a k=1 where ck is any point in the k subinterval. suppose f is con- n a b 3 f1x2 dx = lim a f1ck2 ¢ x n: q k=1 (1) The symbol a 3 f1x2 dx is called a definite integral. Published by Pearson Learning Solutions. we assume the function is non-negative. We use the integral symbol. Walter O. Therefore. as we shall see. if we can integrate f. Inc. In any event. over the interval [a. When the function is negative on part or all of the interval. the notions of area and Riemann sums are nothing more than simple geometry: adding up the areas of many rectangles. Thus. the definite integral is a number which depends on the sign of f. the Fundamental Theorem reduces the summation and limit process into one of integration. b]. We move into the realm of calculus when we consider the limit as n becomes infinite. and April Allen Materowski. we see that if f is non-negative on [a. but since the theorem states we may use any antiderivative. by Warren B.448 * ** Section 5. Another notation will be convenient. we can compute the definite integral. Can we use the calculus to help us compute (1) in a simpler way? The answer to this question results in one of the most beautiful results in the calculus: the so-called Fundamental Theorem of Integral Calculus. then the symbol gives the sum of the signed areas of the appropriate regions.F1a2 a b 3 f1x2 dx = F1x2 b a = F1b2 . Example 1 2 val [a. Thus. Copyright © 2007 by Pearson Education. FUNDAMENTAL THEOREM Suppose f is continuous on [a. Basically. so we choose 3 3 Solution L x2 dx = Applied Calculus for Business. then we have F1x2 = L f1x2 dx Of course there is an arbitrary constant of integration c in the evaluation of this integral. we can choose c = 0. If the function is negative over some or all of the interval. Then b a 3 f1x2 dx = F1b2 . If F is an antiderivative of f. Gordon. Let F be any antiderivative of f. but may have some other interpretation. some remarks and examples. Wang. Published by Pearson Learning Solutions. x3 x3 as the antiderivative we will use. this theorem says that if you can integrate f. the number no longer represents an area. b].F1a2 (2) We shall prove this theorem at the end of the section. b] then f1x2 dx represents the area 3 a of the region bounded by the curve and the x-axis over this interval. then (1) is easily determined. b] then Compute 1 3 x2 dx. we let F1x2 so we may rewrite the Theorem as b b a = F1b2 . and Finance.6 The Definite Integral Fundamental Theorem of Integral Calculus same symbol to represent an infinite sum of rectangles as we used to represent an antib derivative.F1a2 It then follows from the Fundamental Theorem that if f is non-negative on the interf1x2 dx is a positive number representing the area of the region 3 a bounded by the curve and the x-axis. First. . Basically. this number may or may not represent an area. Economics. + c. 6 The Definite Integral * ** 449 We have. 2 1 3 x 2 dx = 1223 1123 x3 2 8 1 7 ` = = = 3 1 3 3 3 3 3 Did you notice that the definite integral in the previous example represents the area of the region bounded by f1x2 = x2 and the x-axis between x = 1 and 2? This is the first example we considered in Section 3. Example 2 (a) Interpret as an area the integral 3 2 3 x3 dx. and April Allen Materowski. Economics. Walter O. Wang. . as the constant will cancel in the calculations. x3 Suppose you chose as the antiderivative + 1 in the previous example. The region is illustrated in Figure 2. we choose c = 0 to make the computations simplest. That s why the theorem states any antiderivative will work. Published by Pearson Learning Solutions. Solution (a) Note that f1x2 = x 3 is positive on the interval [2. Gordon.Section 5. Inc. f(x ) = x 3 A 3 Figure 2: 2 3 x 3 dx representing the Area A (b) We have L x3 dx = 3 x4 x4 and we have + c. (b) Find this area. and Finance.a + 1b = . 3] so the definite integral does indeed represent an area.= 3 3 3 3 3 3 1 The ones cancel. The same would be true for any constant of integration.4. do you 3 see why you get the same answer? 2 1 3 x 2 dx = a 2 1223 1123 x3 8 1 7 + 1b ` = a + 1b . so we choose F1x2 = 4 4 1324 1224 x4 3 81 16 65 ` = = = 4 2 4 4 4 4 4 2 3 x 3 dx = Applied Calculus for Business. by Warren B. Copyright © 2007 by Pearson Education. x 3 Solution An antiderivative of 1/x is ln x . from the Fundamental Theorem. b. . so we have: 5 3 1 dx = ln x x 3 5 3 = ln 5 . Wang. and Finance.450 * ** Section 5. Copyright © 2007 by Pearson Education. g1x22 dx = a a 3 f1x2 dx . then we have a a b 3 f1x2 dx = 0 b (3) a b 3 kf1x2 dx = k b a 3 f1x2 dx b (4) a 3 1f1x2 .F1a2 but b a 3 f1t2 dt = F1t2 b a = F1b2 . Economics. Walter O. then b a 3 f1x2 dx = F1x2 b a = F1b2 . b a 3 g1x2 dx (5) b b 3 f1x2 dx = c a 3 f1x2 dx b (6) a 3 f1x2 dx = a 3 f1x2 dx + c 3 f1x2 dx (7) Applied Calculus for Business. Some of these properties are similar to those for the indefinite integral (see (4) and (5)). c. b b a 3 f1x2 dx = a 3 f1t2 dt Do you see why? Let F be an antiderivative of f. Gordon.F1a2 Basic Properties Note the integration variable is replaced by the limits of integration. The variable of integration in a definite integral is sometimes called a dummy variable. and April Allen Materowski. Suppose f and g are continuous functions on some interval containing a.ln 3 = ln 5 . the result is the same. Thus. and k is a constant. Published by Pearson Learning Solutions. Inc.ln 3 = ln 5 3 It should be noted that the definite integral does not depend on the integration variable. There are some basic properties of the definite integral that we may obtain either from its definition in terms of a Riemann sum. That is. if we replace x by any other letter. by Warren B.6 The Definite Integral Example 3 5 Evaluate 3 1 dx. or by a simple interpretation of the definite integral in terms of an area. there are three cases to be considered: c may be between a and b.6 b The Definite Integral * ** 451 a b b 3 k dx = k1b . The area will then be negative. . rather than from left to right.F1b2 = . by Warren B. Walter O. meaning the width of each rectangle is now a negative number. if we assume the Fundamental Theorem.Section 5. that is. A1 A2 a b c c b b Figure 3: Illustrating a 3 f 1x2 dx = a 3 f 1x2 dx + c 3 f 1x2 dx Applied Calculus for Business. Then if the limits of integration are reversed. and April Allen Materowski. but equal in absolute value to the area of the region. (9) We first prove (3). the constant k may be factored out of the sum using the linearity property. (4) may be proven from a simple consideration of areas.F1a22 = - a 3 f1x2 dx In (7). or to the right of b. Published by Pearson Learning Solutions.a2 (8) a 3 f1x2 dx a 3 g1x2 dx. b n i=1 n i=1 b a 3 kf1x2 dx = lim a kf1ci2 ¢ x = k lim a f1ci2 ¢ x = k n: q n: q a 3 f1x2 dx (5) may be proven from the Riemann sums or by consideration of the additive property of areas. ¢ x = 0. or to the left of a.1F1b2 . a] has length zero. namely the area bounded by kf(x) is k times the area of the region bounded by f(x) (assuming k and f are nonnegative on the interval). the area is being generated from right to left. we can show this result another way. if a 6 b. Alternately.F1a2 but a b b 3 f1x2 dx = F1x2 a b = F1a2 . Alternately. Consider the case where a 6 c 6 b. Consider (6): suppose a 6 b. and the Riemann sum is zero. b]. We leave it as an exercise. Inc. Copyright © 2007 by Pearson Education. Economics. Let F be an antiderivative of f. as in Figure 3. Therefore. since the interval has zero length it has no area so its definite integral is zero. Wang. then b a 3 f1x2 dx = F1x2 b a = F1b2 . where f1x2 g1x2 on [a. and Finance. In terms of Riemann sums it should be clear that the interval [a. Equivalently. if we write the definite integral on the left as a Riemann sum. Gordon. Example 4 Illustrate the properties on 1 3 3 Solution Using the properties.+ 5b dx = 3 x2 dx . Economics.4x + 2ex . letting F be any antiderivative of f.F1c2 = F1b2 . b]. Published by Pearson Learning Solutions.F1c2 adding. .F1a2 = f1x2 dx 3 3 3 c a We leave the proof of (8) as an exercise. Wang.4 x dx + 2 ex dx .F1a2 + F1b2 . The other two cases may be shown in a similar way by indicating the areas of the appropriate regions.6 The Definite Integral The area represented by the definite integral b a 3 f1x2 dx = A 1 + A 2 but c A1 = and a 3 b f1x2 dx A2 = c 3 f1x2 dx proving (7). and Finance. Inc. We illustrate a few of the properties in the next example. x 3 3 3 3 1 1 7 a3x2 . We leave this proof to the exercises as well. Walter O. Copyright © 2007 by Pearson Education.7 dx + 5 dx x x 3 3 3 3 3 3 1 1 1 1 1 * Applied Calculus for Business.F1a2 c 3 f1x2 dx = F1b2 . by Warren B. we have 3 3 a 3x2 . and using property (6). and it has a simple interpretation when f and g are non-negative on [a. (9) may be proven using Riemann sums. Alternately. assuming the Fundamental Theorem. Gordon. we have c b b a f1x2 dx + f1x2 dx = F1c2 . These will be left to the exercises.F1a2 a 3 b f1x2 dx = F1c2 . and April Allen Materowski.452 * ** Section 5.4x + 2ex - 7 + 5 b dx. we have b a 3 c f1x2 dx = F1b2 . and Finance.x 2 . but that would be time-wise foolish.2x2 dx = x4 x3 .1 .123 2 23 . Published by Pearson Learning Solutions. and its sketch is given in Figure 4.2e . A1 A2 x Figure 4: f 1x2 = x 3 .a .x 2 ..x 2 .7 ln 3 + 152 .7 ln x + 5x2 3 = 127 .Section 5.2 + 2e .2x2 + 2ex .7 ln x + 5x b ` = x 3 2 1 3 1 (x3 .2x2 dx.22 b . Walter O.2x and the x-axis between x = . 1x3 .1 and 2.2x Applied Calculus for Business.7 ln 3 Example 5 (a) Evaluate 2 (b) Find the area bounded by f1x2 = x3 .+ 1 = -3 + = 3 4 3 4 4 (b) Observe that f1x2 = x3 .x2 .x 2 .4 .221x + 12.4 + 2ex . It is more convenient to first find an antiderivative of the original integrand and proceed as follows: 3 1 3 a 3x2 .6 The Definite Integral * ** 453 We could then evaluate each integral separately.4x + 2ex - 3 x3 x2 7 + 5 b dx = a 3 . Inc. and April Allen Materowski.2x2 dx = a 2 x4 x3 . Copyright © 2007 by Pearson Education.124 1 .18 + 2e . . by Warren B. Economics. Gordon. 2 -1 3 1x 3 .x2 b ` = 4 3 -1 a 1 .x2 + c 4 3 -1 4 3 1x3 .7 ln 1 + 52 = 20 + 2e3 .2x = x1x . Wang.11 ..122 b = 4 3 4 3 8 1 1 3 9 4 .x 2 . Solution (a) We have L therefore. 1 + x2 taking reciprocals. For example. 5/12 . as it should. Copyright © 2007 by Pearson Education. Observe that 1 + x 2 7 x2 on the interval [1. A 1 + A 2.a .02 b . 0 A1 = -1 3 1x 3 . . and Finance.9/4. Published by Pearson Learning Solutions. Wang. 1 + x2 3 Solution The given integrand does have an antiderivative. Consider the following example.x 2 . Walter O. Inc. the function is positive and for x between 0 and 2 it is negative.2x2 dx will give the negative of A 2. 1 2x2 2 x 2 + x 2 = 2x2 1 1 + x2 2 we again use (9) to obtain 1 1 dx 2 2 3 x 1 1 dx 1 + x2 3 Applied Calculus for Business.x 2 - Do you see why the sum of the areas in (b) is not equal to the result in (a)? However. it follows that 2 2 1 1 1 1 dx 6 dx = 1Verify!2 2 2 2 31 + x 3x 1 On the other hand.122 b = 4 3 4 3 12 2 Now the integral 0 3 1x3 . You can show that 2x2 dx = . we need 2 only compute it and change its sign to find A 2.8/3.1 . Example 6 2 Estimate 1 1 dx. They are considered in second semester calculus. 2].x 2 . We have. however. estimate this integral a variety of ways.x2 b ` = 4 3 -1 a 1 . We could. the required area is 5/12 + 8/3 = 37/12 0 3 1x 3 . on this interval.2x2 dx = a 0 x4 x3 . Gordon. and April Allen Materowski. For x between . we could use Riemann sums and take the midpoint of each interval with a reasonably large value of n to obtain a fairly accurate estimate.123 04 03 5 . Economics. the required area is the sum of the two areas shown. so we have. 1 1 6 2 2 x 1 + x and by (9). therefore. by Warren B.124 1 .1 and 0.454 * ** Section 5. (Why?) The last basic property is helpful in estimating the value of a definite integral in the event we do not know how to compute it. We choose to illustrate (9).8/3 = .6 The Definite Integral Therefore. taking reciprocals. but we have not considered such types. any representative rectangle has height f1x2 = 2x 2. Basically. its value is 1/4. (2.6 The Definite Integral * ** 455 The definite integral on the left may be computed. by Warren B. We then use the Fundamental Theorem to compute the integral. Example 7 Determine the area of the triangular shaped region between the curves and the x-axis in Figure 5. Wang. Copyright © 2007 by Pearson Education.32175. (2. Economics. the integral is equal to 0. we draw an incremental rectangle. every representative rectangle has height g1x2 = .Section 5. 8) f(x) = 2x2 dx f(x) g(x) -4x + 16 dx g(x) Figure 6: Finding the Area of the Triangular Region Applied Calculus for Business. So we have 2 1 4 1 1 1 dx 6 2 2 31 + x (To five decimal places. it tells us that when we want to compute an area. see Figure 6. Inc. Gordon. Observe that if we draw the dotted line x = 2 to the left of this line. .4x + 16. We illustrate this remark in the next example.) Let s go back and re-examine the idea presented in Figure 1. 8) f(x) = 2x2 g(x) -4x + 16 Figure 5: Finding the Area of the Triangular Shaped Region Solution We need to be careful when we draw a representative rectangle. The definite integral then gives the sum of the infinite number of these rectangles. as the curve which determines the height of the rectangle changes at x = 2. and to the right of this line. Published by Pearson Learning Solutions. Walter O. and Finance. and April Allen Materowski. Published by Pearson Learning Solutions.2x2 dx. It can be shown that this limit will also exist when f has a finite number of finite discontinuities on the interval. Wang. n We remark that if f is continuous on the interval a n b x b. Economics. by Warren B. The syntax is Calculator Tips 1expression. Applied Calculus for Business.456 * ** Section 5.6 The Definite Integral The area of the triangular region is the sum of the area of the region to the left on the vertical line x = 2 and the area to its right.4x + 162 dx we leave it for you to verify that A = 40/3. and Finance. Copyright © 2007 by Pearson Education. It s not hard to understand why a finite number of discontinuities are allowable. and April Allen Materowski. as a single point has no width.x 2 . upper limit2 L We use the same integral key as we do for the (indefinite) integral that is. that is. . Walter O. see Figure 7. either from the Catalog or by pressing F3 and scrolling down to the second option and pressing Enter. we see that a discontinuity does not change the area. f is said to be integrable. Whenever this limit exists. see Figure 8. the total area bounded by this curve and the x-axis is c1 c2 b A = A1 + A2 + A3 = a 3 f1x2 dx + c1 3 f1x2 dx + c2 3 f1x2 dx Definite integrals are easily computed with the TI 89. variable. 2 For example. then lim a f1ck2 ¢ x exits. Gordon. Inc. to compute 0 3 1x 3 . 2 4 A = 0 3 2x dx + 2 2 3 1 . n: q k=1 We defined lim a f1ck2 ¢ x = n: q k=1 a 3 f1x2 dx. A3 A1 a c1 A2 c2 b Figure 7: A Function with a Finite Number of Discontinuities In Figure 7 we show a discontinuous function. resulting in no change to the area. lower limit. From the Riemann sums interpretation as an area. (The proof may be easily extended to the case when f is negative over part or all of [a. Walter O. Let us approximate this area by two rectangles.6 The Definite Integral * ** 457 2 Figure 8: Using the TI 89 to compute 0 3 1x 3 . we need only show that A1b2 = F1b2 . b f1x2 dx. A1b2 = Applied Calculus for Business. . Inc. where F is any antiderivative of F. b].A1x2 Let us now consider the area ¢ A. as shown in Figure 9. and Finance. We assume f is a continuous. Economics. Copyright © 2007 by Pearson Education.Section 5. b]. We now consider A1x + ¢ x2. Let f(m) denote the smallest y-value on the curve in interval between x and x + ¢ x. which is the area indicated in Figure 10. Note that when x = a. It is the Area A(x) (to the left of the dotted line) and the additional area ¢ A = A1x + ¢ x2 . Wang. and April Allen Materowski.x 2 . see Exercise 66. and f(M) the largest y-value on this interval. non-negative function over the interval [a. the area 3 a under the curve between a and b. by Warren B. To prove the theorem. note that A1x + ¢ x2 is the area of the region under the curve and above the x-axis between the vertical lines at a and x + ¢ x. Published by Pearson Learning Solutions.2x2 dx We close this section with a proof of the Fundamental Theorem. A1a2 = 0 and when x = b. Gordon. and the x-axis between a and x.) y = f( x ) A (x) x a x b Figure 9: Defining A(x) Let us consider a new function A defined as follows: A(x) is the area bounded by the curve y = f1x2. In Figure 10.F1a2. and Finance. A is an antiderivative f. . must approach x. Economics.A1x2 Then for the area of the region ¢ A.F1a2 Applied Calculus for Business. Gordon.A1x2 lim f1m2 lim lim f1M2 ¢x : 0 ¢x : 0 ¢x : 0 ¢x Now let s keep in mind that both m and M are x-values in the interval between x and x + ¢ x. both m and M. which are sandwiched in the interval between x and x + ¢ x. that is A1x + ¢ x2 .A1x2 f1M2 ¢x We recognize the appearance of the difference quotient in the middle of this inequality. that is. Walter O. so when we let ¢ x : 0. and April Allen Materowski. then we have A1x2 = F1x2 + C. Inc.458 * ** Section 5. and we have f1m2 f1x2 A ¿ 1x2 f1x2 The only way this inequality can be true is if we have the equality f1x2 = A ¿ 1x2. by Warren B.6 The Definite Integral y = f( x ) f( M ) f( m ) A *A x a x x + *x b Figure 10: ¢ A = A1x + ¢ x2 . A1x2 = F1x2 . (Why?) But since A1a2 = 0 we have 0 = F1a2 + C or C = .F1a2 Thus. Copyright © 2007 by Pearson Education. Published by Pearson Learning Solutions. we have f1m2 ¢ x ¢A f1M2 ¢ x We divide by ¢ x and have (assuming ¢ x 7 0) ¢A f1m2 f1M2 ¢x or A1x + ¢ x2 . We next will take the limit as ¢ x : 0. Wang. If F is any other antiderivative. 1 3 4 12x 3 . and April Allen Materowski. k dx. 8 32 3 t2 dt Applied Calculus for Business. 11. 1 0 1 2 31 + x 1 7. we have A1b2 = F1b2 . 4 dx 16. 1t dt 20. Inc. and Finance.32 dx 19. Wang. 0 3 9 12x . Gordon. 1 3 27 13. by Warren B. Published by Pearson Learning Solutions.12 dx 23.2x + 42 dx 22. (a) 3x dx (b) 3x dx (c) 2x 2 dx 18. 5 dx 15. 8 3 2 3 4 e -2t dt 9. (a) 2 3 2 1 5 3x 12 dx (b) 3 3 2 6 1 dx x 3 b 5. 13x2 .2 dt 12. Walter O.6 The Definite Integral * ** 459 and if we let x = b. 0 1 2 1/4 3 11 + x 2 dx 2 3 b 0 3 x13x2 + 121/2 dx 2 3. (a) -1 -1 3 1 3x a k#b dx (b) 1 2 3 1 dx x 3 k# a x dx 6.224 dx x 1 3 7 2.1 1 -2 1 3 t dt 24. (a) 2x2 dx (b) 2 1 1 dx x 3 10 4.1 dt 3 1t 3 w B dw A 1w . . -1 3 3t .Section 5. Copyright © 2007 by Pearson Education. a 3 17. -1 2 3 1 e 1x dx 3 1x 3 10. 1 3 w dw 21. 3 1.3x2 + 4x . k is a constant.6 In Exercises 1 24 evaluate the given definite integral. 1 64 t .F1a2 proving the theorem. EXERCISE SET 5. Economics. 3 1 14. 4 27 3 3 2 et dt 8. 0 215 3 13x . . f(x) 4 . Ex. Ex. [1. lim 1 n 2 a x k.9x A Fig. 2] n: q n k =1 n: q A 33. Wang. 25 26.6 The Definite Integral 29. Ex. Ex. and April Allen Materowski. 5] nk =1 Fig. 3] nk =1 4 n 3 2 a 12xk . Integral in Exercise 1 36. 25. Published by Pearson Learning Solutions. A x= 1 x=3 f(x) = 4x + 1 A x=1 x=2 Fig. and Finance. by a definite integral. 31 In Exercises 32 34. [2. 32. Gordon. Determine the total area between the curve and the x-axis shown in Fig. [1. represent the given sum (which represents a partition over the indicated interval). lim 34. Integral in Exercise 9 Applied Calculus for Business. 35. 28 n: q In Exercises 35 43 sketch the area represented by the given definite integral. 30 x=2 A x=3 31. Fig. Walter O. 26 27. Copyright © 2007 by Pearson Education. Inc. Integral in Exercise 7 37. Ex. f(x) = 4/x1/2 Fig. f(x) = 4/x In Exercises 25 30 (a) determine a definite integral that will determine the area of the region shown and (b) evaluate this integral. Ex. by Warren B. f(x) = x2 + 1 f(x) = e * x A Fig. and then evaluate the integral to determine the value of the sum. 29 30.3x k + 12. Fig.460 * ** Section 5.x2 (x) = x 3 . 31. Ex. Ex. lim 1 n 2+k a e n . Economics. 27 28. Section 5.L be the smallest value f assumes on this interval. 1 3 2 ln x dx 1 and the x-axis 21 + x 2 between x = 1 and x = 2. g1x2 = . Region bounded by f1x2 = and the x-axis. estimate the area bounded by f1x2 = 41. 54. approximate this area using a spreadsheet. 3 + 2x 50. 52.L2 dx = a 3 1f1x2 + L2 dx - a 3 L dx 53. Copyright © 2007 by Pearson Education. (a) Using (9). Using Riemann sums on a spread sheet. (b) using Riemann sums and (c) using the Fundamental Theorem. with a 7 0 and n Z .1. Region bounded by f1x2 = 49. estimate the area bounded by f1x2 = 39. call the y-coordinate h. n + 1 (b) Show both integrands on the right in the result above are non-negative so we may use apply the Fundamental Theorem to them.6 2 The Definite Integral * ** 461 38. Region bounded by f1x2 = 4e .1 and 1. Consider the parabola f1x2 = ax .2x + 8 and the x-axis. 62. between x = 2 and x = 5. Region bounded by f1x2 = 1x and the x-axis between x = 0 and x = 9. between x = 0 and 2. When x = b. Let F be an antiderivative of f. Prove (9) by (a) an area argument. Find b if this area is 1/2. and the x-axis. 1 59. 66. the midpoint of each subinterval. (c) Show an antiderivative of f1x2 + L is F1x2 + Lx and an antiderivative of L is Lx (d) Show that b b b a 3 f1x2 dx = a 3 1f1x2 + L2 dx - a 3 L dx = F1b2 . Wang. b]. Inc. (a) Show b b b b 43. call the y-coordinate h. This exercise proves the Fundamental Theorem when f is negative on part or all of [a. 1 . with L 7 0. Region bounded by f1x2 = 3ex and the x-axis. 51. and April Allen Materowski.2x 2 and the x-axis. Region bounded between f1x2 = 1x. approximate this area using a spread 2 42. 45. Region bounded by f1x2 = 2/x and the x-axis.x2 dx 1 1 3 1 + x3 dx given by your calculator. Applied Calculus for Business. When x = b. Economics. Let . and a large value for n. (c) Compare these results to the value of the 2 integral 29 . Show that the area bounded by the parabola 1 and the x-axis between x = 0 and x = b is A = bh. between x = . Show that the area bounded by 1 the graph and the x-axis between x = 0 and x = b is A = bh. and n = 100 rectangles. Gordon. find an approximate value of the integral estimated in Example 6. where b 7 0. between x = .1 and x = 2. and Finance. between x = 1 and x = 8. Find b if this area is 1/2. -1 3 3 ex dx 1 and the x-axis 1 + x3 between x = 1 and x = 2. -3 3 2 58. b]. (b) using Riemann sums and (c) using the Fundamental Theorem. 1 3 1 22x + 1 dx 57. Consider the area bounded by f1x2 = x2 + 1 and the x-axis. 46. (a) Using (9).F1a2 Proving the Fundamental Theorem for functions which are negative on part or all of [a. Published by Pearson Learning Solutions. 65. 47. with a 7 0. Consider the graph defined by f1x2 = ax .1. 61. (c) Compare these results to the value of the integral dx 3 2 1 + x2 given by your calculator.2x a 3 f1x2 dx = a 3 1f1x2 + L . and n = 100 rectangles. 0 3 3x2 dx + 2 3 In Exercises 44 54. Walter O. Region bounded by f1x2 = 2/x and the x-axis. where b 7 0. between x = . Prove (5) by (a) an area argument. 2 and the x-axis. (b) Using Riemann sums with the midpoint of each subinterval. -1 3 2 2x . and (b) using Riemann sums. 3 n 56. . determine the area of the indicated region 44. between x = 0 and x = b. 55. between x = 2 and x x = 4. Prove (8) by (a) an area argument. Region bounded between f1x2 = 2x. 64. 1 3 x 48. 1 .1 dt 8 1 sheet. Region bounded by f1x2 = 3x and the x-axis. between x = 1 and x = 3. between x = 0 and x = b. Consider the area bounded by f1x2 = x3 + 1 and the x-axis. Prove (7) using an area argument similar to the one used in the text in the case (a) c 6 a and (b) when c 7 b.3 and x = . 63. by Warren B. [Hint: See Example 6] (b) Using Riemann sums with the midpoint of each subinterval. 40. g1x2 = 2 + 4.2x + 162 dx 60. Region bounded by f1x2 = 3x and the x-axis. tal Theorem.163/22 = 1125 . To apply the Fundamen3 x 29x2 + 16 dx. Published by Pearson Learning Solutions. evaluate L We proceed in the usual way. We do this in Table 1. we let 0 u = 9x2 + 16. that is. We redo the 1 problem illustrating how this is done. computed the integral with respect to the variable u. by Warren B. using the substitution method for antiderivatives.7 Substitution and Properties of Definite Integrals 5. Let u = 9x2 + 16.7 Substitution and Properties of Definite Integrals » » » » » Substitution Odd and Even Functions Average Value Derivative of an Integral Calculator Tips 1 Consider the evaluation of the definite integral 0 x 29x2 + 16 dx. Inc. Gordon. Wang. then du = 18x dx and we have L x 29x2 + 16 dx = 1 1 1 1 2 3 29x2 + 16 18x dx = u2 du = u2 + c = 18 L 18 L 18 3 3 1 3 1 u2 + c = 19x 2 + 1622 + c 27 27 We now have 1 0 3 x 29x 2 + 16 dx = 1 3 1 1 1 61 19x 2 + 1622 ` = 1253/2 .462 * ** Section 5. and April Allen Materowski.642 = 27 27 27 27 0 Substitution Notice that we first made a change of variable from x to u. and then switched back to the x variable again (as the limits of integration are given with respect to that variable) to compute the definite integral. It turns out. To evaluate x 29x2 + 16 dx. and Finance. Table 1: Find the Limits of Integration when u = 9x 2 + 16 x 0 1 u = 9x2 + 16 91022 + 16 = 16 91122 + 16 = 24 So we have 1 1 1 1 2 3 25 x 29x 2 + 16 dx = 29x 2 + 16 18x dx = u2 du = u2 ` = 18 3 18 3 18 3 16 3 1 1 25 0 0 16 Applied Calculus for Business. Walter O. Copyright © 2007 by Pearson Education. What needs to be done is to change the limits of integration as well. we must first compute an antiderivative. we begin as be3 fore. . then du = 18x dx in addition we use the transformation u = 9x2 + 16 to determine the new limits of integration in the u-domain. that with an additional substitution we can avoid switching back to the x-variable. Economics. and the limits of integration transform as indicated in Table 2.59726402 4 4 Applied Calculus for Business. Therefore. Inc.G1a2 = b b a 3 G ¿ 1x2 dx = a 3 f1g1x22g ¿ 1x2 dx as required. in the third integral. by Warren B. * This theorem states that for most functions. and Finance. THEOREM 1: SUBSTITUTION THEOREM FOR DEFINITE INTEGRALS Suppose f and g ¿ are continuous on [a. Let u = g1x2.Section 5.7 Substitution and Properties of Definite Integrals * ** 463 1 3 25 1 1 61 u2 ` = 1253/2 . 2 Solution Let u = 2x2. Published by Pearson Learning Solutions.F1a2 = F1g1b22 . we do not have to return to the original variable of integration. by the chain rule.F1a2 = a 3 f1u2 du let G1x2 = F1g1x22. it may be extended to the definite integral. Copyright © 2007 by Pearson Education. note that G1b2 = F1g1b22 and G1a2 = F1g1a22. . then du = 4x dx. then b F1 b 2 . Walter O. Economics. More generally we have the following theorem.e02 = 1e2 . where the substitution method works on an indefinite integral. G ¿ 1x2 = F ¿ 1g1x22g ¿ 1x2 = f1g1x22g ¿ 1x2 Now b a 3 f1u2 du = F1 b 2 . the limits of integration are in the x-domain. provided we substitute for the limits of integration as well. Gordon. as the integral is with respect to x. we are integrating with respect to u and the limits of integration are now given with respect to this variable. Wang. a = g1a2 and b = g1b2 then b b a 3 f1g1x22g ¿ 1x2 dx = a 3 f1u2 du (1) Proof Let F be an antiderivative of f. Moreover. and April Allen Materowski. we have 1 0 3 xe 2x2 2 1 1 1 2 dx = e2x 4x dx = = eu du = a eu b ` = 43 43 4 0 0 0 1 2 1 2 1 1e . once the complete substitution is made.12 L 1.642 = 27 27 27 27 16 Notice that in the first and second integrals. b].F1g1a22 = G1b2 .163/22 = 1125 . Example 1 1 Evaluate 0 3 xe2x dx. f1x22 then a -a 3 f1x2 dx = 0 (2) (b) Suppose f is an even function 1f1 . Gordon. is zero. Economics. Applied Calculus for Business. Solution We let u = x2 + 1. Inc.464 * ** Section 5. a].x2 = f1x22 then a a -a 3 f1x2 dx = 2 0 3 f1x2 dx (3) Proof We illustrate the proof geometrically. Table 3: Find the Limits of Integral when u = x 2 + 1 x -1 1 u * x2 + 1 1 . and April Allen Materowski. a]. Published by Pearson Learning Solutions. consider Figure 1. Copyright © 2007 by Pearson Education. the definite integral. Odd and Even Functions A more interesting question is why the answer in the previous example turned out to be zero. We have the following theorem for continuous functions on [ . Walter O.a.a. we applied basic property (3) from the previous section. THEOREM 2: INTEGRALS OF ODD AND EVEN FUNCTIONS (a) Suppose f is an odd function 1f1 . (a) If f(x) is odd.x2 = . it is being integrated over a symmetric interval to the left and the right of the y-axis. Wang. The general proof uses the previous theorem and is left to the exercises. .7 Substitution and Properties of Definite Integrals Table 2: Find the Limits of Integral when u = 2 x 2 x 0 1 u = 2x2 0 21122 = 2 Example 2 1 Evaluate -1 3 x1x2 + 123 dx. by Warren B. assuming f Ú 0 on [ . Did you notice that the function in the integrand is odd? Moreover. which is the sum of the signed areas. as a result. then dx = 2x dx and the new limits of integration are found in Table 3.122 + 1 = 2 1122 + 1 = 2 So we have 1 2 3 1 2 -1 1 1 x1x + 12 dx = 1x2 + 123 2x dx = u3 du = 0 23 23 3 -1 2 Notice that we didn t have to find the antiderivative here (which is u4/4). and Finance. (b) When f is even. Wang. Theorem 2 is often useful in computing definite integrals when the limits of integration are equal and opposite.A 1 + A 2 where A 1 and A 2 are the areas of the regions bounded by the curve and the x-axis as shown.Section 5. . a a -a 3 f1x2 dx = A1 + A2 = 2A2 or -a 3 f(x) dx = 2 0 3 f1x2 dx. y = f(x ) A1 A2 a a Figure 2: - 3 a f 1x2 dx = 2 0 3 f 1x2 dx when f is Even a Since f is even. Inc. by Warren B. therefore. Published by Pearson Learning Solutions. A 1 = A 2. consider Figure 2. as the next examples illustrate. Gordon. Copyright © 2007 by Pearson Education.7 y = f( x ) Substitution and Properties of Definite Integrals * ** 465 -a A2 a A1 a Figure 1: a 3 a f 1x2 dx = 0 when f is Odd We have -a 3 f1x2 dx = . and April Allen Materowski. the integral is 0. by symmetry. Walter O. Applied Calculus for Business. However. Economics. and Finance. therefore. we have A 1 = A 2. 3x + 4x .3x 2 + 4x . Published by Pearson Learning Solutions.52 dx. Then the average temperature over this time period would be f1t12 + f1t22 + Á + f1tn2 n or written using sigma notation as f1tk2 k=1 n a Clearly the larger n is (that is.52 dx = even integrand 1 0 + 2 21 . b]? We could partition this interval into n subintervals of length ¢ t = 1b . and Finance. Copyright © 2007 by Pearson Education.5x2 1 0 0 3 1 . 2 Example 4 1 Using Theorem 2. . one of which is odd and one of which is even. any polynomial function can be rewritten as a sum of two functions.1022 = . the better the sum represents the average temperature over the given time interval.466 * ** Section 5. evaluate -1 3 12x 3 . Economics. 2 Solution We could evaluate this integral directly using the substitution method. Gordon. Average Value Let y = f1t2 where y is the temperature at time t. that is. but it is simpler to observe that f1x2 = 3xe -x is an odd function (verify that f1 .a2/n and letting tk denote the time used to determine the temperature in the kth subinterval.f1x2). Inc.52 dx = = 21[ .52 dx = 3 2 -1 3 12x + 4x2 dx + odd integrand 3 -1 3 1 . Walter O. our average temperature would be lim n: q a f1tk2 k=1 n n n Applied Calculus for Business. and April Allen Materowski.3x 2 . Solution Observe that we can break the given integral into a sum of two integrals. one of whose integrand is even and the other whose integrand is odd. Since we are dealing with a continuous function.1123 . 1 1 1 -1 3 12x .7 Substitution and Properties of Definite Integrals Example 3 2 Evaluate -2 3 3xe -x dx. the more temperature readings we take).x2 = . How might we go about defining the average temperature over some time interval [a.x3 . f defines a continuous function relating the temperature to the time. 2 2 Therefore it follows that -2 3 3xe -x dx = 0. why not let n become infinite? In this case. by Warren B.5112] . Wang.3x2 .12 As this example illustrates. We have b f1c21b . More generally. and April Allen Materowski.a3 following definition. Published by Pearson Learning Solutions.a3 a (4) Example 5 Find the average value of f1x2 = x2 on the interval [2.a. denoted by f(c).a n: q k=1 k=1 n b n 1 this last term may be replaced by the definite integral f1t2 dt. the average b . the area under the curve f1x2 = x 2 for x between 1 and 2 is equal to the area of the rectangle shown. a DEFINITION The average value of the continuous function f over the interval [a. b]. and Finance.232 = x2 dx = x2 dx = = 13 f1c2 = 33 3 3 2 9 9 5 . Walter O. b]. Inc.a2 = a 3 f1x2 dx This means that the area of the region bounded by the curve and the x-axis over the interval [a.23 2 2 5 5 We can give another interpretation to the average value by rewriting (4) and assuming f Ú 0 on [a. 5]. f( x ) = x 2 f( c ) = 7/3 Figure 3: The Average Value Interpreted as the Height of a Rectangle Applied Calculus for Business. Copyright © 2007 by Pearson Education. we have the b . We can get this as follows: lim n: q a n n f1tk2 1 b .a3 a b 1 temperature over the given time interval is f1t2 dt. by Warren B. Gordon. b] is equal to the area of the rectangle whose height is f(c) and whose width is b . what is missing is ¢ t. .7 Substitution and Properties of Definite Integrals * ** 467 this almost looks like a Riemann sum.Section 5. is b 1 f1c2 = f1x2 dx b . Economics. in Figure 3.a 1 = lim a f1tk2 = lim a f1tk2 ¢ t n b . Solution We have 1 1 1 x3 5 1 117 ` = 153 . Thus. Wang. For example.a n: q k=1 b . we can determine much about their properties. Example 6 Referring to the curve in Figure 3.7 Substitution and Properties of Definite Integrals Therefore. then for any x in this interval. Solution (a) We have 1 x3 2 8 1 7 ` = . We can easily perform the integration and have x G1x2 = 1 3 3t2 dt = t3 x 1 = = x3 . and Finance. and that is certainly a reasonable question. there are functions whose antiderivatives cannot be found. we have c = 27/3 L 1. Inc. in particular. the x-value at which the function assumes its average value (when the top of the rectangle crosses the curve).53 The definite integral provides another method by which we may define functions. by Warren B. one of which we consider below. when multiplied by the length of the interval. consider G1x2 = 1 3 3t2 dt. (Note that the independent variable x is now one of the limits of integration. In fact. the average value would be that y-value.= x2 dx = f1c2 = 3 3 3 3 2 . which. we have x d G ¿ 1x2 = f1t2 dt = f1x2 dx 3 a (6) Proof Let F be any antiderivative of f. their derivatives. Walter O. Gordon. While we may not be able to express such functions in a simple closed form. . Wang. and f1x2 = x2. and April Allen Materowski. For x Derivative of an Integral example. Copyright © 2007 by Pearson Education. THEOREM 3: THE DERIVATIVE OF AN INTEGRAL Suppose f is continuous in some interval containing a. we have c2 = 7/3 and since 1 6 c 6 2. but which are represented as such integrals. However. (a) show the average value is 7/3 and (b) determine c. Economics. produces the area bounded by the curve and the x-axis over the given interval.468 * ** Section 5. if f is any continuous function then we may define the function G as x G1x2 = a 3 f1t2 dt (5) You may ask why anyone would define a function in terms of an integral.) In this example. then Applied Calculus for Business. it turns out that such functions arise in a variety of ways and have important applications.1 More generally. Published by Pearson Learning Solutions.13 1 1 2 (b) Since f1c2 = 7/3. Economics. Note that n is even (symmetric about the y-axis) so it follows that P1 . G 1x2 = . Moreover. .t2 2 dt (7) for different values of x. and April Allen Materowski. Often. Gordon. or equivalently. A very important function in mathematical statistics is the normal distribution.F1a2 d d f1t2 dt = 1F1x2 . and Finance. we need to evaluate x P1x2 = 1 22p 3 0 1 e . then consider the function G1x2 = 1 1 dt (let s pretend we do not rec3t ognize this function).4. Walter O.7 x Substitution and Properties of Definite Integrals * ** 469 G1x2 = and x a 3 f1t2 dt = F1x2 . so tables need only be constructed for x Ú 0. Without explicitly determining the function. it turns out that most problems have to evaluate integrals of the form. and since x 7 0. Most texts give a table which gives the approximate result for any given value of x. by Warren B. most tables stop at x = 3.1/x 2.m s and the limits of integration are found in Table 4 Applied Calculus for Business. (Note that the actual domain for n is 1 . Note that G112 = 0. which means the function is always concave downward. Copyright © 2007 by Pearson Education. since the area between the curve and the x-axis for x 7 3 is negligible (see Figure 5). q 2.x2 = P1x2 (Why?). The problem is easily resolved by the substitution z = then dz = 1 dt s t . to determine the area bounded by the curve and the x-axis from 0 to x. G ¿ 1x2 = 1/x.) Actually. Wang.Section 5. We considered this function in Section 4. so numerical methods are required to compute P for each value of x. sometimes called the Bell Curve or Gaussian distribution. This integrand has no simple closed form for its antiderivative. Inc. x Q1x2 = 1 22ps 3 a e -2 A 1 t-m 2 s B dt Figure 5: n1x2 = 1 22p e. we could draw a sketch of its graph.F1a22 = F ¿ 1x2 = f1x2 G ¿ 1x2 = dx dx 3 a x Suppose x 7 0. Also.2 x2 This poses no problem with a calculator but does with a table. this function is always increasing. its equation is n1x2 = 1 22p ex2 2 Figure 4: A Sketch of G1x2 = 1 dt 3t x and a sketch of its graph is given in Figure 5.q . Published by Pearson Learning Solutions. see Figure 4. 5 e .2 . 1 1y11x2. How might we do this? The easiest t2 way is to enter on the calculator.2 dz z2 Calculator Tips This last integral may now be evaluated using the standard normal distribution tables found in most statistical references. we have 0 1. x. you need to observe that 2 0 2 1 22p 3 -1. and press ENTER. Gordon.m Integral when z = s t a x z = t .52 + P122 t2 P11.t2 2 dt = 1 22p 3 0 e . 2 Figure 6: Computing P(1) with the TI 89 After a few seconds. Using tables.2A 1 t-m 2 s B 1 dt = 1 x-m s s 3 22p a m s e . enter the intethe Y = window.t2 2 dt = 1 22p 3 0 2 e .1. Return to the HOME screen.t2 2 dt = 1 22p 3 .2 dt = P11.m2/s 1x . define y11x2 = 22p gral as indicated in Figure 6. the calculator gives the value as 0. by Warren B.5 e . and April Allen Materowski.341345.5 e .m2/s so we have x x Q1x2 = 1 22ps 3 a e.2 dt t2 and by symmetry.1.5 1 22p 3 -1.52 + P122 are now obtained from the table.5 e .2A 1 t-m 2 s B dt = 1 22p 3 a e. Inc. Walter O. Published by Pearson Learning Solutions.7 Substitution and Properties of Definite Integrals Table 4: Find the Limits of t . 2 1.5.470 * ** Section 5.m s 1a . Suppose we want to compute P(1).1.t2 2 dt + 1 22p 3 0 e .5 e . and Finance.2 dt t2 therefore.t2 2 dt + 1 22p 3 0 e . Wang. -1.2 dt.5 1 22p 3 . 2 Suppose we want to compute 1 22p 3 . This can be done with a calculator as follows: in x 1 e . Economics.910443. . Copyright © 2007 by Pearson Education. Applied Calculus for Business. 22 and press ENTER giving 0. -3 3 2 x dx -2 3 0 x x dx 2. 25x + 4 dx 18. 6 2 10. G1x2 = x dx 0 x 3 x t2 dt 14.Section 5. Published by Pearson Learning Solutions. 1 3 1. -1 3 x2 + 1 35. 2] x + 1 30.x2 and the x-axis. find the average value of the account during the 4 year period. find the average value of the account during the 8 year period. 7. -1 3 2 x2e -2x dx 3 25. -1 3 2 x dx 5 12.3x2 + 2x + 42 dx 4 3 x .1 and 1. Inc.2 and 2.x22 dx 33. The region bounded by f1x2 = x 2 2 x + 1 and the x-axis between x = .1 0 2 1/2 3 12x . 1] x 29. -2 3 3 x 21 . -4 3 1 0 3 x 21 . Copyright © 2007 by Pearson Education. determine the area of the region bounded by the given curves. 23. 0 3 1 11 .1 In Exercises 11 20. by Warren B. 3 24x + 1 0 0 4. If $1000 is deposited into an acocunt for 4 years and earns interest at 5% compounded continuously. 0 3 2 -3 3 2 2x + 3 dx (Hint: let u = 2x + 3.x dx 5 32. Gordon. G1x2 = 1 3 t4 2 3 1 . 2] 31. e 2x 2x 1 3e . 1 1x -2 3 1 13x4 . [0. 3 x 1 2 In Exercises 26 30 (a) determine the average value of the given function over the given interval. Walter O. [0. 2x . 1 dx 19.t2 dt Applied Calculus for Business. f1x2 = xe-x . between x = . and Finance. b] 28. If $5000 is deposited into an account for 8 years and earns interest at 4. In Exercises 33 37 determine G 1x2. [ . . f1x2 = x2. and the x-axis. 21.1x 2 dx In Exercises 21 25. 13. f1x2 = 2x. G1x2 = 13x 2 + 12(2x3 + 2x)5 dx 34. Economics. 26.2x24 dx 17. [a.2 -1 3 17x5 . The region bounded by f1x2 = x 24 . 6.2x 4 + 9x 3 . and April Allen Materowski. x 1 2 3x + 1 1 dx 22. 16. The region bounded by f1x2 = and the x-axis.) 3.2x + 12 dx 5. -1 3 9 x212x3 + 125 dx 20.1. f1x2 = 2 .2x 3 + 4x 2 .2x + 42 1 dx 24.2 and 2. use the properties about odd and even functions to evaluate the given integral.3 and 3.7 In Exercises 1 10 evaluate the given integral. between x = . . 1] dx 27.t2 dt 15. and (b) find the x-coordinate at which the function assumes its average value. ln x dx 9.7 Substitution and Properties of Definite Integrals * ** 471 EXERCISE SET 5. f1x2 = mx + d. -3 3 4 14 . [1. 1 11.75% compounded continuously. The region bounded by f1x2 = x x2 + 1 x3 x +1 4 8. The region bounded by f1x2 = xe-x and the x-axis. Wang. between x = . enter this approximation as y2(x) and then enter 1. 40. x2 41. Use the pervious exercise to determine G ¿ 1x2 if G1x2 = x3 0 3 2x2 + 1 dx. 5. 0. Economics. Gordon. y = T(x) x=a x=b y = B(x) Figure 1: The Region bound by y = B1x2. Show that 1 22p 3 -1. and y = B1x2 (bottom curve) over the interval [a. Suppose you are given G1x2 = generalized. G1x2 = x 2 0 3 t3 dt Solve a x 37. where f is an odd function and g is and even function. and April Allen Materowski.5 e . As an alternative. t.2 dt t2 Solve1y21x2 = 0.25.5 e . x b = 0.t2 2 dt + 1 22p 3 1. 102 (1995) 46-49.. Suppose you want to solve the problem P1x2 = 0.8 Applications of the Definite Integral » » » » » Area Between Curves Consumer and Producer Surplus Continuous Income Flow Probability Calculator Tips Area Between Curves We have already seen that the definite integral represents the area of a region when the function is non-negative. 44. use the approximation for P(x) given in the previous exercise and solve it for x.5 2 2 36. Wang. You may be tempted to use the SOLVE command on your calculator.25.t2 dt = 2 22p 3 0 e .12B + a 7 + x2 b e -x d f 2 30 4 1 42. Show that any polynomial p(x) may be written as p1x2 = f1x2 + g1x2.t2 dt 1 + x2 2 38. b]. u1x2 3 f1t2 dt. That is. and Finance. v1x2 39. Richard. Determine G ¿ 1x2 if G1x2 = x e x2 + 1 3 2x dx. G1x2 = 0 3 24 . it may take some time for the calculator to solve the problem (in a few minutes the calculator does gives as its solution 0. Copyright © 2007 by Pearson Education. b] Applied Calculus for Business. y = T1x2 on [a. where we want to determine the area between the curves y = T1x2 (top curve). Let us assume you have n(x) stored as y1(x) then you would enter x2 43. Consider Figure 1. American Mathematical Monthly.472 * ** Section 5.8 x Applications of the Definite Integral 1y11t2. x2 L While this will yield an approximate solution for x. Calculating Normal Probabilities. . It is shown that the integral in (7) may be accurately approximated by 2 1 2 2 2 1 p P1x2 L e 1 c 7e -x /2 + 16e -x A2 .67449). by Warren B.25 for x. Published by Pearson Learning Solutions. Walter O. Compare this approximation with results obtained by your calculator for various values of x between 0 and 3. J. We can go one step further and show that it may be used to evaluate the area between two or more curves in a very simple way. Show how Theorem 3 may be See Bagby. Inc. Example 1 Find the area between f1x2 = x2 . we draw a representative thin rectangle over the given interval. by Warren B. dx x=a y2 =T(x) x=b y = T(x) T(x) . the width of the representative rectangle is dx and its area is 1T1x2 .y1 = T1x2 . Inc. y2 . and April Allen Materowski. We also include a representative rectangle in the sketch.B(x) y = B(x) y1 = B(x) Figure 2: Area of Representative Rectangle is 1T1x2 .1 to 2 Applied Calculus for Business. Wang. .B1x22 dx We need to sum this infinite number of thin rectangles. The height of the rectangle is the difference between the two y-values.B1x2. see Figure 3. Copyright © 2007 by Pearson Education.B1x22 dx (1) We illustrate with some examples.8 Applications of the Definite Integral * ** 473 The method by now should be fairly straight-forward.3 and g1x2 = x + 4 over the interval from . Economics.Section 5.3 and g 1x2 = x + 4 Over the Interval from . Walter O. Published by Pearson Learning Solutions. T(x) = x + 4 x=2 x =-1 B(x) = x2 -3 Figure 3: Area Between f 1x2 = x 2 .1 to 2. That means we use the definite integral. and Finance. Gordon. see Figure 2. Solution We first sketch the curves.B1x22 dx. and we obtain the area A b A = a 3 1T1x2 . note that the top and bottom curves do not change over this interval. 4 and g 1x2 = 2x We need to determine the two points of intersections of the curves.7b = + + 7x b ` = a .474 * ** Section 5. their y-values are the same.1x . f( x ) = 2 x 2 . the curve on the top is the line g1x2 = x + 4 and the bottom curve is the parabola f1x2 = x2 .2 = 0 factoring. Wang.2x . x2 .a + .4 (2.4 and g1x2 = 2x.x . Gordon. Copyright © 2007 by Pearson Education. Applied Calculus for Business. 4) g (x) = 2 x (-1. We also indicate a representative rectangle. f1x2 = g1x2 or 2x2 . Economics. Walter O. When the curves intersect.8 Applications of the Definite Integral Everywhere on this interval. . -2) Figure 4: The area of the Region Bounded by f 1x2 = 2x 2 .322 dx = -1 3 1 . Inc. as the representative rectangles begin and end at these points.+ 3 2 3 2 3 2 2 -1 Example 2 Determine the area between the curves f1x2 = 2x 2 . so we have at the intersections. see Figure 4. Published by Pearson Learning Solutions.3.x2 + x + 72 dx = a- 2 x3 x2 8 4 1 1 39 + 14 b . Solution It is always useful to sketch the graph.4 = 0 dividing the equation by 2. by Warren B. and Finance. and April Allen Materowski.4 = 2x or we have the quadratic equation 2x2 . where we see the line is the top curve and the parabola is the bottom one. The area is 2 2 2 A = -1 3 11x + 42 . 2x 2 + 2x + 42 dx = a- 2x 16 2 + 4 + 8b . x = . we could select an x-value between these intersections and compute the corresponding y-values on each curve. We first find the three vertices. B h ( x) = 4 x f(x) = x2 g(x) = 2x A O Figure 5: Region bound by f 1x2 = x 2. You can better appreciate the need for visualization with the next example.4b = 9 + x 2 + 4x b ` = a 3 3 3 -1 We should remark that if there are two intersections. and Finance. by Warren B.1 and 2 at the intersections (the corresponding y-coordinates are . the larger y-value is the curve on the top. . then they define one region over which we need to integrate in order to determine the area.a + 1 . Economics.221x + 12 = 0 Thus. and without drawing the graph. Copyright © 2007 by Pearson Education. We could actually solve the previous problem by first determining the points of intersection. g1x2 = 2x and h1x2 = 4x. However. the smaller the bottom curve.12x .8 Applications of the Definite Integral * ** 475 1x .Section 5. and we have 2 2 A = -1 3 3 112x2 . as we did above. Published by Pearson Learning Solutions. Gordon. and April Allen Materowski. Inc. Walter O.422 dx = 2 2 -1 3 1 . and h 1x2 = 4x Observe the triangular region is bounded by the three curves with vertices are O. Wang. Solution We first give a sketch showing the region in Figure 5. f 1x2 = 2x. usually it is an easy matter to sketch the graphs (or have the calculator do it for us). A.2 and 4 respectively). Example 3 Determine the area of the triangular region bounded by the three curves f1x2 = x 2. To find A we find the intersection of f(x) with g(x). or solve x2 = 2x Applied Calculus for Business. so we can better visualize the region. and B. Applied Calculus for Business. and Finance. x2 . Copyright © 2007 by Pearson Education.16) f(x) = x2 h ( x) = 4 x g ( x) = 2 x A(2. We also note that if we draw representative rectangles in this region.4x = 0 or x1x .a8 .b b = b ` = 14 .x22 dx = 0 3 2x dx + 2 3 14x .476 * ** Section 5. Wang. the x-coordinate at B. we have that the required area is 2 4 2 4 A = x2 0 3 2 0 14x . We draw the line x = 2 to emphasize this important point.02 + a a 32 3 2 3 3 3 Note that we could have also found the area bounded by h and f from O to B and subtracted the area bounded by g and f from O to A. but the curve on the bottom is now f. 4) O(0. We leave it as an exercise for you to show this gives the correct area. by Warren B. and x = 2. B(4 . Gordon. . we set x2 = 4x solving. for x-values between 0 and 2 the curve on the top will be h and the curve on the bottom will be g. Therefore. the curve on the top is still h. we proceed the same way. for x between 2 and 4. the x-coordinate at A.2x2 dx + + a 2x 2 3 4 2 3 14x . To find B. Inc.2x = 0 x1x . Walter O.42 = 0 which yields x = 0 (the origin) and x = 4. Thus.x22 dx = x 64 8 28 b .22 = 0 this gives x = 0. See Figure 6. the x-coordinate at O. 0) Figure 6: Illustrating the Different Representative Rectangles The area of the triangular region is the sum of the areas of the abutting regions to the left and right of the vertical line through A. Economics.8 Applications of the Definite Integral or x2 . Published by Pearson Learning Solutions. and April Allen Materowski. the bottom curves are different to the left and right of vertex A. For example. violating the definition of a function. We shall sum horizontal rectangles between the two point A and B.B1x2 which is always positive.Section 5. consider the equation y2 = x. Example 4 Find the area bounded by the x-axis and the curve y = x2 . so we will consider the problem in terms of y. by Warren B.2 and 2 (verify). Economics.422 dx = 2 -2 3 1 . However. This is useful in cases when the equation of one or more of the curves is not a function of x. and Finance. The height of the rectangle is the T1x2 . . this equation does not define y as a function of x because to each x-value other than 0.x = 0. the area is determined directly via (1).8 Applications of the Definite Integral * ** 477 If all or part of a region is below the x-axis.) Sometimes. Published by Pearson Learning Solutions. then this does define a function of y. we sum horizontal rectangles. Wang. But because the rectangle s height is represented by dy.y = 6 and y2 . Observe that if we draw any horizontal rectangle the height of the rectangle is dy and the width of the rectangle is the difference between the x-value on the right curve and the xvalue of the left curve. if we think of x in terms of y. we will be Applied Calculus for Business. Solution See Figure 6. we obtain two different y-values. Example 5 Find the area of the region bounded by x . The next example illustrates this. Solution Note that y 2 = x is not a function of x. namely the curve is written as x = y 2 and the line as x = y + 6. y=0 y = x2 . Gordon. y = 0. and April Allen Materowski. Copyright © 2007 by Pearson Education. That means instead of summing vertical rectangles.1x . Inc. Walter O. The graph intersects the x-axis at . it is easier to integrate with respect to y to compute an area. We illustrate the idea in the next example. but can be written as a function of y. we leave the details to you. even with two negative y-values this difference will be positive.x 2 + 42 dx = 32 3 (The easiest way to compute this integral is to observe that it is an even function over a symmetric interval. so we have 2 2 A = -2 3 10 . We give a sketch of the region in Figure 7.4.4 Figure 6: Computing an Area Below the x-axis Observe that the top curve is now the x-axis. Economics. but note that x = y2 has to be considered as two functions namely y = 1x and y = . so we have 3 3 A = -2 3 1y + 62 . Walter O. . So the x-value on the right curve is x = y + 6. Gordon. B (9. by Warren B.2 and continue until the highest. y = .1x.y22 dy The rectangles begin at the lowest y-value. Consider Figure 8.478 * ** Section 5. Mathematically. so the problem transforms into finding the area of the region bounded by x .y = 6 and x 2 . and the x-value on the left curve is x = y2 so the area of the representative rectangle is 11y + 62 . -2) Figure 7: Area by Horizontal Rectangles Notes: (1) Here is another way to solve the previous example. interchange x with y everywhere.y2 + y + 62 dy = 125 6 B (9. the area does not change. Thus. and April Allen Materowski. Inc. (2) The given example can also be done using vertical rectangles. 3) x = y2 dy x=y+6 A (4.y = 0. this means we are rotating the coordinate system. Wang. y = 3.8 Applications of the Definite Integral integrating with respect to y. We leave the details to you as an exercise. 3) y=x-6 A (4. and Finance. Under this rotation.y 2 dy = 2 -2 3 1 . we need to write the two x-values that represent the width of the rectangles in terms of y. -2) Figure 8: Using Vertical Rectangles to Find the Area in Example 5 Applied Calculus for Business. Copyright © 2007 by Pearson Education. Published by Pearson Learning Solutions. Now y is a function of x and you can find the area using vertical rectangles. If the units are small (in the limit. Gordon. which shows the demand curve the item and the price per item. causing an oversupply and forcing the producers to lower their price. their intersection E1xe. the total value to the consumer for xe units of the items is p112 + p122 + Á + p1xe2 which is the shaded area shown under the demand curve (note that each rectangle has width 1 so the area of each rectangle is the indicated price).622 dx We leave the details for you to complete. The area is then the sum of the areas of the two regions. see Figure 9.1x . To sell two units of the item. Economics. Inc.. p (1) p (2) p (3) Consumer and Producer Surplus p = D (x ) pe . While pe is the market price for a given item.Section 5. but actually pays pe. then the shaded area. Published by Pearson Learning Solutions. Thus. by the fundamental theorem Applied Calculus for Business. Recall.8 Applications of the Definite Integral * ** 479 Note that the bottom curve changes to the right of A. the price would then be p(2). We can continue this process until we reach equilibrium. Walter O.1x22 dx + 4 3 1 1x . that when supply and demand equations are plotted on the same set of axes. if the price is below the equilibrium price. since the consumer is willing to pay the price p(1) (the value the consumer attached to the item). and April Allen Materowski. If the price goes above this value. the first region to the left of A and the second to its right. pe2 is called the point of market equilibrium. sold at the price p(1). and Finance. the total value to the consumer. supply and demand will be equal. especially if it is an item that is greatly desired by the consumer. it often happens that the consumer is willing to pay a higher price for the item. a person would have been willing to pay the price p(2) for that item (the value the consumer attaches to the item). there will be a shortage followed by an upward adjustment in the market. the equilibrium price is the price that the market determines for any given commodity. the supply will increase and the demand decrease. Suppose only one item is available to the consumer. the price paid by the consumer is actually pe. xe Figure 9: The Demand Curve and the Value to the Consumers at Each Price Thus. the difference between the two prices 1p112 . for the second unit.1 . Copyright © 2007 by Pearson Education. Similarly. as the width of the rectangles approaches zero). so when one unit is sold. We have 4 9 A = 0 3 1 1x . by Warren B. However. Wang.. It is characterized by the fact that at the equilibrium price. and so on. Similarly.pe2 represents a surplus for the consumer. . will become. thus generating a (smaller) surplus. three units of the item would have a price of p(3). p e) Figure 10: Consumer Surplus Example 6 Given the Demand and Supply Functions p = 12 . the consumer actually pays xepe for xe units of the item. (Why?). Gordon. p = D (x) A = Consumer Surplus p = pe E ( x e. Economics.480 * ** Section 5. we find that market equilibrium is at (2. Inc. 8). by Warren B. Copyright © 2007 by Pearson Education.xepe Thus. and April Allen Materowski. geometrically.x 2 and the Market Equilibrium Point (2.122182 = 16/3 Applied Calculus for Business. the consumer surplus is the area A below the demand curve and above the line p = pe shown in Figure 10. .x22 dx . 8) Figure 11: The Demand Function p = 12 . Therefore. Published by Pearson Learning Solutions. Wang. 8) The consumer surplus is 2 0 3 112 . (Note the demand function is p = 12 .8 Applications of the Definite Integral xe L 0 D1x2 dx However. and Finance. the total surplus to the consumer is given by the difference xe 0 3 D1x2 dx . determine the consumer surplus. Solution Solving the two equations simultaneously.x2 and p = 4 + x2.) p = 12 . Walter O.x 2 (2.x 2. Inc. b] is n b lim f1xk2 ¢ x = n: q a k=1 a 3 f1x2 dx Continuous Income Flow where xk is any point in the kth subinterval. and the value t years in the future is Rert. Copyright © 2007 by Pearson Education. determine the producer surplus. 8) p = 4 + x2 122182 - 0 3 14 + x22 dx = 32/3 It is coincidental that the value of the producer surplus in this example was equal to the consumer surplus in the previous example. economists define producer surplus. Walter O. Figure 13: The Supply Function p = 4 + x 2 and the market Equilibrium Point (2. We observe that if f1x2 = g1x2h1x2.8 Applications of the Definite Integral * ** 481 Just as we defined consumer surplus as the total value to the consumer derived by paying a lower price (the equilibrium price) for the items. then the value of R dollars t years earlier is Re -rt.Section 5. Solution We have the producer surplus is (see Figure 13) 2 p=8 (2. Gordon. Economics. Wang.3 that if interest is compounded continuously at a rate r. p = S(x) p = pe = producer surplus E ( x e. Applied Calculus for Business. Published by Pearson Learning Solutions. The derivation is very similar to the one given above for consumer surplus and we leave it to the exercises. the total value derived by the producer by charging a lower price (than the equilibrium price) for the items. see Figure 12. 8) Recall that the definition of the definite integral for a continuous function over the interval [a. then we have n b a n: q k =1 lim g1xk2h1xk2 ¢ x = a 3 g1x2h1x2 dx We also recall from Section 2. and April Allen Materowski. by Warren B. If p = S1x2 is the supply function. . p e) Figure 12: Producer Surplus Example 7 In the previous example. and Finance. then the producer surplus is defined by xe xepe - 0 3 S1x2 dx It is the area above the supply curve and beneath the line p = pe. . Walter O. Copyright © 2007 by Pearson Education. by Warren B. We partition this interval in the usual way and then the product T1tk2 ¢ t represents the total value of the flow over the interval ¢ t. (b) What is its final value? Solution Since the flow of money is uniform. That is. we wanted the final value S of a continuous flow at time t. Wang.8 Applications of the Definite Integral Let T(t) represent a continuous flow of money per year over the interval [0. therefore. we may write t A = 0 3 T1u2e -ru du (2) We note there is a special case. if instead. after integration.000 per year flowing uniformly over a 10 year period if it earns 4% interest compounded continuously.04 10 = $ 12. Gordon.e-rt d r (3) Furthermore. and April Allen Materowski.295. therefore. T1tk2 ¢ te -rtk The approximate present value of the flow over the given interval is then A L a T1tk2 ¢ te -rtk k=1 n We let n become infinite and we have n A = lim a T1tk2 ¢ te n: q k=1 -rtk n t -rtk = lim a T1tk2e n: q k=1 ¢t = 0 3 T1t2e -rt dt The integral follows from the generalization of the definition of the definite integral given above. we have # S = 8242e. then we have t S = Ae = e rt rt 0 3 T1u2e -ru du (4) Example 8 (a) Find the present value of $1.482 * ** Section 5. t] with continuously compounded interest rate r. T = 1000. A = Tc 1 . Not to confuse the dummy variable of integration with the time t in the upper limit. (2) becomes.04 (b) Using (4). and Finance. (a) We.04 10 A = 1000 c d = $ 8242 .e -. have # 1 . Inc. when the flow is uniform (constant).62 Applied Calculus for Business. The present value of this sum is. Published by Pearson Learning Solutions. Economics. T1t2 = T. b]. b]. b] is one. the normal distribution defined by n(x) in the last section is one such pdf whose interval of definition is 1 . Wang. Economics.8 Applications of the Definite Integral * ** 483 Example 9 A continuous money flow begins at $10. Then one question that might be asked is what is the probability that the lightbulb will fail during the third hour of operation? If we are given a pdf defined by the equation y = f1x2 over the interval [a. Copyright © 2007 by Pearson Education. and April Allen Materowski. We shall not examine infinite intervals here. b] b Probability 1b2 a 3 f1x2 dx = 1 (We remark that the given interval may be infinite.02u du = 10. T1t2 = 10.1/ 2 x + 1 on the interval [0. Walter O. and Finance.) Thus.Section 5.581.1/ 2 x + 1 on the interval [0. Gordon. Published by Pearson Learning Solutions. We substitute into (2) and we have 5 5 A = 0 3 10. d] is contained within the domain [a. sometimes the distribution of data is best described by a continuous function called a probability density function or pdf for short. 2].02u 5 d = $ 47. see Figure 14.000 and increases exponentially at 3% per year for 5 years. However.3. we could consider it defined over [ . Solution We have. for a function to represent a pdf on an interval [a.03t. Inc.000e0. q 2. Figure 14: f 1x2 = . In fact.03ue-0. suppose we are studying the lifetime of a lightbulb. b] 1a2 f1x2 Ú 0 on [a. Example 10 Suppose a pdf is given by the equation f1x2 = . For example. then the probability that x will lie between c and d is given by d Pr1c x d2 = c 3 f1x2 dx (5) Observe that since the interval [c.000 0 3 e-0. (a) Verify the function satisfies the two conditions given above and (b) find Pr A 1/ 2 x 1B. 2] Applied Calculus for Business.) For a pdf.02 0 q T1u2 In the study of probability theory.000 c e-0.29 .000e0.q .0. Find the present value if money is worth r = 5% compounded continuously. for all practical purposes. by Warren B.05u du = 10. . Solution (a) A sketch of the function shows that it is non-negative on the given interval. (This corresponds to the case of a discrete distribution where the sum of all the probabilities is one. These non-negative functions have the property that the total area bounded by the curve and the x-axis over its domain of definition [a. the question that is asked is to determine the probability that an event will occur during a specified interval. 3]. the value of the integral in (4) (the probability) is always a number between 0 and 1. C A B Figure 16: Graphing the Two Functions We see from Figure 16 that there are three intersections. Between A and B. and April Allen Materowski. we set y11x2 = x3 . Copyright © 2007 by Pearson Education. We find the three intersections. Applied Calculus for Business. we leave the verification A -21 x + 1 B dx = 16 to you. (Do you need 3 to integrate to show this?) 0 1 (b) Pr A 1/ 2 x 1B = 3 1 2 5 . Economics.484 * ** Section 5.4. which we labeled A. Figure 15: Defining the Two Functions We plot the graph on the calculator. see Figure 16. and Finance. Solution In the Y = Screen. by Warren B. the cubic curve y1(x) is on top and between B and C the parabola y2(x) is on top.3x 2 + 1 and y21x2 = x2 . Inc. B. see Figure 15. The calculator can sketch these curves easily (you may need to set the appropriate window) and also determine the intersection points of the curves.8 Applications of the Definite Integral 2 We leave it as an exercise for you to verify that A -21 x + 1 B dx = 1. (Once again. Published by Pearson Learning Solutions. Example 11 Find the area of the region bound by the curves f1x2 = x3 . Wang. a graph is very useful in determining which curve is on top or bottom. see Figure 17. We illustrate with the following example.3x 2 + 1 and g1x2 = x2 . and C. Walter O.4. Gordon. . making sure our window shows the appropriate intersections.) Calculator Tips We saw that in determining the area of more complicated regions. 1 and 9x . 12.1 and g1x2 = 5x . Inc. 1.000 and increases exponentially at 6% per year for 12 years. 2. determine the area of the region between the given curves. . x = . In Exercises 16 20.3x2 + 2x + 1 and g1x2 = 1/ 4120x 2 + 4x .x + 8.384 + 12p = 0. (b) What is its final value? 23. y = 0. x2 + 6x . 3. 2p + x . .000 is a continuous uniform stream over the year. Gordon. The winner of a $1. 9.1 and g1x2 = .000 Lotto Jackpot receives 50.x + 3y = 2.4. .x = .152.11p = 0 21. We obtain EXERCISE SET 5. (a) Find the present value of $10. and Finance. x.25 5 + 25 b + 1 a y21x2 .2 and y = 11. 8p .1 and x = 2. 24. x = . Find the (a) present value and (b ) final value.2 and x = 2. determine the (a) consumers surplus and (b) producers surplus for the given demand and supply curves.25 5 . p = 216 . x + 3y2 = 4 and . f1x2 = x2 . 10. 14.000 a year for 20 years.y21x2. 5. 4.x 20. f1x2 = x and . Applied Calculus for Business. 5 .1. Wang. they 5 .3x + 2x . 25.000 and increases exponentially at 2% per year for 15 years. If money earns 4% interest compounded continuously. f1x2 = 2x + 1.8 In Exercises 1 15. if money is worth r = 4% compounded continuously.300 = 0.8 Applications of the Definite Integral * ** 485 Figure 17: Finding the Intersections The calculator gives the three intersections (you need to scroll to see the third one). A pdf is given by the equation f1x2 = 16 18 . We now have the calculator compute the area 2 2 for us. 13.5y = 17. . A continuous money flow begins at $12. f1x2 = x3 + 3.000 per year flowing uniformly over a 20 year period if it earns 6% interest compounded continuously. Copyright © 2007 by Pearson Education. and (b) determine Pr11 x 32.. y = 0.200 = 0 17.3x 2 + 4x .2x . Published by Pearson Learning Solutions.x .1 and x = 2. 8 24 and press ENTER.4x2 + 7 and g1x2 = 2x2 + 1 8.490 = 0 3 2 2 2 18. x . . f1x2 = 2x3 . (b) What is its final value? 22.1. 16. 6. x. f1x2 = . x = . f1x2 = x3 and y = 8x.x = 2 and y . Walter O.000. We enter 1 a y11x2 . Assume the $50.x = 0. x = . f1x2 = 2x3 . Economics.x + 5y = 12. y2 . f1x2 = x2 . 5p + 14x . and April Allen Materowski. x2 + 8x + 220 .5p + 2x + 10 = 0. (a) Find the present value of $8.7p + 2x + 7 = 0.3x + 4 and g1x2 = .1. f1x2 = 2x . p = 2x + 14. 15.2y2 = 1 and x + 2y = 5. 11.1. f1x2 = . (a) Verify that f is a pdf. A continuous money flow begins at $1. f1x2 = 2x2 .1. 4]. if money is worth r = 5% compounded continuously. what is the lump sum amount that is paid off immediately? 1 26.3x2 + 2x .2x2 .4x . b 2 2 2 107 35 25 + .Section 5.y11x2. x = . Find the (a) present value and (b ) final value.000 per year flowing uniformly over a 7 year period if it earns 3% interest compounded continuously. by Warren B.25 5 + 25 are x = . y = 0. p = 236 .2x2 over the interval [0.x 19. 7. 34.S1x22 dx. Show that the area between the demand and supply functions is xe 0 3 1D1x2 . The pdf in Exercise 26.12 1u du = 1u .52. The pdf in Exercise 28. A continuous money flow begins at $ T0 and increases exponentially at a continuous rate j per year for t years. 38. Gordon. and we have x 2x + 1 dx = 1u . (a) Determine the value of k if the function represented by the graph in Figure Ex.5 x 12. (d) Pr1 x 12. and April Allen Materowski. 41.12u1/2 du = 1u3/2 . Copyright © 2007 by Pearson Education. Determine the formula for the producer surplus.1. 1]. Given f1x2 = 32 14 . Suppose we try the substitution u = x + 1. (a) Show that f is a pdf and . (a) Show that f is a pdf and determine (b) Pr11 x 22.u1/22 du = L L L 21x + 125/2 21x + 123/2 2u5/2 2u3/2 + C = + C 5 3 5 3 L We can simplify this expression by noting that we may factor 21x + 123/2 from the expression as follows: Applied Calculus for Business. (c) Pr1 . k 3 27. Economics.2. Check your result by multiplying (3) by ert. a constant. Find the present value if money is worth r compounded continuously. 39. also note that x = u . 5. Suppose the lowest x-value for a demand function is x = a. a number m such that Pr1a x m2 = 0. 29 40. Show (2) becomes (3) when T1t2 = T. 28. (d) Pr1x 0. by Warren B. . 30.3 and 5. The pdf in Exercise 29.x22 over the interval [ . x 2x + 1 dx. then L du = dx. compute (4). Determine the formula for the consumer surplus.e -2 determine (b) Pr10 x over the interval [0. b]. 29. 35.52.9 Two Integration Techniques » » » Substitution Integration by Parts Tabular Integration We saw in Sections 5. Walter O. 37. (f) Pr1x Ú 02. 36. Published by Pearson Learning Solutions. analogous to (3). If T1t2 = T is constant. (e) Pr1x 12. 29 is a pdf.0. 33.5 is called the median of the distribution. (b) Find Pr1 . The pdf in Exercise 27. Figure Ex. 2]. Wang. 31.1 x 02. 32. obtaining a specific formula for S. Solve Example 3 by the alternative method suggested at the end of its solution.5 x 12. However. In Exercises 30 33 find the median. for example. where an appropriate substitution reduces the integral to a form that can be easily integrated. Substitution Consider. Inc. there are integrals that are not of this form. and Finance. Suppose the lowest x-value for a supply function is x = a. (c) Pr10.486 * ** Section 5. Solve Example 5 by first interchanging x and y as suggested in the note following the solution.9 Two Integration Techniques For a pdf defined over the interval [a.7 that integrals of the form f1g1x22g ¿ 1x2 dx were easily L solved by the substitution u = g1x2. Given f1x2 = 2e -2x 1 . 32614x + 12 12112x .3 giving du = 2 dx will work. We have Evaluate L x12x . and x = u + 2 . Example 2 x12x . Example 3 1 Evaluate 0 x dx. 3x + 1 Solution We let u = x + 1. We illustrate in the next example. Copyright © 2007 by Pearson Education. the new integrand contains only the new variable. Wang. L Solution We use the same substitution.22 + C 15 15 The substitution works if when we are done. and the expression in the new integrand is one that we can easily integrate. we must always return to the original variable.b + C = + C = 21x + 123/2 a 5 3 5 3 2 2 1x + 123/2[31x + 12 . but the substitution u = 2x .Section 5. We also note from Table 1 the new limits of integration. That is. and April Allen Materowski.3252 dx = 1 2 L u5 du = 1 u6 2 6 + C = 1 6 1 u + c = 12x .3.326 + C 12 12 Let s change the previous example a little so it is not in the form considered in Section 5.1.3252 dx = 1 2 L 1u + 32 5 u 2 du = 1 4 L 1u6 + 3u52 du = 7 + 3u6 u6 u 1 u6 2u + 7 a + b + C = a b + C = b + C = 6 4 7 2 4 14 12x . However.325 dx.3.325 dx = 1 2 12x . . then du = dx. when evaluating a definite integral. and x = u . Published by Pearson Learning Solutions.325 dx. Economics. also note that 3 2x = u + 3. Walter O. Applied Calculus for Business. L L 12x .9 Two Integration Techniques * ** 487 21x + 125/2 21x + 123/2 1x + 12 1 . and Finance. we substitute for the limits of integration as well.5] + C = 1x + 123/213x . Example 1 Evaluate L Solution This is actually of the form we considered in Section 5. Inc.3 giving du = 2 dx. by Warren B.32 + 72 + C = + C 56 56 You should note that when evaluating an indefinite integral. We have 12x . making it unnecessary to return to the original variable. We could expand the integrand.325 dx = 1 u 4a 7 1 2 L x12x . Gordon.326 12x . we may use a complete substitution. u = 2x . 12/2. 1 x + 4 L Applied Calculus for Business. Often the choice is obvious.ln 22 . Walter O.1 2 1/3 u 2 1 2 3 8 x 212x + 121/32 dx = 8 1 3 A B du = 10 3 u2 .1 u 1 1 dx = du = a . 1 2 2 2 0 x u . 8 3 = A 2 3 8 B N = 2N somewhat simplified the calculations. Example 4 Evaluate 3 7/2 x2 2 3 2x + 1 dx.11 . Published by Pearson Learning Solutions.ln 12 = 1 . our objective is to choose u to be the expression which best simplifies the integrand into one which can be easily integrated.1/1x + 12. Gordon.b du = a 1 . giving du = 2 dx.2u + 1 1/3 u du = 4 3 1 7 4 1 3 1u 7/3 .ln 2 (Note that since the degree of the numerator is the same as the degree of the denominator. by Warren B.488 * ** Section 5. Economics.9 Two Integration Techniques Table 1: New Limits of Integration x 0 1 u = x + 1 u = 0 + 1 = 1 u = 1 + 1 = 2 We have. One final example on the substitution method: Example 5 Evaluate dx . From Table 2.2u 4/3 + u 2 du = 1/3 1 3u 8a 6u3 3u3 8 1 7332 27 29301 b = + b ` = 8a 10 7 4 35 140 1120 1 N The arithmetic was somewhat messy in the last example. and April Allen Materowski. Wang.ln u 2 2 1 = 12 . and x = 1u . we obtain the limits of integration Table 2: New Limits of Integration x 0 7/2 u = 2x + 1 u = 2102 + 1 = 1 u = 217/22 + 1 = 8 We have.b du = u u u u 3 3 3 3x + 1 1 1 1 1u . 0 Solution Let u = 2x + 1. and Finance. 7/2 7/2 1 2 0 1 2 8 1 8 3 x2 2 3 2x + 1 dx = 0 u . fortunately. we could have performed long division in the integrand to obtain 1 .) When applying substitution. Inc. Copyright © 2007 by Pearson Education. . Economics. Doing so. and April Allen Materowski. yet we shall see that it is most effective in integrating expressions which are the product of two different types of functions. and Finance. In the event that we are evaluating a definite integral. it is nothing more than the reformulation of the product rule. Gordon.4 ln u 2 + C1 = 21 1x + 42 . We make the observation that if u and v are differentiable functions of x. (Note that to evaluate 4 1 du requires the substitution u + 4 = w.4 ln u + 4 2 + c2 = du = 2 a 1 u + 4 Lu + 4 L 21 1x . giving 2u du = dx.) Another approach to this problem is to let u = 1x. we have 2 u 4 b du = 21u . or equivalently.4 ln1 1x + 422 + C1 = 2 1x . Published by Pearson Learning Solutions. . Wang. d dv du 1uv2 = u + v . 1 dx 1 u-4 4 dx = 21u .Section 5. u2 = x. Inc. then 1x = u . Walter O. giving u dv = uv v du (1) Integration by Parts L L (1) is called the integration by parts formula.) There are many different techniques of integration.8 ln1 1x + 42 + 8 + c1 = 2 1x . which is one of the most powerful and is based only on the product rule for derivatives. Copyright © 2007 by Pearson Education.8 ln1 1x + 42 + C (Note that we set 8 + C1 = C. it is called integration by parts.42 du = 2 du = 2 a 1 . so we must divide. but in an introductory text like this we shall consider only one more method.4.9 Two Integration Techniques * ** 489 1 Solution Let u = 1x + 4. and du = dx or dx = 2 1x du = 2 1x 21u .4 ln1 1x + 422 + c2 which is the same result obtained using the other substitution. the corresponding integration by parts formula is Applied Calculus for Business. dx dx dx or multiplying by dx (writing in differential format) d1uv2 = u dv + v du or rewriting u dv = d1uv2 . So we have.b du = = u Lu L u L L 1x + 4 L 1x + 4 21u .42 du. by Warren B. Integration by parts does not appear to be a deep result.v du We integrate each side of this equation. or perhaps you observed du = Lu + 4 Lu + a ln u + a + C. We then have 1 dx 1 u dx = 2u du = 2 du = u + 4 u 1 x + 4 1 x + 4 L L + 4 L L Now note that the degree of the numerator is the same as the denominator. How do we know if made the proper identification. 2. Consider p1x2q1x2 dx. then seeing how they cancel and/or combine into one constant at the end. Walter O. then we have x2 x dx = .) xe2x dx = e2x a x2 x2 2x x 2e2x 2e dx = b x 2e2x dx 2 2 L2 L L Note that the resulting integral is more complicated and higher order in x than the original integral. and April Allen Materowski. we remind you that eax dx = eax a where the constant a Z 0.9 Two Integration Techniques b b a 3 u dv = uv b a - a 3 v du (2) To use integration by parts effectively we must examine the integrand and identify one of its factors as u and its remaining factor as dv. Then we usually let u = p1x2 and dv = q1x2 dx in (1). When we cannot easily integrate q(x). p1x2 = x and q1x2 = e2x. Inc. We now try the first choice: u = x then du = dx Substitution into (1) gives 2x xe2x dx = x A 1 2e B x 2x 2x A1 2 e B dx = 2 e 1 2 2x 2x . rather than carrying around additional constants of integration (when computing v. In the previous example. (and the constant of integration is omitted). then we may try reversing u and dv as the next example illustrates. where p(x) is a polynomial and q(x) is another function L which we can integrate. Applied Calculus for Business. Here are the most obvious choices for u and dv u = x u = e2x dv = e2x dx or dv = x dx Suppose we use the second choice. Gordon. as there are often several possibilities? 1. that is. Published by Pearson Learning Solutions. for example). Copyright © 2007 by Pearson Education.1 + C e2x dx = x 2e 4e dv = e2x dx v = L e2x dx = 1 2x e . by Warren B. so obviously this choice for u and dv is will not work. u = e2x and dv = x dx. Economics. The resulting product vdu should usually be a simpler product (or lower order) than the original product udv. L Consider L xe2x dx. Wang. . (We indicate below why we may ignore the condu = 2e2x dx and v = 2 L stant of integration. Whichever expression is identified as dv we should be able to integrate. 1 Before going any further.490 * ** Section 5. 2 L L L Note that in the previous example. and Finance. You should convince yourself why this is permissible by carrying along all constants of integration. to simplify our work. we added one constant of integration at the last step. Gordon. Published by Pearson Learning Solutions. and Finance. v = x L L 1 dx = x ln x dx = x ln x . Solution Note the polynomial expression is x. Walter O. and April Allen Materowski. Inc. Copyright © 2007 by Pearson Education.x 2 + C 2 4 The next example is somewhat unusual in that there is a single function in the integrand. L We now let L u = ln x then du = 1/x dx and we have using (1): ln x dx = 1ln x2x x# dv = 1 dx. Instead we reverse the choices as indicated above. We will illustrate how this happens in the next example and then learn how to set up a simple table that will keep track of the multiple use of integration by parts for us. we do not (yet) know how to integrate ln x. u = ln x Then du = 1/x dx We now have x ln x dx = 1ln x2 # x 2 2 dv = x dx v = x 2/2 L L 1 dx = A x2 B x 2 x2 2 ln x - 1 2 L x dx = x2 1 ln x . Example 7 Evaluate L ln x dx.Section 5. Economics. but the process may need to be repeated. We make the following identification Applied Calculus for Business. but if we let dv = ln x dx. Wang. Example 8 Evaluate L x2e2x dx. Solution This problem is very similar to the introductory example. by Warren B. and we proceed the same way.9 Two Integration Techniques * ** 491 Example 6 Evaluate L x ln x dx. .x + C x L It often turns out that integration by parts works. Solution The trick here is to rewrite the integral as follows: ln x dx = ln x # 1 dx. rather than a product of functions. 12x2 A 1/ 4e2x B + L even easier. namely x2e2x dx = x2e2x x2e2x x 2x xe2x dx = . Let s continue the process.492 * ** Section 5. Also note that the signs alternate. 1 + 2x2 A 1/ 2e2x B . differentiation and integration. Inc. We did this before so let s just write down what you get when you evaluate by parts. but why stop here? One more row (integration) makes the evaluation Applied Calculus for Business.e2x + 1 4e 2 2 L Tabular Integration Let s analyze the last problem a little more to see how we can automate the process a bit. Table 4: Continuing the Table Sign + + Differentiations x2 2x 2 Integrations e2x 1 2x 2e 1 2x 4e 2x If we stop right now then we obtain 1 + 2x2 A 1 2e B - 2x A1 2 e B 12x dx2. We have. and Finance. by Warren B. Gordon. Published by Pearson Learning Solutions. then we obtain from Table 4.1 4e b + C = 2 2 2 L x2e2x x 2x + C . Wang.a e2x .9 Two Integration Techniques u = x2 then du = 2x dx substituting into (1) gives 2x x2e2x dx = x2 A 1 2e B - dv = e2x dx 2x v = 1 2e L L 2x A1 2 e B 12x dx2 = x 2e2x xe2x dx 2 L Notice the last integral on the right needs to be evaluated by parts again. and April Allen Materowski. 2x 122 A 1 4 e B dx. Lets keep track of what was done in the previous problem in a tabular format (we omit the differential dx). Economics. Table 3: The Start of Tabular Integration Sign + Differentiations x2 2x Integrations e2x 1 2x 2e L we multiply expressions along the diagonal and integrate across the columns in Table 3. Copyright © 2007 by Pearson Education. inserting the signs as shown. You ll note that in a repeated integration by parts. Walter O. This means If we stop now. . we are doing two operations. Section 5. Published by Pearson Learning Solutions. and the cross integration term vanishes. a column of derivatives. when we get to the last row. by Warren B. Economics.e2x + 1 4e 2 2 L Table 6 gives the generalization. We have from Table 5. p1x2q1x2 dx. and the signs continue to alternate. and April Allen Materowski.u2v3 + u3v4 + Á .9 Two Integration Techniques * ** 493 Table 5: Completing the Table Sign + + Differentiations x2 2x 2 0 Integrations e2x 1 2x 2e 1 2x 4e 1 2x 8e Now that we have a 0 in the last row. a column of alternating signs (beginning with + ). can always be quickly evaluated using tabular integration (integration by parts). Whenever we stop. where p(x) is a polynomial L and q(x) is a function all of whose integrals are easily determined. beginning with dv (with all dx terms omitted). Moreover.1vn + 1 . Gordon. The table structure is just a short hand method of doing repeated integration by parts a bookkeeping method and works well on most problems which require integration by parts. We remark that integrals of the form Applied Calculus for Business. Walter O. . Table 6: Tabular Integration row 1 2 3 o n Sign + + o 1 . Wang. v1 v2 v3 o vn o Basically. the objective is to continue the process by adding rows until the last row has a zero in the derivative column (as in Table 5). we integrate the product (including the sign) of the derivative and integral columns (following the horizontal arrow). u1 u2 u3 o un Int.12nun .12x2 A 1/ 4 e2x B + 122 A 1 8e B We now add the constant of integration.12n + 1 Deriv. repeated integration by parts has the form + u1v2 . and a column of the integrals. Copyright © 2007 by Pearson Education. the last term is the integral of the product of the last row with the appropriate sign in the alternation process.12n + 1 L unvn dx Note that the subscripts correspond to the row number in the table. the integration by parts result is obtained by multiplying the entries in the derivative column (along with the associated sign) with the entry on the next row of the integral column (follow the slanted arrow). and have that x 2e2x x 2x x2e2x dx = + C . In summary. Inc. and Finance. 2x 1 + 2x2 A 1/ 2 e2x B . we form three columns. + 1 . It is especially convenient for other products of functions not considered in this text. beginning with u. 1 3 x ln1x2 dx 12. L -2 5. . Exercise 2 24. Copyright © 2007 by Pearson Education. 0 3 2x 2 3 x + 1 dx x2 22x . 16.1 dx 3x3 2 4 x + 3 dx In Exercises 16 17. Economics. (c) Pr1x 22. its mean.1 7 1 -1 3 x2e -2x dx dx 15. 0 3 m = xe xe dx 8. In the evaluation of 2x xe2x dx. we 2e L carry along the intermediate constant of integration). 4] is a pdf. by Warren B. 2] is a pdf. (c) Pr11 x 22. Given a pdf defined over the interval [a. and Finance. f1x2 = x ln x and y = 0. b]. Exercise 4 26. Exercise 3 25. Walter O. 3xe2x dx 2xe -3x dx 2 . Published by Pearson Learning Solutions.9 In Exercises 1 5. L x n ln1x2 dx 3. 17. 22. 21. find the mean.1 and 1. Exercise 5 27.2x L 1 7. is defined by b 4. Wang.494 * ** Chapter Review EXERCISE SET 5.x over the interval 128 [0. between x = . and April Allen Materowski. between x = 1 and 2. -3 3 In Exercises 6 15. evaluate the given integral. Inc. Find (b) Pr12 x 32. L x 2x + 1 dx 3x + 2 L 2x . Gordon. suppose we use v = 1 + c (that is. 10. n 3 4x dx x4 dx 2x Le 2 11. 4e4x2e-2x 18. and then use integration by parts to evaluL ate the integrals in Exercises 22 26. a 3 xf1x2 dx. m. f1x2 = x2e -2x and y = 0. L L ln12x + 32 dx x1ln x2 dx 2 CHAPTER REVIEW Key Ideas Antiderivative Integration Theorems Simple Power Rule Simple Logarithmic Rule The Simple Exponential Rule Particular Solutions Equations of Motion Marginal Functions Separable Differentiable Equations Reversing the Chain Rule Generalized Power Rule Generalized Logarithmic Rule Generalized Exponential Rule Areas by Rectangles Left Endpoints Right Endpoints Midpoints Sigma Notation Applied Calculus for Business. Find (b) Pr10 x 12.13 [0. The pdf defined in Exercise 19. evaluate the integral by substitution. 1. (Note: you do not need to integrate to solve (c). Evaluate the integral 2ax + b dx. (a) Verify that the function defined by f1x2 = 4 over the interval e . find the area of the region bounded by the graphs of the given equations. 13. L L 20. The pdf defined in Exercise 18. 14. In Exercises 20 21.) 15x 19. (a) Verify that the function defined by f1x2 = 24 . 9. 6. Exercise 1 23. Show that it is unnecessary. 2. 2x. 6. Solve the differential equation = 2 . . An object is thrown vertically upward with an initial velocity of 48 ft/sec off the ledge of a building 160 ft above ground. find G ¿ 1x2. 7. by Warren B. if money is worth r = 6% compounded continuously. 3t3 21 + 4t2 dt 2x 2 3 8 + x2 dx e . (b) right endpoint and (c) midpoint of each subinterval. and the x-axis between x = . (a) How high does it go. (a) Evaluate lim a a 2 a b + 4 a b + 7 a b . Evaluate L 4 . (b) right endpoint and (c) midpoint of each subinterval. Evaluate dt 2 1 n i 3 i 2 i 2. (a) Verify that the function defined by f1x2 = e 8 . 17.12. (b) Determine Pr A .t2 dt. Economics.5% interest compounded continuously. Find the area of the region bounded by f1x2 = 2x4. 5.x dy xy3 9.1/2 (d) Pr1x Ú 3.000 and increases exponentially at 4. x 11. Using Riemann sums. Evaluate 0 3 12. Inc. Find the (a) present value and (b) final value. Published by Pearson Learning Solutions. Using four rectangles. Evaluate 1ln x22 x 1 3 x2 t 21 . and Finance. and p = 28 . x = 0 and x = 1. g1x2 = 2x and h1x2 = 16x.4i2 i=1 15 3 t1ln t2 1 3 15.2i 2 + 4i . The supply and demand curves for some commodity are given by the equa2 tions p = 1 4 x + 1. Determine the (a) consumer surplus and (b) producer surplus. Evaluate a 13i3 . dx x + 1 10. 18. and April Allen Materowski. 2].1 b .72 i =1 n 3 14. 1 3 dx 24. The marginal revenue for some commodity is given by the equation 18 . 2 29 .Chapter Review * ** 495 Linearity Property Summation Formulas Reimann Sums Areas by Riemann Sums Definite Integral Fundamental Theorem of Integral Calculus Basic Properties Substitution Odd and Even Functions Average Value Derivative of an Integral Area Between Curves Consumer and Producer Surplus Continuous Income Flow Substitution Integration by Parts Tabular Integration 1.5u 21.5u du 23. (c) Pr1 .12. and (b) with what speed does it impact with the ground? 8. (b) Represent n: q n i=1 n n n (a) as a definite integral. determine the demand equation. Compare these approximations with the exact area. Copyright © 2007 by Pearson Education. approximate the area bounded by f1x2 = 2x3 + 1 and the x-axis for 1 x 5. Gordon. (a) Find the average value (b) of the function whose equation is f1x2 = xe-3x over the interval [1. find the area of the region bounded by f1x2 = 3x2 + 2. (a) Find the present value of $20.2e 2 . using the (a) left endpoint.1 and x = 2. For the region described in the previous exercise. Wang. 3.x. Evaluate a 14i + 1 .2% per year for 20 years. Walter O. y = 0.6x + 20 -1 3 x x 3 10 3 is a pdf. Find the area of the region bounded by f1x2 = x 3 21 . 1 2x + 1 20.3x R ¿ 1x2 = . Applied Calculus for Business. Evaluate 0 3 3x 5e -2x dx 4. given y102 = .1. (b) What is its final value? 22. If G1x2 = 1 3 t 21 . Evaluate L 1 16. x 5/2 B .t2 dt.000 per year flowing uniformly over a 30 year period if it earns 4.1 x 3. (b) Determine (approximately) the x-value at which the function assumes its average value. use a spreadsheet with n = 100 rectangles to approximate the area using the (a) left endpoint. 19. find G ¿ 1x2. A continuous money flow begins at $10. If G1x2 = 13. Published by Pearson Learning Solutions. Economics. Inc. and Finance. Copyright © 2007 by Pearson Education. and April Allen Materowski. Gordon. Wang. .Applied Calculus for Business. by Warren B. Walter O. multiple integrals and their applications. and Finance. Copyright © 2007 by Pearson Education.6 An Introduction to Functions of Several Variables This chapter extends the notions of the previous chapters to functions of two or more variables. Inc. Economics. Published by Pearson Learning Solutions. . Walter O. Wang. It also introduces the method of LaGrange multipliers. by Warren B. Gordon. and April Allen Materowski. Applied Calculus for Business. The definition extends is a natural way to any number of variables. Example 1 1 3 Suppose the productivity z is given by the equation z = f1x. as the next two examples illustrate.2 3 1 3 1 Example 2 Given the function defined by the equation w = f1x. 1Recall 814 = 2 and that 164 = A 2 4 16 B 3 = 23 = 8. Gordon. Wang. y2 = 100x4y4. and y is the number of units of capital available. It follows that in realistic applications. Computations with functions of two or more variables are handled the same way as with a function of a single variable. The notions of limits and continuity may be extended to a function of several variables in a very natural way. 3. 22. productivity is a function of the two variables. capital and labor.1. 162 = 1001812411624 = 100132182 = 2400. Consider first the case of a function of two variables. While we are not examining these concepts in details. as our objective is to provide an elementary introduction to the subject. (x. The idea of the difference quotient also can be generalized. 3. Copyright © 2007 by Pearson Education. y).1 Functions of Several Variables » » » » » Functions of Several Variables Difference Quotients Three-Dimensional Coordinate System Surfaces Calculator Tips Function of Several Variables In Chapter 1. z2 = xy . that is. that is. A more detailed treatment of these concepts is left to a more advanced class.1 . written z = f1x. shortly. . y. Economics. f181. we introduced the notion of a function of two or more variables. we associate a unique value z. Published by Pearson Learning Solutions. The definition of function extends in a very natural way. Solution 4 81 = 3. We shall not consider these generalizations here. we shall use them as needed. Productivity may depend on the amount of available capital as well and the number of units of labor available. You substitute for each variable as it appears in the equation.498 * ** Section 6. 16). where x is the number of units of labor available. Applied Calculus for Business.1 Functions of Several Variables 6. and April Allen Materowski. 22 = 1 .1.12132 . Inc. cost of steel. The cost of an automobile may depend on the labor. we may often need to deal with functions of more than a single variable. Walter O. and Finance. Solution f1 . y2. To each pair.12122 + 3132122 = 17. the cost is a function of four variables. Determine f(81. we first review what was done there and then continue with a visualization of such functions.xz + 3yz. determine f1 . fiberglass and rubber. We remark that nice functions have limits and are continuous. we shall give meaning to theses difference quotients. by Warren B. y2 = = h h h12x + h + 2y2 2xh + h2 + 2hy = = 2x + 2y + h h h (b) f1x. assuming neither h nor k is zero. Economics.f1x. Wang. Gordon. Published by Pearson Learning Solutions. y2. that these difference quotients have simple interpretations.f1x. y2 = x2 + 2xy + y2 1x2 + 2xh + h2 + 2xy + 2hy + y22 . assuming neither h nor k is zero. y2 f1x. y2 = x + 3y2. and may be obtained using differentiation. y2 .1 Functions of Several Variables * ** 499 Example 3 Given z = f1x. y2 = x2 + 2xy + y2 f1x. y2 = 2x + 2y + k. y2 . by Warren B. y. h:0 k:0 h k Solution From the previous example. y + k2 . we found that f1x. Copyright © 2007 by Pearson Education. y + k2 . y + k2 . (b) . we computed the corresponding z value.1x2 + 2xy + y22 f1x + h. Walter O. y2 . 1. y2 .f1x. then each point (x. and April Allen Materowski.1x2 + 2xy + y22 = k k k12x + 2y + k2 2xk + 2yk + k2 = = 2x + 2y + k k k Difference Quotients We now take the previous example one step further. therefore. Suppose. z = f1x. k f1x. y2 (a) . y2 = lim 12x + 2y + k2 = 2x + 2y lim k:0 k:0 k We shall see. Suppose we consider a function of two variables. determine f1x + h. to each pair (x. and Finance. y. y2 f1x.f1x. that is. z) represents a point in three-dimensional Three Dimensional Coordinate System Applied Calculus for Business. y2 1x2 + 2xy + 2xk + y2 + 2yk + k22 . y2 = x2 + 2xy + y2. Example 4 Given z = f1x. 1) we find f12. That is. z). y2 . then at (2.f1x. determine f1x + h.f1x. therefore h h:0 lim f1x + h. . and we have the triple (2. y) there corresponds a unique z.f1x. y2 = lim 12x + 2y + h2 = 2x + 2y h:0 h (b) similarly.Section 6. For example. if z = f1x. in the next section. to each pair (x. 12 = 2 + 31122 = 5. y + k2 . y2 = 1x + h22 + 21x + h2y + y2 = x2 + 2xh + h2 + 2xy + 2hy + y2 f1x.f1x. Inc. y2 (a) = 2x + 2y + h.f1x. we have the triple (x. y2 = x2 + 2xy + y2. y + k2 = x2 + 2x1y + k2 + 1y + k22 = x2 + 2xy + 2yk + y2 + 2yk + k2 f1x. h k Solution (a) f1x + h. 5). y) in the domain of the function. (b) lim . we found that f1x + h. y + k2 . y2 (a) lim . by computing a limit for each of the indicated difference quotients.f1x. z y x Figure 1: Three-Dimensional Coordinate System The dotted portion of the axes represent the negative part of the axes.500 * ** Section 6. we will need to locate such points. 0) is the x-y plane (the usual twodimensional plane). y. and Finance. Copyright © 2007 by Pearson Education. A two dimensionalcoordinate system divides the plane into four quadrants. For example. in any detail in this text. Surfaces may have extreme points which may be the maximum and minimum values of the function. but will illustrate such surfaces as they prove useful to us. giving the eight octants). there are two choices for z. We should. Inc. Gordon. Economics. however. z) is the two-dimensional yz plane. we obtain the graph of the surface. z) is the two-dimensional xz plane and set of all points of the form (0.coordinate system takes the two-dimensional coordinate plane. Note that the set of all points of the form (x. We draw the coordinate system in Figure 1. Surfaces Figure 2: A Surface with a Minimum Applied Calculus for Business. the drawing of three dimensional surfaces. and April Allen Materowski.dimensional. the set of all points of the form (x. y. A three-dimensional surface illustrates such points. The positive x-axis should be visualized as coming out of the page. 0. by Warren B. above or below the xy plane. We will not consider. consider the surface in Figure 2. In order to optimize a function. Published by Pearson Learning Solutions.1 Functions of Several Variables space. and adds a third axis. By plotting the points determined by the equation z = f1x. y2 on a threedimensional coordinate system. . Walter O. a three-dimensional coordinate system divides space into eight octants (to each of the four quadrants in the usual xy plane. to produce the three-dimensional coordinate system. and the totality of all these points produce a graph which is called a threedimensional surface. the z-axis. Wang. note that the three. be visualized in three dimensions for fixed values of one of its variables). however. so we clearly cannot draw its graph (it could. see Figure 4. Figure 4: Z = Screen Suppose z = f1x. its minimum z-value is at the bottom (at the origin). Gordon. Figure 3: A Surface Resembling A Mountain Range with Valleys Calculator Tips Our objective is now clear. and we want to evaluate f(81. y2 = 100x1/4y3/4. where we use z1 in place of z.Section 6. clearly. Wang. we need to determine how to locate its maximum and minimum values. y. with many (relative) extrema. and Finance. consider w = f1x. Inc. z2. 16). What about a function of three or more variables? For example. This surface looks like a mountain range along with its valleys. Walter O. Once this is done then the Y = screen becomes a Z = screen. You must first press the MODE key and change FUNCTION to 3D (three-dimensional) by scrolling down. by Warren B. Published by Pearson Learning Solutions. y. w) is fourdimensional. Applied Calculus for Business. Consider the surface shown in Figure 3. we shall see that the methods considered will work higher dimensional problems as well. the set of all points (x. z. and April Allen Materowski. . we need to learn how to locate such points and classify them. Given a function of two variables. Copyright © 2007 by Pearson Education. Economics. Computations with functions of two variables can also be performed on your calculator. we illustrate in Figure 5.1 Functions of Several Variables * ** 501 This is a bowl-shaped surface. Inc. and Finance. For example. Figure 7: z1 = x2 + 2y2 While the graph is not as clear as we might prefer. Economics. (To insert the axis press * (this opens the graph format window). We next return to the home screen and enter z1(81.dimensions. Copyright © 2007 by Pearson Education. press MODE and change GRAPH back to FUNCTION. Wang.1 Functions of Several Variables Figure 5: Defining z = f1x. Figure 6: Computing f(81.502 * ** Section 6. Published by Pearson Learning Solutions. Walter O. a discussion may be found in the TI manual on page 75. Applied Calculus for Business. Remember to return to a function of a single variable. 16). The WINDOW settings are more detailed in three. Gordon.16) The calculator can also graph three-dimensional surfaces. see Figure 6. change the axes option from OFF to axes. suppose we define z1 = x2 + 2y2 then the calculator. gives Figure 7. y2 = 100x1/4y3/4. and April Allen Materowski. it provides a good visualization of the surface as well as indicating a minimum. by Warren B.) Note that you can use the trace option (F3) in the graph window as well as finding z-values on the graph (F5 then option 1). . after adjusting the WINDOW and zooming in. f1x. 3) 3.2 Partial Derivatives * ** 503 EXERCISE SET 6.f1x. 0). f1x. In Exercise 8. y2 . Copyright © 2007 by Pearson Education. determine (a) f11. 1. x).f1x. (b) f(16. (b) f10. 8. 1 + k2 . y + k2 . . y2 h:0 . . k:0 k h:0 f12 + h. 4. y2 = 75x1/3y2/3. (Let x be the length. 22 7. 1). y2 = 25 ln1x + y22.y2 . (b) f(16.1 1. z2 h . determine (a) lim (b) lim (c) lim h:0 f1x + h. each of the sides cost $2 per square foot and the top costs $1 per square foot.f1x. y2 = 2x2 . .1 b determine S(10. (b) f1 . (b) f(2. 17. n2 = R a r 14. z2 = 2x2 .2. 1). 2 . determine (a) f(2. 2). f1x.1 4xh + 2h2 = lim lim = h:0 h:0 h h h14x + 2h2 lim = lim 14x + 2h2 = 4x h:0 h:0 h lim Applied Calculus for Business. determine (a) lim (b) lim k f1x + h. y2 = 2x2 . y.3x y. The cost of manufacturing a rectangular box is as follows: The base costs $4 per square foot.y .f1x. z + l2 . y. by Warren B. y. f1x. 2). . y2 consider lim for this particular function. z2 = 100x1/2y1/3z1/6.3y3 + 12 . 1). then we have h:0 h f1x + h. .12. (b) f(2. f1x.12. Walter O. and Finance. suppose we f1x + h. and April Allen Materowski. f1x.f1x. Inc. (b) f(2.3y3 + 12 = lim = h:0 h:0 h h 21x2 + 2xh + h22 . (c) f1x.f1x.3y3 + 1 . h:0 f1x. determine (a) f(3.) 6. determine (a) f( 1. y2 . h f1x.3xy2 . y2 121x + h22 .f1x. f1x. Economics. 12 (d) lim .f12. determine (a) f (3. Determine the cost function for the manufacture of this box. 3). 1). y2 . 0. f1x.3xy2 . y. y the width and z the height of the box. 64) 6.f1x.2 Partial Derivatives » » » » » » » Partial Derivative Visualization of the Partial Derivative Level Curves Contours Cobb-Douglas Production Function Utility Functions and Indifference Curves Higher Order Partial Derivatives Calculator Tips Given the function defined by the equation z = f1x.f1x.f12. (b) f(x. y2 15. S1R. y2 = x/y. y2 . In Exercise 1. y2 k f12 + h. determine (a) f(1.f1x. f1x. (c) f1x + h. x) 12. determine (a) f(1.3y3 + 1.3z. (c) f(x. f1x. 1). y2 10. 11. 27. y2 h . y. f1x. f1x. y2 .2y3. r. y2 = 100x1/4y3/4. y. 8) 5. k:0 k h:0 . z2 . y. y + k2 . h f1x. y + k. y2 = x . 3 + k2 . y2 . (b) f(27. determine (a) f(5.f1x. Published by Pearson Learning Solutions. 81). determine (a) f12. (b) f(0. y + k2 16. y2 = 10ex 2 2 2 3 (c) lim . 5) 8. y2 = 4x3y2. y2 = 2xy . (b) f(1. Gordon. .32 11 + r2n . determine (a) lim (b) lim (c) lim k:0 f1x + h. z2 .1.2y3z2 + z2. 4. .3x2y3.4. Wang. . f1x. 2. 2). 16). 32 (d) lim . y2 k:0 . 24) 13. determine (a) f(4. y2 h .000. z2 l k:0 l:0 .Section 6. f1x. 32.2x2 + 3y3 .0025. determine (a) f(8. determine (a) f11. 0). In Exercise 2. (b) f142.22 9. y.12x2 . z2 k f1x. z2 = 100 2x + 2y . suppose we compute lim lim f1x. y2 computing lim . the k:0 k first limit is nothing more than the ordinary derivative of f with respect to x. to remind us that the differentiation is partial. we differentiate with respect to y in the usual way.9y2 . These limits are call partial derivatives. . a partial derivative has alternate representations. y2 . we keep the other variables constant during the differentiation process. y + k2 . assuming x is treated like it is a constant. namely. to find the partial derivative with respect to y. While there is a second variable.9yk .3k22 = lim 1 . denoted by or or zx1x. y2 or fy1x.f1x.9yk . assuming y is treated like it is a constant. using the rules we have already developed.f1x. Gordon. y + k2 . namely by using ordinary differentiation rules. we differentiate with respect to x in the usual way. Therefore. Applied Calculus for Business. for example f1x + h.9y2k . 0 0 The notation means to differentiate with respect to x.3y3 + 12 = lim = k:0 k:0 k k 12x2 .3k3 lim = lim = k:0 k:0 k k k1 .f1x. assuming this limit exists. y2 . treating x as a constant wherever it appears.f1x. the first with respect to the variable x. the variable y does not change.f1x. y2 . similarly. More formally. Wang. y2 or fx1x. y2 when computing lim . k:0 k You should note.31y + k23 + 12 . and Finance. by Warren B. As a result of our previous observations.3k22 = . y2 .9y2 . to find the partial derivative with respect to x. and when h:0 h f1x. Copyright © 2007 by Pearson Education. and April Allen Materowski. y2 0f 0z (a) the partial derivative with respect to x. and means to differenti0x 0y ate with respect to y.504 * ** Section 6.f1x.9yk . but there is a slight difference in each difference quotient. y + k2 . that just like an ordinary derivative. Walter O.2 Partial Derivatives Similarly. and. notice that the second variable does not change in the quotient. the variable x does not change. then we have k:0 k f1x. that is. the symbol 0 resembles a partial d. y2 is 0y 0y f1x.12x2 . treating y as a constant wherever it appears. y + k2 . y2 12x2 . we may replace z with f in the various notations. and the second limit is the ordinary derivative of f with respect to y. we have the following definition: Partial Derivative DEFINITION 1 Given the function defined by the equation z = f1x. y2 lim . Thus. lim h:0 h 0f 0z (b) the partial derivative with respect to y. denoted by or or zy1x.2x2 + 3y3 + 1 . assuming this limit exists.31y3 + 3y2k + 3yk2 + k32 + 12 . they involve a second variable.9y2 lim k:0 k:0 k Each of these limits resemble the ordinary derivative. Economics. Inc. we can compute the partial derivative without having to determine the limit (assuming they exist). Published by Pearson Learning Solutions. y2 is 0x 0x f1x + h. it is sometimes called the partial or round d. that is. and the second with respect to the variable y. determine (a) fx1x. so we apply the constant multiplier rule. to evaluate a 0 x 1a. the last two terms are also constants in this differentiation. y2 = 2x3 + 3y4 + 4x5y3 + 2x . Applied Calculus for Business.b2 partial derivative with respect to y at (a. We then have fy1x. similarly. 0 x 11. b) is indicated as follows 0f 0z ` ` using the alternate notations: or or fx1a. Similarly. b) is indicated by: 0f 0z ` ` or or fy1a.Section 6.0 + 0 0x fx1x. y2 = 0 0 0 0 0 0 12x32 + 13y42 + 14x5y32 + 12x2 13y2 + 192 = 0y 0y 0y 0y 0y 0y 0 + 12y3 + 4x5 0 3 1y 2 + 0 . b2. In the third term. Copyright © 2007 by Pearson Education.3 = 12y3 + 12x5y2 .b2 0 y 1a. y2. y2 = 12y3 + 4x5 0 3 1y 2 . Gordon. 22 = 61122 + 2011241223 + 2 = 168.22 Remember.b2 0f ` = fx11. Solution 0 12x3 + 3y4 + 4x5y3 + 2x . y2 = 6x2 + 4y3 (b) 0 1x52 + 2 = 6x2 + 4y3[5x4] + 2 = 6x2 + 20x4y3 + 2 0x 0 12x3 + 3y4 + 4x5y3 + 2x . Published by Pearson Learning Solutions. 4 and y3 are each treated as multiplicative constants in the third term.b2 0 x 1a.3 + 0 0y The first term is zero since x is treated as a constant in this differentiation. and similarly for the fourth and sixth terms.2 Partial Derivatives * ** 505 Example 1 Given f1x. and April Allen Materowski. Wang. Economics. Inc. In Example 1. we have at our disposal all the derivative rules.3y + 9. 4x5 is a constant multiplier. by Warren B. 0 y 1a. so we have fx1x. y2 and (b) fy1x.3y + 92 = 0y fy1x. and Finance.3 = 12y3 + 4x5[3y2] . since a partial derivative is a derivative with respect to a specified variable while the other variables are treated as constants. Walter O. y2 = (a) Note that the second term is 0 because y is being held constant when we differentiate with respect to x so the product 3y4 is constant.3 0y The evaluation of a partial derivative with respect to x at (a.3y + 92 = 0x 0 0 0 0 0 0 12x32 + 13y42 + 14x5y32 + 12x2 13y2 + 192 = 0x 0x 0x 0x 0x 0x 0 6x2 + 0 + 4y3 1x52 + 2 . b2. . suppose we con- Applied Calculus for Business. y2 = 1ln12x2 + 3y422 = 12x2 + 3y42 = 2 4 0y 2x + 3y 0 y 2 (b) 12y3 1 0 0 1 a 12x22 + 13y42 b = 10 + [12y3]2 = 4 0y 2 4 0y 2x + 3y 12x2 + 3y4 2x + 3y 2 Example 3 0f . the two partial derivatives have interpretations that are not surprising. Copyright © 2007 by Pearson Education. Published by Pearson Learning Solutions.3x22 1y42 = 0y 0y 0y 0y e-3x y 1 .3x22[4y3] = .12x2y3e-3x y 2 4 2 4 Example 4 Given f1x. y2 define a smooth surface. y2 and (b) fy1x.506 * ** Section 6.2 Partial Derivatives Example 2 Find (a) fx1x. Let the equation z = f1x. -12 Solution We need to use the product rule. We have 0f 0 0 0 12x513x2 + 2y32102 = 2x5 13x2 + 2y3210 + 13x2 + 2y3210 12x52 = = 0x 0x 0x 0x 0 2x5[1013x2 + 2y329] 13x2 + 2y32 + 13x2 + 2y3210[10x4] = 0x 0 0 12y32b 2 + 13x2 + 2y3210[10x4] = 20x513x2 + 2y329 a 13x22 + 0x 0x 20x513x2 + 2y3291[6x + 0]2 + 10x413x2 + 2y3210 = 120x613x2 + 2y329 + 10x413x2 + 2y3210 and 0f ` = 1120x613x2 + 2y329 + 10x413x2 + 2y32102 11.12329 + 101124131122 + 21 .3x2y42 = e-3x y 1 . Gordon. . 0 0 1 (a) fx1x. using the generalized exponential rule f1x. y2 = 1ln12x2 + 3y422 = 12x2 + 3y42 = 0x 2x2 + 3y4 0 x 1 0 0 1 4x a 12x22 + 13y42b = 1[4x] + 02 = 4 0x 2 4 2 0x 2x + 3y 2x + 3y4 2x + 3y 0 0 1 fy1x. Solution We have. and April Allen Materowski. y2 if f1x. 0y Solution We have. Inc. Economics. y2 = e -3x y . y2 = ln12x2 + 3y42. by Warren B. using the chain rule (generalized logarithmic rule).123210 = 1201121129 + 1011211210 = 120 + 10 = 130 For a function of two variables. y2 = 2x513x2 + 2y3210. and Finance. -12 1201126131122 + 21 . Walter O. determine 0f ` 0 x 11. Wang. -12 = 0 x 11. determine 2 4 0f 0 -3x2y4 0 2 4 0 2 4 = 1e 2 = e-3x y 1 . . The slope of the tangent line to this curve at P is 0f ` .) You might have observed that we used the term smooth when describing the surface above. When we used the term smooth for a function of a single variable. where the blade intersects the surface defines another curve. Published by Pearson Learning Solutions. meaning the function has all partial derivatives. has a value that is a linear combination of fx1a.b2 face at this point P. Economics. by Warren B. and in Figure 2. Wang. This intersection defines another curve. Inc. Walter O. the slope 0f ` of the tangent line to this curve at P is . The intersection of the blade and the surface results in a curve. What about the slope of the tangent line to this curve at P? It can be shown that the slope of the tangent line at P on this curve. Gordon. if a blade parallel to the yz-plane cutting through this sur0 x 1a.b2 Visualization of the Partial Derivative Blade ( y = b ) Figure 1: The Intersection of the Blade (the Plane y = b ) with the Surface z = f1x. and Finance. called a directional derivative. the slope of tangent line at P to the curve which is formed from the intersections is the partial derivative with respect to x. it meant that the function was differentiable everywhere.plane cutting through this surface at this point P. Copyright © 2007 by Pearson Education. we use the term now in a similar sense. b2 We will not consider this derivative in this course. Applied Calculus for Business. the slope of the tangent line at P to the curve which is formed from the intersections is the partial derivative with respect to y. y2 at P Blade ( x = a ) Figure 2: The Intersection of the Blade (the Plane x = a ) with the Surface z = f1x. b2 and fy1a. see Figure 2. (You might wonder what if we placed the blade at any angle perpendicular to the xyplane and cut through the surface at P. and April Allen Materowski. 0 y 1a.Section 6.2 Partial Derivatives * ** 507 sider the point P on the surface where x = a and y = b. see Figure 1. Imagine a large blade (a plane) parallel to the xz. Similarly. y2 at P In Figure 1. y.1.22 formed when the surface is cut by the plane (blade) (a) y = . We could then make the same observation that this corresponds to the derivative with respect to the given variable. -12 0f ` = 120. . as we shall see in the next section. (a) fx1x. by Warren B. and April Allen Materowski. y. Economics. those points where the derivative is zero proved to be very significant in the analysis of the function. Inc. 0f ` = 130. Solution 0 0 1x2y3z42 = y3z4 1x22 = y3z4[2x] = 2xy3z4 0x 0x Note that y and z are treated as constants so we applied the constant multiplier rule when taking the derivative. Applied Calculus for Business. to find the partial derivative with respect to a given variable (assuming the partial derivative exists) use the ordinary rules of differentiation with respect to this given variable. determine (a) fx1x. y. in terms of a limit of a difference quotient. determine the slope of the tangent line to the curve at the point 11. (b) x = 1. We know that for a function of a single variable.508 * ** Section 6. Gordon. 0 0 2 3 4 (c) fz1x. we can define the partial derivatives in the same way as we did above. (b) fy1x.1. z2. . therefore. z2 = x2y3z4. keeping the other variables constant. -12 Solution (a) From Example 4. z2 = The next example is really nothing more than finding the partial derivatives and then solving two equations in two unknowns. Wang. (b) fz1x. Copyright © 2007 by Pearson Education. and Finance. We illustrate finding partial derivatives of a function of three variables. y. y. We take this observation as the generalized definition of a partial derivative. For functions of more than two variables. z2. (b) We leave it to you to show that slope of the tangent line is 120. the 0 y 11. treating all other variables as if they are constants. keeping all variables constant.2 Partial Derivatives Example 5 Given the surface defined by f1x. y2 = 2x513x2 + 2y3210. Walter O. z2 = 1x y z 2 = x2y3 1z42 = x2y3[4z3] = 4x2y3z3 0z 0z Now x and y are treated as constants. y. y. Therefore the slope of the 0 x 11. 0 0 (b) fy1x. Published by Pearson Learning Solutions. z2 = 1x2y3z42 = x2z4 1y32 = x2z4[3y2] = 3x2y2z4 0y 0y Now x and z are treated as constants. DEFINITION 2 Given a function of two or more variables. Example 6 Given w = f1x. we have tangent line is 130. z2. except the one we are differentiating with respect to. The same will be true for functions of two or more variables. 4. y2 for fixed values of z.6x .6y + 72 = 0 The first of these equations gives y = x2. fx1x.Section 6. . 6522 and(3. Wang. Economics.6xy . For example. Walter O. Geometrically. y2 = 0. Another way of thinking about this is to consider z the altitude of the surface above (or below) the xy-plane. Since y = x2. y2 = 6x2 . and the points on the surface where both partial derivatives are zero are 1 .32 = 0 yielding x = . and April Allen Materowski. we must solve the system 6x2 . Applied Calculus for Business. Thus. What this means is to consider the surface z = f1x. Gordon. substitution into the second equation gives .6y and fy1x. see Figure 4. y2 = 0 and fy1x. Level Curves Contours Figure 3: The Surface z = 2x 2 + y 2 Now suppose we cut the surface as described above by three blades (horizontal planes).2 Partial Derivatives * ** 509 Example 7 Find the points at which both partial derivatives are zero if f1x. this means to cut the surface by blades (planes) parallel to the xy plane. 162 and (3.12 = 0 or 1x + 421x . the two partial derivatives are zero at 1 .6y = 0 . we have the corresponding y values of 16 and 9. 9). Sometimes it is convenient to represent a three dimensional surface by level curves or contour maps. consider the surface z = 2x2 + y2. y2 = .4. Solution We are asked to find the simultaneous solution to fx1x. 309). 9. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education.4. This is a right circular cone and its surface is drawn in Figure 3.4 or 3. At each altitude (for fixed z) the equation of the surface becomes an equation in x and y. a two-dimensional curve. and Finance.61x22 + 72 = 0 or x2 + x . 92 = 309.6x . We find that (verify!). Inc. 16.3y2 + 72y + 12.6y + 72 Therefore.6x . 162 = 652 and f13. by Warren B. y2 = 2x3 . The corresponding z-values are f1 . Wang. The term level is used because the z-values. we obtain Figure 5. Economics. are constant at the fixed altitude the surface is level. and Finance. we have 5 = 2x2 + y2 or x + y2 = 25. the altitudes. y2 = e-2x .2y 2 Applied Calculus for Business. . Each of these circles is a level curve or contour of the surface. Gordon.2y . Published by Pearson Learning Solutions. if z = 5. and April Allen Materowski. The term contour map means the set of level curves. using other values for z. by Warren B. y2 = e-2x 2 . Inc. 2 z = 10 z=7 z=5 Figure 5: Selected Level Curves (Contours) of z = 2x 2 + y 2 Example 8 2 2 Determine the level curves of the surface z = f1x. For example. Solution Figure 6 is a graph of the actual surface. Copyright © 2007 by Pearson Education. and drawing them in the xy-plane. Figure 6: The Surface z = f 1x. Walter O.2 Partial Derivatives z = 10 z=7 z=5 Figure 4: The Surface z = 2x 2 + y 2 Cut by Three Planes Each of these intersections is a circle. the two terms are often used interchangeably.510 * ** Section 6. y2 = e-2x 2 . is given by z = f1x. Then the number of units produced. the equation of each level curve is a circle centered at the origin with radius 1 2 . Economics. (Why?) All other values of z will be less than 1.2 Partial Derivatives * ** 511 Notice the maximum value z = 1.1*2 ln k Note that since k 1. we may interpret this derivative as a marginal function. Figure 7: Selected Level Curves of z = f 1x. To determine the level curves.ln k or x2 + y2 = . the partial derivative of the production function with respect to y (capital) measures the rate of change of production with respect to capital when labor is kept fixed. this partial derivative is equal to the change in production when one additional unit of capital is available (for constant labor) and is known as the marginal product of capital.2y2 or 2x2 + 2y2 = . Cobb-Douglas Production Function Applied Calculus for Business. and Finance. Walter O. Let x represent the number of units of labor available. Similarly. Gordon. The partial derivative of the production function with respect to x (labor) measures the rate of change of production with respect to labor when capital is kept fixed. Wang. Thus.2y2 taking the logarithm of each side of this equation gives ln k = . and is known as the marginal product of labor. See Figure 7. y2 = Cxayb where C 7 0 is a constant and the constants a and b are chosen to represent the characteristics of the particular model. equal to the change in production when one additional unit of labor is available (for constant capital). and y the number of units of available capital.Section 6. that is. occurs when x = y = 0. z.2 ln k. and April Allen Materowski. As in the case of a single variable. Published by Pearson Learning Solutions. Inc.2x2 . and the right hand side of this equation is non-negative.2y 2 The Cobb-Douglas production function is used by economists to relate productivity to labor and capital. ln k 0. by Warren B. Copyright © 2007 by Pearson Education. . we choose z to be any constant value k 1 and we have k = e-2x 2 . Inc. (b) if a + b = 1 then f1gx. y2 + yfy1x. y2 = f1x. Gordon. Wang. then (a) f1gx. we obtain the curve xy3 = 1. then we have k 2 1 3 k 4 A x4 y4 B 4 = a b 2 4 k xy3 = 16 x4 y4 = 1 3 For example. Figure 8: Selected Isoquants of the Cobb-Douglas Production 1 3 Function z = f 1x. (b) the marginal product of capital and (c) its level curves (The level curves for a production function are called isoquants. gy2 = C1gx2a1gy2b = Cgaxagbyb = ga + bCxayb = ga + bf1x. In Figure 8 we give a few such level curves. Solution (a) f1gx.) Solution 3 0f y4 1 3 3 0 1 3 3 0 1 (a) = A 2x4y4 B = 2y4 A x4 B = 2y4 4 x 4 = 3 0x 0x 0x 2x 4 1 0f 1 3 1 0 3 1 1 0 3x 4 3 (b) = A 2x4y4 B = 2x4 A y4 B = 2x4 4 y 4 = 1 0y 0y 0y 2y 4 (c) Let k be any constant value for z.512 * ** Section 6. . by Warren B. Economics. y2 = 2x 4y 4 Example 10 Show that for the Cobb-Douglas Production function z = f1x. y2 Applied Calculus for Business. Published by Pearson Learning Solutions.2 Partial Derivatives Example 9 1 3 Given the Cobb-Douglas production function whose equation is z = f1x. when k = 2. These level curves may be sketched using our knowledge of curve sketching for a function of a single variable. gy2 = gf1x. and April Allen Materowski. and Finance. y2 = Cxayb. y2. Determine (a) the marginal product of labor. y2. Walter O. y2 = 2x4y4. gy2 = ga + bf1x. Copyright © 2007 by Pearson Education. y2. and xfx1x. the given production function. Solution (a) If u = c. we have. then the consumer will prefer the bundle with the greater value. Published by Pearson Learning Solutions. (For example. y2 = f1x. as drawn in Figure 9a. which have utility u1x2. y2 = xC b xayb . the statement f1gx. we have xfx1x. should be rewarded at a level equal to its marginal product. y2 = aCxayb + b Cxayb = 1a + b 21Cxayb2 = f1x.2 Partial Derivatives * ** 513 (b) when a + b = 1. by Warren B. y2 + yfy1x. Walter O. x gallons of gasoline and y gallons of milk together are called a bundle. and x2 gallons of gasoline and y2 gallons of milk. Wang. y) is a function of two variables and its value represents how much the consumer values x gallons of gasoline and y gallons of milk. That is.1yb = aCxayb yfy1x. y2 is the marginal product of labor and fy1x. then the level curves of x + 2y = c are straight lines. and Finance. Therefore. y) thus creating a function of two variables. y2 is the marginal product of capital. That is. It should make perfect sense that each factor. which provides a simple economic interpretation. and April Allen Materowski. y2 = f1x. . if we compare two bundles. Similarly. y2. Economics. y2 is the corresponding reward to capital. yfx1x. gy2 = gf1x. y2 = 2x4y4. The result xfx1x. xfx1x. y2 In (a). y2 + yfy1x. the contour is called an indifference curve. y2 xfx1x. Since u is fixed on each contour. A utility function is a function of two variables. two points on the same contour have the same utility. y2 is interpreted by economists as indicating they are constant returns to scale. The result in (b) of the previous example is a special case of a theorem due to Euler which we shall examine more fully in the exercises. labor and capital. Example 11 1 3 Given the utility function (a) u1x. Gordon. Thus u(x. Level curves arise in a variety of applications. y2. y2 = xCaxa . gy2 = ga + bf1x. y22. y2 = 3x2y. (b) u1x. that is. y12. y2 = x + 2y. This function is known as the utility function and can be thought of as the value the consumer places on the bundle. draw its indifference curves. Inc. For example. the consumer is indifferent to the specific choice of x and y on a contour. y2 + yfy1x. y2 says that these rewards total exactly to the amount produced. its graph is a three dimensional surface. x1 gallons of gasoline and y1 gallons of milk which have utility u1x1. Economic theory often uses the property that more is better . a proportional increase in x and y results in the same increase in f. y2 = gf1x. so we may consider their contours level curves. Thus. the total reward to labor is xfx1x. We can assign a numerical value to each pair (x. tripling both labor and capital results in a tripling of production.1 = b Cxayb adding. Suppose we have two consumption items such as x gallons of gasoline and y gallons of milk. Utility Functions and Indifference Curves Applied Calculus for Business. In other words.Section 6. In this particular example. We consider another example from economic theory. Recall that fx1x. Copyright © 2007 by Pearson Education. if increasing labor by one unit results in an increase of production of 100 units per month the meaning of marginal product of labor then the new laborer should receive a monthly wage equivalent to the 100 units. f1gx. It is the Utility Function. (c) u1x. and April Allen Materowski. If z = T1x. Copyright © 2007 by Pearson Education. by Warren B. Figure 10. y2 represents the temperature at location (x. The notion of a level curve arises in many applications. See Figure 10. . called isotherms. y2 = 3x 2y (c) This is the Cobb-Douglas production function given in the previous example. Economics. For example. Published by Pearson Learning Solutions.html) Applied Calculus for Business. Weather Map Showing Isotherms (from: http://vathena. Gordon. Figure 9b: Selected Indifference Curves of u 1x. weather maps often show curves of constant temperature.arc. y2 = x + 2y (b) Here we have 3x2y = c. whose level curves are given in Figure 9b.gov/curric/weather/hsweathr/isotherm. Inc. and Finance. its level curves were drawn in Figure 8.2 Partial Derivatives Figure 9a: Selected Indifference Curves of u 1x.nasa. Walter O. Wang.514 * ** Section 4. y) then the level curves are the locations along which the temperature is constant. Copyright © 2007 by Pearson Education. where the level curves are labeled with the elevation at the given position. and Finance. These are detailed drawings. which shows the elevation as a function of position. Weather Map Showing Isobars (from: http://vathena.arc. Typically. and April Allen Materowski. Walter O.topozone. by Warren B.nasa. Published by Pearson Learning Solutions. SD. these maps are produced by geological surveys.html) There are many other applications of level curves that you may come across. .2 Partial Derivatives * ** 515 Similarly if z = B1x.Section 6. y2 represents the barometric pressure at position (x. Figure 11. Gordon. y) then the isobars are the levels curves along which the pressure is constant. Wang. USGS Topographical Map (from: http://www. One other is the notion of a topographical map. see Figure 11. Economics. Figure 12 is such a topographical map.com/) Applied Calculus for Business. Inc.gov/curric/weather/hsweathr/isobar. Figure 12: Portions of Hells Canyon. then 0x Suppose now we take the partial derivative of fx1x. y2 = we have 1fx1x. 0f Similarly. and Finance. y2 = 0 2f 0y 0x 0 0f a b 0y 0x Note that the alternate notations have a different ordering with respect to the differentiation variables. we have two options. Walter O. The situation is a little different in this case. y2 = Higher Order Partial Derivatives fy1x.516 * ** Section 6. Economics.2 Partial Derivatives We remark that for a function of three variables. y2 = Applied Calculus for Business. y2 = fxx1x. y. y2 = 0 2f 0 x2 0f with respect to y. We represent this second partial derivative as This is the partial derivative of fy1x. y2. y22x = 0f with respect to x a second partial de0x rivative. For example. . Consider 1fy1x. y2 = 0f and 0x 0f . y2 = with respect to y or with respect to 0y x. That is for each value of w. suppose z = f1x. z2. the equation defines a surface. The notation means to first differentiate with respect to the variable on the right. 0 2f y. we can differentiate with respect to x 0y 0 0f a b 0x 0x or y. y22y = 0 0f a b 0y 0y 0f with respect to y. and then differentiate with respect to the variable on the left. we can consider the notion of level surfaces. Consider 1fx1x. Copyright © 2007 by Pearson Education. In each case. y22y = We represent this as fxy1x. We shall not consider these level surfaces here as that would require a greater examination of the sketching of three-dimensional surfaces. x. we could differentiate fy1x. Published by Pearson Learning Solutions. and we have the two first partial derivatives. and April Allen Materowski. Wang. a second partial de0y rivative. Gordon. Using the subscript notation fxy means to first differentiate with respect to the left variable. y. As with a function of a single variable. by Warren B. We represent this second partial derivative as This is the partial derivative of fx1x. Inc. and then differentiate with respect to the variable on the right. say w = f1x. 0y 0x x. as we have different choices to differentiate with respect to. we may consider taking higher order derivatives of a function of two or more variables. fx1x. defined in an analogous way.3y + 9. that is. and Finance. and then differentiate with respect to the variable on the left. y2 = 0y 0x 0y 0 x2 = 0 2f 0y 2 0f 0f = 6x2 + 20x4y3 + 2 and fy1x. y.32 = 36y2 + 24x5y 0y Applied Calculus for Business. y2. y2 = we have 1fy1x.Section 6. y2 = 0 2f 0 y2 0f with respect to x. . x2 = . Gordon. two of which 0 2f 0 2f are mixed.3 0x 0y 0 2f fyy1x. and 0 2f then differentiate with respect to the variable on the right. We may also replace f with z in the above notations. y2 = fyx1x. x. Note that we may use 0y 0x 0x 0y either the subscript or the derivative notations interchangeably. then 0y Suppose now we take the partial derivative of fy1x. Economics. y2 = 0 16x2 + 20x4y3 + 22 = 12x + 80x3y3 0x 0 2f 0 = 16x2 + 20x4y3 + 22 = 60x4y2 fxy1x. y2 = Therefore fxx1x. determine the four second partial derivatives. these two mixed partial derivatives are always the same as indicated in the following theorem.2 Partial Derivatives * ** 517 fyy1x. It turns out that for smooth functions. and leave them to the exercises. namely fxy1x. for a function of two variables. Wang. y2 = = 12y3 + 12x5y2 . y. Copyright © 2007 by Pearson Education. Solution In Example 1 we found that fx1x. Walter O. we have four second derivatives. THEOREM For the smooth function z = f1x. and April Allen Materowski. we have fxy1x. Published by Pearson Learning Solutions. by Warren B. y2 The smoothness condition requires that the second partial derivatives all be continuous in a given region. x. The proof of this theorem may be found in more advanced calculus texts. Example 12 Given f1x. the subscript notation fyx means to first differentiate with respect to the left variable. y22x = we represent this as fyx1y. y2 = and fyx1y. y2 = 2x3 + 3y4 + 4x5y3 + 2x . Thus. Inc. We can also consider third and higher order derivatives. x2 = 0 2f 0x 0y 0 0f a b 0x 0y Note once again that the alternative notations have a different ordering. y2 = = 0 112y3 + 12x5y2 . The notation means to 0x 0y first differentiate with respect to the variable on the right. y2 gives . 0z ` For example. 3x2y = c. Applied Calculus for Business. the level curves in Examc ple 11. as asserted by the theorem.4xy 2x1 . let y1 = . 22 Indifference curves can also be plotted using the calculator. y4 = .121x2 + y2 + z22-22[2y] = 2 1x + y2 + z222 Calculator Tips Given. Economics. 2) results in 0x 0y 0x 2 0 2z z 0 and d(d(z1(x. To 0y evaluate a partial derivative at a point is accomplished using the with symbol (the bar). may be written as y = . For example. Gordon. 3x2 1 2 3 4 that is. y. If you are in the 3D mode then you may define the function in the Z = screen. x. y). For example. 0 x 11. Choose a window 2 2 2 3x 3x 3x 3x2 where x 7 0 (why?) and then have the calculator plot these graphs. determine wxy1x.518 * ** Section 6. Wang. computing its partial derivatives with a calculator is done the same way as computing the ordinary derivative. y2 = 0 2f 0 = 112y3 + 12x5y2 . y). Walter O. z2. z2 = ln1x2 + y2 + z22. . y) gives . y). d(z1(x. x). z = f1x. and so on. y). Copyright © 2007 by Pearson Education. If you are in the Function 0z mode then you may enter the equation directly. and now we need only choose values for c.2 Partial Derivatives and fyx1x. by Warren B. You ll note that the calculator does not distin0 y 0x 0 y2 guish. y2. producing the indifference (level) curves. Then d(z1(x. and Finance. Suppose we let z1 = 5x2y3. y). y3 = . d15x2y3. Solution We have wx = 0 0 1 2x 1ln1x2 + y2 + z222 = 2 1x2 + y2 + z22 = 2 2 2 0x 0x x + y + z x + y2 + z2 wxy = 0 0 2x a 2 b = 2x 1x 2 + y 2 + z22-1 = 2 2 0y x + y + z 0y 0 2x[ .11x2 + y2 + z22-2 1x2 + y2 + z22] = 0y . Inc. especially in the case when we may solve for y explicitly in terms of x. d(z1(x. Example 13 Given w = f1x. y. in its notation between an ordinary or partial derivative. y. y) results in . and April Allen Materowski. d(z1(x. d15x2y3. x2 x = 1 and y = 2 gives and so on. x) results in 0 2z 0z 0z . 2) results in 2 .32 = 60x4y2 0x 0y 0x Observe that the two mixed partial derivatives are equal. Published by Pearson Learning Solutions. y2 = . Gordon. find the point(s) at which fx1x. y2 = ln12x2 + 3y22 14. fy1x. Exercise 8 52. 16. f1x. . z2 = 4x 3 y2 z2 + 4x2 + 2y3 + 5z5 + 3x . y2 = ln1x2 + y22 The partial differential equation c2uxx . If f1x. f1x. c = 2. y. find fx1 . y2 for the function defined in the given Exercises. In Exercises 59 62.12.2 In Exercises 1 15.2y326 13. by Warren B. fy12. .2x + 2y + 2z. y2 = 2x + 3y 37. y.18x + 4y2 .2y + 11z + 12. y2 = 3x2y3e-4x y 3 2 11. y2 at the indicated point. 1. find (a) fx1x.y2 4. u1x. f1x. y2. 44. 2.40z + 25.Section 6.3y . find the point(s) at which fx1x. Exercise 3 49. (b) y and (c) z.16x + 3y2 . 28. z2 = 0.1. u1x. y. Inc.2y + 11z + 12 17.5 3. f1x. y2 = 2x y ln1x + y 2 In Exercises 16 21. In Exercises 56 . . y2 = 2 x . . For f1x. f1x. Exercise 2 48. y2. z2 = ln1x + y + z 2 20. y2 = 4x2/5y3/5 42. 12. z2. y) is said to be harmonic if it is a solution to the partial differential equation uxx + uyy = 0. . 1) x2 . and fz1 . z2 and fxzy1x. y2 = 4x1/3y2/3 In Exercises 36 41. 36. . z2 = 2x3y2z2 + 2x2 + 2y2 + 3z2 + 2xy + 3xz + 5yz . z2 = 2x2 . If f1x. f1x. . 27.1. . f1x. u(x. f1x. y2 = 2y2113x2 . find the partial derivatives with respect to (a) x. y2. 12. 59. z2 = 2x y z e 2 4 3 3x4y2z2 In Exercises 22 25. 1. y. f1x. 22. z2 = 0. .12. find fx11. find the point(s) at which fx1x.7x2y5 + 8x2 . z2 = 0.12 25. y. y. find fzxy1x. For f1x. y2 = 4y312x2 + 3y425 8.1. 30. Exercise 4 50.12. z2 = 4x2 + 2y3 + 5z5 + 3x . . Exercise 11 53. Wang.3y425 12. y2 = 3x3 . and Finance. y2 = 3x2 . f1x.7x2y5 + 8x2 . z2 = 3e2x y z 2 3 4 2 3 2 2 2 2 2 21. f1x. y2 = 0 and fy1x. y2 = 8x2e-2y 2 32.4x4 + 5x3y5 .12.2. z2 = 0. z2 = 3e2x y z . Determine the slope of the tangent line to the surface defined by f1x. y2 = 0. (c) fxy1x. and fz1x. 46. f1x. z2 = ln1x2 + y2 + z22. y. y2 = 2 x + y2 24.2y3 .7x + 9y2 + 11 x2 . t2 = 4e21x . y) is a utility function. If f1x. f1x.3y5 + 15z4 + 8x2 + 2y3 . f1x. f1x.2. 29.5 (2. y. y. y2 = 2x2y3 ln1x2 + y22 11. y2 = 2x2 . y.2x2y3z4 + 5z + 2 18. z2 = 0. For f1x.y2 6. z2 = 2x2 + 2y2 + 3z2 + 2xy + 3xz + 5yz . y2 = 3x + 4y 38. (b) fy1x. y. show that u(x.2x . For f1x. y2 = 2xye3xy 10.1. and fz11. y. z2 = 4x2 + 2y3 + 5z5 + 3x . y2 = 3x4 . u1x. Exercise 1 47. y2 = 2x1/2y1/2 35. y. y2. 22 23. y2 = 2y212x3 .8x + 3y2 + 12y + 2. f1x. 12. f1x. y. z2 = 2x4 . Walter O.2. and fyx1x. .2t2 9. f1x.12 formed when cut by the plane (a) y = . y2 = 0 and fy1x. u1x.y 33. f1x.22 26. y2 = 2xy3 40. z2 = ln12x2 + 3y2 + 4z22 19.2y + 11z + 12. (b) x = 1. 2) formed when cut by the plane (a) y = 2. y.y2 57. 2. 2. Exercise 7 51. Copyright © 2007 by Pearson Education. y. y. (b) fyy1x. 45.24y + 4z2 . y. u1x. find the partial derivatives with respect to (a) x and (b) y. y2 = 3x4 . y. y. u1x. fy1x. f1x. u1x.utt = 0 where c is a constant is called the wave equation. (b) x = 1. u1x. Sketch some of its indifference curves. 31. y. 1.2y3 .y 2 3 4 3 2 Applied Calculus for Business. y2 = 3x31x2 + y224 7. y2 = x2 . z2. Exercise 13 54. and April Allen Materowski. f1x. y2 = 2 x + y2 x2 + y2 5. z2 = 0. y. Determine the slope of the tangent line to the surface defined by f1x.2 Partial Derivatives * ** 519 EXERCISE SET 6.2y + 7 2.58 show the given function is harmonic. y2 = 0. y2 = x2 . y2 = 2x2 + 3y2 + 4x3y4 + 9x . sketch some of the level curves (contours) or the surface defined by f. y2 = 4x3y215x4 . f1x. u1x. For f1x. and fz12. f1x. y.3y225 at the point (1. y2 = 2xy 58. f1x.3y . f1x. y2 = 2x1/2y1/2 41. 56. u1x. y2 = x2 + y2 34. Published by Pearson Learning Solutions. f1x. find the point(s) at which fx1x. f1x. y2 = 3x2y3e-4x y 12. t) satisfies the wave equation for the given value of c. z2 and fxyz1x.y2 1 . and fz1x. For f1x.12.2x + 2y + 2z.24y + 10. 1. where c is a constant. 55. f1x. y2 = 14x2 + 3y327 at the point 11. . y2 = ln 2x3 + 2y2 15.1. u(x. 43. Economics. . In Exercises 31 35. f1x. y2 = 4x3y 39. In Exercises 46 53 find (a) fxx1x. fy11.12. fy1 . f1x. find fxzy1x. y2 = x2 . find fx12. b2 Ú f1x. where f and g are smooth functions. Hint: Use the limit definitions of the partial derivative. f1x. 64. y2 = x3y3 x2 + y3 70. z2 + zfz1x. y. u1x. when a + b + g = 1. Wang. y2 = 100x4 y4 . z2 = 1a + b + g2f1x. (a) Suppose a company has a production function f1x. In Exercises 67 .3t2 3 62. Exercise 67 76. y2 = 3x2y3. z2 = 3xy2z3 73. and April Allen Materowski. y. y. Applied Calculus for Business. xfx1x. 02 60. 02 and a similar limit for fyx10. y2 = Cxayb 72. b)) if f1a. u1x. y2 = x3y3 71. where H 1 and H 2 approach zero as ¢ x and ¢ y both approach zero. 02 = 1. y) near the point P. give its degree. Let f1x.1 and (b) fyx10. z2 = gf1x. but at a slower rate. y. c = 3. f1x. y2. b. f1x. z2. Hint: Use fx10. z2 = Cxaybzg. by Warren B. y2 = . b)) if f1a. 02. y. y. c = 2. t2 = f1x . t2 = 4e21x + 2t2 61. t2 = 51x . that for a function of two variables. (b) a relative minimum at the point P(a.y2 x2 + y2 0 b if 1x.y for all y. dy.74.3 Extrema Euler s theorem states that for smooth homogenous functions of degree g. f(a. y2 + yfy1x. y. 65. y2 = x for all x. z2 + yfy1x.) 66. and if it is. A function is said to be homogeneous of degree n if f1gx. determine which of the following functions is homogenous. z2 (c) in particular. Copyright © 2007 by Pearson Education. (b) xfx1x. z2 + yfy1x. y2 = gf1x. y2 for all points (x.f1x. z2. is said to have (a) a relative maximum at the point P(a. y. b. Consider the function f defined by f1x2 = c xy a x2 . z2 = Cxaybzg n 1 3 (a) Show that fx10. y2 for all points (x. dz2 = da + b + gf1x. (This may be interpreted that as labor increases. y2 ¢ y + H 1 ¢ x + H 2 ¢ y. w = f1x.utt = 0. . The next definition deals with relative and (absolute) extrema. Similarly. Show that the marginal product of labor is positive and that the marginal product is a decreasing function of labor for constant capital. k (c) What does this imply about the partial derivatives? k:0 fxy10. y + ¢ y2 . y2 = 10. y2 Z 10. 02 if 1x. Show that u1x.3 Extrema » » » » Extrema Critical Points Second Partial Derivative Test Calculator Tips In this section we generalize the notions of extrema considered in Chapter 3 to functions of two or more variables. Walter O. DEFINITION 1 The function defined by the equation z = f1x. f1x. Inc. y. gz2 = gnf1x. Consider the Cobb-Douglas production function of three variables. y2 = x + y 68. f1x. For the function defined in the previous exercise. 67. y2 = fx1x. Published by Pearson Learning Solutions. z2 In Exercises 75 77. y2 and for functions of three variables. gy. 02 = . 0 + k2 . y. z2 = 3xy2z3 2x2 + 2y2 + 3z2 74. y. 75. y. 02 = lim 6. show that (a) fxy10. u1x. b2 f1x. Show that the mixed second derivatives for the Cobb-Douglas Production function are equal. give an economic interpretation of these results. y) near the point P. Gordon. y2. verify it satisfies Euler s theorem. y2 ¢ x + fy1x. z2 + zfz1x. (b) fy10. f1x.fx10.520 * ** Section 6. Economics. f(a. Exercise 68 77. Show that (a) f1dx.ct2 + g1x + ct2 is a solution to the wave equation c2uxx . f1x. 79. 80. t2 = 51x + 3t23 63. y2 = 2x + y x2 + y2 69. Exercise 72 78. f1x. y. xfx1x. Show that f1x + ¢ x. c = 3. production increases. y. a function of three variables is homogenous of degree n if f1gx. if the function is homogeneous. gy2 = g f1x. and Finance. y2 for all points (x. the z-value is smaller than all other nearby z-values. for a function of two variables. f(a. b. Wang. b2 + f1x. and at a minimum it is smaller than any other z-value on the surface. b)) if f1a. then the slopes of the tangent lines we obtain are precisely the partial derivatives with respect to x and y. but they all are going up to the peak and descend after the peak. . b)) if f1a. Walter O. (d) a minimum at the point P(a. Gordon. b. then the tangent line at the peak is horizontal to each of these curves. if we were descending a valley (approaching a relative minimum point). and April Allen Materowski. Economics. the z-value is higher than any other z-value on the surface. and Finance. therefore. The notion of relative extrema and extrema generalize in a similar way to a function of any number of variables.3 Extrema * ** 521 (c) a maximum at the point P(a. by Warren B. Copyright © 2007 by Pearson Education. Notice that at a relative maximum. Inc. and at a relative minimum. At a maximum. there are many different paths to the peak. see Figure 2. the z-value is higher than all other nearby z-values. Thus.Section 6. y) in the domain of f. Consider Figure 1. f(a. b2 * f1x. the same would be true. y) in the domain of f. P Figure 2: Illustrating a Relative Minimum Applied Calculus for Business. we have that both fx and fy are zero at the relative maximum. we have the same distinction between relative extrema and extrema as we have for a function of a single variable. Published by Pearson Learning Solutions. In particular. Clearly the peak is an example of a relative maximum. How do we determine the relative extrema? Visualize a mountain and suppose you are hiking to the peak. Similarly. y2 for all points (x. if we cut this smooth surface at the peak with planes parallel to the xz and yz-planes. Extrema P Figure 1: Illustrating a Relative Maximum If you think of each path through the mountain peak as a curve. 3 Extrema Critical Points We know that for a function of a single variable there were points where the derivative vanished which were not relative extrema. y2 = 0 For a smooth function of three variables. For the most part. so such critical points will not arise often. z2 = 0 fz1x. See Figure 4. yet the point may still be a relative extremum. we give a sketch of a surface that is called a saddle because it looks similar to a saddle you would place on a horse. we will deal with smooth functions. it must be the lowest point in all directions. Copyright © 2007 by Pearson Education. consider Figure 3. Inc. in another a minimum. We now generalize that definition. it most be the highest point in all directions. Gordon. y. Thus the point P in one direction is a maximum. The same is true for a function of two or more variables. Recall that for a function of a single variable. This means. z2 = 0 fy1x. Wang. to be a relative minimum. The point P is a sharp point: neither partial derivative exists at this point. by Warren B. yet the function has a relative maximum there. to find the critical point for a smooth function of two variables. Suppose you sat in the saddle. and similarly. Walter O. y2 = 0 fy1x. P Figure 3: A Critical Point Where the Partial Derivatives Do Not Exist It is possible for a function of two variables to have both partial derivatives be zero at a point P and yet the point is neither a relative maximum or relative minimum. we solve fx1x. . we defined the notion of critical points and then examined them to determine what they were. Published by Pearson Learning Solutions. for a function of any number of variables. and April Allen Materowski. but if you were mounting the saddle. y. But for it to be a relative maximum. as a result. there are surfaces at which one or more of the partial derivatives does not exist. similarly. z2 = 0 and. the point at which you were pulling your leg over (from side to side) is a maximum. we included in our definition of a critical point those points in the domain of the function at which the derivative failed to exist.522 * ** Section 6. and. To visualize this. Applied Calculus for Business. The point at which your bottom rests is a minimum in one direction (from the front to the back of the saddle). y. DEFINITION 2 Any point in the domain of a function at which any first partial derivative does not exist or at which all first partial derivatives are simultaneously zero is called a critical point. we solve the system of equations fx1x. and Finance. Economics. 6y + 18 = 0 yielding x = . 3. y2 = 6x2 . We next clear fractions in the second equation and then x2 4 2 + + xy.3y2 + 24x + 18y + 10. Example 2 Determine the critical points for f1x. Inc. y2 = Solution We have -4 + y x2 -2 fy1x. Gordon.2. Wang. Copyright © 2007 by Pearson Education. the point 1 .3 Extrema * ** 523 P Figure 4: A Saddle Point at P Example 1 Determine the critical points for f1x.2 and y = 3. f1 . We shall later show that this point is a saddle point.2 + xy2 = 0 4 .6y + 18 so we solve the equations 12x + 24 = 0 . Walter O. . Published by Pearson Learning Solutions. and April Allen Materowski. 132 is the only critical point of the given function. by Warren B. Solution We have fx1x.2. we have . and Finance. y2 = . y2 = so we must solve -4 + y = 0 x2 -2 + x = 0 y2 The first equation gives y = substitute for y Clearing fractions.Section 6. Economics. y2 = 2 + x y fx1x. 32 = 13. therefore. y2 = 12x + 24 fy1x. x y Applied Calculus for Business. the three critical points are 1 . Applied Calculus for Business. we substitute into y = .2. by Warren B. . Walter O.x9 4 a 3 b + 16x = 0 4 . 22 = f12. which gives x = . To find the corresponding yx3 values.2 or 2. y2 = 4y3 + 16x We solve the equations 4x3 + 16y = 0 4y3 + 16x = 0 If we solve the first equation for y.2. 2.22 = . either x = 0 or x8 = 256. . Economics.2.3 Extrema Substituting and solving. Wang.x9 + 16x = 0 16 .32.2562 = 0 Thus. . we have y = x3 4 We substitute this value for y in the second equation to obtain 4a .x9 + 256x = 0 . 0) and 12. (0. Solution We have fx1x. y2 = 4x3 + 16y fy1x.322.2. . Copyright © 2007 by Pearson Education. and April Allen Materowski. and we have y = 1. y2 = x4 + y4 + 16xy. Gordon.x3 3 b + 16x = 0 4 . -2 + xa 4 2 b = 0 x2 16 -2 + 3 = 0 x 3 .and find the corresponding y-values are 0. . To find y we substitute x = 2 into y = Example 3 Find the critical points for f1x. Inc. 02 = 0.x1x8 . and Finance. 6) is the only critical point of the function.524 * ** Section 6. 0. 4 We have f10. thus. thus the x2 point (2. 1. Published by Pearson Learning Solutions.2x + 16 = 0 x3 = 8 x = 2 4 .322. 2 and . 12 = 6. f12. f1 . 5. b2fyy1a. the curves are concave downward. b2 have the same sign. but remember for smooth functions the two mixed partials are the same. the critical point is a relative minimum. y2fyx1x. It gives us a simple test by which we may classify the critical points. b2 6 0 then f has a saddle point at P. y2]2 You may note the absence of the mixed partial derivative fyx1x. that means the product fxx1a. Published by Pearson Learning Solutions. y2 the discriminant.[fxy1x. b2 7 0 means that curves in the surface passing through the point P are concave upward. called the discriminant. implying that fxx1a. 2.3 Extrema * ** 525 Example 4 Find the critical points for f1x. Gordon. . Case II: If D1a. however. THEOREM . . Solution fx1x. b2 and fyy1a.5 = 0 These equations have as their solution x = 3. that there is a theorem for a function of two variables that is analogous to the second derivative test for a function of a single variable. It turns out. at the point P(a. Wang. Since D1a. y. and April Allen Materowski. y) is defined by D1x. Case III: If D1a. b2 is positive and greater than [fxy1a. therefore the only critical point of this function is 13. b2 6 0 then f has a relative maximum at P. y. 2. Economics. some remarks are in order. If we could easily draw the sketch of a surface. (1) Note in Case I. a relative maximum. y2 = fxx1x. denoted by D(x. . therefore. b2]2. b. z2 = xy + xz + yz + 3x + 2y . we could have just as well used fyy1a. Case I: (a) If D1a. b2 7 0 and fxx1a. z2 = x + y . y2.Section 6. z2 = y + z + 3 fy1x. b)). (b) If D1a. we define another function used in this test. y. (Verify!). y2 has a critical point when x = a and y = b. Applied Calculus for Business. y = 2. z2 = x + z + 2 fz1x. b2 7 0 and fxx1a. Inc. Copyright © 2007 by Pearson Education.5z. It is simpler to write one of them squared. Second Partial Derivative Test DEFINITION 3 For the smooth function z = f1x. Before we state it. y2. (2) While the conditions in Case I use fxx1a. b2 = 0 then the test is inconclusive. b2 6 0. fxx1a. z = . then visually. Walter O. so we could replace the squared term by the product fxy1x. y2fyy1x. b2 7 0 in its place.5 and f13.52 = 19. y. b2 7 0 then f has a relative minimum at P. b2 7 0. we could determine if the critical points are relative maxima or minima.SECOND PARTIAL DERIVATIVE TEST Suppose the smooth function defined by the equation z = f1x. and Finance. Similarly. While the proof of this theorem is beyond the scope of the text. b2 7 0. y2 . 132. by Warren B.5 We must solve the three linear equations y + z + 3 = 0 x + z + 2 = 0 x + y . f(a. if fxx1a. it is called the Second Partial Derivative Test. We found.62 .2. y2 = Solution 1. y2 = 0.3y2 + 24x + 18y + 10. 4.526 * ** Section 6. We illustrate this theorem on each of the three examples considered in this section. 1. solve fx1x. y2fyy1x. 3. 3. Classify each critical point. In Example 1. Evaluate fxx at each critical point.6 fxy1x. . y2. If D 6 0. Walter O. y2 = 3 x 4 fyy1x. Evaluate D at each critical point. Published by Pearson Learning Solutions. y2 = 3 y fxy1x. Find the critical points.[fxy1x. y2]2 = 11221 . if D = 0. y2 and fxy1x. Solution We have 1. y2. by Case II we may conclude that the critical point 1 . Example 6 Classify the critical points for f1x. y2 = . y2 = 12 fyy1x.2. y2 = . Gordon. fx1x. y2. fy1x. Copyright © 2007 by Pearson Education. fyy1x. and April Allen Materowski. -4 + y x2 -2 fy1x. y2 = 12x + 24 fy1x. Example 5 Classify the critical points for f1x. 4 2 + + xy. y2 2. D1x. Economics. we found that the only critical point occurred when x = . y2 . 6). y2 = 2 + x y 8 fxx1x.1022 = . y2 = 0 and fy1x. by Warren B. in Example 2. y2 = fxx1x. y2 = 1 fx1x. and Finance. 5. that the critical point was (2. Evaluate fx1x. y2 = 0 2. 132 is a saddle point. that is.72. Inc. if D 7 0 proceed to the next step. then critical point is a saddle.6y + 18 fxx1x. so in particular D1 . fxx1x. Wang. y2 = 6x2 .3 Extrema Applying The Second Partial Derivative Test 1. 32 6 0 We do not have to proceed any further. 3. y2 = 2.2 and y = 3. test fails. x y Applied Calculus for Business. Example 7 Classify the critical points for f1x. if needed. 22 = 48 7 0 not required fxx12. fxx12. We also recall that if a function of a single variable is continuous on a closed interval then it assumes both its maximum and minimum values. the relative maximum in Figure 1 is in fact the maximum of the function.1. 02 = . Consider the next example which illustrates the determination of the minimum of a function. b) 1 . we will be more concerned with computing their critical points and.2. y2 = x 4 + y 4 + 16xy (a. 12 = 1 7 0. Economics. . by Warren B. Thus far. therefore. y2 = 12y2 fxy1x. 0. Wang. 0) and 12. y2 = 4y3 + 16x fxx1x.2.2. y2 = 16 2. the extrema at interior points of the domain or on the domain s boundary. Solution We have 1. y2 . y2fyy1x. y2 = fxx1x. 0) 12. We shall show another method of solving such problems involving constraints in the next section. y2 * 144x 2y 2 + 256 D1 . for example. 3. D1x.22 D1x. therefore D12.[fxy1x. that is. classify them other ways. y2 = 4x3 + 16y fy1x.22 = 2048 7 0 f xx1x.Section 6. Copyright © 2007 by Pearson Education. 12 = 3 7 0 x3y 3 4. y2 .3 Extrema * ** 527 3. we will try to justify our conclusions by means of a graph. It will happen that this often will be the case for functions of two or more variables. 32 . what about the extrema themselves? We saw when we studied functions of a single variable that very often the relative extrema turned out to be the extrema. y2 = 12x2 fyy1x. 2.22 = 48 7 0 Classification relative minimum saddle point relative minimum While the second partial derivative test does generalize to functions of three or more variables.322. . . When this happens. and the relative minimum of the function in Figure 2 is its minimum. D1x. Applied Calculus for Business. .322. Note. fx1x. we will not do so here. y2fyy1x. Published by Pearson Learning Solutions. . Gordon. y2 = fxx1x. y2 * 12x 2 fxx1 . and we will leave its examination to the exercises.2. We compute D and fxx at the x and y coordinates of the critical points in Table 1.2. (0. Instead.256 6 0 D12. It illustrates the generalization of the optimization problems we considered in Chapter 1. . y2 = x4 + y4 + 16xy. the critical point is a relative minimum. We found in Example 3 the following critical points: 1 . And 4. y2]2 = so we are in Case I. 22 = 2048 7 0 D10.[fxy1x. we have considered only relative extrema. This theorem generalizes to functions of two or more variables as well. and Finance. Inc.256. Table 1: Classifying the Critical Points of f1x. and April Allen Materowski. by geometrical or physical considerations. y2]2 = 144x2y2 . 22 (0. . Walter O. which now is a function of x and y. . width and height as indicated in Figure 5.3 Extrema Example 8 An open rectangular box is to have a volume of 4 cubic feet.2 = 0 y y - Applied Calculus for Business. so we first rewrite the expression and then find its partial derivatives. We use the constraint to replace one of the variables in the function to be minimized. Gordon. that is. it follows that x. and April Allen Materowski. Note that each face of the box is a rectangle. for a combined sum of 2yz square feet. The area of the front and back faces are each yz. Thus we are asked to minimize the total surface area of the box which is f1x. for a combined sum of 2xz square feet. y2 = x . y.2 y We must solve 8 = 0 x2 8 x . labeling the length. Wang. the dimensions of the box must each be a positive number. y and z. by Warren B. z 7 0 This is very similar to the two dimensional problems that we studied in Chapter 3. Solution z x y Figure 5: An Open Box We first draw a picture of the box. Determine the dimensions of the box that uses the least amount of material. y. y2 = y . Using the least amount of material means we want to minimize the surface area of the box. and Finance. f1x. and the area of the bottom is xy square feet. z2 = xy + 2xz + 2yz subject to the constraint that xyz = 4 with x.528 * ** Section 6. Inc. Walter O. The area of the left and right faces are each xz. the box is required to have a volume of 4 cubic feet. We have f1x.2 x 8 fy1x. Economics. y2 = xy + 8y-1 + 8x-1 8 fx1x. y. y2 = xy + 2x a 4 4 8 8 b + 2y a b = xy + + xy xy y x We want to determine the critical points. We have from the constraint z = 4 xy substitute for z in f. xyz = 4. From the physical nature of the problem. So the objective is to minimize f1x. Copyright © 2007 by Pearson Education. z2 = xy + 2xz + 2yz in addition. Published by Pearson Learning Solutions. We now indicate how this can be combined with the solve command to find the critical points.8 = 0 8 2 xa 2 b . If we can show that f1a + h. 5x. Often.Section 6. by Warren B. then f(a. y2. Similarly for a maximum. Recall that d(z1(x. Substitution of x = 2 into y = When physical or geometrical reasoning are not conclusive. y2 = xy + has + 2 y x x exactly one critical point when x = y = 2. How do we conclude that this is the minimum of f? From a practical point of view. the mode must be 3D and you press F1 to get the Z = Screen).f1a. Using the second partial derivative test. Suppose f has a relative minimum at (a.8 = 0 x 64 = 8 x3 8x3 = 64 x3 = 8 x = 2 8 8 8 . y62 This calculation determines when both first partial derivatives are simultaneously equal to 0. For example.3 Extrema * ** 529 The first equation gives y = stitute for y. Recall that to solve two equations in x and y on the calculator. 5x. y) gives fy. This approach may be difficult to apply when the function is complicated. y2 = 0. y2 as z1 in the Z = Screen (remember. Published by Pearson Learning Solutions. and Finance. see Exercise 23. b) is the minimum of the function (Why?). there must be a way of constructing the box that uses the least amount of material. where we have f1a + h. Thus. b2 Ú 0 for every choice of h and k. x) gives fx and d(z1(x. f1x. y). b + k2. y2.f1a. y). you will see that the point (2. when working with functions of two or more variables on your calculator. From the constraint equation. If we sketch the graph of the surface (the calculator can provide a rough sketch). when entering products. We saw in the previous section how to use the calculator to find the partial derivatives. Calculator Tips Applied Calculus for Business. 8 . and the only candidate for solution is the relative minimum. you should verify that this critical point is a relative minimum. Wang. So the box has dimensions 2 feet by 2 feet by 1 foot. b + k2 . gives y = 2. b) then any other point in the domain may be written as 1a + h. otherwise the calculator will consider the xy as a single variable. We clear fractions in the second equation and then subx2 xy2 . (Make sure that you use the curly parenthesis to indicate the variable x and y. Gordon. it then follows that z = 1. and April Allen Materowski. xyz = 4.) Also. Inc. some algebraic manipulations are required with this approach. Copyright © 2007 by Pearson Education. xy should be entered as x * y. which therefore is the minimum. Walter O. . Economics. we may try to use the definition of an extreme value. the syntax is solve1equation 1 and equation 2. x2 = 0 and d1z11x. use the multiplication symbol *. y62 Suppose we define the function z = f1x. so we need only enter solve1d1z11x. b + k2 . b2 0. 2) gives the minimum. y2 = e-2x 2 2 3 3 3 2 4 .3y2 2 22.3xy2 + 48xy 17. The cost of producing a rectangular box is as follows: the sides cost $2 per square foot. 35. 30. f1x. (b) Prove that this is in fact a minimum of the function by k2 showing that f10 + h. y2 = 12x2y3. the joint revenue derived from the sale of these two commodities is given by the equation R1x.xy + 3x . y.3 Extrema EXERCISE SET 6. f1x2 L f1a2 + f ¿ 1a21x .it is. y. y2 = 12x y .49z 27. 32. y2 = x4 + y4 + 16xy 20.x y 18. In Exercise 32. thereby proving R is maximized at the critical point.2y2 + 6xy + 6x + 8y. yn2. f1x. y2 = 2x2 .3 Find and classify. Gordon. If we sum all of the squared errors.ln xy2 21. using the second partial derivative test.5xz .9xy 19. we approximated it by the equation of its tangent line at a.The equation of a hemisphere is given by f1x. y2 = . 3). satisfying this requirement. b2 . y2 = x2 + y2 . f1x.R1a + h. 1x2.24y + 20 2. f1x.2y2 + 16x . this linearization represents a plane that is tangent to the surface at the point (a. f1x.b2 (1) 2 4 + + xy x y 1 8 + + 2xy x 2y 2 4 + xy x y 12. xi. y.24y + 10 6. the critical points of the function defined in Exercises 1 22. Suppose the box in Example 8 is to have a top as well. f1x. b2 = a 1yi . f1x. In Exercises 24 26 Find the critical points.18x + 4y2 . y2. y2 = xy + 9.8x + 3y2 + 12y + 2 4.value. and the value of y predicted by the regression line. y2 near the point (a. It can be shown that geometrically. y2 = 2x4 + 2y2 . y2 = + + cxy occurs when x y 2 1/3 b a x = a b and y = x. . y2 = 6xy + 4y . y2 = xy . *1). f1x. let y = ax + b be the equation of the least square (regression) line. Find its critical points. f1x. 2 2 2 38. f1x. b) be the critical point.2x + 2y + 2z 26.y2. y2 = 4x2 . y2 = x3 + y3 . we linearized the function.1xn. In the previous example give a symmetry argument that yields the solution by inspection. f1x. 1x3. 3 + k2 . For each x. (1) is called the linearization of the function z = f1x. Published by Pearson Learning Solutions. f1x. y12. The United Postal Service requires that for any rectangular package. Find three numbers whose sum is 12 if their product is to be as large as possible. let (a. determine the dimensions that minimize the cost. b)? You could almost guess at the approximation that generalizes one dimensional linearization. 1. y2 L f1a. y2 = 6x2 + 4y2 + 30x . b + k2 = 61h . z2 = x2 + 2y2 + 13z2 + 3xy . Copyright © 2007 by Pearson Education.12xy + 5 13. Á . by Warren B. and April Allen Materowski. z2 = 2x2 .f10. f1x. b2 + fx1a. y22. Determine the dimensions of the package of largest volume that may be sent. f1x. y2 = 8 8 + y x 33. there is the observed y-value. In Exercises 37 39.k22 + k2/2. This should not surprise us.24y + 4z2 . Determine the number of each commodity that should be sold to maximize the revenue. Suppose the Revenue derived from the sale of these two commodities is given by the equation R1x.12yz + 7x + 25y . 34.x y .y4 5.3y + 6 8. Show that R1a. a b 39. f1x. y2 = 11. the sum of its length and girth (cross sectional distance) not exceed 108 inches. f1x. If the volume of the box is to be 10 cubic feet. y2 = x3y3 + x2y4 + 5000. y2 = x4 + y4 + 16xy near (1. yi. f1x. f1x. Find three numbers whose sum is S if their product is to be as large as possible. b21y . f1x.20y . near this point the values on the plane approximate the values of the surface. as we are approximating a function in two variables. Linearization Recall that when approximating a function defined by y = f1x2 near x = a. y2 = 12x2 + y22e-1x + y22 Notice that the linearization involves both partial derivatives. (a) Show that f1x. y2 = 10. f1x.1axi + b2 is the error at x = xi. y2 = 2xy2 + 2x2y . y2 = 3x2 . 29. x units of one commodity and y units of the other.8x + 4 14. and Finance. f1x.1axi + b222 i=1 n Applied Calculus for Business. Economics. b) and. that is.15 3.x2 . Show that the critical point for f1x. x y 23. determine the maximum profit. y2 = x4 . y2 = x3 + 6xy + y2 15.2x . Inc. and the cost function is given by C1x. y32. y2 = 4 2 + + xy near (1.64x2 + 8y + 3 7. we have f1a. the top $1 per square foot and the base $4 per square foot.4x2y . f1x. f1x. y2 = 24 . y2 = . Verify the critical point found in Example 8 is a relative minimum. bc a 40. 31. 28. y2 = 2x3 + 2y3 . A producer sells x units of one commodity and y units of another. y2 = 2x2y + 3xy2 . Determine its dimensions.a2 + fy1a. z2 = 2x + 2y + 3z + 2xy + 3xz + 5yz . 36. linearize the given function. Wang. Determine a and b so as to minimize the sum of the squares of the errors. f1x.2). f1x. A producer sells two commodities. 37. This difference yi . f1x. f1x.40z + 25 25.6y + 5 has a relative minimum at (0.36xy 16. (Why?) Given n data points 1x1. therefore.a2 This was a good approximation near a because the tangent line is very close to the curve near a. What about approximating the function defined by z = f1x.530 * ** Section 6. Walter O. b21x . 32 = h2 + k2 + 34 Ú 0. f1x. 24.16x + 3y2 .6x2 . This theorem generalizes: for a continuous function z = f1x. 43. 5) x+y=5 n Note that in Section 1.Section 6. y2 = 4xy + 21 . show that the constants a. and substituting for b in the first equation Recall the Extreme Value Theorem for a function of a single variable states that continuous function on a closed interval attain their extrema either at critical points or at their endpoints. defined on a closed region.8 we wrote this expression as a = sxy . and its extrema on the boundary are now found by the methods of Chapter 1.4 f is a function of a and b. Published by Pearson Learning Solutions.y2 defined over the region bounded by the circle x2 + y2 1.x2 .x2 . and x + y = 5. implementation requires nothing more than finding partial derivatives and solving a system of equations.6y + 2 defined over the region bounded by y = 0. x = 0 and x + y = 8. Show that the second equation in the previous exercise gives b = y . Following the previous exercise. see Figure Ex. b2 = 0 The Method of Lagrange Multipliers * ** 531 41. the function will attain its extrema at either critical points (within the region) or at boundary points of the region. y2. 6. (5. Such two variable optimization problems were solved by using the constraint equation to solve for one variable in terms of other and then reducing the problem to a single variable. Find the extrema of the function f1x. and Finance. b and c for the quadratic regression curve y = ax2 + bx + c may be found by solving the equations i=1 n i=1 n i=1 2 a yi = na + b a xi + c a xi i=1 i=1 2 3 a xiyi = a a xi + b a xi + c a xi 2 2 3 4 a xi yi = a a xi + b a xi + c a xi i=1 i=1 i=1 i =1 n i=1 n i=1 n n n n n n n Figure Ex. reducing the function on the boundary to one of a single variable. b2 = 0 fb1a. x = 0. y2 = x2 + y2 . Walter O. 43. by Warren B. and solving for a gives n n n i =1 a = i=1 a xiyi n i=1 a xi a yi n n 2 i=1 2 a xi - a a xi b i=1 (0. For example. To find the critical points of this function solve the equations fa1a. Find the extrema of the function f1x. 0) 42. To find the appropriate boundary point. Gordon. This was a three dimensional minimization problem that was reduced to a two dimensional one by using the constraint to solve for one variable in terms of the other two.x # y sx . and April Allen Materowski.4y + 28 defined over the region bounded by y = 0. Copyright © 2007 by Pearson Education.4 The Method of Lagrange Multipliers » » Method of Lagrange Calculator Tips In Chapter 3 we considered optimization problems subject to constraints. 43 44. Economics. y2 = x2 + y2 . n n where x = i =1 a xi n and y = i=1 a yi n . Wang.ax. we may use the equation of the boundary to eliminate one variable. While the proof of his theorem requires more advanced mathematics. In the last section. . Inc. The critical points are found by the methods of this section. 45. Method of Lagrange Applied Calculus for Business. Find the extrema of the function f1x.4x . in finding the dimensions of the rectangle of largest area with a given perimeter. The determination of the critical points of such optimization problems can be found using a method developed by the mathematician Joseph Louis Lagrange (1736 1813). we considered the problem of minimizing the surface area of a box if it was required to have a given volume. for a. b and c.4x . The symbol l is the Greek letter lambda and is called the Lagrange multiplier. Inc. x.10 = 0 There are many ways to do the algebra. Wang. y. z2 then the critical values of x. Copyright © 2007 by Pearson Education. This does not effect the critical points of f (our objective) in any way.10l We now find the critical points of L. and l. define the function L1x. y. y. Why?) We solve the system y + 2l = 0 x + 2l = 0 2x + 2y .2l Applied Calculus for Business. introduce a new variable l and define the function L1x. Given the function defined by w = f1x. z2 = 0. This method generalizes to any number of variables. y. as indicated in the previous section to make this determination. Example 1 Maximize f1x. y. y. y. We form the function L1x. l2 = x + 2l = 0 Ll1x.10 = 0 (Note that Ll1x. y. and l. z2. Lx1x. Gordon. x. y2 subject to a constraint c1x. y. l2 = f1x. Economics. y. then c1x. Published by Pearson Learning Solutions.10 = 0. y. l2 = xy + l12x + 2y . note that the first equation gives y = .10 = 0. It will be the negative of the value we find. . l2 = 0 always yields the constraint equation. y. y. l2 = xy + 2xl + 2yl . by Warren B. y2 Then the critical values of x and y of f subject to c = 0 will be among the critical values of L with respect to the three variables. While the method of Lagrange multipliers yields the critical points of f. z2 + lc1x. y.lc1x. z. The only difference in this form is the value for l at the critical point of L. z. y and z of f will be among the critical values of L with respect to the three variables.2l and the second gives x = . z2 subject to the constraint c1x. y2 + lc1x. l2 = f1x. y2 = xy subject to 2x + 2y = 10 Solution The constraint 2x + 2y = 10 needs to be written in the form 2x + 2y . y2 = 2x + 2y . We note that some other texts write the equation of the form L1x. y. l2 = f1x.4 The Method of Lagrange Multipliers METHOD OF LAGRANGE Given the function defined by z = f1x.532 * ** Section 6. Walter O. y2 = 0. l2 = 2x + 2y . We will need to use other methods. z2 . and April Allen Materowski. z. y. y. l2 = y + 2l = 0 Ly1x. it does not identify which of those critical points are its maxima and minima. We may also have additional constraints as we shall see below.102 or L1x. and Finance. y2 = xy.42 = xy + 2xz + 2yz + lxyz .1x + 2z2 = l xz (2) (1) Applied Calculus for Business.4 = 0 and we have L1x.4l We next find all partial derivatives and set them equal to zero. z2 = xy + 2xz + 2yz subject to the constraint xyz = 4. as we did above. We could always find l. we have y = 5/2 as well. In the next example. Economics. l2 = xy + 2xz + 2yz + l1xyz . Note that the problem in Example 1 is asking you to find the maximum area of the rectangle whose perimeter is 10.1y + 2z2 = l yz We solve for l in the next two equations the same way and obtain . Thus. Walter O. 2x + 21x2 . Lx = y + 2z + lyz = 0 Ly = x + 2z + lxz = 0 Lz = 2x + 2y + lxy = 0 Ll = xyz . y. y = . Example 2 Minimize f1x. and April Allen Materowski. it must be here. y. we have z = 25/4. 5/2. Gordon. we consider the method of Lagrange multipliers to solve the minimization problem considered in the last section. and since z = f1x. and Finance. Note that we could also have used x = . not L. f subject to the constraint c = 0 has as its only critical point (5/2. z. It s easy to show. yielding x = y = 5/2. Published by Pearson Learning Solutions. that this point is a relative maximum. but we really want the critical points of f.Section 6. We could also find l.4 The Method of Lagrange Multipliers * ** 533 this means that y = x we substitute for y in the third (constraint) equation. and if f has a maximum. The best way of solving this last example and of solving most constrained optimization problems in two variables.5/4. where we also have an easy way of justifying the maximum. but we really do not need it.10 = 0 4x = 10 x = 5/2 since y = x. Inc. we solve each of the first three equations for l. we often may express the variables in terms of one variable and then substitute into Ll = 0 to solve for this variable. Copyright © 2007 by Pearson Education. is by using the methods of Chapter 1. Alternately.lyz or . using the second partial derivative test.2l. 25/4). We rewrite the first equation as y + 2z = . Solution We rewrite the constraint xyz = 4 in the form xyz .2l and substitute into the constraint equation to find that l = . One way of solving the system of equations is to solve for all the variables in terms of l. as we want the critical points of f. Wang. by Warren B. and then substitute for them in the equation Ll = 0 and solve for l. .4 = 0 As suggested above. thus the dimensions of the box are 2 * 2 * 1. In fact. these variables represent the dimensions of the box and must all be positive. y and z cannot be zero by the constraint equation xyz = 4. z2 = 1x . Then we could have used y = x to obtain z in terms of x. and Finance. . and April Allen Materowski. to obtain x x1x2a b = 4 2 x3 = 8 x = 2 since y = x. y = 2 and z = x/2 = 1. Since each of these equations equal l. from the statement of Example 8 in the last section. Wang. so we shall set equation 112 = equation (2) and then equation 112 = equation (3).1x + 2z2 = yz xz We clear fractions by multiplying each side of the equation by xyz to obtain x1y + 2z2 xy + 2xz 2xz y We now set equation 112 = equation (3) . Gordon. Applied Calculus for Business. Copyright © 2007 by Pearson Education. Now that we have y = x and z = x/2. this would give z in terms of y (verify this!).322 if x2 + y2 + z2 = 1. y and z at the critical points of f1x. by Warren B.1y + 2z2 . Published by Pearson Learning Solutions. we substitute into the fourth equation (the constraint) xyz .12x + 2y2 = l xy (3) Note that we need not be concerned about dividing by zero when solving each of these equations for l because x.222 + 1z . Walter O.1y + 2z2 . Inc. multiply by xyz to obtain x1y + 2z2 xy + 2xz xy x z12x + 2y2 2xz + 2yz 2yz 2z x z = 2 = = = = = = = = y1x + 2z2 xy + 2yz 2yz x Note that we solved for y and z in terms of x.4 The Method of Lagrange Multipliers and . Economics. y.12x + 2y2 = yz xy Once again.122 + 1y . We could have just as easily set equation (2) equal to equation (3). Example 3 Find the values of x.534 * ** Section 6. Setting equation 112 = equation (2) .4 = 0. they are equal to each other. -2 -3 114 . y. by Warren B. We remark that geometrically. and Finance.1 = 0 and L1x. Walter O. 114 2 2 2 1 2 3 b + a b = 1 b + a 1 + l 1 + l 1 + l B and A 1114.32 Ll = x2 + y2 + + 2lx = 0 + 2ly = 0 + 2lz = 0 z2 . the preceding example determines the points on the unit sphere centered at the origin nearest and farthest from the point (1. 1 -1 2 -2 3 -3 x = = . y and z at the critical points are A 1 14 .y = = . 2 3 114 . In the first of these equations 21x . we find y = 2 1 + l 3 z = 1 + l 1 1 + l We substitute for x.214.y = = .222 + 1z . .12 Ly = 21y . Similarly. l2 = 1x . x = = .12 then Lx = 21x .322 + l1x2 + y2 + z2 . 214 Thus.122 + 1y . y and z in the constraint equation to obtain a or 14 = 1 11 + l22 11 + l22 = 14 1 + l = .1 = 0 We shall solve for each of the variables in terms of l. and April Allen Materowski. 1 + l 1 + l 1 + l 214 214 214 -1 Thus.4 The Method of Lagrange Multipliers * ** 535 Solution We rewrite the constraint as x2 + y2 + z2 .Section 6. so the two critical points found are the extrema of f. and z = = . Gordon. 114 B. Copyright © 2007 by Pearson Education. the values of x. 2. for each value of l we obtain a critical point. Inc. Applied Calculus for Business. z.22 Lz = 21z . when 1 + l 1 + l 1 + l 214 214 214 1 1 2 2 3 3 1 + l = 214. Wang.12 + 2lx = 0 we divide by 2 to obtain x . Published by Pearson Learning Solutions.1 + lx = 0 or 1l + 12x = 1 or x = Similarly. and z = = . Economics. When 1 + l = . 3). y. then we consider L1x. y. z.4 The Method of Lagrange Multipliers Additional constraints are handled in a similar way. y. z2 = x2 + z2 . y. z2 = xy . m2 = f1x. y. z2 + lc11x. Wang. equation (8) so it is rejected. We have L1x.z -z 2x (9) We substitute into (7) for x to obtain z2 + z2 . .z2 = 0 x2 = z2 x = . Economics. y. Inc. z.8 = 0 Lm = xy .8 = 0 and c21x. and m. z2 = 0 and c21x. Walter O. Copyright © 2007 by Pearson Education.42 Lx = z + y + 2xl + my = 0 Ly = x + mx = 0 Lz = x + 2zl = 0 Ll = x2 + z2 . y. l. and Finance. y. y and z for f will be among the critical values of x. x + 2a -z bz = 0 2x z2 x = 0 x x2 .8 = 0 2z2 = 8 Applied Calculus for Business. z. y. y. The values of x. We substitute m = . z2 + mc21x. Example 4 Determine the critical points of w = f1x. We illustrate with an example. by Warren B. y.1 into (4) and obtain z + y + 2xl .1. which yields either x = 0 or m = . z2 = xz + xy subject to the constraints x2 + z2 = 8 and xy = 4. Suppose we want to optimize the function f(x.1. thus m = .4 = 0. z) subject to the constraints c11x.82 + m1xy . Solution We rewrite the two constraint equations in the form c11x. l. l. Gordon. m2 = xz + xy + l1x2 + z2 . z2 and find its critical points with respect to the five variables x. y. z. and April Allen Materowski.4 = 0 (4) (5) (6) (7) (8) From (5) we have x11 + m2 = 0. z2 = 0.536 * ** Section 6. x = 0 contradicts the second constraint. The only difference with the additional constraint is the complexity of the algebra in determining the critical points of L. y. and m of L.y = 0 or solving for l yields l = We substitute for l in (6). l. Published by Pearson Learning Solutions. Maximize f1x. We can easily compute w = f1x.z = 0. Find the critical points of f1x.x = 2. Find the critical points of f1x. Maximize f1x. y2 = 0 and d1L1x. y2 = 2x2 + y2 such that 3x + 4y = 12. y2 = 2xy such that y = 27 . For ease. 82 The method of Lagrange reduces.2. 2. y. We may use the solve command on the calculator to solve these equations. Maximize f1x. we store this function into memory. Wang. Minimize f1x. 2.2. Maximize f1x. . y. y2 = xy such that x2 + y2 = 16. when x = . 20. y. Minimize f1x. 2. 19. y. y2 = x + y such that xy = 8. Minimize f1x. y. y. z2 = 2x2 + y2 x + y + 2z = 7 and y . x = . 16. 13. Minimize f1x. z2 = xyz such that xy + 2xz + 2yz = 8. .2.4 The Method of Lagrange Multipliers * ** 537 z2 = 4 z = . 2.2. 9. z2 = xz + xy for each set of critical values. by Warren B. z2 = xyz such that 36x2 + 9y2 + 4z2 = 1. 23.42 STO L(x. to the solution of a system of equations.Section 6. and press the backspace key to delete this function).2. 2. Find the critical points of f1x. 2. t2 = 0. 22. An athletic field is a rectangular region with a semicircle on each end. .x + 2y . t). y2 = x + 4xy such that x y = 4. Maximize f1x. 18.x . Copyright © 2007 by Pearson Education.3z = 21 and 2x . The four critical points are therefore 1 .22 and when z = 2. From (8). y.42 + y such that y = 2 1x. Walter O. Find the critical points of f1x. t62 The calculator will produce the solution after a few seconds. Minimize f1x. z2 = xyz such that 3x + 2y + 5z = 27. enter solve1d1L1x. 12. 5x.42. 17 2 z such that 2 Applied Calculus for Business. y.2. Find three numbers whose sum is 12 if their product is to be as large as possible. z. y. z. 4x 10. 02 and 12. 14. y. y. t2. z. as the key-strokes are simplified. when z = . 2 2 2 2 2 2 2 15. z. we use t. Gordon. after the partial differentiations.3 11. 5. y. . y2 = 1x . Inc. y2 = 2x + 3y such that xy = 24 8. y. and t for L.2. z2 = xyz such that 2x + 3y + 4z = 36. so y = 4/x Thus. we have x = . 1 . l2 = xy + 2xz + 2yz + l1xyz . Consider Example 2 where we found the critical points of L1x. 2. . 4. determine its dimensions if the field if its rectangular portion is to have maximum area. 21.2 Thus. Calculator Tips EXERCISE SET 6.2 and when x = 2. z2 = xy + 2xz + 2yz such that xyz = 32. y. t2. subject to the given constraint(s). Maximize f1x. y2 = 1x . z2 = x + z such that x2 + y2 + z2 = 8. and Finance. z2 = 2xy + 2xz + yz such that xyz = 108.3y + 4z = 15.22 and 12. If the perimeter of the field is to be 250 yards. y = . z2 = 0 and d1L1x. Minimize f1x. 25. y. y. z2 = x2 + y2 + z2 such that x . y2 = xy such that 2x + 6y = 18. y. so we have 1 . 2) where the four y-values need to be determined. so we get 1 . Find the critical points of f1x. that is. Published by Pearson Learning Solutions. 17. we have xy = 4. 22 and (2. . Minimize f1x. 7. . y2 = xy such that 4x + 2y = 12. Maximize f1x.2. y. use the method of Lagrange multipliers to optimize f as indicated. Minimize f1x. remember to delete L from memory (VAR-LINK then scroll down to 1. z2 = 3x2 + 4y2 + 2z2 such that . y = 2. y. y2 = xy such that y = . x2 = 0 and d1L1x. Economics.4 In Exercises 1 29. Maximize f1x. 6. 3. Maximize f1x. y. y. 12. z. When you finish. To determine the critical values of x. We suggest that instead of using l as the symbol representing the Lagrange multiplier. y.2y + 3z = 6 and z .2. t2. z2 = x2yz3 such that x + y + z = 12.2. z. 02. z. we enter x * y + 2 * x * z + 2 * y * z + t * 1x * y * z . where x 7 3 x . t2. 24. z2 = x + y + z such that x2 + y2 + z2 = 36. and April Allen Materowski. Maximize f1x. 1. y. y. z.22 + y such that y = 1x. y. 82. y2 and p2 = G1x. Suppose we have two products. and y the number of units of capital. all the variables are non-negative.000.538 * ** Section 6. we could write p1 = F1x. Copyright © 2007 by Pearson Education. For example. if a decrease in demand for one product (due to an increase in its price) corresponds to an increase in demand for the other product. we must have 7 0. Inc. If each unit of labor costs $500 and each unit of capital $200. p1. 6. The United Postal Service requires that for any rectangular package. and April Allen Materowski. Let us also assume that the price for each of these products is related. this could be done in this case as well. Production is modeled by the Cobb Douglas production function 2 1 f1x. Walter O. the price of an automobile depends on the prices of the products used in its construction. if two 0y 0x products are substitutes. p22 and y = g1p1. Gordon. 30. p2 Ú 0. the top $1 per square foot and the base $4 per square foot. where x represents the number of units of labor.) For (1) to represent a demand function we must have for fixed p2. the demand for y may increase or decrease. For example. then the products are said to be substitutes. 27. 28. Recall that a demand function was one relating price per item to the demand for the item. and the amount allocated to labor and capital is $300. Find the critical points of f1x. 0 p2 Consider (1). y.5 Economic Applications » » » » Demand Equations Complementary and Substitute Products Joint Revenue. Thus. demand for the item decreased. that is. similarly. aspirin and acetaminophen. 0 p1 0 p2 Applied Calculus for Business. Suppose when the demand for x decreases the demand for y increases. so we have. one important economic application was the study of demand functions. Economics. and p1 is the per unit price when x units of the first product are demanded. and. y2. 26. for constant p2. Determine the dimensions of the package of largest volume that may be sent satisfying this requirement. x as a decreasing 0x function of p1. use Lagrange multipliers to determine the number of units of labor and capital which maximizes the level of production. and p2 is the per unit price when y units of the second product are demanded. Let us write the demand equations in the form x = f1p1. and Finance. the sum of its length and girth (cross sectional distance) not exceed 108 inches. Show that Ll = 0 always yields the constraint equation c = 0. Cost and Profit Marginal Rate of Substitution Demand Equations When we studied functions of a single variable. and similarly. determine the dimensions that minimize the production cost of the box. y2 = 200x3 y3 . sugar and honey. as p1 increases x decreases (since f is a demand function). If the volume of the box is to be 10 cubic feet. we must have 7 0. The cost of producing a rectangular box is as follows: the sides cost $2 per square foot. Published by Pearson Learning Solutions. We now generalize the notion to a function of two variables. p22 (2) (1) Complementary and Substitute Products In a realistic economy. 31. however. by Warren B. that means that 6 0. for (2) to represent a demand 0 p1 0y function we must have 6 0. and when the price increased. y2 = x2 + 24xy + 8y2 such that x2 + y2 = 25. butter and margarine are examples of substitute products. x. (When we studied the one variable problem.5 Economic Applications 29. Wang. That is. . we wrote the price as a function of demand. Find three numbers whose sum is S if their product is to be as large as possible. by Warren B. we note that 0y 0x = . Copyright © 2007 by Pearson Education.0. 2 p1p2 p1p2 0x 0x -8 = 6 0. . 3 0 p2 0 p2 p1p2 0y -4 = 3 6 0. Walter O. and April Allen Materowski. we have 0y 0x = 2 7 0 and = 4 7 0.3p2 + 12. Cost and Profit Applied Calculus for Business.8. therefore. and since both p1 and p2 are positive. Determine if the products are substitutes. and 6 0.1 and p2 = . we have that 0 p1 p1p2 The notion of revenue generalizes as well. Differentiating. if two products are complementary. y2 The next example considers the maximization of the profit. Published by Pearson Learning Solutions. Example 2 The demand equations for two related products are defined by x = Determine if the products are substitutes.6y + 10. y2 = R1x. 0y 0x Thus. the products are complementary. and y = 4p1 . and Finance.0. Similarly. bread and butter. and bicycles and helmets are examples of complementary products. tires and gasoline. For example. Joint Revenue. 0 p 0 p 2 1 0 y 0x # in which case 6 0. so the products are substitutes.5 Economic Applications * ** 539 On the other hand. we must have 6 0. Solution First. Inc. y2 . complementary or neither.Section 6. Gordon. R1x.3 6 0 as they 0 p1 0 p2 both should be if the equations represent demand functions. y2 = xp1 + yp2 If we let C(x. We have the joint revenue function derived from the two products. Two 0 p1 0 p2 0y 0x products are neither substitutes nor complementary if and are of opposite signs.6 6 0 and = . two products are said to be complementary if a decrease in demand for one results in a decrease in demand for the other.2y + 5. Solution We have that 4 2 and y = 2 . We leave it as an exercise for you to show that p1 = .6p1 + 2p2 + 9. Wang. complementary or neither.0. 0 p2 0 p1 Note in the previous problem we could give p1 and p2 as functions of x and y.4x . 0 p2 0 p1 Example 1 The demand equations for two related products are defined by x = .0.C1x. y) represent the joint cost function in producing the two products then the profit derived from the two products is given by P1x.3x . Economics. Wang. y.540 * ** Section 6.7y + 622 .C1x. and the production level of 240. and April Allen Materowski. y2 = R1x.7y2 + 56y Px1x. where x is the number of units of labor. y2 = xp1 + yp2 .2x + 6y + 252 + y1 .P15 + h. y2 = 5x + 6y.2x + 6y + 25 and p2 = . $6 to produce each unit of Product 2.6y or P1x. and the joint revenue function R1x.000 units is to be achieved. If each unit of labor costs $4. y2 = 0. 4) produces a maximum as well.14 and Pxy1x. y2 = . We leave the first method to the exercises and proceed with the Lagrange multiplier approach.12 = 0 We solve the second and third of these equations for l.4x + 20 = 0 Py1x. y2 . P = 162. Walter O. Economics. Find the demands and prices for the two products if profit it to be maximized.4. each unit of capital $1. 42 = 56 7 0 and it follows from the second partial derivative test that the profit has a relative maximum at x = 5 and y = 4. . by Warren B. where p1 and p2 are the per unit prices in dollars for Product 1 and Product 2 respectively. y2 = 4x + y such that 1 2 20000x3y3 = 240000. or we can use the method of Lagrange multipliers.6x .12 B 3 3 Lx1x. y2 = . We could solve this constraint equation for y in terms of x and then substitute for it into the cost equation and set its partial derivatives to zero. (a) determine the number of units of labor and capital that minimize the cost. y2 = 1 + 2 3 lx y = 0 1 -1 -2 2 1 2 Ll = x3y3 . we have L1x. y2 = x 3y 3 . y2 = 4 + 1 3 lx y = 0 3 3 Ly1x.6y = x1 . First. l2 = 4x + y + l A x3y3 . Therefore. Inc.5x . Solution We have the joint cost function C1x. y2 = xp1 + yp2 and the profit is then P1x. Pyy1x.2x2 + 20x . we have x = 5 and y = 4. y2 = 20000x 3y 3. we find that p1 = $ 39 and p2 = $ 4. It is not difficult to show that (5. (b) Determine the minimum cost. Thus. Example 4 1 2 Assume the Cobb-Douglas production function f1x. 4 + k2 = 2h2 + 7k2 7 0 therefore at the point where x = 5 and y = 4. y2 = . Published by Pearson Learning Solutions. obtaining 1 2 Applied Calculus for Business. 42 . D15. Solution (a) We need to minimize the joint cost C1x. We leave it as an exercise for you to show that P15. Substituting. y2 = . and Finance. and the profit is indeed a maximum. Pxx1x. we may divide the constraint equation 1 2 by 20000 and rewrite it as C1x.14y + 56 = 0 Solving. Copyright © 2007 by Pearson Education. y2 = .7y + 62.6x .5 Economic Applications Example 3 Suppose it costs $5 to produce each unit of Product 1. and y is the number of units of capital. Gordon. x is the demand for Product 1 and y is the demand for Product 2.12 = 0. and the two products have demand equations p1 = .5x . Copyright © 2007 by Pearson Education.3 3 which gives the condition y = 8x. giving x = 3. in a more realistic situation. the ratio of the marginal productivity of labor to the marginal productivity of capital is equal to the corresponding ratio of their costs. Therefore y = 8132 = 24.5 Economic Applications * ** 541 l = l = Therefore. we have .12x3 y3 which gives 2 2 .a. Economics.3y3 2x3 . In the previous example. Consider the dual problem of maximizing the productivity of the Cobb-Douglas productivity function where each of the x units of labor costs $a. to minimize cost and achieve the given production level. by Warren B. we would obtain 1 2 2 20000 x . then we can show using Lagrange multipliers that for the total joint cost to be minimized at a fixed production level. 242 = 4132 + 11242 = $ 36.12 = 0 or 4x = 12.12x3 y3 . which is what we found in the above solution. Wang. If we consider the general Cobb-Douglas production formula. (Note. where each of the x units of labor costs $a and each of the y units of capital costs $b. Inc. and April Allen Materowski. each unit of y units of capital costs $b and the total cost of labor and capital cannot exceed some given fixed value. Gordon. It is shown in a more advanced calculus course that the marginal rate of substitution is nothing more than the negative of the slope of the tangent line to the level curve.Section 6. (b) The minimum cost is C13. x and y may represent the number of units of labor and capital in thousands. This is known in Economics as the Marginal Rate of Substitution. . we must have 0f 0x a = 0f b 0y 1 2 Marginal Rate of Substitution Applied Calculus for Business.3y3 2x3 1 1 1 1 2 2 = y = 8x We now substitute into the constraint equation to obtain x318x23 . and then the cost would be in thousands of dollars. f1x.3y 3 3 4 = 2 1 1 1 20000 x 3y . Walter O. Thus. we require 3 units of labor and 24 units of capital. and Finance. we must have 0f 0x a = 0f b 0y That is. Published by Pearson Learning Solutions. We can show that for productivity to be maximized. y2 = Axay1 . 542 * ** Section 6. given a constraint on the items x. Gordon.5 Economic Applications That is. Economics. and April Allen Materowski. L1x.2xy/5 = l y2 2xy = 2 5 Since y = 0 is not a possible solution (Why?) we have y = 4x 5 and substitute for y in the constraint equation to solve for x 2x + 514x/52 = 60 6x = 60 x = 10 y = 41102/5 = 8 Thus. l2 = xy2 + l12x + 5y . the ratio of the marginal productivity of labor to the marginal productivity of capital is equal to the corresponding ratio of their costs.602 Lx = y2 + 2l = 0 Ly = 2xy + 5l = 0 Ll = 2x + 5y . by Warren B. We first write the constraint in the form c1x. once again. y2 = xy2. Published by Pearson Learning Solutions. For a utility function. to maximize the utility. Thus. We note that this example could also be solved by using the marginal rate of substitution to find the relationship between x and y. Applied Calculus for Business. Inc. and Finance. y. 8). We may use Lagrange multipliers to solve this problem. Walter O. What is his best buy? Solution By best buy. we may determine the bundle that maximizes the utility function.60 = 0. Example 5 Given the utility function u1x. Copyright © 2007 by Pearson Education. .60 = 0 We have . the best bundle is (10. we mean to maximize the utility function subject to the constraint 2x + 5y = 60. y2 = 2x + 5y . as illustrated in the next example. y. suppose the price for each item x is $2 and for each item y is $5 and the consumer has at most $60 to spend. Wang.y2/2 = l . x = . 12.5 In Exercises 1 7 the demand equations for two related products are given. .3p2 + 15. and April Allen Materowski. Show that in Example 3.0.2p2 + 9. x2. x = . Wang. Assume the Cobb-Douglas production function f1x. 11. Á .3p1 + 2p2 + 20. and y = . where x is the number of units of labor. x2. If 1 3 1 4 1 3 0f ai 0 xi = 0f aj 0 xj at the optimal point. Copyright © 2007 by Pearson Education. Á . (b) Find the minimum cost. Á . xn2 = a1x1 + a2 x2 + Á + an xn + l f1x1 . and y = . Solve Example 4 by the alternate method suggested and verify that the critical point is indeed a minimum. (b) Find the maximum productivity.p1 . Economics.5y + 12. (a) determine the number of units of labor and capital that minimize the cost. x = . 14. xn2 = f1x1. suppose the price for each item x is $4 and each item y is $2 and the consumer has at most $2.Chapter Review * ** 543 EXERCISE SET 6.2p1 . 1. and Finance. In the previous example show that the demand levels produce a maximum at the critical values x = a and y = b. each unit of capital $16.4x . What is his best buy? 15. 6. y2 = 16200x5 y5 . and the production level of 80. 4.1 and p2 = . and y = .7y + 62? 18. Applied Calculus for Business. and y is the number of units of capital.3x . y2 and p2 = G1x. p1 = . What conditions on the partial derivative determine when the products are complementary. x = . and the production level of 540. 9. complementary or neither. 4 + k2 = 2h2 + 7k2 7 0.3p2 + 12.0.4y0. Suppose the two products have demand equations p1 = .3p1 .000 to spend.3p2 + 12. substitute or neither? (b) What is the relationship between the products whose demand equations are p1 = . each unit of capital $3.p1 . x = 7. Determine if the products are substitutes. y2 = 10000x4 y4 . that is given (a) L1x1.000. xn2 + l1a1x1 + a 2x2 + Á + anxn2 then 8.0. b2 . by Warren B. Show in Example 1. and y is the number of units of capital. and y is the number of units of capital. and the fixed overhead cost is $10. and y = 5p1 . Published by Pearson Learning Solutions. where p1 is the per unit price demanded for x items of the first product and p2 is the per unit price when y units of the second product are demanded.0. y2 = 10000x0. determine the number of units of labor and capital that maximize productivity. Inc. (b) Find the minimum cost. and y = .5p1 + 2p2 + 9. Given the utility function u1x. if we solve for p1 and p2 in terms of x and y. If each unit of labor costs $8.6y + 10. x n2 then 0f ai 0 xi = 0f aj 0 xj at the optimal point. If each unit of labor costs $27. each unit of capital $243. where p1 and p2 are the per unit prices in dollars. by showing that P1a.P1a + h. x2 .2x + y + 20 and p2 = 3x . x = 4 6 . (a) determine the number of units of labor and capital that minimize the cost.8. Assume the Cobb-Douglas production function f1x. Find the demand levels for the two products if profit it to be maximized. x units of Product 1 and y units of Product 2 are produced. 5. 2. 3. x = . Assume the Cobb-Douglas production function f1x. p1p2 p1p2 2 each unit of labor costs $1. and y = p1 . b + k2 = 21h .3p1 + 2p2 + 8.2x + 6y + 25 and p2 = . and then explain why this proves P has a maximum at x = 5 and y = 4.6x . y2. Á . Prove the marginal rate of substitution more generally.4p1 . where x is the number of units of labor. 10. where x is the number of units of labor. Suppose it costs $8 to produce each unit of Product 1 and $12 to produce each unit of Product 2.2y + 5. (b) L1x1.P15 + h. 13. P15. and if the total cost may not exceed $10. Walter O.000 units is to be achieved. (a) Suppose the related demand equations are written in the form p1 = F1x. 16. 17. y2 = 10000x4 y4 .6.2p2 + 4. x2.2p2 + 3. Gordon. p1p2 p1p2 2 12 . 42 . and y = .4p2 + 8.000 units is to be achieved.k22 + 3k2. not the function whose partial derivative with respect to x is 2x + 3y3.2x + 3y2dx 12x3y3 + 5 . by Warren B. treated as if it were a constant in this integration. Example 1 Evaluate each of the following integrals: (a) (b) L L 12x3y3 + 5 . Economics. and it is understood that the variable y is 0 x.2x + 3y2dy Applied Calculus for Business. 3y2.544 * ** Section 6. Published by Pearson Learning Solutions. . We indicate this by writing Partial Integration 12x + 3y32dx L You might ask why do we write the differential symbol as dx. it was treated as a constant. find a function whose partial derivative with reSuppose we have f1x. Wang. Similarly. there is always a constant of integration. y2 = x2 + 3xy3 + 2y + 3. You should have observed that we asked to find a function. Walter O. since we are integrating with respect to the variable x. then spect to x is 2x + 3y3. Inc. and April Allen Materowski. the convention is to use dx. we really should write While this is true. Copyright © 2007 by Pearson Education.6 Double Integrals 6. Now suppose 0x we want to reverse the question. Gordon. we have 12x + 3y22dy = 2xy + 3 y3 + C1x2 = 2xy + y3 + C1x2 3 L Notice here that the 2x term in the integrand is treated as a constant when the integration is with respect to y. that is.6 Double Integrals » » » » » » » » Partial Integration Double Integrals Iterated Integrals Areas Volume Probability Density Function Average Value Calculator Tips 0f = 2x + 3y3. We have 12x + 3y22dx = 2 x2 + 3y2x + c1y2 = x2 + 3xy2 + c1y2 2 L Note that when we integrated the second term. and Finance. Of course. but in this case. since y is treated as a constant with respect to the integration. the constant of integration is a function of y. 2x + 3y2dx = a x=y x4y3 = + 5x .a + 5y . by Warren B.y2 + 3y3 b = . .y4 + 3y3 .2x + 3y2dx = 2 x4 3 x2 y + 5x . since we integrated with respect to x.x2 + 3xy b ` 2 x=y 2 a 1y224y3 1y24y3 + 51y22 .2x + 3y2dy = 2x3 x3y4 3y2 + 5y .a + 5112 .2xy + + c1x2 2 2 y4 y2 + 5y . As in the case of an ordinary definite integral.y4 + 6y2 . Economics.2xy + 3 + c1x2 4 2 L = We can also consider a definite integral of a function of two variables. Published by Pearson Learning Solutions.x2 + 3xy + C1y2 2 12x3y3 + 5 . Example 2 1 Evaluate (a) 0 3 y2 12x3y3 + 5 .1y22 + 31y22y b = 2 2 y11 y7 y11 y7 + 5y2 .5y 2 2 2 2 Applied Calculus for Business.x2 + 3xy b ` 2 x=1 = a 1224y3 1124y3 + 5122 .1122 + 3112y b 2 2 = 8y3 + 6 + 6y . Copyright © 2007 by Pearson Education.2 + 3yx + C1y2 4 2 x4y3 + 5x .2x + 3y2dx (b) Solution y 3 12x3y3 + 5 .6 Double Integrals * ** 545 Solution (a) L = (b) 12x3y3 + 5 . to remind us that we substitute for x when doing the computation. 2 (a) 1 3 12x3y3 + 5 . Inc. and April Allen Materowski.1222 + 3122y b . Gordon. we need only any antiderivative. and Finance.a y3 15y3 + 4 + 3y b = + 3y + 2 2 2 Note that when we evaluated the integral we indicated the limits of integration as x = 1 and x = 2. Wang. y2 (b) y 3 12x3y3 + 5 .2x + 3y2dx. Walter O.Section 6. we choose the one where the constant of integration is zero.1y222 + 31y22y b .a + 51y2 . so for simplicity.2x + 3y2dx = a x=2 x4y3 + 5x . Such an integral is an example of an iterated integral.a xe0 .x21x2 + 1x23 1023 2x3 b . observe that by indicating the limits of integration as x = y2 and x = y. not a function of x or y.x2102 + b = xex . by Warren B. we emphasize which variable is being replaced by the limits of integration.x2 + y22dy = a xey . this is most common in applications. With this example in mind.x2y + y3 y = x = b` 3 y=0 a xex .546 * ** Section 6. Inc.x 3 3 3 2 x Double Integral 13x2 + 2y2dy dx. Gordon. Economics. the iterated integral is a number.6 Double Integrals Once again. Let us compute this particular iterated integral.a + b = + 4 3 x=1 3 4 3 12 Notice. Published by Pearson Learning Solutions. first we evaluate the inner integral and then the outer. . we define a double integral in terms of an iterated integral as follows: Applied Calculus for Business. Walter O. Wang.x2 + y22dy. called a double 33 1 0 integral. as the next example illustrates. How shall we interpret it? It makes perfectly good sense to consider this as two combined problems. namely Consider the following expression 2 x 2 x 1 33 0 13x + 2y2dy dx = 2 1 3 3 0 + 13x2 + 2y2dy * dx that is. and April Allen Materowski. Solution x 0 3 1xey . We proceed the same way when we integrate with respect to y. Example 3 x Evaluate 0 3 1xey . We have x 0 3 13x2 + 2y2dy = 13x2y + y22 ` y=x y=0 = 3x3 + x2 therefore. Copyright © 2007 by Pearson Education. and Finance. 2 x 2 x 2 Iterated Integral 1 33 0 13x + 2y2dy dx = 2 1 3 3 0 + 13x + 2y2dy * dx = 2 1 3 13x3 + x22dx = 3x4 x3 x = 2 8 3 1 163 ` = a 12 + b . in this example. That happens because the limits of integration in the second integral were constants. Inc. y2dx dy = c 2 x2 3 + L1y2 3 f1x.Section 6. we obtain b T1x2 b T1x2 b y = T1x2 b a 3 3 B1x2 dy dx = a 3 + B1x2 3 1 dy * dx = a 3 y` y = B1x2 dx = a 3 1T1x2 . y2dx * dy (2) Example 4 Evaluate the double integral (a) Solution 2 x2 2 0 1 33 0 15x + 6y2dy dx (b) 2 1 2 x2 33 y 110xy2 . Copyright © 2007 by Pearson Education. that is. and April Allen Materowski.6 b T1x2 b T1x2 Double Integrals * ** 547 a 3 3 B1x2 f1x. Wang.2x2 ` * dy = 2 1 3 10 .02dx = 2 0 1 3 8x4 dx = 248 5 (b) 1 33 y 110xy2 . Economics.22dx * dy = 1 2 3 a15x2y2 . .28 Consider the double integral we obtain from (1) if we let f1x. by Warren B.B1x22 dx Applied Calculus for Business. Gordon. Walter O. y2dy * dx (1) and d R1y2 d R1y2 c 3 3 L1y2 f1x. y2dy dx = a 3 + B1x2 3 f1x.151y2 y .22dx dy (a) 1 33 0 15x2 + 6y2 dy dx = 1 2 3 3 0 + 15x2 + 6y2 dy * dx = 1 2 3 + 5x2y + 3y2 ` y = x2 y=0 * dx = 2 1 2 0 3 15x21x22 + 31x222 . and Finance. y2 = 1.21y222dy = 2 2 1 3 1 . b T1x2 a 3 3 B1x2 dy dx What does this represent? If we evaluate this integral.5y4 + 2y2dy = . Published by Pearson Learning Solutions.22dx dy = 2 1 3 3 y x=0 x=y + 110xy2 . dy dx = a 3 3 B1x2 dy dx = A1R2 (3) y = T(x) x=a x=b R y = B(x) Figure 1: The Region R Bounded by y = B1x2. Published by Pearson Learning Solutions.548 * ** Section 6. Wang. it represents the area of the region R bounded by the curves y = T1x2 and y = B1x2 between x = a and x = b. and Finance.3 and g1x2 = x + 4 on the Interval from . See Figure 1. Solution We first sketch the curves. Walter O.1 to x = 2. . Gordon. y = T1x2 on [a.3 and g1x2 = x + 4 over the interval from x = . Therefore if R represents the region indicated in Figure 1. Copyright © 2007 by Pearson Education. we may define b T1x2 L RL as the area of the Region R. and April Allen Materowski. by Warren B. Inc. We also include a representative rectangle in the sketch. Economics. that this last integral has a very simple geometric interpretation.6 Double Integrals Areas We saw in Chapter 5. T(x) = x + 4 x=2 x =-1 B(x) = x2 -3 Figure 2: Area Bounded Between f1x2 = x2 . see Figure 2.1 to 2 Applied Calculus for Business. b] Example 5 Find the area of the region bounded between f1x2 = x2 . Wang. d Therefore.3.1x2 . Example 6 Find the area of the region bounded by x . we may define d RL L R1y2 dx dy = c 3 3 L1y2 dx dy = A1R2 (4) as the area of the Region R. and April Allen Materowski. x = R1y2 on [c.x2 + x + 72 dx = 39 2 We can similarly interpret (2). Applied Calculus for Business. Walter O. The area is 2 RL L 2 x+4 2 dy dx = -1 3 23 x -3 dy dx = -1 3 11x + 42 .x = 0. Inc. and Finance. y2 = 1. See Figure 3.6 Double Integrals * ** 549 Everywhere on this interval. the upper curve is the line g1x2 = x + 4 and the lower curve is the parabola f1x2 = x2 . .322 dx = -1 3 1 . Copyright © 2007 by Pearson Education. by Warren B.y = 6 and y2 .Section 6. x = R (y) y=d x = L (y) R y=c Figure 3: The Region R Bounded by x = L1y2. Gordon. we have d R1y2 d R1y2 d c 3 3 L1y2 dx dy = c 3 + L1y2 3 dx * dy = c 3 1R1y2 . Published by Pearson Learning Solutions. Economics. if R represents the region indicated in Figure 3. If we let f1x.L1y22 dy This last integral represents the area of the region R bounded by the curves x = L1y2 and x = R1y2 between y = c and y = d. Note.y22 dy = 125 6 B (9. and April Allen Materowski.8. Example 7 Evaluate 112 .2y2 dA over the region R bounded by the x-axis. Of course.2 and the highest is d = 3. we may write (5) dA RL L where dA is either dydx or dxdy. and when we integrate in the order dxdy.4x . Copyright © 2007 by Pearson Education. and Finance. we have.550 * ** Section 6. Therefore.6 Double Integrals Solution This is Example 5 in Section 5. Walter O. we are covering the region with horizontal rectangles. we define f1x. when we integrate in the order dydx we are covering the region R with vertical rectangles. 3) x = y2 x=y+6 A (4.y2 dA to be the value of the iterated double integral evaluated over the region RL L R. Applied Calculus for Business. Economics. 3 y+6 3 y+6 A = 3 RL L dx dy = -2 3 3 2 y dx dy = -2 3 + y 3 2 dx * dy = -2 3 1y + 62 . We illustrate this definition in the next example. . the line L RL y = 6 . by Warren B. -2) Figure 4: Area of the Region Bound by x = y + 6 and x = y2 Since the area of a region may be computed by either the use of (3) or (4). Wang. Gordon.2x between the y-axis and x = 3. A1R2 = We can now generalize the notion of a double integral. Inc. Published by Pearson Learning Solutions. We see from Figure 4 that the curve on the right R1y2 = y + 6 and the curve on the left is L1y2 = y2. the lowest y value is c = . however. that is. we must be careful to make sure we indicate the limits appropriately when we use (5). 4x . and R1y2 = .Section 6.y x = L1y2 = 0.2x dx = y=0 0 3 111216 .4xy .02 dx = 3 0 3 14x2 . by Warren B.2x between the y-axis and x = 3 Second.2x Figure 5: The Region Bounded by the x-axis. Copyright © 2007 by Pearson Education. Gordon. We evaluate this double integral two ways. Wang. Economics.4x .2x222 . and Finance.16 .4x16 . Published by Pearson Learning Solutions. Note that 6 . the line y = 6 .4x .y22 ` y = 6 . First we evaluate the integral by 112 .2y2 dydx = 0 3 3 112y . 2 Applied Calculus for Business. we evaluate the integral by reversing the order of integration.6 Double Integrals * ** 551 Solution A sketch of the region is given in Figure 4.2y2 dy dx = 3 0 3 3 0 112 .4x .24x + 362 dx = 36 y = 6 .2x2 . and April Allen Materowski. we have RL L 3 6 .2y2 dy dx RL L L RL From Figure 5.2x2 . Inc. .2y2 dA = 112 . Walter O.2x 112 . 4) x = y /2 or y . We illustrate in the next example. Gordon. as y ranges from 0 to 4. Copyright © 2007 by Pearson Education. we have. RL L 6 6-y 2 112 . Solution While there are methods that can be used to integrate 24 . This means we must first examine the region y R. Walter O. Example 8 Evaluate the double integral 4 2 0 3 3y 2 24 . Published by Pearson Learning Solutions.2y2 dx dy = 0 33 0 112 .4x .2y2 dA = L RL 112 . Inc. and Finance.x2.6 Double Integrals Therefore. this region is bounded by the lines x = and x = 2. Instead.4x . we shall evaluate this integral by reversing the order of integration. x = 2. by Warren B. for x between 0 and 4 Applied Calculus for Business. Economics. (2. they are beyond the scope of this text (they require trigonometric substitution).2xy22 ` 2 y x=62 dy = x=0 0 3 a 18y .2x x=2 R Figure 6: The Region bounded by x = y/2.2y2 dx dy = 6 6 0 3 112x .552 * ** Section 6. See 2 Figure 6.2x .x2 dx dy.4x . Wang.6y + y2 b dy = 36 2 The notion of reversing the order of integration is sometimes very useful in the computation of a double integral. and April Allen Materowski. . y2 = 12 .10 . RL L f1x. and April Allen Materowski. y)dA is then the volume of this representative tube. Their product f(x. Then the volume of the solid bounded below by the region R and above by the surface z = f1x. y2 be a non-negative continuous function. yi.x2 dx = 2 2 2 2 16 . . y2 dA L RL Example 9 Find the volume of the solid bounded above by z = f1x. so we have.14 . Let the length of the longest diagonal of all these rectangles is denoted by 7 ¢ 7 .82 = 3 3 3 3 0 We remark that just as a single definite integral may be represented as a Riemann sum. 1xi. y2 is V = f1x. Then it can be shown. Copyright © 2007 by Pearson Education. Applied Calculus for Business. In fact. and ¢ Ai the area of this rectangle. the area of each of the base tubes is dA and the height of each of these representative tubes is z = f1x.4x . Inc.2y and below by the region in the x-y plane bounded by the coordinate axes and the line y = 6 . a double integral may be also be represented as such a sum.6 Double Integrals * ** 553 Note that x = y/2 may be written as y = 2x. by Warren B. which is the volume of the solid is then the double integral V = following definition. y2 dA (6) Volume You recall that with a single integral. Gordon.2x. suppose the region R is partitioned into n rectangles by drawing lines parallel to the coordinate axes that cover the closed and bounded region R.x223/2 ` = . and let z = f1x. y2. and Finance. In fact we considered a very thin representative rectangle of width dx and multiplied it by its height f(x) to obtain an element of area b f(x)dx. y2 dA. Wang. 4 2 2 2x 0 3 3y 2 24 .1 .x2 dy dx = 2 0 3 A y 24 .x B ` 2 dx = y=0 0 3 2x 24 . Walter O. We may use this sum analogy to realize that a volume is the infinite sum of infinitesimally thin rectangular tubes.43/22 = .yi2 ¢ Ai = lim 7¢7 : 0 a i=1 L RL f1x.2 is any point in the ith rectangle. We summarize with the DEFINITION 1 Let R be a bounded closed region in the x-y plane. the infinite sum of all these rectangles became the area A = a 3 f1x2 dx. for a continuous function f defined over R that n f1xi. Economics.Section 6. area was nothing more then the limit of the infinite sum of infinitesimally small rectangles. Published by Pearson Learning Solutions.x dx dy = 2 0 y = 2x 33 0 2 24 . The infinite sum of the volume of all these tubes. 4) y = 2x y = x2 Figure 8: Region Bounded by the Intersection of the Curve y = x 2 and the Line y = 2x. 0) and (2. by Warren B. x2 = 2x. Economics. Published by Pearson Learning Solutions. yielding (0. . Inc.554 * ** Section 6. Solution The region R in the x-y plane is shown in Figure 8. and below by the region formed by the intersection of the curve y = x2 and the line y = 2x. where we found the value of this integral to be 36. Applied Calculus for Business. the volume of the solid is 36 cubic units. Walter O. Wang.2y2 dA This is exactly the integral we computed in Example 7. Gordon. Note when the curves intersect. and April Allen Materowski.4x .2x Example 10 Find the volume of the solid bounded above by the surface z = f1x. y2 = 4 + x2 + y2. Note that we have V = L RL f1x. y2 dA = L RL 112 . y = 6 -2 x Figure 7: The Region R bounded by the Coordinate Axes and the Line y = 6 . 4) as indicated (verify!) (2. and Finance.6 Double Integrals Solution The region R in the x-y plane is indicated in Figure 7. Thus. Copyright © 2007 by Pearson Education. Draw a representative rectangle to determine which are the bottom and top (or left and right) curves. . by Warren B. In Section 5. b] to be non-negative with unit area on this interval. We illustrate in the next example. Gordon. We have. and Finance. Economics. 2 2x V = 2 RL L 14 + x + y 2 dA = y3 y = 2x dx = b` 3 y = x2 2 2 2 0 33 2 x 14 + x2 + y22dy dx = 0 3 a 4y + x2y - 0 3 a- x6 14 3 1208 x . Wang. y2 dA = 1 Probability Density Function RL L If the event (set) E is a subset of R.4x2 + 8x b dx = . Our objective in this section is to introduce the concept of volume as an application of a double integral.8. y2 Ú 0 on R and (b) f1x. y2 dA as an iterated integral using the information (3) Evaluate the iterated integral We could generalize Definition 1 to determine the volume between two surfaces. We generalize this notion to two dimensions as follows: Let R be defined as above. clearly indicating the curves bounding the region and its endpoint. Inc. then the probability that the event E occurs is given by Pr1E2 = f1x. y2 and below in the x-y plane by the region R is as follows: (1) Sketch the region R in the x-y plane. and April Allen Materowski. Published by Pearson Learning Solutions. Applied Calculus for Business. then f(x. RL L f1x.x4 + 3 3 105 The procedure for determining the volume of the solid. Walter O. y2 dA (7) EL L Thus. bounded above by the continuous function z = f1x.Section 6. (2) Write the integral V = from (1). we leave as an exercise for you to do the evaluation in the reverse order. y) is said to be a joint pdf on R if (a) f1x.6 Double Integrals * ** 555 Since the solid is bounded above by z = f1x. we have V = RL L 14 + x2 + y22 dA We choose to evaluate this integral with dA = dy dx. Copyright © 2007 by Pearson Education. we considered a probability density function (pdf) for a function of a single variable defined on an interval [a. y2 = 4 + x2 + y2. the notion of a two-dimensional probability is nothing more than the evaluation of a double integral. We do this in the exercises for the interested reader. Inc. 4) y = x2 x=2 Figure 9: The Region R Bounded by y = x 2.6 Double Integrals Example 11 27 2 2 Given f1x. f is indeed a pdf over the region R. Gordon. the x-axis and the line x = 2. y2 = 512 x y over the region R in the x-y plane bounded by y = x2. (b) The subset E is the subset of R below y = 1 as shown in Figure 10. Copyright © 2007 by Pearson Education. Solution (a) The region R is shown in Figure 9. the x-axis and the Line x = 2 We need only compute 2 x2 27 2 2 27 2 2 x y dA = x y dy dx = R L 512 3 3 512 L 0 0 0 27 2 y3 y = x 9 9 x9 2 ` ` = 1 a x b dx = x8 dx = 512 3 512 9 0 3 512 3 y = 0 0 2 2 2 Therefore. (2. y = 1. x = 2 and the x-axis. Published by Pearson Learning Solutions. (a) verify that f is a pdf over the region R. Wang. determine Pr(E). y = x2 (or ) x=2 y=1 E Figure 10: The Subset E Applied Calculus for Business. . Walter O. by Warren B. and Finance. and April Allen Materowski. Economics.556 * ** Section 6. (b) Let E be the subset of R bounded by y = x2. y2 over R = A1R2 Average Value (7) Example 12 Determine the average value of the function f1x. . and Finance. Inc. Pr1E2 = 1 2 27 2 2 512 x y 27 2 2 27 2 2 x y dA = x y dx dy = 512 512 EL 3 3 L 0 1y 0 27 x3 2 x = 2 9 11 y ` dy = 18y2 . Walter O.Section 6. EL L that is. We could use a double integral to compute the area of the region R. x = 2 and the x-axis. but it is most easily found by realizing it is a right triangle with height 8 and base 2. its geometric interpretation turned out to be that y-value which when multiplied by the length of the interval yields the area of the region. Economics. by Warren B. Wang. Similarly. We have. that is. Copyright © 2007 by Pearson Education.6 Double Integrals * ** 557 dA. it is simplest to compute this integral using dA = dx dy. Gordon. b]. (2. we consider the notion of the average value of a function of two variables over the region R. the x axis and the Line x = 2 Applied Calculus for Business. y2 dA RL L Average Value of f1x. with horizontal rectangles. we define f1x. 8) y = 4x x=2 Figure 11: The Region R bounded by y = 4x.y7/22 dy = 512 3 512 256 x = 1y 3 3 0 1 1 We remark that reversing the order of integration results in two double integrals that need to be evaluated. see Exercise 49.7 we considered the average value of a function of a single variable over the interval [a. Published by Pearson Learning Solutions. therefore A1R2 = 1*2 122182 = 8. In Section 5. As our last example of an application of a double integral. Solution The region R is shown in Figure 11. y2 = y 29x2 + y2 over the region bounded by y = 4x. we may think of the average value of a function of two variables as that z-value which when multiplied by the area of the region yields its volume. and April Allen Materowski. 2ey + 2x . y2 . g1x. 0. 22 and pressing enter (remember. x. 3. Published by Pearson Learning Solutions. We remark that the notion of a double integral can be extended to any number of variables. 2 33 0 1 2 1x2 + y2 + 22 dy dx 0 33 1x2 + y2 + 22 dx dy 0 4 2x 9. y. Consider the integral in the Example 12.3y + 52 dx 13ex . or intuitively from its interpretation as a volume. EXERCISE SET 6.x2 . in particular the linearity properties. and Finance. 0 6. we could easily define a triple integral as an iterated integral. 4x22. 5. kf1x. 4.x2 . 0 33 x 1x2 + 3y2 + 22 dy dx Applied Calculus for Business. Copyright © 2007 by Pearson Education.558 * ** Section 6. L 1 xy 21 . L L L L L 12x2 . Wang. y2 dA . Walter O. Gordon.3y + 52 dy xy 21 . we need only enter * ¿ ¿ 11 1 1 1 y *19x 2 + y 22.y2 dy 8. y2 dA = k f1x. Economics.3xy3 + y2 + 32 dy 12x2 . and April Allen Materowski. g1x. 2. y22 dA = f1x. 0. by Warren B. 13ex . y2 dA (7) RL L L RL and 1f1x. .y2 dx In Exercises 7 10 evaluate the given integral. y2 dA (8) RL L L RL L RL Calculator Tips The validity of these properties follows from the Riemann sum representation of the double integral. Inc. the integration symbol is above the number 7).6 In Exercises 1 6 evaluate the given integral 1.2ey + 2x . Double integrals may be easily evaluated on the TI 89 as an iterated integral. see Exercise 60.6 Double Integrals 2 4x Volume = 2 RL L y 29x + y dA = 2 2 0 2 33 0 y 29x2 + y2 dy dx = 0 y = 4x 1 98 392 a 19x2 + y223/2 ` b dx = x3 dx = 33 3 y=0 3 3 0 and Average Value of y 29x + y over R = 2 2 392 3 8 = 49 3 The double integral has many properties in common with the ordinary single integral that are useful in evaluating the double integral.3xy3 + y2 + 32 dx 7. so for example. y x . (a) f1x. Walter O. 17 Applied Calculus for Business. (b) f1x. 15 R y = x2 16. evaluate RL L f1x. (b) f1x. (b) f1x. Wang. y2 = 2x2 + 3y2 + 4 1x y . 13 Figure Ex. (a) f1x. y = x2 11. y2 = x2 . y2 = 1. (b) f1x. 0) Figure Ex. (a) f1x. y2 dA for R and f as given. 11 12. Copyright © 2007 by Pearson Education.x2 R Figure Ex. y2 = 1. (a) f1x. y2 = 1. y2 = 2x2y y=8 Figure Ex.6 3 y2 Double Integrals * ** 559 14. . (b) f1x. y2 = 4x2y y = 3x 2 y = 4 . (a) f1x. Gordon. (b) f1x. Economics. Inc. by Warren B. 2) R (4. and Finance.2x + 3y + 12 dx dy 2 2 10. 16 17. y2 = y R y = 2x2 R Figure Ex. y2 = 1.y2 Figure Ex.Section 6. 14 15. (a) f1x. y2 = 1. y2 = 1. y2 = 2x2y3 R (0. 2) (4. y2 = x. y2 = xy + 2x2 + 3 y=x y = 6x y = 4-x R Figure Ex. 2 33 2y In Exercises 11 18. and April Allen Materowski. (b) f1x. (a) f1x. Published by Pearson Learning Solutions. 12 13. y2 = h. f1x. determine Pr(E) if E is the region x2 + y2 = 1/4.560 * ** Section 6. f1x. R: 3x + 4y = 24. and April Allen Materowski. y = 4x and x = 3. 0 33 2 f1x. 0 3 3 2 y x f1x. Wang. y2 and below by the region R in the x-y plane. reversing the order of integration. Copyright © 2007 by Pearson Education. 0 3 3 f1x. . y2 = 4xy + 2x + 1. y2 = 6x2y. 43. 50. 2 18. y = 4x R y=x 32. R is the region bounded by y = 2x.x2 in the first quadrant. y2 dx dy y 2 4-y 25. f1x. R: rectangle determined by 1 39. y2 dx dy 2 4 .y2. y = x and y = 2. y2 = x. 0 33 2 f1x. Sketch the region R determined by this integral. 36.2. y2 = x. 19. (b) f1x. 38. f1x. 2 1. y2 dy dx 55. y = 2x 34. (b) Determine Pr(E) if E is the subset of R bounded by y = 6x. for the given function f. R is the region bounded by y = 2x2 and y = 2 1x. 0 33 0 f1x. y2 = 1/p.4 RL L f1x. f1x. 0 3 3 y 23 1 3x2 f1x. R is the region in Exercise 15. f1x. R is the region bounded by y = 0. find the volume of the solid bounded above by the surface z = f1x. y2 = 3x2 + 2y. R: rectangle determined by 0 40. 46. determine k. y2 = 1/16. 0 y 2. x = 1 and x = 2. (a) If f1x. Published by Pearson Learning Solutions. y2 = 12 . evaluate the double integral cated region R. y = 4x and x = 2. and Finance.y 29. f1x. 42. 35. 0 3 3y 6 f1x. y2 dy dx + 2 3 3 0 f1x. y2 = 2xy. y = 2x and x = 1. 0 y 4. R: y = x.3 2 x . f1x. y2 dx dy 0 1 24 . x x 3.2y. by Warren B. Let R be the square 0 x 4. In Exercises 50 . (Do you need to integrate in this exercise?). Do not integrate to determine the volume. 0 x2 36 1y 33 f1x. f1x. x = 0. If E is the square. y2 dy dx 20. f1x. y2 dy dx 48. R: x2 + y2 = a2. y2 = 3x2y2. f1x. 51. 8 27. R is the region bounded by y = 0. 54. f1x. 0 y y 5. Let R be the circle x2 + y2 = 1 and the joint pdf over R is f1x. y2 = y. determine Pr(E). Redo Example 11(b). 2. f1x. R is the region bounded by x = 0. y2 = 1. assume a < b and c < d. 47. 0 x 2. 33. suppose E was the region bounded by the quarter circle y = 21 . f1x. 0 3 3 2 2x 8 2 f1x. (a) f1x. y) over the region R. y2 = 3x2 . 41. recognize the required area. 3 4 1 . y2 dx dy 1 1x 23. -1 2 2 3 33 f1x.3y2. y2 dy dx x 1 1y 24. f1x. R: y = 2x.4 and y = 2x . Given the integral 33 a c dy dx. R is the region in Exercise 17. from x = 0 to x = 1. 44. y2 = kx2y2 is a joint pdf over the region R in Exercise 12. y2 = x2exy. and without doing any integration. f1x. y = x2. y2 = 2x + 3y . Gordon. y2 = In Exercises 38 43. determine Pr(E). y2 = 2x + 3y . 45. y2 = 2x + 3y + 2.54. R is the region in Exercise 12. y2 = kxy is a joint pdf over the region R in Exercise 11.x2 b d 30. determine the average value of the function f(x. Figure Ex. R is the region bounded by y = 6x2 and y = x. -1 2 6 6x 33 f1x. f1x. R is the region in Exercise 16. Economics. 0 33 0 2 f1x. determine k. R is the region in Exercise 11. y2 dy dx + 1 33 0 f1x. 18 In Exercises 19 30 determine the region R determined by the given double integral. y2 dA over the indi- 31. 53. y2 = 6.6 Double Integrals In Exercises 31 37. (a) If f1x. y2 = 4 . f1x. determine its value. y2 dy dx 1 2y 28. Walter O. x = 0 and y = 0. y = x. In Exercise 44. y2 = 4x2y. 1 + y2 37. Applied Calculus for Business.1. y2 dx dy 21. (b) Determine Pr(E) if E is the subset of R defined by 2 x 3 and 1 y 2. 49. 52. Inc. and y = 2. y2 dx dy 4 4-x 26.x2 . R is the region bounded by y = x2 . and the joint pdf over R is f1x. y2 dy dx 22. z2. y2 = 4x2/3y1/3. (b) Determine the minimum cost. (b) A dart board that is one foot in diameA(R) ter has a one inch bull s-eye in its center. Published by Pearson Learning Solutions. y. Draw the indifference curve for the utility function u1x. Evaluate Exercise 9 by reversing the order of itnegraiton. suppose the price for each item of x units of Product 1 is $6 and for each of y units of Product 2 is $2. z2. z2 = 2x y z . Copyright © 2007 by Pearson Education. z2 = 4x2 + y2 + z2 if 2x + 3y + 4z = 104.2y2 + 42 dy dx 3 3 1 2x 19. determine the probability it lands within the bull s-eye. 18. and y is the number of units of capital. Economics. 0x 2 f 0 4. Draw the indifference curve for the utility function u1x. y. z2. determine (a) f12. x y 3 2 5 Extrema Critical Points Second Partial Derivative Test Method of Lagrange Critical Points Second Partial Derivative Test Demand Equations Complementary and Substitute Products Joint Revenue. . and the consumer has at most $40 to spend. (b) lim h:0 h 2. . y.5p2 + 12. y2 = . 17. 8. y. 57. The demand equations for two related products are defined by x = . 1) by the plane y = 1. and the production level of 80. A(E) . (b) Prove it is an extremum. and y = 2p1 . what is his best buy? 16. 0y 5.2x2 . y2 = 8xy2. z2. by Warren B. Given the utility function u1x. y. each unit of capital $3. . Gordon. complementary or neither. 7.y2 where it is cut through the point (2.f12. where R is the region bounded by the x-axis. Cost and Profit Marginal Rate of Substitution 11. 6. Find three positive numbers whose product is to be as large as possible if the sum of the first. and April Allen Materowski. Determine the equation of the tangent line to the surface defined by z = x2 . f12.f12. f1x.2x3y2 + 4x2y2 + 9. Find the critical points of f1x. 12 . 59. 12.2. Assume the Cobb-Douglas production function f1x. . Using a double integral.Contours Cobb-Douglas Production Function Utility Functions and Indifference Curves Higher Order Partial Derivatives 1. (d) fzyx1x.3z) dz dy dx the line x * 4 and the line y * 8x. 12. f1x. Evaluate the double integral 1 dxdy. 3 4 .Chapter Review 1 2x 2x + 3y * ** 561 56. f12 + h. y2 = 500x5 y5 . Classify the critical points of the function defined by 8 3 f1x.y .x2 2 3 13x2 . .12. If each unit of labor costs $64. (a) Show that A(R) 58.12 . y2 = 3x2 . y2 = 10x3y2. Inc.2. z2 = 2x2y3z2 + ln1x2 + y2 + z22. 14. Given the integral 3 3 R dA. determine (a) f12. find the area of the region R bounded by the curves 2 x .2 + k. (c) fzxy1x. y) is said to be a uniform pdf over R if f(x. Wang. f(x. 15. y2 = 3x + 2y. y2 = 3x3ex . y2 = ln 2x3 + y3 find 2 . A dart is thrown randomly at the target. Evaluate the double integral 0 x 1 1 3 3 e dydx. 60. why?) 2 2 y2 61. 13. y. y2 = + + 9xy. 9. (b) fxzy1x. How many different order of integratins does a triple integral have? 62. Given f1x. (a) Classify the critical point of the function defined by f1x. Evaluate Applied Calculus for Business.7p1 + 3p2 + 12. Pr(E) = 1 . y. S the region R and determine the value of this integral without doing any integration. determine (a) fxyz1x. and Finance.4y = 0 and 4y + x = 8. (There will be two integrals. 12 .000 units is to be achieved.5y2 + 2xy + 6x + 12y. twice the second and 9 times the third is 648. (a) determine the number of units of labor and capital that minimize the cost. f1x. 10. Draw the indifference curve for the utility function u1x. . find 2 . 1 + x2 33 0 y CHAPTER REVIEW Key Ideas Functions of Several Variables Difference Quotients Three-Dimensional Coordinate System Surfaces Partial Derivative Visualization of the Partial Derivative Level Curves . y) = for a uniform pdf. (b) lim k:0 k 0 2f 2 2 3. where x is the number of units of labor. . y2 = x3 + 12xy + 4y2. Classify the critical points of the function defined by f1x. Walter O. Determine if the products are substitutes. Given f1x. Evaluate the triple itnegral 33 3 0 x 0 (x + 2y .12 . and April Allen Materowski. Evaluate the integral 0 33 4x e-y dy dx. Wang. (b) if E is the region bounded by x = 0.4y and below by the region R in the x-y plane bounded by x = 0. y2 = kxy is a joint pdf over the region R bounded by x = 0. (a) Determine k. Gordon. y = 0 and x + 2y = 4. Walter O. 20.562 * ** Chapter Review 22. Copyright © 2007 by Pearson Education. 2 8 21. y2 = x2 + y2 over the region R defined in Exercise 20. f1x. y = 0 and x2 + y2 = 4. by Warren B. 2 Applied Calculus for Business. Find the average value of f1x. and Finance. find Pr(E). . 23. y2 = 8 .2x . Published by Pearson Learning Solutions. y = 0 and x2 + y2 = 1. Economics. Inc. Find the volume of the region bounded above by z = f1x. therefore an understanding of their properties is useful. Copyright © 2007 by Pearson Education. . It then examines the relationship between matrices and linear systems of equations. by Warren B. Wang. Applied Calculus for Business. and April Allen Materowski. and Finance. Economics. Published by Pearson Learning Solutions. Gordon. Walter O.A Matrices and Linear Systems This Appendix introduces the elementary properties of matrices. Matrices are an important tool in many different disciplines. Inc. D is a column vector. C is an example of a row vector. Thus. Walter O. 2. In particular. In most programming languages they are called arrays. Economics. In a square matrix. DEFINITION 1 A matrix is a rectangular array of numbers. Gordon. and April Allen Materowski. -3 0 1 C = 11. they may be used to represent relations. any matrix consisting of one column is called a column vector. the 3-1 entry of D is 0 and the 1-1 entry of E is 5. or systems of equations. that is. the 3-2 entry of B is 34. In other words. row * column. The numbers in the matrix are called entries or elements.2 5 4 1 4 *2 2 1 3 b 2 5 *4 .1 Basic Operations » » » » » Matrices Addition of Matrices Scalar Multiplication Zero Matrix Calculator Tips Matrices appear in a wide variety of applications. Some additional examples are: 1 0 B = 2 8 3 D = +2* 0 2 -1 34 4 3 11 0 0. C is a 1 * 4 matrix. 2-2. Inc. Note that the dimension is given in the form. The commas between the entries in C were put in for readability and could have been omitted.1 Basic Operations A. and Finance. 42 E = a B is a 4 * 5 matrix. Any matrix with the same number of rows as columns is called a square matrix. Copyright © 2007 by Pearson Education. D is a 3 * 1 matrix and E is a 2 * 2 matrix. The matrix E has the same number of rows and columns. the 2-3 entry of B is 11. Thus. 3. the main diagonal consists of those elements whose row and column positions are the same. etc. A matrix is used whenever there is a need to store information. by Warren B. entries 1-1. We denote matrices by upper case letters. When identifying a particular entry in a matrix we always give its row location before its column location.) The dimension or size of this matrix is 3 * 4. the main di- Applied Calculus for Business. . Published by Pearson Learning Solutions. Wang. 1 A = +3 2 Matrices 2 0 -4 -1 8 1 0 5* 9 is a matrix with three rows and four columns. (The rows are horizontal while the columns are vertical. 3-3. Any matrix consisting of one row is called a row vector. A matrix may have any number of rows and any number of columns. For example. Similarly. We begin by defining a matrix.564 * ** Appendix A. the 1-2 entry of C is 2. collections of vertices and edges (called graphs). Observe that each row of A can be thought of as a row vector and each column of A may be considered a column vector. Each entry in the sum is obtained by adding the corresponding entries of the matrices summed. a 2n2 and the third column of A as the column vector a13 a . Copyright © 2007 by Pearson Education. we need to define certain basic terms. Wang. 3. More generally. Thus (1) and (2) mean exactly the same thing. the second row of A may be written as the row vector 1a 21. We start with equality. 2. 1 5 . is indicated on the outside. For example. Applied Calculus for Business. we can abbreviate the matrix A by writing A = 1a ij2m * n (2) where aij represents any arbitrary element in A and the dimension of A. by Warren B. and 2. that is. 2. Gordon. That is. m * n.j). That is. 8. DEFINITION 2 1. 7 3 2 2 1 3 3 1 8 1 4 0 + 9 2 the diagonal entries are 1. it has m rows and n columns. where cij = a ij + bij for each pair (i. and scalar multiplication. in the following square matrix. Any entry in A has a subscript indicating its position in the matrix. a 22. Inc. and April Allen Materowski. aij is the entry in A that is in the i-th row and j-th column. . 23 + o a m3 In order to work with matrices. and Finance. Two matrices of the same size are equal if and only if their corresponding entries are equal. addition. their sum A + B = C = 1cij2m * n. if A = 1a ij2m * n and B = 1bij2m * n then A = B if and only if a ij = bij for each pair (i. In general. In fact.1 Basic Operations * ** 565 agonal of a square matrix begins in the upper left hand corner and proceeds down to the lower right hand corner of the matrix. a 23.Appendix A. Two matrices are said to have the same size if and only if they have the same dimension. Published by Pearson Learning Solutions. if A = 1aij2m * n and B = 1bij2m * n. Two matrices of the same size may be added. Economics. Á .j). For example. we may represent an arbitrary matrix by a 11 a 21 A = * a 31 o am1 a 12 a 22 a 32 o am2 Á Á Á o Á a 1n a 2n a 3n o a mn (1) A as given in (1) is a m * n matrix. Walter O. Thus a 23 is the entry in the second row third column. Their sum is a matrix with the same dimension. Thus x . and Finance.4 3 . 3 To see that the definition of matrix addition is a natural one. Published by Pearson Learning Solutions. 5. and April Allen Materowski. 3 3A = + 0 9 -6 27 15 9 39 * . and B represents its profits on the same items during February. in thousands of dollars. by Warren B. These examples will also show that the definitions for equality and multiplication by a scalar are the natural choices. That is. Let us illustrate the definitions with examples. Gordon. Example 1 Addition of Matrices Given that A = B. and y = 1. if k is any scalar and A = 1aij2m * n. and the 2-3 entry indicates a net profit of thirteen thousand dollars for that product. 3 Applied Calculus for Business. the 1-1 entry indicates a net loss of four thousand dollars for the product represented by that entry. A + B = B + A and A + 1B + C2 = 1A + B2 + C. Walter O. you should convince yourself that when addition is defined. Observe that for addition to be performed the matrices must be the same size. consider the following. (Verify this!) Example 2 Determine the sum of the two matrices A and B.y 7 5 9 b and B = a x + y 7 5 b . if the entry is negative) that a company makes on the sale of each of 9 products during January. However. (loss. Z.3 -4 13 + 0 * = + 2 1 + 3 13 2 21 1 0 13 * 4 -2 9 5 3 -5 13 * and B = + 2 1 10 4 12 -4 -3 0*. Then the company s combined profits for the two months are given by C.566 Appendix A. and x + y = 11.A. Scalar Multiplication Example 3 Compute 3A. To multiply a matrix by a number. Thus. Their sum 1 . where A = a x . That is. Since the two matrices are equal we know that the corresponding entries are equal. 11 Solution. To multiply A by 3 means that we multiply each entry in A by 3. Thus.5 C = +0 + 2 3 + 10 -2 + 4 9 + 12 5 . it satisfies the commutative and associative laws. Solution. Wang.y = 9. Economics.12A is denoted by the symbol . Inc. a zero matrix is a matrix all of whose entries are zero. means to multiply each entry in the matrix by the number. then kA = 1kaij2m * n. The matrix 1 . in the given matrices. where 1 A = +0 3 Solution. .1 Basic Operations 4. These two linear equations may be solved simultaneously to yield x = 10. Suppose the matrix A represents the profits. called a scalar. Copyright © 2007 by Pearson Education. where A is given in Example 2. determine x and y. If two matrices do not have the same dimension then addition is not defined. Walter O. since to add Z to A means that A and Z have the same dimension. These observations mean that we can manipulate matrices like numbers in many algebraic operations. 3 4 12 -4 6 3 -9 -3 1 0* . type [-5. Wang. Find the matrix X. or finally.26 * . -1.B.+0 3 3 -6 .B. in this example. we have A . Gordon. if A represents the profits of a company. . and accessed by first pressing the 2nd function key. Inc. Also note that with this definition.13 * = + 4 2 14 12 6 .18 . A + A + A = 3A. Published by Pearson Learning Solutions. (This means that all the entries in Z are 0. A zero matrix Z has the special property that if A is any matrix with the same dimension as Z. by Warren B. (The last entry in the last row is followed by the right bracket. A = C . and we assume that the profits tripled. Figure 1: A Matrix in Memory Applied Calculus for Business. For example. 2.12 . then adding . we proceed as follows: on the entry line in the HOME screen. if A + B = C.A = A + 1. respectively.A2 = Z. The elements on each row of the matrix are separated by commas. then the new profits are represented by 3A. and Finance.B = C . Suppose we wanted to place the matrix B from Example 2 into the calculator s memory and store the matrix as b.) Note that there are an infinite number of zero matrices. or A + Z = C . except the last entry in the row where a semicolon (located above the 9 in orange) is used to indicate the start of the next row. 4 Calculator Tips It is very straightforward to define a matrix and perform basic matrix operations with the TI 89 calculator. Given any matrix A.12 .B to each side of the matrix equation yields A + B .Appendix A. 12.B. as b. Economics. 3] press STO b Note the left and right bracket symbols are in orange above comma and division keys.13 * 2 -2 3 9 13 * = 5 1 Solution. then A + Z = A.) The STO : key tells the calculator to store the matrix in its memory. 0. as one would expect. -3. Example 4 1 Given that *2 X + + 0 3 1 Zero Matrix -2 9 5 3 -5 13 * = + 2 1 10 4 12 -4 -3 0 * . Copyright © 2007 by Pearson Education. we have 1*2 X = + 2 10 4 12 -4 -3 1 0* . 4. -5 As indicated above. and April Allen Materowski. 10.+0 3 3 -2 9 5 3 -6 13 * = + 2 1 7 1 -5 X = 2 + *2 10 Multiplying each side of the equation by 2 gives -6 X = 2 + 2 7 6 3 -9 -6 .1 Basic Operations 567 Again. Solve for x. S Ms. Walter O. 7. then we may have the calculator compute 2A + 3B by just entering these expressions and adding. January Sales Houses Lots Mr. Each sells houses and lots. subtraction and scalar multiplication of matrices are performed similarly to the way they are done with real numbers on the calculator. Addition. and their sum was used to represent the combined profit. 6. Gordon. Economics. and Ms.2 . (c) Write the third and fourth columns as column vectors and determine their sum.11 54 21 210 9 0. A + B = a 1 2 3 2 -1 b. A real estate brokerage employs three salespersons: Mr. z. press ENTER again. Redo exercise 8 for rij = 3i + 2j.000 $ 3. (d) Show that the combined commission matrix Tc is the sum of the two commission matrices given in (c) and that Tc = 0. and w: a 1 0 -x 1 b + 2a 2 2 -1 1 b .000 $10. 3. Published by Pearson Learning Solutions.67 .500 $9. (ii) 2-1 entry. H $70. (d) What is the dimension of A? 3 47 .9 0.2B (c) 1*2 B.06T.9 1. 12. Ms. (iv) 1-4 entry.4B = a 12.2 9 *4 1 5 11 43 57 (a) What is its dimension? (b) Determine the sum of the 1-2 and 4-3 entries. For example. Find matrices A and B so that the following equations are satisfied simultaneously. Wang.000 $8.7 3. . for this example.000 $12. 32 + y1 . 2A + 3B = a 3A . assume we have the two matrices A and B stored in the memory of the calculator.06J + 0. 10. by Warren B. -5 1 0 if i = j . Let R = 1rij24 * 2 with rij = i . Gold. (e) Show.000 $45. The ABC and DEF hardware companies report their annual profits for the following items in thousands of dollars: 5.06F. Silver. Let I be the n * n matrix defined by I = 1dij2n * n where dij = e find I if (a) n = 2.0. (iii) 1-3 entry. Determine the matrices representing the commissions for each of the two months. that multiplication of a sum of matrices by a scalar is distributive. if i Z j (a) Represent each company s profits as a separate 1 * 6 matrix.000 $30. pressing ENTER and then to confirm.B = a 2 1 0 9 0 b. 4.j. that is. (c) Write the fourth column as a column vector. where k is any scalar.1 1.568 * ** Appendix A. 12 = 1 . y. -7 2A . Let A and B be given in exercises 1 and 2.0. both defined as in Example 2. how would the answer differ from the result in (b)? 11. 2.2 8. (b) Write the first row as a row vector.1. where A and B are defined in Example 2 We can now quickly perform operations on this matrix from the entry line on the HOME screen.31 5 12 . (c) Suppose that each broker receives a 6% commission on every sale. representing the total sales for the two months.5.4 . Call these matrices Jc and Fc. Deletion of all one letter variables is accomplished by pressing F6 (2nd F1) and choosing clear a-z. Their sales records for January and February are shown below.000 $80. Copyright © 2007 by Pearson Education. 13. 0. (v) 3-1 entry. 1 4 b.1 0. k1A + B2 = kA + kB. Hernandez. Inc. (b) .1 1 .000 $40. determine the combined profits of the two companies.2 2. see Figure 2. Determine (a) 3A.3 45 3 1. We note that we often give one letter names to matrices so we can erase them quickly from the calculator s memory (as they take up valuable memory space). EXERCISE SET A. 9.000 (a) Represent the sales in January and the sales in February by matrices J and F respectively. (b) Using the two 1 * 6 matrices.3a 1 y -1 z b = a 6 11 0 b.8 DEF 3. Applied Calculus for Business. and Finance.061J + F2 = 0. w ABC Nuts Bolts Hammers Saws Washers Nails 5. Show that for any two matrices of the same size. (c) Suppose a 6 * 1 matrix was used for each company s profits. Given the matrix A = + . (b) n = 3. (b) Determine the matrix T. and April Allen Materowski. Given the matrix B = 1 0 37 .2 0 0.5 8* 9 1 81 1*2 (a) Determine the (i) 1-2 entry. Find R. (c) n = 4 (d) describe I for any n.000 $7.2. -2 1 b. Find matrices A and B so that the following equations are satisfied simultaneously. G Ms.1 Basic Operations Figure 2: 2A + 3B. Determine x and y if x11.5 2.000 February Sales Houses Lots $35. Published by Pearson Learning Solutions. 4 It may first appear strange that the product of two vectors should be a number. Thus. Once vector multiplication is understood. Economics. Many students wonder why the product of two matrices is not simply defined to be the matrix obtained by multiplying the corresponding entries of the two matrices in a manner analogous to finding the sum of two matrices. 2. 5.VECTOR MULTIPLICATION To multiply a row vector positioned on the left with the same number of entries as a column vector on its right. To start with. but as various applications are discussed throughout the text. 72 represent the poundage ordered of apples. Copyright © 2007 by Pearson Education. the product of two matrices is easily obtained. We shall see that the only restriction imposed by our definition is that the number of columns of the matrix on the left must equal the number of rows of the matrix on the right. The product is obtained by multiplying the corresponding entries and summing. the following example may indicate why the definition is a natural one. Let r = 14. . it would limit the definition to matrices of the same size. In fact. Solution. The definition will at first seem unusual. by Warren B. It is important to note that in accordance with Definition 1. and April Allen Materowski. Wang. potatoes and tomatoes respectively. there are mathematical systems in which such a definition is used but it has very limited application. 6 Solution. you will better understand why the definition is chosen. of tomatoes at 78 cents per pound.Appendix A. of apples at 89 cents per pound.2 Matrix Multiplication » » » » » » » Vector Multiplication Matrix Multiplication Identity Matrix Inverse Matrix Matrix Form for a Linear System of Equations Transpose of a Matrix Calculator Tips We are almost ready to define the product of two matrices. 32 # + 5 * . of potatoes at 32 cents per pound and 7 lb. the product of a row vector with a column vector is a number. Represent the poundage order by a row vector. Inc. Example 1 Compute the vector product of 11.2 Matrix Multiplication 569 A. and the corresponding prices by the column vector Applied Calculus for Business. 5 lb. the product is 112142 + 122152 + 132162 = 32. the prices as a column vector and use vector multiplication of the two vectors to determine the total cost. the corresponding entries are each multiplied together and the individual products are then summed. Vector Multiplication DEFINITION 1 . However. Example 2 Suppose an order of produce reads 4 lb. Gordon. and Finance. This requirement leads directly to the definition of vector multiplication which follows. Walter O. The definition indicates that the matrix on the left must have the same number of columns as the matrix on the right has rows. We shall use.782 = $10. Economics. 12 Solution. and if it can. and the same number of columns as the matrix on the right (seven). the vector multiplication of a row vector (on the left) by a column vector (on the right). that matrix multiplication is not commutative.892 + 1521. Gordon. Thus the product is a 2 * 7 matrix. Matrix Multiplication We are now ready to define multiplication of two matrices. Thus C has the form C = a 11 b . . as you shall see.2 Matrix Multiplication 0. is not defined.MATRIX MULTIPLICATION Let A = 1aij2M * K and B = 1bij2K * N. Since the number of columns in the matrix on the left (five) is the same as the number of rows in the matrix on the right.322 + 1721. It will have the same number of rows as the matrix on the left (two). and the same number of columns as the matrix on the right. This will also imply. in our definition. by Warren B. find the product C = AB. The ij entry of the vector product is the product of the i-th row of the matrix on the left with the j-th column of the matrix on the right. Observe that the product of the matrices in the reverse order.62. the 5 * 7 matrix on the left and the 2 * 5 matrix on the right. This will mean that we can only multiply matrices in which the number of columns in the matrix on the left is the same as the number of rows in the matrix on the right. c21 c22 Applied Calculus for Business.32 * 0.89 c = + 0. We need only determine the four entries. the product is defined. Inc.78 then the total price for the order is rc = 1421. C will have 2 rows (the same as A) and two columns (the same c c12 as B). The product matrix has the same number of rows as the matrix on the left.570 Appendix A. and April Allen Materowski. and Finance. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education. what is the dimension of the product? Solution. Wang. Example 3 Can a 2 * 5 matrix be multiplied on the right by a 5 * 7 matrix. Since the number of columns in A is the same as the number of rows in B. Example 4 Let A = a 1 4 2 5 7 3 b and B = + 9 6 11 8 10 * . DEFINITION 3 . matrix multiplication is defined. Then the product AB = C = 1cij2M * N where cij = the vector product of the i-th row of A with the j-th column of B. Walter O. Notice that the multiplication of the two vectors corresponds exactly to the way the total price would ordinarily be computed. Economics. thus we write c11 C = + c21 c31 c12 c22 * . and. In general matrix multiplication is not commutative. in the above example. 1 5 Solution.. (See. 2. even when both products have the same dimension. c22 = 14. Gordon. and Finance. c11 = 11. Published by Pearson Learning Solutions.Appendix A. we find the product BA.) Example 5 Compute the product 2 + -1 1 1 3 -1 3 0 2 1 4 2 2* # 1 -2 3 -1 0 . Wang. C = a 58 139 64 b. The matrix on the left is 3 * 4 and the one on the right is 4 * 2. Inc. which is not the matrix we obtained from the product AB. The product is defined and has dimension 3 * 2. c32 We will compute one entry from each row and leave the remaining as exercises for you. by Warren B. 154 It is important that C be labeled correctly. Copyright © 2007 by Pearson Education. Exercises 3 and 4. for example. If. it will be a 3 * 3 matrix. Applied Calculus for Business. 2. 5. that is AB Z BA. Walter O.2 Matrix Multiplication 571 We have. We find (after computing the remaining three entries). c21 = 14. 7 32 # + 9 * = 112172 + 122192 + 1321112 = 58 11 8 32 # + 10 * = 112182 + 1221102 + 1321122 = 64 12 7 62 # + 9 * = 142172 + 152192 + 1621112 = 139 11 8 62 # + 10 * = 142182 + 1521102 + 1621122 = 154 12 c12 = 11. 5. as the subscripts on each entry give the row and column from A and B (respectively) which are to be multiplied together. . and April Allen Materowski. Given the matrix A M * N does there exist a matrix which when multiplied by A leaves A unchanged? Rephrasing the question. 3.12102 + 122112 + 1 . that is. if we multiply A on the left. (why?) Additionally. 3.12 + 1 . Economics. such that A-1A = AA-1 = I? Applied Calculus for Business. IU * VA M * N = A M * N. 1 1 I3 = + 0 0 0 1 0 0 0*. Gordon. thus we have S = N.1. If the matrix is on the right of A. aa -1 = a -1a = 1. Inc. we would have. -9 Identity Matrix Recall that for any non-zero real number a. = 1 .22152 = . In order for multiplication to be defined. but this matrix is not always the same. 1 I2 = a 0 0 b.1. and April Allen Materowski. we write IN instead of IN * N. we shall omit the subscript N.2 Matrix Multiplication c12 = 12. . c32 -1 0 = 11. Therefore I is an N * N (square) matrix. we must have R = N.) Thus. It should be clear what IN looks like for any N. The question that still remains is what are the entries in I? It turns out that I has a particularly simple form. . Having the notion of an identity naturally poses the following question: does a matrix have a reciprocal. that is.22 # . and I is a M * M square matrix. (The proof of this may be found in the exercises. we find that U = V = M. can we divide by a matrix? We know that if a is a nonzero number the product of a with its reciprocal 11/a = a -12 is 1. A M * NIR * S = A M * N. The matrix IN is called the identity matrix. Wang.. the number of columns in the product comes from the number of columns in the matrix on the right. All its diagonal entries are 1 and its non-diagonal entries are 0. 1a = a1 = a.12112 + 132122 + 102112 + 122132 = 11 1 3 c21 = 1 . When there is no ambiguity. Does there exist a matrix. respectively. Similarly. 0. similarly for IM. = 1121 . As above. is there a matrix that behaves like the number one? The answer is yes. and Finance.572 Appendix A. = 1221 .. now call the matrix IU * V. Copyright © 2007 by Pearson Education. Walter O. then calling it IR * S. .9 1 5 19 C = + 11 -5 21 11 * . by Warren B. Published by Pearson Learning Solutions. You should try some simple examples to convince yourself that this is correct. 1 Inverse Matrix where I2 and I3 are the 2 * 2 and 3 * 3 identity matrices. Since I is a square matrix. 2.. 1. -1 0 42 # . For each value of N there is a different identity.12 + 112102 + 132112 + 142152 = 21 1 5 1 2 22 # . and we have. which could be denoted by A-1. It is an N * N matrix with ones along the main diagonal and zeros elsewhere. there is an entire class of matrices that is analogous to the number zero in this respect. only zero has no reciprocal. if A is a square matrix and A-1 is a left inverse. However. as required. If a matrix A has an inverse it is said to be invertible or non-singular.Appendix A. but will assume familiarity with them. It is understood that all multiplications and additions indicated in Table 1 are defined. A singular matrix is one that has no inverse.) For the moment we shall not be concerned about the process used to find an inverse. among the real numbers. Example 7 Perform the indicated matrix multiplications to rewrite the given matrix equation without matrices. Table 1: Matrix Properties Properties of Matrices ADDITION A + B = B + A A + 1B + C2 = 1A + B2 + C A + Z = A MULTIPLICATION AB Z BA 1AB2C = A1BC2 AI = A A1B + C2 = AB + AC 1B + C2D = BD + CD A1kB2 = kAB INVERSE AND IDENTITY AI = IA = A A-1A = AA-1 = I Commutativity Associativity Zero matrix exists Non-Commutativity Associativity Identity matrix exists Distributivity over addition Associativity over scalar multiplication The next example demonstrates the relationship between matrix multiplication and linear systems of equations. Table 1 is a summary of the various properties satisfied by matrices. . Economics. then it is also a right inverse! (See Exercise 33 for an example of a partial inverse. A-1 does not always exist. by Warren B. 1 -1 2 -1 b is the inverse of A = a 2 1 1 b. and if AA-1 = I. Wang. This relationship will be examined in greater detail in succeeding sections. it is not necessary to do both calculations in order to verify that A-1 is the inverse of A. It may also be shown that if A is a square matrix. Published by Pearson Learning Solutions. Thus. As you know. Copyright © 2007 by Pearson Education. Applied Calculus for Business. and Finance. However. it is called the inverse. Such matrices are called singular. We will leave their verification to the exercises. a 1 -1 -1 # 2 b a 1 2 1 c b = a 11 1 c21 c12 b c22 Calculating we find that c11 = c22 = 1 and c12 = c21 = 0. A must be a square matrix (why?). That is. Inc. 1 We need only show that the product (in either order) of these matrices is I. Instead let us see how it may be used if it is known. Example 6 Show that A-1 = a Solution. Walter O. In order for both of these multiplications to be defined. Thus A-1A = I.2 Matrix Multiplication 573 When such a matrix exists. and April Allen Materowski. then A-1A = I. Gordon. even for a square matrix. and Finance. Published by Pearson Learning Solutions. 3x + 5y = 6 2x + 7y = 9. The first matrix on the left is a 2 * 2 matrix and the column vector on the right is a 2 * 1 matrix. Solution. then the column vector of unknowns and finally the vector of numbers on the right hand side of the equations. X is a column vector whose entries are the unknowns. and April Allen Materowski. can be rewritten as a very simple matrix equation. therefore their product is a 2 * 1 matrix. Wang.2 Matrix Multiplication a 3 2 5 x 6 b a b = a b. y Thus we have. their corresponding entries are equal and we have. We first write down coefficient matrix. 3 z -1 9 11 -1 Applied Calculus for Business. by Warren B. . 7 y 9 Solution. (and is therefore called the coefficient matrix). (This had to be the case. Economics. otherwise how could the product equal the 2 * 1 matrix on the right of the equal sign?) Thus. It will be just AX = B. and B is a column vector of the numbers to the right of the equal sign. 72a b = 2x + 7y. Example 8 Write the system of linear equations 3x + 4y . the product is a where x c11 = 13. where A is the matrix of the coefficients of the variables.574 Appendix A. Walter O. y and x c21 = 12. a 6 3x + 5y b = a b 9 2x + 7y c11 b c21 If two matrices are equal. 52a b = 3x + 5y. Inc. Matrix Form for a Linear System of Equations It is not hard to generalize from the previous example that any system of linear equations.7z = 2x + y + 5z = 2y + 3z = In the matrix form AX = B. Copyright © 2007 by Pearson Education. 3 +2 0 4 1 2 -7 x 9 5 * + y * = + 11 * . Gordon. Thus we see that the given matrix equation is equivalent to the above system of equations. However. by Warren B. 2 Solution. Wang. Copyright © 2007 by Pearson Education. By associativity. the unknowns in each equation must appear in the same relative positions: All x s in one column. exists. It is important that we line up the unknowns. we have X = A-1B. can we solve this equation for X? The answer is sometimes. -1 Example 9 Solve the system of equations 2x + y = 21 x + y = -4 using the fact that for A = a 2 1 1 1 b . where A. then X = A-1B.2 Matrix Multiplication 575 Note that the third equation has no x term so we consider it 0x. (where A. Remember how to solve the corresponding algebraic equation. thus 1A-12AX = 1A-12B. Gordon. Of course. Published by Pearson Learning Solutions. which in this example is Applied Calculus for Business. and Finance. all y s in another. The given system may be written in the form AX = a 21 b -4 x where A is as given and X = a b . A-1. Walter O. If AX = B. ax = b for x? We divide by a (multiply by a -1) to obtain the solution a -1b. That is. Suppose we are given the matrix equation AX = B. B and X are described above. Thus. and all z s in a third. .Appendix A. Inc. since there are some matrices (singular ones) that behave like the number zero. A-1 = a -1 1 -1 b. we have the following theorem. when the inverse does exist. multiply on the left by A . PROOF The proof follows immediately from the definition of the inverse and identity. a cannot be zero. according to Theorem 1. and April Allen Materowski. 1A-1A2X = A-1B or IX = A-1B or X = A-1B. B and X are described above) if the inverse of A. we cannot expect to always be able to solve our matrix equation for X. Economics. y Therefore. THEOREM 1 Given the matrix equation AX = B. It is this relationship that motivates the examination of linear systems in the next section. 9 Solution. and April Allen Materowski. that is. However. we mean to take the product of k A s. if A is invertible. when it exists. For example. 1 A = +3 4 t 5 7*. once the inverse of a coefficient matrix is known. the system of equations is easily solved by matrix multiplication. Economics. We know that not every linear system of equations has a unique solution. DEFINITION 3 The transpose of a matrix. suppose we name them A and B. Making the first row of A the first column of At and the second row of A the second column of At we have.) The TI 89 calculator can do matrix multiplication once the matrices are stored in its memory. Example 10 Transpose of a Matrix Determine the transpose of A = a 1 5 3 7 4 b. Notice that this definition implies that AkAm = AmAk. Walter O. Copyright © 2007 by Pearson Education. the graphs of these two equations are parallel lines. Clearly. A is a 2 * 3 matrix so At will be a 3 * 2 matrix. Inc. consider the equations 2x + y = 6 4x + 2y = 9 As you know. A3 = A2A = AAA and in general.29 As the preceding exercise clearly indicates. In fact. so that matrices which are powers of the same matrix must commute under multiplication. the lines have no points in common and this system has no solution. once the inverse is calculated. Ak + 1 = AkA. and Finance. Similarly. after examining a procedure for determining the solution to a linear system of equations. 9 Calculator Tips If A is a square matrix then we define A2 = AA. Published by Pearson Learning Solutions. we have yet to determine the amount of labor involved in finding the inverse matrix.576 Appendix A. denoted by At. k.2 Matrix Multiplication x 1 a b = a y -1 or x = 25 and y = . Hence. if A = 1aij2M * N then At = 1a ji2N * M. After all. which never intersect. the remaining labor is minimal. is the matrix obtained from A by interchanging its rows with its columns. To multiply the matrices given in Example 4. Recall that for a real number a. the coefficient matrix corresponding to this system cannot have an inverse and must be a singular matrix.29. There is one additional matrix that we define. Gordon. (Note that we can also define negative integral powers. it is convenient to define A0 = I. Wang. Therefore. by Warren B. A-n = 1A-12n = 1An2-1. Thus. we shall discover a procedure for finding the inverse of a matrix. we must ask ourselves if this is an efficient method. a0 = 1. To determine the transpose we need only interchange rows with columns. . when we raise a matrix to an integral power. 2 -4 . Applied Calculus for Business. It is clear that there is relationship between the existence of a unique solution to a linear system of equations and the existence of the inverse of its coefficient matrix. -1 21 25 ba b = a b. Verify that 1DE2F = D1EF2 8. a¿5 .2F2 11. 1 -1 -2 -1 2 -1 3 * show that A-1 = 1*2 + 7 2 5 0 -4 -2 1 -1*. Solve for x: 1x. BC2 13. C1B . Wang. ). 2. (c) Does DE = ED? 4.A2 7. Suppose the matrix A as given in Example 5 is stored in the calculator.5u . the calculator can quickly determine its inverse (if the matrix is not too large. If A = + 2 -5 2 5 . 7.. y and z. 1 -1 1 b 3 2 -1 B = -3 5 E = a 2 1 1 3 . (a) DE.3EF 15. then you need only type on the entry line of the home screen.55u .2z = 3 2x + 5y + z = . -1 1 A = 3 -4 D = a 0 -2 0 2 . If A = + 1 3 1 1 1 b show that K 2 .11 for u. (b) k = 4.2v . 12A . Let A = a 0.5x .2 (e) What do you think will happen as k becomes larger and larger? Applied Calculus for Business. Let K = a 1 24. Published by Pearson Learning Solutions. Walter O. 1 26. 8FE + 2CF 1 19. 3C2F .2 Matrix Multiplication * ** 577 and enter them into the calculator s memory. and April Allen Materowski. then we need only enter a*b and the product is given. 23. but that -2 2 2 b and B = a -2 -1 F = a 1.2a 4 -3 2 3 b f a b = 33 -1 x 4 b . 12 and H = . and Finance. v. show that 1AB2-1 = B-1A-1. Compute GH and HG where G = 1 . (a) Write the system of equations x . (b) Use the result of Exercise 26 to solve this system for x. 1 5 25. 5 3 2 20.4 b compute Ak for (a) k = 2.w = . (b) Use the previous exercise to solve this system for x. (a) CF. A1D + 3E . Inc. Copyright © 2007 by Pearson Education.2K = Z.10y + 9z = 12 in the form AX = B.10 -2 . . If a matrix is invertible.2 In Exercises 1 18 use the following matrices to perform the given operations. 29. (b) FC. (c) Does CF = FC? 5.6 0. 1A + B21F + 2C2 12.y + 2z = 0 in the form AX = B. 4CD . 3.3BC 14.Appendix A. by Warren B.8 0. and w. see Figure 1. y and z.2DE 18. -1 1 -1 2 -5 3 1 b . (a) AB + C. 0.3B2C 6. (b) 1A + B2C 2. 2A .y + 2z = 3 x . 22e a 22. Show that 1D + E21D + E2 Z D2 + 2DE + E 2. (a) 1AC2t (b) CtAt (c) Any observations? 17.1. (b) ED.55 1 * show that A-1 = + 23 9 -5 -2 1 0 . show that AB = 0. If A and B each have an inverse. -1 27.4DC 16. solve the system .2 . (c) k = 8. (c) Using the result of Exercise 26. 3C . 42 and H = + 3 * .1. Economics.12 5*. Given A = a BA Z 0. 28.12w = 7 23u + v + 5w = 3 . Figure 1: Illustrating Matrix Multiplication EXERCISE SET A. We shall see how to determine when the inverse exists and how to find it in the next section. 0 8 -1 b -2 1 C = a -3 -3 1 2 b 5 2 b 4 21. Gordon. Why doesn t it? 10. Verify that D1E + F2 = DE + DF 9.16 and press Enter to obtain A-1.2y + 3z = 5 3x . (a) A + BC. (b) AC + B 3. Compute GH and HG where G = 1 . when they are well defined. (a) Write the system of equations x + 2y . . we indicate the separation between the coefficient matrix and the right hand side by a vertical line. called the augmented matrix. which corresponds to the right hand side of the system. Although is not necessary. when it exists. Thus. 31. it is convenient to use subscripts and name them x1. we store all the important information in a matrix. if a system of equations is written in the matrix form AX = B. It will prove useful to examine systems of linear equations in some detail before we return to the question of finding the inverse of a matrix.1 -1 1 1 4 . call them B and C. Thus. the augmented matrix corresponding to (1) is a 11 a 21 am1 a 12 a 22 am2 a 13 a 23 o a m3 Á Á Á a 1n a 2n amn b1 4 b2 ¥ o bm § (2) The augmented matrix is a shorthand representation for the system. is unique. by Warren B. y and z. Hint: Refer to Exercise 13 in Exercise Set 1. Compute BAC in two different ways. When a system has two unknowns. but should be thought of as being there. where I is n * n. Each row of the augmented matrix represents the corresponding equation of the system. x4.578 * ** Appendix A. Economics. and equal signs have been omitted in the augmented matrix. and Finance. Published by Pearson Learning Solutions. Gordon. 32.3 Gauss-Jordan Reduction 2 2 -1 1 1 1 b and A-1 = 11/32 £ . For A = a but BA Z I3. x2. when it exists. when there are three. Show that the inverse matrix. Wang. Hint: Assume that A has two inverses. x3. we usually call them x and y. Note that the unknowns. and April Allen Materowski.3 Gauss-Jordan Reduction » » » » Augmented Matrix Gauss-Jordan Reduction Matrix Inversion Calculator Tips In the previous section we indicated that there is a connection between solutions of systems of linear equations and the existence of the inverse of a matrix. Inc. Show that the n * n identity matrix is unique for each n. Copyright © 2007 by Pearson Education. Thus. Walter O. we may denote them x. 33. Show that AB = I2 3 30. A. However. . when there are more than three. Á . which is formed by entering the coefficient matrix and then adding an extra column. Show that AI = A for any m * n matrix A. a system of m linear equations in n unknowns may be written as a 11x1 + a12x2 + a 13x3 + Á + a1nx1n = b1 a21x1 + a22x2 + a 23x3 + Á + a2nx2n = b2 o am1x1 + a m2x2 + a m3x3 + Á + amnxmn = bm (1) Augmented Matrix Rather than work directly on this set of m equations in n unknowns. then the augmented matrix has the form 1A B2 Applied Calculus for Business. say n unknowns. xn. We shall see that it is more convenient to perform the calculations on the augmented matrix rather then on the system itself. Thus. we have. in which case we say that the system is inconsistent. and Finance. Solution. Proceeding as in the previous example. Copyright © 2007 by Pearson Education. In particular. that is. It is important to keep in mind that the number of equations in the system should remain the same. There are several ways to solve systems of equations but all come down to the same thing.2 2x . 3x + 9y = 23 Solution.Appendix A. x + 4y = . The augmented matrix is obtained by augmenting an additional column to the coefficient matrix. and April Allen Materowski. In this section we shall only consider systems possessing unique solutions. the additional column being the right hand side of the system. by Warren B. Gordon.3y = 7 Applied Calculus for Business. (3) Adding a multiple of one equation to any other equation. we have to understand exactly what types of operations are permitted on the augmented matrix. (2) Multiplication of every term on both sides of an equation by a nonzero constant. Since the augmented matrix is shorthand for the system. Thus. Wang. It may happen that the system has no solution. there are three operations that are allowed when working with a system of equations: (1) Swapping (Interchanging) any two equations. from which we may more easily determine the solution. Published by Pearson Learning Solutions. operation (3) means that we can replace any equation by a new equation obtained by adding to the original equation any multiple of another equation in the system. Economics. any operation that may be performed on the system may be performed on the matrix. the system is said to be consistent. our objective is to determine its solution. Other cases will be examined in the next section.) Example 3 Solve the following system of equations. a 2 5 -3 7 4 41 b ` -9 34 Given a linear system of equations. If a solution exists. the augmented matrix is a 2 3 5 11 b ` 9 23 Example 2 Represent the system 2x .9z = 34 as an augmented matrix. (To be precise. .3y + 4z = 41 5x + 7y . Walter O. Inc.3 Gauss-Jordan Reduction 579 Example 1 Represent the system 2x + 5y = 11 as an augmented matrix. to perform operations on the given system that will replace it by an equivalent system (one with the same solution set). by Warren B.2 and add it to the second equation. the allowable operations could just as easily have been performed on the entries in the augmented matrix used to represent this system of equations. the solution will be Applied Calculus for Business. Our objective is to reduce it to a form that is so simple that we are able to read off the unique solution directly from the matrix just as we did for the two by two system in Example 3. the augmented matrix was a 1 0 0 2 b ` 1 -1 Gauss Jordan Reduction In this matrix. we see that the coefficient side is exactly the Identity matrix and the right hand side gives the solution. In other words.2 .11y = 11 We now see that we can divide the second equation by . Copyright © 2007 by Pearson Education. Walter O. Once we do this.) Thus. our goal shall be to transform the left hand side of the augmented matrix to the Identity matrix. .42e2.2e1 + e2. take a look back at that solution. we try to eliminate one of the unknowns from one of the equations.3 Gauss-Jordan Reduction Solution.2 + y = -1 We could substitute y = . Gordon. The question that naturally arises is what is our objective? If we are to perform elementary operations to the augmented matrix. (e2 represents the modified second equation. we can indicate this step by writing e2 = . DEFINITION 1 . (3) Adding a multiple of one row to the corresponding entries of any other row. Now. thus eliminating x from the second equation. we replace the second equation by the new one. In symbols.4 times the second equation to the first.580 Appendix A.ELEMENTARY ROW OPERATIONS The following operations are called elementary row operations: (1) Swapping any two rows of the augmented matrix.11 and solve for y. In other words add . Thus. Thus we have the following definition. Performing these operations on the augmented matrix produces a new matrix. our system of equations is now x + 4y = . Wang. to what do we want to transform it? There are several possibilities.1 in the first equation and solve for x. Published by Pearson Learning Solutions. the modified equation e1 = e1 + 1 . Denoting the two equations as e1 and e2. Inc. Instead. One way to do this is to multiply the first equation by . Using what is usually referred to in preliminary algebra courses as the addition method. and Finance. (2) Multiplying every entry in a row by a non-zero constant. All of the operations performed on the system effected only the coefficients and the right hand side constants. Economics. and April Allen Materowski. In matrix form. let us continue to use the addition method to eliminate y from the first equation. Actually. The solution set of the system represented by this modified matrix is identical to the original solution set. This leaves x = 2 y = -1 There is no equivalent system for which it could be any easier to read off the solutions! You should realize that the unknowns x and y played no role in the method of solution. in general. x + 4y = . There are many ways to achieve the desired reduction.11. convert and eliminate. the algorithm that we choose is the following: Gauss-Jordan Reduction Algorithm 1. we represent the procedure as 1 A B2 : 1 I C2 C is the resulting right hand side after the operations have been performed. Walter O. x + 4y = . This is accomplished by multiplying the entire second row by the reciprocal of . To perform either of these actions. Choose any diagonal entry from the coefficient matrix portion of the augmented matrix not yet selected and convert it to 1.11 r2 We now go back to Step 2 of the algorithm. if we select the 1 in the 1-1 position. (2) The elimination of the non-zero entries in the pivotal column may be accomplished by multiplying the row containing the pivot (leading one) by the negative of the entry we wish to eliminate. we may use the elementary operations described above. In short.2r1 + r2 There is no need to perform any preliminary conversions to obtain a pivot (leading one).) To eliminate the 2 in the pivotal column. we multiply row 1 by .2 . called the pivotal column. If the left hand side of the modified augmented matrix is the Identity matrix the reduction is complete.11 1 r2 = .3y = 7 Solution. yielding Applied Calculus for Business. Use the pivot (leading one) from the preceding step to eliminate all non-zero entries in its column. Copyright © 2007 by Pearson Education. Gordon.3 Gauss-Jordan Reduction * ** 581 on the right. Pictorially.Appendix A. we have already used the 1-1 entry. its augmented 1 ~ 2 4 -2 b ` -3 7 a r2 = . 3. and use elementary operations on it so that A is transformed into I.2 and add the modified elements 1 . This kind of reduction is called Gauss Jordan Reduction. (1) To convert any non-zero entry to a 1. This 1 is called a leading one or pivot. matrix is This is the same system we solved in the previous example. we may multiply every entry in its row by the reciprocal of the selected entry. Construct the augmented matrix from the given system of equations.8 42 to the corresponding elements in row 2. It is necessary to explain two terms used in the above algorithm. This is indicated by the expression r2 = . (We indicate the pivot by circling the 1. and April Allen Materowski. We illustrate the algorithm by the next few examples. Economics. and Finance. Inc. by Warren B. since. we select the 2-2 entry and convert it to a 1. 2.2 2x . otherwise go to Step 2. a 1 0 -2 4 b ` 11 . We obtain. Example 4 Solve the following system using Gauss-Jordan Reduction. and then add this modified row to the row containing that entry. as indicated on the right of the above matrix. we begin with the augmented matrix.2r1 + r2 to the right of the above matrix. . Published by Pearson Learning Solutions. Wang. 2 and add this modified row to the corresponding entries in the second row. and April Allen Materowski. Economics. we find that x = 2 and y = . the instruction to the right of the matrix shows the operation to be used to get the first pivot (leading one).1. it never changes. So we multiply the second row by the reciprocal of this entry. x + 0y = 2 or x = 2 and 0x + y = . . . or y = . Published by Pearson Learning Solutions.2 3 r2 + r1 Applied Calculus for Business.2r1 becomes 1 . observe that once a 1 is obtained in a pivotal position. Thus. The row containing the pivot also does not change during the sequence of operations in which it is used to eliminate the entries in the pivotal column. Since we have already used the 1-1 entry. Gordon. 3x + 2y = 7 2x . and Finance.3 Gauss-Jordan Reduction a 1 0 ~ -2 4 b ` 1 -1 r 1 = . we must add a multiple of the first row to the second so that the 2-1 entry in the augmented matrix is changed to 0.1.14/32. by Warren B. We need only multiply the first row by .4 42 and add it to row 1. Inc. which has the Identity matrix on the left. Thus.2 . Copyright © 2007 by Pearson Education. We begin by writing the augmented matrix a 3 2 7 2 b ` -3 -4 r1 = 1 3 r1 Note. Note that .4/3 . each entry in the first row is multiplied by 1/3. we shall divide the first row by 3 to obtain the modified first row with a 1 in the diagonal position. by using row reductions on the augmented matrix. This results in a 1 0 ~ 1 2 3 ` 7 3 2 b r1 = . In using the algorithm. Thus. we multiply row 2 by . Example 5 Solve the following system of equations by Gauss Jordan Reduction.2r1 + r2 We shall now use the pivot (leading one) to eliminate all other entries in its column. Specifically.582 Appendix A. we must select the 2-2 entry. Similarly.1.3y = .4 Solution. This instruction is indicated to the right of the above matrix. In this case. Walter O. we have a 1 ~ 2 2 3 -3 ` 7 3 -4 b r 2 = . That is. obtaining a 1 0 2 0 b ` -1 1 The solution is now easily read off from this matrix.4 (the modified row 2 is now 10 . once an entry in the pivotal column becomes zero it too never changes.13 r2 Step 2 of the algorithm tells us to select another diagonal entry of the coefficient matrix and convert it to a 1.3/13 as indicated to the right of the above matrix. we have a 1 0 2 3 13 -3 ` 7 3 26 b -3 3 r 2 = . Wang.4r2 + r1 To eliminate the 4 above the pivot. by Warren B.15 . Similarly.z = . obtaining 1 . obtaining 1 .3r1 + r2 : r3 = .r3 + r1 : r2 = r3 + r2 1 £0 0 0 1 0 0 1 3 0 -2 1 3 ~ Applied Calculus for Business. We use the instruction written to the right of the last matrix.3 Solution. So. In this case. and add these entries to the corresponding entries in row 3. we must use row one twice. We are done and we read off the solution: x = 1 and y = 2.50 .13 . Gordon.8 5x + 7y + 2z = . Walter O. . Now we use this pivot element to clear all the non-zero entries from the second column as shown below. 1 £0 0 0 1 0 1 4 3 -1 -5 1 3 r1 = . (the entry in the 2-1 position) we multiply each entry in row 1 by . 1 £0 0 -2 1 17 ~ 3 14 -1 3 -5 . and add these entries to the corresponding entries in row 2.9 . we begin with the augmented matrix and perform the necessary row reductions to obtain the Identity matrix. Therefore. we obtain a 1 0 0 1 ` b 1 2 We now have reduced the left hand side of the augmented matrix to the Identity.702. and April Allen Materowski. Published by Pearson Learning Solutions.5. the 1 is obtained by multiplying row 3 by 1/4. the 2/3 must be eliminated.10 3 . Inc.73 r2 = We select the 2-2 entry to become the new pivot and make it a 1 by multiplying row 2 by 1/10. Copyright © 2007 by Pearson Education. Example 6 Solve the following system of equations. this time we must eliminate the two non-zero entries in the pivotal column.2/3 . multiply row 2 by .5 10 . As before.2/3 and add the entries to the corresponding entries in row 1.13 .5r1 + r3 1 10 r2 1 £0 0 3 14 . as indicated to the right of the above matrix. x . However.3 Gauss-Jordan Reduction 583 We use the new pivot (leading one) to eliminate all entries in its column. to eliminate the 5 in row 3 we multiply each entry in row 1 by .Appendix A.17r2 + r3 0 2r2 + r1 0 1 0 1 4 -1 3 -5 4 12 r3 = 1 4 r3 The 3-3 entry is the next pivot. the modified entries in row 2 then become 10 . and Finance.3. Wang.422. To eliminate the 3 in the pivotal column. 1 £3 5 ~ -2 4 7 3 14 -1 3 -8 2 -3 -2 10 17 r2 = .2y + 3z = 14 3x + 4y .73 r1 = 1 : £0 r 3 = . Economics.3 6 . As indicated above.4/32. Note. The reductions are again indicated to the right of each matrix. y = 1.39 r1 r 2 = . Solution. The row reductions are indicated at the right of each matrix. y = . assuming it exists. 3 .10 0 . Inc. The solution is now easily read from the last matrix: x = 1. This was accomplished by multiplying row 3 by .2y + 7z = 25 4x + y .4z = .5 11 0 5 37 4 1 . by clearing the third column. by Warren B. we have the following two systems of linear equations. Wang. Let A-1 = a b . the algorithm allows us to proceed in any order when we are selecting the pivot. 3x + 5z = 1 2x + 3z = 0 3y + 5w = 0 2y + 3w = 1 Applied Calculus for Business.584 Appendix A. Economics.23 A quick look at the augmented matrix reveals that one of the diagonal entries is a 1. We multiply 2 3 z w -1 A by A and set it equal to I.5r3 + r1 r2 = r3 + r2 0 2 3 0 29 1 23 : .3 Gauss-Jordan Reduction The process is completed. Example 7 Solve the system of equations 3x . The following example will be used to illustrate the above remarks.1 .1.r3 . Matrix Inversion We return to the question of finding the inverse to a given matrix. we selected the 2-2 entry as our first pivot as it is the most convenient diagonal entry to work with. obtaining a 3x + 5z 2x + 3z 3y + 5w 1 b = a 2y + 3w 0 0 b 1 equating the corresponding elements in the two matrices. . obtaining 10 0 . Therefore.2 7 25 1 -1 3 6 £4 2 3 -4 -5 ~ r 1 = 2r2 + r1 r 3 = . we have that AA-1 = I.39 £ 14 10 : 0 1 0 0 .10r1 + r3 0 2 3 0 1 1 3 The problem is now solved. we should be able to find the inverse.1 3 6 r 3 = . thus. Walter O. 3 5 x y Given A = a b our objective is to find A-1. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education. and multiplying row 3 by + 1 and adding to the corresponding entries in row 2.3r2 + r3 :£ 11 £ 4 10 0 1 0 5 37 -1 3 6 1 23 1 £ 14 10 ~ ~ 0 1 0 r 1 = .1 . Gordon.32.78 0 3 29 1 23 1 £0 0 0 1 0 1 r 1 = .z = 6 2x + 3y . and April Allen Materowski. Any other choice would immediately involve us in computation with fractions. by equating the elements in the product of A with its inverse to the corresponding entries in I. x = 2. From its definition. and Finance. as indicated in the above matrices. and adding to the corresponding elements in row 1. and z = 3.14r1 + r2 r 3 = . Remember.2 and z = 3. and then generalized to develop an algorithm for finding the inverse. and April Allen Materowski. the left hand side of which is the above coefficient matrix. the first column on the right (column 3) gives the solution for x and z. Walter O. but with two columns. a 3 2 5 1 ` 3 0 0 b 1 Observe that the right hand side of this matrix is precisely the Identity matrix. obtaining a 1 0 ` 0 b r = . but it is immediately apparent that to reduce the left hand side of each of the augmented matrices to the identity would require exactly the same operations. Thus.Appendix A. Given the square matrix A. it will also be obvious (at some point in the reduction process) if a matrix is singular. When this happens. Thus. First. We now perform row operations on this matrix. the algorithm is -1 1A I2 : 1I A 2 We know that not every matrix is invertible. we form another matrix.3 Gauss-Jordan Reduction * ** 585 Observe that the coefficient matrices for the two systems are identical. and Finance. we have the following matrix. the right hand side of this matrix becomes the inverse. the right hand side will then be the inverse matrix. by Warren B. Gordon. each column representing the right hand side of one of the above systems. Thus. Example 8 Find A-1 if A = a Solution. Inc. 3 We perform row reductions (which are indicated to the right) on 5 1 ` 3 0 1 3 -2 3 0 b r = 1r 1 1 3 1 : a 1 ~ 2 5 3 3 5 3 ` 1 3 0 1 3 0 b r = . Pictorially. we generalize the above discussion. a 3 2 5 3 -1 3 3 2 5 b.2r1 + r2 1 2 We use the pivot in the 1-1 position as indicated to clear that column. We shall actually perform the reductions in Example 8. Copyright © 2007 by Pearson Education. The augmented matrix needed to find x and z is a 3 2 5 1 ` b 3 0 and the augmented matrix needed to find y and w is a 3 2 5 0 ` b 3 1 Of course. Economics. so that the left hand side becomes the Identity.3r2 1 2 : a 1 0 1 ~ ` 2 0 b r = -5 3 r2 + r1 -3 1 Applied Calculus for Business. Instead. from this algorithm. to find its inverse form the matrix 1A I2 Using Gauss Jordan Reduction reduce the left hand side of this matrix to the Identity. Published by Pearson Learning Solutions. We will be unable to get the identity on the left hand side. and the second column on the right (column 4) gives the solution for y and w. Suppose we are given a singular matrix and we apply this algorithm. . What will appear instead will be a row containing zeros. we could perform separate row reductions on each of these augmented matrices. It occurred because the product of A and its inverse yield I. Wang. This is no coincidence. Walter O.36 = £ 14 3 2 -1 0 . We leave the matrix multiplication as an exercise for you. A-1 if A = £ 2 -3 2 3 -6 3 10 . we use the pivot as indicated to obtain a 1 0 -3 0 ` 2 1 5 b -3 The left hand side is the Identity.11 4 . We leave the calculations as an exercise. 4 1 9 2 -5 . The next example is similar to the previous one with the exception that the arithmetic is a bit messy.11 14 . 1 To check this result it is only necessary to show that either AA-1 or A-1A is I.3 Gauss-Jordan Reduction To complete the solution. Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education.2r2 + r1 : £ 0 1 . Economics. . and Finance. Thus. therefore the right hand side is the inverse. Inc. 2 3 £ 2 3 10 -3 -6 -8 1 2 3 £0 1 -4 3 0 0 1 1 ~ 1 0 3 0 1 0 0 1 0 2 -1 3 0 0 1 2 3 1 0 r 2 = . Example 10 3 Find the inverse if A = £ 2 3 Solution. We have.r2 1 0 r = .4 3 2 -1 1 0 0 1 3 0 1 0 0 . -8 Solution. Why? Applied Calculus for Business. The reductions are indicated to the right of each matrix. Gordon. Wang. 0 It is convenient to choose the 2-2 entry as our first pivot. A -1 . multiplying A by its inverse must yield I.586 Appendix A. with the pivots (leading ones) circled.36 2 . therefore the right hand side is the inverse. we have A-1 = a -3 2 5 b -3 It is a simple matter to check our result. by Warren B. and April Allen Materowski.2r1 + r2 3 0 : £0 -1 4 -2 1 r 3 = 3r1 + r3 1 0 0 1 3 0 0 1 0 11 -3 2 0 r 1 = . Example 9 1 Find the inverse of A.1 4 £0 1 0 3 0 0 1 3 0 1 0 0 r 2 = .11r3 + r1 0 1 r = 4r3 + r2 1 2 ~ ~ The left hand side of this last matrix is the Identity matrix. 1/7 Since the left hand side of the last matrix is the Identity.5 0 22 1 .22r3 + r1 : r2 = 5r3 + r2 7 3 £ 1 3 -1 3 22 0 0 1 2 5 . . a row of zeros will appear at some stage bringing the reduction process to a halt. we have found the inverse.1/3 1 ~ 0 1 0 0 3/7 3 0 0 1 0 1 £0 0 0 1 0 0 3/7 3 0 .21 . we factor out 1/105 and write the inverse in the more convenient form A-1 = 45 1 £ .45 1 1 0 3 0 0 9 1 0 1 0 -1 5 45 r1 = 3 7 r1 : 6/35 0 .35 -1 7 . making it impossible to select a pivot element with which to continue the process.22/105 19/105 . performing the indicated reductions. 3 10 11 -2 . Inc.1/7 1 1/7 6/35 . For simplicity.5 0 22 1 -4 0 2 1 -5 3 0 1 0 . Wang.Appendix A.1/5 . Economics.22/105 r = -1 3 r1 + r2 1/9 2 : r3 = 1 3 r1 + r3 1/45 .15 . As you will see a row of zeros appears on the left hand side.15 0 45 0 -9 1 r3 = 1 45 r3 : .2/35 . and Finance. If it is not invertible. Gordon.1/21 £ 1/3 . given any matrix. where A = £ 8 6 Solution.36 .4 £ 2 1 -5 3 0 1 -1 1 0 -1 3 0 5 ~ r1 = . Published by Pearson Learning Solutions.22 We proceed in the usual fashion.22 105 19 105 1 .9r2 + r3 0 0 1 45 .22 19 -5 When you check your result using this last form for the inverse. if it is invertible the reductions illustrated above result in the Identity appearing on the left and the inverse on the right.4r2 + r1 : £ r 3 = . multiply A by A-1.15 105 15 18 -6 . you will appreciate the advantage of factoring out 1/105. Our next example illustrates what happens when one tries to find the inverse of a singular matrix by using Gauss Jordan Reduction. Walter O. that is.3 Gauss-Jordan Reduction 587 3 4 2 1 0 0 £2 1 -5 3 0 1 0 3 9 0 0 0 1 ~ r 1 = . Example 11 1 Determine if the matrix A is invertible. Copyright © 2007 by Pearson Education. and April Allen Materowski. . Applied Calculus for Business. Thus. it is A-1 = 3 7 £ -1 7 1 7 6 35 2 . by Warren B. Essentially. Therefore.22 0 0 1 0 0 0 1 0 1 0 0 1 . We shall illustrate. m. it will be necessary to rename and store the modified matrix after each operation is performed. Published by Pearson Learning Solutions. You ll need to confirm the deletion by pressing Enter twice. and Finance. renamed as a2. We cannot select the 3-3 element as a pivot and we cannot clear the third column. a. Gordon. We illustrate the three commands by reducing the augmented matrix of the system given in Example 7. Inc. adds the modified row to row J. mRow1m. These three operations can be done by the calculator.3r2 + r1 : r3 = 7r2 + r3 0 2 Observe in the matrix on the left the third row is all zeroes.14 -1 2 0 0 0 1 r1 = . If you have used two or more characters to name the matrix using Clear from a-z option.10 3 -5 7 4 7 -6 3 14 1 . J2 : a3 multiplies row I by the constant m. press F1 and choose the delete option. and adds a multiple of one row to a multiple of another row. Instead.8r1 + r2 : r3 = . you will need to use the Var-Link key.10 -6 -2 10 7 1 3 4 7 ~ 16 7 10 7 . . we conclude that A is a singular matrix.14 r2 : -2 1 3 . Calculator Tips Gauss Jordan Reduction performs three operations on a matrix. Economics.6r1 + r3 0 0 1 0 0 1 1 r2 = .588 Appendix A.) Our present goal is to emphasize an understanding of the reduction process without getting bogged down with the intermediate arithmetic calculations.14 -7 3 1 -7 0 1 0 -2 1 . r2 : a2 multiplies row r of matrix a by the constant m and stores this new matrix. swaps two rows. In performing these reductions on a matrix. I. with their syntax that is to be used to perform these reductions.20 -8 . that we can further automate this process to a single calculator command. Copyright © 2007 by Pearson Education. and stores this new matrix as a3.3 Gauss-Jordan Reduction 1 £8 6 1 £0 0 1 £0 0 1 £0 0 ~ 3 10 11 3 . and April Allen Materowski.36 3 0 .14 r2 = . n2 : a1 interchanges row m and row n and stores this new matrix as a1. Wang. but first indicate the three corresponding commands. Suppose we have stored the matrix named a in the calculator s memory. Scroll down to each named matrix. Note that after you have finished the reduction you will need to clear the stored matrices from memory. The command (found in the Catalog. Walter O. multiplies any row by a non-zero constant. doing all the algebraic manipulations for us. Applied Calculus for Business. by Warren B.( F6) will not work. these three commands automate the reduction process. (We shall see in the next section. a. to delete these stored matrices from memory. located above the key. mRowAdd1m. or may be entered by hand) RowSwap1a. Wang.5 We proceed as follows. 4. Inc. Figure 2: Result When .2y + 7z = 25 4x + y . -2.3 Times Row 2 Is Added to Row 3 (Note. See Figure 1. -4. Walter O. 2.z = 6 2x + 3y . we enter [3. and April Allen Materowski. Figure 1: Result When 2 Times Row 2 Is Added to Row 1 We clear the column in Figure 2. and Finance. Gordon. by Warren B. 1. .1 in the third row to a one. 6.Appendix A. Published by Pearson Learning Solutions.4z = .1 and Renaming the Matrix Applied Calculus for Business. see Figure 3.3 Gauss-Jordan Reduction * ** 589 3x . On the entry line on the HOME screen. Figure 3: Multiplying Row 3 by . 3. 7. -1. Economics. Copyright © 2007 by Pearson Education. -5] : a As in Example 7 we proceed as follows. we are constantly renaming the matrix as we perform these operations!) We need to convert the . 25. Figure 7: Clearing Row 2 Figure 8: Clearing Row 3. Applied Calculus for Business. however. by Warren B. Published by Pearson Learning Solutions. Figure 6: Obtaining the Leading 1 in Row 1 To finish. . Copyright © 2007 by Pearson Education.5x. The reduction is now completed. Wang. we need only clear column 1. Gordon. Economics. on the entry line of the HOME screen as follows: solve13x . the curly braces used to enclose the variables are above the parentheses keys. see Figures 7 and 8.2y + 7z = 25 and 4x + y . doing so means you cannot backtrack and recheck your operations. y. Walter O. x = 2. z62 Note the difference between the subtraction operation and the negation symbol. also. Figure 4: .5 Times Row 3 Added to Row 1 Figure 5: 1 Times Row 3 Added to Row 2 We obtain the leading one in the first row as indicated in Figure 6. you need only use the clear a z option. All that is left to do is to clear the memory of the calculator.4z = -5. and Finance. y = 1 and z = 3.z = 6 and 2x + 3y . It should be pointed out that at each reduction you can rename the matrix a.590 * ** Appendix A. and April Allen Materowski. Consider the previous example. We remark that linear systems can be solved directly with the use of the solve command. Inc. but makes cleaning up calculator memory much easier.3 Gauss-Jordan Reduction We next clear the third column see Figures 4 and 5. we enter each equation connected by the use of an and. Inc. and April Allen Materowski.z = 2 + w = 9 + 2w = 8 = -1 .y + z = .t = 1 = . nickels and dimes. The price of a quart of milk. The price of three quarts of milk. 4 of ball points.3 In Exercises 1 14 solve the given system of equations 1.z + 2w = 4 + z .2y + 6z = .00 per case.z + y 2z 3z 3x z 2w 14.2w = . x x x -x 2x x 2x + 6y . two dozen eggs and two loaves of bread is $3. three dozen eggs. x .3x + 4y .y . 22. Walter O. A stationery supplier wishes to clear his inventory of 124 boxes of pencils. .4 x + 2y . 19.770.4y + 5z = 16 9. and one dozen eggs is $1. 3x .2x + 3y + 2z = . A package B order is 4 boxes of pencils. the expense is $3. the middle digit is the sum of the end digits.x + t .46 in pennies. the vitamins at $10.w .00 per case. A = + 2 1 1 27.2z = 4 2x .2y + 2z = .16 x + 2y .05. For a certain three-digit number. the new number is 99 less than the original number. where a is the overhead.z = 3 2x + 5y + 2z = .5y = 23 2x + 7y = . if it exists. 20. Josh had $4. where h is the hundreds digit.4z = . the sum of the digits is 16.y 12. 2x . and 2 of erasers. When there are 30 patients and 1 physician. and u is the units digit (for example. A = a 23. and y is the number of resident physicians. Published by Pearson Learning Solutions. Find the number.w + t + 2t + 2w . When there are 40 patients and 2 physicians. . + 2 1 3 28. Applied Calculus for Business. A = a 24.60.z = 9 3x + 7y + 5z = . Economics.1 10. x + 3y = 6 x .00 per case. . 345 = 100132 + 10142 + 5). He finds that his major customers order supplies as follows: a package A order is 5 boxes of pencils.3z .x + y + z = .840.050. What are the prices of the three commodities? In Exercises 22 30 find the inverse of A. and a loaf of bread is $4. How many packages of each type should he make up in order to exactly clear his inventory? 18. 17. and ten more nickels than dimes. x y x x .y + 3z = 1 3x + z = -2 4. Wang. b. A druggist fell victim to a fire the day after receiving a small shipment from the warehouse.6 x + 2y + 3z = 5 2x .2y .3y .13 11. 3 of ball points. 2 of ball points.y + y .4 2y + 9z = .2 . Find the number.3 Gauss-Jordan Reduction * ** 591 EXERCISE SET A.z = . Copyright © 2007 by Pearson Education. 72 boxes of ball point pens. The shipment consisted of 15 cases of cold tablets. A package C order is 2 boxes of pencils. x + 2y . 3x + 7y .3y + z = 7 x + y .3y = 12 and aspirins having a total value of $92. 5. The Board of Trustees of a small hospital assumes that the daily operating expenses E can be expressed by the formula E = a + bx + cy.y = . For a certain three-digit number. + 1 2 1 1 0 1 0 -2 15.3 8. b is the cost per patient.13 3. how many cases of each type were in the shipment? 16. He had 94 coins in all. How many coins of each type did he have? 21. vitamins. and if there was one fewer case of aspirins than of the other two combined.15 = -1 + y + z + 2y . x is the number of patients.7y + 3z = 21 5x + y + 3z = . and 32 boxes of erasers. and the aspirins at $3.3z + 6w = 3 + 4z = 9 . the expense is found to be $6. and Finance. by Warren B. When there are 50 patients and 5 physicians in the hospital.3 x + y . t is the tens digit. If cold tablets were valued at $7.5 . 3x .Appendix A.y = y = 2t + 2w = x + z = 1 26.t + 3w + 4t + w + t .5 .26 = 23 = . A = a 1 2 0 1 -1 3 1 5 3 b 2 1 b 0 -2 b 6 4 b 11 1 1 2 -1 1* 2 1 -2* 1 2 0* 2 2.2 6x . and 1 of erasers. the tens digit is equal to the sum of the other two.4y + 3z = . Gordon.8 .3z + 2z . A = a 25.4z = 22 7.3z = . 2x .y .2 6.20.y . and interchanging the units and tens digit will create a number 27 less than the original one. and c. c is the cost per resident physician. 2x x 2x x 13.6 x . A three-digit number may be expressed as 100h + 10t + u. Find a. The price of two quarts of milk. the sum of the digits is 18.z = 0 x . and if the digits are reversed. the expense is $4. and 1 of erasers. However.10 9 (b) Use the result of (a) to solve the system x + 2y . -2 1 31.7 b . Inc.3 0.3z = 6 2x + 3y + 4z = 1 3x + 2y + 5z = 0 1 2 -2 5 1*. find 1 -1 -1 4 b. 4 -1 0 1 -1 1 0 2 3 1*. e1T . 43. Given A = a -1 2 -1 3 b.I2-1 = Z1T . Published by Pearson Learning Solutions. Copyright © 2007 by Pearson Education.x .bc Z 0.I2-1 exists. 1 3 2 -4 (c) A-1 and B-1. Solve for X and Y. 33. (d) Show that 1AB2-1 = B-1A-1.5 . the equations AX = B and YA = B where 1 A = +2 1 2 40. This variation selects a new diagonal entry of the coefficient matrix as a pivot at each step. Where is the contradiction? 45.3+2 -2 3 -1 3*. . d a d -c -b b. Z is the matrix of all zeroes). B = +1 1 2 1 0 -2 1 -2* 1 -1 1 3 2 0* 1 3 1 4 * find B if (a) A = + 2 1 0 46.I2-1 = Z. eT .2y . -1 Z A -1 + B .2r2 132 r1 = 2r1 .2x .5x . and T = a 0.1 2 r3 Observe that the ones are obtained during the set of reductions numbered (4). (b) 1AB2-1. Gordon. (a) Find the inverse of A = + 2 3 the system -3 4 * . but does not immediately convert it to a one. Show 1A + B2 1 34.11 Applied Calculus for Business. (ii) Exercise 23. by Warren B. A = 3 4 1 -1 30. Wang.592 * ** Appendix A. Now if 1T . In the algorithm for solving a linear system of equations. This could be written as eT = eI. Simplify into a single matrix + 0 2 3 13 -2* a 2 4 2 3 -1 b .2z = .y + 2z = .5 2x + 5y + z = 12 . This exercise illustrates a modification of the Gauss Jordan Reduction Algorithm. 0 eT = e. (b) Use the result of (a) to solve 5 x + 2y . (b) Using the method illustrated in (a) Find the inverse of the matrix given in (i) Exercise 22. Given that A = a and A-1 = 1 ad . which avoids the introduction of fractions until after the last pivot is used to clear its column.r3 122 r1 = r1 . Economics. Solve for X and Y. it is easy to check that e = 19. and Finance.I21T . (a) Given the matrix 1 A = +0 1 Proceed as follows: 112 r3 = r1 . (Remember. show that the inverse exists when ad . (a) Find the inverse of A = + 2 . What should you do when there is no acceptable non-zero pivot? Consider the following cases: (a) Solve the system of equations: x + 2y .9 0.I2 = Z. the equations AX = B and YA = B where A = a 1 2 -1 1 b.3 -1 -2 0 3 2 -3 2 2 3 2 0 1 2 1 . 36. what is C? 42. we could write e1T . and April Allen Materowski. a 1 2 . -2 1 -1 4 3 1 Gauss-Jordan Reduction 41. B = a -3 2 1 35.7r3 r2 = 2r2 + 3r3 r1 = 1 2 r1 142 r2 = 1 2 r2 r3 = . 1A B2 : 1I C2. (iii) Exercise 25. Show that if A and B are any two invertible matrices. (a) Find AB.eI = Z. Given A = a 2 A-1B + 3C. and conclude that the only solution is e = 0.z = 7 .10y + 9z = 21 1 37. Given AB = + 0 -1 1 (b) A = + 2 1 -1 2 -2 1 2 2 2 0*. y2. we would like to find e so that 0. 1AB2-1 = B-1A-1. 44.2z = 2 . (a) Given the system of equations ax + by = e cx + dy = f under what conditions on the coefficients can we be guaranteed a unique solution? (b) Determine the slope of each of the lines in (a) and interpret the condition found in (b) in terms of the slopes. Walter O. Given e = 1x. and C = a 3 -1 0 2 1 b .bc 1 2 29. 72 is a solution. B = a -3 2 -5 b 2 39. -1 1 2 3 2 2 2 b .1 32. B = a b . A = 2 3 a c b b . 2 1 2 -1 3* 1 38. Walter O.4 Inconsistent Linear Systems and Systems with Infinitely Many Solutions » » » » » Alternative Theorem Inconsistent Systems Infinitely Many Solutions Reduced Row Echelon Form Calculator Tips In the previous section.X2 satisfies the equation AX = Z. To get a non-zero acceptable pivot interchange rows 2 and 3. we saw that the system has a unique solution if and only if the coefficient matrix is non-singular. Gordon. In that case. If the system does not have a unique solution there are again two possibilities: Either there is no solution.4 Inconsistent Linear Systems and Systems with Infinitely Many Solutions * ** 593 If you start with the entry in the 1-1. First. (2) the system has no solution. PROOF Either the system has a unique solution or it does not.X22 = A1X12 .B = Z.position as a pivot. We write the system in the matrix form AX = B.Appendix A. Wang. where Z is the zero column vector. indicates all the possibilities. Published by Pearson Learning Solutions. Economics. we considered only systems with as many equations as unknowns. (b) Find the inverse of the matrix 1 A = + -1 -2 2 -2 -1 -1 -2* 2 47. there are three alternatives: (1) the system has a unique solution. Applied Calculus for Business. To see this. . or there is more than one solution. Thus. after the first step the 2-2 and the 3-3 entries should both be zero. That is. In this exercise you must extend the procedure used in Exercise 46. Inc. Let X1 and X2 be two solutions to the system. That leaves two questions. what about systems for which the coefficient matrix is singular? Second. systems for which the coefficient matrix is square. consider A1X1 . we have AX1 = B and AX2 = B Observe that X1 .2y = 16 (b) Find the inverse of the matrix 0 A = +1 1 1 0 -2 2 3* 0 A. and Finance. by Warren B. (3) the system has an infinite number of solutions. called the Alternative Theorem. and April Allen Materowski.A1X22 = B . To complete the proof we must show that the alternative that there is more than one solution implies that there are an infinite number of solutions. That means that each satisfies the equation AX = B. (a) Solve the system of equations: + y + 2z = 16 x + 3z = 37 x . Alternative Theorem Alternative Theorem Given any system of linear equations. Copyright © 2007 by Pearson Education. what about systems in which the number of equations and unknowns are not the same? The following theorem. the system has no solution. Consider AW = A1k1X1 . Certainly.4r1 + r3 0 ~ 2 1 1 3 9 .2r1 + r2 : £0 r3 = . For example. the equations represent two parallel lines. Let us write the above system as an augmented matrix and perform the usual reductions on its rows. Performing the reductions indicated above. We proceed in the usual manner. We form the augmented matrix and use Gauss Jordan reductions 1 £2 4 ~ 2 5 9 3 9 4 3 3 10 7 1 r2 = . However. it means that the system is inconsistent. W is a solution to the system. Example 1 Show that the following system is inconsistent. If at any time you arrive at a row which has all zeros on the left and a non-zero number on the right. x + y cannot be equal to both 7 and 9 at the same time! Thus.594 * ** Appendix A. we are interested in recognizing when any linear system is inconsistent. Geometrically. However.X22 + X1 (note that there are an infinite number of choices for W. you might have observed that the equations represented by the last two rows of the matrix are inconsistent. proceed in the usual manner. so we shall take the above example a little further. the original system is inconsistent. Walter O. it has an infinite number of solutions. by Warren B. a 1 ~ 1 1 7 1 ` b r = . (Why?) Therefore. Gordon. As parallel lines do not intersect. consider the system x + y = 7 x + y = 9 Clearly. Copyright © 2007 by Pearson Education.X22 + A1X12 = kZ + B = B. This is an absurdity. * We remarked in the last section that if a system has no solution it is said to be inconsistent. the information (the two equations) is contradictory. The immediate question is what makes a system inconsistent? Think of each equation in a system as a piece of information. each k produces another W). it has no solution. which means either a unique solution or an infinite number of solutions. we proceed on the assumption that you might have missed this observation. and a consistent system is one which has a solution. Inc. a clue to the identities of the unknowns.2 3 . Now we may answer that the system will be inconsistent when the clues are contradictory. Wang. Published by Pearson Learning Solutions.15 -2 . yields Applied Calculus for Business. and Finance. given any linear system. or 0 = 2. For any k. this system is inconsistent.r1 + r2 : a 1 9 2 0 7 1 ` b 0 2 Inconsistent Systems The last row means that 0x + 0y = 2.X222 + A1X12 = kA1X1 . and we must conclude that the system is inconsistent.2r2 + r1 : r 3 = . In general. that is.X22 + X12 = A1k1X1 . and April Allen Materowski. .r2 + r3 Before performing the last set of reductions shown above. Economics. Thus.4 Inconsistent Linear Systems and Systems with Infinitely Many Solutions Let k be any real number and let W = k1X1 . the case of an inconsistent system with two unknowns requires no detailed examination. we have shown that if a linear system has more than one solution. x + 2y + 3z = 9 2x + 5y + 4z = 3 4x + 9y + 10z = 7 Solution.29 r 1 = . Economics. obtaining x = 3 . We know that a line contains an infinite number of points and the coordinates of any point on the line satisfy the equation and give a solution. Geometrically. you may omit these rows as they occur. we write the solution as y = k and x = 3 .2 3 . which becomes the free variable.15 0 . the first member of the ordered pair is x and the second is y.r1 + r2 : a 2 3 2 0 3 2 ` b 0 0 Infinitely Many Solutions Unlike the inconsistent case. (Of course. and in this case. and Finance. We solve the equation for x.14 It is now obvious from the last row of the above matrix that the system is inconsistent. There is a simple point to be underscored with this example.) Notice that the x variable is represented by a leading one in the diagonal position of this last matrix (the 1-1 entry). Sometimes. using the augmented matrix.2k. Published by Pearson Learning Solutions. Here we are asked to solve a system containing the single equation x + 2y = 3. It turns out that this is a fairly straight-forward problem.) We may think of y as being a free variable. (If you wish. Now x is thought of as being determined by y. Summarizing. .Appendix A. how to describe the solutions. Copyright © 2007 by Pearson Education. or keep them together as the bottom rows of the matrix. obtaining leading ones in the diagonal positions on the left hand side of the Applied Calculus for Business. It is a simple matter to describe all the solutions. To emphasize this interpretation. we could also have chosen to solve for y in terms of x. This is typical of systems of equations with infinitely many solutions. k). a 1 ~ 1 2 3 1 ` b r = . it is convenient to write the answer as an ordered pair in the form 13 . Consider the system of equations x + 2y = 3 x + 2y = 3 where we have the same equation written twice. it was studied in algebra. The next case to consider is how to determine when a linear system has an infinite number of solutions.4 Inconsistent Linear Systems and Systems with Infinitely Many Solutions 595 1 £0 0 0 1 0 7 39 . sometimes called a parameter. we have. our object is to perform Gauss Jordan Reduction on the augmented matrix.2k. by Warren B.2y Now we are free to choose any number for y and substitute it into the above equation to find the corresponding x-value. in fact we really have only one piece of information. If we solve the above problem with the two equations. Inc. systems with an infinite number of solutions may often have entire rows (both the left and right hand side) with only zero entries. Gordon. One of the equations is redundant. but it is impossible to obtain a 1 for the variable y. Walter O. The system reduces to one in which the equations are consistent and the number of unknowns exceeds the number of equations. Wang. this single equation represents a line. If we again think of each equation as a piece of information. Where of course. This case is not new to you. and April Allen Materowski. then the above example indicates that the two pieces of information are not independent. 4 Inconsistent Linear Systems and Systems with Infinitely Many Solutions non-zero rows. Published by Pearson Learning Solutions. Walter O. From the non-zero rows in the last matrix we have the equations x + 7z = 39 y . . we may write our solution as x = 39 . Inc.2 3 . When we can go no further. We shall solve for x and y in terms of z.7z y = .15 Solving for x and y in terms of z. by Warren B. z will be the free variable. and April Allen Materowski. The next example illustrates this case.15 0 0 It is impossible to obtain a 1 in the 3-3 position which is the diagonal position representing the variable z. The procedure is still the same. Locate as many leading ones as possible. and solve for the variables represented by the leading ones in terms of the free variables. those variables for which we cannot obtain a 1 in their diagonal positions will be the free variables. 1 r2 = .15 .2r2 + r1 : r 3 = . and Finance.15 r 1 = .15 + 2z If we call the free variable k.7k y = .r2 + r3 7 39 3 . we have x = 39 . Economics. Example 3 Solve the following system x x 3x 2x + + + y y 2y 5y + + + z 7z 6z 7z + + + 5w 9w 8w 31w = 2 = 30 = 20 = . The variables represented by 1 s will be determined in terms of these free variables. We form the augmented matrix and perform the reductions.4r1 + r3 0 1 0 £0 1 0 0 1 2 3 9 £2 5 4 3 3 4 9 10 21 ~ ~ 2 1 1 3 9 .38 Applied Calculus for Business. Example 2 Solve the system x + 2y + 3z = 9 2x + 5y + 4z = 3 4x + 9y + 10z = 21 Solution.2z = .2r1 + r2 : £0 r3 = .2 . Copyright © 2007 by Pearson Education.596 Appendix A. Wang. Gordon. Hence.15 + 2k z = k It may well happen that there may be more then one free variable.2 . We now have two non-zero rows in the last matrix.4k + 2l = .7l = k = l The solution to this system involves the two free variables. we have x y z w = 16 . Copyright © 2007 by Pearson Education.38 5 2 .42 0 The original augmented matrix has 4 rows.3r2 + r4 21 . Economics. We may mentally add a row of all zeroes to the bottom of the matrix. Therefore. Inc.y + z + w = 6 2x .4 Inconsistent Linear Systems and Systems with Infinitely Many Solutions 597 1 ~ Solution. § 1 3 2 1 -1 2 5 1 -2 -1 3 1 7 6 -7 1 6 3 -9 5 2 r2 = .z .14 + 3z . We have four unknowns and three equations. We have. Example 4 Solve the system of equations x .r1 + r2 -9 4 30 ¥ r3 = .14 0 4 r2 + r3 : § ¥ r3 = -7 14 0 r4 = . So. there can only be two 1 s in the diagonal positions on the left. we let z = k and w = l.3r1 + r3 : 8 20 r4 = .2w = 16 y . Gordon. .14 Solving for x and y in terms of the free variables z and w we have. Sometimes we describe this by saying the system has a two parameter family of solutions. we obtain another solution. and April Allen Materowski. x + 4z . k and l. Wang.Appendix A. we have completed the row reductions. giving us four equations and a square coefficient Applied Calculus for Business.4z + 2w y = .3z + 7w = . The coefficient matrix will not be square and our algorithm instructs us to use elements along the diagonal of the coefficient matrix as the pivots.3w = 4 Solution.3y + 3z + w = 7 x + y . Published by Pearson Learning Solutions. x = 16 .14 4 ¥ 0 0 0 0 1 0 § 0 0 1 0 § 0 0 1 1 -1 3 ~ 1 -3 3 -9 5 2 1 r2 + r1 r1 = 7 . Walter O.1 2 r 2 = r2 : -7 14 21 .7w As before.42 0 1 0 0 4 -3 0 0 -2 16 7 . Thus.2r1 + r4 31 .14 4 28 ¥ . and Finance.14 + 3k . what do we do? Consider the following example. For each choice of k and l. Suppose we have a system in which the number of unknowns is greater than the number of equations. by Warren B. There is no acceptable pivot in the 3-3 position. we see a nonzero entry in the 4-4 position.2r3 + r1 : £0 r 2 = . but now the only non-zero entry in the last row is the 3-4 entry. Published by Pearson Learning Solutions. -1 1 1 6 1 r2 = .598 Appendix A.2r2 + r3 0 2 -2 -4 -2 3 0 1 ~ -1 1 1 6 3 -1 1 .4 Inconsistent Linear Systems and Systems with Infinitely Many Solutions matrix. and April Allen Materowski.1 6 r3 : 0 0 . then we have x = 7 y = 3 + k z = k w = 2 0 1 0 0 -1 0 2 11 3 1 5 1 2 1 r 1 = . Continuing with this pivot with the zero row removed.r1 + r3 1 1 -1 -3 4 3 0 1 -1 1 1 6 1 r = r2 + r1 1 -1 1 3 5 1 : £0 £0 r = .z = 3 w = 2 or x = 7 y = 3 + z w = 2 if we name the free variable z = k. 1 0 § 0 0 0 1 0 0 0 -1 0 0 2 11 1 4 5¥ 0 0 1 2 However. if we interchange the imaginary row of zeroes with the third row. there will be infinitely many solutions (why?). this row gives us no information but neither does it cause any harm.2r1 + r2 3 3 1 7 : £0 £ 2 -3 r = . The first two pivots(leading ones) were the 1-1 and 2-2 entries. Walter O. which we select as our pivot (leading one).12 ~ Something unusual is about to occur. Inc. However.r2 : 2 -2 -4 -2 0 0 2 11 1 -1 1 3 5 r3 = . our last pivot is simply not a diagonal entry on the left hand side. Now. it is really not necessary to insert the row of zeroes. and Finance.5 r 2 = . by Warren B.r3 + r2 0 0 1 0 0 -1 0 0 7 3 0 3 1 2 ~ Note that the above example illustrates that the method we have been using all along to obtain the leading ones (pivots)and zeros may be generalized as follows: define the leading Applied Calculus for Business. Copyright © 2007 by Pearson Education. In essence.6 . Economics. we begin to perform row reductions on the augmented matrix. if the equations are consistent. Gordon.1 . Wang. we have 1 £0 0 We now have x = 7 y . Without actually writing the extra row. . Of course. the leading one of the lower row will always be to the right of the leading one of the upper row. Five times the second plus 21 is equal to twice the first. Example 5 Find two numbers such that the first is six more than twice the second. the system would have been inconsistent. we shall work on all the rows. Published by Pearson Learning Solutions. Thus.12 and y = . the last row would reduce to zeroes on the left hand side but not a zero on the right. but how do we choose a pivot? We use the diagonal of the first two equations to determine the pivots. and April Allen Materowski. That is. but performing Gauss-Jordan Reduction on the augmented matrix (that is.2r1 + r3 -5 21 3 0 -1 1 -2 6 1 r1 = r1 + 2r2 3 1 -9 : £0 £0 r = . The objective is now to put the matrix in rref. Inc. obtain the rref of the augmented matrix) will soon determine the proper alternative. Now that we know that Gauss-Jordan Reduction is nothing more than finding the reduced row echelon form of a matrix. Gordon. In addition. each must be in the form ax + by = c. Any matrix having this property is said to be in row echelon form. and five times the second plus 21 is equal to twice the first. given any system of equations. means that 5y + 21 = 2x. In summary.Appendix A. then the matrix is said to be in reduced row echelon form. This generalization enables us to solve problems like the last example without the insertion (or consideration) of imaginary rows containing all zero entries. as we do our row reductions. It may not be obvious by inspection. Let the first number be x and the second be y. However.1r3 9 3 0 . the first is six more than twice the second. Economics.1r2 r = . the second is three more than the first.2y = 6 -x + y = 3 2x . we first rewrite them in the usual order. Now. we can use a command from the calculator that will automatically Calculator Tips Applied Calculus for Business. In order to use our standard method to solve the system.9. Copyright © 2007 by Pearson Education. In any two rows.12 3 1 -9 0 0 The last row is all zeroes and we can read the solution from the first two rows: x = .1r2 + r3 0 1 -9 3 0 ~ 6 3 9 r2 = . by Warren B. . Solution. You should convince yourself that had the last number in the third equation been anything other than 21. What about a system in which the number of equations exceeds the number of unknowns? Let us look at such a case. 1 £ -1 2 rref ~ -2 6 1 -2 r2 = r1 + r2 1 3 3 : £0 -1 r = . we know that there are three alternatives regarding its solution: there is a unique solution. x . no solution.5y = 21 We now set up our augmented matrix. all rows consisting entirely of zeros shall be placed together at the bottom of the matrix. means that y = x + 3. We now have three equations and two unknowns. Hence. Wang. Walter O. or an infinite number of solutions. The second is three more than the first. which is the case.4 Inconsistent Linear Systems and Systems with Infinitely Many Solutions 599 one (pivot) to be the first non-zero entry of a row (assuming the row s entries are not all zero). or rref for short. means that x = 2y + 6. and Finance. if all the other entries in the column containing the leading one are zero. not all zero. If it has a total inventory of 7. (c) an infinite number of solutions.6x .6z 2x + 3y .1 3x + y .2y + z = 8 x + 2y + 3z = 11 2.2y . $2 ties. (b) a unique solution.2z + 6w = 9 x + 5y + 4z .x + 7y + 3z = 2 16.3y .x + 7y = 6 5 7 18.z = 1 . and Finance.3z = 7 x .4y = . rref(A) results in this form of the matrix.y + z + w = 4 x + 2y .2y + z = 8 2x + y .4z = 7 .y = .y + z = 3 2x + 2y .z = 8 x + 2y = 3z + 6 3x + 2y = .2y . x + 2y + 3z .x + y . the matrix is transformed into its equivalent reduce row echelon form. Gordon. Wang.3y + z = .z x + y 14. and $4 ties.8y + 6w + 2v = 41 3x . x . x + y + z + w = 2x + 2y 3x + 3y + 4z + w = 16 9 . Published by Pearson Learning Solutions. x .6w = 33 13. (b) a unique solution. EXERCISE SET A. stored in the calculator s memory. (d) Describe the solution set in (c).9 2x + 4y = .4 Solve the following systems whenever possible.2 3x . (c) an infinite number of solutions.z + w + v = 19 2x .z = 2 7. Given any matrix A. .000.3y + z = . Walter O.1 Applied Calculus for Business. x + 2y + 2z = 5 3x + 8y + cz = 8 x + 4y + 3z = k 20. Determine for which values of c and k the following system has (a) no solution.3z = 7 x .3z = 7 x .y + z = 3 2x + 2y .2y + z = 8 x + 2y + 11z = 49 3. x .3z + w = 3 x .16z + 17w = 35 x .3 3x + 5y = .8y + 3z = .3 . 2y .2 2x . x .6w = 30 11. 2x 7y = 3 = 7 . x . 10.5y + 2z = .y + z = 3 2x + 2y .y + z = 6 x + 2y + 3z . x . 8 x .000 how many of each type are there? 19. (d) Describe the solution set in (c). Then Figure 1.2z + 6w = 9 x + 5y + 4z .3w = .6w = 12 . Economics. x + y + 2z = 4 x + 2y + 3z = 6 x + 3y + cz = k 2x . and we may immediately determine its solution from this form.600 * ** Appendix A.2z = 312 + w2 x + 2y .y = -1 3x + y = 2x + 2y = 4.5 x + 2y + 5z = .5y + 6z = 2 11x . Ties Unlimited stocks $1 ties.4y = . shows the effect of entering rref(a).5z + 2w + v = 36 6x . and April Allen Materowski. Inc.5 x + 4y = 5.4 Inconsistent Linear Systems and Systems with Infinitely Many Solutions provide this reduction. Suppose the augmented matrix in Example 4 is stored in the calculator s memory as a. 8. 1.7y + 6z = 34 6.1 4y + 4z = 0 17. Copyright © 2007 by Pearson Education. x + y = 3 4 2 12.y + z = 3 2x + 2y . Determine for which values of c and k the following system has (a) no solution.6w = 12 .5z 15.x + y .y + z + 9w = 14 5x + 2y .000 ties valued at $12. by Warren B.8z .3z = 7 x . x . Figure 1: rref(a) With one command. 9. x + 3y = 3x .9z + 11w = 22 x . How many ties of each type may it possibly have? If the number of $2 ties is 2. If A = + . (a) Using the rref command on your calculator. Determine the solution(s) to 3x . Gordon.4B = a 2.bc Z 0.10y + 9z = .5y + 2z = 9 5x . If A = + 4 5 -3 2 8 5 3 * .5A2 + 8A . Determine for which values of c and k the following system has (a) no solution.12 1 2 -2 4.4I = Z (Z is a -4 3 4 1 b. (b) Give a geometric interpretation of this condition.55 . 2x .5x .A + 8I b = I (c) Use (b) to find A-1. by Warren B. 3x + 2y + cz = 4 5x + y . Copyright © 2007 by Pearson Education. determine the conditions on the coefficients in order for there to be a unique solution. If A and B are each 2 * 3 matrices and if 2A + 3B = a 3A . (b) Visualizing each equation as being a three dimensional plane.10 9 AB.4y + 3z = . y and z: x + 2y . Consider the system of equations ax + by + cz = A dx + ey + fz = B gx + hy + jz = C * ** 601 (a) Write down the augmented matrix for this system and show there is a unique solution if and only if ad . and April Allen Materowski.Appendix A. give a geometric interpretation of this condition. .2 5. (d) describe the solution set in (c).3y + 3z = . Walter O.10y + 2z = 20 10.2z = k 26x + 5y + 3z = 12 3.5 . Determine the solution(s) to 3x . Consider the system of equations ax + by = e cx + dy = f Inconsistent Linear Systems and Systems with Infinitely Many Solutions 22.3C.17y + 4z = 10 9. find A-1 use it to solve Exercise 5. -6 4 -6 -6 2 * (a) Show that A3 . and Finance.2 5x .4 21. 6 5 -2 2 0 -1 5 3 b and 2 5 7.1 3 zero matrix). y and z. (b) Use the result from (a) to solve the following system for x. Find BBt where B is given in the previous exercise. (b) a unique solution and (c) and infinite number of solutions. .2 2x . CHAPTER REVIEW Key Ideas Matrix Addition of Matrices Scalar Multiplication Zero Matrix Vector Multiplication Matrix Multiplication Inverse Matrix Matrix Form for a Linear System Transpose Augmented Matrix Gauss-Jordan Elimination Matrix Inversion Alternative Theorem Inconsistent Systems Infinitely Many Solutions 1. 9 Applied Calculus for Business.3y + 5z = 28 4x + 2y + 3z = 11 5x + 8y + 9z = 13 2 6. Wang. Given A = + 23 1 5 * and B = + 2 5 1 * (a) compute -5 0 -1 .2z = 5 2x + 5y z = 9 . Inc. B = a 2 -1 3 mine 2AB . Use Gauss Jordan reductions to solve for x. 4 4 8. If A = a 5 -3 2 1 1 b determine A and B.4y + 2z = 4 5x .2 . Published by Pearson Learning Solutions. Economics. -2 3 b and C = a 4 -1 2 2 1 b deter0 1 5 (b) show A a A2 . Walter O. and Finance. Inc. and April Allen Materowski. Wang. Copyright © 2007 by Pearson Education. Economics. Published by Pearson Learning Solutions. Gordon. by Warren B.Applied Calculus for Business. . by Warren B. Walter O. Copyright © 2007 by Pearson Education. .Answers to Exercises and Pre/PostTests Applied Calculus for Business. Gordon. and Finance. Published by Pearson Learning Solutions. Wang. Economics. Inc. and April Allen Materowski. 2 22 3 8.1 1.92/4 10. Economics. 26 4 Applied Calculus for Business.2/9 8. 10 17.33 18. 45 9. 2/9 8.3 4. 3/7 2. and April Allen Materowski. 12x . 5 . 26/11 9.1 1. M = 21 32. . 15x . 120 mi 38. 13x . Walter O. 65/84 21. 119 + 2y2/5 Pretest 0. 5/4 2. M = 19. 12 cm 35.604 * ** Section 0. 84 33. 755/9 2. 47. by Warren B. Inc. . 2.2 Answers to Exercises and Pre/Post Tests 23. 4 hrs 40. 17/2 6.322 27. 6 ft Exercise Set 0. . L = 12 31. 8/3 6. . . 2 25 2 Posttest 0. . 23 2 5. 0 4. 0. 15 5. (a) 2 hrs (b) 160 mi. 9/4 4. 531. V/1pr 22 28. 3 yrs. 5/14 44. 5/91F . 2 3. 216 37. . 150* 42. l = 13 34. .2 1.14/11 13. . 3 23i 10. 4/3 10. 25* 41.ax2/b 26. 336 36. 16 2.by2/a 25. and Finance. 517. 20 8. 12 14. RW/1W . 0. 1 2. 1 6. . 1 2. . 4 3. w = 5.192/2 24.37342 7. 9/2 3. 4 hrs 39. 3 4. 4% Pretest 0. 0. 2 22i 7. Published by Pearson Learning Solutions.13/6 6. $6. .270/109 L . . 4 7.1 0. J = 16. . 2 25i 7. 8/15 16. Wang. 32. 15 15.R2 30. . 36 43.2 1. .47706 9. 12A . 5 . 2 11.52/3 Posttest 0. . . 12/15 45. . 19/18 20. Gordon.2.6 3.2/3 6. 7/3 4.3 3.ah2/h 29. . 3/2 5. 7 5.3/7 12. 19. .32/7 7. 0 5. 3 22 9. 29/7 19. 59/21 10.1 1. 59/158 L 0. 1c . 1c . Copyright © 2007 by Pearson Education.480 46. 0. 5 . 26. 2 25i 2.4/3. .2. 2 215 9 15 . 2 22 3 42. 7i 19. 3 2 2i 3 2 . 5 . . . . . 5i 16. 2 210 4.449 15. . 1x + 5/222 .25/4 10.551. 5 32. 2 23i 10. 3 10.3 . 2 26 15. 2 25 5 3 . . .449 17. . . 211. 25/4 3.464 30.4. 2. 1x . 51. 4i 24. . 2 4. . . 1x . Copyright © 2007 by Pearson Education.7 . 0. 2 2 6i 2 5 . 2 22 13. 2230 10 50. 2 23i 45.1 .317. 3 11.317 16.4 . 8 6.4. 3 23i 26. 5 2. 2 2 5i 2 54. .5 . . . 53.3 . 211. 2 23 27. . 2 23 14.3 . 2 215i 6 Pretest 0. Gordon. . 5. 5 . .322 .36 7. 5 33. . Wang. 3 25 2 5. 3 22 .0.4 . 1x + 322 . . 2 26i 31.522 .1. 48. . 5.3 1. 46. 2 26 9. . 2 23 3 . . 47.3 1.9 4. Walter O. . 3 22 12. 26. 2 . 36 2.6. . 5 37. 3i 43. 16 2. 2 2 3i 3 Exercise Set 0. .551. .3 1. 1x . . 4i Applied Calculus for Business. and Finance. 2 . 3 7. 52. 2i 18. 26 Exercise Set 0.8.49/4 12. 1x + 9/222 . 2 22i 23. . 6 3. 3 . 1x + 3/222 . 39.7 .5/222 . .3/222 .3 . 4. 2 26 2 1 . 1 . 3 22i 25. 5 .317 14.4 .2 1. . 6i 4 3 . 0. . .9/4 9. 1x + 222 . 3i 20. 3 25 2 6 . 3 22 2 .2 0.5/2.3 . .9 3.25/4 11.464.3 . . .81/4 13. 8 9. 5 25i 10 3 . 2170 10 . 5 8. 2 27 4.4 2. 2 23 5 * ** 605 44. 1x . 4i 21. . 49. 56. by Warren B. . 57. 6i 17. . Published by Pearson Learning Solutions. 4 23 3 Posttest 0. 3 22 4 . 1x + 622 .2.16 5. . 1x . 3 22i 2 15 . 7 .3 Answers to Exercises and Pre/Post Tests 8.5. 5i 3 . 1x .7 . 29. 2 35.422 . 3 34. . 2 23.317.7/222 . Economics. . 9/4 3. 40. . 10 36. .Section 0. . 6 . . 4 38. 7 5.2. 55. . 4 22i 28. 41.25 6. . . 4 . Inc. . . and April Allen Materowski.9/4 8. 0. 8.2. 40. . 241 8 1 .9. 25 5.0. . 3i 27.6. 0. .303.1 . 223i 8 18. 255i 4 5 . Posttest 0. 7. Inc. 2 27. 4. 50.583. . 21. .5/2. . . 4 .871.472.0. .132 13.222 14. 3 . Gordon.464 38. . . . .5 .171 10 46.0. . 26.1.743. 17.2.4. 1. . . .6. 4.2.129 17. .854. 214 .317.464. 5. 16.854.0. . 8.3/4. 2 23i 39. 2 23i 35. 7.2. .303 2 5.0.5 .82 3. 2 210.323.0. .303 2 40. 211.5 . c) 1.449 18. 2321 . and April Allen Materowski. Exercise Set 0. .854 2 3 . 2 3i 2 .171. . .902 12 . by Warren B. .0. 2.2 . 12. 27. 6 . 215i 2 . 1. .323 2 5 .5 . 12.102 9. . 213 . 1 .3 . 0.336 sec.2 m Pretest 0.5 . 4/3 43.3. 5 . 27 . 712 15. 4.0.313.0.449 20.303. 38.3. . 2 25i 37. 2 . 2 23i 33. . . 214i 34.4 .0. 2103i 8 9 .325. . 4. 239i 2 .3.871.5 . 5i 34. 3 25 . Published by Pearson Learning Solutions.402. 213 .129 42. . 18. 5i 36.472 19. 6 . 0.5 . 19. 5. 41. .464.6. 2 23. 14.1. 6 .2. 3.7. 221.2.464 36.854 2 5 .3 . 15. 2 . 2 27. 3/2 41. 23. .1 .1.29 29. 2 3i 3 33. 234 9 3 . 3 . . . 3 25 .152. 32. 3.0. 11) 10. .5 . 160. 24. 2 23i 37. . 2393 .3 0.325. 13.3/2. Economics.664 or 198.4 .102 6. . 30.464 21. 215i 2 . 48.0. 45.325 23. 2119i 12 31. . 2.303 2 .551. 214 2 4 . 5 27. . .325 20. . 02 12. . .606 * ** Section 0.1 . 2 25i 35. 5/4 4. .102 5. 0. 3 25 .3 . 26. 0.8.3 . 49.317 24.1 . .142 8.4/3 44. 13.317 19. . . b. 11. . 5. h = 5 52. .243 12 15 . 1. . . 5. 02 11. .4 1.5 . 2 25.9. 1/6 30. 2 2. 213 . 3 25 .7.112 2. 3/2 Applied Calculus for Business. . . 3 .5.1 . .854 2 3 . . 12. 5 28. 13. 1. 15. 2. 7. b = 8.5.14. . .122 4. 0. . 10. 233 4 .417 26.1 . . 5 .2. 25. and Finance.34 sec 53.5. 239i 2 . 2 . 39.3 . . 0.1 . . 4.4.464. .29. 0. Walter O.3. 214i 32. 2 23. 211. 3 . 5 29. 47. 3. . 2/3 42. 3i 25. 42 7. Wang. 3 .129 sec. 4 . 2 25.4 Answers in 1 15 are in the form (a. 4 . . 2. 4 .29.472 2. 2 .0. 2.212 16.551. 5.472.0. . .2 m by 6. . 14. 15. 214 .854. 2. .4 1. 6 .854 2 5 . 2 210. (5.4 Answers to Exercises and Pre/Post Tests 3. 3 . 2 . . 5 26. 3 25 . 5. . Copyright © 2007 by Pearson Education. 2 23. 23i 2 . 1/6 28.303. .854. 2119i 12 51.29 31. 5 1. 8 67. 4 69. x 5.0. Walter O. .4 or x Ú 2/3 3/2 or x 7 5 x 2.4/3 23.1 . Published by Pearson Learning Solutions. x 6 1 or x 7 3 Exercise Set 0. . 49.832ft by 16.062 sec (b) 3. 221.323.5 54. x 3. x .Section 0. 2 6 x 6 5 20. x . . 19683 72.2 or x Ú 3 14. 2. 2393 .402. . x 1 or 2 x 6 4 39. x 2/3 or x Ú 3/4 0 or x Ú 5/2 48. (a) 3. and Finance.195 sec or 4.0. 82 and 94 miles 61. x .5 1. 50. . x 11. 13. Gordon. 6/13.953 sec 59.4 6 x 6 3 6. 2321 .7.5 or x Ú 3/5 8. x 6 .17/6 or x Ú 2 . 6.0. x 6 .9. Inc. 0 6 x 6 3/2 or 3/2 6 x 6 4 45. x 6 . . . . or .171 10 * ** 607 3.4/3. 3/4 6 x 6 1 3. 3 6 x 8 47. . . 27 .243 12 15 .3 6 x 6 4 49.3 or x Ú 4 48. Wang.2.4/3 46. (a) 50. . . . 1 6 x 3 50. 2/3 44.4 or x Ú 3 x 7/4 4.3/2. 32 Pretest 0.4 or 0 38. 47. 15. x 6 .17/6 6 x 6 2 36. . x 6 0 4.805 sec (b) 5 sec (c) 5. .6. 0.5 31. (a) 18.4/3 or x 7 3/2 .971 in 56.5/2.5 1.3 19. Economics. 2103i 8 9 . x 27.5 or 0 6 x 6 5 32. 52.2 6 x 6 1 or x 7 3 41. x 25.416 in 57.4 = . 4ft 60. x 15. and April Allen Materowski.171.1 or x 7 1 33. x 6 3 or x 7 8 46.5 6 x 6 3/5 9. 255i 4 .2 or 1 x 6 3 71. . x 6 .5/3 7.3/2 or x Ú 4/5 . 512. 16.3/4. (a) 0. x 6 . x 3/2 Applied Calculus for Business.3/2 6 x 5.5/2 or x Ú 4 34. . . 13. 0 6 x 6 5/2 13.3/2 or x Ú 2/5 5/2 0 or 2/3 x 6 5/3 or x Ú 5/2 2.15.4 6 x 6 5/2 10. . x = .323 2 5 . . 11. x 18. 10 and 12 mph 63.5 . by Warren B. .743.2 .902 12 . 7. 3 68. .583.5/3 6 x 4. 1.5/2 6 x 6 4 35. .5/3 or x 7 7/4 .5/4 or x Ú 5/6 x x x 0 or x Ú 4 0 or x Ú 5 4 24.3/2 6 x 6 4/5 26. . 4/3 45. x 6 .584ft (b) 12. 2/3 43. . . . .5/4 6 x 6 5/6 28. x 37. x . .4 29. 5.417 12.536 sec (c) 3. . 3 25 .3 or 0 x x 2 x 5/3 . 3 36 2 3 12 2 2 . x 3/2 . .3/2 or . (a) 3.4 or x Ú . x 6 2/3 or 5/2 6 x 6 4 or x 7 4 42. x x 5/2 0 or x = 3/2 or x 7 4 Posttest 0.11/20 6 x 6 1 40. . .5 17. . .832 ft 62.4 or 0 6 x 6 4 30. .17/2. x 7 3/2 2. x 16. .7 or x 7 6 21.13. x 6 .807 sec 58.5 Answers to Exercises and Pre/Post Tests 43. 2 70. or . 34 + 2 210 2 x 3/2 2. 51. 47 (b) 15. 150 (b) 100 (c) 5000 65. 3 + 110 2 . 2. Copyright © 2007 by Pearson Education. . . 0.11/4 6 x 6 4/3 5. .5 .11/20 or x Ú 1 . x .4/5 6 x 6 3 44.11 55.5 .1 or x Ú 2 53. 4 hr 64. 50 66. 6 5.12.1 1. .9/2 4. 5/2 3.19/42 11. . 10.3) 9. 19/2. Published by Pearson Learning Solutions. . Posttest 1. y = .3 or x = 0 or 1*2 6 x 54. 02. 9/5 5. x + 11y = 47 9.11 6. y = 2x + 5 2.3 25 65. . x = 12 20.112/13.1 Answers to Exercises and Pre/Post Tests 8. Gordon. 10.2/7 3.1 1. (0. .4/3 or .3x + 11 2 1 51. 2x + 5y = 5 Exercise Set 1.22 6 x 6 1 + 22 62.1 1. 3/4. y = .2 22 x x 6 2 or x = 0 58. 5x . x 6 3 53. 0.6 or . and April Allen Materowski.q 6 x 6 q 12. y = 3x .2 4. x 1 . 12/7 4.3 2. y = 6 7. . . 02.2 22 or x 7 2 22 61. .3x + 2y = 38 6.22 or x Ú 1 + 22 x 2 + 3 25 63. . ¤ 66. 1 . y = 14x + 143/10 15.4/3 6 x 6 0 or 0 6 x 6 *2 or x 7 2 57. y = 1 4x + 3 6. and Finance. . . 02. Walter O. 1 . y = . Inc.1 17. x = 5 7. y = 1*2 x + 1 13. Copyright © 2007 by Pearson Education. Pretest 1.6 6 x 6 .3 16. y = 3x .608 * ** Section 1.2y = 8 10.112/472 10. 5/4. Wang. 1 .32 7.9/2 or 7/4 59. y = 2x + 4 2. y = 2x + 4 19. . . Applied Calculus for Business.3 25 or x 7 2 + 3 25 64. 14.1x . 10.6 18.9/2 6 x 6 . 02. by Warren B. y = 1 6x 14. . x 6 . 2 .12/5.1/2 or . x 7 3 52. y = . 3x . y = 4. x = 5 8.13/47. 10.2y = 16 9. 2x + 3y = 27 2. y = 1*2 x + 1/4 5. x = .1*2.3 6 x 6 . y = 2 8.5 or 5 6 x 6 6 56. x 6 . 9/42 23. x . x 55. Economics.5 6 x 6 5 or x Ú 6 . y = 8 21.3x 3. x 6 2 .1/2 6 x 6 0 or 0 6 x 6 7/4 or x 7 2 2 22 60. Gordon. 30. Applied Calculus for Business. Inc.1 Answers to Exercises and Pre/Post Tests 24.Section 1. 31. Wang. * ** 609 25. y = *19/4 27. and April Allen Materowski. . 29. 28. Published by Pearson Learning Solutions. by Warren B. 26. and Finance. Walter O. Economics. Copyright © 2007 by Pearson Education. 34. Published by Pearson Learning Solutions. 35. Wang. and Finance. Gordon. Copyright © 2007 by Pearson Education. 0) x Applied Calculus for Business. y x = 12 (12. Inc. and April Allen Materowski. * ** Section 1.610 32. x = *1 37.1 Answers to Exercises and Pre/Post Tests 36. Walter O. 38. 39. by Warren B. . Economics. 33. Inc.6/52 (b) y = 2 5x * ** 611 65. 82.7 46. (3. y = 24x . and April Allen Materowski. y = 4 63.1 Answers to Exercises and Pre/Post Tests 40. 5y + 2x = 20. 1 . t = . . no slope 52. (c) (a) (b) Applied Calculus for Business.9 67. . x = 1 60. y = . 4/7 59. Gordon.6 (b) x = . 130 81. 166 43. 24 55. x = 12 64. no slope 58.Section 1. y = 4 7x + 4 71. 5y . no slope 48. 62. .14.0). y = . y = .0.7 (d) They are horizontal translations of each other.62 (b) y = . (a) 3x + 7y = . y = 1.5 69. y = 1*2 x + 3 41.05x + 4000 80. . Each graph may be obtained from (a) by moving it 3 units to the left or right.8/5 50.6/5 (b) x = 3 44.2 56. Copyright © 2007 by Pearson Education.2x 68. x = .3 7x . 1*2 53.5y = 16 (b) 5x + 2y = .14 45. 10. y = 3x .3/7. (b) no (c) examine the slopes 79. 7/5 49. Walter O. 0 51. (a) 2/5. 76. (a) y = . y = .2x = 20.2 (b) 75.5x . Published by Pearson Learning Solutions. 10. (a) 2x . (a) y = .18 (0. 14x + 2y = 3 66. (a) 3 C2 2 AB 42. x = 1*2 (0.5y = 59 61. (a) 291/5 (b) ` C ` 2A2 + B2 AB 78.18 73. -6/5) (b) 7x .4 47. 54. Economics. 7x . (a) . x = 8. .3y = 16 74.8 72. 37/5) + 37 5 70. 0 57. 8x + 5y = 0. . and Finance. 02. 20 77. Wang. by Warren B. 5. 116. 116 14. Is a function 7. 3. -2 x1x + h2 -4 1x . Is a function 9. 76. * ** Section 1. 84. m 24. Is not a function 8. Copyright © 2007 by Pearson Education. (d) (e) (b) 0 (c) 4/3 (d) 12x + 12/14x . Is a function 4. D = 53. Gordon. The second line is obtained from the first by moving it h units horizontally and k units vertically.22. 31. R = 5 . 5. 32.2 Answers to Exercises and Pre/Post Tests 11. (a) . 5. 5. 96 R = 5 . (25/14. 10x + 5h . 196 R = 53. 7.1 (c) (f) They are each obtained from (a) by moving it 4 units up or down and simultaneously moving it 3 units to the left or right. 5. 116 33. D = 54. 4.32 20.1 1. 3 2. 46 R = 556 17. 2. 2x + h + 3 26.5.321x + h . R = 53. 76 13. 5. 2x + 2h . 11.5.1.1 (b) 7 (c) 1 (d) 18x2 . and Finance. 9. (a) .5.8 (c) 1 (d) 6x . and April Allen Materowski. 3.1 + 12x .16 15.32 1 1x + h + 3 + 1x + 3 2 12x + 2h . 9. by Warren B. D = 52.q 6 x 6 q (b) (c) .612 83. D = 5 . 5 23. 85. Inc. D = 52. 236 10. 2. 3. Each graph may be obtained from (a) by moving it 4 units Up or down. 87.b 2m2 + 1 Exercise Set 1. Wang. 7. 66 R = 5 . 1. 2.3. D = 53. 4. Is a function 3. Economics. .q 6 y 6 q Applied Calculus for Business. Is not a function 6. 2. 3x2 + 3xh + h2 29. (a) . Published by Pearson Learning Solutions. 3.32 (e) 1x + h + 12/12x + 2h .5 25. Walter O.5 (e) 3x + 3h . B . 76 16.1 (e) 2x2 + 4xh + 2h2 .2 1. 196 (b) (a) 12.5 18. Is a function 2. D = 51. 2. 4x + 2h + 3 27. R = 5 . 5.4 28. D = 5 . 126. (a) 2x + y = 6 (b) 2x . 46/7) 89. (a) . 30.1 19.5 (b) . Is a function 5. (a) 0 (a) (b) (b) 0 (c) 12 (d) 48 (e) 4x 22x + 1 21. 96 R = 5 .y = 4 (c) 4x + 3y = 2 88.1/3 (c) (d) They are vertical translations of each other. 2 4x + 7 x Z -7 4 12 a 6 12 a 19 a 7 19 1 h 6 1 h 6 6 h Ú 6 1. Is 38. Is 50.14 (d) 3x .q 6 x 6 q (b) (c) y Ú 0 36. by Warren B. (a) x Ú 0 (b) (c) y Ú 0 47. (a) C1h2 = c 25 + 55h 67 + 48h (b) $245 (c) $403 16 + 3a 57. and April Allen Materowski. 3 46.5/2 44. Economics.75 1. Is 48. (a) x Ú .q 6 x 6 q (b) (c) y Ú 0 * ** 613 35. h1a2 = c 28 + 2a 66 (b) 43 in (c) 58 in 58.25 80 56. Inc.9 (c) 12x2 + 13x . Not 37. . Gordon. Copyright © 2007 by Pearson Education. (a) .2 Answers to Exercises and Pre/Post Tests 34.q 6 x 6 q (b) (c) y Ú 5 41. and Finance. Is 52. x Z 3/2 43. Is 51. x Z . (a) . (a) . x Z .4/3. 0. Wang.q 6 x 6 q (b) (c) y 4 42. (a) x Ú 4/3 (b) (c) y Ú 0 Applied Calculus for Business.1/2 (b) (c) y Ú 0 55. x Z .25 + 0.3. Published by Pearson Learning Solutions.q 6 x 6 q (b) (c) . (a) . Is 53.x .50d d 6 1 d Ú 1 39. Is 54. 3/2 45.q 6 y 6 q 40. Walter O.Section 1. (a) 7x + 5 (b) . (a) C1d2 = e (b) $6. Is 49. (a) . (a) x (b) x 68. (a) 12 . Economics. Wang.5 78.32 (b) x Z . 63. (a) x 69. (a) 5 (b) . (a) x + 2 x Ú 1 (b) x2 + 2 .3 (b) . (a) x2 + 1*2 x + 11/2 (d) (b) x2 .x + 2x .2 75. f1d2 = g1b2 72.3x2 x Z 0 x2 60. Gordon. Walter O.4 62. . (a) x (b) x 67.4 .q 6 x 6 q (b) 4x2 + 12x + 12 . (a) x (b) x -q 6 x 6 q -q 6 x 6 q -q 6 x 6 q -q 6 x 6 q -q 6 x 6 q -q 6 x 6 q x Ú .1 70.2.1*2 x + 9/2 (c) 1*21x3 + x 2 + 5x + 52 2x2 + 10 x Z -1 x + 1 7x .3 (c) 2x3 . Inc. (a) 2x + 9 . (b) x x Ú . and Finance. 2 2x2 . 3 x + 2 66. (a) 5 (b) . Copyright © 2007 by Pearson Education. 3 2 1 1 2 3 Applied Calculus for Business.614 * ** Section 1. (a) 2x2 + 3x + 4 (b) .2 74. (a) x2 + 2x . 5x x is in the domain of f and f(x) is in the domain of g6 71.3 61. and April Allen Materowski.4 (d) (c) (d) 12x + 6 1x + 2221x .3x 121x .q 6 x 6 q 64.21 3x + 7 26 x Z .1x + 82 (b) x2 .9x . x Z .q 6 x 6 q 12 + 2x x Z 0. Published by Pearson Learning Solutions.2x2 + 3x + 10 (c) 6x3 + 14x2 .3/2 76. 59.6 77.5 (c) 6 (d) .2 Answers to Exercises and Pre/Post Tests 73.22 3x .6 4x + 2 2 2 (d) 2x .5 (c) 5 (d) . (a) 2 x . 4 65. by Warren B. 48x3 80.3 2 2 13.2y2 (h) .22 = 1. 89. f1x2 = c 14. f1x2 = x54. $33 6.3 86. R1x2 = 45x.4xy .000 . g1x2 = 3x2 .950t + 10. v = x + 1r. and Finance.2000.9k .90 per item 10. f152 = 8 2x . R1x2 = 1. (b) .2 87.2y2.4xy . f102 = . Land = $ 10. C1x2 = 0.4x .8 (c) 2x + h . p = . Copyright © 2007 by Pearson Education.9kyz2 .5. v = x + 2x + 1x. x = 2500 7. u = 4x + 3x + 2x + 1x. 9.2x + 23 84. $166. (a) 3. 92. For non integer values. (a) C1x2 = 2500 + 30x. . 12. 96y2 (h) 16x3y.9yk .) 1 -x 2x .000.75x + 4500.2 1.4xy . f1x2 = 2 3 x.11x5 .9 15. $30 and $3800. x L 4 miles. . (a) 864 (b) 1600 (c) 24x2y2 + 24xy2 + 8y2h2 (d) 16x3y2 + 8x3k (e) 96y2 + 48hy2 + 8y2h2 (f) 48x 3 + 8x3k (g) 24x2y2. Economics. Wang.7x 2 + 3x .8x + 1000. by Warren B.9 .3. C1x2 = 0.67 82.3y3l + 2z + l * ** 615 Exercise Set 1.3 Answers to Exercises and Pre/Post Tests 79.2y2 + h (d) . House = $ 90.26 yrs.9y2.3k2 (g) 2x .3 1.9y2z2 . 8 w = x + 1v.6y3z . f11. x L 16 miles 91.4x . g1x2 = x4 . 93. Walter O.2y2 (d) . f1x2 = d 6x . f142 = 3 (Either end points may have equality in 14 and 15. Inc.2xk .22 = 11/2.2y2 (f) .80 8. overhead = $ 1000.3z2k2 (e) . P1x2 = 0.7x . Marginal cost = $ 0.3 -1 x -1 6 x 1 f1 .52 = 1. Gordon.2xk . A1t2 = .2x + 19 85. It reduces to $80 3.2kx .000.25x 83.006x + 22.1 1 x .3/2.9y2 .Section 1. f11.1. C1x2 = 0.4x + 2000. g1x2 = x . (a) 218 (b) . and April Allen Materowski.0.8x .2500 (c) $266. 5.22 = 2. f1x2 = 1x.15 (c) 2x . f1x2 = x18. f1u2 = 1u.2x. f122 x Ú 3 f1 . $27 5. f1x2 = c 2 (b) $3. $500 4. u = 1w w = x + 3x + 2x + 1x.52 = .8 11.67 (b) 15x . R1x2 = 2. Published by Pearson Learning Solutions. f152 = 7 x 7 1 Applied Calculus for Business. g1x2 = 2x8 .3k2 (e) 4 + h .12 = . 2. r = x + 1x 88. Ceil1x2 = flr1x2 + 1 81.1 x 1 1 6 x 2 x 7 2 x 6 1 1 x 6 3 f1 . 4 . . (b) 1 (c) 1 x Applied Calculus for Business.4 1. C1x2 = c 2 + 0.000 20 6 t 0. (7/3. T1x2 = f 0. 1 (b) 2 (c) 2 3.700 6 x 311.18.000t + 800.25x .554.50 0. 3.3 x 100 100 6 x 1000 x 7 1000 C(x) 5. (b) . 1.000t + 1. 19.33x .08x 23.000 6 x 56.6. T1x2 = d 6. 18. T120. by Warren B.1 (c) .5x .51 rather earn $19.2.99. Walter O.700 0. 7/3) t .793.5x .800 6 x 114.65 6.24.650 6 x 174. Copyright © 2007 by Pearson Education.13 24. Inc.1 2. and April Allen Materowski. 1) (b) two intersections at (0. 5/3). 2.700 174.28x .616 * ** Section 1. . (11/3.35x .4 Answers to Exercises and Pre/Post Testss 6. and Finance. 1.300. 3. 2) and (10.3 (c) . Wang.650 114.1. Three.800 56.15x . Economics. (1/5. 0 10 20 T119.06x 22 + 0.6.50 0.0012 = $ 123.9992 = $ 116. 3/5). Published by Pearson Learning Solutions.819.50 0.2. Gordon.380 22. 6) 20.3 1.04x 20 40 (b) 3 (c) 3 0 6 x 14.9.3.19.5 6.999 Exercise Set 1. V1t2 = e .000. . 17.5x . (a) one intersection at (2.10x 0.000 0 21.5x x 6 10 x 6 20 x 6 30 x Ú 30 16.950 x 7 311.6 (b) .000 14.950 4.20. Published by Pearson Learning Solutions. (b) . Gordon.6 (c) 6 . Economics. (b) . 1 11. (b) None (c) None 9. 1.5. 3. 5 7. 214 2 17.4 Answers to Exercises and Pre/Post Tests 6. by Warren B. (b) . and Finance. 210 8. Walter O.2. 25 Applied Calculus for Business.Section 1. (b) .2 (c) 1 . 210 2 16.2. 0 (c) . 5 (c) .5.5. . 4.9.1. (b) .1 (c) .1. (b) 1.2 .2.1 12. Inc. 222 2 10.3.2 .4 (c) . (b) None (c) None 15. 1 (c) .2 (c) . (b) 1 (c) 1 13.2. 0 14. 4. * ** 617 (b) .1.8 .4. Wang. and April Allen Materowski.2.5. (b) . Copyright © 2007 by Pearson Education. (b) . .1 (c) . 1. h22 = 4p1y .8yh = 0 20. . (a) 100 (b) 31. (b) . y = 2x2 . and Finance. and April Allen Materowski. Economics. Published by Pearson Learning Solutions. Inc.5.000 (c) $1400 31.65 sec 27.5 sec (b) 6.4ac 6 0 (b) b2 . y = . Copyright © 2007 by Pearson Education. y = 500 ft 29. Wang. (a) 276 ft. 2145 4 26. by Warren B.4p1y . (a) 250 (b) $350.1. x2 + y2 . y = 363.2 (c) .000 32. 2. (b) None (c) None (3. x = 360 ft.7.5 Answers to Exercises and Pre/Post Tests (b) .4ac 6 0 and a 6 0 34. 1.k2 (b) 1x . C13.k2 38.8xh . Walter O. 0) 2. (b) None (c) None (4. C14. 4. (b) None (c) None Exercise set 1.3x 2 + 5x + 8 36.618 18.4ac 6 0 and a 7 0 (c) b2 . 22 r = 2 21.3x + 10 35. * ** Section 1. (b) None (c) None 25. x = 375 ft. (b) 71 (c) $1185 33. (a) x2 = 4py (b) x 2 = .3. Gordon.2. 1. (a) 1x .h22 = . -1) Applied Calculus for Business.5 1. 2217 8 2. (b) .4 1.5 . (a) b2 . y = 900 ft 28.2 (c) 5 .332 ft 30.8 (c) 7 .501 ft. 297 4 24.4py 37. 22 r = 3 23.3.2xy . x = 777. (a) 144 ft (b) 6 sec 19. and Finance. . Gordon.4. Economics. C13. C10.22 r = 5 7. Walter O. Published by Pearson Learning Solutions. 32 r = 4 6. 02 r = 5 Applied Calculus for Business.3. 02 r = 2 (3. .22 r = 3 * ** 619 (4. Copyright © 2007 by Pearson Education. C10.42 r = 2 8.-1) 4. . by Warren B.1.Section 1. C1 . and April Allen Materowski. Inc. C1 . .5 Answers to Exercises and Pre/Post Tests 3. -7) 5. C1 . Wang. Gordon. 52 r = 5 10.5 Answers to Exercises and Pre/Post Tests 12. . Inc. Walter O. C11. .22 r = 2 13. C1 . . Published by Pearson Learning Solutions. . 5/22 r = 2 9. Wang.4. Copyright © 2007 by Pearson Education. C1 .1/42 r = 2 Applied Calculus for Business. C13/2.3. Economics. C13/2. C12/3. by Warren B. 32 r = 4 14. and April Allen Materowski.4/32 r = 11/3 11. and Finance.620 * ** Section 1. (a) y = 35. Contradiction 19. Gordon. (b) 35 (b) the y-values on the tangent line 5 5 and on the circle near the point of tangency are almost the same 36.4ad 6 0 27. by Warren B. Point 22. point 24. Contradiction 18. Walter O.2 25 (c) 3 .4 (c) . (a) . (a) y = 3 . Copyright © 2007 by Pearson Education. Economics. Circle 20. Applied Calculus for Business. * ** 621 (c) (a) (b) (d) They are translations of the same circle 16.4ad 7 0 (b) b2 + c2 . Inc.4ad = 0 (c) a = 0.5 Answers to Exercises and Pre/Post Tests 15. Wang.Section 1. bc Z 0. Point 25. Contradiction 23. (a) 5 + 2 26 (b) 5 . and Finance.215 (b) 3 .5/62 r = 3 32. 12/3.2 23 29. and April Allen Materowski. Published by Pearson Learning Solutions. (a) y = 5/12xz + 34. Contradiction 17. (a) 3 + 215 (b) 3 + 2 25 (c) 3 + 27 30.2 26 (c) 8 (d) 2 33. (a) 2 23 (b) 4 (c) 2 23 28.25 x 4 4 169 12 12 169 x 33 (b) 34.2 23 (b) .27 31. Circle 26. . (d) b2 + c2 . 36. 37. . (a) b2 + c2 . Point 21. (a) 3 . Applied Calculus for Business.6 1. by Warren B. 44.5 1. Walter O. x = 16. Exercise Set 1. 40.622 38. Wang. 39. and Finance. Gordon. . p = 4 41. * ** Section 1. (a) Demand. (a) 3 221 5 3 221 5 (b) (c) 6 26 5 6 26 5 (d) 43. and April Allen Materowski. Copyright © 2007 by Pearson Education. Published by Pearson Learning Solutions. Inc. Economics.6 Answers to Exercises and Pre/Post Tests 42. p = 8 6. (a) Supply. and Finance. (a) Demand. Published by Pearson Learning Solutions. (a) Demand. (a) Supply. p = 4 8. Gordon. 94) (d) R1x2 = .444 Applied Calculus for Business. 2p +x *300 = 0 4. p = 10 * ** 623 3. (a) Demand x = 22. Inc. 50). Economics.6 Answers to Exercises and Pre/Post Tests 2.Section 1. p = 40 p +14x -490 = 0 Demand Supply -5p + 2x + 70 = 0 (c) (105/4. . by Warren B. Copyright © 2007 by Pearson Education. x = 110/7. x = 12. (d) R1x2 = . Wang. (a) Supply.14/5x2 + 98x (e) 2. p = 6 7.000 9. and April Allen Materowski.1*2 x2 + 150x (e) 10. p = 21 Demand Supply 8p *x *200 = 0 (c) (200. 5. Walter O. 14. 894 30. (a) 2300 (c) (11. (e) 146. 7x .491. 48) 26. (d) R1x2 .5x + 110. and Finance. Economics. 35/3) Demand 25. and April Allen Materowski. p = .18 (c) c1x2 = 100 + 2000/x (d) 100.1/121x3 + 6x 2 . 11940 32. Inc.80 (e) 4. (a) R1x2 = x 22000 . . Walter O. * ** Section 1. 2x + 4p = 21 27.5x (b) 0 x 400 (c) 35.33 million. n = 400 .600p + 2400 = 0 (d) 25. 73. 5) (d) R1x2 = x 236 . 350) 24.000 (c) 196 (d) c1x2 = x . Copyright © 2007 by Pearson Education. by Warren B. 2x .20/3x + 80.5.5p + 15 = 0 16. $345 33.742).500 31. x + 500p = 20000 18.75 x2 +8x +220 -11p = 0 Supply 29. (a) 60 cords (b) $25 15.624 10.x (e) 21 11. (a) 5040 (b) 99. 107/8) 21. (2. 26. (140/9.3/20p + 12 19.1. Published by Pearson Learning Solutions. (3. Gordon. 3) (d) R1x2 = x 216 . $4800 28.73 (d) 267. (a) R = . (50.20x2 + 1000x (b) 0 (c) $220 x 50 Demand x2 + 6x *384 + 12p = 0 (c) (5. Wang. which would exceed population of NYC (and capacity of system) Supply *7p + 2x + 7 Demand =0 (c) (7. (13/4. 345.5. p = . 20.x 3 Applied Calculus for Business.84x2. $12. (a) R1x2 = 20 1x (b) P1x2 = 20 1x .84 13.6 Answers to Exercises and Pre/Post Tests 17. 760/9) Supply 2.10 (c) 4 (d) 10 34. 2297. (17/3.3 + 2300/x (e) 120 (f) 1. (a) $40 (b) $65 (c) Perhaps everyone has a radio.x (e) 55 12. 9) 23. x = .5x . (b) 12. Economics. Copyright © 2007 by Pearson Education. and Finance. . 2 2. by Warren B. Published by Pearson Learning Solutions. 0. . Wang. 3/2. 3 3. .7 Answers to Exercises and Pre/Post Tests * ** 625 Exercise Set 1. 4 5. 2 Applied Calculus for Business. and April Allen Materowski. .3. .2.Section 1. No zeros 8. 3 6.7 1. 0.3. Walter O. 2 4. Inc. Gordon. 0.3/2. 4.3 7. . .2. 0. Published by Pearson Learning Solutions.3. Odd 20. x = . x = . and April Allen Materowski. . Inc. Neither 17. Copyright © 2007 by Pearson Education. . Even 14. Applied Calculus for Business.626 * ** Section 1. Wang. Walter O. x = 2 37. Neither 18. Odd 24. y = 0 41. Even 21. 39. Both 28. y = 0 47.3 34. . y = 0 42. Both 26. y-axis 27. y = 0 46. None. x = 3 33. x = . Origin 48. Odd 16. (a) violates vertical line test (b) Exercises 25 and 27 31. Even 23. 3 36. by Warren B. x = 5 32. y = 2 45. Odd 22. None 44. 0. Origin 30. Even 13. Origin 29.2 38. x = 5/2 9. 1 12. Neither 15. Odd 19.3 10. Gordon. 11. y = 3/2 43. 22 35. y = 2 40. Economics. and Finance.7 Answers to Exercises and Pre/Post Tests 25. 7 Answers to Exercises and Pre/Post Tests 49. Copyright © 2007 by Pearson Education. * ** 627 50. Gordon. 54.Section 1.1106 and 6. 52. Inc. 53.4. by Warren B. 55. x L . 56. Wang. Applied Calculus for Business. Economics.2186 51. and Finance. and April Allen Materowski. . Published by Pearson Learning Solutions. (b) twice. Walter O. by Warren B. and April Allen Materowski. .7 Answers to Exercises and Pre/Post Tests (b) 57. Economics. Published by Pearson Learning Solutions. (a) (c) (d) (e) (f) Applied Calculus for Business.628 * ** Section 1. Wang. Copyright © 2007 by Pearson Education. and Finance. Inc. Walter O. Gordon. Economics.Section 1. by Warren B. Published by Pearson Learning Solutions. and Finance. Walter O. (a) (b) * ** 629 (c) (d) (e) (f) Applied Calculus for Business. and April Allen Materowski. Gordon. . Inc. Copyright © 2007 by Pearson Education. Wang.7 Answers to Exercises and Pre/Post Tests 58. . and April Allen Materowski. Inc.7 Answers to Exercises and Pre/Post Tests (b) 59. (a) (c) (d) (e) (f) Applied Calculus for Business. Copyright © 2007 by Pearson Education. and Finance. Published by Pearson Learning Solutions. Gordon. Wang. by Warren B. Economics. Walter O.630 * ** Section 1. (a) (b) * ** 631 (c) (d) (e) (f) Applied Calculus for Business. Wang. Copyright © 2007 by Pearson Education. Economics. Gordon. Walter O. Inc. . by Warren B.7 Answers to Exercises and Pre/Post Tests 60.Section 1. and Finance. Published by Pearson Learning Solutions. and April Allen Materowski. Observe 2x . Wang.8/252 62.7 1.4. f1x2 = 0 65. . (a) 64. Gordon.7 2. Walter O. Theorem 2.810 Applied Calculus for Business.8 1. 3/2 (b) Exercise Set 1. 67. Yes. Inc. (c) y = 1108x + 10442 y = 4x*2 x = *3 y = 688x + 6092 x=3 (d) Widening (f) $9. Economics. Copyright © 2007 by Pearson Education.1.8/25 or y 7 0 (d) 11/4. 68. Published by Pearson Learning Solutions. No 69. . See Section 3. (a) x Z .632 * ** Section 1. x = *5/2 y = 2x *11 (b) 70. and Finance. also symmetry with respect to the x-axis. and April Allen Materowski. (a) (b) (c) y .8 Answers to Exercises and Pre/Post Tests 61.4 10 = 2 x + 3 x + 3 (c) 4267. by Warren B. there is certainly some correlation between number of wins and bating average. 8. and Finance.8 Answers to Exercises and Pre/Post Tests 3. Published by Pearson Learning Solutions. (a) 7. Gordon. (b) R = 0. Inc.976633 (d) 7. Copyright © 2007 by Pearson Education. From the data. and April Allen Materowski.608 (c) It is widening 4. (c) It is unclear. (a) * ** 633 (b) 1. (a) (b) 1077. For 1977 to 1982.985237 (c) It was unattractive in the 1980s.3 (c) 1997 (b) 9. by Warren B. (a) 5. Walter O. . Economics. (a) (b) The money spent on each form of advertising grows changes the same way. For 1983 to 1986 Applied Calculus for Business. the different forms of advertising could be reaching different consumers. Wang.Section 1.2 6. (c) 0. Published by Pearson Learning Solutions. (There is a lot of information on the web regarding this topic. Walter O. 16.75 x Ú 5 (b) Something in the country resulted in the decrease.067857x + 5. 14. and Finance. (b) f1x2 = e 12.44 x . (b) f1x2 = e 11. or some other reason. Gordon. .) 15.28x + 137. (a) (d) 10 0. Wang.0. (a) f1x2 = e 4. (b) Applied Calculus for Business.8 Answers to Exercises and Pre/Post Tests (c) 10. Inc.6 x 4 .75x + 179.5.996 x Ú 11 (c) The divorce rate in the 1980s decreased. 13. (a) It changes after 1980 1x = 212 36x + 428 if x 21 1100 if x Ú 23 (c) Riding became safer (perhaps because of helmet usage.634 * ** Section 1. and April Allen Materowski. by Warren B. (a) Chapter Review 1. perhaps better health care. Copyright © 2007 by Pearson Education. Economics.207x + 3. y = . 0.1 .2x 2 17. Gordon. .6x 2 + x + 12 (c) 3/2. (a) x 3/4 (b) 23.3x 5 .5/2. 14. by Warren B. (a) (7/2.3y = 6 8. 2x .1*2 x .3 (d) same as before. Walter O. (a) V1 . . 4) or [4. All x except 2/3 and 3/2 * ** 635 y=4 x = *2 x=4 15.3 5. 8] (c) A 5. y = 5/4x 1 2 7.4 (e) f1x .2. Economics.2x x + 2 2.5/3 (b) 7 (c) 3 11. Published by Pearson Learning Solutions. NA = not available x f(x) (a) f1x .8x 3 (b) .2.2. No.1 . Copyright © 2007 by Pearson Education.22 + 3 (f) f1x + 22 + 4 -2 3 NA -2 5 -1 NA 2 -1 4 NA -3 6 0 NA 1 0 -2 3 1 0 -6 6 5 1 -3 4 4 -1 -7 7 8 2 1 -2 7 3 -3 1 11 3 4 -3 9 6 0 0 13 4 7 1 NA 9 3 4 NA 5 9 4 NA 11 5 7 NA .1. 13. and Finance. . 1.Chapter 1 Review Answers to Exercises and Pre/Post Tests 2. (a) 9 + 2h (b) 5 + 2h (c) 4x + 2h .4/3 (d) . 102 (b) . x Z 3 y = *1 2 16 3.4. same x corresponds to different y 10.22 (b) f1x + 22 (c) f1x2 + 2 (d) f1x2 .7/32 (b) 6. q 2 5 . 32 or (3. 3/2 12. 24 18. y = x + 3 3 4. Inc.2x 3 16.4y = 17 (b) 4x + 3y = 6 9. 23 B 20. . (a) 3x . (a) . 0).2. (a) The first is a supply and the second a demand function (b) [2. and April Allen Materowski. Wang. x Z 4 4 . . 10. 25 (c) y Ú 0 (c) . [ . (a) $790 (b) $330 21. (a) (b) 3x2 + 10x + 8 3 .3. x Z 3 . 22 Applied Calculus for Business. 105 19. (a) .34. Copyright © 2007 by Pearson Education.52 34. 11. (a) 194 (b) 2x . y2 x2 + = 1 16 25 36. Inc.94 26. x 2 + y2 .2.52 (b) Turning points are indicated in sketch by T. it is a point. . Walter O. T T T Note: y-values are positive on 1 .4x + 6y . by Warren B.3 = 0. C12.3. Gordon. (a) (d) odd (e) even 31. 1.200 + 600/x. 29.636 * ** Chapter 1 Review Answers to Exercises and Pre/Post Tests 33. 3/2 35. Economics. (a) y = 5/12 x + 169/12 (b) on Tan. and April Allen Materowski.32 r = 5 x = -1 y=2 x=1 Zeros: x = *4. 100 25. . Wang. 50. . C12.2. (c) Applied Calculus for Business. 27. and Finance. 32 r = 0. (a) 24.3.995844 30. (a) odd (b) even (c) even 28.995833 on Circle 11. . (There is also one in the interval 1 . (b) 32. Published by Pearson Learning Solutions. (a) (c) 0 (1.4/3 4. (a) 12. (c) . Inc. Connect the segments. .166710 4. 1/4 17.01 .99 0. 16) 30.6x + 7 21. 1 2 1x (2. Gordon.001 2. (4.166710 5. 0 14. 6 13. 1 + 2/x2 2 1x (3.0001 1.30000 0.9999 1. Wang. 162. (a) Q mPQ (b) 1/6 11. 8) (c) 12 8. Copyright © 2007 by Pearson Education.7 19. 4) (c) 2 5. 2/x2 25.3/x 2 24. and April Allen Materowski.00001 . (a) 1 . .5 16. . .999 . 28. (a) Q mPQ 0.162 (b) 1 . . .1 Answers to Exercises and Pre/Post Tests * ** 637 Exercise Set 2. (a) Q mPQ 1.2.1 2. 92.299895 1.0001 . (a) 0 (b) The TL is the line itself which has slope 0.166662 5.2) (c) 1/4 7. 2x . 0) (b) does not exist 31. (a) 10. (a) 26.42 29. 3/4 3.001 .300011 0. 13. 0) y=0 9. 2 15.92 (c) 1 . (a) (2.9999 1.4. .999 1. and Finance.99 1.166713 4.Section 2.5 6.01 2. (a) (0. 12 18. (a) (b) 0.1666204 (1.3001051 4.9999 . . Walter O.42 (c) 11. and 27. Published by Pearson Learning Solutions. (b) 2 Applied Calculus for Business. . 1) 20. .3.999 0.99 .1 1. by Warren B. .162.999 .001 1.9999 .001 . Economics. 22.72 (b) 13.3 12. . 12. m 23.166710 5.01 (b) 4 (c) 1 (d) does not exist.0001 2. 14x6 + 7. 38. 1 .3. y = 5 -5 . (a) (64.6416 . (b) 2x (c) 1 (d) Does not exist 33.2 1. (a) 2. y = . by Warren B. . 28.2154 1/3 in f1x2 = x . .708 26.9995 34. 0) 22.132. 2 x 1/3 30.162 (c) x = .12x 3 .16 100.0001 .1 (b) (0.642 and (2.16 464. .0. Economics.866667 (c) 15. Wang. 1092. . y = 7 2x + 1 16.544 4. Gordon.00 464.1 4.00 21.16x + 81.177 (e) 0.638 * ** Section 2.2 (c) 2 34. Area = 2k 23. Inc.4642 (d) 0.4642 . a . y = 14. (b) 2x + 7 (c) 9 (d) 9 35.544 . (a) 0 (b) m (c) . Copyright © 2007 by Pearson Education.6x + 16 (b) y = .. and April Allen Materowski. 8) 36. (a) 4 4.828 (b) 0. 15x4 .01 . 2x 12. y = 4x + 1 25. 10 2.001 .4.. 1/12 4 .132 18.6 21.32t + 256 10.2 12 u 11. .0.b2 . 57x56 27. Published by Pearson Learning Solutions.001 . 12 x5 21 4x7/4 32.0. 0) (b) . the slope is becoming infinite as h approaches zero for f1x2 = x4/3 it approaches 0.3x 1/2 + 5.6416 . y = - 20. 29.2154 0.1 .0001 .2 x 5 x6 32. h mPQ . 114. 2 3x1/3 -2 x3 1 2 1x mPQ ..6x + 13.01 21.0464 0.p 22 2 + 3 22 100. 2 3.1 2. (a) 13. y = 6x + 13 24.0464 . b 4a 2a 1 79 x + 11 11 19..1072 17. . (a) 1 1 + 7 + 128x 3 42 4 x3 32 3 x2 8 32 8. y = 88x . (a) y = 4 16 x + 3 3 (b) 15. 24). 3.866387 33. 1 .1 0. 112 31. (a) (5/2.6 6 x 6 0 or 0 6 x (b) x 6 . y = . 1. Walter O.2. 12. (a) . (0. and Finance.2 Answers to Exercises and Pre/Post Tests 6. 13.b 4ac .3/x2 (d) 2/x 2 Exercise Set 2..6 3x4 Applied Calculus for Business. 3). (a) y = 6x + 4. 3/4 + 3 + 14v v v 9. 1 .129 575 145 x 3 3 25 56 x 3 12 15. 6x5 + 7x 2 . 6 21. (b) f1x2 = x is one such example.121x . Copyright © 2007 by Pearson Education. (a) 7x2 .252 f1x2 = 1x . (a) * ** 639 Exercise Set 2. 9/13 17. and Finance.3 Answers to Exercises and Pre/Post Tests (b) 4 (c) thy are the same 35. 31. Removable discontinuity at x = 3. (a) 25 (b) 25 (c) 49 (d) 49 (e) 59 (f) 60 (g) removable discontinuity at x = 5. 0 22.121x . 7 7. and April Allen Materowski.1 16.1 (d) 1 27. (a) 1 (b) . 5 9. (a) yes (b) no 32. 3 6. Published by Pearson Learning Solutions. for example.1 (d) 1 35. Function is undefined and therefore discontinuous at x = 5 30. (a) 0 (b) 0 36. 1 28. (a) 0. . (a) 7 (b) (2.3 (b) 2h . by Warren B. (b) 6 12.3 1.3 5.Section 2. 34.52 (b) x = 1 (c) removable 33. (a) 37. Wang. r + s 2 29.1 (c) . 0.9/2 4. . 5) 36. 5 2. 2x 24.1*2 19. 9/2 14. 1/8 13. (h) (b) nonremovable discontinuity at x = 0.4 11. (a) 0 (b) 1 (c) .52 1x . Non removable discontinuity at x = 2 and 3 Applied Calculus for Business. Gordon. 9 3.2 2. non-removable one at x = 9. Walter O. (a) Any graph without holes or jumps will do. . 16/25 23. (a) 1 (b) . Continuous function 38.1 (c) . Economics. . . There are many possibilities.15 8.3 15. (a) 1 (b) 0 (c) 1 26. f1x2 = or 2 2 1 x 1 21 x . Inc. 2 23 20. 1 2 21 + x 25. 18 10. 1/6 18. discont at x = 2 not diff. f102 6 0. (a) 0 (b) 0 10. Inc. 54. there either. and Finance.1 6 s 6 1 61. (a) 6x + 5 (b) 4y .q 9. Removable discontinuity at x = 1 and 2. . Wang. (i) Rem disc at x = 0 (ii) Not diff at x = 0 (iii) d102 = 2 (iv) not diff. (a) (b) x = 1 (c) f112 = 1 (d) no. at x = 3 (c) and (d) cont. Non-rem. min at x = 0 77. everywhere except x = 3 (iii) f132 = 6 (iv) yes (c) everywhere cont.121x 2 . everywhere not diff. No 68.1 (c) max at x = 3. Everywhere discontinuous 70. 68 (b) it is possible. (b) x = 27 71. 0 12. Published by Pearson Learning Solutions.01. Continuous 46. (a) yes (b) yes (b) possible but not guaranteed 74. Economics. (a) 6xy2 (b) 6x y 64. 68 76. Continuous 41. Continuous 45. Non rem.q 6. disc at x = 3 (ii) not diff at x = 3 (iii) L132 = 2 (iv) not diff.11 65.001 x 75 so Ex 75 is applicable 39. Non remov. (a) 1/1000 (b) q Applied Calculus for Business. elsewhere. 60. by Warren B. (b) 0.3 6 P 82.4 1. not diff. 3/2 2. (a) 0 (b) q 18. (a) and (b) (i rem. (a) 0. Gordon. (a) 8 (b) .121x . P/32 among other choices Exercise Set 2. f132 = 27. at x = 3 and 4 63. (a) q (b) q 15. if . 58. except at x = and 4 55. (c) x . 8 11. At x = 3 57.22 1x2 . Disc at x = 5. diff everywhere except at x = 5. 73.121x . . differentiable 53. f122 = . 5/6 3. (i) Cont. Continuous and diff. Continuous 47. (a) f1x2 = (b) f1x2 = 1 1x . (a) q (b) q 16. At x = 2 and 4 (ii) diff. f122 = 4 not diff. 62.1. non removable 50. Copyright © 2007 by Pearson Education. 59. 24 + PB or x .640 * ** Section 2. for x 7 1 (ii) not diff. Yes. Walter O. C is continuous on 2 81. but not guaranteed by the theorem. f122 7 0 72. (a) and (b) (i) disc at x = 3 (ii) diff.4 Answers to Exercises and Pre/Post Tests 66. differentiable 49.42 1x . and April Allen Materowski. . (a) . 48. it follows from Ex.2 6 min A 24 .2 42. f122 = 4. 56. (a) Follows from Ex. 5/2 8.2 6 min11. Non-rem.q 17. 8/9 13. 51. (a) (i) continuous everywhere (ii) not diff. Non-removable discontinuity at x = . (i) non-remov disc. Cont and diff for all x except x = 0. non-removable at x = 3 40. At x = 5 44. (a) q (b) q (c) q (d) . q 5.q (b) q 7. (a) q (b) . at x = 1 (b) (i) non-remov. disc. At x = 0 43.22 2 (c) x .P. 52. 0 4.q 14.3 2. disc. disc at x = 0 continuous and diff. and diff. 2 (b) x = 4 (c) y = .Section 2. Inc.2 (d) 21. Economics. (a) x = .1 (d) 25. (a) x = 0 (b) x = 3/2 (c) y = 1*2 (d) 22. (a) x = 0 (b) x = . by Warren B. (a) x = . Copyright © 2007 by Pearson Education.1*2 (b) x = 3 (c) y = 2 Applied Calculus for Business. Published by Pearson Learning Solutions. Walter O. (a) x = 0 (b) x = 1 (c) y = 2 (d) (d) * ** 641 24. . Wang. 1 (c) y = 0 (d) 23. (a) x = 2 (b) x = . and Finance. 1/4 20. (a) x = 3/2 (b) x = 1 (c) y = 2 (d) 26.4 Answers to Exercises and Pre/Post Tests 19.1 (c) y = . and April Allen Materowski. Gordon. 642 * ** Section 2. 2 (d) 29.2 32. (a) 1 (b) . (a) x = 0 (b) x = . and Finance.2. Copyright © 2007 by Pearson Education. (a) x = .1. . 2/3 (b) x = 0. 1.1 33. Published by Pearson Learning Solutions. Economics. 3 (b) x = . 3/5 (c) y = 12/5 (d) 27.1 (c) y = 3 (d) 36. (a) x = 0 (b) x = .3/2. and April Allen Materowski. Wang. (a) 2 (b) . 1 (c) y = . Walter O. (a) x = 0 (b) none (c) y = 4 (d) 34. 2 (d) 30. by Warren B. 2 (c) y = 4 (d) 35. (a) x = 0 (b) x . Gordon. 3/2 (c) y = 1/8 Applied Calculus for Business. (a) x = 0 (b) none (c) y = .4 Answers to Exercises and Pre/Post Tests 31. (a) x = . Inc. (a) x = 0 (b) none (c) y = 0 (d) 28. x 5120x 4 . (a) A 1. 02. and April Allen Materowski.577.23297 12.21x + 242 5. by Warren B.221x + 22 25. Wang. 3/17 14. Inc. 2. 1 .9851.07/x25x. 0.17. and Finance.92 2. 2) 18. x L .x2 1x2 + 122 (b) y = x. (c) y = 0 (d) 215 5 (b) x = . 1*2 B (b) (1. (a) x = 0. y = .33. 2x/1x 2 + 122 7. y = 52/3x . Published by Pearson Learning Solutions. (c) y = 2 (d) 1 . 21x217x 4 . 46. y = 3x . 22 (b) A .4 2.2 (b) x = .13/13x . 4x 5 .6/x 3 10. 19. .6738 47. .22. Let f1x211 + 0.84/5 15.1.3x 2 + 16 3x5 5 39.2 (c) x1x .21x .72 4.0. x L .3 4x2 2x . (a) f1x2 = 21x + 321x .12x + 0. (a) x = . 0) 1 .08x + 0.1. Copyright © 2007 by Pearson Education. Economics.1*2 B . .64. y = 2x + 3 42. 625/16). (a) 12 (b) 4 Applied Calculus for Business. (a) y = . 2x3150x 6 + 135x 2 + 82 3.5 Answers to Exercises and Pre/Post Tests (d) 44. 0. . 1*2 B 41.121x . 217 8 45. 12x2 + 722 9. * ** 643 38.1.Section 2. Gordon.18/1x + 322 4x12x3 .1.7590.5 1. . A 1. (4. 11.83585 (b) x = 1. y = 6x . Walter O. 11. .2 43.1. (3/2.54 23.52 8. (a) 1 . evaluate for large values of x.3x + 1 13. 2x13x4 + 4x 2 .424 Exercise Set 2.2x 2 . y = 1*2. (a) 17.2 16.x . y = 0. .52 28.222 6. y = 0. 262 3 2r2 .3 . 0) (c) [f1x2]N has horizontal tangents at the zeros of f and at the horizontal tangents of f. g1x2 = 5x3 . f1x2 = x + 5x + 5. Gordon.42102-1/2131x2 . g1x2 = 3x . f1x2 = x . Exercise Set 2. f and g have derivative zero at the same x-values.422 + 21x 2 .272 12x11x2 .5 2. by Warren B.296x + 8 29.59t2 + 4t .1 12. (a) x + 2 1x + 1 (b) x + 1 3 . (a) x (b) x 8.644 * ** Section 2.2 2 2 26. and Finance. 10. 412x + x 4211 + 2x 32 3x2 3 + 2 + 44.72 14. y = 1*2 x 28.321x 2 + 2x + 82715x 3 + 6x 2 . 41x2 .4x .12 33. -1 1x . Wang. (1.115r 2 .100u 17. 13/2.221/3 4 40. 2 22x3 + 3x + 2 .1/4152. (a) 7y6 (b) 7y6 3. g1x2 = 2x . 2. f1x2 = x3. y = .4 (b) 5x551x2 .13 x + 16 6 dy dx dy dx + 5x4y7 30.f ¿ 1x 2 .2x 4.7 9. (a) 3 . Economics.1 x6 + 1 b 4 34. 415x2 . 2x x2 + 1 36. 15u2 + 923/2 18. Inc. and April Allen Materowski. 1 .122 21. 39. y = . g1x2 = 3 9 7 (c) 71x5 + 12y6 31. y = 108x . f1x2 = x3/5.6 1. (2. 2xS21x22 37. 1x 2 + 124 20. . 313t2 + 121t3 .13t + 12 15. 02 2 23x + 2 1x 2 32x + 23x + 2 1x 16689645 26 45. 13t2 + 2t23/2 312x2 + 12 16. (a) x + 5 (b) 1x + 52 2.3x + 723 (b) 2x9 . 2x51x2 . (a) 3y2 + 6y (b) 13y2 + 6y2 dy dx dy dx (c) 30y1y + 221y3 + 3y2 + 429 32. Published by Pearson Learning Solutions. (a) 10. . 1 1x 2x + 3 2x . 2x1x . g1x2 = x 2x 2 + 4 13. g ¿ 1x2 = 1f1x22 y-values are reciprocals.6 Answers to Exercises and Pre/Post Tests . h1x2 = x2 .1/42. 22. (a) f1x2 = x1/2. 3x2 a x6 . 13x 3 + 923 38. 32 Applied Calculus for Business. g1x2 = 5x3 .142 24. 1712x + 321x2 + 3x + 1216 3110x . Copyright © 2007 by Pearson Education.1x + 42 (b) 6 5.4292 (c) f ¿ 1g1h1x22g ¿ 1h1x22h ¿ 1x2 42. .927111x2 . 4102.122 19.x 10.423 + 1x2 . Walter O.3 31t2 .7x + 3 10. 5t2 .2x x 1 (b) 3x . (a) 1/x (b) 1/x 6. 6x2S212x 32 43. 25.7x + 221/4 . and the 2 23. 41. (a) 4x . (a) 12x3 .123/2 2x + 1 r135r2 .42. 12 4 1x + 5213 13 1 6 1x + 5211 11 (b) 13/2. 0). 2xE1x22 35.4 11.6t + 126125t4 . f1x2 = . .189 27.122/3 82 3 4t1t + 32 31t + 421/3 29.3x3 + 7 7. 8 million 1 5. R ¿ 1x2 = x + 7 4 2 (c) 6.002 (d) 10. 5 21. and Finance.181. 16. (a) (i) 1060.32t .16t2 + 176t (b) 112 ft/sec (c) . 3 19.Section 2. Copyright © 2007 by Pearson Education. and April Allen Materowski. (a) s = .20) -5 4.6 2.48 9.5/48 22. Published by Pearson Learning Solutions. m 20.1/10x + 30 (d) 35 5(e) 5(f) 300.9 At x = 14/5.10 (ii) 1070 (b) (i) 9.66 sec. Walter O. (b) 640.7 Answers to Exercises and Pre/Post Tests * ** 645 Exercise Set 2. 2.16h + 176 R(x) = -5/4x2 +7x R' = -5/2x + 7 (e) .02 ft/sec 14. (a) (i) 240 (ii) 241.996 ft/sec 13. by Warren B.16h2 + 176h (d) . 654.5 sec. 3 17. Gordon. (a) P1x2 = .500 20 (b) 2000 (c) P ¿ = . Each answer is 3.90 (b) 1. (a) p = x + 7 4 -5 2 -5 (b) R1x2 = x + 7x. Inc. . Profit 7. 5.32t + 176 (f) (I) 144 ft/sec (ii) 80 ft/sec (g) 5.208 ft/sec 11. 15. time to reach maximum height (h) 484 ft 10.90 (ii) 9. .50. max.80 (or 2. m see 14. (a) 80 (b) 16 (c) 64 8.000ft 12.90 (c) 1. maximum profit Applied Calculus for Business.00) (d) 0. (a) 576 ft (b) 12 sec (c) .1/5x + 40 (f) 200.90 (c) 0.10 (or 0.70 2.32th . (a) in 200 sec.x2 + 30x . . Wang. . 656 (or 652) 3. the rate is the slope.1000 10 (c) P ¿ = . 20 18. (a) 112 (b) 16 (c) . .7 1. R is maximized (d) $3.128. (a) P1x2 = (b) 2000 (d) 50 (e) 1 2 x + 40x . Economics. 4in.5 ft/sec 30. 42 9. 5 7.17/22 (b) 1 . Economics. (a) 3x2 (b) x/4 x 34. Gordon. (a) y = 4 1 5 29 1 30 x + .2 + 25 (b) 1 2 25 . 64 22 L 90.a k 1/3 b 1x .12 (b) t 7 2 24. (a) .6xy 11.1 or t 7 3 26. 14.9 Answers to Exercises and Pre/Post Tests 13.12 (c) x2 + 1 33. 1 3.12/m . (a) 23 2 x (b) 23 6 35. 4.2y2 . . 2.1221x3 .3/2.5/72 39. . 3/16 37.8xy . 4in 13. . 13/2) (b) 1 . Wang. 0 4. 3in. (a) . g = 120° 3.2x2y32 Exercise Set 2. 80 26 L 195. therefore f1x2 = 0 at least once on (3. (a) y = 3 5 x + 4 4 217 b 3 (b) a 2. . 15 22 8 22.7 2. a = b = 110°.96 ft/sec 31. 54in 8.h2 + k h (b) x1 = a2/3h1/3. .x + 4 4 3 25 (ii) y = x 4 4 1 7 20. (a) (b) + S0 29.6x 3y2 2y211 + x3y22 12.6t . b = d = h = 65° 4.8 (b) t 7 1 25.3x 21x 2 .y/x or . 2 x 13 . (c) When tan line is vertical.s2/t2 or s 1 (b) .2xy3 4xy + 3x2y2 + 5y4 + 2y y1/31216x 5/4 + 36x1/4 . Copyright © 2007 by Pearson Education.x + . by Warren B.1/3 x 6x2y3 + 4y2 . 3x2 . a 28. (d) v2 0 64 2 15. 3. .162 211 .t (c) it becomes infinitely large 17. but f13) 6 0 and f142 7 0. (a) (i) y = . (a) 3t . f1x2 = x3 . Inc. (a) 4.9 (b) t 6 . (a) 9x 23x2 + 9 . 15/2.y 8. 80ft 14.3/8 2.v32 3sv2 3sv2 211 . y = .7/x 2 2x . (a) 3x21x3 .12 (b) 6x21x3 . . x x15 + 6y22 9. (a) 125/2.32 3 13 10.8 1. . 1/4 5. and April Allen Materowski. Walter O.t2/s2 or 18. g = 70° 2.3x1x2 .12x . (a) y = .y = . a = u = 115°.6x22 y1/3 10. 2x . y11 . (a) y = . .8 ft (b) 12.6x2 . (a) (2.646 * ** Section 2.9 1.272 x1/41240y10/3 + 168y4/3 .1 = 0. 9y 7. .20 36.9/2. a = b = 60°. and Finance.9 38. 9.1/3 16.8 ft 12. (a) 8t . y = 79 362 x + 373 373 3 25 19.y = .x + 5 5 2 2 7 7 Applied Calculus for Business. y1 = a 2/3k1/3 distance = a Exercise Set 2.x + 2 8 11.x2/y2 4x 6. Published by Pearson Learning Solutions.122 23x2 + 9 (b) x2 + 1 32. . (a) 6t . 4) 23. .v32 1 s 1 t t 1 23. 12 5.x + 3 3 7 1 (b) y = x 3 3 21. 11019. 0.1. by Warren B. 17/3) 18.x + 3 6 2 3 (b) (5. 1/6 7.532089. 2. 828/3721. 1. . 72 2. 0.174802 11. (a) 8x3 . 1.x + 3 3 3.87x . Inc. .258259 2.334.x + 5 5 Exercise Set 2. y = .56% Chapter Review 1.0. 6 3.8155/18 19. dx/dt = . 11/3) 17. (a) dy dy dx = dt dx dt * ** 647 (b) - 2x 3y2 8 11 8. 5/2 6.22072 8. 80.146121x .203564x .3. ym = m + n m + n 1 2 7 13 21.1989 ft/sec 16p Exercise Set 2. . (a) 3p/5000 cubic km/day (b) It will increase by 10 km 21. 3312/21025 million per year 18. Both increase. (a) 2 2x 3 + 1 (b) y = 20x .879385 6.31.4526269.443357 (e) 2. f ¿ 132 = 0. 5 22 (c) (13/3. (a) decreasing at 1/6 lb per week (b) $0 per week 17. 32/3 ft/sec 24. (a) .6x + 5.814321x + 0. dy/dt = 3 dy dx + y x dt dt dz 8.3399. larger at 2 m/sec. . .2. 3.1461.1.2. method will not work. 9680/3 (b) 4690.2732 ft/sec 23.0. If x0 7 2. 1. x = 5. shorter side at 4 m/sec. . (a) 21x3 + 1023/2 Applied Calculus for Business.347296.1.164035 5.3 (c) y = 13.x + 4 4 13. 256/767 5. 229. Economics. . .23. 2.9 2.sec 26.10 1. (a) 1*2.2207 (b) f1x2 L 51x + 1.28 in/hr 10. (a) 8p L 25.1179 in/min 12 12.0. 9.11 Answers to Exercises and Pre/Post Tests (b) 241.2 in/sec 11. . 48p L 150. 8/3 ft/sec. (16/5.2 (b) y = . f ¿ 122 = 0. 24/13 ft/sec 14. (a) 1/12 in/min 22 (b) L 0.796 sq ft/sec 16. 320 (ii) 410.4/3 x17x 3 + 42 7.259921 13. xm = 27. Copyright © 2007 by Pearson Education. f13) 7 0 so a root is in (1. 121/12 million (b) 92/25. (a) = z dt (b) 50mph 9. in/hr (b) in either case at 2p L 6. shorter 25. and Finance. (a) .11 1. Wang. and April Allen Materowski. 36/5) ny1 + my2 nx1 + mx2 .626577 3. 620. . 49/6.339921x . 10 L 0. They are the same. Longer side is increasing at 14/3 cm/sec. 4/p L 1. r = 7. Walter O.5 3x2/3 x 68 95 (b) y = .000 psi/sec 20. 2. (a) y = . 10. (a) 2 (b) 25/6 6. (a) . 2.2511896 (d) y = 13.29.28 x1x3 + 402 8. If x0 6 2 they converge to smaller root. Gordon. decr. (a) (i) 820.35 units/sec 22. 1 10 + 3 2 1x x 25 35 (b) y = x 2 2 20 8 4. Published by Pearson Learning Solutions. iterates converge to larger root.903654 4.133 sq.4x + 7 (c) 1 . y = .x + . . 27 2.Section 2. (a) f112 6 0.2x . (a) 1*2 (b) . shorter decreasing at 5/3 cm. f ¿ 122 = 0. 3) (b) y = 10x . 15190/3. (a) -4 1x + 122 1 7 (b) y = .8143.8174032.4 4. . 5/6 ft/min 15.1. 2. (a) .3. (100/13. s = 61*4 (c) 0 6 t 6 1*2 13. (a) 3.292 ft/sec 14. 5 ft/sec 18. y = 7 25 x + 8 24 23.3. $24 3 11. 3 (d) f1 .1/6 20. (a) 6t . and Finance. Inc.648 * ** Chapter 2 Review Section 3. 3 ft/min 17.1 Answers to Exercises and Pre/Post Tests 13 7 x 18 18 9x2 21. .25 ft (c) .25307 Exercise Set 3. Walter O. Published by Pearson Learning Solutions. Economics. Wang.92 10. 16 ft 15.32 = . y = 19.x 2 + 12x 5 (b) $3. 3 (c) . (a) 212x3 + 523/2 2x3 + 1 (b) 213x + 4251x2 .1 1. (a) 1320 ft (b) 1332.12t2 (b) t = 0. (a) (b) y = 9. 172/13) 16. Copyright © 2007 by Pearson Education. 1*2. 4. 1. Gordon. (b) 1 (c) 1 (d) f112 = 1 Applied Calculus for Business. by Warren B. (b) .60 12. (a) 5 9 x 4 2 (b) everywhere continuous (c) not differentiable at x = 1 2. and April Allen Materowski.124124x 2 + 20x . 2) 11. . Dec on 1 6 x 6 3/2. (1. F (d) C. m = . m = 0 10. (a) D (b) C. m = 3 (c) M = 6. 32 20. 0 B . inc on t 7 . None 22. Dec on x 6 1/3.9/7.13 6. 1 . and Finance. 1 . 1 . Yes 1 (b) they are the same 9.1/42. everywhere except when x = .q . Dec on x 6 0.1.5112. 137. 11. D. 02 A 0. 11. 2) 19.899/1282 17.b2 . G (c) B. 2) 12.2 M = relative Maximum. 1 .1 13.121 31. Exercise Set 3. 4ac 4a B 14. m = . 1/42 23. 732. m = . m = . 52. m(1. inc on x 6 1 or x 7 3 8. inc on x 7 0 10.82. inc on x 7 1/27 12. (a) M = (b) none 32 4 3 . Inc.2702 15. (a) no Max. inc on t 7 0 5. Gordon. (a) M = 18.4/9 B 24. (2/3. Always decreasing 2. m = 1/4 39. m = 2 (b) M = 6. 1 . (a) M = 1*2. (3/2. M = 0. 335/27) 16.1.1. (a) M = 14. Dec on x 6 2. 13. (a) No (b) No 36.2 23. . (b) M = 39. inc on 1 . 02 1 Applied Calculus for Business. 0) 13. (a) M = 3.3/2. no min 30. inc on x 7 0 6. (0. Dec on 1 . . inc on x 7 2 11. m = 0 (b) M = 1*2. Published by Pearson Learning Solutions.2ba. . 10. m = 2 32. Dec on its domain 1 .2.Section 3. Dec on x 6 0. (a) M = 34. by Warren B. Dec on 1 6 x 6 3. m = relative minimum. Dec on t 6 .1 3. 0 B 18.3.2. (a) always increasing (b) always decreasing 3. (a) (b) M = 18. m = 2 34. and April Allen Materowski. A .2 Answers to Exercises and Pre/Post Tests 5. . M = 4 26 9 . 25. (0. 02 21. Wang. m = 2 (b) none (c) No Max. A *2. 42 14. 11/4. no min (d) M = 1*2. m = 9 7. * ** 649 m = 0 35. 0) 27. inc on x 6 1 or x 7 3/2 9. Inc. m = 6 (b) None 29. 4) 28. m = . 1 . 15. A . (0. E 8. A 2 23.2 7. Dec on t 6 0. N = neither 1.72. 1/27). 13/2. (a) No (b) No (c) Min at (1. 2 23 B . q 2 15. 02. A 0. (0. Walter O.1*2 (c) M = 1*2.1*2 40. . 3 2 3 3 B . 4) 26. Copyright © 2007 by Pearson Education. 1 .3. (8. Economics. 1).1/27 6 x 6 02 or ((0. 45/4).1*2 33. 37. 10. inc on x 7 1/3 4. . (a) M = 18. 32. m11. 1.3. . and Finance. Walter O. m1 . m111/4. M10. 45/4). . and April Allen Materowski.899/1282. Copyright © 2007 by Pearson Education.1. 2 23 B 22. b2 16. M(0.82 21. 137. m(3/2. 02. A . m21 . 26. Inc. M1 . 28. Published by Pearson Learning Solutions. 335/27) 20. 32 24.2 Answers to Exercises and Pre/Post Tests 29. m(0. Applied Calculus for Business. 32. m15. 2) 23. Gordon. 30. 33. . 5). M1 .9/7. 02 25. M(2/3. by Warren B. 27. 31.650 * ** Section 3. Wang. 732 19.2702. N(0. m A 0.2. Economics.72.5112. 0) 17. N1 .2ba . . 4ac 4a B is a m if a 7 0 is a M if a 6 0 18. . 0 for n Ú 5 2. 0. 12a4x2 + 6a3x + 2a2. f (0)) (b) yes 19. (50. 15.8. 4a4x3 + 3a3x2 + 2a2x + a1. f(2)) (-1.3 23 2 .1 30. otherwise CU Applied Calculus for Business. v = 15t2 . CD if . 4x3.12t2 (b) 0. Gordon. Walter O. 28) 29. Published by Pearson Learning Solutions. -9 3 2y -9 3 2y 3/2 4 3 2 3 2 27.15t4. . a = 30t .Section 3. 8 B 33. (a) .3 1. -1 3/2 411 . I3 I2 M I3 I1 m 21. a = 8 . 70 cents (b) yes. CU if x 6 . 1 B 26.1 6 x 6 1. 12x2. and Finance.4t3. 3/x 12. the rest are all 0 9. Inf 14. See 19(a) 22.1 6 x 6 1. CU if x 6 16. and April Allen Materowski. a = 18t A . 5a5x + 4a4x + 3a3x + 2a2x + a1. (a) 20x3 + 6 (b) 24x2 + 6 3. they are all 0 8.2.60t3 (b) 0. CD if x 6 . (a) (2.1. CD if x 7 . CU if x 6 5/6.2 or 0 6 x 6 1 I1 .3 23 2 or 0 6 x 6 3 23 2 . CU if x 7 1*2 I A 1*2. CD if . v = 9t2 . 1542. (a) 0 (b) 0 (c) 18 (d) 0 (e) 0 7. 26 3 * ** 651 (0.12 8 2 5 I1 m 20. Wang. Copyright © 2007 by Pearson Education. 3 -2 8 B . (a) CU (b) CD 25. Always CD 24. (c) (5. 22 2 (-3. CU if . CD if x 6 .24 3 13x + 12 2 6. 0. 50) 37. . it can have both. 02 and A 3 23 23 2 . by Warren B. 6 * 12 38. I1 .24/x5 (b) 4. 10.2 3.3 23 2 6 x 6 0 or x 7 3 23 2 . Inc.3 Answers to Exercises and Pre/Post Tests 34. 302 and (1. 120a5x + 24a4. 8 2 3 B 11. . 3/x3/2 10.4. 10. (a) 22 2 26 6 (1. 20a5x + 12a4x + 6a3x + 2a2. CD if x 6 1*2.f (5)) M I4 I5 Exercise Set 3. f(1)) (0. 24.1/y3 32. Always CU 23. 22. CD if x 7 5/6 I2 I4 (c) 6 (b) 201x + 12 119x + 12 . (a) v = 8t . 24a4x + 6a3. 9 ) 2/3 (c) 26 3 (d) 0 17. 120a5. a = 30t 16.1.162 and (1. CD everywhere 3 31. (3. (a) v = 15t2 . 2) 28. 60a5x2 + 24a4x + 6a3. CU if . 13. f(-1)) 36. CU if x 7 1. 24a4. otherwise CD I A 16. 24x. 1.t2 213x2 + 521x4 + 10x2 + 12 2 1 x . (a) 5.2 6 x 6 0 or x 7 1.0) 35. Economics. 0) (c) (d) 18. Economics. 2) 41. 162.3 10 . 9 3 B 51. Published by Pearson Learning Solutions. . Walter O. CD on . . 54. 335/27) 42.3. CU if x Z 0 40.2 23 16 23 . . and Finance. 2 23. M1 .3 Answers to Exercises and Pre/Post Tests 49. 22 567 22 22 I1 A . m(1.3. Additional x-intercepts at . m11.1. Always CU 39. mA6 + 2 23 . Gordon.144 B 52. 2 23.272 43. 215 5 Applied Calculus for Business.16 23 . M1 .1/42. and April Allen Materowski. 02 44. 02 and (243. MA6 50. m at A . 9 3 B. m11. m1 . . . .652 * ** Section 3. 1012. Always CU 37. CD if x 6 0 or x 7 243 CU if 0 6 x 6 243 I10. 2 23 48. CD on x 6 3 35. 1082. 46. m1 . 02. Copyright © 2007 by Pearson Education. m at A . I3 A 3 10 .119 B 53. Wang. 1/42 45. 5000 B . m1 .2. Inc. 45/4). 27) 38. by Warren B. 47. . M(2/3.3/2. m(3/2.3 6 x 6 3 36. M13/2. 34.567 22 5000 B other zeros x = . Walter O. . Copyright © 2007 by Pearson Education. 0). Note: Vertical tangent line at (0.3223 . 8 . 59. by Warren B. and April Allen Materowski. 56. * ** 653 m A . 2 22 58.Section 3. Inc. 61. 27) I1 A . zero at x = 32 Inflection at (243. 62. 8 B Applied Calculus for Business. . Wang.3. 60. Published by Pearson Learning Solutions. I3 A 3 2 2 . Economics.3 Answers to Exercises and Pre/Post Tests 55. I2(4/5. Gordon. 2048/3125) other zeros: x = . and Finance.16 23 9 B 57.23 8 3 23 B . 3ac 7 0 66. 0). (7. y = 24 12.5) (-4. a2 = 10. f(3)) 18. 0). 2400 ft. Published by Pearson Learning Solutions. a5 = 1 Exercise Set 3. radius = height of rectangle. y = 40 11.5)) (. 71. 0) (d) neither at (1. 5 * 5 * 2 15.4 Answers to Exercises and Pre/Post Tests 69.5112 73.1) (5. r = 16. f(1)) (-. d = 17. f(-1)) 75. (6. a0 = 1. and April Allen Materowski. f(1)) y = 60 p + 4 (2. m1 .3) (-1. f(5)) 67. (a) neither (b) m(2.137.5)) m (0. h = 24 1/3 p m (3. x = 20. . Copyright © 2007 by Pearson Education. x = 13. (a) 100 * 200 (b) 100 * 100 8. f(0)) 65. and Finance. Economics. (a) h = 2r. 0). (0.7) (0. . M (2.4 1. a1 = 5. 64. a4 = 5.-2) 68. x = 60.f (-.3. p 3 2 3p 30 p + 4. (50.5. Wang. 50) (5. 36 * 36 3. 15 * 15 7. Walter O. f(2)) (c) doesn t look sleek. f(-1)) M (1. f(0)) m (-2. 0) (c) M(2. a3 = 10. m (-2.3 3. Inc. 3 80. M (-1.f(. 10. Gordon. 5. 2 cm 3 1/3 . f(-2)) 70.654 63. 6 * 6 * 3 14. 02. (a) f 1x2 (b) 2f 1x2 76. M (1. 10) Applied Calculus for Business. (a) 20 * 40 (2.2) (b) 20 22 * 20 22 9. f(2)) (0. (0. by Warren B.5/7. f (7)) (-1. * ** Section 3. (a) b2 . M1 .5. Wang.4 6. 0.1000x2 + 8000x . 350 trees 12.09.60 (c) 1406. (a) dy = 10x dx (b) 10x ¢ x + 51 ¢ x22 (c) 5 ¢ x (d) (i) 100.05 is better because the 1 is closer to 1.7 2. 6 23 .09. 9 213 213 . 11 3.14. 0. and April Allen Materowski. 213 213 6 . Published by Pearson Learning Solutions.006 large 11.21100p .08 using (b) 7.1005. (a) A 24.05 (b) 2.4 3. $32/9. 17. 0.0. B.1 (b) . (a) 2 10.12 . 75 11.3. .00125 large. (a) y = . (a) and (b) 3 miles (c) 1 mile from A 22. . (a) 2 22 * 22 (b) 2 22 * 27. (a) 6. by Warren B. (1. (61/25. (a) 253.44 small (b) 8. -9 213 213 B + 3 22 4 26. 0.015 small 10.1.037 large (b) 2.1 9. *2/25) 23.1 29. (a) (4.9975 large 8.25 (d) 50 (e) not a demand equation (f) 25 5.93 (c) 6. 60.x22 or . (a) No max (b) 33. 0). (a) (b) p . (a) . Copyright © 2007 by Pearson Education. (a) p = .15 3.6 Answers to Exercises and Pre/Post Tests 19.1000x + 10.8/5 14. 5 (ii) 0. (a) y = 5 4x + 2 (b) y = 5 6x + 13 6 22 4 x (c) A 25. 02 12 (b) A 2 . 93 (ii) 0. p12 + 4 in for circle yes.1 (e) .963 large 9.6 1. 48 * 24 * 24 21.5 1.01 than is 0 6. 1) B.Section 3. . $256/27 (c) . (a) 9x2 dx (b) 9x2 ¢ x + 9x1 ¢ x22 + 31 ¢ x23 (c) 9x ¢ x + 31 ¢ x22 (d) (i) 900.000 (b) C = 2000x + 2500.02 x (b) 8/3.090903.2500 (c) 4 7. 3 22 2 p for square. 993.p2 2 2x 1x 5 2p or p-+ 1 (c) 3.089103.0.4 . 1 . 16. $4. 80 8. A 2 B 10 210 . Economics.9963 large (b) . (a) 3.12 . 48 p + 4 in. (a) 256.0. y = 1 3x + 1 4.1402. Decreases by 4750 Applied Calculus for Business.2x + 5 (b) y = 10x . Gordon.000 per coat. A 2 B 13 213 * ** 655 16. (b) 3. P = . (a) 4. Walter O.23/2 (c) 0 (d) . (a) 2. 212 4. B.256 small (b) 1. 10 4 210 8 12 . 17. (a) y = 1 + 1*2 x (b) y = (c) 5. (a) 25 (b) 97/4 2.8 .3 (b) y = .0903 (iii) .200 + x or 200 4+ 4p 4 1100 .2 + 13. (a) y = 12x .50 per thous. $49.00741 large 12. 105.8 thousands 15.0897 14.4. 2 (ii) 0. 15 * 10 20. A -6 .05 4 13. f is not differentiable at . (a) 6x21x3 + 52 dx (b) 6x21x3 + 52 ¢ x + 130x + 15x421 ¢ x22 + 20x31 ¢ x23 + 6x1 ¢ x25 + 101 ¢ x23 + 1 ¢ x26 + 15x21 ¢ x24 (c) 130x + 15x421 ¢ x2 + 20x31 ¢ x22 + 6x1 ¢ x24 + 101 ¢ x22 + 1 ¢ x25 + 15x21 ¢ x23 15. Inc.1*2 (f) . 0. (a) dy = 14x + 22 dx (b) 14x + 22 ¢ x + 21 ¢ x22 (c) 2 ¢ x (d) (i) 14. 125 ft. and Finance. 0. Exercise Set 3. Exercise Set 3. . (a) dy = (b) .21x2 + 3x + 22 1 x . 1/3) 7.2 0. (a) dy = 5. 0).5) (b) 11x + ¢ x23 + 521.212x + 12-0.72 dx (b) x5120x4 . (a) . f(-2)) 3 (2. (a) M = . Inc.12x + 120. 322 (c) 12. . 22.1x3 + 521.100u 2 3/2 du 1 5u + 92 4 2 7 ¢x + P ¢x dy dx (c) D = 28. m = . f (-4)) M (3.7 .2. See Exercise 16 11. 1 . .42 (b) (0. (2/5.1/27. 02 5.4 dx 12x + 102 0.1x3 + 521. 1 . f (7)) Applied Calculus for Business. f (½)) M (-1. Walter O.125 (b) M = 0. M = 155 9.6 ¢x 2 3 0. m = .32 312x 3 2 2 2 11x + ¢ x23 + 521.2) (-4.4 2. 108/3125). (a) 3s2 ¢ s + 3s1 ¢ s22 + 1 ¢ s23 (b) 3s ¢ s 31. 4) (4. .92 + 1 40 12. 0). (a) none (b) 14. 162 4. (b) T1x2 = (c) 8. 2 3 . .7 17 12 x 1 12 1x 2 . . (a) 1 .1x2 + 423 (b) D 2 32x + 3x + 2 . m = .3. (a) 2s ¢ s + 1 ¢ s22 (b) 2s ¢ s (c) and (d) s s2 *s s* s (* s)2 8. (a) 12.4 10.7 ¢x 3 .6 Chapter 3 Review Answers to Exercises and Pre/Post Tests 1.20 9 B (-3. Published by Pearson Learning Solutions. (a) 13. At most 0. .3/42. m = . 1 . (1/27. 2/9).1997% 22. 11x + ¢ x22 + 423 . (0. 0) Chapter Review 1. f (2)) (b) 12x + 2 ¢ x + 12 (c) 17. f (-3)) (7. 0). and Finance. m = .648 (c) m = 0. (1.5. . (b) T = 2 + 32. 0).4 B .22 (d) M = 8.10. f (4)) dx m ((0. 0). 3/42 (b) (0.24p L 32. f (-2)) 15.1282 (c) (0.22 2x 2 2 dx 1 x + 12 -7 2 dx 1 2x . (a) 2x13x5 + 4x2 .656 * ** Section 3. 0.4.6 .92 17x2 . (a) 10. (a) M (-2. 60%/r 23. 10. Gordon. . 0) + 12 2 (b) 3x 1x .12x + 120. A .7 m (½.21x + 242 dx 24. (a) dx (. (a) 11. (a) (0. Economics.1699 cm (b) small 19. and April Allen Materowski. Copyright © 2007 by Pearson Education. .1.152 dx 27. -5) 3.122 6 (b) A . m = .75 (b) 0.1) (b) M (1.30 (b) M = 0. 12.6 16. See Exercise 27 14.½ f(½)) (2.2.0001 in 21.142 (b) (0. 1 .1x 1x + 12 (c) 12x + 2 ¢ x + 120. (a) (b) 25.4 (c) M = 2. by Warren B.2. (a) 5.2/92 (b) (1. (a) M = 16.6 . .7 (-2.1x21x3 + 520.7 18.5.62 dx (b) (0.7 . f (2)) m (0. See Exercise 17 s s* s 29. Wang.24 in 20. (a) (b) 26. .102. See Exercise 28 13. 0) 6.162. 13. and April Allen Materowski. . (a) .01862 B 17. Inc. .4 B inflection points A . Walter O. Copyright © 2007 by Pearson Education. 23 9 (c) Inflection points at A 4 18. .4 B . Gordon. . Published by Pearson Learning Solutions.0145 B and A 4 + 26 . Wang.20 9 B Applied Calculus for Business.22. (a) (c) * ** 657 (b) Zeros at x = (d) . by Warren B. 2 3 . 10 0. and Finance.Chapter 3 Review Answers to Exercises and Pre/Post Tests 16. (a) (b) (b) (c) 6 m1 A . m2 A 22. Economics.26 . 10 0. 3 2 8 B Inflection points A .96 (b) 0. 3. Inc.60x-6 (b) . Published by Pearson Learning Solutions. . No I1 A . (a) 250 (b) 325/4 27. t = 0.13x22 dx (c) . 2 (c) 16 (d) 3 21. Copyright © 2007 by Pearson Education.122 1 x + 62 2 2 dx 31. 5ft. Yes 27. Decreasing 13.30586 B 29. 2323 . (b) .15x22 dx (b) x411 .4t3 a = 24t . Increasing 21. . Increasing 12.3 23 8 3 B . by Warren B. 6) 25. Wang. Yes 3. Yes 4. (a) v = 12t2 .42155 B .658 * ** Chapter 3 Review Section 4. 02 I3 A 2 23. (a) 4. and April Allen Materowski. (a) v = . Yes 9. and Finance. Increasing 20. No 25. No 22. Increasing 18.1 1. Economics. I210. No 6. No 7.0015 30. Yes 24. Yes 2. a = . 20/3 (ii) none 23. Yes 23. Yes Applied Calculus for Business. 10 (ii) none (b) (i) 40/3. Decreasing 15. Yes 5. I2 A 1 + 3 25 5 . (a) (i) 10.32 (b) 256 20. Increasing 14. (a) (d) 19.32t + 64.x22315 .212x2 + 3x . Walter O. (a) *2 x + 1 (b) 1. high. 12. 0. Neither 16. 3 1/3 ft. (6. (a) 114x . (a) Exercise Set 4.2496 22. wide 24. Yes 8. 1 4B 28. (a) 12x . Yes 10. Neither 17.2 23.768 5 1 2x + 12 (b) I1 A 1 1 3 25 5 .0. Gordon.1 Answers to Exercises and Pre/Post Tests 28. 64 26. Decreasing 19. Yes 11.12t2 (b) t = 0. Yes 26. Copyright © 2007 by Pearson Education. (a) 1 .q . and Finance.x2/3 34.1 Answers to Exercises and Pre/Post Tests 29. q 2 (f) 1 . by Warren B. q 2 (c) 1 . (a) x Ú 0 (c) y Ú 1 (f) x Ú 1 (g) y Ú 0 (h) f-11x2 = 22x . q 2 (h) f-11x2 = 1x + 32/2 33. Gordon. q 2 (c) 1 . . (a) x Ú 0 (c) y Ú 0 (f) x Ú 0 (g) y Ú 0 (h) f-11x2 = 1x * ** 659 (e) (b) (b) v. (a) x 3 (b) (c) y Ú 0 (f) x Ú 0 (g) y 3 (h) f-11x2 = 16 .q . Published by Pearson Learning Solutions. Walter O. 30.Section 4.q .q .x 2 (h) f-11x2 = - (e) (b) 32. q 2 (f) 1 . (a) 1 . Wang. q 2 (g) 1 .q . q 2 (g) 1 . and April Allen Materowski.q .x22/2 (e) (b) (e) Applied Calculus for Business. (a) x Ú 4 (c) y Ú 0 (f) x Ú 0 (g) y Ú 4 (h) f 1x2 = 1x + 82/2 -1 2 35. (a) x (b) y (f) x (g) y 0 7 7 0 27 .2 2 (b) (b) (e) (e) 31. q 2 (h) f-11x2 = 17 . Inc.q . Economics.q . see Exercise 57 73.5x 3 + 2x -1 16 . Copyright © 2007 by Pearson Education. (a) 1 . . 72. (a) f ¿ 1x2 = 15x4 + 6x2 7 0 (b) 1/21 69. (a) 7/3 (b) 4/3 41.q . q 2 (c) 1 .q . by Warren B. q 2 (c) 1 . q 2 (f) 1 .1*2 (b) . (a) 7/3 (b) 14/3 43. q 2 (h) f-11x2 = x 8 3 x2 . f-11x2 = 52.q . f-11x2 = 49.q .q . q 2 (g) 1 .3 . f-11x2 = x + 9 5 x . f ¿ 1x2 = 6 0 63. f ¿ 1x2 = .q .x2 .6 2 215 . q 2 (g) 1 . q 2 (c) 1 . No. see exercises 69 and 70.q . (a) .8x 6 0 2 2/3 7 0 3x 19 2 12x + 52 2 7 0 66. It is one-to-one (e) (b) 70. f-11x2 = 48. Wang. f-11x2 = 50.1 Answers to Exercises and Pre/Post Tests 42. Walter O.q .x 2 5 . f-11x2 = 47.1 2 22x . (a) No (b) only if they are one-to-one Applied Calculus for Business. Published by Pearson Learning Solutions.15/7 45. f ¿ 1x2 = 64.q .2x (b) (e) 61.2 11 . Inc. (a) 1 . q 2 (f) 1 . Yes.q . q 2 1 3 x (h) f-11x2 = 2 (b) (e) 46. and Finance.3/5 44. Economics.9/11 (b) .x 4 3x + 7 9x .5x 5 0 0 36. q 2 (g) 1 . (a) f ¿ 1x2 = 6x + 3 7 0 38. It is one-to-one 39. f ¿ 1x2 = 65. (a) 3 (b) 4 40. q 2 (f) 1 . (a) 1 . f-11x2 = 37.q . f-11x2 = 51. (a) . (a) 1 (b) 5 2 71. and April Allen Materowski. f ¿ 1x2 = 2 7 0 62. q 2 (h) f-11x2 = x3 + 1 1 (b) 1/27 67.660 * ** Section 4. Gordon.q . 21. and Finance.06156706343 10. f ¿ 1x2 = 76. 0. . 14. See Exercise 15.94 8. 1.92 2 2 * ** 661 6 0 16.003607 13.3912.632527 11. See Exercise 14.00179239 12. . and April Allen Materowski.242273 4. Economics. 0. f is increasing increasing decreasing decreasing f is Concave Upwards Concave Downwards Concave Upwards Concave Downwards g is Concave Downwards Concave Upwards Concave Upwards Concave Downwards 17.1 4. 15. 19. by Warren B. Published by Pearson Learning Solutions. . Copyright © 2007 by Pearson Education. Inc.0069593 3.9874 7. Wang. 0.41x2 + 92 1x .67789 2.99142 5. 22. 20. . Gordon. 0. Applied Calculus for Business. 18. 0. it is not one-to-one Exercise Set 4. 7. (b) No.2 1. 7.10299 9.Section 4. Walter O. 1.126485 6.2 Answers to Exercises and Pre/Post Tests 74. 0.32165728. 36.3/4 38. Economics. Gordon. 30.2 . 35.2 28. by Warren B. 2/3 32. y = 3x + 2 26. . Walter O.2 Answers to Exercises and Pre/Post Tests 33. Copyright © 2007 by Pearson Education. 41. y = 3x . .3 34. 4/21 42. y = 3x . 7/13 40. . 31. Wang. 13/22 39.2 25. 19/2 43.18/7 44. y = 3x + 2 27. They are symmetric about the y-axis Applied Calculus for Business. 37. and April Allen Materowski. * ** Section 4. Inc. 24. and Finance. 11/4 45. y = 3x .2 + 3 29.662 23. Published by Pearson Learning Solutions. y = 3x . (a) 664.9062511. (a) 296.0855 2.35 (c) 8028.062t (b) 396. (a) 10. They are symmetric about the y-axis 64.4142x (c) 3x 61.14774711. 508. Linear(or power) function 63. among other reasons.4257 * ** 663 (c) (i) 67438. 138.88710. 23.88710.167.611.9837092x 67.95 (b) 8029. (a) 1261 (b) 530 (c) 223 50.646. (a) 2.8615 lb 55. (a) 736. (a) 2 (b) 2 59.55552x (b) 119.9952 (b) 9. 0.92 (c) 2. Exercise Set 4.529410.166.09 13.02 9.9062513. They are symmetric about the y-axis 49.17 12. 11.32x/5 L 9.33 (b) $70. and Finance. (a) 8041.123352x 65. (a) 9.89 (b) $3386.90 (b) 2.973 52.722x/20 L 138.6974 4. (a) 3x 0. Inc.645.66 (b) 2112.Section 4.3 Answers to Exercises and Pre/Post Tests 46. (a) 2x + 3 (b) 25 7 x 7 5 t 1.67 6.2307711. 4. 4. A B (c) y = 5x 60. Gordon.2307711. Published by Pearson Learning Solutions. and April Allen Materowski.6875 lb (b) 1. (a) 366. 20.17 (c) 662. better information.64 (b) 20. Economics. 9.2 4.918 (c) 263.07983 lb (c) 4.99 (c) 2113.92 10.13910. (a) $52. (a) 20.22x/10 L 4. contraception.64 (c) 716.50 (b) 736.07 7.1133 5.96 8.84 11.642. .53 53.18 (c) 736.00971 lb 57.9203 lb (b) 4.485427 (b) 3122x/2 L 311.081. 10.35 58.88 (c) 1479. Wang.1353 3.3 47. (a) 31.839 million (c) 10. (a) 1140 (b) 289 (c) 19 51.92 (b) 1479.07111.529410.02 (b) 716.01007 lb 56.541 million 68.979892x 66. Copyright © 2007 by Pearson Education. (a) 716. Power function 62. better treatment.3533.53710.241 million 54.0538742x Applied Calculus for Business.108.83842x (ii) 55732.12673 lb (b) 8.33 (b) 662. by Warren B.07 (c) 20. (a) 2110.168.2227092x. They are symmetric about the y-axis 48.852x/8 L 23.983x2. Walter O. (a) 1480.2088x-0. (a) $1628. (a) 27. Published by Pearson Learning Solutions. . 18. 15. Gordon.664 14. 22. Economics. and Finance. 20. and April Allen Materowski. 16. 23. Wang. 17. Applied Calculus for Business. by Warren B. * ** Section 4.3 Answers to Exercises and Pre/Post Tests 19. Walter O. 21. Inc. Copyright © 2007 by Pearson Education. 1 2 x11 + ex 2 y11 + e 2 9x + 2 . 10. 2xe2x1x + 12 6.4 Answers to Exercises and Pre/Post Tests 24. 4e-1/3) 28.09861 33. y = e-11 .2xe2x y + 2ex . 4e-1/32.12 2ex x 2 1e + 12 4 x -x 2 1e + e 2 2 2 2 2x[12x2 + 12e2x .693147 * ** 665 Exercise Set 4. and Finance.4 1. 2x . y = x + 1 19.Section 4. x3e-3x1 .5 21. (d) 29. Copyright © 2007 by Pearson Education. 8e-2/32 Applied Calculus for Business. . 2axeax 2 5. 11. xex1x + 22 8.2x + 32 22.12ex . inflection at 12/3.9 gm (e) Max at 11/3. dec on x 7 1/3 (b) CD on x 6 2/3.8 gm (c) 4. y = e316x . 4x ye 5 2x y 3 2 + 6e 2y + 8y + 3 18. axeax 14. 1x2 .2x 1 x . (a) 479 gm (b) 326 gm (c) 6. . Gordon. 16. y = x 23. 26.3x + 42 7.e-x 4.0.ex + 2y x + 2y 2e . y = . y = 2x . .9 gm 32.3e-3x 3.2x .52 24. 1. Inc. Economics. 17. 2. .1] 1 x + 12 2 2 12.2x + 62e2x 15. 12x2 . 9. (a) 98. CU on x 7 2/3 (c) (1/3. Wang. 2e2x 25. by Warren B. and April Allen Materowski.6x4y2e2x y . Walter O.8 gm (b) 29.3e2y 2 3 2 3 2 y 2 27.x + 1 20. Published by Pearson Learning Solutions. (a) inc on x 6 1/3. xex 13. 30.3 4. log2 16 = 4 4.0717 b . log16 8 = 3/4 7. * ** Section 4.1/3 5.3 21. 4 20. b0 = 1 12. (b) inc for all x (c) CU for x 7 0. I3 a .2/3 M(3/2. by Warren B. 2 A ae Exercise Set 4. log6 1 = 0 6. 0. e-2) 26. log3 81 = 4 2. e-2) *I2 *I1 (f) none (0.293 b . Economics. 12/324 = 16/81 17. e-1/2) I(½. 23 = 8 10. 1) (f) (0. 1) is the minimum point. 16. 33 = 27 15. 27e /8) * I3 *I2 I1 (0. 8-4/3 = 1/16 11. and Finance. 1/16 Applied Calculus for Business. 11*223 = 1/8 3 . 2 26. . . 2 18. 1. 10-2 = 1/100 13. 22 22 . log9/16 27/64 = 3/2 9. and April Allen Materowski. Copyright © 2007 by Pearson Education. . 102 = 100 14. 0. 3/2 25. log1/3 1/9 = 2 8. 0) Note: I1 a 1 27. . 0) -3 (e) 22. Wang. 29.1167 b 2 2 28. Inc. 3 24. 0) is an inflection point 30.23 3 + 23 Note: I2 a .5 Answers to Exercises and Pre/Post Tests (b) dec for x 6 0. Published by Pearson Learning Solutions. log8 1*2 = . Walter O. CD for x 6 0 (d) none (e) M(1/4.2 3.5 m(0. 5 23. 1/a 32.4 4. Gordon. 2/3 19.707 b 2 2 1. inc for x 7 0 (c) CU for all x (d) rel min at (0. log3 1/9 = . I2 a 1 + . M(1.666 25. 2 base. 8 29. (a) 7. (a) 11.1 acid 33. * ** 667 42. . Economics. 1/x 48.Section 4. 81 28. 5/x 50. 1/x 46.5 Answers to Exercises and Pre/Post Tests 27.3/x Applied Calculus for Business. Walter O. 40. by Warren B. 39. almost neutral (b) 6. (a) 7. and April Allen Materowski. and Finance. 43.6 acid 32. (a) 10. .3 base (b) 4. Copyright © 2007 by Pearson Education.1 base. Gordon. 1/8 31. almost neutral 34. almost neutral (b) 8.5 base (b) 3.1 acid 36.4 acid (b) 5. 38. 1/x 47. 41.9 acid. 37. 3/x 49. (a) 4. Inc. 44. Wang. Published by Pearson Learning Solutions.6 base 35. 5 30. 45. 2x ln x1e2x + x2 x1e 1 x ln x 2x Exercise Set 4.96076 51. log 28.075. 2-2ln 2) 62. 53. 1*2 ln x + 2 ln y 9. by Warren B.5 log6 v . y = 2x .3010 16. Gordon. 1. m (2. Wang. 3. -1/ e ) 61.398 17. m (1/e . Economics. log5 16 22.023609. 6x 2 4 + 3x 5x4 5 x + 12 51ln x24 x e. 10 log4 x + 1/3 log4 y . 57.A.2 59.2x ln x2 x e2x + x2 + 1 . 3 ln x + 4 ln y + 5 ln z 8. x = 0 is V.7632.552. y = 0 is H. 1. log6 45 21. y = e-21 . 0. Walter O. 32. log3 2 . 1 15. log 3 z 25. 3 ln x + 4 ln y .log3 y 3.77124 33. 1 + 2 log x + 3 log y 6.6 Answers to Exercises and Pre/Post Tests 64. ln 63. 2 4 1x + 323 2 4x + 1 M (1. . 5 log6 x + 1/3 log6 y + 5 log6 z .1.9 log6 r 13. log 3 + log x 2. 56.729716 34.6020 14.0. log3 8 20.128 17 + x + 12 2 2 58. .075 . ln x3y2 23.2.1/4 log4 z 12. 2.6020 19. and April Allen Materowski. Inc.11866 65. log x4 1z y2 2 3 w2 w4 2 5 v2 x5 1 3y z 2 5 w2 2x + 5 x7 B 1x + 127 x + 1 1x . 1. 26.023609 + 1. log x + 2 log y 5. . 1*2 log x + 3/2 log y 10.030215 ln x (b) .5 4.1*2 log6 w . (a) .204 18. ln 1 4 x2 3 y2z3 27. log2 x + 2 log2 y 4. 55. Copyright © 2007 by Pearson Education. A.58496 x = 0 is V. 5 ln x + 3/4 ln y 11. 1.63093 Applied Calculus for Business. log3 31. 0.07299) 29. ln 30.5 ln z 7. ln x2 y4 1xy2 24. 2.668 * ** Section 4. and Finance. Published by Pearson Learning Solutions.6 1.1.7x + 112 60. 54.0.0973) * I (2. n/x 52.4599 ln x (b) 0.2x11 . (a) 0. . 1.A. 1 2x + 1 5 315x + 72 2 x ln 3 5 x ln 7 2x * ** 669 64.635777 52.Section 4. x242x12x ln 4 + 32 80. and April Allen Materowski.3/7 41. 2. .26178 45. 15y 3m 22d 53. 86. 83. 76. 7. 1. Inc.0.15465 46.228 24x . 13y 3m 7d 54.2x4. 6.231544 47. 3. 2 38.6 Answers to Exercises and Pre/Post Tests 35.2x 121x2 + 12 ln 4 + 12 1 x + 12 31 7x + 22 2x13x + 12 4111x22 + 452 5x12x + 92 1 + ln x x ln x .25x x 5x + 4 7 60x 2x + + 2 2 x 3x + 8 2x + 5 2 2 2 2 2 2 - 50x4 5 2x + 9 - 3x 2 21 3x + 12 63. 3. Published by Pearson Learning Solutions. 211 44. 77. 6/x 70. 81.53 x + Applied Calculus for Business. 3 50. 0.10% 62. 2 39.74901 49. and Finance. . 88. A 3x 24 - 5 + + 5 21 5x + 72 2 4x . 1/x 69. 1 51. y = 1 8 95. 89. 5/x 73.62% 61.18x2 + 92 x ln 214x + 32 3 + 2 x . 3 37. 85. 74.22x x x + 4 3x + 7 3 + 2x 8+ 1 . ex11 + x ln x2xe x . 1/x 68. Walter O. 5 43.04317 48. 6x ln 2123x 2 79. 84. by Warren B.86135 36. Economics. y = 15 4 92. (a) 6y 5m 21d (b) 6y 4m 20d (c) 6y 3m 29d (d) 6y 3m 19d (e) 6y 3m 19d 57.7 69 53 94. 10y 6m 16d 55. y = . Gordon. 2 ln 313 2 78.1 2x x B A 1x2 + 524 25x + 7 B 2x 2 x + 1 2 -1 BA 13x .1 42. 67. Wang. . Copyright © 2007 by Pearson Education. . A 2 8x x + 90. 3/4 40. 9/x 71.44% 59. 82. 6/x 72.1 x + 1 2 B 91. 75. 65.112x2 + 252 x13x + 52 . 211 + ln x2x 93. 3y 10m 18d 56. . (a) 13y 8m 24d (b) 17y 3m 29d 58.65% 60. 5. 87. 0. 1. 02 billion 6.68 billion (d) 2.468 billion (b) 4.0155109t (b) 2016 5.16 * 1017 ergs 30. 2535.549 hours 2.69 7. and Finance. Gordon.82 * 1019 ergs 27. 79.4 1/3 2 B (III) not 1 . 5. After 10.1 (b) .096 years 19.96 years 8. 4.9 * 109 billion 38.4 days 21. 9. (a) 10. inc when x 7 0 (IV) not 1 . Walter O. 28.1796 1grams (b) 1620 years 11. 3.1 (b) . and April Allen Materowski.01 * 1018 ergs 31.220 years 17.670 * ** Section 4.01 billion (c) 5. 3.7 1. (a) 170.74 hours 3. .51 billion (d) 1. 19. 4.24 billion Chapter Review 1. 5.1 (c) 1/8 1 2 2x + 2 2.45% 13.2 (d) (e) (b) . (a) f ¿ 1x2 7 0 Applied Calculus for Business.1. (a) 5040.98 years 18.q 6 x 6 q .59 billion (b) 4. Inc. dec for x 6 0.980 years 15. (I) (a) 1 .6 kilograms 10.62 * 1022 ergs 28. 2.630 (b) 0. 5. 11.919.08 * 1021 ergs 29.1.1.1 (b) x Ú 0. 5. 5. 5. 11. 8.1 26.828 times 36. Copyright © 2007 by Pearson Education. 1000 times 34.55 * 1024 ergs 32.527 billion (f) 7.q 6 y 6 q (d) (e) f-11x2 = A x . (a) 9.85% 16. inc for x 6 0 (V) not 1 . dec for x 7 0. 43.46 billion (c) 2.7 Chapter 4 Review Answers to Exercises and Pre/Post Tests (e) 0.5/2 (II) (a) 1 . .54% 9. by Warren B. 6. .3 24.03 weeks 22.6 23.3 minutes Exercise Set 4.6 25. y Ú .030 years 14. inc when x 7 0 (VI) (a) 1 .60 (b) 5.q 6 y 6 q (d) (e) f-11x2 = 1*2x . dec when x 6 0. 7. Economics.8°F (b) 23.3 weeks 20. Published by Pearson Learning Solutions.q 6 x 6 q . Wang. (a) P1t2 = 149000e0. 5623 times 35. (a) 3. 178 times 33.01474781 (b) 17. 1. (a) 3. 177.04 billion 4.0. (a) . 3e + x .2ex + 7x + c 29.3x + 7 2.e-x + c .1 1.4x5 x 2 . Published by Pearson Learning Solutions. 20.9e + 8 ln x . 23. Call 2004t = 0. 1 a + bx 13. x3 3 ln 5 x 17.892789 14. 7ex + c 27.3x + 3x + 3x 27x4 4 + c 4x3 2 3 + 2x + x + c t5 2t3 5 + 3 + t + c 2 22 3/2 + c 3 x -4 7 1 x + 3x3 . 2et 4 3t5 5 x + 7 ln t . -1 2 + c 2x 2 3/2 + 3x 2 3/2 + 3t 3 4/3 + 4x 4 5/4 t + 5 * ** 671 c c c c 8.224742x 6. 5.4x3/2 + 2x + 2 4. 3x . and April Allen Materowski. x .380 yrs 19. 3x212x2 + 13x22 + 12ln13x2 + 122 3x + 1 2 e-x 5x 5+ 2 . ln 8 L 2. They differ by a constant Exercise Set 5. . 12 + ln x2 18. e -1/8) -3/2 I ( e /2 . 9. by Warren B. 14.12 + c 2 4t c 24. Inc.079442 15. (a) 941. 12. x + 2 8.3546511. 21.x2 + 9x + 2 x 2 x 2 3. 1.211. Copyright © 2007 by Pearson Education. (a) basic (11. and Finance. . 4 e -2/3) t8 4 + c 1 2 4w 15.5x + c 32.94 7. 3x + c 2. 19. + c y = 0 is H. 2et + c 28.4x + c (c) . 11.95 (b) 2341.2 ln x - 19 6 + 14 3 x - 199 24 + 2 ln 2 10. (a) ln x (b) ln x 34.6x .2x ln12 + 5x2 r4 3 2 + r + 4r + c 5 3 4 3/2 . A 3x 12 16.Chapter 4 Review Section 5. x2 . 16. 7. 11. Gordon. 10.½ /2.12t + c 30.42) (b) acidic (3. 2x + 5 ln x + 1 6. 3x + c M (1/3.3e2 x3 3 5. x4/4 + c . 4 ln x + c A B 25. 2 ln x + c 26. 10. Wang. (a) 1*2 e2x + c (b) . 6.1 4e I m ( e . 2 e -1/3) I (2/3. 5 ln x .89 1x . f1x2 = 4.-3 e -3/8) 33. 5. 5.2x4 + c 7 4 2t . 3ex + + x + 4 .2 Applied Calculus for Business. 3.2x 2 3x .39 5. 17. 18.A. Economics. 2e . 12. y = .22 B A 12 B 2 4x + 9 4 35. 4.3e-21x + 22 9. 2 1x + c 4 7/4 + c 7x 7 2/7 w + c 2 2 5/2 s + c 5 4 9/4 + c 9x 1/3 13. .2 1 .022 ¿ (b) 8. 22. Walter O.54) + 5 3 Exercise Set 5.2 Answers to Exercises and Pre/Post Tests 3.2t . $1057.95 (c) 6.61 4.21x2 + 92 .2 1. x3 6 2 20.x2 + 3x x 2 1 2 6x 9. 4. (a) P1t2 = 7.21% 8.2 ln x + 7. 3 Answers to Exercises and Pre/Post Tests 20.84 ft/ sec (c) gravity acts the same way in both cases. 10.26 ft/sec (b) 62.2x2 + 8x (b) .33 (b) 143.4y2 + cx3y2 12. (a) 3. 12x3 + 322 + c 2 1 3/2 6 14x + 142 13x5 + 129 + c 9 12x + 1210 + c 10 2x2 + c 5.n = k11 . Gordon. N Z .53 19. 16. y = 1 3 x + c 4 x 24. M = 1. Inc. y1 . N Z 1. x3y3 . 2.768 sec 18.M = 1 + N axN + 1 + c 31. P1x2 = .16x3 = 2x3y ln x . .2x 1 2 + c 41 e + 12 1 2x + 12 + c 2 ln1e c + 8x + 5 + c Applied Calculus for Business. 24. + c 6.672 11.2gs102 13.2 5.2y + cx3y 29. its acceleration will be zero Exercise Set 5. N = .1 y1 . y = 4 3x 2 3 7 2 2x 2 axN + 1 34. x2 + 50x + 5000 21. 1 .2210 + c 180 1 3/2 + c 3 12x + 12 1 3 3/2 + c 6 14x + 32 4 1 -3 x 3e 11. 18. y = x 2 3 + 5x 35. (a) 2 . Walter O.M = a11 .84 ft/sec (b) 282. and April Allen Materowski. 3. 96 ft/sec 16.2x + 3x 1 2/3 2 4 12 + 3x 2 + c 1 4x3 + c 12 e 2 3x5 + 7 + c 3e 1 4x3 + 2x2 + 1 + c 4e 1 2x 3/2 1 1 + 3 e 2 + c 9 1 *2 ln1x2 + 12 + c ln153x + 22 + c 3 ln 5 .x + c2 36. 17. 14. 19. 28. 8. 15. .21 .x + c + c -1 3/2 4 + c 813x + 4x + 12 1 2 2 ln 3x + 4x + 1 + 1 5 3 2 2 ln 4x . y1 .e-kt2 k 37. (a) 78.9x3 = . (b) v = g11 . If n = 1. y2 = 8e + c -5 5 9x + c x 27. 17.1 y = c xa M Z 1. 2y = . M = 1.800 23. 9.76 ft 15. ln 2x + 1 + c 1 2 6 36 13x + 22 + c 13x6 . * ** Section 5. .2x + 8 22.M M Z 1. (a) 62. Copyright © 2007 by Pearson Education. and Finance.x2 + 2xy + y2 = c 3/2 33. 3x3y4 . e 7. y = cekx otherwise. 4.953 sec (b) 1. by Warren B. y = 3 5 52 + c 25.1 y = ce N + 1 . 20. N = . 21. 14.M2 ln x + c. (a) .05x2 + 830x . 22.3 1.0. 23. 12. y = 2 3 26. 13.3xy2 .n2x + c 30. If it has constant velocity. (a) 282. Economics. Published by Pearson Learning Solutions. Wang. 2 31. (a) 2. 7.953125 (b) 18. 28.0833 (c) 0. 31.5659156 (c) 2.21875 (b) 21.333008 23.02 2.42 i=1 Applied Calculus for Business.9375 (c) 2 22. 10.5 5 8 2x x5 5 26. 1. 4. a 2 i=1 i 5 i=2 19 4.3 5.75 (c) 2 13.5 20 1 1.4531596 (c) 6.2 (c) 0.32779589 (b) 0.40625 (b) 3.5 12.052734 (b) 20.46875 (c) 20. and April Allen Materowski. 10. Wang.3125 (b) 3. (a) 2. Published by Pearson Learning Solutions. (a) 0.32172789 24. Walter O. (a) 2. Economics.1 8 A3 + xB + c * ** 673 16.1 and am Z 0 38.333325 25.5243268 (b) 2. 36. a i=1 i 23 2. 10 29.x + 1 a ln ax + 1 ax + c ae 1ln x24 + c 4 42x + c 2 ln 4 2 54x + c 4 ln 5 1 2 2 ln1x + 12 + c ln x + 1 + c b + c 33. (a) 0.75 6.703125 (c) 16. (a) . 34.0625 (b) 1.898438 (c) 20. 11 11. 8 9.75 7. If n = m . (a) 0. 35.875 (c) 2 18. (a) 13.328125 14.03 3. 0. Copyright © 2007 by Pearson Education. 50.08335 26. (a) 19.210938 15. x2 2 . . (a) 3.25 (b) 1.125 (b) 3. a 12i + 12 1 3.5 17. (a) 20.5. (a) 20.ln11 + e-x2 + c1 (b) x .288350 (b) 20. (a) 0.75 (b) 0. 4. 1. 53. (a) 3.5. 26 4. 6.625 8. (a) 6.31579589 (c) 0. ln ln x + c 1ln x2N + 1 N + 1 + c when N 3 2 (a) 4 3 x + 6x + 9x 1 (b) 6 12x + 323 + C 625 17 17 x 250 14 7 x (c) They differ by a constant. a 15i . 30.332031 Z -1 + c 19. 4x 27.5 Answers to Exercises and Pre/Post Tests 2 4 25.59375 (c) 3. . 29. 3.3 (c) 0. 56 5.6875 20.875 (c) 3. (a) 3.75 Exercise 5.3889496 27.5 (b) 0. by Warren B.125 (b) 1.773438 (b) 20. 10 10. Inc. 20.5452056 28. 19.0833 (b) 0. Gordon.4 1.ln1e + 12 + c2 (c) they differ by a constant x Exercise Set 5.6875 (c) 3.5. and Finance. 32.34375 37.Section 5. + + 150 11 11 x + + + c 21.5. 3.615234 (c) 20.37835 (c) 20.5 30.3253785 (b) 6. (a) 2. 5. 1 21.1 i 5. x = 0 and x = 1 31. k1b .674 * ** Section 5.12 12 47. (a) 3/4 i 1/3 1 (b) lim a A n B n n: q i=1 n Exercise Set 5. an + 1 . 105 10.745 17. Gordon.1 xi 8.ln 22 20.4 2 2 23. 1*21ln 5 . 8/3 12. x = 1 and x = 2 34. a ari .192. 1/4 37. a or a i = 2 2i + 1 i = 1 2i + 3 6.159.6 6.a1 + an .1 7. y = 0.910 18. (a) 50 (b) 500 (c) n + 1 2 i=1 n i=1 n 12 44. 1/10.a0 22. . 20/3 29. y = 0. 3 14.2 20. 11/2 11.150 (c) 2n + 1 .e. . Wang.358. 75/4 30. Á . and Finance.2 . 11/5 15. Published by Pearson Learning Solutions. Region bounded by y = 4x3. (a) 6 (b) . x = 0 and x = 1 32. Copyright © 2007 by Pearson Education. 7/9 17. 10 2. (a) 62 (b) 2. 22. by Warren B.2 7. lim a ln A 1 + n Bn n: q i=1 i=1 n n 6 5 i .320 19. (a) 11/2 (b) 5/6 (c) 10/3 38. a 2i . 38/3 8. . 16 48. (a) 15/2 (b) . (a) 15/4 (b) 1/4 (c) 1*2 (d) 17/4 Applied Calculus for Business.e2 23. and April Allen Materowski. 228 15.022 26.6 1. Region bounded by y = 3x2. Inc.12 b 30 n21n + 12212n2 + 2n . 18 10. x = 0 and x = 1 33. Region bounded by y = 6 + 2x2. 1 27. Consider the interval [0.19. 1].5 5. with the partition 50. e . a i=1 i 9. 1/10 + 2 # 9/110n2. y = 0. y = 0. 195/4 9.872 25. (a) ln 2 (b) ln 2 (c) ln 2 18. (a) a (b) (c) n1n + 12 2 2 n1n + 1216n3 + 9n2 + n . 32 11. 1/10 + 3 # 9/110n2.640 24. 21e . 14/3 16. 8 35. 3457/840 12. 1793/12 13. 2/3 36. 240 14.305. Walter O.15/2 5. an + 1 . (a) ln b/a (b) ln b/a 19. (a) 2/3 i 1/2 1 (b) lim a A n B n n: q i=1 n 49. 850. 791. 20 3. 13/2 24. 13/3 e. 1 28. lim a ei/n n n: q i 1 41. 8270 16. Economics.a1 21. . 14/3 39. 292. 81. Region bounded by y = 5x3.961.097. 1/10 + 9/110n2.6 Answers to Exercises and Pre/Post Tests 1 40. 8 13.a2 4. e2 34. Economics. 32. (a) 14 . 14x + 12 dx 25. Inc. 35. (a) 3 1 (b) 7 3 26. A x =3 x =1 A A 41. Published by Pearson Learning Solutions. 36. 2 3 5 ex dx = e3 . Walter O. 1 3 3 x dx = 7/3 2 33. x=4 A x=9 A Applied Calculus for Business. 3x 3 1/2 (b) 8 A 22 . (a) 2 3 2 1x2 + 12 dx (b) 22/3 27. by Warren B.1 B 29. Wang.6 Answers to Exercises and Pre/Post Tests 2 * ** 675 37. 1 3 12x3 . . Gordon. (a) 3 2 4 x dx 1 (b) 4 ln 3 30. and April Allen Materowski.Section 5.e-2 31. (a) 4 1 A dx 38. and Finance. 81/2 2 A 39.x22 dx -2 3 2 (b) 32/2 28. (a) e-x dx 1 3 (b) e-1 . Copyright © 2007 by Pearson Education.3x2 + 12 dx = 192 y=5 40. x2 3 1 .674537 11 + x222 60. 6 14. Gordon. and April Allen Materowski.t2 dt b Exercise Set 5.x2 35.56226116 (c) ln A A 25 + 2 B A 22 .562262 ln 2 a11 + x 2 24 .2 6 42.262997 or 1. 0. $6082. (a) (b) (b) 29. 0 8. 3 5 x 2 x 2 2 + p 23 18 L 0. 10. 0. 9 2.01 33. . 0 43. (a) 0 (b) 0 27. 21e2 . 9 Applied Calculus for Business. 0 5. 32 3. 9 17. 38/15 3.517999 65. 26/5 21. 1*2 ln 5/2 7. 0 18. . by Warren B.2x 37.x .254353 6 A 6 1 22 (b) 0. e2 .69 32. Wang. ln 10 24. Economics.f1u1x22u ¿ 1x2 43. (a) 30.466221 61. 2 ln 3 49.7 1. and Finance. Walter O. 3x5 6 x + 1 - e 4x 2e4x + 1 Exercise Set 5. . 0 13. 8 54.8 Answers to Exercises and Pre/Post Tests 9.e-22 53.0.2 ln 3 6. x4 2 36. . 1 . 9 48. (a) m1a + b2 + 2d 2 1 *21a + b2 28. Inc. 9 46.52846 1 . Copyright © 2007 by Pearson Education. 16/3 22. (a) ln13/42 6 A 25. 63/2 45.e-4 4 (b) 0. 1ln 222 2 1 ln1e2 2 + 12 11.e-1 23. 39. f1v1x22v ¿ 1x2 .6 5. 2x 2x4 + 1 44. . Published by Pearson Learning Solutions. 21 .32174914 (using n = 100) 0 3 24 . 9/2 19. 2 ln 3 50. 1 4. 18 47. ln 5 44. * ** Section 5. 1*2 ln 17 26. 0. 1/5 2. (a) 1/3 1 13 1 *2 ln 5/2 (b) 1. 0 12. 956/15 20. x2 34. 32/3 4.e .676 42.12 52. (a) 3/16 6 A 6 3/8 (b) 0.24535037 (c) 58. 3 ln 2 + 4 51.287403 31.8 1. 16. 0 15. $1107. 31e2 .1 B B L 0. 28/3 57. Inc.174. (a) $93.12e 211 . (a) $63. 416 .29 25.268941 (d) 0.22 (b) $278. Published by Pearson Learning Solutions.4 12x2 + 2x + 12e-2x 1 3 2 128 132x . (a) 16. 1. Walter O.0592 (b) 320.24x + 12x 1 . 6. (a) 11/3 (b) 7 19.a2pe - a 3 S1x2 dx Exercise Set 5.r rt . 2 15 1x + 123/213x . (b) 0.40224/1105 3 2x + c 412x .9 1.892.5e 4 xn + 1 n + 1 ln 2 -2 A x- 1ln x22 2 + c 1 n + 1 B +c if n = .58 (b) $77. 8 8.4e-32 9 1 . 9n4 + 10n3 + 21n2 . 9 6. 16/7 Chapter Review 1.000 (b) 2500 17.1771 16. 1.75 (b) 21.02 21.8 Chapter 5 Review Answers to Exercises and Pre/Post Tests 5. (a) 13/2 (b) 3 12x3 + 4x2 + 7x .3/4 ln12x + 32 . Wang.9792 (c) 315.613. 6 12. (a) $12. 1 105 12x .996. 1331/750 7. 1209/14 4. 12. . T0 c j .r2t .348.x ln x + 1 2B + c A e . 14. Copyright © 2007 by Pearson Education. by Warren B. 0 32.1 * ** 677 xe 40. (a) 1260 (b) 180 18.17157 31.338. (a) 311. 1 11. 27/2 15. (a) 204 (b) 452 (c) 310 5.69 22. (b) 0.731059 (c) 0. (b) 0. 9.1 17.09 (b) $23.00 24. and Finance.9904 exact area is 316 Applied Calculus for Business.11x + 42 + c 3.969.381282 20. 8. A = T C e r D e1j . 13. Economics.32e4x + c 6x + 32e-2x + c 11.12 dx 0 3. . a 3 D1x2 dx . 25.79 26.22 + c 2.4 4. 0 37. (a) 1*2 (b) 19/32 30. 2 2x . $688.11537 21. 9 14. 7.92 23.777.138. and April Allen Materowski.424367 (c) 0.575633 19.Section 5.4 12x4 + 4x3 + 6x2 + 1 2 12x + 32 x2 2 2 1ln x2 2 -2 + c .1 d 38. e .64n 12 1 2.a2pe xe 41.x + c . 10. Gordon.59 20.1 if n Z .3/4 18. (a) $152.353093 (c) 0. e 16.123/2115x2 + 6x + 22 5.11 (b) $309. (b) 1*2 27.731059 29.28311 33. 2 ln 2 . 1xe . 2 ln 2 . 32 9. (b) 5/32 (c) 11/32 (d) 11/16 (e) 27/32 (f) 1*2 28. 0. (a) 17/3 (b) 6.5e 4 15. 9/2 13. (a) 10. 37/6 10.1xe . 6xy3 (b) 2x .3y42412x3 .20 2.3x3 (c) 4 .16 4 92 1 ln 4 .3x3 16. (a) 143 (b) .36y (d) . by Warren B. (a) 196 ft (b) . 4x + 12x2y4 + 9 (b) 6y + 16x3y3 . y = . (a) 432 (b) 800 8. (a) 0 + c (b) 0 (c) 25 ln 2x2 12.3y2 (b) .957.0202 (b) 1.ln11 + x 2 1 1 1 + 4t223/216t2 .9x2y2 (c) 2y . 12x3 . (a) 16xe-2y 2 2 1x .2 1.9x2 15. 247. 14.x2 24.22 48 (c) 0. (a) 150 (b) 900 5. Walter O.678 * ** Chapter 5 Review Section 6.028 14. 9x2 . x 21 .163333 21. (a) 3x 1x + y223111x2 + 3y22 (b) 24x3y1x2 + y223 7.8758057 (b) 4/3 20. 32/315 19.3y424 (b) 4y12x3 . 15. (a) 60x2y212x3 .32x2ye-2y 9.9x2y (b) .35x2y4 .k2 31 .2y .4xy 2 2 2 2 2 2 2 2 2 2 2 Exercise Set 6.y2 .966. Published by Pearson Learning Solutions. Inc.112 ft/sec 8.4x2y 1x + y 2 .14xy5 + 16x (b) . (a) 4x . (a) 400 (b) 2400 6.7 (b) 25x3y4 + 18y 4.6x3y22 (b) 3x2y2e-4x y 13 .x 9.y 2 4x y 2 6. (a) 2ye3xy13xy + 12 (b) 2xe3xy13xy + 12 10.2e-5u + c 10 1 3 3 1ln 22 1ln 322 . (a) 6xy3e-4x y 11 . (a) 2x . C1x. (b) 7 214 . z2 = 5xy + 4xz + 4yz 2 . (a) 3 (b) .2 2. 10. (a) 4 (b) . 15 7. (a) .33y42 3 2 3 2 Applied Calculus for Business.836667 (d) 0. (a) 100 213 (b) 100 243 13. (a) (b) 4xy2 1x + y 2 . 12.55 (b) $1.8 (b) .2 7. . and April Allen Materowski. 93/5 18.6xy . (a) $167. (a) 2y .60 (b) $557.16x3 + 15x2y5 .6y2 .3 3. and Finance. 1 2 10. Copyright © 2007 by Pearson Education.12y3 (d) 2x . Wang. (a) $329. 2x3 21 . Gordon. (a) 0.4y3z + 2z 17.1ln 222 A B 21ln 32 1ln 22 45 . Economics.6 . (a) 3/2 (b) 2/3 1x .2 Answers to Exercises and Pre/Post Tests (c) x + h y 6.226. (a) 0. (a) 80xy312x2 + 3y424 (b) 12y212x2 + 3y42412x2 + 23y42 8.x4 Exercise Set 6.8x3y22 11.4 9. (a) (b) 5.41683 17. (a) 800 (b) 5400 4.638.y 2 2 2 2 2 (b) . p = 29 .12 40 3 3 9 .2yk .77 3.269. 11. y.1 1.327e.6y2z2 (c) .90 22.2 8 2 2 16.86 23. 13. (a) 10 (b) 10 (c) 10ex 11. and April Allen Materowski. . (b) 2x2y2 A 16.2 (c) 25z4 + 11 17.2y325145x4 .96e16.732e-32 25. (a) 8x3 + 16x . Applied Calculus for Business. 36 27. by Warren B.472 30.32 ln 5 (b) 64/5 + 24 ln 5 26. (a) 100 (b) . (a) . 72e16. .Section 6. (a) (b) (c) 4x 2 2 2 2x + 3y + 4z 6y 2x + 3y + 4z 8z 2 2 2 2x + 3y + 4z 2x 2 2 2 x + y + z 2y 2 2 2 2 2 2 33. (a) . (a) 124x5y6z5 + 4xy4z32e3x y z (b) 112x y z + 8x y z 2e 22. (a) (b) 4x 2 2 2x + 3y 6y 2x + 3y 2 2 * ** 679 31. (a) 1040 (b) . (a) 12xy z e x + y + z 2z 2 2 2 x + y + z 3 4 2x2y3z4 2 3 4 (b) 18x2y2z4e2x y z (c) 24x2y3z3e2x y z 6 5 5 2 3 4 4 2 2 21.564e-32 (b) . (c) 112x6y6z4 + 6x2y4z22e3x y z 35.1/3 28. Copyright © 2007 by Pearson Education.2 Answers to Exercises and Pre/Post Tests 12. and Finance.4xy3z4 (b) . . Economics.15y4 + 6y2 . (a) 8x + 3 (b) 6y2 .64 2 3 3 3x4y2z2 4 2 2 34. 192e16 29. . (a) (b) (c) 19. (a) y = 56x .8x2y3z3 + 5 18. Gordon.16/25 (b) . (a) 12x2y215x4 . 11. 1/3.8/25 24. 22.32/5 . 20. (a) 4xy3 A x2 2 2 x + y 2y 2 x + y 2 2 + ln1x2 + y22 B + 3 ln1x2 + y22 B 32. Wang. (a) (b) 14. Published by Pearson Learning Solutions.4y 2 13. 3 4 3 5 4 3 3x2 3 2 21x + 2y 2 2y x + 2y 3 2 15. 2/3. Walter O.57 (b) y = 63x .6x2y2z4 (c) 60z3 .2y 2 1x . Inc.2y32 (b) 40x y15x .149 23. . Inc.432x3y4 + 4x62 (c) 120x2y12x3 . Walter O.1*2. 12.48x2 + 30xy5 (b) 100x3y3 + 18 (c) . (a) 36x2 .64xye-2y 2 52. 41. 11*2. 38. 4. (a) 18x .27y42 Applied Calculus for Business.14y5 + 16 (b) .75x2y4 49.140x2y3 39. (4. by Warren B. and April Allen Materowski. 37. Wang. . and Finance. . 5) 45. (a) 16e-2y 2 2 (b) 1 .32x2 + 128x2y22e-2y (c) .3y42312x3 . (c) .4y213x2 . Economics. 42.70xy4 48. Published by Pearson Learning Solutions. (3.y22 1x + y 2 1x + y 2 2 2 2 3 2 2 3 4x213y2 .22 43. (a) 4 + 24xy4 (b) 6 + 48x3y2 (c) 48x2y3 47.3y423114x3 .12y . * ** Section 6.3y42312079y8 . Copyright © 2007 by Pearson Education. .1/32 46.2 Answers to Exercises and Pre/Post Tests 40. (a) 240y 12x + 3y42316x2 + y42 (b) 24y12x2 + 3y42314x4 + 192x2y4 + 759y82 (c) 240xy212x2 + 3y42312x2 + 19y42 51.x22 1x + y 2 3 2 2 3 50. 3) 44.8xy1y2 . (a) (b) (c) .x22 . Gordon.3y42 (b) 412x3 . (a) 120x12x3 .680 36. Copyright © 2007 by Pearson Education.322 20. sp = saddle point. . 8 9. 0) for 0 6912) 18.62 12. 1 .5172 sp 10. Comp 2. Published by Pearson Learning Solutions. 4. Inc. 6 23 B 21. 18) 29. radius = 25. 4.1 70.Section 6. 0. Not homogeneous 71. 2. rm (2. 0). 02. 18. A 2 3 4. TF (0. 2. y2 = 8 . TF = test fails.182 4. 500) 125 2p yd. .2. 0. 6) 34. 1 .4 1. . A 2 23. 0. 6) 11. (0. 0) 8. 12 12. 2 3 4. 4252.5. 0). rm 12/3. 6 73. . 4. sp (0. 108 17. . 18. 48 11. rM(4.2002 30. 0). sp (0. rm 12. f1x. sp (0. 6.412x2 . 2 23.6 23 B . min = 20 45. s/3) 30. 16/3. 0). 3. A .3. TF (0. 158. .5 1. 1 68. 1 . (2.2002. $1912 at (4.107/22 2. 0. 12.2/3. . . 132182 22.1082 15. .18.52 7. . 4. 0). rm 16. and April Allen Materowski. 0. 11) 33. 2. 0. 5 74.3. 0. 0.12x + 12y 43.5.2. (6. 3. 5 3 4B 22 35. . 192 3. (36. 4) 29.3 rm = relative minimum. 425). S/3) 27. S/3. rM 1 . 12 5.2 23. 02. rm 12.35. Wang. Max = 60. 1024/3) 17. sp 1 . . (s/3.11 sp10. 48 16. . 0. by Warren B. 27/4 3. Both are 24x2yz 67. 1 . 2) 37.24 8. Walter O. 11. 1. . TF (x. . . 2/e2. . 5/2) 28. . .3y22 1 2x + 3y 2 12x + 3y 2 . . 2 + ln 242 21.4 10.2. rm 1 . s/3. 7/4 4. 1.1. 1/4. (S/3. sp 16. 0. . .532 6. Economics.5/2. . 0. 2. 2. 522 13. All variables appear symmetrically. 0). 4) 20. .2 6. 2.22 26. . sp (0. 10/32 23. 1 . rm 1 .4 22 7. rm (4.2802 16. 72 13. (2.1/3. 0).2 23. . Both are 48xy2z 55. 2. 16. 1) 22. 42. . . . 0. 23 324 18. 5/3. . (400. Sub 3. 0. (36. 4.2. sp (0. . f1x. 4. . 2. . rm 13. 1. 6912 19.2 612x2 .13. 0. 5. sp (18. . sp (3.452 28. 4. 4. rm (1. TF (0. 0). Max = 2. 4. 0. y.2. (3. 10. . 3. 1/4.11 44. 0). 3. 1 . rm 12. (4. (4. and Finance. 2. 0. Max = 7 min = . . 42 14. .702 24.2 23. 9/2 2. a + b 72. 108 6. rM = relative maximum. 2. sp 10. 0) for all x. a + b + g Exercise Set 6. 0.322. . 11.3y22 54.4. 4) 26. Gordon. sp (0. 0) 5.1. x = y = z = 2 3 4 32. 14. y2 = 10 . Length = 125/2 yd.5 Answers to Exercises and Pre/Post Tests 53. 0. rM (4. 2. .2. rm 13. rM 1 . y 12 at other y it is a sp. Exercise Set 6. 243/10 14. rm (0. Sub Applied Calculus for Business. 18) 36. rM (0. 2 23. 0).24xy 12x + 3y 2 2 2 2 2 2 2 2 2 2 * ** 681 31. 15. min = . 3.2.2x + 1 2y 38. sp ((12. .1552 25.272 19. rm 1 . (a) (b) (c) . 1/e2 24. 6) 10. sp 10. 14. 1*2 69. 2. 12) 9. 8 26 9 Exercise Set 6. 2.2. 1 x x 3. see Fig.x2 and y = 0. (a) 8 (b) 512/3 12. 256/3 54. see Fig.x . (a) 21 (b) 4 21. Comp 9.12 21 . (a) 1/16 (b) 15/64 45. 59. (a) (25. (5. Same as 23 25. Published by Pearson Learning Solutions. . 1/4 48. Same as 29 31. and April Allen Materowski. 3exy . (a) (16. 15 26. (a) 36 (b) 14148 5 13. ( a.5 Chapter 6 Review Answers to Exercises and Pre/Post Tests 22. 216 34. 224/3 36. Bounded by y = x.e .000. Bounded by y = 6x and y = x see Fig. 6 4 62. y = 4 . 96 42. Ex. Same as 25 27. (a) (2500.000 14. (a) 8/3 18.682 * ** Section 6. see Fig. Neither 5. 22/3 9. 2 2048 21 1504 105 3016 15 55. 600) 0 p1 0 p1 0 p2 0 p2 17. 14 24.2xe .6 1. 528 10. 1/144 3 and . . 81) (b) $1728 12. 64 32. Ex. 800/243) (b) $1.a )( d . Same as 27 29.3xy + 5x + c1y2 4. 1 + 3y + c1x2 + 3x + c1y2 + 5y + c1x2 + c1x2 + c1y2 40. (200. 1/4 47. (a) 32/3 (b) (b) (b) 20. 1/128 50.1 2 1 2 ln 2 17.y2 3 y1x2 + y2 . 64 58.4 + 2x y 3 2x 3x2y3 + xy2 3 2 x 2 y 38. 12. Bounded by y = x2. (a) 1/3 (b) 37/21 15. Copyright © 2007 by Pearson Education. and 2 y y 2 2 60. 0x 0x 0 p1 0 p2 if 0 y # 0 x 6 0 then neither (b) neither 33. and Finance. 3 52. 2500). 17 30. 36 39. e4 . Wang. 7/6 43. y3 3xy4 2 3 . x + y = 4 and y = 0. 37. Inc. 2) 11.1 Chapter Review 1. (a) 4 (b) 16 16. c ) 56. d) (b. (a) . 3y2 2 2 2 2 x1x + y . Comp 7.000 13.x2 . if 0 y 6 0 and 6 0 then comp.19/2 61. (a) 28/3 19. (a) if 0 y 7 0 and 7 0 then sub. Ex. Bounded by y = 2x2. 6. 16 28.2ey + 2xy 5. (a) 9 (b) 9 14. Same as 21 23. 1 30240 4.3 2 Exercise Set 6. by Warren B. 3e + x . (a) 1/186624 (b) 323/60279552 46. (b) 25. Comp 6. see Fig. Walter O. Bounded by y = 3x2. Applied Calculus for Business. pa2h 44. 4/3 41. 13897 168 11.c ) ( b. 2. 9/7 35. 1 2 ln 5 e4 . d ) Value =( b . Ex. 188 35 3. Ex. y = 8 and x = 0. and y = 1x.y2 3 7. 18 51. Gordon. c ) ( a. Economics. 4 53.12 21 . 22/3 8. 1 Answers to Exercises and Pre/Post Tests 2.9.1082 12. . . by Warren B. 0. 12. 35 54 3 3. 5/3 + k2 = h2 + 1h .6. 0.11 2 27 5 2 . .22 . . 640) (b) $3200 15.14 Applied Calculus for Business.86 21 2 .1. 70 125 4.62 . 45.82.3y12x3 . rel min 14/3.5 * 9 17 11. 0.e. Gordon.9.k22 + k2 Ú 0 13. (216. 182 11. (a) 64 (b) . rel min 16. 10/3 Exercise Set A. -9 0 3 135 1. .1.0.3 (ii) . 4. 0). Wang.3.108 . 16.y2 20.5 243 9 24 * 3 2 .5 * 21 600 180 * 720 2. 70 5. (a) (20. 0.10 22 . 0.Chapter 6 Review Section A. y = . Inc.3. 4.2.155 37 2 11 2 6 9 2 57 2 -1 43 2 / 4.67.400 2. 5/3. 0. 2. .7 10.114 (c) + 105 0 1 2 0. (a) rel max (7/3. 8) 16.f17/3 + h. (a) (i) . x = 1.7/3.42 . 1 .10 4 . saddle point (0.31 (c) . (4.1 1. y = 4x . (a) J = . 30 80 7 8. . (a) 4 * 5 (b) 58 43 5.y32 21 x + y 2 3 3 2 2 * ** 683 18.24 .200 (c) Jc = . and April Allen Materowski. 32 7.2. . 32/3 21. *2. (2.74 . 2. Economics. Walter O. (a) . z = 0. and Finance. 416) 1 (b) (8. w = .43. 18 . 2.9. 40 45 105 (b) T = . 108.700 10 3* 12 35 F = . 1. 1. (a) 15. Copyright © 2007 by Pearson Education.7.18 . 6x12x4 + 7x2 + 32ex 4.800 420 510 * 540 7.18 -1 2 . 3 (c) .2 (iii) 45 (iv) 3 (v) 9 (b) 11. 16xyz 1x + y + z 2 2 2 2 3 (b) 1/16 23.420 (b) .6 27 -6 .100 Fc = .1.7. . 24xy2z + 6.0. 5/32 . Substitutes 14.800 4. Published by Pearson Learning Solutions. 17) (b) f17/3.64 8 (a) 1*2 5.22 9.3.5. 3.4.8002/105 19. 22. 8 * 1 2 (d) 3 * 4 2.6) (c) the corresponding entries are the same. 0 -2 3 2 47 2 9 2 1 8 5 2 8. 24) 17.64 3. 13. 1.94 .0. . 33333504 . 15 49 14.47 -6 .3328 b .12 2. and Finance. 1 2 7 10 .36 . -5 .164 29. GH = 25. 3. a 10 b .42 8. by Warren B.36 212 (b) a (c) a 6. Walter O. 196/31. . Wang. a 18. 5 37 9 15 .36 .27 (c) u = 35. 1 . -3 -2 20.3/2. . .2. 7. a 15. v = .28 .25 93 263 -1 -2 1 -7 . HG = -1 -5 21. 19. a B = a 0 1 -3 1 17 a 2 4b 3 . t) unless otherwise indicated. .66666752 . x = .1. 13 16 -3 b -1 3 35 1 0 1 8.64 . Economics.66666496 2 1 3 1b 3 (d) a 3 2 3 Exercise Set A. 35 .23 3. .15 .305. . GH = 15.68 .2.12 10 10. 20 .1.28 .3344 . (a) a (b) a 4. Copyright © 2007 by Pearson Education. (a) + 1 3 1 27.1. z = .3 -5 2 6 10 9 6 3 15 4 12 * 20 21 14 7 35 3 2 .15 (b) -3 .152 28 b 66 94 b 76 -5 12 10 b 23 12 b .49 12. R = 11 14 10. R = 2 3 5 8 9.169 1 -1 2 4 -1 14 -9 . y = 1*2.1 A. 42 5. (a) + 2 -5 -1 -2 -1 2 x 3 3* +y* = +5* 2 z 0 -2 x 3 1* + y* = + -2* 9 z 12 (b) x = .10 (b) x = . 1 .1/32 6. 11.32 b . 9 55 . y = 16/7 Exercise Set A.32 7.35 . w.25 11. A = a 11.27 .20 -2 -9 2. Gordon.6656 . z = 5/2 2 5 . 1 .11 7 25 .11 17. . 1/2. Inc. (a) a (b) a 1 20 5. a . 40 . 22 Applied Calculus for Business.34 76 16.3/7. (a) 0 .43 B = 9 .13 5 .176 6 . z.z -5 49 -2 b -4 -1 b -5 10 b 14 4 b 22 19. y = 127. a 1 12 17 a 19 b. .6672 . 1 .33333248 b .12 12. 15 .85/312 3.2 1. A = 1 1 1 32 13.18 w = .31 .44 -1 HG = + . 11. and April Allen Materowski. 1. Undefined . y. 2. (a) Undefined -2 .684 * ** Section A. Published by Pearson Learning Solutions.29 35 20 70 . 14 104 43 45 . 1 5 0 11 -1 b . 1. . x = 3 or 9 1 25. 22 4. .3 Answers are in the order (x.3 Answers to Exercises and Pre/Post Tests -1 0 . (a) 7 1 + 2 26 -5 x 1 (b) + y * = + 1 * z -1 Applied Calculus for Business.3. 3) 1 -9 10 a (c) 1 5a 34.12 1 -7 -2 11 8 6 0 4 4 Y = a -8 4* 8 6 3 -9 -8 10 7 -7 10 4 b -4 6 7 6 2 -5* -6 2 1 b 0 3 b -1 -2 1 39.16 + 98 138 38. 6.14 18 14 -8 8 5 . Copyright © 2007 by Pearson Education. a 0 1 40.24 . (1. 1. Wang. (a) 24. 3/2) 4. Gordon. 1. 1/22 11. 50 43 . a2 = 13. 12. 594 19.11 b -9 -5 11 b 5 2 b. 1c. . Inconsistent 2. 9.55 36. Published by Pearson Learning Solutions. 1 .10 * -1 1 4 2a 26. X = 4 + 2 8 1 15 + -2 Y = 1 4 + -5 2 23.12 * . 3. 21/10) 8. None 25.I is singular. 1 . 20 type A. 46. c = 70 17. (8. a 16 3 -3 1 -4 3 2 35. 7.4 1. (a) 12. 1 4+ 2 -4 3 20 + 20 (b) 1 4 .7k/3 + 6l. 5k/4 + 1/4. Walter O. None 4 1 +5 28.11 9a 4 -8 9 29 2* 9 20 . Inconsistent 3 b 1 3. 24 dimes.3 2 -1 0 -1 4 2 -1 -2 1* 0 -3 -2* 1 Exercise Set A. 3.10 1 -2 (b) X = 1 3+ 2 -1 47. 1 6 b 4 .17 b . and Finance. 72 16. 90¢/doz.42 14. 5/22 13. 7. . 22. k. k. 1 . 35¢/qt. . k2 6.2.Section A. (1/2. 1 -2 4a * ** 685 . 7 .16 41. 8 type C 18. (a) a (b) 5 5 -2 5 21 . A-1B 43. 3.1.35 30 . 12 15. 1 . 4.bc Z 0 (b) .28 . 112 -6 (b) X = + .6.milk. . Inconsistent 11. Economics. 113/4 + k/4. . . eggs. 45¢/loaf bread. (a) 14. Show T . . 19 2 . 0) 12.2 . l2 13. 44. 1. .k/3.1/3.12 6 12 12 . (a) + 23 -5 -2 1 0 . -9 12 43 . 22 10.9/2. 1 . .12 10. Inc. 5.30 131 * 23 . 4k/3 .16 14 4 17 . Inconsistent 7. a = 700. X = a 1 0 . . 2.c/d (c) there will be a unique solution as long as the lines are not parallel.12 5* -1 x -1 (b) + y * = + 2 * z 4 92 1 37. 385 20.25 . .3 A. by Warren B. v.22 12. b = 100.a/b.16 15 1 29.1.47 51 1 30. 16.27 . (13/10. 66 14 25 31. .10 5 4 b -1 2 0 -1 3 -1 -8 -2 4* -1 2 -7* 1 20 .12 9. 34 nickels. 36 pennies 21.32 27. and April Allen Materowski. . 117/3 + k/3. 12.2. 1.4 Answers to Exercises and Pre/Post Tests 9. (a) ad . Inconsistent 5. 2 type B. 11. Copyright © 2007 by Pearson Education. . 68/37 + 67z. 32 7. k. Gordon. then x = 4000 and z = 1000. l2 Chapter Review 1.6 + 2l/3. 11. . A = 2. Wang. and Finance. l2 20. 232 16.95 (c) c = .l. z be the number of $1. 96. (a) AB = I (b) 1 . k. z2 Applied Calculus for Business.7/2 . 1666.9 1 6 7 b -8 . No solution 9.4 Chapter Review Answers to Exercises and Pre/Post Tests 14.232 5.l. Economics. 114. by Warren B. 19.15 b 29 2 23 15 b 22 B = 1 -4 a 17 6 -7 13 7 b . k. . 2 .l. a 20 6 26 1 z. 2. 1 . 123/2 + 2k .k. and $4 ties respectively.12 15. y. x = 2000 + 2k. .95 and k Z 88/37 (b) c Z .3. (a) c = 7 and k Z . Inc.l/2.A + 2I = + 1 4 4 -3 6 -1 6 6 -2* 7 8. z b 11 11 11 11 10. 12. (a) c = . Walter O. Á . . A-1 = -2 1 2 5 A . . where k = 0. 1. If y = 2000. a 3.5l.15 4.5 + 3k + 4l. y = 5000 .269. a 6 2 1 23 a 17 .10 22 . (a) c = 4 and k Z 8 (b) c Z 4 (c) c = 4 and k = 8 (d) 12 .2 (d) 112 .3k.103 . and April Allen Materowski. z = k.95 and k = 88/37 (d) 14/37 . 19 + 2k .686 * ** Section A. $2. l2 18.27/4 + k/2 + 2l/3. + z.2 (b) c Z 7 (c) c = 7 and k = . . Let x.13z. Published by Pearson Learning Solutions.l/3. . 1. l2 17. 242 Conjugate Expression. 217 Constant Multiplier Rule. 120 Critical Number. 256 Continuous Compounding. 18. 181 General Power Rule. 349 Continuous Income Flow. 175. Copyright © 2007 by Pearson Education. 323 325 Definite Integral. 79 Bundle. 520. 593 Antiderivative. 121. 548 Using Left Endpoints. 511 Column Vector. 51 Absolute Value Function. 182 Derivative of Inverse. 42 43 Corner. 238 Cost.Index Absolute Value. 261. 307 Applied Calculus for Business. Published by Pearson Learning Solutions. 14 15 Decay Rate. 162 ex. 348. 422 423 Riemann Sums. 425 Asymptotes Horizontal. 538 Demand Function. 160. 91 Bell Curve. 481 483 Contour Map. 564 Complementary Product. 179 ex. and April Allen Materowski. 331 Difference Quotients. 564 Discontinuity. 278 Algebraic Function. 509 Coordinate System. Wang. 135. 312 Differentiate. 513 Consumer Surplus. 193. by Warren B. 207 Augmented Matrix. 256 Distance Formula. 218 of a Linear Function. 79. 224 Average Value. 37 Chain Rule. 440 Using Double Integrals. 174 Correlation Coefficient. 380 Circle. 342 Decomposition. 226 Axis of Symmetry. 104 Cobb-Douglas Production Function. 131. 538 539 Completion of the Square. 499 Differentiable. . 348 Concave Downward. 80 Derivative. 513 Carbon Dating. 136. 105 Composition Property of Inverses. 116 Dependent variable. 315 Differential. 174 Acceleration. Gordon. 447 448 Demand Equation. 467. 469 Break Even Point. Economics. 66 Depreciation. 103 Equation of. 227 Average Cost. 176 Dimension of a Matrix. 64 e. 557 Average Velocity. 522 525 Cubic Regression. 211 Simple Power Rule. 180 Sum Rule. 116. 193. 143 Alternate Interior Angles. 62 Consistent System. 329 Compound Interest. 479 481 Continuity. 509 Contours. 579 Constant Returns to Scale. 62. 155 Corresponding Angles. 472 By Rectangles. Inc. 279 280 Concavity. 279 Congruence. 261 Critical Point. 266. 426 Using Right Endpoints. 176 Differentiable Function. 77 Change of Base. 424 Using Midpoints. 350 Elastic Demand. Walter O. and Finance. 70 Decreasing Function. 205 Vertical. 179 Product Rule. 210 Quotient Rule. 279 280 Concave Upward. 398 Area Between Curves. 54 Domain. 238 Alternative Theorem. 578 Average Rate of Change. 389 Ceiling Function. 60 Composition. Walter O.688 * ** Index Elasticity of Demand. 539 540 Joint Revenue. 520 Even Function. 581 Level Curves. 464 Generalized Exponential Rule. 464 Simple Exponential Rule. 455. 323 325 Indefinite Integral. 521 Extreme Value Theorem. 42 General Equation. 277 Homogeneous Function. 386 387 Exponential Functions. 307 309 Elementary Row Operation. 207 Inflection Point. 462. 201 ex. by Warren B. 448. Gordon. 501. 82 of Several variables. 124. 546 Even Functions. 359 Extrema. 354. 419 Generalized Logarithmic Rule. 539 541 Joint Profit. 489 Intermediate Value Theorem. 153 Lines. 201 ex. 407 Euler s Theorem. 456 Integral. 354 355 Growth Rate. 215 Piecewise-linear. 325 334 Irregular Partition. Economics. 516 Higher Order Derivatives. Inc. 51 Parallel Lines. 55 Logarithm. 322. 468 Double. 61 Function of Several Variables. 115. 279 Instantaneous Rate of Change. 572 Implicit Differentiation. 69. 405. 322 323 Hydronium Ion. 307 308 Infinite Limits. 205 Horizontal Line Test. 75. 45. 222 223 Marginal Function. 70 Functional Notation. 364 Identity Matrix. 203 206 Linearization. 398 399 Integration by Parts. 515 Isoquants. 457 Gauss-Jordan Reduction. 250 Joint Cost. 258 First Derivative Test. 452. 557 Derivative of. 130 Exponential Domination. 512 Isotherm. 45 Vertical. 358 Generalized Logarithmic Rule. Copyright © 2007 by Pearson Education. 580 Ellipse. 580 Gaussian Distribution. 78. 498 Fundamental Theorem of Integral Calculus. 48 Slope-Intercept Form. 2 Linear Regression. 401 Simple Power Rule. 362 Logarithmic Differentiation. 249. 539 540 Lagrange Multipliers. 415 Odd Functions. 390 Marginal Cost. 266 Floor Function. 572 Invertible Matrix. 266. 153 Horizontal. 373. 227 Integrable Function. 310 320 Line of Best Fit. 514 Iterates. 520 Horizontal Asymptote. 32 Inverse Matrix. 513. . 541 Applied Calculus for Business. 66 Indifference Curve. 357 Exponential Growth. 573 Inverse Function. 399 Integrals by Substitution. 43 Linear Equation. 111 Equations of Motion. 398 Average Value. 342 Harmonic Function. 579. 519 Higher Order Partial Derivatives. 509 515. 402 Simple Logarithmic Rule. Wang. and Finance. and April Allen Materowski. 418 Generalized Power Rule. 136. 518. 43 Point-Slope Equation. 77 Free Variable. 231 Inconsistent System. 446 Isobar. 383 Logistic growth. 69 Interval Notation. 595 Function. 520. 467. 257. 497. Published by Pearson Learning Solutions. 124 Marginal Rate of Substitution. 398 Independent Variable. 237 Perpendicular Lines. 594 Increasing Function. 531 537 Leading One. 518 Inelastic Demand. 486 Integrand. 519 Limits at Infinity. 368 Generalized Exponential Rule. 513 514. 16 Quadratic Equation. Copyright © 2007 by Pearson Education. 263 Optimization of Quadratics. 566 Scaling. 507 Particular Solutions. 196 Revenue. 223 Market Equilibrium. 223 Marginal Product of Labor. 364 Piecewise Function. 257. 435 Sign Analysis. 511 Marginal Product of Capital. 44 Slope of a Tangent Line. 522 523 Scalar Multiplication. 162 ex. 392 Riemann Sums. 275 ex.Index * ** 689 Marginal Revenue. 331 334 One Sided Limit. 79. 243 ex. 566 Transpose. Published by Pearson Learning Solutions. 358 359. 168 Slope of a Curve. 500 Odd Function. 13. 261. 117 Matrix Inversion. 196 Non-singular Matrix. 516. 73. and Finance. 275 ex. 174 Sigma Notation. 503 504. 520 522 Newton s Method. 261. 160. 86 ax2 b = 0 Applications. Wang. 42 Row Vector. 483 Probability Density Function. 124. 14 Midpoint Formula. 174 Smooth Function. 119 Richter Scale. 132. 40 Regular Partition. 276. 82 Pivot. 202 Rationalization. 260 Saddle Point. 307 Perpendicular Bisector. 307 Relative Extrema. 249 252 Non-linear Regression. 439 440 Rolle s Theorem. 564 Row Echelon Form. 599 rref. 436 Sigma Notation. 494. 343 Probability. 14. 521 522 Relative Minimum. 64 Rational Function. 151 Secant Method. Gordon. 166 Smooth. 79. 280 282 Second Partial Derivative. 261. 584 Matrix. 257. 469 Octant. Economics. 169 Second Derivative. 573 Slant (Oblique) Asymptote. 322 329. Walter O. . 599 Regression. 259. 10 pH. 259. 43 44 Median. 564 Addition. 354 Simple Logarithmic Rule. 521 522 Relative Maximum. 573 Scalar Multiplication. 243 ex. 160. 520 522 Marginal Profit. 31 40 Non-removable Discontinuity. 63 Reduced Row Echelon Form. 411 Sharp Point. 244 248 Relative Change. 595 Partial Derivative. 367 Singular Matrix. 386 387 Power Function. 131 One-to-One Function. 156 Non-linear Inequalities. 122 Quadratic Regression. 86 Quartic Regression. 576 Mean Value Theorem. 14. Linearity Property. 500. 281 Separable Differential Equations. 481 Profit. 24 Quadratic Function. 573 Normal Distribution. and April Allen Materowski. 239 Simple Exponential Rule. 14 Secant Line. 151 Slope. 88 Scatter Plot. 242 Minimum. 259. 388 Range. 440 Related Rates. 16 Radioactive Decay. 405 406 Percent Change. 269 Similarity. 162 ex. 26 Formula. 68. 599 Rule of Thumb. Inc. 252 ex. 530 ex. 286 Applied Calculus for Business. by Warren B. 190 Only Critical Point Test. 560 Producer Surplus. 566 Form of Linear Systems. 32 Sign Diagram. 511 Maximum. 574 Multiplication. 97 Parameter. 483 486. 570 Properties. 500. 525 Second Derivative Test for Relative Extrema. 521 522 Removable Discontinuity. 555 556. 581 Population Growth. 94 Vertical Line Test. 144 Transversal. 95 Wave Equation. 500 501 Symmetry about y-axis. 237 Volume. 225 226. 500 Topographical Map. 278 Vertex of a Parabola. 91. 95. . 307 Utility Function. 238 Unitary Demand. and April Allen Materowski. 127 y-intercept. 109. 171 Three-dimensional Coordinate System. Inc. 278 Substitute Product. and Finance. Copyright © 2007 by Pearson Education. 143 Translations. 538 539 Summation Formulas. 166 Equation of. 519 Zero of a Function. by Warren B. 87. 127 Zero Matrix. Gordon. 513 Vector Multiplication. Wang.690 * ** Index Speed. Published by Pearson Learning Solutions. 515 Transcendental Function. 135. Economics. 90. Walter O. 115 Surface. 436 Supply Function. 553 x-intercept. 130 Symmetry about Origin. 131 Tabular Integration. 569 Velocity. 67 Vertical Asymptote. 492 Tangent Line. 207 Vertical Angles. 567 Applied Calculus for Business.