UNIVERSITY
CALCULUS EARLY TRANSCENDENTALS Second Edition
Joel Hass University of California, Davis Maurice D. Weir Na...
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UNIVERSITY
CALCULUS EARLY TRANSCENDENTALS Second Edition
Joel Hass University of California, Davis Maurice D. Weir Naval Postgraduate School George B. Thomas, Jr. Massachusetts Institute of Technology
Editor in Chief: Deirdre Lynch Senior Acquisitions Editor: William Hoffman Sponsoring Editor: Caroline Celano Senior Content Editor: Elizabeth Bernardi Editorial Assistant: Brandon Rawnsley Senior Managing Editor: Karen Wernholm Associate Managing Editor: Tamela Ambush Senior Production Project Manager: Sheila Spinney Digital Assets Manager: Marianne Groth Supplements Production Coordinator: Kerri McQueen Associate Media Producer: Stephanie Green Software Development: Kristina Evans and Marty Wright Executive Marketing Manager: Jeff Weidenaar Marketing Coordinator: Kendra Bassi Senior Author Support/Technology Specialist: Joe Vetere Rights and Permissions Advisor: Michael Joyce Image Manager: Rachel Youdelman Manufacturing Manager: Evelyn Beaton Senior Manufacturing Buyer: Carol Melville Senior Media Buyer: Ginny Michaud Design Manager: Andrea Nix Production Coordination, Composition, and Illustrations: Nesbitt Graphics, Inc. Cover Design: Andrea Nix Cover Image: Black Shore III—Iceland, 2007. All content copyright © 2009 Josef Hoflehner For permission to use copyrighted material, grateful acknowledgment is made to the copyright holders on page C1, which is hereby made part of this copyright page. Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book, and Pearson Education was aware of a trademark claim, the designations have been printed in initial caps or all caps. Library of Congress CataloginginPublication Data Hass, Joel. University calculus: early transcendentals/Joel Hass, Maurice D. Weir, George B. Thomas, Jr.—2nd ed. p. cm. Rev. ed. of: University calculus. c2007. ISBN 9780321717399 (alk. paper) 1. Calculus—Textbooks. I. Weir, Maurice D. II. Thomas, George B. (George Brinton), 1914–2006. III. Title. QA303.2.H373 2011 515—dc22
2010035141
Copyright © 2012, 2007, Pearson Education, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. For information on obtaining permission for use of material in this work, please submit a written request to Pearson Education, Inc., Rights and Contracts Department, 501 Boylston Street, Suite 900, Boston, MA 02116, fax your request to 6176713447, or email at http://www.pearsoned.com/legal/permissions.htm. 1 2 3 4 5 6—CRK—14 13 12 11
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ISBN 13: 9780321717399 ISBN 10: 0321717392
CONTENTS
1
Preface
ix
Functions
1 1.1 1.2 1.3 1.4 1.5 1.6
2
14
Limits and Continuity 2.1 2.2 2.3 2.4 2.5 2.6
3
Functions and Their Graphs 1 Combining Functions; Shifting and Scaling Graphs Trigonometric Functions 21 Graphing with Calculators and Computers 29 Exponential Functions 33 Inverse Functions and Logarithms 39
Rates of Change and Tangents to Curves 52 Limit of a Function and Limit Laws 59 The Precise Definition of a Limit 70 OneSided Limits 79 Continuity 86 Limits Involving Infinity; Asymptotes of Graphs QUESTIONS TO GUIDE YOUR REVIEW 110 PRACTICE EXERCISES 111 ADDITIONAL AND ADVANCED EXERCISES 113
52
97
Differentiation 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11
116 Tangents and the Derivative at a Point 116 The Derivative as a Function 120 Differentiation Rules 129 The Derivative as a Rate of Change 139 Derivatives of Trigonometric Functions 149 The Chain Rule 156 Implicit Differentiation 164 Derivatives of Inverse Functions and Logarithms Inverse Trigonometric Functions 180 Related Rates 186 Linearization and Differentials 195 QUESTIONS TO GUIDE YOUR REVIEW 206 PRACTICE EXERCISES 206 ADDITIONAL AND ADVANCED EXERCISES 211
170
iii
iv
Contents
4
Applications of Derivatives 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8
5
289 Area and Estimating with Finite Sums 289 Sigma Notation and Limits of Finite Sums 299 The Definite Integral 305 The Fundamental Theorem of Calculus 317 Indefinite Integrals and the Substitution Method 328 Substitution and Area Between Curves 335 QUESTIONS TO GUIDE YOUR REVIEW 345 PRACTICE EXERCISES 345 ADDITIONAL AND ADVANCED EXERCISES 349
Applications of Definite Integrals 6.1 6.2 6.3 6.4 6.5 6.6
7
230
Integration 5.1 5.2 5.3 5.4 5.5 5.6
6
Extreme Values of Functions 214 The Mean Value Theorem 222 Monotonic Functions and the First Derivative Test Concavity and Curve Sketching 235 Indeterminate Forms and L’Hôpital’s Rule 246 Applied Optimization 255 Newton’s Method 266 Antiderivatives 271 QUESTIONS TO GUIDE YOUR REVIEW 281 PRACTICE EXERCISES 281 ADDITIONAL AND ADVANCED EXERCISES 285
214
353
Volumes Using CrossSections 353 Volumes Using Cylindrical Shells 364 Arc Length 372 Areas of Surfaces of Revolution 378 Work 383 Moments and Centers of Mass 389 QUESTIONS TO GUIDE YOUR REVIEW 397 PRACTICE EXERCISES 397 ADDITIONAL AND ADVANCED EXERCISES 399
Integrals and Transcendental Functions 7.1 7.2 7.3
The Logarithm Defined as an Integral 401 Exponential Change and Separable Differential Equations Hyperbolic Functions 420 QUESTIONS TO GUIDE YOUR REVIEW 428 PRACTICE EXERCISES 428 ADDITIONAL AND ADVANCED EXERCISES 429
401 411
Contents
8
Techniques of Integration 8.1 8.2 8.3 8.4 8.5 8.6 8.7
9
448
563
Parametrizations of Plane Curves 563 Calculus with Parametric Curves 570 Polar Coordinates 579 Graphing in Polar Coordinates 583 Areas and Lengths in Polar Coordinates 587 Conics in Polar Coordinates 591 QUESTIONS TO GUIDE YOUR REVIEW 598 PRACTICE EXERCISES 599 ADDITIONAL AND ADVANCED EXERCISES 600
Vectors and the Geometry of Space 11.1 11.2 11.3 11.4 11.5 11.6
486
Sequences 486 Infinite Series 498 The Integral Test 507 Comparison Tests 512 The Ratio and Root Tests 517 Alternating Series, Absolute and Conditional Convergence 522 Power Series 529 Taylor and Maclaurin Series 538 Convergence of Taylor Series 543 The Binomial Series and Applications of Taylor Series 550 QUESTIONS TO GUIDE YOUR REVIEW 559 PRACTICE EXERCISES 559 ADDITIONAL AND ADVANCED EXERCISES 561
Parametric Equations and Polar Coordinates 10.1 10.2 10.3 10.4 10.5 10.6
11
431
Infinite Sequences and Series 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10
10
Integration by Parts 432 Trigonometric Integrals 439 Trigonometric Substitutions 444 Integration of Rational Functions by Partial Fractions Integral Tables and Computer Algebra Systems 456 Numerical Integration 461 Improper Integrals 471 QUESTIONS TO GUIDE YOUR REVIEW 481 PRACTICE EXERCISES 481 ADDITIONAL AND ADVANCED EXERCISES 483
v
ThreeDimensional Coordinate Systems 602 Vectors 607 The Dot Product 616 The Cross Product 624 Lines and Planes in Space 630 Cylinders and Quadric Surfaces 638 QUESTIONS TO GUIDE YOUR REVIEW 643 PRACTICE EXERCISES 644 ADDITIONAL AND ADVANCED EXERCISES 646
602
vi
Contents
12
VectorValued Functions and Motion in Space 12.1 12.2 12.3 12.4 12.5 12.6
13
Curves in Space and Their Tangents 649 Integrals of Vector Functions; Projectile Motion 657 Arc Length in Space 664 Curvature and Normal Vectors of a Curve 668 Tangential and Normal Components of Acceleration 674 Velocity and Acceleration in Polar Coordinates 679 QUESTIONS TO GUIDE YOUR REVIEW 682 PRACTICE EXERCISES 683 ADDITIONAL AND ADVANCED EXERCISES 685
Partial Derivatives 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8
14
649
686 Functions of Several Variables 686 Limits and Continuity in Higher Dimensions 694 Partial Derivatives 703 The Chain Rule 714 Directional Derivatives and Gradient Vectors 723 Tangent Planes and Differentials 730 Extreme Values and Saddle Points 740 Lagrange Multipliers 748 QUESTIONS TO GUIDE YOUR REVIEW 757 PRACTICE EXERCISES 758 ADDITIONAL AND ADVANCED EXERCISES 761
Multiple Integrals 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8
763 Double and Iterated Integrals over Rectangles 763 Double Integrals over General Regions 768 Area by Double Integration 777 Double Integrals in Polar Form 780 Triple Integrals in Rectangular Coordinates 786 Moments and Centers of Mass 795 Triple Integrals in Cylindrical and Spherical Coordinates Substitutions in Multiple Integrals 814 QUESTIONS TO GUIDE YOUR REVIEW 823 PRACTICE EXERCISES 823 ADDITIONAL AND ADVANCED EXERCISES 825
802
Contents
15
Integration in Vector Fields 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8
16
17
828
Line Integrals 828 Vector Fields and Line Integrals: Work, Circulation, and Flux 834 Path Independence, Conservative Fields, and Potential Functions 847 Green’s Theorem in the Plane 858 Surfaces and Area 870 Surface Integrals 880 Stokes’ Theorem 889 The Divergence Theorem and a Unified Theory 900 QUESTIONS TO GUIDE YOUR REVIEW 911 PRACTICE EXERCISES 911 ADDITIONAL AND ADVANCED EXERCISES 914
FirstOrder Differential Equations 16.1 16.2 16.3 16.4 16.5
Online
Solutions, Slope Fields, and Euler’s Method 162 FirstOrder Linear Equations 1610 Applications 1616 Graphical Solutions of Autonomous Equations 1622 Systems of Equations and Phase Planes 1629
SecondOrder Differential Equations 17.1 17.2 17.3 17.4 17.5
vii
Online
SecondOrder Linear Equations 171 Nonhomogeneous Linear Equations 178 Applications 1717 Euler Equations 1723 Power Series Solutions 1726
Appendices
AP1 A.1 A.2 A.3 A.4 A.5 A.6 A.7 A.8 A.9 A.10 A.11
Real Numbers and the Real Line AP1 Mathematical Induction AP6 Lines, Circles, and Parabolas AP10 Conic Sections AP18 Proofs of Limit Theorems AP26 Commonly Occurring Limits AP29 Theory of the Real Numbers AP31 Complex Numbers AP33 The Distributive Law for Vector Cross Products AP43 The Mixed Derivative Theorem and the Increment Theorem Taylor’s Formula for Two Variables AP48
AP44
Answers to OddNumbered Exercises
A1
Index
I1
Credits
C1
A Brief Table of Integrals
T1
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PREFACE We have significantly revised this edition of University Calculus, Early Transcendentals to meet the changing needs of today’s instructors and students. The result is a book with more examples, more midlevel exercises, more figures, better conceptual flow, and increased clarity and precision. As with the previous edition, this new edition provides a briefer, modern introduction to calculus that supports conceptual understanding but retains the essential elements of a traditional course. These enhancements are closely tied to an expanded version of MyMathLab® for this text (discussed further on), providing additional support for students and flexibility for instructors. In this second edition, we introduce the basic transcendental functions in Chapter 1. After reviewing the basic trigonometric functions, we present the family of exponential functions using an algebraic and graphical approach, with the natural exponential described as a particular member of this family. Logarithms are then defined as the inverse functions of the exponentials, and we also discuss briefly the inverse trigonometric functions. We fully incorporate these functions throughout our developments of limits, derivatives, and integrals in the next five chapters of the book, including the examples and exercises. This approach gives students the opportunity to work early with exponential and logarithmic functions in combinations with polynomials, rational and algebraic functions, and trigonometric functions as they learn the concepts, operations, and applications of singlevariable calculus. Later, in Chapter 7, we revisit the definition of transcendental functions, now giving a more rigorous presentation. Here we define the natural logarithm function as an integral with the natural exponential as its inverse. Today, an increasing number of students become familiar with the terminology and operational methods of calculus in high school. However, their conceptual understanding of calculus is often quite limited when they enter college. We have acknowledged this reality by concentrating on concepts and their applications throughout. We encourage students to think beyond memorizing formulas and to generalize concepts as they are introduced. Our hope is that after taking calculus, students will be confident in their problemsolving and reasoning abilities. Mastering a beautiful subject with practical applications to the world is its own reward, but the real gift is the ability to think and generalize. We intend this book to provide support and encouragement for both.
Changes for the Second Edition CONTENT In preparing this edition we have maintained the basic structure of the Table of Contents from the first edition, yet we have paid attention to requests by current users and reviewers to postpone the introduction of parametric equations until we present polar coordinates. We have made numerous revisions to most of the chapters, detailed as follows:
•
Functions We condensed this chapter to focus on reviewing function concepts and introducing the transcendental functions. Prerequisite material covering real numbers, intervals, increments, straight lines, distances, circles, and the conic sections is presented in Appendices 1–4.
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• •
•
•
•
•
•
•
Limits To improve the flow of this chapter, we combined the ideas of limits involving infinity and their associations with asymptotes to the graphs of functions, placing them together in the final section of Chapter 2. Differentiation While we use rates of change and tangents to curves as motivation for studying the limit concept, we now merge the derivative concept into a single chapter. We reorganized and increased the number of related rates examples, and we added new examples and exercises on graphing rational functions. L’Hôpital’s Rule is presented as an application section, consistent with our early coverage of the transcendental functions. Antiderivatives and Integration We maintain the organization of the first edition in placing antiderivatives as the final topic of Chapter 4, covering applications of derivatives. Our focus is on “recovering a function from its derivative” as the solution to the simplest type of firstorder differential equation. Integrals, as “limits of Riemann sums,” motivated primarily by the problem of finding the areas of general regions with curved boundaries, are a new topic forming the substance of Chapter 5. After carefully developing the integral concept, we turn our attention to its evaluation and connection to antiderivatives captured in the Fundamental Theorem of Calculus. The ensuing applications then define the various geometric ideas of area, volume, lengths of paths, and centroids, all as limits of Riemann sums giving definite integrals, which can be evaluated by finding an antiderivative of the integrand. Series We retain the organizational structure and content of the first edition for the topics of sequences and series. We have added several new figures and exercises to the various sections, and we revised some of the proofs related to convergence of power series in order to improve the accessibility of the material for students. The request stated by one of our users as, “anything you can do to make this material easier for students will be welcomed by our faculty,” drove our thinking for revisions to this chapter. Parametric Equations Several users requested that we move this topic into Chapter 10, where we also cover polar coordinates. We have done this, realizing that many departments choose to cover these topics at the beginning of Calculus III, in preparation for their coverage of vectors and multivariable calculus. VectorValued Functions We streamlined the topics in this chapter to place more emphasis on the conceptual ideas supporting the later material on partial derivatives, the gradient vector, and line integrals. We condensed the discussions of the Frenet frame and Kepler’s three laws of planetary motion. Multivariable Calculus We have further enhanced the art in these chapters, and we have added many new figures, examples, and exercises. As with the first edition, we continue to make the connections of multivariable ideas with their singlevariable analogues studied earlier in the book. Vector Fields We devoted considerable effort to improving the clarity and mathematical precision of our treatment of vector integral calculus, including many additional examples, figures, and exercises. Important theorems and results are stated more clearly and completely, together with enhanced explanations of their hypotheses and mathematical consequences. The area of a surface is still organized into a single section, and surfaces defined implicitly or explicitly are treated as special cases of the more general parametric representation. Surface integrals and their applications then follow as a separate section. Stokes’ Theorem and the Divergence Theorem continue to be presented as generalizations of Green’s Theorem to three dimensions. A number of new examples and figures have been added illustrating these important themes.
EXERCISES AND EXAMPLES We know that the exercises and examples are critical components in learning calculus. Because of this importance, we have updated, improved, and increased the number of exercises in nearly every section of the book. There are over 750 new exercises in this edition. We continue our organization and grouping of exercises by
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topic as in earlier editions, progressing from computational problems to applied and theoretical problems. Over 70 examples have been added to clarify or deepen the meaning of the topic being discussed and to help students understand its mathematical consequences or applications to science and engineering. The new examples appear throughout the text, with emphasis on key concepts and results. ART Because of their importance to learning calculus, we have continued to improve existing figures in University Calculus, Early Transcendentals, and we have created a significant number of new ones. We continue to use color consistently and pedagogically to enhance the conceptual idea that is being illustrated. We have also taken a fresh look at all of the figure captions, paying considerable attention to clarity and precision in short statements.
y y 1x
No matter what positive number is, the graph enters this band at x 1 and stays. y
N – 1 0 y –
z
M 1
x
–
No matter what positive number is, the graph enters this band at x – 1 and stays.
FIGURE 2.50, page 98 The geometric explanation of a finite limit as x : ; q .
y x
FIGURE 15.9, page 835 A surface in a space occupied by a moving fluid.
MYMATHLAB AND MATHXL The increasing use of and demand for online homework systems has driven the changes to MyMathLab and MathXL® for University Calculus, Early Transcendentals. The MyMathLab and MathXL courses now include significantly more exercises of all types.
Continuing Features RIGOR The level of rigor is consistent with the first edition. We continue to distinguish between formal and informal discussions and to point out their differences. We think starting with a more intuitive, less formal, approach helps students understand a new or difficult concept so they can then appreciate its full mathematical precision and outcomes. We pay attention to defining ideas carefully and to proving theorems appropriate for calculus students, while mentioning deeper or subtler issues they would study in a more advanced course. Our organization and distinctions between informal and formal discussions give the instructor a degree of flexibility in the amount and depth of coverage of the various topics. For example, while we do not prove the Intermediate Value Theorem or the Extreme Value Theorem for continuous functions on a … x … b, we do state these theorems precisely, illustrate their meanings in numerous examples, and use them to prove other important results. Furthermore, for those instructors who desire greater depth of coverage, in Appendix 7 we discuss the reliance of the validity of these theorems on the completeness of the real numbers.
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WRITING EXERCISES Writing exercises placed throughout the text ask students to explore and explain a variety of calculus concepts and applications. In addition, the end of each chapter contains a list of questions for students to review and summarize what they have learned. Many of these exercises make good writing assignments. ENDOFCHAPTER REVIEWS In addition to problems appearing after each section, each chapter culminates with review questions, practice exercises covering the entire chapter, and a series of Additional and Advanced Exercises serving to include more challenging or synthesizing problems. WRITING AND APPLICATIONS As always, this text continues to be easy to read, conversational, and mathematically rich. Each new topic is motivated by clear, easytounderstand examples and is then reinforced by its application to realworld problems of immediate interest to students. A hallmark of this book has been the application of calculus to science and engineering. These applied problems have been updated, improved, and extended continually over the last several editions. TECHNOLOGY Technology can be incorporated in a course using this text, according to the taste of the instructor. Each section contains exercises requiring the use of technology; these are marked with a T if suitable for calculator or computer use, or they are labeled Computer Explorations if a computer algebra system (CAS, such as Maple or Mathematica) is required.
Text Versions UNIVERSITY CALCULUS, EARLY TRANSCENDENTALS, Second Edition Complete (Chapters 1–15), ISBN 0321717392  9780321717399 Single Variable Calculus (Chapters 1–10), ISBN 0321694597  9780321694591 Multivariable Calculus (Chapters 9–15), ISBN 0321694600  9780321694607 University Calculus, Early Transcendentals, introduces and integrates transcendental functions (such as inverse trigonometric, exponential, and logarithmic functions) into the exposition, examples, and exercises of the early chapters alongside the algebraic functions. Electronic versions of the text are available within MyMathLab (www.mymathlab.com) or at CourseSmart.com.
Instructor’s Edition for University Calculus, Early Transcendentals, Second Edition ISBN 0321717473  9780321717474 In addition to including all of the answers present in the student editions, the Instructor’s Edition includes evennumbered answers for Chapters 1–11.
Print Supplements INSTRUCTOR’S SOLUTIONS MANUAL Single Variable Calculus (Chapters 1–10), ISBN 0321717481  9780321717481 Multivariable Calculus (Chapters 9–15), ISBN 032171749X  9780321717498 The Instructor’s Solutions Manual by William Ardis, Collin County Community College, contains complete workedout solutions to all of the exercises in University Calculus, Early Transcendentals.
STUDENT’S SOLUTIONS MANUAL Single Variable Calculus (Chapters 1–10), ISBN 0321694627  9780321694621 Multivariable Calculus (Chapters 9–15), ISBN 0321694546  9780321694546 The Student’s Solutions Manual by William Ardis, Collin County Community College, is designed for the student and contains carefully workedout solutions to all of the oddnumbered exercises in University Calculus, Early Transcendentals.
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JUSTINTIME ALGEBRA AND TRIGONOMETRY FOR EARLY TRANSCENDENTALS CALCULUS, Third Edition ISBN 0321320506  9780321320506 Sharp algebra and trigonometry skills are critical to mastering calculus, and JustinTime Algebra and Trigonometry for Early Transcendentals Calculus by Guntram Mueller and Ronald I. Brent is designed to bolster these skills while students study calculus. As students make their way through calculus, this text is with them every step of the way, showing them the necessary algebra or trigonometry topics and pointing out potential problem spots. The easytouse table of contents has algebra and trigonometry topics arranged in the order in which students will need them as they study calculus.
CALCULUS REVIEW CARDS The Calculus Review Cards (one for Single Variable and another for Multivariable) are a student resource containing important formulas, functions, definitions, and theorems that correspond precisely to the Thomas’ Calculus series. These cards can work as a reference for completing homework assignments or as an aid in studying, and are available bundled with a new text. Contact your Pearson sales representative for more information.
Media and Online Supplements TECHNOLOGY RESOURCE MANUALS Maple Manual by James Stapleton, North Carolina State University Mathematica Manual by Marie Vanisko, Carroll College TIGraphing Calculator Manual by Elaine McDonaldNewman, Sonoma State University These manuals cover Maple 13, Mathematica 7, and the TI83 Plus/TI84 Plus and TI89, respectively. Each manual provides detailed guidance for integrating a specific software package or graphing calculator throughout the course, including syntax and commands. These manuals are available to students and instructors through the University Calculus, Early Transcendentals Web site, www.pearsonhighered.com/thomas, and MyMathLab.
WEB SITE www.pearsonhighered.com/thomas The University Calculus, Early Transcendentals Web site contains the chapters on FirstOrder and SecondOrder Differential Equations, including oddnumbered answers, and provides the expanded historical biographies and essays referenced in the text. Also available is a collection of Maple and Mathematica modules and the Technology Resource Manuals.
Video Lectures with Optional Captioning The Video Lectures with Optional Captioning feature an engaging team of mathematics instructors who present comprehensive coverage of topics in the text. The lecturers’ presentations include examples and exercises from the text and support an approach that emphasizes visualization and problem solving. Available only through MyMathLab and MathXL.
MyMathLab Online Course (access code required) MyMathLab is a textspecific, easily customizable online course that integrates interactive multimedia instruction with textbook content. MyMathLab gives you the tools you need to deliver all or a portion of your course online, whether your students are in a lab setting or working from home.
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Interactive homework exercises, correlated to your textbook at the objective level, are algorithmically generated for unlimited practice and mastery. Most exercises are freeresponse and provide guided solutions, sample problems, and tutorial learning aids for extra help.
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“Getting Ready” chapter includes hundreds of exercises that address prerequisite skills in algebra and trigonometry. Each student can receive remediation for just those skills he or she needs help with.
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Personalized homework assignments that you can design to meet the needs of your class. MyMathLab tailors the assignment for each student based on his or her test or quiz scores. Each student receives a homework assignment that contains only the problems he or she still needs to master.
•
Personalized Study Plan, generated when students complete a test or quiz or homework, indicates which topics have been mastered and links to tutorial exercises for topics students have not mastered. You can customize the Study Plan so that the topics available match your course content.
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Multimedia learning aids, such as video lectures and podcasts, Java applets, animations, and a complete multimedia textbook, help students independently improve their understanding and performance. You can assign these multimedia learning aids as homework to help your students grasp the concepts.
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Homework and Test Manager lets you assign homework, quizzes, and tests that are automatically graded. Select just the right mix of questions from the MyMathLab exercise bank, instructorcreated custom exercises, and/or TestGen® test items.
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Gradebook, designed specifically for mathematics and statistics, automatically tracks students’ results, lets you stay on top of student performance, and gives you control over how to calculate final grades. You can also add offline (paperandpencil) grades to the gradebook.
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MathXL Exercise Builder allows you to create static and algorithmic exercises for your online assignments. You can use the library of sample exercises as an easy starting point, or you can edit any courserelated exercise.
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Pearson Tutor Center (www.pearsontutorservices.com) access is automatically included with MyMathLab. The Tutor Center is staffed by qualified math instructors who provide textbookspecific tutoring for students via tollfree phone, fax, email, and interactive Web sessions.
Students do their assignments in the Flash®based MathXL Player, which is compatible with almost any browser (Firefox®, Safari™, or Internet Explorer®) on almost any platform (Macintosh® or Windows®). MyMathLab is powered by CourseCompass™, Pearson Education’s online teaching and learning environment, and by MathXL®, our online homework, tutorial, and assessment system. MyMathLab is available to qualified adopters. For more information, visit www.mymathlab.com or contact your Pearson representative.
MathXL Online Course (access code required) MathXL is an online homework, tutorial, and assessment system that accompanies Pearson’s textbooks in mathematics or statistics.
•
Interactive homework exercises, correlated to your textbook at the objective level, are algorithmically generated for unlimited practice and mastery. Most exercises are freeresponse and provide guided solutions, sample problems, and learning aids for extra help.
•
“Getting Ready” chapter includes hundreds of exercises that address prerequisite skills in algebra and trigonometry. Each student can receive remediation for just those skills he or she needs help with.
•
Personalized Homework assignments are designed by the instructor to meet the needs of the class, and then personalized for each student based on his or her test or quiz scores. As a result, each student receives a homework assignment in which the problems cover only the objectives for which he or she has not achieved mastery.
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•
Personalized Study Plan, generated when students complete a test or quiz or homework, indicates which topics have been mastered and links to tutorial exercises for topics students have not mastered. Instructors can customize the available topics in the study plan to match their course concepts.
•
Multimedia learning aids, such as video lectures, Java applets, and animations, help students independently improve their understanding and performance. These are assignable as homework, to further encourage their use.
•
Gradebook, designed specifically for mathematics and statistics, automatically tracks students’ results, lets you stay on top of student performance, and gives you control over how to calculate final grades.
•
MathXL Exercise Builder allows you to create static and algorithmic exercises for your online assignments. You can use the library of sample exercises as an easy starting point or use the Exercise Builder to edit any of the courserelated exercises.
•
Homework and Test Manager lets you create online homework, quizzes, and tests that are automatically graded. Select just the right mix of questions from the MathXL exercise bank, instructorcreated custom exercises, and/or TestGen test items.
The new, Flashbased MathXL Player is compatible with almost any browser (Firefox, Safari, or Internet Explorer) on almost any platform (Macintosh or Windows). MathXL is available to qualified adopters. For more information, visit our Web site at www .mathxl.com, or contact your Pearson representative.
TestGen TestGen (www.pearsonhighered.com/testgen) enables instructors to build, edit, print, and administer tests using a computerized bank of questions developed to cover all the objectives of the text. TestGen is algorithmically based, allowing instructors to create multiple but equivalent versions of the same question or test with the click of a button. Instructors can also modify test bank questions or add new questions. Tests can be printed or administered online. The software and testbank are available for download from Pearson Education’s online catalog. (www.pearsonhighered.com)
PowerPoint® Lecture Slides These classroom presentation slides are geared specifically to the sequence and philosophy of the Thomas’ Calculus series. Key graphics from the book are included to help bring the concepts alive in the classroom.These files are available to qualified instructors through the Pearson Instructor Resource Center, www.pearsonhighered/irc, and MyMathLab.
Acknowledgments We would like to express our thanks to the people who made many valuable contributions to this edition as it developed through its various stages:
Accuracy Checkers Rhea Meyerholtz Tom Wegleitner Gary Williams
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Reviewers for the Second Edition Robert A. Beezer, University of Puget Sound Przemyslaw Bogacki, Old Dominion University Leonard Chastofsky, University of Georgia Meighan Dillon, Southern Polytechnic State University Anne Dougherty, University of Colorado Said Fariabi, San Antonio College Klaus Fischer, George Mason University Tim Flood, Pittsburg State University Rick Ford, California State University—Chico Robert Gardner, East Tennessee State University Christopher Heil, Georgia Institute of Technology David Hemmer, SUNY—Buffalo Joshua Brandon Holden, RoseHulman Institute of Technology Alexander Hulpke, Colorado State University Jacqueline Jensen, Sam Houston State University Jennifer M. Johnson, Princeton University Hideaki Kaneko, Old Dominion University Przemo Kranz, University of Mississippi John Kroll, Old Dominion University Glenn Ledder, University of Nebraska—Lincoln Matthew Leingang, New York University Xin Li, University of Central Florida Maura Mast, University of Massachusetts—Boston Val Mohanakumar, Hillsborough Community College—Dale Mabry Campus Aaron Montgomery, Central Washington University Yibiao Pan, University of Pittsburgh Christopher M. Pavone, California State University at Chico Cynthia Piez, University of Idaho Brooke Quinlan, Hillsborough Community College—Dale Mabry Campus Rebecca A. Segal, Virginia Commonwealth University Andrew V. Sills, Georgia Southern University Edward E. Slaminka, Auburn University Alex Smith, University of Wisconsin—Eau Claire Mark A. Smith, Miami University Donald Solomon, University of Wisconsin—Milwaukee John Sullivan, Black Hawk College Maria Terrell, Cornell University Blake Thornton, Washington University in St. Louis David Walnut, George Mason University Adrian Wilson, University of Montevallo Bobby Winters, Pittsburg State University Dennis Wortman, University of Massachusetts—Boston
FPO
1 FUNCTIONS OVERVIEW Functions are fundamental to the study of calculus. In this chapter we review what functions are and how they are pictured as graphs, how they are combined and transformed, and ways they can be classified. We review the trigonometric functions, and we discuss misrepresentations that can occur when using calculators and computers to obtain a function’s graph. We also discuss inverse, exponential, and logarithmic functions. The real number system, Cartesian coordinates, straight lines, circles, parabolas, ellipses, and hyperbolas are reviewed in the Appendices.
1.1
Functions and Their Graphs Functions are a tool for describing the real world in mathematical terms. A function can be represented by an equation, a graph, a numerical table, or a verbal description; we will use all four representations throughout this book. This section reviews these function ideas.
Functions; Domain and Range The temperature at which water boils depends on the elevation above sea level (the boiling point drops as you ascend). The interest paid on a cash investment depends on the length of time the investment is held. The area of a circle depends on the radius of the circle. The distance an object travels at constant speed along a straightline path depends on the elapsed time. In each case, the value of one variable quantity, say y, depends on the value of another variable quantity, which we might call x. We say that “y is a function of x” and write this symbolically as y = ƒ(x)
(“y equals ƒ of x”).
In this notation, the symbol ƒ represents the function, the letter x is the independent variable representing the input value of ƒ, and y is the dependent variable or output value of ƒ at x.
DEFINITION A function ƒ from a set D to a set Y is a rule that assigns a unique (single) element ƒsxd H Y to each element x H D. The set D of all possible input values is called the domain of the function. The set of all values of ƒ(x) as x varies throughout D is called the range of the function. The range may not include every element in the set Y. The domain and range of a function can be any sets of objects, but often in calculus they are sets of real numbers interpreted as points of a coordinate line. (In Chapters 12–15, we will encounter functions for which the elements of the sets are points in the coordinate plane or in space.)
1
2
x
Chapter 1: Functions
f
Input (domain)
Output (range)
f (x)
FIGURE 1.1 A diagram showing a function as a kind of machine.
x a D domain set
f (a)
f (x)
Y set containing the range
FIGURE 1.2 A function from a set D to a set Y assigns a unique element of Y to each element in D.
Often a function is given by a formula that describes how to calculate the output value from the input variable. For instance, the equation A = pr 2 is a rule that calculates the area A of a circle from its radius r (so r, interpreted as a length, can only be positive in this formula). When we define a function y = ƒsxd with a formula and the domain is not stated explicitly or restricted by context, the domain is assumed to be the largest set of real xvalues for which the formula gives real yvalues, the socalled natural domain. If we want to restrict the domain in some way, we must say so. The domain of y = x 2 is the entire set of real numbers. To restrict the domain of the function to, say, positive values of x, we would write “y = x 2, x 7 0.” Changing the domain to which we apply a formula usually changes the range as well. The range of y = x 2 is [0, q d. The range of y = x 2, x Ú 2, is the set of all numbers obtained by squaring numbers greater than or equal to 2. In set notation (see Appendix 1), the range is 5x 2 ƒ x Ú 26 or 5y ƒ y Ú 46 or [4, q d. When the range of a function is a set of real numbers, the function is said to be realvalued. The domains and ranges of many realvalued functions of a real variable are intervals or combinations of intervals. The intervals may be open, closed, or half open, and may be finite or infinite. The range of a function is not always easy to find. A function ƒ is like a machine that produces an output value ƒ(x) in its range whenever we feed it an input value x from its domain (Figure 1.1). The function keys on a calculator give an example of a function as a machine. For instance, the 2x key on a calculator gives an output value (the square root) whenever you enter a nonnegative number x and press the 2x key. A function can also be pictured as an arrow diagram (Figure 1.2). Each arrow associates an element of the domain D with a unique or single element in the set Y. In Figure 1.2, the arrows indicate that ƒ(a) is associated with a, ƒ(x) is associated with x, and so on. Notice that a function can have the same value at two different input elements in the domain (as occurs with ƒ(a) in Figure 1.2), but each input element x is assigned a single output value ƒ(x).
EXAMPLE 1 Let’s verify the natural domains and associated ranges of some simple functions. The domains in each case are the values of x for which the formula makes sense. Function y y y y y
= = = = =
x2 1>x 2x 24  x 21  x 2
Domain (x)
Range ( y)
s  q, q d s  q , 0d ´ s0, q d [0, q d s  q , 4] [1, 1]
[0, q d s  q , 0d ´ s0, q d [0, q d [0, q d [0, 1]
The formula y = x 2 gives a real yvalue for any real number x, so the domain is s  q , q d. The range of y = x 2 is [0, q d because the square of any real number is nonnegative and every nonnegative number y is the square of its own square root, y = A 2y B 2 for y Ú 0. The formula y = 1>x gives a real yvalue for every x except x = 0. For consistency in the rules of arithmetic, we cannot divide any number by zero. The range of y = 1>x, the set of reciprocals of all nonzero real numbers, is the set of all nonzero real numbers, since y = 1>(1>y). That is, for y Z 0 the number x = 1>y is the input assigned to the output value y. The formula y = 1x gives a real yvalue only if x Ú 0. The range of y = 1x is [0, q d because every nonnegative number is some number’s square root (namely, it is the square root of its own square). In y = 14  x, the quantity 4  x cannot be negative. That is, 4  x Ú 0, or x … 4. The formula gives real yvalues for all x … 4. The range of 14  x is [0, q d, the set of all nonnegative numbers. Solution
1.1
3
Functions and Their Graphs
The formula y = 21  x 2 gives a real yvalue for every x in the closed interval from 1 to 1. Outside this domain, 1  x 2 is negative and its square root is not a real number. The values of 1  x 2 vary from 0 to 1 on the given domain, and the square roots of these values do the same. The range of 21  x 2 is [0, 1].
Graphs of Functions If ƒ is a function with domain D, its graph consists of the points in the Cartesian plane whose coordinates are the inputoutput pairs for ƒ. In set notation, the graph is 5sx, ƒsxdd ƒ x H D6. The graph of the function ƒsxd = x + 2 is the set of points with coordinates (x, y) for which y = x + 2. Its graph is the straight line sketched in Figure 1.3. The graph of a function ƒ is a useful picture of its behavior. If (x, y) is a point on the graph, then y = ƒsxd is the height of the graph above the point x. The height may be positive or negative, depending on the sign of ƒsxd (Figure 1.4). y
f (1)
y
f (2) x
yx2
0
1
x
2 f(x)
2 (x, y) –2
x
y x2
2 1 0 1
4 1 0 1
3 2
9 4
2
4
x
0
FIGURE 1.3 The graph of ƒsxd = x + 2 is the set of points (x, y) for which y has the value x + 2 .
EXAMPLE 2
FIGURE 1.4 If (x, y) lies on the graph of ƒ, then the value y = ƒsxd is the height of the graph above the point x (or below x if ƒ(x) is negative).
Graph the function y = x 2 over the interval [2, 2].
Make a table of xypairs that satisfy the equation y = x 2. Plot the points (x, y) whose coordinates appear in the table, and draw a smooth curve (labeled with its equation) through the plotted points (see Figure 1.5).
Solution
y
How do we know that the graph of y = x 2 doesn’t look like one of these curves? (–2, 4)
(2, 4)
4
y
y
y x2 3 ⎛3 , 9⎛ ⎝2 4⎝
2 (–1, 1)
1
–2
0
–1
1
2
y x 2?
y x 2?
(1, 1) x
FIGURE 1.5 Graph of the function in Example 2.
x
x
4
Chapter 1: Functions
To find out, we could plot more points. But how would we then connect them? The basic question still remains: How do we know for sure what the graph looks like between the points we plot? Calculus answers this question, as we will see in Chapter 4. Meanwhile, we will have to settle for plotting points and connecting them as best we can.
Representing a Function Numerically We have seen how a function may be represented algebraically by a formula (the area function) and visually by a graph (Example 2). Another way to represent a function is numerically, through a table of values. Numerical representations are often used by engineers and scientists. From an appropriate table of values, a graph of the function can be obtained using the method illustrated in Example 2, possibly with the aid of a computer. The graph consisting of only the points in the table is called a scatterplot.
EXAMPLE 3 Musical notes are pressure waves in the air. The data in Table 1.1 give recorded pressure displacement versus time in seconds of a musical note produced by a tuning fork. The table provides a representation of the pressure function over time. If we first make a scatterplot and then connect approximately the data points (t, p) from the table, we obtain the graph shown in Figure 1.6. p (pressure)
TABLE 1.1 Tuning fork data
Time
Pressure
Time
0.00091 0.00108 0.00125 0.00144 0.00162 0.00180 0.00198 0.00216 0.00234 0.00253 0.00271 0.00289 0.00307 0.00325 0.00344
0.080 0.200 0.480 0.693 0.816 0.844 0.771 0.603 0.368 0.099 0.141 0.309 0.348 0.248 0.041
0.00362 0.00379 0.00398 0.00416 0.00435 0.00453 0.00471 0.00489 0.00507 0.00525 0.00543 0.00562 0.00579 0.00598
Pressure 0.217 0.480 0.681 0.810 0.827 0.749 0.581 0.346 0.077 0.164 0.320 0.354 0.248 0.035
1.0 0.8 0.6 0.4 0.2 –0.2 –0.4 –0.6
Data
0.001 0.002 0.003 0.004 0.005 0.006
t (sec)
FIGURE 1.6 A smooth curve through the plotted points gives a graph of the pressure function represented by Table 1.1 (Example 3).
The Vertical Line Test for a Function Not every curve in the coordinate plane can be the graph of a function. A function ƒ can have only one value ƒ(x) for each x in its domain, so no vertical line can intersect the graph of a function more than once. If a is in the domain of the function ƒ, then the vertical line x = a will intersect the graph of ƒ at the single point (a, ƒ(a)). A circle cannot be the graph of a function since some vertical lines intersect the circle twice. The circle in Figure 1.7a, however, does contain the graphs of two functions of x: the upper semicircle defined by the function ƒ(x) = 21  x 2 and the lower semicircle defined by the function g (x) =  21  x 2 (Figures 1.7b and 1.7c).
1.1 y
y
y
–1 –1
0
1
x
–1
yx
0
1
2
x
3
x, x,
x Ú 0 x 6 0,
First formula Second formula
y f (x) 2 y1
1 –1
(c) y –1 x 2
whose graph is given in Figure 1.8. The righthand side of the equation means that the function equals x if x Ú 0, and equals x if x 6 0. Piecewisedefined functions often arise when realworld data are modeled. Here are some other examples.
y
–2
x
Sometimes a function is described by using different formulas on different parts of its domain. One example is the absolute value function ƒxƒ = e
FIGURE 1.8 The absolute value function has domain s  q , q d and range [0, q d .
y –x
1 0
PiecewiseDefined Functions
1 –1
x
y x
3 2
–3 –2
1
FIGURE 1.7 (a) The circle is not the graph of a function; it fails the vertical line test. (b) The upper semicircle is the graph of a function ƒsxd = 21  x 2 . (c) The lower semicircle is the graph of a function gsxd =  21  x 2 .
y y –x
0
(b) y 1 x 2
(a) x 2 y 2 1
5
Functions and Their Graphs
EXAMPLE 4
y x2
0
1
The function x, ƒsxd = • x 2, 1,
x
2
FIGURE 1.9 To graph the function y = ƒsxd shown here, we apply different formulas to different parts of its domain (Example 4).
x 6 0 0 … x … 1 x 7 1
First formula Second formula Third formula
is defined on the entire real line but has values given by different formulas, depending on the position of x. The values of ƒ are given by y = x when x 6 0, y = x 2 when 0 … x … 1, and y = 1 when x 7 1. The function, however, is just one function whose domain is the entire set of real numbers (Figure 1.9).
y yx
3
EXAMPLE 5
The function whose value at any number x is the greatest integer less than or equal to x is called the greatest integer function or the integer floor function. It is denoted :x; . Figure 1.10 shows the graph. Observe that
2 y ⎣x⎦
1 –2 –1
1
2
3
x
:2.4; = 2, :2; = 2,
:1.9; = 1, :0.2; = 0,
:0; = 0, : 0.3; = 1
: 1.2; = 2, : 2; = 2.
–2
FIGURE 1.10 The graph of the greatest integer function y = :x; lies on or below the line y = x , so it provides an integer floor for x (Example 5).
EXAMPLE 6 The function whose value at any number x is the smallest integer greater than or equal to x is called the least integer function or the integer ceiling function. It is denoted < x = . Figure 1.11 shows the graph. For positive values of x, this function might represent, for example, the cost of parking x hours in a parking lot which charges $1 for each hour or part of an hour.
6
Chapter 1: Functions
Increasing and Decreasing Functions
y yx
3
If the graph of a function climbs or rises as you move from left to right, we say that the function is increasing. If the graph descends or falls as you move from left to right, the function is decreasing.
2 y ⎡x⎤
1 –2 –1
1
2
x
3
DEFINITIONS Let ƒ be a function defined on an interval I and let x1 and x2 be any two points in I.
–1
1. If ƒsx2) 7 ƒsx1 d whenever x1 6 x2 , then ƒ is said to be increasing on I. 2. If ƒsx2 d 6 ƒsx1 d whenever x1 6 x2 , then ƒ is said to be decreasing on I.
–2
FIGURE 1.11 The graph of the least integer function y = < x = lies on or above the line y = x , so it provides an integer ceiling for x (Example 6).
It is important to realize that the definitions of increasing and decreasing functions must be satisfied for every pair of points x1 and x2 in I with x1 6 x2 . Because we use the inequality 6 to compare the function values, instead of … , it is sometimes said that ƒ is strictly increasing or decreasing on I. The interval I may be finite (also called bounded) or infinite (unbounded) and by definition never consists of a single point (Appendix 1). The function graphed in Figure 1.9 is decreasing on s  q , 0] and increasing on [0, 1]. The function is neither increasing nor decreasing on the interval [1, q d because of the strict inequalities used to compare the function values in the definitions.
EXAMPLE 7
Even Functions and Odd Functions: Symmetry The graphs of even and odd functions have characteristic symmetry properties.
DEFINITIONS
A function y = ƒsxd is an even function of x if ƒs xd = ƒsxd, odd function of x if ƒs xd = ƒsxd,
for every x in the function’s domain. y y x2 (x, y)
(–x, y)
x
0 (a) y y x3
0
(x, y) x
(–x, –y)
The names even and odd come from powers of x. If y is an even power of x, as in y = x 2 or y = x 4 , it is an even function of x because s xd2 = x 2 and s xd4 = x 4 . If y is an odd power of x, as in y = x or y = x 3 , it is an odd function of x because s xd1 = x and s xd3 = x 3 . The graph of an even function is symmetric about the yaxis. Since ƒs xd = ƒsxd, a point (x, y) lies on the graph if and only if the point s x, yd lies on the graph (Figure 1.12a). A reflection across the yaxis leaves the graph unchanged. The graph of an odd function is symmetric about the origin. Since ƒs xd = ƒsxd, a point (x, y) lies on the graph if and only if the point s x, yd lies on the graph (Figure 1.12b). Equivalently, a graph is symmetric about the origin if a rotation of 180° about the origin leaves the graph unchanged. Notice that the definitions imply that both x and x must be in the domain of ƒ.
EXAMPLE 8 (b)
FIGURE 1.12 (a) The graph of y = x 2 (an even function) is symmetric about the yaxis. (b) The graph of y = x 3 (an odd function) is symmetric about the origin.
ƒsxd = x 2
Even function: s xd2 = x 2 for all x; symmetry about yaxis.
ƒsxd = x 2 + 1
Even function: s xd2 + 1 = x 2 + 1 for all x; symmetry about yaxis (Figure 1.13a).
ƒsxd = x
Odd function: s xd = x for all x; symmetry about the origin.
ƒsxd = x + 1
Not odd: ƒs xd = x + 1 , but ƒsxd = x  1 . The two are not equal. Not even: s xd + 1 Z x + 1 for all x Z 0 (Figure 1.13b).
1.1
y
Functions and Their Graphs
7
y
y x2 1
yx1
y x2
yx 1
1 x
0
(a)
–1
x
0
(b)
FIGURE 1.13 (a) When we add the constant term 1 to the function y = x 2 , the resulting function y = x 2 + 1 is still even and its graph is still symmetric about the yaxis. (b) When we add the constant term 1 to the function y = x , the resulting function y = x + 1 is no longer odd. The symmetry about the origin is lost (Example 8).
Common Functions A variety of important types of functions are frequently encountered in calculus. We identify and briefly describe them here. Linear Functions A function of the form ƒsxd = mx + b , for constants m and b, is called a linear function. Figure 1.14a shows an array of lines ƒsxd = mx where b = 0, so these lines pass through the origin. The function ƒsxd = x where m = 1 and b = 0 is called the identity function. Constant functions result when the slope m = 0 (Figure 1.14b). A linear function with positive slope whose graph passes through the origin is called a proportionality relationship.
m –3
y m2 y 2x
y –3x m –1
m1
y
yx
1 m 2 1 y x 2 x
y –x 0
2 1 0
(a)
y3 2
1
2
x
(b)
FIGURE 1.14 (a) Lines through the origin with slope m. (b) A constant function with slope m = 0.
DEFINITION Two variables y and x are proportional (to one another) if one is always a constant multiple of the other; that is, if y = kx for some nonzero constant k.
If the variable y is proportional to the reciprocal 1>x, then sometimes it is said that y is inversely proportional to x (because 1>x is the multiplicative inverse of x). Power Functions A function ƒsxd = x a , where a is a constant, is called a power function. There are several important cases to consider.
8
Chapter 1: Functions
(a) a = n, a positive integer. The graphs of ƒsxd = x n , for n = 1, 2, 3, 4, 5, are displayed in Figure 1.15. These functions are defined for all real values of x. Notice that as the power n gets larger, the curves tend to flatten toward the xaxis on the interval s 1, 1d, and to rise more steeply for ƒ x ƒ 7 1. Each curve passes through the point (1, 1) and through the origin. The graphs of functions with even powers are symmetric about the yaxis; those with odd powers are symmetric about the origin. The evenpowered functions are decreasing on the interval s  q , 0] and increasing on [0, q d; the oddpowered functions are increasing over the entire real line s  q , q ). y
y
yx
1 –1
y
y x2
1 0
–1
FIGURE 1.15
1
x
–1
y
y x3
1
0
1
x
–1
–1
0
y y x5
y x4
1 x
1
–1
–1
1
0
1
x
–1
–1
0
1
x
–1
Graphs of ƒsxd = x n, n = 1, 2, 3, 4, 5, defined for  q 6 x 6 q .
(b) a = 1
or a = 2 .
The graphs of the functions ƒsxd = x 1 = 1>x and gsxd = x 2 = 1>x 2 are shown in Figure 1.16. Both functions are defined for all x Z 0 (you can never divide by zero). The graph of y = 1>x is the hyperbola xy = 1, which approaches the coordinate axes far from the origin. The graph of y = 1>x 2 also approaches the coordinate axes. The graph of the function ƒ is symmetric about the origin; ƒ is decreasing on the intervals s  q , 0) and s0, q ). The graph of the function g is symmetric about the yaxis; g is increasing on s  q , 0) and decreasing on s0, q ). y y y 12 x
y 1x 1 0
1
x
Domain: x 0 Range: y 0
(a)
1 0
x 1 Domain: x 0 Range: y 0 (b)
FIGURE 1.16 Graphs of the power functions ƒsxd = x a for part (a) a = 1 and for part (b) a = 2.
(c) a =
1 1 3 2 , , , and . 2 3 2 3
3 The functions ƒsxd = x 1>2 = 2x and gsxd = x 1>3 = 2 x are the square root and cube root functions, respectively. The domain of the square root function is [0, q d, but the cube root function is defined for all real x. Their graphs are displayed in Figure 1.17, along with the graphs of y = x 3>2 and y = x 2>3 . (Recall that x 3>2 = sx 1>2 d3 and x 2>3 = sx 1>3 d2 .)
Polynomials A function p is a polynomial if psxd = an x n + an  1x n  1 + Á + a1 x + a0 where n is a nonnegative integer and the numbers a0 , a1 , a2 , Á , an are real constants (called the coefficients of the polynomial). All polynomials have domain s  q , q d. If the
1.1
9
Functions and Their Graphs
y y
y
y
y
y x
y x 23
3
y x
1 1 0
x 32
1 Domain: 0 x Range: 0 y
x
0
1 Domain: – x Range: – y
Graphs of the power functions ƒsxd = x a for a =
FIGURE 1.17
1
1 x
0
x
x
0 1 Domain: – x Range: 0 y
1 Domain: 0 x Range: 0 y
2 1 1 3 , , , and . 2 3 2 3
leading coefficient an Z 0 and n 7 0, then n is called the degree of the polynomial. Linear functions with m Z 0 are polynomials of degree 1. Polynomials of degree 2, usually written as psxd = ax 2 + bx + c, are called quadratic functions. Likewise, cubic functions are polynomials psxd = ax 3 + bx 2 + cx + d of degree 3. Figure 1.18 shows the graphs of three polynomials. Techniques to graph polynomials are studied in Chapter 4. 3 2 y x x 2x 1 3 2 3 y
4
y y
2
–2
16
2 –1
–4
y (x 2)4(x 1)3(x 1)
y 8x 4 14x 3 9x 2 11x 1
0
2
x
4
1
–2
x
2
–4
–1
–6 –2
1
x
2
–8 –10
–4
–16
–12 (a)
FIGURE 1.18
0
(c)
(b)
Graphs of three polynomial functions.
Rational Functions A rational function is a quotient or ratio ƒ(x) = p(x)>q(x), where p and q are polynomials. The domain of a rational function is the set of all real x for which qsxd Z 0. The graphs of several rational functions are shown in Figure 1.19. y y
8 2 y 5x 2 8x 3 3x 2
y
4 2 2 y 2x 3 2 7x 4
–4
2
4
x
–5
0
5
2 10
–1
–2
y 11x3 2 2x 1
4 Line y 5 3
1
–2
6
x
–4
–2 0 –2
2
4
6
x
–4
–2
NOT TO SCALE
–4
–6 –8 (a)
FIGURE 1.19 of the graph.
(b)
(c)
Graphs of three rational functions. The straight red lines are called asymptotes and are not part
10
Chapter 1: Functions
Algebraic Functions Any function constructed from polynomials using algebraic operations (addition, subtraction, multiplication, division, and taking roots) lies within the class of algebraic functions. All rational functions are algebraic, but also included are more complicated functions (such as those satisfying an equation like y 3  9xy + x 3 = 0, studied in Section 3.7). Figure 1.20 displays the graphs of three algebraic functions. y x 1/3(x 4)
y
y x(1 x)2/5
y y 3 (x 2 1) 2/3 4 y
4 3 2 1
1
–1 –1 –2 –3
x
4
x
0
0
5 7
x
1
–1
(c)
(b)
(a)
FIGURE 1.20 Graphs of three algebraic functions.
Trigonometric Functions The six basic trigonometric functions are reviewed in Section 1.3. The graphs of the sine and cosine functions are shown in Figure 1.21. y
y
1
3 0
–
2
1
– 2 x
0 –1
–1 (a) f (x) sin x
3 2
5 2
x
2
(b) f(x) cos x
FIGURE 1.21 Graphs of the sine and cosine functions.
Exponential Functions Functions of the form ƒsxd = a x , where the base a 7 0 is a positive constant and a Z 1, are called exponential functions. All exponential functions have domain s  q , q d and range s0, q d, so an exponential function never assumes the value 0. We discuss exponential functions in Section 1.5. The graphs of some exponential functions are shown in Figure 1.22. y
y y 10 x
y 10 –x
12
12
10
10
8
8 y
6 y 3x
4 2 –1
–0.5
FIGURE 1.22
0 (a)
1
6 4
y 2x 0.5
3 –x
2
y 2 –x x
Graphs of exponential functions.
–1
–0.5
0 (b)
0.5
1
x
1.1
11
Functions and Their Graphs
Logarithmic Functions These are the functions ƒsxd = loga x, where the base a Z 1 is a positive constant. They are the inverse functions of the exponential functions, and we discuss these functions in Section 1.6. Figure 1.23 shows the graphs of four logarithmic functions with various bases. In each case the domain is s0, q d and the range is s  q , q d. y
y
y log 2 x y log 3 x
1
0
1
–1
x y log5 x
1
y log10 x –1
FIGURE 1.23 Graphs of four logarithmic functions.
0
1
x
FIGURE 1.24 Graph of a catenary or hanging cable. (The Latin word catena means “chain.”)
Transcendental Functions These are functions that are not algebraic. They include the trigonometric, inverse trigonometric, exponential, and logarithmic functions, and many other functions as well. A particular example of a transcendental function is a catenary. Its graph has the shape of a cable, like a telephone line or electric cable, strung from one support to another and hanging freely under its own weight (Figure 1.24). The function defining the graph is discussed in Section 7.3.
Exercises 1.1 Functions In Exercises 1–6, find the domain and range of each function. 1. ƒsxd = 1 + x 2
2. ƒsxd = 1  2x
3. F(x) = 25x + 10 4 5. ƒstd = 3  t
4. g(x) = 2x 2  3x 2 6. G(t) = 2 t  16
In Exercises 7 and 8, which of the graphs are graphs of functions of x, and which are not? Give reasons for your answers. 7. a.
b.
y
y
8. a.
0
y
b.
x
0
x
Finding Formulas for Functions 9. Express the area and perimeter of an equilateral triangle as a function of the triangle’s side length x.
y
10. Express the side length of a square as a function of the length d of the square’s diagonal. Then express the area as a function of the diagonal length.
0
x
0
x
11. Express the edge length of a cube as a function of the cube’s diagonal length d. Then express the surface area and volume of the cube as a function of the diagonal length.
12
Chapter 1: Functions
12. A point P in the first quadrant lies on the graph of the function ƒsxd = 2x . Express the coordinates of P as functions of the slope of the line joining P to the origin.
y
31. a.
b.
(–1, 1) (1, 1) 1
2
13. Consider the point (x, y) lying on the graph of the line 2x + 4y = 5. Let L be the distance from the point (x, y) to the origin (0, 0). Write L as a function of x. 14. Consider the point (x, y) lying on the graph of y = 2x  3. Let L be the distance between the points (x, y) and (4, 0). Write L as a function of y.
15. ƒsxd = 5  2x
16. ƒsxd = 1  2x  x 2
17. g sxd = 2ƒ x ƒ
18. g sxd = 2 x
19. F std = t> ƒ t ƒ
20. G std = 1> ƒ t ƒ
21. Find the domain of y =
x + 3 4  2x 2  9
22. Find the range of y = 2 +
.
27. F sxd = e
4  x , x 2 + 2x ,
28. G sxd = e
1>x , x,
x 6 0 0 … x
0
2
0
1
2
3
4
y 3
(2, 1) 5
x Ú 0 x 6 0.
38. y = 
1 39. y =  x
40. y =
1 x2
1 ƒxƒ 42. y = 2 x 44. y = 42x 46. y = s xd2>3
Even and Odd Functions In Exercises 47–58, say whether the function is even, odd, or neither. Give reasons for your answer.
b.
2
:x;, <x=,
37. y = x 3
48. ƒsxd = x 5
47. ƒsxd = 3
x
y
30. a.
ƒsxd = e
45. y = x 3>2
2
2
36. Graph the function
y
b. (1, 1)
1
35. Does < x = =  :x ; for all real x? Give reasons for your answer.
43. y = x 3>8
y
b. < x = = 0 ?
34. What real numbers x satisfy the equation :x; = < x = ?
41. y = 2ƒ x ƒ
Find a formula for each function graphed in Exercises 29–32. 29. a.
t
Increasing and Decreasing Functions Graph the functions in Exercises 37–46. What symmetries, if any, do the graphs have? Specify the intervals over which the function is increasing and the intervals where it is decreasing.
x … 1 x 7 1
2
–A
x
T 3T 2T 2
Why is ƒ(x) called the integer part of x?
PiecewiseDefined Functions Graph the functions in Exercises 25–28.
0 … x … 1 1 6 x … 2
T
a. :x; = 0 ?
b. ƒ x + y ƒ = 1
1  x, 2  x,
T 2
T 2
The Greatest and Least Integer Functions 33. For what values of x is
b. y 2 = x 2 a. ƒ y ƒ = x 24. Graph the following equations and explain why they are not graphs of functions of x.
26. g sxd = e
y
b. (T, 1)
0
23. Graph the following equations and explain why they are not graphs of functions of x.
0 … x … 1 1 6 x … 2
y
32. a.
0
x . x2 + 4
x, 25. ƒsxd = e 2  x,
(–2, –1)
x 1 (1, –1) (3, –1)
A
2
a. ƒ x ƒ + ƒ y ƒ = 1
x
3
1
Functions and Graphs Find the domain and graph the functions in Exercises 15–20.
y
2
x
1 –1
–1 –2 –3
1
x 2 (2, –1)
t
49. ƒsxd = x + 1
50. ƒsxd = x 2 + x
51. gsxd = x 3 + x
52. gsxd = x 4 + 3x 2  1
2
x x2  1
53. gsxd =
1 x2  1
54. gsxd =
55. hstd =
1 t  1
56. hstd = ƒ t 3 ƒ
57. hstd = 2t + 1
58. hstd = 2 ƒ t ƒ + 1
Theory and Examples 59. The variable s is proportional to t, and s = 25 when t = 75. Determine t when s = 60.
1.1 60. Kinetic energy The kinetic energy K of a mass is proportional to the square of its velocity y. If K = 12,960 joules when y = 18 m>sec, what is K when y = 10 m>sec?
66. a. y = 5x
Functions and Their Graphs
b. y = 5x
c. y = x 5
y g
61. The variables r and s are inversely proportional, and r = 6 when s = 4. Determine s when r = 10. 62. Boyle’s Law Boyle’s Law says that the volume V of a gas at constant temperature increases whenever the pressure P decreases, so that V and P are inversely proportional. If P = 14.7 lb>in2 when V = 1000 in3, then what is V when P = 23.4 lb>in2?
h
x x
14 x
x x
b. Confirm your findings in part (a) algebraically.
x
64. The accompanying figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units long. a. Express the ycoordinate of P in terms of x. (You might start by writing an equation for the line AB.)
y
70. Three hundred books sell for $40 each, resulting in a revenue of (300)($40) = $12,000. For each $5 increase in the price, 25 fewer books are sold. Write the revenue R as a function of the number x of $5 increases.
B
P(x, ?)
A –1
0
x
1
x
In Exercises 65 and 66, match each equation with its graph. Do not use a graphing device, and give reasons for your answer. b. y = x 7
T 68. a. Graph the functions ƒsxd = 3>sx  1d and g sxd = 2>sx + 1d together to identify the values of x for which 3 2 6 . x  1 x + 1 b. Confirm your findings in part (a) algebraically. 69. For a curve to be symmetric about the xaxis, the point (x, y) must lie on the curve if and only if the point sx, yd lies on the curve. Explain why a curve that is symmetric about the xaxis is not the graph of a function, unless the function is y = 0 .
b. Express the area of the rectangle in terms of x.
65. a. y = x 4
f
T 67. a. Graph the functions ƒsxd = x>2 and g sxd = 1 + s4>xd together to identify the values of x for which x 4 7 1 + x. 2
22 x
x
0
63. A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 14 in. by 22 in. by cutting out equal squares of side x at each corner and then folding up the sides as in the figure. Express the volume V of the box as a function of x.
x
13
71. A pen in the shape of an isosceles right triangle with legs of length x ft and hypotenuse of length h ft is to be built. If fencing costs $5/ft for the legs and $10/ft for the hypotenuse, write the total cost C of construction as a function of h. 72. Industrial costs A power plant sits next to a river where the river is 800 ft wide. To lay a new cable from the plant to a location in the city 2 mi downstream on the opposite side costs $180 per foot across the river and $100 per foot along the land.
c. y = x 10
2 mi P
x
Q
City
y 800 ft
g h
Power plant NOT TO SCALE
0 f
x
a. Suppose that the cable goes from the plant to a point Q on the opposite side that is x ft from the point P directly opposite the plant. Write a function C(x) that gives the cost of laying the cable in terms of the distance x. b. Generate a table of values to determine if the least expensive location for point Q is less than 2000 ft or greater than 2000 ft from point P.
14
Chapter 1: Functions
1.2
Combining Functions; Shifting and Scaling Graphs In this section we look at the main ways functions are combined or transformed to form new functions.
Sums, Differences, Products, and Quotients Like numbers, functions can be added, subtracted, multiplied, and divided (except where the denominator is zero) to produce new functions. If ƒ and g are functions, then for every x that belongs to the domains of both ƒ and g (that is, for x H Dsƒd ¨ Dsgd), we define functions ƒ + g, ƒ  g , and ƒg by the formulas sƒ + gdsxd = ƒsxd + gsxd. sƒ  gdsxd = ƒsxd  gsxd. sƒgdsxd = ƒsxdgsxd. Notice that the + sign on the lefthand side of the first equation represents the operation of addition of functions, whereas the + on the righthand side of the equation means addition of the real numbers ƒ(x) and g(x). At any point of Dsƒd ¨ Dsgd at which gsxd Z 0, we can also define the function ƒ>g by the formula ƒ ƒsxd a g bsxd = swhere gsxd Z 0d. gsxd Functions can also be multiplied by constants: If c is a real number, then the function cƒ is defined for all x in the domain of ƒ by scƒdsxd = cƒsxd.
EXAMPLE 1
The functions defined by the formulas ƒsxd = 2x
and
g sxd = 21  x
have domains Dsƒd = [0, q d and Dsgd = s  q , 1]. The points common to these domains are the points [0, q d ¨ s  q , 1] = [0, 1]. The following table summarizes the formulas and domains for the various algebraic combinations of the two functions. We also write ƒ # g for the product function ƒg. Function
Formula
Domain
ƒ + g ƒ  g g  ƒ ƒ#g
sƒ + gdsxd = 2x + 21  x sƒ  gdsxd = 2x  21  x sg  ƒdsxd = 21  x  2x sƒ # gdsxd = ƒsxdgsxd = 2xs1  xd ƒ ƒsxd x g sxd = gsxd = A 1  x gsxd g 1  x sxd = = ƒ ƒsxd A x
[0, 1] = Dsƒd ¨ Dsgd [0, 1] [0, 1] [0, 1]
ƒ>g g>ƒ
[0, 1) sx = 1 excludedd (0, 1] sx = 0 excludedd
The graph of the function ƒ + g is obtained from the graphs of ƒ and g by adding the corresponding ycoordinates ƒ(x) and g(x) at each point x H Dsƒd ¨ Dsgd, as in Figure 1.25. The graphs of ƒ + g and ƒ # g from Example 1 are shown in Figure 1.26.
1.2
y
y
yfg
g(x) 1 x
8 6
15
Combining Functions; Shifting and Scaling Graphs
y ( f g)(x) y g(x)
4 2
g(a)
y f (x)
1 2
f (a) g(a)
yf•g
f (a) x
a
0
f(x) x
1
FIGURE 1.25 Graphical addition of two functions.
0
1 5
2 5
3 5
4 5
1
x
FIGURE 1.26 The domain of the function ƒ + g is the intersection of the domains of ƒ and g, the interval [0, 1] on the xaxis where these domains overlap. This interval is also the domain of the function ƒ # g (Example 1).
Composite Functions Composition is another method for combining functions.
DEFINITION If ƒ and g are functions, the composite function ƒ ⴰ g (“ƒ composed with g”) is defined by sƒ ⴰ gdsxd = ƒsgsxdd. The domain of ƒ ⴰ g consists of the numbers x in the domain of g for which g(x) lies in the domain of ƒ. The definition implies that ƒ ⴰ g can be formed when the range of g lies in the domain of ƒ. To find sƒ ⴰ gdsxd, first find g(x) and second find ƒ(g(x)). Figure 1.27 pictures ƒ ⴰ g as a machine diagram and Figure 1.28 shows the composite as an arrow diagram. f g f (g(x)) x g
x
g
g(x)
f
f (g(x))
FIGURE 1.27 A composite function ƒ ⴰ g uses the output g(x) of the first function g as the input for the second function f.
f
g(x)
FIGURE 1.28 Arrow diagram for ƒ ⴰ g . If x lies in the domain of g and g(x) lies in the domain of f, then the functions f and g can be composed to form (ƒ ⴰ g)(x).
To evaluate the composite function g ⴰ ƒ (when defined), we find ƒ(x) first and then g(ƒ(x)). The domain of g ⴰ ƒ is the set of numbers x in the domain of ƒ such that ƒ(x) lies in the domain of g. The functions ƒ ⴰ g and g ⴰ ƒ are usually quite different.
16
Chapter 1: Functions
EXAMPLE 2
If ƒsxd = 2x and gsxd = x + 1, find
(a) sƒ ⴰ gdsxd
(b) sg ⴰ ƒdsxd
(c) sƒ ⴰ ƒdsxd
(d) sg ⴰ gdsxd.
Solution
Composite
Domain [1, q d
(a) sƒ ⴰ gdsxd = ƒsg sxdd = 2g sxd = 2x + 1 (b) sg ⴰ ƒdsxd = g sƒsxdd = ƒsxd + 1 = 2x + 1 (c) sƒ ⴰ ƒdsxd = ƒsƒsxdd = 2ƒsxd = 21x = x
1>4
(d) sg ⴰ gdsxd = g sg sxdd = g sxd + 1 = sx + 1d + 1 = x + 2
[0, q d [0, q d s  q, q d
To see why the domain of ƒ ⴰ g is [1, q d, notice that gsxd = x + 1 is defined for all real x but belongs to the domain of ƒ only if x + 1 Ú 0, that is to say, when x Ú 1. Notice that if ƒsxd = x 2 and gsxd = 2x, then sƒ ⴰ gdsxd = A 2x B 2 = x. However, the domain of ƒ ⴰ g is [0, q d, not s  q , q d, since 2x requires x Ú 0.
Shifting a Graph of a Function A common way to obtain a new function from an existing one is by adding a constant to each output of the existing function, or to its input variable. The graph of the new function is the graph of the original function shifted vertically or horizontally, as follows.
Shift Formulas Vertical Shifts y
y x2 2 y x2 1 y x2
y = ƒsxd + k
Shifts the graph of ƒ up k units if k 7 0 Shifts it down ƒ k ƒ units if k 6 0
Horizontal Shifts
y = ƒsx + hd
y x2 2
Shifts the graph of ƒ left h units if h 7 0 Shifts it right ƒ h ƒ units if h 6 0
2 1 unit
EXAMPLE 3
1 –2
0 –1
2
x
2 units
–2
FIGURE 1.29 To shift the graph of ƒsxd = x 2 up (or down), we add positive (or negative) constants to the formula for ƒ (Examples 3a and b).
(a) Adding 1 to the righthand side of the formula y = x 2 to get y = x 2 + 1 shifts the graph up 1 unit (Figure 1.29). (b) Adding 2 to the righthand side of the formula y = x 2 to get y = x 2  2 shifts the graph down 2 units (Figure 1.29). (c) Adding 3 to x in y = x 2 to get y = sx + 3d2 shifts the graph 3 units to the left, while adding 2 shifts the graph 2 units to the right (Figure 1.30). (d) Adding 2 to x in y = ƒ x ƒ , and then adding 1 to the result, gives y = ƒ x  2 ƒ  1 and shifts the graph 2 units to the right and 1 unit down (Figure 1.31).
Scaling and Reflecting a Graph of a Function To scale the graph of a function y = ƒsxd is to stretch or compress it, vertically or horizontally. This is accomplished by multiplying the function ƒ, or the independent variable x, by an appropriate constant c. Reflections across the coordinate axes are special cases where c = 1.
1.2 Add a positive constant to x.
Add a negative constant to x.
y
y y x – 2 – 1
4 y (x 3) 2
y x2
17
Combining Functions; Shifting and Scaling Graphs
y (x 2) 2 1
1 –3
–4 1
0
–2
2
–1
4
6
x
x
2
FIGURE 1.30 To shift the graph of y = x 2 to the left, we add a positive constant to x (Example 3c). To shift the graph to the right, we add a negative constant to x.
FIGURE 1.31 Shifting the graph of y = ƒ x ƒ 2 units to the right and 1 unit down (Example 3d).
Vertical and Horizontal Scaling and Reflecting Formulas For c 7 1, the graph is scaled:
y = cƒsxd
Stretches the graph of ƒ vertically by a factor of c.
1 y = c ƒsxd
Compresses the graph of ƒ vertically by a factor of c.
y = ƒscxd y = ƒsx>cd
Compresses the graph of ƒ horizontally by a factor of c. Stretches the graph of ƒ horizontally by a factor of c.
For c = 1, the graph is reflected:
y = ƒsxd y = ƒs xd
EXAMPLE 4
Reflects the graph of ƒ across the xaxis. Reflects the graph of ƒ across the yaxis. Here we scale and reflect the graph of y = 2x.
(a) Vertical: Multiplying the righthand side of y = 2x by 3 to get y = 32x stretches the graph vertically by a factor of 3, whereas multiplying by 1>3 compresses the graph by a factor of 3 (Figure 1.32). (b) Horizontal: The graph of y = 23x is a horizontal compression of the graph of y = 2x by a factor of 3, and y = 2x>3 is a horizontal stretching by a factor of 3 (Figure 1.33). Note that y = 23x = 232x so a horizontal compression may correspond to a vertical stretching by a different scaling factor. Likewise, a horizontal stretching may correspond to a vertical compression by a different scaling factor. (c) Reflection: The graph of y =  2x is a reflection of y = 2x across the xaxis, and y = 2 x is a reflection across the yaxis (Figure 1.34). y
y y 3x
5
compress
y x
stretch
2
2
1
compress
1 0
y 3 x
3
3
–1
y x
4
4
1
2
3
y 3 x 4
x
FIGURE 1.32 Vertically stretching and compressing the graph y = 1x by a factor of 3 (Example 4a).
stretch
1 –1
0
y
y –x
1
2
3
4
y x y x3 x
FIGURE 1.33 Horizontally stretching and compressing the graph y = 1x by a factor of 3 (Example 4b).
1
–3
–2
–1
1
2
3
x
–1 y –x
FIGURE 1.34 Reflections of the graph y = 1x across the coordinate axes (Example 4c).
18
Chapter 1: Functions
Given the function ƒsxd = x 4  4x 3 + 10 (Figure 1.35a), find formulas to
EXAMPLE 5
(a) compress the graph horizontally by a factor of 2 followed by a reflection across the yaxis (Figure 1.35b). (b) compress the graph vertically by a factor of 2 followed by a reflection across the xaxis (Figure 1.35c). y 20
y
y 16x 4 32x 3 10 y f (x)
x4
4x 3
10
y – 12 x 4 2x 3 5
20 10
10 –1
0 –10
10 1
2
3
4
x
–2
–1
0 –10
x
1
–1
0
1
2
3
4
x
–10 –20
–20 (b)
(a)
(c)
FIGURE 1.35 (a) The original graph of f. (b) The horizontal compression of y = ƒsxd in part (a) by a factor of 2, followed by a reflection across the yaxis. (c) The vertical compression of y = ƒsxd in part (a) by a factor of 2, followed by a reflection across the xaxis (Example 5). Solution
(a) We multiply x by 2 to get the horizontal compression, and by 1 to give reflection across the yaxis. The formula is obtained by substituting 2x for x in the righthand side of the equation for ƒ: y = ƒs 2xd = s 2xd4  4s 2xd3 + 10 = 16x 4 + 32x 3 + 10 . (b) The formula is y = 
1 1 ƒsxd =  x 4 + 2x 3  5. 2 2
Exercises 1.2 Algebraic Combinations In Exercises 1 and 2, find the domains and ranges of ƒ, g, ƒ + g , and ƒ # g. 1. ƒsxd = x,
g sxd = 2x  1
2. ƒsxd = 2x + 1,
g sxd = 2x  1
In Exercises 3 and 4, find the domains and ranges of ƒ, g, ƒ>g, and g >ƒ. 3. ƒsxd = 2,
g sxd = x 2 + 1
4. ƒsxd = 1,
g sxd = 1 + 2x
Composites of Functions 5. If ƒsxd = x + 5 and g sxd = x 2  3 , find the following. a. ƒ( g (0)) c. ƒ( g (x))
b. g(ƒ(0)) d. g(ƒ(x))
e. ƒ(ƒ(5))
f. g (g (2))
g. ƒ(ƒ(x))
h. g (g (x))
6. If ƒsxd = x  1 and gsxd = 1>sx + 1d , find the following. a. ƒ(g (1>2))
b. g (ƒ(1>2))
c. ƒ(g (x))
d. g (ƒ(x))
e. ƒ(ƒ(2))
f. g (g (2))
g. ƒ(ƒ(x))
h. g (g (x))
In Exercises 7–10, write a formula for ƒ ⴰ g ⴰ h. 7. ƒ(x) = x + 1,
gsxd = 3x , hsxd = 4  x
8. ƒ(x) = 3x + 4, 9. ƒsxd = 2x + 1, 10. ƒsxd =
x + 2 , 3  x
gsxd = 2x  1, hsxd = x 2 1 , x + 4
gsxd = gsxd =
x2 , x + 1 2
1 hsxd = x hsxd = 22  x
1.2
Combining Functions; Shifting and Scaling Graphs
x . Find a function y = g(x) so that x  2 (ƒ ⴰ g)(x) = x.
Let ƒsxd = x  3, g sxd = 2x , hsxd = x 3 , and jsxd = 2x . Express each of the functions in Exercises 11 and 12 as a composite involving one or more of ƒ, g, h, and j.
19. Let ƒ(x) =
11. a. y = 2x  3
20. Let ƒ(x) = 2x 3  4. Find a function y = g(x) so that (ƒ ⴰ g)(x) = x + 2.
b. y = 2 2x
c. y = x 1>4
d. y = 4x f. y = s2x  6d3
e. y = 2sx  3d
3
12. a. y = 2x  3
Shifting Graphs 21. The accompanying figure shows the graph of y = x 2 shifted to two new positions. Write equations for the new graphs.
b. y = x 3>2
c. y = x 9
d. y = x  6
e. y = 22x  3
f. y = 2x 3  3
19
y
13. Copy and complete the following table. g(x)
ƒ(x)
(ƒ ⴰ g)(x)
a. x  7
2x
?
b. x + 2 c.
?
3x
?
2x  5
2x 2  5
d.
x x  1
x x  1
?
e.
?
1 1 + x
x
f.
1 x
?
x
–7
Position (a)
0
4
y –x 2
x
Position (b)
22. The accompanying figure shows the graph of y = x 2 shifted to two new positions. Write equations for the new graphs. y Position (a)
14. Copy and complete the following table. g(x)
ƒ(x)
(ƒ ⴰ g)(x)
ƒxƒ
?
a.
1 x  1
b.
?
x  1 x
x x + 1
c.
?
2x
ƒxƒ ƒxƒ
d. 2x
?
y x2 3
x
0 Position (b)
15. Evaluate each expression using the given table of values: –5
2
1
0
1
2
ƒ(x)
1
0
2
1
2
g(x)
2
1
0
1
0
x
a. ƒsgs 1dd
b. gsƒs0dd
c. ƒsƒs 1dd
d. gsgs2dd
e. gsƒs 2dd
f. ƒsgs1dd
23. Match the equations listed in parts (a)–(d) to the graphs in the accompanying figure. a. y = sx  1d2  4
b. y = sx  2d2 + 2
c. y = sx + 2d2 + 2
d. y = sx + 3d2  2 y
Position 2
Position 1
16. Evaluate each expression using the functions ƒ(x) = 2  x,
g(x) = b
x, x  1,
2 … x 6 0 0 … x … 2.
a. ƒsgs0dd
b. gsƒs3dd
c. gsgs 1dd
d. ƒsƒs2dd
e. gsƒs0dd
f. ƒsgs1>2dd
In Exercises 17 and 18, (a) write formulas for ƒ ⴰ g and g ⴰ ƒ and find the (b) domain and (c) range of each.
3 (–2, 2) Position 3
2 1
–4 –3 –2 –1 0
(2, 2) 1 2 3 Position 4
(–3, –2)
1 17. ƒ(x) = 2x + 1, g (x) = x 18. ƒ(x) = x 2, g (x) = 1  2x
(1, –4)
x
20
Chapter 1: Functions
24. The accompanying figure shows the graph of y = x 2 shifted to four new positions. Write an equation for each new graph.
55. The accompanying figure shows the graph of a function ƒ(x) with domain [0, 2] and range [0, 1]. Find the domains and ranges of the following functions, and sketch their graphs.
y (1, 4)
y
(–2, 3) (b)
(a) (2, 0)
1
y f (x)
0
2
x
(–4, –1)
(c)
(d)
Exercises 25–34 tell how many units and in what directions the graphs of the given equations are to be shifted. Give an equation for the shifted graph. Then sketch the original and shifted graphs together, labeling each graph with its equation. 25. x 2 + y 2 = 49
Down 3, left 2
26. x 2 + y 2 = 25
Up 3, left 4
27. y = x
3
Right 1, down 1
29. y = 2x
Left 0.81
30. y =  2x
Right 3
31. y = 2x  7
Up 7
1 32. y = sx + 1d + 5 2
e. ƒsx + 2d
f. ƒsx  1d
g. ƒs xd
h. ƒsx + 1d + 1
56. The accompanying figure shows the graph of a function g(t) with domain [4, 0] and range [3, 0] . Find the domains and ranges of the following functions, and sketch their graphs.
–2
y g(t)
Down 5, right 1
0
t
–3
a. g s td
b. g std
c. g std + 3
d. 1  g std
e. g s t + 2d
f. g st  2d
g. g s1  td
h. g st  4d
Left 2, down 1
35. y = 2x + 4
36. y = 29  x
37. y = ƒ x  2 ƒ
38. y = ƒ 1  x ƒ  1
39. y = 1 + 2x  1
40. y = 1  2x
41. y = sx + 1d
42. y = sx  8d2>3
43. y = 1  x 2>3
44. y + 4 = x 2>3
3 x  1  1 45. y = 2
46. y = sx + 2d3>2 + 1
2>3
1 x  2
1 49. y = x + 2
53. y =
d. ƒsxd
–4
Graph the functions in Exercises 35–54.
51. y =
c. 2ƒ(x)
y
Up 1, right 1
34. y = 1>x 2
47. y =
b. ƒsxd  1
Left 1, down 1
28. y = x 2>3
33. y = 1>x
a. ƒsxd + 2
x
1 sx  1d2 1 + 1 x2
1 48. y = x  2 50. y = 52. y = 54. y =
1 x + 2 1  1 x2 1 sx + 1d2
Vertical and Horizontal Scaling Exercises 57–66 tell by what factor and direction the graphs of the given functions are to be stretched or compressed. Give an equation for the stretched or compressed graph. 57. y = x 2  1,
stretched vertically by a factor of 3
58. y = x 2  1,
compressed horizontally by a factor of 2
59. y = 1 +
1 , x2
compressed vertically by a factor of 2
1 , stretched horizontally by a factor of 3 x2 61. y = 2x + 1, compressed horizontally by a factor of 4 60. y = 1 +
62. y = 2x + 1,
stretched vertically by a factor of 3
63. y = 24  x 2,
stretched horizontally by a factor of 2
64. y = 24  x ,
compressed vertically by a factor of 3
2
65. y = 1  x ,
compressed horizontally by a factor of 3
66. y = 1  x ,
stretched horizontally by a factor of 2
3 3
1.3 Graphing In Exercises 67–74, graph each function, not by plotting points, but by starting with the graph of one of the standard functions presented in Figures 1.14–1.17 and applying an appropriate transformation. 68. y =
69. y = sx  1d3 + 2
70. y = s1  xd3 + 2
1 71. y =  1 2x
2 72. y = 2 + 1 x 74. y = s 2xd2>3
3 73. y =  2 x
A
1 
Combining Functions 77. Assume that ƒ is an even function, g is an odd function, and both ƒ and g are defined on the entire real line ⺢ . Which of the following (where defined) are even? odd?
x 2
67. y =  22x + 1
21
Trigonometric Functions
a. ƒg
b. ƒ>g
c. g >ƒ
d. ƒ 2 = ƒƒ
e. g 2 = gg
f. ƒ ⴰ g
g. g ⴰ ƒ
h. ƒ ⴰ ƒ
i. g ⴰ g
78. Can a function be both even and odd? Give reasons for your answer. T 79. (Continuation of Example 1.) Graph the functions ƒsxd = 2x and g sxd = 21  x together with their (a) sum, (b) product, (c) two differences, (d) two quotients.
75. Graph the function y = ƒ x 2  1 ƒ . 76. Graph the function y = 2ƒ x ƒ .
2 T 80. Let ƒsxd = x  7 and g sxd = x . Graph ƒ and g together with ƒ ⴰ g and g ⴰ ƒ .
Trigonometric Functions
1.3
This section reviews radian measure and the basic trigonometric functions.
Angles
B'
Angles are measured in degrees or radians. The number of radians in the central angle A¿CB¿ within a circle of radius r is defined as the number of “radius units” contained in the arc s subtended by that central angle. If we denote this central angle by u when measured in radians, this means that u = s>r (Figure 1.36), or
s B 1 C
C ir
A' A r
it cir cl
e
Un
θ
cle of ra diu
sr
s = ru
FIGURE 1.36 The radian measure of the central angle A¿CB¿ is the number u = s>r. For a unit circle of radius r = 1, u is the length of arc AB that central angle ACB cuts from the unit circle.
(u in radians).
(1)
If the circle is a unit circle having radius r = 1, then from Figure 1.36 and Equation (1), we see that the central angle u measured in radians is just the length of the arc that the angle cuts from the unit circle. Since one complete revolution of the unit circle is 360° or 2p radians, we have p radians = 180°
(2)
and 180 1 radian = p ( L 57.3) degrees
1 degree =
or
p ( L 0.017) radians. 180
Table 1.2 shows the equivalence between degree and radian measures for some basic angles.
TABLE 1.2
Angles measured in degrees and radians
Degrees
180
135
90
45
0
30
45
60
90
120
135
150
180
270
360
U (radians)
p
3p 4
p 2
p 4
0
p 6
p 4
p 3
p 2
2p 3
3p 4
5p 6
p
3p 2
2p
22
Chapter 1: Functions
An angle in the xyplane is said to be in standard position if its vertex lies at the origin and its initial ray lies along the positive xaxis (Figure 1.37). Angles measured counterclockwise from the positive xaxis are assigned positive measures; angles measured clockwise are assigned negative measures. y
y Terminal ray Initial ray x Positive measure
Initial ray
Negative measure
Terminal ray
x
FIGURE 1.37 Angles in standard position in the xyplane.
Angles describing counterclockwise rotations can go arbitrarily far beyond 2p radians or 360°. Similarly, angles describing clockwise rotations can have negative measures of all sizes (Figure 1.38). y
y
y
y
3 x
– 5 2 x
9 4 hypotenuse
opposite
adjacent opp hyp adj cos hyp opp tan adj
hyp opp hyp sec adj adj cot opp csc
y
y r x
O
r
x
Nonzero radian measures can be positive or negative and can go beyond 2p.
The Six Basic Trigonometric Functions
FIGURE 1.39 Trigonometric ratios of an acute angle.
P(x, y)
x
Angle Convention: Use Radians From now on, in this book it is assumed that all angles are measured in radians unless degrees or some other unit is stated explicitly. When we talk about the angle p>3, we mean p>3 radians (which is 60°), not p>3 degrees. We use radians because it simplifies many of the operations in calculus, and some results we will obtain involving the trigonometric functions are not true when angles are measured in degrees.
sin
FIGURE 1.38
– 3 4
x
You are probably familiar with defining the trigonometric functions of an acute angle in terms of the sides of a right triangle (Figure 1.39). We extend this definition to obtuse and negative angles by first placing the angle in standard position in a circle of radius r. We then define the trigonometric functions in terms of the coordinates of the point P(x, y) where the angle’s terminal ray intersects the circle (Figure 1.40). y sine: sin u = r x cosine: cos u = r y tangent: tan u = x
r cosecant: csc u = y r secant: sec u = x x cotangent: cot u = y
These extended definitions agree with the righttriangle definitions when the angle is acute. Notice also that whenever the quotients are defined,
FIGURE 1.40 The trigonometric functions of a general angle u are defined in terms of x, y, and r.
sin u cos u 1 sec u = cos u tan u =
1 tan u 1 csc u = sin u cot u =
1.3
2
3
2 1
4
As you can see, tan u and sec u are not defined if x = cos u = 0. This means they are not defined if u is ;p>2, ;3p>2, Á . Similarly, cot u and csc u are not defined for values of u for which y = 0, namely u = 0, ;p, ;2p, Á . The exact values of these trigonometric ratios for some angles can be read from the triangles in Figure 1.41. For instance,
6
4
3
2
1
2
p 1 = 4 22 p 1 cos = 4 22 p tan = 1 4 sin
1
sin
23 2p = , 3 2
p 1 = 6 2
sin
23 p = 3 2
cos
2p 1 =  , 3 2
cos
tan
2p =  23. 3
⎛cos 2 , sin 2 ⎛ ⎛– 1 , 3 ⎛ ⎝ 3 3⎝ ⎝ 2 2 ⎝ y
y
A all pos
P 1
3 2
x T tan pos
sin
23 p p 1 = cos = 6 2 3 2 p p 1 tan = tan = 23 6 3 23 The CAST rule (Figure 1.42) is useful for remembering when the basic trigonometric functions are positive or negative. For instance, from the triangle in Figure 1.43, we see that
FIGURE 1.41 Radian angles and side lengths of two common triangles.
S sin pos
23
Trigonometric Functions
C cos pos
2 3 x
1 2
FIGURE 1.42 The CAST rule, remembered by the statement “Calculus Activates Student Thinking,” tells which trigonometric functions are positive in each quadrant.
FIGURE 1.43 The triangle for calculating the sine and cosine of 2p>3 radians. The side lengths come from the geometry of right triangles.
Using a similar method we determined the values of sin u, cos u, and tan u shown in Table 1.3. TABLE 1.3
Values of sin u, cos u, and tan u for selected values of u
Degrees
180
u (radians)
p
135 3p 4
90 p 2
45 p 4
0 0
30 p 6
45 p 4
60 p 3
90 p 2
120 2p 3
135 3p 4
150 5p 6
22 2
1 2
 22 2 1
sin u
0
 22 2
1
 22 2
0
1 2
22 2
23 2
1
23 2
cos u
1
 22 2
0
22 2
1
23 2
22 2
1 2
0

tan u
0
1
0
23 3
1
23
1
1 2
 23
180
270 3p 2
360
0
1
0
 23 2
1
0
1
 23 3
0
p
2p
0
24
Chapter 1: Functions
Periodicity and Graphs of the Trigonometric Functions
Periods of Trigonometric Functions Period P : tan sx + pd = tan x cot sx + pd = cot x Period 2P :
sin sx + 2pd = sin x cos sx + 2pd = cos x sec sx + 2pd = sec x csc sx + 2pd = csc x
When an angle of measure u and an angle of measure u + 2p are in standard position, their terminal rays coincide. The two angles therefore have the same trigonometric function values: sin(u + 2p) = sin u, tan(u + 2p) = tan u, and so on. Similarly, cos(u  2p) = cos u, sin(u  2p) = sin u, and so on. We describe this repeating behavior by saying that the six basic trigonometric functions are periodic.
DEFINITION A function ƒ(x) is periodic if there is a positive number p such that ƒ(x + p) = ƒ(x) for every value of x. The smallest such value of p is the period of ƒ. When we graph trigonometric functions in the coordinate plane, we usually denote the independent variable by x instead of u. Figure 1.44 shows that the tangent and cotangent functions have period p = p, and the other four functions have period 2p. Also, the symmetries in these graphs reveal that the cosine and secant functions are even and the other four functions are odd (although this does not prove those results). y y
y y cos x
– – 2
2
0
y sin x
3 2 2
x
y
coss xd = cos x secs xd = sec x
3 2 2
2
0
y
y sec x
1 – 3 – – 0 2 2
– – 2
3 2 2
x
y sinx Domain: – x Range: –1 y 1 Period: 2 (b)
Domain: – x Range: –1 y 1 Period: 2 (a)
Even
y tan x
x
0 3 2 2
x
Domain: x , 3 , . . . 2 2 Range: – y Period: (c) y
y csc x
1 – – 0 2
– 3 – – 2 2
y cot x
1 2
3 2 2
x
– – 0 2
2
3 2 2
x
Odd
sin s xd tan s xd cscs xd cots xd
= = = =
Domain: x , 3 , . . . 2 2 Range: y –1 or y 1 Period: 2 (d)
sin x tan x csc x cot x
Domain: x 0, , 2, . . . Range: y –1 or y 1 Period: 2
Domain: x 0, , 2, . . . Range: – y Period: (f)
(e)
FIGURE 1.44 Graphs of the six basic trigonometric functions using radian measure. The shading for each trigonometric function indicates its periodicity.
y P(cos , sin )
Trigonometric Identities
x 2 y2 1
The coordinates of any point P(x, y) in the plane can be expressed in terms of the point’s distance r from the origin and the angle u that ray OP makes with the positive xaxis (Figure 1.40). Since x>r = cos u and y>r = sin u, we have
sin cos O
x 1
x = r cos u,
y = r sin u.
When r = 1 we can apply the Pythagorean theorem to the reference right triangle in Figure 1.45 and obtain the equation
FIGURE 1.45 The reference triangle for a general angle u .
cos2 u + sin2 u = 1.
(3)
1.3
25
Trigonometric Functions
This equation, true for all values of u, is the most frequently used identity in trigonometry. Dividing this identity in turn by cos2 u and sin2 u gives
1 + tan2 u = sec2 u 1 + cot2 u = csc2 u
The following formulas hold for all angles A and B (Exercise 58).
Addition Formulas cossA + Bd = cos A cos B  sin A sin B sinsA + Bd = sin A cos B + cos A sin B
(4)
There are similar formulas for cos sA  Bd and sin sA  Bd (Exercises 35 and 36). All the trigonometric identities needed in this book derive from Equations (3) and (4). For example, substituting u for both A and B in the addition formulas gives
DoubleAngle Formulas cos 2u = cos2 u  sin2 u sin 2u = 2 sin u cos u
(5)
Additional formulas come from combining the equations cos2 u + sin2 u = 1,
cos2 u  sin2 u = cos 2u.
We add the two equations to get 2 cos2 u = 1 + cos 2u and subtract the second from the first to get 2 sin2 u = 1  cos 2u. This results in the following identities, which are useful in integral calculus.
HalfAngle Formulas cos2 u =
1 + cos 2u 2
(6)
sin2 u =
1  cos 2u 2
(7)
The Law of Cosines If a, b, and c are sides of a triangle ABC and if u is the angle opposite c, then
c 2 = a 2 + b 2  2ab cos u.
This equation is called the law of cosines.
(8)
26
Chapter 1: Functions
We can see why the law holds if we introduce coordinate axes with the origin at C and the positive xaxis along one side of the triangle, as in Figure 1.46. The coordinates of A are (b, 0); the coordinates of B are sa cos u, a sin ud. The square of the distance between A and B is therefore
y B(a cos , a sin )
c 2 = sa cos u  bd2 + sa sin ud2
c
a
= a 2scos2 u + sin2 ud + b 2  2ab cos u ('')''* 1
C
b
A(b, 0)
x
= a 2 + b 2  2ab cos u. FIGURE 1.46 The square of the distance between A and B gives the law of cosines.
The law of cosines generalizes the Pythagorean theorem. If u = p>2, then cos u = 0 and c 2 = a 2 + b 2.
Transformations of Trigonometric Graphs The rules for shifting, stretching, compressing, and reflecting the graph of a function summarized in the following diagram apply to the trigonometric functions we have discussed in this section. Vertical stretch or compression; reflection about y = d if negative
Vertical shift
y = aƒ(bsx + cdd + d Horizontal shift
Horizontal stretch or compression; reflection about x = c if negative
The transformation rules applied to the sine function give the general sine function or sinusoid formula ƒ(x) = A sin a
2p (x  C)b + D, B
where ƒ A ƒ is the amplitude, ƒ B ƒ is the period, C is the horizontal shift, and D is the vertical shift. A graphical interpretation of the various terms is given below. y
(
Horizontal shift (C)
Amplitude (A)
This axis is the line y D.
D
DA
)
y A sin 2 (x C) D B
DA
Vertical shift (D) This distance is the period (B).
0
x
Two Special Inequalities For any angle u measured in radians,  ƒ u ƒ … sin u … ƒ u ƒ
and
 ƒ u ƒ … 1  cos u … ƒ u ƒ .
1.3
To establish these inequalities, we picture u as a nonzero angle in standard position (Figure 1.47). The circle in the figure is a unit circle, so ƒ u ƒ equals the length of the circular arc AP. The length of line segment AP is therefore less than ƒ u ƒ . Triangle APQ is a right triangle with sides of length
y P
O cos
sin
1
Q
27
Trigonometric Functions
QP = ƒ sin u ƒ ,
AQ = 1  cos u.
From the Pythagorean theorem and the fact that AP 6 ƒ u ƒ , we get
x A(1, 0)
sin2 u + (1  cos u) 2 = (AP) 2 … u2.
1 – cos
(9)
The terms on the lefthand side of Equation (9) are both positive, so each is smaller than their sum and hence is less than or equal to u2: sin2 u … u2
FIGURE 1.47 From the geometry of this figure, drawn for u 7 0, we get the inequality sin2 u + (1  cos u) 2 … u2.
(1  cos u) 2 … u2.
and
By taking square roots, this is equivalent to saying that ƒ sin u ƒ … ƒ u ƒ
and
ƒ 1  cos u ƒ … ƒ u ƒ ,
 ƒ u ƒ … sin u … ƒ u ƒ
and
 ƒ u ƒ … 1  cos u … ƒ u ƒ .
so
These inequalities will be useful in the next chapter.
Exercises 1.3 Radians and Degrees 1. On a circle of radius 10 m, how long is an arc that subtends a central angle of (a) 4p>5 radians? (b) 110°? 2. A central angle in a circle of radius 8 is subtended by an arc of length 10p . Find the angle’s radian and degree measures. 3. You want to make an 80° angle by marking an arc on the perimeter of a 12in.diameter disk and drawing lines from the ends of the arc to the disk’s center. To the nearest tenth of an inch, how long should the arc be? 4. If you roll a 1mdiameter wheel forward 30 cm over level ground, through what angle will the wheel turn? Answer in radians (to the nearest tenth) and degrees (to the nearest degree). Evaluating Trigonometric Functions 5. Copy and complete the following table of function values. If the function is undefined at a given angle, enter “UND.” Do not use a calculator or tables. U
P
2P>3
0
P>2
3P>2
U
6. Copy and complete the following table of function values. If the function is undefined at a given angle, enter “UND.” Do not use a calculator or tables.
P>6
P>4
5P>6
sin u cos u tan u cot u sec u csc u In Exercises 7–12, one of sin x, cos x, and tan x is given. Find the other two if x lies in the specified interval. 7. sin x =
3 , 5
p x H c , pd 2
9. cos x =
1 , 3
x H c
p , 0d 2
10. cos x = 
11. tan x =
1 , 2
x H cp,
3p d 2
1 12. sin x =  , 2
3P>4
sin u cos u tan u cot u sec u csc u
P>3
x H c0,
8. tan x = 2,
5 , 13
p d 2
p x H c , pd 2 x H cp,
3p d 2
Graphing Trigonometric Functions Graph the functions in Exercises 13–22. What is the period of each function? 13. sin 2x
14. sin ( x>2)
15. cos px
16. cos
17. sin
px 3
19. cos ax 
px 2
18. cos 2px p b 2
20. sin ax +
p b 6
28
Chapter 1: Functions
21. sin ax 
p b + 1 4
22. cos ax +
2p b  2 3
Graph the functions in Exercises 23–26 in the tsplane (taxis horizontal, saxis vertical). What is the period of each function? What symmetries do the graphs have? 23. s = cot 2t
24. s = tan pt
pt 25. s = sec a b 2
t 26. s = csc a b 2
T 27. a. Graph y = cos x and y = sec x together for 3p>2 … x … 3p>2 . Comment on the behavior of sec x in relation to the signs and values of cos x. b. Graph y = sin x and y = csc x together for p … x … 2p . Comment on the behavior of csc x in relation to the signs and values of sin x. T 28. Graph y = tan x and y = cot x together for 7 … x … 7 . Comment on the behavior of cot x in relation to the signs and values of tan x.
Solving Trigonometric Equations For Exercises 51–54, solve for the angle u, where 0 … u … 2p. 3 51. sin2 u = 52. sin2 u = cos2 u 4 53. sin 2u  cos u = 0
54. cos 2u + cos u = 0
Theory and Examples 55. The tangent sum formula The standard formula for the tangent of the sum of two angles is
Derive the formula. 56. (Continuation of Exercise 55.) Derive a formula for tan sA  Bd . 57. Apply the law of cosines to the triangle in the accompanying figure to derive the formula for cos sA  Bd . y
29. Graph y = sin x and y = :sin x; together. What are the domain and range of :sin x; ?
1
30. Graph y = sin x and y = < sin x = together. What are the domain and range of < sin x = ?
A
Using the Addition Formulas Use the addition formulas to derive the identities in Exercises 31–36. 31. cos ax 
p b = sin x 2
32. cos ax +
p b = sin x 2
33. sin ax +
p b = cos x 2
34. sin ax 
p b = cos x 2
35. cos sA  Bd = cos A cos B + sin A sin B (Exercise 57 provides a different derivation.) 36. sin sA  Bd = sin A cos B  cos A sin B 37. What happens if you take B = A in the trigonometric identity cos sA  Bd = cos A cos B + sin A sin B ? Does the result agree with something you already know? 38. What happens if you take B = 2p in the addition formulas? Do the results agree with something you already know? In Exercises 39–42, express the given quantity in terms of sin x and cos x. 39. cos sp + xd
40. sin s2p  xd
41. sin a
42. cos a
3p  xb 2
43. Evaluate sin
7p p p as sin a + b . 12 4 3
44. Evaluate cos
11p p 2p b. as cos a + 12 4 3
p 45. Evaluate cos . 12
p 12
0
x
1
58. a. Apply the formula for cos sA  Bd to the identity sin u = cos a
p  u b to obtain the addition formula for sin sA + Bd . 2
b. Derive the formula for cos sA + Bd by substituting B for B in the formula for cos sA  Bd from Exercise 35. 59. A triangle has sides a = 2 and b = 3 and angle C = 60° . Find the length of side c. 60. A triangle has sides a = 2 and b = 3 and angle C = 40° . Find the length of side c. 61. The law of sines The law of sines says that if a, b, and c are the sides opposite the angles A, B, and C in a triangle, then sin C sin A sin B a = b = c . Use the accompanying figures and the identity sin sp  ud = sin u , if required, to derive the law. A
5p . 46. Evaluate sin 12
50. sin2
B
1
3p + xb 2
Using the DoubleAngle Formulas Find the function values in Exercises 47–50. 5p p 47. cos2 48. cos2 8 12 49. sin2
tan A + tan B . 1  tan A tan B
tansA + Bd =
3p 8
c
B
h
a
A
c
b
b C
B
a
h
C
62. A triangle has sides a = 2 and b = 3 and angle C = 60° (as in Exercise 59). Find the sine of angle B using the law of sines.
1.4 63. A triangle has side c = 2 and angles A = p>4 and B = p>3 . Find the length a of the side opposite A. T 64. The approximation sin x L x It is often useful to know that, when x is measured in radians, sin x L x for numerically small values of x. In Section 3.11, we will see why the approximation holds. The approximation error is less than 1 in 5000 if ƒ x ƒ 6 0.1 . a. With your grapher in radian mode, graph y = sin x and y = x together in a viewing window about the origin. What do you see happening as x nears the origin? b. With your grapher in degree mode, graph y = sin x and y = x together about the origin again. How is the picture different from the one obtained with radian mode?
Graphing with Calculators and Computers
69. The period B
29
Set the constants A = 3, C = D = 0 .
a. Plot ƒ(x) for the values B = 1, 3, 2p, 5p over the interval 4p … x … 4p . Describe what happens to the graph of the general sine function as the period increases. b. What happens to the graph for negative values of B? Try it with B = 3 and B = 2p . 70. The horizontal shift C Set the constants A = 3, B = 6, D = 0 . a. Plot ƒ(x) for the values C = 0, 1 , and 2 over the interval 4p … x … 4p . Describe what happens to the graph of the general sine function as C increases through positive values. b. What happens to the graph for negative values of C ? c. What smallest positive value should be assigned to C so the graph exhibits no horizontal shift? Confirm your answer with a plot.
General Sine Curves For ƒsxd = A sin a
2p sx  Cdb + D, B
identify A, B, C, and D for the sine functions in Exercises 65–68 and sketch their graphs. 1 1 65. y = 2 sin sx + pd  1 66. y = sin spx  pd + 2 2 2pt p 2 1 L 67. y =  p sin a tb + p 68. y = sin , L 7 0 L 2 2p COMPUTER EXPLORATIONS In Exercises 69–72, you will explore graphically the general sine function ƒsxd = A sin a
71. The vertical shift D Set the constants A = 3, B = 6, C = 0 . a. Plot ƒ(x) for the values D = 0, 1 , and 3 over the interval 4p … x … 4p . Describe what happens to the graph of the general sine function as D increases through positive values. b. What happens to the graph for negative values of D? 72. The amplitude A
Set the constants B = 6, C = D = 0 .
a. Describe what happens to the graph of the general sine function as A increases through positive values. Confirm your answer by plotting ƒ(x) for the values A = 1, 5 , and 9. b. What happens to the graph for negative values of A?
2p sx  Cdb + D B
as you change the values of the constants A, B, C, and D. Use a CAS or computer grapher to perform the steps in the exercises.
1.4
Graphing with Calculators and Computers A graphing calculator or a computer with graphing software enables us to graph very complicated functions with high precision. Many of these functions could not otherwise be easily graphed. However, care must be taken when using such devices for graphing purposes, and in this section we address some of the issues involved. In Chapter 4 we will see how calculus helps us determine that we are accurately viewing all the important features of a function’s graph.
Graphing Windows When using a graphing calculator or computer as a graphing tool, a portion of the graph is displayed in a rectangular display or viewing window. Often the default window gives an incomplete or misleading picture of the graph. We use the term square window when the units or scales on both axes are the same. This term does not mean that the display window itself is square (usually it is rectangular), but instead it means that the xunit is the same as the yunit. When a graph is displayed in the default window, the xunit may differ from the yunit of scaling in order to fit the graph in the window. The viewing window is set by specifying an interval [a, b] for the xvalues and an interval [c, d] for the yvalues. The machine selects equally spaced xvalues in [a, b] and then plots the points (x, ƒ(x)). A point is plotted if and
30
Chapter 1: Functions
only if x lies in the domain of the function and ƒ(x) lies within the interval [c, d]. A short line segment is then drawn between each plotted point and its next neighboring point. We now give illustrative examples of some common problems that may occur with this procedure. Graph the function ƒsxd = x 3  7x 2 + 28 in each of the following display or viewing windows:
EXAMPLE 1
(a) [10, 10] by [10, 10]
(b) [4, 4] by [50, 10]
(c) [4, 10] by [60, 60]
Solution
(a) We select a = 10, b = 10, c = 10, and d = 10 to specify the interval of xvalues and the range of yvalues for the window. The resulting graph is shown in Figure 1.48a. It appears that the window is cutting off the bottom part of the graph and that the interval of xvalues is too large. Let’s try the next window.
10
10 –4 –10
60 4 –4
10
–10
–50
(a)
(b)
10
–60 (c)
FIGURE 1.48 The graph of ƒsxd = x 3  7x 2 + 28 in different viewing windows. Selecting a window that gives a clear picture of a graph is often a trialanderror process (Example 1).
(b) We see some new features of the graph (Figure 1.48b), but the top is missing and we need to view more to the right of x = 4 as well. The next window should help. (c) Figure 1.48c shows the graph in this new viewing window. Observe that we get a more complete picture of the graph in this window, and it is a reasonable graph of a thirddegree polynomial.
EXAMPLE 2
When a graph is displayed, the xunit may differ from the yunit, as in the graphs shown in Figures 1.48b and 1.48c. The result is distortion in the picture, which may be misleading. The display window can be made square by compressing or stretching the units on one axis to match the scale on the other, giving the true graph. Many systems have builtin functions to make the window “square.” If yours does not, you will have to do some calculations and set the window size manually to get a square window, or bring to your viewing some foreknowledge of the true picture. Figure 1.49a shows the graphs of the perpendicular lines y = x and y = x + 3 22, together with the semicircle y = 29  x 2 , in a nonsquare [4, 4] by [6, 8] display window. Notice the distortion. The lines do not appear to be perpendicular, and the semicircle appears to be elliptical in shape. Figure 1.49b shows the graphs of the same functions in a square window in which the xunits are scaled to be the same as the yunits. Notice that the scaling on the xaxis for Figure 1.49a has been compressed in Figure 1.49b to make the window square. Figure 1.49c gives an enlarged view of Figure 1.49b with a square [3, 3] by [0, 4] window.
If the denominator of a rational function is zero at some xvalue within the viewing window, a calculator or graphing computer software may produce a steep nearvertical line segment from the top to the bottom of the window. Example 3 illustrates this situation.
1.4 8
Graphing with Calculators and Computers 4
4
–4
–6
4
31
6
–3 –6 (a)
–4 (b)
3 0 (c)
FIGURE 1.49 Graphs of the perpendicular lines y = x and y = x + 322 , and the semicircle y = 29  x 2 appear distorted (a) in a nonsquare window, but clear (b) and (c) in square windows (Example 2).
1 . 2  x Solution Figure 1.50a shows the graph in the [10, 10] by [10, 10] default square window with our computer graphing software. Notice the nearvertical line segment at x = 2. It is not truly a part of the graph and x = 2 does not belong to the domain of the function. By trial and error we can eliminate the line by changing the viewing window to the smaller [6, 6] by [4, 4] view, revealing a better graph (Figure 1.50b).
EXAMPLE 3
Graph the function y =
10
4
–10
10
–6
6
–10 (a)
–4 (b)
1 . A vertical line may appear 2  x without a careful choice of the viewing window (Example 3). FIGURE 1.50 Graphs of the function y =
Sometimes the graph of a trigonometric function oscillates very rapidly. When a calculator or computer software plots the points of the graph and connects them, many of the maximum and minimum points are actually missed. The resulting graph is then very misleading.
EXAMPLE 4
Graph the function ƒsxd = sin 100x.
Solution Figure 1.51a shows the graph of ƒ in the viewing window [12, 12] by [1, 1]. We see that the graph looks very strange because the sine curve should oscillate periodically between 1 and 1. This behavior is not exhibited in Figure 1.51a. We might 1
–12
1
12
–1 (a)
–6
1
6
–1 (b)
–0.1
0.1
–1 (c)
FIGURE 1.51 Graphs of the function y = sin 100x in three viewing windows. Because the period is 2p>100 L 0.063 , the smaller window in (c) best displays the true aspects of this rapidly oscillating function (Example 4).
32
Chapter 1: Functions
experiment with a smaller viewing window, say [6, 6] by [1, 1], but the graph is not better (Figure 1.51b). The difficulty is that the period of the trigonometric function y = sin 100x is very small s2p>100 L 0.063d. If we choose the much smaller viewing window [0.1, 0.1] by [1, 1] we get the graph shown in Figure 1.51c. This graph reveals the expected oscillations of a sine curve. 1 sin 50x. 50 Solution In the viewing window [6, 6] by [1, 1] the graph appears much like the cosine function with some small sharp wiggles on it (Figure 1.52a). We get a better look when we significantly reduce the window to [0.6, 0.6] by [0.8, 1.02], obtaining the graph in Figure 1.52b. We now see the small but rapid oscillations of the second term, s1>50d sin 50x, added to the comparatively larger values of the cosine curve.
EXAMPLE 5
Graph the function y = cos x +
1.02
1
–6
6
–0.6
0.6 0.8 (b)
–1 (a)
FIGURE 1.52 In (b) we see a closeup view of the function 1 sin 50x graphed in (a). The term cos x clearly dominates the y = cos x + 50 1 second term, sin 50x , which produces the rapid oscillations along the 50 cosine curve. Both views are needed for a clear idea of the graph (Example 5).
Obtaining a Complete Graph Some graphing devices will not display the portion of a graph for ƒ(x) when x 6 0. Usually that happens because of the procedure the device is using to calculate the function values. Sometimes we can obtain the complete graph by defining the formula for the function in a different way.
EXAMPLE 6
Graph the function y = x 1>3 .
Some graphing devices display the graph shown in Figure 1.53a. When we 3 compare it with the graph of y = x 1>3 = 2 x in Figure 1.17, we see that the left branch for Solution
2
–3
2
3
–2 (a)
–3
3
–2 (b)
FIGURE 1.53 The graph of y = x 1>3 is missing the left branch in (a). In x # 1>3 (b) we graph the function ƒsxd = ƒ x ƒ , obtaining both branches. (See ƒxƒ Example 6.)
1.5
Exponential Functions
33
x 6 0 is missing. The reason the graphs differ is that many calculators and computer software programs calculate x 1>3 as e s1>3dln x . Since the logarithmic function is not defined for negative values of x, the computing device can produce only the right branch, where x 7 0. (Logarithmic and exponential functions are introduced in the next two sections.) To obtain the full picture showing both branches, we can graph the function ƒsxd =
x # 1>3 ƒxƒ . ƒxƒ
This function equals x 1>3 except at x = 0 (where ƒ is undefined, although 0 1>3 = 0). The graph of ƒ is shown in Figure 1.53b.
Exercises 1.4 Choosing a Viewing Window T In Exercises 1–4, use a graphing calculator or computer to determine which of the given viewing windows displays the most appropriate graph of the specified function. 1. ƒsxd = x 4  7x 2 + 6x a. [1, 1] by [1, 1]
b. [2, 2] by [5, 5]
c. [10, 10] by [10, 10] d. [5, 5] by [25, 15] 2. ƒsxd = x 3  4x 2  4x + 16 a. [1, 1] by [5, 5]
b. [3, 3] by [10, 10]
c. [5, 5] by [10, 20]
d. [20, 20] by [100, 100]
3. ƒsxd = 5 + 12x  x 3 a. [1, 1] by [1, 1]
b. [5, 5] by [10, 10]
c. [4, 4] by [20, 20]
d. [4, 5] by [15, 25]
19. ƒsxd =
x2 + 2 x2 + 1
20. ƒsxd =
x2  1 x2 + 1
21. ƒsxd =
x  1 x2  x  6
22. ƒsxd =
8 x2  9
24. ƒsxd =
x2  3 x  2
6x 2  15x + 6 4x 2  10x 25. y = sin 250x 23. ƒsxd =
27. y = cos a 29. y = x +
26. y = 3 cos 60x
x b 50
28. y =
x 1 sin a b 10 10
1 sin 30x 10
30. y = x 2 +
1 cos 100x 50
31. Graph the lower half of the circle defined by the equation x 2 + 2x = 4 + 4y  y 2 . 32. Graph the upper branch of the hyperbola y 2  16x 2 = 1 .
4. ƒsxd = 25 + 4x  x 2 a. [2, 2] by [2, 2]
b. [2, 6] by [1, 4]
33. Graph four periods of the function ƒsxd =  tan 2x .
c. [3, 7] by [0, 10]
d. [10, 10] by [10, 10]
34. Graph two periods of the function ƒsxd = 3 cot
Finding a Viewing Window T In Exercises 5–30, find an appropriate viewing window for the given function and use it to display its graph. x3 x2  2x + 1 5. ƒsxd = x 4  4x 3 + 15 6. ƒsxd = 3 2 5 4 3 4 7. ƒsxd = x  5x + 10 8. ƒsxd = 4x  x 9. ƒsxd = x29  x 2
10. ƒsxd = x 2s6  x 3 d
11. y = 2x  3x 2>3
12. y = x 1>3sx 2  8d
13. y = 5x
14. y = x 2>3s5  xd
2>5
 2x
15. y = ƒ x 2  1 ƒ 17. y =
x + 3 x + 2
1.5
16. y = ƒ x 2  x ƒ 18. y = 1 
1 x + 3
x + 1. 2
35. Graph the function ƒsxd = sin 2x + cos 3x . 36. Graph the function ƒsxd = sin3 x . Graphing in Dot Mode T Another way to avoid incorrect connections when using a graphing device is through the use of a “dot mode,” which plots only the points. If your graphing utility allows that mode, use it to plot the functions in Exercises 37–40. 1 1 37. y = 38. y = sin x x  3 x3  1 39. y = x :x; 40. y = 2 x  1
Exponential Functions Exponential functions are among the most important in mathematics and occur in a wide variety of applications, including interest rates, radioactive decay, population growth, the spread of a disease, consumption of natural resources, the earth’s atmospheric pressure, temperature change of a heated object placed in a cooler environment, and the dating of
34
Chapter 1: Functions
fossils. In this section we introduce these functions informally, using an intuitive approach. We give a rigorous development of them in Chapter 7, based on important calculus ideas and results.
Exponential Behavior
Don’t confuse the exponential function 2x with the power function x 2. In the exponential function, the variable x is in the exponent, whereas the variable x is the base in the power function.
When a positive quantity P doubles, it increases by a factor of 2 and the quantity becomes 2P. If it doubles again, it becomes 2(2P) = 22P, and a third doubling gives 2(22P) = 23P. Continuing to double in this fashion leads us to the consideration of the function ƒ(x) = 2x. We call this an exponential function because the variable x appears in the exponent of 2x. Functions such as g(x) = 10 x and h(x) = (1>2) x are other examples of exponential functions. In general, if a Z 1 is a positive constant, the function ƒ(x) = a x,
a0
is the exponential function with base a.
EXAMPLE 1 In 2010, $100 is invested in a savings account, where it grows by accruing interest that is compounded annually (once a year) at an interest rate of 5.5%. Assuming no additional funds are deposited to the account and no money is withdrawn, give a formula for a function describing the amount A in the account after x years have elapsed. If P = 100, at the end of the first year the amount in the account is the original amount plus the interest accrued, or Solution
P + a
5.5 bP = (1 + 0.055)P = (1.055)P. 100
At the end of the second year the account earns interest again and grows to (1 + 0.055) # (1.055P) = (1.055) 2P = 100 # (1.055) 2.
P = 100
Continuing this process, after x years the value of the account is A = 100 # (1.055) x. This is a multiple of the exponential function with base 1.055. Table 1.4 shows the amounts accrued over the first four years. Notice that the amount in the account each year is always 1.055 times its value in the previous year.
TABLE 1.4
Year 2010 2011 2012 2013 2014
Savings account growth Amount (dollars) 100 100(1.055) 100(1.055) 2 100(1.055) 3 100(1.055) 4
= = = =
105.50 111.30 117.42 123.88
Increase (dollars)
5.50 5.80 6.12 6.46
In general, the amount after x years is given by P(1 + r) x, where r is the interest rate (expressed as a decimal).
1.5
y 10 x
a n = 1442443 a # a # Á # a.
12 10
n factors
If x = 0, then a 0 = 1, and if x = n for some positive integer n, then
8 6 y 3x
4 2
a n =
y 2x
–1 –0.5 0 0.5 1 (a) y 2 x, y 3 x, y 10 x
n
a 1>n = 2a, which is the positive number that when multiplied by itself n times gives a. If x = p>q is any rational number, then
y 10 –x
q
a p>q = 2a p =
12 10 8 6 4 2
y 2 –x
n
1 1 = aa b . an
If x = 1>n for some positive integer n, then
x
y
y
35
For integer and rational exponents, the value of an exponential function ƒ(x) = a x is obtained arithmetically as follows. If x = n is a positive integer, the number a n is given by multiplying a by itself n times:
y
3 –x
Exponential Functions
–1 –0.5 0 0.5 1 (b) y 2 –x, y 3 –x, y 10 –x
x
FIGURE 1.54 Graphs of exponential functions.
A 2a B . q
p
If x is irrational, the meaning of a x is not so clear, but its value can be defined by considering values for rational numbers that get closer and closer to x. This informal approach is based on the graph of the exponential function. In Chapter 7 we define the meaning in a rigorous way. We displayed the graphs of several exponential functions in Section 1.1, and show them again here in Figure 1.54. These graphs describe the values of the exponential functions for all real inputs x. The value at an irrational number x is chosen so that the graph of a x has no “holes” or “jumps.” Of course, these words are not mathematical terms, but they do convey the informal idea. We mean that the value of a x, when x is irrational, is chosen so that the function ƒ(x) = a x is continuous, a notion that will be carefully explored in the next chapter. This choice ensures the graph retains its increasing behavior when a 7 1, or decreasing behavior when 0 6 a 6 1 (see Figure 1.54). Arithmetically, the graphical idea can be described in the following way, using the exponential function ƒ(x) = 2x as an illustration. Any particular irrational number, say x = 23, has a decimal expansion 23 = 1.732050808 . . . . We then consider the list of numbers, given as follows in the order of taking more and more digits in the decimal expansion,
TABLE 1.5 Values of 213 for
rational r closer and closer to 23 r
2r
1.0 1.7 1.73 1.732 1.7320 1.73205 1.732050 1.7320508 1.73205080 1.732050808
2.000000000 3.249009585 3.317278183 3.321880096 3.321880096 3.321995226 3.321995226 3.321997068 3.321997068 3.321997086
21, 21.7, 21.73, 21.732, 21.7320, 21.73205, . . . .
(1)
We know the meaning of each number in list (1) because the successive decimal approximations to 23 given by 1, 1.7, 1.73, 1.732, and so on, are all rational numbers. As these decimal approximations get closer and closer to 23, it seems reasonable that the list of numbers in (1) gets closer and closer to some fixed number, which we specify to be 223. Table 1.5 illustrates how taking better approximations to 23 gives better approximations to the number 213 L 3.321997086. It is the completeness property of the real numbers (discussed briefly in Appendix 7) which guarantees that this procedure gives a single number we define to be 213 (although it is beyond the scope of this text to give a proof ). In a similar way, we can identify the number 2x (or a x, a 7 0) for any irrational x. By identifying the number a x for both rational and irrational x, we eliminate any “holes” or “gaps” in the graph of a x. In practice you can use a calculator to find the number a x for irrational x, taking successive decimal approximations to x and creating a table similar to Table 1.5. Exponential functions obey the familiar rules of exponents listed on the next page. It is easy to check these rules using algebra when the exponents are integers or rational numbers. We prove them for all real exponents in Chapters 4 and 7.
36
Chapter 1: Functions
Rules for Exponents If a 7 0 and b 7 0, the following rules hold true for all real numbers x and y. ax 1. a x # a y = a x + y 2. y = a x  y a 3. (a x ) y = (a y ) x = a xy 4. a x # b x = (ab) x 5.
a ax = a b b bx
EXAMPLE 2 1. 2. 3. 4. 5.
x
We illustrate using the rules for exponents to simplify numerical expressions.
31.1 # 30.7 = 31.1 + 0.7 = 31.8
A 210 B 210
3
=
A 210 B
A 522 B 22 = 522 p
7
#8
p
4 a b 9
= (56)
1>2
=
31
# 22
=
2 A 210 B = 10
= 52 = 25
p
41>2 2 = 1>2 3 9
The Natural Exponential Function e x The most important exponential function used for modeling natural, physical, and economic phenomena is the natural exponential function, whose base is the special number e. The number e is irrational, and its value is 2.718281828 to nine decimal places. (In Section 3.8 we will see a way to calculate the value of e.) It might seem strange that we would use this number for a base rather than a simple number like 2 or 10. The advantage in using e as a base is that it simplifies many of the calculations in calculus. If you look at Figure 1.54a you can see that the graphs of the exponential functions y = a x get steeper as the base a gets larger. This idea of steepness is conveyed by the slope of the tangent line to the graph at a point. Tangent lines to graphs of functions are defined precisely in the next chapter, but intuitively the tangent line to the graph at a point is a line that just touches the graph at the point, like a tangent to a circle. Figure 1.55 shows the slope of the graph of y = a x as it crosses the yaxis for several values of a. Notice that the slope is exactly equal to 1 when a equals the number e. The slope is smaller than 1 if a 6 e, and larger than 1 if a 7 e. This is the property that makes the number e so useful in calculus: The graph of y e x has slope 1 when it crosses the yaxis.
y
y
y 2x m 0.7
y
y ex
m 1.1
m1
1
1 0 (a)
x
y 3x
1 0
(b)
x
0
x
(c)
FIGURE 1.55 Among the exponential functions, the graph of y = e x has the property that the slope m of the tangent line to the graph is exactly 1 when it crosses the yaxis. The slope is smaller for a base less than e, such as 2x, and larger for a base greater than e, such as 3x.
1.5
37
Exponential Functions
Exponential Growth and Decay The exponential functions y = e kx, where k is a nonzero constant, are frequently used for modeling exponential growth or decay. The function y = y0 e kx is a model for exponential growth if k 7 0 and a model for exponential decay if k 6 0. Here y0 represents a constant. An example of exponential growth occurs when computing interest compounded continuously modeled by y = P # e rt, where P is the initial monetary investment, r is the interest rate as a decimal, and t is time in units consistent with r. An example of exponential 4 decay is the model y = A # e 1.2 * 10 t, which represents how the radioactive isotope carbon14 decays over time. Here A is the original amount of carbon14 and t is the time in years. Carbon14 decay is used to date the remains of dead organisms such as shells, seeds, and wooden artifacts. Figure 1.56 shows graphs of exponential growth and exponential decay. y y 20 1.4
15
1
10
y e1.5x
y e –1.2 x
0.6
5 0.2 –1
– 0.5
0
0.5
1
1.5
2
x
– 0.5
0
0.5
1
1.5
2
2.5
3
x
(b)
(a)
FIGURE 1.56 Graphs of (a) exponential growth, k = 1.5 7 0, and (b) exponential decay, k =  1.2 6 0.
Investment companies often use the model y = Pe rt in calculating the growth of an investment. Use this model to track the growth of $100 invested in 2000 at an annual interest rate of 5.5%.
EXAMPLE 3
Let t = 0 represent 2010, t = 1 represent 2011, and so on. Then the exponential growth model is y(t) = Pe rt, where P = 100 (the initial investment), r = 0.055 (the annual interest rate expressed as a decimal), and t is time in years. To predict the amount in the account in 2014, after four years have elapsed, we take t = 4 and calculate
Solution
y(4) = 100e 0.055(4) = 100e 0.22 = 124.61.
Nearest cent using calculator
This compares with $123.88 in the account when the interest is compounded annually from Example 1.
EXAMPLE 4
Laboratory experiments indicate that some atoms emit a part of their mass as radiation, with the remainder of the atom reforming to make an atom of some new element. For example, radioactive carbon14 decays into nitrogen; radium eventually decays into lead. If y0 is the number of radioactive nuclei present at time zero, the number still present at any later time t will be y = y0 e rt,
r 7 0.
38
Chapter 1: Functions
The number r is called the decay rate of the radioactive substance. (We will see how this formula is obtained in Section 7.2.) For carbon14, the decay rate has been determined experimentally to be about r = 1.2 * 10 4 when t is measured in years. Predict the percent of carbon14 present after 866 years have elapsed. If we start with an amount y0 of carbon14 nuclei, after 866 years we are left with the amount Solution
y(866) = y0 e (1.2 * 10 L (0.901)y0.
4
)(866)
Calculator evaluation
That is, after 866 years, we are left with about 90% of the original amount of carbon14, so about 10% of the original nuclei have decayed. In Example 7 in the next section, you will see how to find the number of years required for half of the radioactive nuclei present in a sample to decay (called the halflife of the substance). You may wonder why we use the family of functions y = e kx for different values of the constant k instead of the general exponential functions y = a x. In the next section, we show that the exponential function a x is equal to e kx for an appropriate value of k. So the formula y = e kx covers the entire range of possibilities, and we will see that it is easier to use.
Exercises 1.5 Sketching Exponential Curves In Exercises 1–6, sketch the given curves together in the appropriate coordinate plane and label each curve with its equation. 1. y = 2x, y = 4x, y = 3x, y = (1>5) x 2. y = 3x, y = 8x, y = 2x, y = (1>4) x 3. y = 2t and y = 2t
4. y = 3t and y = 3t
5. y = e x and y = 1>e x
6. y = e x and y = e x
In each of Exercises 7–10, sketch the shifted exponential curves. 7. y = 2x  1 and y = 2x  1 8. y = 3x + 2 and y = 3x + 2 9. y = 1  e x and y = 1  e x 10. y = 1  e x and y = 1  e x Applying the Laws of Exponents Use the laws of exponents to simplify the expressions in Exercises 11–20. 11. 162 # 161.75 44.2 43.7 4 15. A 251>8 B 13.
17. 2
23
19. a
# 723
2 22
b
4
12. 91>3 # 91>6
35>3 32>3 16. A 1322 B 22>2 14.
18.
1>2 A 23 B # A 212 B
20. a
1>2
26 2 b 3
Composites Involving Exponential Functions Find the domain and range for each of the functions in Exercises 21–24. 1 21. ƒ(x) = 22. g(t) = cos (e t ) 2 + ex 3 23. g(t) = 21 + 3t 24. ƒ(x) = 1  e 2x Applications T In Exercises 25–28, use graphs to find approximate solutions. 25. 2x = 5
26. e x = 4
27. 3x  0.5 = 0
28. 3  2x = 0
T In Exercises 29–36, use an exponential model and a graphing calculator to estimate the answer in each problem. 29. Population growth The population of Knoxville is 500,000 and is increasing at the rate of 3.75% each year. Approximately when will the population reach 1 million? 30. Population growth The population of Silver Run in the year 1890 was 6250. Assume the population increased at a rate of 2.75% per year. a. Estimate the population in 1915 and 1940. b. Approximately when did the population reach 50,000? 31. Radioactive decay The halflife of phosphorus32 is about 14 days. There are 6.6 grams present initially. a. Express the amount of phosphorus32 remaining as a function of time t. b. When will there be 1 gram remaining?
1.6
Inverse Functions and Logarithms
39
32. If John invests $2300 in a savings account with a 6% interest rate compounded annually, how long will it take until John’s account has a balance of $4150?
35. Cholera bacteria Suppose that a colony of bacteria starts with 1 bacterium and doubles in number every half hour. How many bacteria will the colony contain at the end of 24 hr?
33. Doubling your money Determine how much time is required for an investment to double in value if interest is earned at the rate of 6.25% compounded annually.
36. Eliminating a disease Suppose that in any given year the number of cases of a disease is reduced by 20%. If there are 10,000 cases today, how many years will it take
34. Tripling your money Determine how much time is required for an investment to triple in value if interest is earned at the rate of 5.75% compounded continuously.
b. to eliminate the disease; that is, to reduce the number of cases to less than 1?
a. to reduce the number of cases to 1000?
Inverse Functions and Logarithms
1.6
A function that undoes, or inverts, the effect of a function ƒ is called the inverse of ƒ. Many common functions, though not all, are paired with an inverse. In this section we present the natural logarithmic function y = ln x as the inverse of the exponential function y = e x, and we also give examples of several inverse trigonometric functions.
OnetoOne Functions A function is a rule that assigns a value from its range to each element in its domain. Some functions assign the same range value to more than one element in the domain. The function ƒsxd = x 2 assigns the same value, 1, to both of the numbers 1 and +1; the sines of p>3 and 2p>3 are both 13>2. Other functions assume each value in their range no more than once. The square roots and cubes of different numbers are always different. A function that has distinct values at distinct elements in its domain is called onetoone. These functions take on any one value in their range exactly once. y
y y x
y x3 x
0
x
0
DEFINITION A function ƒ(x) is onetoone on a domain D if ƒsx1 d Z ƒsx2 d whenever x1 Z x2 in D.
EXAMPLE 1
Some functions are onetoone on their entire natural domain. Other functions are not onetoone on their entire domain, but by restricting the function to a smaller domain we can create a function that is onetoone. The original and restricted functions are not the same functions, because they have different domains. However, the two functions have the same values on the smaller domain, so the original function is an extension of the restricted function from its smaller domain to the larger domain.
(a) Onetoone: Graph meets each horizontal line at most once. y
y x2
Same yvalue y Same yvalue
1 –1 0
0.5 1
x
6
5 6
x
y sin x (b) Not onetoone: Graph meets one or more horizontal lines more than once.
FIGURE 1.57 (a) y = x 3 and y = 1x are onetoone on their domains s  q , q d and [0, q d. (b) y = x 2 and y = sin x are not onetoone on their domains s  q , q d .
(a) ƒsxd = 1x is onetoone on any domain of nonnegative numbers because 1x1 Z 1x2 whenever x1 Z x2 . (b) gsxd = sin x is not onetoone on the interval [0, p] because sin sp>6d = sin s5p>6d. In fact, for each element x1 in the subinterval [0, p>2d there is a corresponding element x2 in the subinterval sp>2, p] satisfying sin x1 = sin x2, so distinct elements in the domain are assigned to the same value in the range. The sine function is onetoone on [0, p>2], however, because it is an increasing function on [0, p>2] giving distinct outputs for distinct inputs. The graph of a onetoone function y = ƒsxd can intersect a given horizontal line at most once. If the function intersects the line more than once, it assumes the same yvalue for at least two different xvalues and is therefore not onetoone (Figure 1.57).
40
Chapter 1: Functions
The Horizontal Line Test for OnetoOne Functions A function y = ƒsxd is onetoone if and only if its graph intersects each horizontal line at most once.
Inverse Functions Since each output of a onetoone function comes from just one input, the effect of the function can be inverted to send an output back to the input from which it came.
DEFINITION Suppose that ƒ is a onetoone function on a domain D with range R. The inverse function ƒ 1 is defined by ƒ 1sbd = a if ƒsad = b. The domain of ƒ 1 is R and the range of ƒ 1 is D.
The symbol ƒ 1 for the inverse of ƒ is read “ƒ inverse.” The “ 1” in ƒ 1 is not an exponent; ƒ 1sxd does not mean 1> ƒ(x). Notice that the domains and ranges of ƒ and ƒ 1 are interchanged. Suppose a onetoone function y = ƒsxd is given by a table of values
EXAMPLE 2 x
1
2
3
4
5
6
7
8
ƒ(x)
3
4.5
7
10.5
15
20.5
27
34.5
A table for the values of x = ƒ 1s yd can then be obtained by simply interchanging the values in the columns (or rows) of the table for ƒ: y 1
ƒ s yd
3
4.5
7
10.5
15
20.5
27
34.5
1
2
3
4
5
6
7
8
If we apply ƒ to send an input x to the output ƒ(x) and follow by applying ƒ 1 to ƒ(x), we get right back to x, just where we started. Similarly, if we take some number y in the range of ƒ, apply ƒ 1 to it, and then apply ƒ to the resulting value ƒ 1syd, we get back the value y with which we began. Composing a function and its inverse has the same effect as doing nothing. sƒ 1 ⴰ ƒdsxd = x,
for all x in the domain of ƒ
sƒ ⴰ ƒ 1 dsyd = y,
for all y in the domain of ƒ 1 sor range of ƒd
Only a onetoone function can have an inverse. The reason is that if ƒsx1 d = y and ƒsx2 d = y for two distinct inputs x1 and x2 , then there is no way to assign a value to ƒ 1syd that satisfies both ƒ 1sƒsx1 dd = x1 and ƒ 1sƒsx2 dd = x2 . A function that is increasing on an interval so it satisfies the inequality ƒsx2 d 7 ƒsx1 d when x2 7 x1 is onetoone and has an inverse. Decreasing functions also have an inverse. Functions that are neither increasing nor decreasing may still be onetoone and have an
1.6
41
Inverse Functions and Logarithms
inverse, as with the function ƒ(x) = 1>x for x Z 0 and ƒ(0) = 0, defined on ( q , q ) and passing the horizontal line test.
Finding Inverses The graphs of a function and its inverse are closely related. To read the value of a function from its graph, we start at a point x on the xaxis, go vertically to the graph, and then move horizontally to the yaxis to read the value of y. The inverse function can be read from the graph by reversing this process. Start with a point y on the yaxis, go horizontally to the graph of y = ƒsxd, and then move vertically to the xaxis to read the value of x = ƒ 1syd (Figure 1.58).
y
DOMAIN OF
y 5 f (x)
f RANGE OF
f –1
y
y
0
x
x DOMAIN OF
x 5 f –1(y) y
0
f
RANGE OF
f
–1
–1 is
(a) To find the value of f at x, we start at x, go up to the curve, and then over to the yaxis.
(b) The graph of f the graph of f, but with x and y interchanged. To find the x that gave y, we start at y and go over to the curve and down to the xaxis. The domain of f –1 is the range of f. The range of f –1 is the domain of f.
y
f –1
x y5x x5f
–1(y)
(b, a) 0
y DOMAIN OF
(c) To draw the graph of f –1 in the more usual way, we reflect the system across the line y 5 x.
f –1
y 5 f –1(x)
f –1
(a, b)
RANGE OF
RANGE OF
x
x
x
0 DOMAIN OF
f –1
(d) Then we interchange the letters x and y. We now have a normallooking graph of f –1 as a function of x.
FIGURE 1.58 The graph of y = ƒ 1(x) is obtained by reflecting the graph of y = ƒ(x) about the line y = x.
We want to set up the graph of ƒ 1 so that its input values lie along the xaxis, as is usually done for functions, rather than on the yaxis. To achieve this we interchange the x and yaxes by reflecting across the 45° line y = x. After this reflection we have a new graph that represents ƒ 1 . The value of ƒ 1sxd can now be read from the graph in the usual way, by starting with a point x on the xaxis, going vertically to the graph, and then horizontally
42
Chapter 1: Functions
to the yaxis to get the value of ƒ 1sxd. Figure 1.58 indicates the relationship between the graphs of ƒ and ƒ 1 . The graphs are interchanged by reflection through the line y = x. The process of passing from ƒ to ƒ 1 can be summarized as a twostep procedure. 1. 2.
y
y 2x 2
yx
Solve the equation y = ƒsxd for x. This gives a formula x = ƒ 1s yd where x is expressed as a function of y. Interchange x and y, obtaining a formula y = ƒ 1sxd where ƒ 1 is expressed in the conventional format with x as the independent variable and y as the dependent variable.
EXAMPLE 3
1 x + 1, expressed as a function of x. 2
Find the inverse of y =
Solution y 1x1 2
1.
Solve for x in terms of y:
1 –2
1 x + 1 2 2y = x + 2
x
1
FIGURE 1.59 Graphing ƒsxd = s1>2dx + 1 and ƒ 1sxd = 2x  2 together shows the graphs’ symmetry with respect to the line y = x (Example 3).
The graph is a straight line satisfying the horizontal line test (Fig. 1.59).
x = 2y  2. 2.
–2
y =
Interchange x and y:
y = 2x  2.
The inverse of the function ƒsxd = s1>2dx + 1 is the function ƒ 1sxd = 2x  2. (See Figure 1.59.) To check, we verify that both composites give the identity function: 1 ƒ 1sƒsxdd = 2 a x + 1b  2 = x + 2  2 = x 2 ƒsƒ 1sxdd =
1 s2x  2d + 1 = x  1 + 1 = x. 2
y
EXAMPLE 4
y x 2, x 0
Find the inverse of the function y = x 2, x Ú 0, expressed as a function
of x. yx
For x Ú 0, the graph satisfies the horizontal line test, so the function is onetoone and has an inverse. To find the inverse, we first solve for x in terms of y: Solution
y x
y = x2 2y = 2x 2 = ƒ x ƒ = x 0
x
FIGURE 1.60 The functions y = 1x and y = x 2, x Ú 0 , are inverses of one another (Example 4).
ƒ x ƒ = x because x Ú 0
We then interchange x and y, obtaining y = 1x. The inverse of the function y = x , x Ú 0, is the function y = 1x (Figure 1.60). Notice that the function y = x 2, x Ú 0, with domain restricted to the nonnegative real numbers, is onetoone (Figure 1.60) and has an inverse. On the other hand, the function y = x 2, with no domain restrictions, is not onetoone (Figure 1.57b) and therefore has no inverse. 2
Logarithmic Functions If a is any positive real number other than 1, the base a exponential function ƒ(x) = a x is onetoone. It therefore has an inverse. Its inverse is called the logarithm function with base a.
DEFINITION The logarithm function with base a, y = loga x, is the inverse of the base a exponential function y = a x (a 7 0, a Z 1).
1.6 y
y 2x yx
y log 2 x
2 1 0
x
1 2
Inverse Functions and Logarithms
43
The domain of loga x is (0, q ), the range of a x. The range of loga x is ( q , q ), the domain of a x. Figure 1.23 in Section 1.1 shows the graphs of four logarithmic functions with a 7 1. Figure 1.61a shows the graph of y = log2 x. The graph of y = a x, a 7 1, increases rapidly for x 7 0, so its inverse, y = loga x, increases slowly for x 7 1. Because we have no technique yet for solving the equation y = a x for x in terms of y, we do not have an explicit formula for computing the logarithm at a given value of x. Nevertheless, we can obtain the graph of y = loga x by reflecting the graph of the exponential y = a x across the line y = x. Figure 1.61 shows the graphs for a = 2 and a = e. Logarithms with base 2 are commonly used in computer science. Logarithms with base e and base 10 are so important in applications that calculators have special keys for them. They also have their own special notation and names:
(a)
loge x is written as ln x. log10 x is written as log x.
y 8
y ex
The function y = ln x is called the natural logarithm function, and y = log x is often called the common logarithm function. For the natural logarithm,
7 6
ln x = y 3 e y = x.
5 4
In particular, if we set x = e, we obtain e
(1, e)
2
y ln x
ln e = 1
1 –2
–1
0
1
2
e
4
x
because e 1 = e.
Properties of Logarithms (b) x
FIGURE 1.61 (a) The graph of 2 and its inverse, log2 x. (b) The graph of e x and its inverse, ln x.
HISTORICAL BIOGRAPHY* John Napier (1550–1617)
Logarithms, invented by John Napier, were the single most important improvement in arithmetic calculation before the modern electronic computer. What made them so useful is that the properties of logarithms reduce multiplication of positive numbers to addition of their logarithms, division of positive numbers to subtraction of their logarithms, and exponentiation of a number to multiplying its logarithm by the exponent. We summarize these properties for the natural logarithm as a series of rules that we prove in Chapter 3. Although here we state the Power Rule for all real powers r, the case when r is an irrational number cannot be dealt with properly until Chapter 4. We also establish the validity of the rules for logarithmic functions with any base a in Chapter 7.
THEOREM 1—Algebraic Properties of the Natural Logarithm For any numbers b 7 0 and x 7 0, the natural logarithm satisfies the following rules: 1. Product Rule:
ln bx = ln b + ln x
2. Quotient Rule:
b ln x = ln b  ln x
3. Reciprocal Rule:
1 ln x = ln x
4. Power Rule:
ln x r = r ln x
Rule 2 with b = 1
*To learn more about the historical figures mentioned in the text and the development of many major elements and topics of calculus, visit www.aw.com/thomas.
44
Chapter 1: Functions
EXAMPLE 5
Here we use the properties in Theorem 1 to simplify three expressions.
(a) ln 4 + ln sin x = ln (4 sin x)
Product Rule
x + 1 = ln (x + 1)  ln (2x  3) (b) ln 2x  3 1 (c) ln = ln 8 8
Quotient Rule Reciprocal Rule
= ln 23 = 3 ln 2
Power Rule
Because a x and loga x are inverses, composing them in either order gives the identity function. Inverse Properties for ax and loga x 1. Base a: a loga x = x, 2. Base e: e
ln x
= x,
loga a x = x,
a 7 0, a Z 1, x 7 0
ln e = x,
x 7 0
x
Substituting a x for x in the equation x = e ln x enables us to rewrite a x as a power of e: x
a x = e ln (a )
Substitute a x for x in x = e ln x.
= e = e (ln a) x. x ln a
Power Rule for logs Exponent rearranged
Thus, the exponential function a is the same as e kx for k = ln a. x
Every exponential function is a power of the natural exponential function. a x = e x ln a That is, a x is the same as e x raised to the power ln a: a x = e kx for k = ln a. For example, 2x = e (ln 2) x = e x ln 2,
and
5  3x = e (ln 5) (  3x) = e  3x ln 5.
Returning once more to the properties of a x and loga x, we have ln x = ln (a loga x) = (loga x)(ln a).
Inverse Property for a x and loga x Power Rule for logarithms, with r = loga x
Rewriting this equation as loga x = (ln x)>(ln a) shows that every logarithmic function is a constant multiple of the natural logarithm ln x. This allows us to extend the algebraic properties for ln x to loga x. For instance, loga bx = loga b + loga x. Change of Base Formula Every logarithmic function is a constant multiple of the natural logarithm. ln x loga x = (a 7 0, a Z 1) ln a
Applications In Section 1.5 we looked at examples of exponential growth and decay problems. Here we use properties of logarithms to answer more questions concerning such problems.
EXAMPLE 6 If $1000 is invested in an account that earns 5.25% interest compounded annually, how long will it take the account to reach $2500?
1.6
Inverse Functions and Logarithms
45
From Example 1, Section 1.5 with P = 1000 and r = 0.0525, the amount in the account at any time t in years is 1000(1.0525) t, so to find the time t when the account reaches $2500 we need to solve the equation
Solution
1000(1.0525) t = 2500. Thus we have (1.0525) t = 2.5
Divide by 1000.
ln (1.0525) = ln 2.5 t
Take logarithms of both sides.
t ln 1.0525 = ln 2.5 ln 2.5 L 17.9 t = ln 1.0525
Power Rule Values obtained by calculator
The amount in the account will reach $2500 in 18 years, when the annual interest payment is deposited for that year.
EXAMPLE 7
The halflife of a radioactive element is the time required for half of the radioactive nuclei present in a sample to decay. It is a remarkable fact that the halflife is a constant that does not depend on the number of radioactive nuclei initially present in the sample, but only on the radioactive substance. To see why, let y0 be the number of radioactive nuclei initially present in the sample. Then the number y present at any later time t will be y = y0 e kt . We seek the value of t at which the number of radioactive nuclei present equals half the original number: 1 y 2 0 1 e kt = 2 1 kt = ln = ln 2 2 ln 2 t = . k
y0 e kt =
Reciprocal Rule for logarithms
(1)
This value of t is the halflife of the element. It depends only on the value of k; the number y0 does not have any effect. The effective radioactive lifetime of polonium210 is so short that we measure it in days rather than years. The number of radioactive atoms remaining after t days in a sample that starts with y0 radioactive atoms is
Amount present y0 y y0 e–5 10
3
–3t
y = y0 e 5 * 10 t.
1y 2 0 1y 4 0
The element’s halflife is ln 2 k ln 2 = 5 * 10 3 L 139 days.
Halflife =
0
139
278
t (days)
Halflife
FIGURE 1.62 Amount of polonium210 present at time t, where y0 represents the number of radioactive atoms initially present (Example 7).
Eq. (1) The k from polonium’s decay equation
This means that after 139 days, 1/2 of y0 radioactive atoms remain; after another 139 days (or 278 days altogether) half of those remain, or 1/4 of y0 radioactive atoms remain, and so on (see Figure 1.62).
Inverse Trigonometric Functions The six basic trigonometric functions of a general radian angle x were reviewed in Section 1.3. These functions are not onetoone (their values repeat periodically). However, we can restrict their domains to intervals on which they are onetoone. The sine function
46
Chapter 1: Functions
increases from 1 at x = p>2 to +1 at x = p>2. By restricting its domain to the interval [p>2, p>2] we make it onetoone, so that it has an inverse sin1 x (Figure 1.63). Similar domain restrictions can be applied to all six trigonometric functions.
y x sin y y sin –1x Domain: –1 x 1 Range: –/ 2 y / 2
2
–1
Domain restrictions that make the trigonometric functions onetoone
x
1
– 2
– 2
sin x 1
1 0
–1
FIGURE 1.63 The graph of y = sin1 x.
2
x
tan x
cos x 2
0 –1
y = sin x Domain: [p>2, p>2] Range: [ 1, 1] y
y
y
y
x
– 2
2
x
y = tan x Domain: s p>2, p>2d Range: s  q , q d
y = cos x Domain: [0, p] Range: [1, 1] y
sec x
0
y
csc x
cot x 1 0
y y sin x, – 2 x 2 Domain: [–/2, /2] Range: [–1, 1] 1 – 2
x
2
0 –1 (a) y
x sin y y sin –1x Domain: [–1, 1] Range: [–/2, /2]
2
1 2
x
– 2
2
0
x 0
y = csc x Domain: [p>2, 0d ´ s0, p>2] Range: s  q , 1] ´ [1, q d
y = sec x Domain: [0, p>2d ´ sp>2, p] Range: s  q , 1] ´ [1, q d
2
x
y = cot x Domain: s0, pd Range: s  q , q d
Since these restricted functions are now onetoone, they have inverses, which we denote by y y y y y y
= = = = = =
sin1 x cos1 x tan1 x sec1 x csc1 x cot1 x
or or or or or or
y y y y y y
= = = = = =
arcsin x arccos x arctan x arcsec x arccsc x arccot x
These equations are read “y equals the arcsine of x” or “y equals arcsin x” and so on. –1
0
1
x
– 2
(b)
FIGURE 1.64 The graphs of (a) y = sin x, p>2 … x … p>2 , and (b) its inverse, y = sin1 x . The graph of sin1 x , obtained by reflection across the line y = x , is a portion of the curve x = sin y .
Caution The 1 in the expressions for the inverse means “inverse.” It does not mean reciprocal. For example, the reciprocal of sin x is ssin xd1 = 1>sin x = csc x. The graphs of the six inverse trigonometric functions are obtained by reflecting the graphs of the restricted trigonometric functions through the line y = x. Figure 1.64b shows the graph of y = sin1 x and Figure 1.65 shows the graphs of all six functions. We now take a closer look at two of these functions.
The Arcsine and Arccosine Functions We define the arcsine and arccosine as functions whose values are angles (measured in radians) that belong to restricted domains of the sine and cosine functions.
1.6 Domain: –1 x 1 Range: – y 2 2 y 2
y
–1
x
1
2
y cos –1x 2
– 2
–1
–2
–1
x
1
2
y sec –1x
2
–2 –1
1
2
–1
Domain: – ∞ x ∞ Range: 0 y y
y csc–1x 1
2
2
x
– 2
x
(d)
FIGURE 1.65
(c)
Domain: x –1 or x 1 Range: – y , y 0 2 2 y
Domain: x –1 or x 1 Range: 0 y , y 2 y
y tan –1x x 1 2
– 2
(b)
(a)
–2
Domain: – ∞ x ∞ Range: – y 2 2 y
Domain: –1 x 1 0 y Range:
y sin –1x
47
Inverse Functions and Logarithms
–2
–1
(e)
y cot –1x
1
2
x
(f )
Graphs of the six basic inverse trigonometric functions.
DEFINITION y sin1 x is the number in [p>2, p>2] for which sin y = x. y cos1 x is the number in [0, p] for which cos y = x. The “Arc” in Arcsine and Arccosine For a unit circle and radian angles, the arc length equation s = r u becomes s = u , so central angles and the arcs they subtend have the same measure. If x = sin y , then, in addition to being the angle whose sine is x, y is also the length of arc on the unit circle that subtends an angle whose sine is x. So we call y “the arc whose sine is x.”
The graph of y = sin1 x (Figure 1.64b) is symmetric about the origin (it lies along the graph of x = sin y). The arcsine is therefore an odd function: sin1s xd = sin1 x.
(2)
The graph of y = cos1 x (Figure 1.66b) has no such symmetry.
EXAMPLE 8
Evaluate (a) sin1 a
23 1 b and (b) cos1 a b. 2 2
Solution
y
(a) We see that x2 1 y2 5 1
sin1 a
Arc whose sine is x
Arc whose cosine is x
Angle whose sine is x 0
x Angle whose cosine is x
1
x
23 p b = 2 3
because sin (p>3) = 23>2 and p>3 belongs to the range [p>2, p>2] of the arcsine function. See Figure 1.67a. (b) We have 2p 1 cos1 a b = 2 3 because cos (2p>3) = 1>2 and 2p>3 belongs to the range [0, p] of the arccosine function. See Figure 1.67b.
48
Chapter 1: Functions
y
Using the same procedure illustrated in Example 8, we can create the following table of common values for the arcsine and arccosine functions.
y cos x, 0 x Domain: [0, ] Range: [–1, 1]
1
2
0 –1
x
23>2 22>2 1> 2 1>2  22>2  23>2
(a) y x cos y
y cos –1 x Domain: [–1, 1] Range: [0, ]
2 –1 0
sin x 1 x
x
cos1 x
p>3 23>2 p>4 22>2 p>6 1> 2 p>6 1>2 p>4  22>2 p>3  23>2
p>6 p>4 p>3 2p>3 3p>4 5p>6
x
1
y
(b)
2 3 0 1
FIGURE 1.66 The graphs of (a) y = cos x, 0 … x … p , and (b) its inverse, y = cos1 x . The graph of cos1 x , obtained by reflection across the line y = x , is a portion of the curve x = cos y .
y
sin –1 3 2 3
cos–1 ⎛– 1⎛ 2 p 3 ⎝ 2⎝ 2
3
3 x
sin 3 2 3 (a)
FIGURE 1.67 (Example 8).
Chicago
–1 0
2p 3 x
cos ⎛ 2 p⎛ – 1 ⎝3 ⎝ 2 (b)
Values of the arcsine and arccosine functions
179 Springfield
180
61
12 62 b
St. Louis
Plane
EXAMPLE 9
During an airplane flight from Chicago to St. Louis, the navigator determines that the plane is 12 mi off course, as shown in Figure 1.68. Find the angle a for a course parallel to the original correct course, the angle b, and the drift correction angle c = a + b.
a
c
From Figure 1.68 and elementary geometry, we see that 180 sin a = 12 and 62 sin b = 12, so 12 a = sin1 L 0.067 radian L 3.8° 180 12 b = sin1 L 0.195 radian L 11.2° 62
Solution FIGURE 1.68 Diagram for drift correction (Example 9), with distances rounded to the nearest mile (drawing not to scale).
c = a + b L 15°.
y –1(–x)
cos
cos–1x –1 –x
0
x
1
x
FIGURE 1.69 cos1 x and cos1s xd are supplementary angles (so their sum is p).
Identities Involving Arcsine and Arccosine As we can see from Figure 1.69, the arccosine of x satisfies the identity cos1 x + cos1s xd = p, or cos1 s xd = p  cos1 x. Also, we can see from the triangle in Figure 1.70 that for x 7 0, sin1 x + cos1 x = p>2.
(3) (4)
(5)
1.6
cos–1x
1
x
sin–1x
FIGURE 1.70 sin1 x and cos1 x are complementary angles (so their sum is p>2).
49
Inverse Functions and Logarithms
Equation (5) holds for the other values of x in [1, 1] as well, but we cannot conclude this from the triangle in Figure 1.70. It is, however, a consequence of Equations (2) and (4) (Exercise 76). The arctangent, arccotangent, arcsecant, and arccosecant functions are defined in Section 3.9. There we develop additional properties of the inverse trigonometric functions in a calculus setting using the identities discussed here.
Exercises 1.6 Identifying OnetoOne Functions Graphically Which of the functions graphed in Exercises 1–6 are onetoone, and which are not? 1.
2.
y
y
y 3x 3
Graphing Inverse Functions Each of Exercises 11–16 shows the graph of a function y = ƒsxd . Copy the graph and draw in the line y = x . Then use symmetry with respect to the line y = x to add the graph of ƒ 1 to your sketch. (It is not necessary to find a formula for ƒ 1 .) Identify the domain and range of ƒ 1 . 11.
x
0
–1
0 1 y x4 x2
12. y
y
x
y f (x) 2 1 , x 0 x 1 1
3.
4.
y
1 0
y f (x) 1 1x , x 0 x
1
y y int x 0
x
1
y 2x
13.
14. y
y
x
y f (x) sin x, –x 1 2 2
5.
6.
y y 1x
y y x1/3
– 2
2
0
– 2
x
15.
16. y
6
8. ƒsxd = e
2x + 6, x + 4,
10. ƒsxd = e
x … 3 x 7 3
x , 2
x … 0
x , x + 2
x 7 0
2  x 2, x 2,
x … 1 x 7 1
1 9. ƒsxd = d
x 6 0 x Ú 0
f (x) f (x) 6 2x, 0x3
In Exercises 7–10, determine from its graph if the function is onetoone. 3  x, 3,
x
–1
y
7. ƒsxd = e
2
0
x
x
0
y f (x) tan x, –x 2 2
0
3
x
1 –1 0
x 1, 1 x 0 2 x, 0 x 3 3
2 3
x
–2
17. a. Graph the function ƒsxd = 21  x 2, 0 … x … 1 . What symmetry does the graph have? b. Show that ƒ is its own inverse. (Remember that 2x 2 = x if x Ú 0 .) 18. a. Graph the function ƒsxd = 1>x . What symmetry does the graph have? b. Show that ƒ is its own inverse.
50
Chapter 1: Functions
Formulas for Inverse Functions Each of Exercises 19–24 gives a formula for a function y = ƒsxd and shows the graphs of ƒ and ƒ 1 . Find a formula for ƒ 1 in each case. 19. ƒsxd = x 2 + 1,
x Ú 0
20. ƒsxd = x 2,
x … 0
y
y
y f (x)
1
y f –1(x)
0
0
21. ƒsxd = x  1
x
1
x
1
2
x Ú 1
y
y y f (x)
1 x
1
–1
0
40. a. Find the inverse of ƒsxd =  x + 1 . Graph the line y = x + 1 together with the line y = x . At what angle do the lines intersect?
c. What can you conclude about the inverses of functions whose graphs are lines perpendicular to the line y = x ?
y f (x)
1
c. What can you conclude about the inverses of functions whose graphs are lines parallel to the line y = x ?
b. Find the inverse of ƒsxd = x + b (b constant). What angle does the line y = x + b make with the line y = x ?
y f –1(x)
y f –1(x)
39. a. Find the inverse of ƒsxd = x + 1 . Graph ƒ and its inverse together. Add the line y = x to your sketch, drawing it with dashes or dots for contrast. b. Find the inverse of ƒsxd = x + b (b constant). How is the graph of ƒ 1 related to the graph of ƒ?
y f –1(x)
22. ƒsxd = x  2x + 1,
3
–1
b. What can you conclude about the inverse of a function y = ƒsxd whose graph is a line through the origin with a nonzero slope m? 38. Show that the graph of the inverse of ƒsxd = mx + b , where m and b are constants and m Z 0 , is a line with slope 1> m and yintercept b>m .
y f (x)
1
Inverses of Lines 37. a. Find the inverse of the function ƒsxd = mx , where m is a constant different from zero.
x
1
Logarithms and Exponentials 41. Express the following logarithms in terms of ln 2 and ln 3. b. ln (4> 9)
a. ln 0.75 23. ƒsxd = sx + 1d2,
x Ú 1 24. ƒsxd = x 2>3,
y
1
y f (x) 1
–1 0
x
1
Each of Exercises 25–36 gives a formula for a function y = ƒsxd . In each case, find ƒ 1sxd and identify the domain and range of ƒ 1 . As a check, show that ƒsƒ 1sxdd = ƒ 1sƒsxdd = x . 25. ƒsxd = x 5
26. ƒsxd = x 4,
27. ƒsxd = x + 1
28. ƒsxd = s1>2dx  7>2
3
29. ƒsxd = 1>x 2, 31. ƒsxd =
x 7 0
x + 3 x  2
33. ƒsxd = x 2  2x,
x Ú 0
30. ƒsxd = 1>x 3, 32. ƒsxd =
x … 1
f. ln 213.5
e. ln 322
y f –1(x)
x
3 d. ln 2 9
42. Express the following logarithms in terms of ln 5 and ln 7.
y f –1(x)
1 0
x Ú 0
y y f (x)
–1
c. ln (1> 2)
x Z 0
2x 2x  3
34. ƒsxd = s2x 3 + 1d1>5
(Hint: Complete the square.) x + b , b 7 2 and constant 35. ƒsxd = x  2 36. ƒsxd = x2  2bx, b 7 0 and constant, x … b
a. ln (1> 125)
b. ln 9.8
c. ln 727
d. ln 1225
e. ln 0.056
f. sln 35 + ln s1>7dd>sln 25d
Use the properties of logarithms to write the expressions in Exercises 43 and 44 as a single term. sin u 1 b 43. a. ln sin u  ln a b. ln s3x 2  9xd + ln a b 5 3x 1 c. ln s4t 4 d  ln b 2 44. a. ln sec u + ln cos u b. ln s8x + 4d  2 ln c 3 2 t  1  ln st + 1d c. 3 ln2
Find simpler expressions for the quantities in Exercises 45–48. 45. a. e ln 7.2 46. a. e
ln sx 2 + y 2d
b. e ln x b. e
c. e ln x  ln y
2
ln 0.3
c. e ln px  ln 2
47. a. 2 ln 2e
b. ln sln e e d
48. a. ln se sec u d
b. ln se se d d
x
c. ln se x
2
 y2
d
c. ln se 2 ln x d
In Exercises 49–54, solve for y in terms of t or x, as appropriate. 49. ln y = 2t + 4
50. ln y = t + 5
51. ln s y  bd = 5t
52. ln sc  2yd = t
53. ln s y  1d  ln 2 = x + ln x 54. ln s y 2  1d  ln s y + 1d = ln ssin xd
1.6 In Exercises 55 and 56, solve for k. 55. a. e 2k = 4 b. 100e 10k = 200 1 56. a. e 5k = b. 80e k = 1 4 In Exercises 57–60, solve for t. 1 57. a. e 0.3t = 27 b. e kt = 2 1 58. a. e 0.01t = 1000 b. e kt = 10 59. e 2t = x 2
c. e k>1000 = a c. e sln 0.8dk = 0.8 c. e sln 0.2dt = 0.4 c. e sln 2dt =
1 2
60. e sx de s2x + 1d = e t 2
Simplify the expressions in Exercises 61–64. 61. a. 5log5 7 d. log4 16 62. a. 2log2 3 d. log11 121 63. a. 2log4 x log5 s3x 2d
64. a. 25
b. 8log822 e. log3 23
c. 1.3log1.3 75 1 f. log4 a b 4
b. 10 log10 s1>2d
c. plogp 7
e. log121 11
1 f. log3 a b 9
x
b. log e se d
69. a. arccos (1) 70. a. arcsin (1)
c. log4 s2e
c. shifting left 1, up 3 units.
x
sin x
d
23 1 b c. cos1 a b 2 22 b. arccos (0) b. arcsin a
75. Find a formula for the inverse function ƒ 1 and verify that (ƒ ⴰ ƒ 1)(x) = (ƒ 1 ⴰ ƒ )(x) = x. 50 100 a. ƒ(x) = b. ƒ(x) = 1 + 2x 1 + 1.1x 76. The identity sin1 x + cos1 x = P>2 Figure 1.70 establishes the identity for 0 6 x 6 1 . To establish it for the rest of [1, 1] , verify by direct calculation that it holds for x = 1 , 0, and 1 . Then, for values of x in s 1, 0d , let x = a, a 7 0 , and apply Eqs. (3) and (5) to the sum sin1s ad + cos1s ad .
a. shifting down 3 units.
Arcsine and Arccosine In Exercises 67–70, find the exact value of each expression.  23 1 1 b c. sin1 a b 67. a. sin1 a b b. sin1 a 2 2 22 b. cos1 a
74. If a composite ƒ ⴰ g is onetoone, must g be onetoone? Give reasons for your answer.
b. shifting right 1 unit.
Express the ratios in Exercises 65 and 66 as ratios of natural logarithms and simplify. log x a log 2 x log 2 x 65. a. b. c. log 3 x log 8 x log x2 a log210 x log 9 x log a b 66. a. b. c. log 3 x log b a log22 x
1 68. a. cos1 a b 2
51
77. Start with the graph of y = ln x. Find an equation of the graph that results from
c. log2 se sln 2dssin xd d
b. 9log3 x
Inverse Functions and Logarithms
1 22
b
Theory and Examples 71. If ƒ(x) is onetoone, can anything be said about gsxd = ƒsxd ? Is it also onetoone? Give reasons for your answer. 72. If ƒ(x) is onetoone and ƒ(x) is never zero, can anything be said about hsxd = 1>ƒsxd ? Is it also onetoone? Give reasons for your answer. 73. Suppose that the range of g lies in the domain of ƒ so that the composite ƒ ⴰ g is defined. If ƒ and g are onetoone, can anything be said about ƒ ⴰ g ? Give reasons for your answer.
d. shifting down 4, right 2 units. e. reflecting about the yaxis. f. reflecting about the line y = x. 78. Start with the graph of y = ln x. Find an equation of the graph that results from a. vertical stretching by a factor of 2. b. horizontal stretching by a factor of 3. c. vertical compression by a factor of 4. d. horizontal compression by a factor of 2. T 79. The equation x 2 = 2x has three solutions: x = 2, x = 4, and one other. Estimate the third solution as accurately as you can by graphing. T 80. Could x ln 2 possibly be the same as 2ln x for x 7 0? Graph the two functions and explain what you see. 81. Radioactive decay The halflife of a certain radioactive substance is 12 hours. There are 8 grams present initially. a. Express the amount of substance remaining as a function of time t. b. When will there be 1 gram remaining? 82. Doubling your money Determine how much time is required for a $500 investment to double in value if interest is earned at the rate of 4.75% compounded annually. 83. Population growth The population of Glenbrook is 375,000 and is increasing at the rate of 2.25% per year. Predict when the population will be 1 million. 84. Radon222 The decay equation for radon222 gas is known to be y = y0 e 0.18t , with t in days. About how long will it take the radon in a sealed sample of air to fall to 90% of its original value?
2 LIMITS AND CONTINUITY OVERVIEW Mathematicians of the seventeenth century were keenly interested in the study of motion for objects on or near the earth and the motion of planets and stars. This study involved both the speed of the object and its direction of motion at any instant, and they knew the direction at a given instant was along a line tangent to the path of motion. The concept of a limit is fundamental to finding the velocity of a moving object and the tangent to a curve. In this chapter we develop the limit, first intuitively and then formally. We use limits to describe the way a function varies. Some functions vary continuously; small changes in x produce only small changes in ƒ(x). Other functions can have values that jump, vary erratically, or tend to increase or decrease without bound. The notion of limit gives a precise way to distinguish between these behaviors.
2.1
Rates of Change and Tangents to Curves Calculus is a tool that helps us understand how a change in one quantity is related to a change in another. How does the speed of a falling object change as a function of time? How does the level of water in a barrel change as a function of the amount of liquid poured into it? In this section we introduce the ideas of average and instantaneous rates of change, and show that they are closely related to the slope of a curve at a point P on the curve. We give precise developments of these important concepts in the next chapter, but for now we use an informal approach so you will see how they lead naturally to the main idea of the chapter, the limit. The idea of a limit plays a foundational role throughout calculus.
Average and Instantaneous Speed HISTORICAL BIOGRAPHY Galileo Galilei (1564–1642)
In the late sixteenth century, Galileo discovered that a solid object dropped from rest (not moving) near the surface of the earth and allowed to fall freely will fall a distance proportional to the square of the time it has been falling. This type of motion is called free fall. It assumes negligible air resistance to slow the object down, and that gravity is the only force acting on the falling body. If y denotes the distance fallen in feet after t seconds, then Galileo’s law is y = 16t 2, where 16 is the (approximate) constant of proportionality. (If y is measured in meters, the constant is 4.9.) A moving body’s average speed during an interval of time is found by dividing the distance covered by the time elapsed. The unit of measure is length per unit time: kilometers per hour, feet (or meters) per second, or whatever is appropriate to the problem at hand.
52
2.1
EXAMPLE 1
Rates of Change and Tangents to Curves
53
A rock breaks loose from the top of a tall cliff. What is its average speed
(a) during the first 2 sec of fall? (b) during the 1sec interval between second 1 and second 2? The average speed of the rock during a given time interval is the change in distance, ¢y, divided by the length of the time interval, ¢t. (Increments like ¢y and ¢t are reviewed in Appendix 3.) Measuring distance in feet and time in seconds, we have the following calculations:
Solution
(a) For the first 2 sec:
¢y 16s2d2  16s0d2 ft = = 32 sec 2  0 ¢t
(b) From sec 1 to sec 2:
¢y 16s2d2  16s1d2 ft = = 48 sec 2  1 ¢t
We want a way to determine the speed of a falling object at a single instant t0, instead of using its average speed over an interval of time. To do this, we examine what happens when we calculate the average speed over shorter and shorter time intervals starting at t0. The next example illustrates this process. Our discussion is informal here, but it will be made precise in Chapter 3.
EXAMPLE 2
Find the speed of the falling rock in Example 1 at t = 1 and t = 2 sec.
We can calculate the average speed of the rock over a time interval [t0 , t0 + h], having length ¢t = h, as
Solution
¢y 16st0 + hd2  16t0 2 . = h ¢t
(1)
We cannot use this formula to calculate the “instantaneous” speed at the exact moment t0 by simply substituting h = 0, because we cannot divide by zero. But we can use it to calculate average speeds over increasingly short time intervals starting at t0 = 1 and t0 = 2. When we do so, we see a pattern (Table 2.1).
TABLE 2.1
Average speeds over short time intervals [t0, t0 + h] Average speed:
¢y 16st0 + hd2  16t0 2 = h ¢t
Length of time interval h
Average speed over interval of length h starting at t0 1
Average speed over interval of length h starting at t0 2
1 0.1 0.01 0.001 0.0001
48 33.6 32.16 32.016 32.0016
80 65.6 64.16 64.016 64.0016
The average speed on intervals starting at t0 = 1 seems to approach a limiting value of 32 as the length of the interval decreases. This suggests that the rock is falling at a speed of 32 ft> sec at t0 = 1 sec. Let’s confirm this algebraically.
54
Chapter 2: Limits and Continuity
If we set t0 = 1 and then expand the numerator in Equation (1) and simplify, we find that ¢y 16s1 + 2h + h 2 d  16 16s1 + hd2  16s1d2 = = h h ¢t =
32h + 16h 2 = 32 + 16h. h
For values of h different from 0, the expressions on the right and left are equivalent and the average speed is 32 + 16h ft>sec. We can now see why the average speed has the limiting value 32 + 16s0d = 32 ft>sec as h approaches 0. Similarly, setting t0 = 2 in Equation (1), the procedure yields ¢y = 64 + 16h ¢t for values of h different from 0. As h gets closer and closer to 0, the average speed has the limiting value 64 ft >sec when t0 = 2 sec, as suggested by Table 2.1. The average speed of a falling object is an example of a more general idea which we discuss next.
Average Rates of Change and Secant Lines Given an arbitrary function y = ƒsxd, we calculate the average rate of change of y with respect to x over the interval [x1 , x2] by dividing the change in the value of y, ¢y = ƒsx2 d  ƒsx1 d, by the length ¢x = x2  x1 = h of the interval over which the change occurs. (We use the symbol h for ¢x to simplify the notation here and later on.)
y y f (x) Q(x 2, f(x 2 ))
Secant y
P(x1, f(x1))
DEFINITION The average rate of change of y = ƒsxd with respect to x over the interval [x1 , x2] is
x h 0
x2
x1
FIGURE 2.1 A secant to the graph y = ƒsxd . Its slope is ¢y>¢x , the average rate of change of ƒ over the interval [x1 , x2] .
x
ƒsx2 d  ƒsx1 d ¢y ƒsx1 + hd  ƒsx1 d = = , x2  x1 h ¢x
h Z 0.
Geometrically, the rate of change of ƒ over [x1, x2] is the slope of the line through the points Psx1, ƒsx1 dd and Qsx2 , ƒsx2 dd (Figure 2.1). In geometry, a line joining two points of a curve is a secant to the curve. Thus, the average rate of change of ƒ from x1 to x2 is identical with the slope of secant PQ. Let’s consider what happens as the point Q approaches the point P along the curve, so the length h of the interval over which the change occurs approaches zero. We will see that this procedure leads to defining the slope of a curve at a point.
Defining the Slope of a Curve P L O
FIGURE 2.2 L is tangent to the circle at P if it passes through P perpendicular to radius OP.
We know what is meant by the slope of a straight line, which tells us the rate at which it rises or falls—its rate of change as a linear function. But what is meant by the slope of a curve at a point P on the curve? If there is a tangent line to the curve at P—a line that just touches the curve like the tangent to a circle—it would be reasonable to identify the slope of the tangent as the slope of the curve at P. So we need a precise meaning for the tangent at a point on a curve. For circles, tangency is straightforward. A line L is tangent to a circle at a point P if L passes through P perpendicular to the radius at P (Figure 2.2). Such a line just touches the circle. But what does it mean to say that a line L is tangent to some other curve C at a point P?
2.1
Rates of Change and Tangents to Curves
55
To define tangency for general curves, we need an approach that takes into account the behavior of the secants through P and nearby points Q as Q moves toward P along the curve (Figure 2.3). Here is the idea: 1. 2. 3.
HISTORICAL BIOGRAPHY
Start with what we can calculate, namely the slope of the secant PQ. Investigate the limiting value of the secant slope as Q approaches P along the curve. (We clarify the limit idea in the next section.) If the limit exists, take it to be the slope of the curve at P and define the tangent to the curve at P to be the line through P with this slope.
This procedure is what we were doing in the fallingrock problem discussed in Example 2. The next example illustrates the geometric idea for the tangent to a curve.
Pierre de Fermat (1601–1665) Secants
Tangent
P
P Q
Tangent
Secants
Q
FIGURE 2.3 The tangent to the curve at P is the line through P whose slope is the limit of the secant slopes as Q : P from either side.
Find the slope of the parabola y = x 2 at the point P(2, 4). Write an equation for the tangent to the parabola at this point.
EXAMPLE 3
We begin with a secant line through P(2, 4) and Qs2 + h, s2 + hd2 d nearby. We then write an expression for the slope of the secant PQ and investigate what happens to the slope as Q approaches P along the curve: Solution
Secant slope =
¢y s2 + hd2  22 h 2 + 4h + 4  4 = = h h ¢x =
h 2 + 4h = h + 4. h
If h 7 0, then Q lies above and to the right of P, as in Figure 2.4. If h 6 0, then Q lies to the left of P (not shown). In either case, as Q approaches P along the curve, h approaches zero and the secant slope h + 4 approaches 4. We take 4 to be the parabola’s slope at P. y y x2
Secant slope is
(2 h) 2 4 h 4. h
Q(2 h, (2 h) 2) Tangent slope 4 Δy (2 h)2 4 P(2, 4) Δx h 0
2
2h
x
NOT TO SCALE
FIGURE 2.4 Finding the slope of the parabola y = x 2 at the point P(2, 4) as the limit of secant slopes (Example 3).
Chapter 2: Limits and Continuity
The tangent to the parabola at P is the line through P with slope 4: y = 4 + 4sx  2d
Pointslope equation
y = 4x  4.
Instantaneous Rates of Change and Tangent Lines The rates at which the rock in Example 2 was falling at the instants t = 1 and t = 2 are called instantaneous rates of change. Instantaneous rates and slopes of tangent lines are intimately connected, as we will now see in the following examples.
EXAMPLE 4
Figure 2.5 shows how a population p of fruit flies (Drosophila) grew in a 50day experiment. The number of flies was counted at regular intervals, the counted values plotted with respect to time t, and the points joined by a smooth curve (colored blue in Figure 2.5). Find the average growth rate from day 23 to day 45. There were 150 flies on day 23 and 340 flies on day 45. Thus the number of flies increased by 340  150 = 190 in 45  23 = 22 days. The average rate of change of the population from day 23 to day 45 was Solution
Average rate of change:
¢p 340  150 190 = = L 8.6 flies>day. 45  23 22 ¢t
p 350 Q(45, 340) Number of flies
56
300 p 190
250 200 P(23, 150)
150
p 8.6 flies/day t t 22
100 50 0
10
20 30 Time (days)
40
50
t
FIGURE 2.5 Growth of a fruit fly population in a controlled experiment. The average rate of change over 22 days is the slope ¢p>¢t of the secant line (Example 4).
This average is the slope of the secant through the points P and Q on the graph in Figure 2.5. The average rate of change from day 23 to day 45 calculated in Example 4 does not tell us how fast the population was changing on day 23 itself. For that we need to examine time intervals closer to the day in question.
EXAMPLE 5
How fast was the number of flies in the population of Example 4 growing
on day 23? To answer this question, we examine the average rates of change over increasingly short time intervals starting at day 23. In geometric terms, we find these rates by calculating the slopes of secants from P to Q, for a sequence of points Q approaching P along the curve (Figure 2.6). Solution
2.1
57
Rates of Change and Tangents to Curves
p
Slope of PQ ≤p/≤t (flies / day) 340 45 330 40 310 35 265 30
(45, 340) (40, 330) (35, 310) (30, 265) FIGURE 2.6

150 23 150 23 150 23 150 23
L 8.6 L 10.6 L 13.3 L 16.4
Q(45, 340)
300 Number of flies
Q
B(35, 350)
350
250 200 P(23, 150)
150 100 50 0
10 20 30 A(14, 0) Time (days)
40
50
t
The positions and slopes of four secants through the point P on the fruit fly graph (Example 5).
The values in the table show that the secant slopes rise from 8.6 to 16.4 as the tcoordinate of Q decreases from 45 to 30, and we would expect the slopes to rise slightly higher as t continued on toward 23. Geometrically, the secants rotate about P and seem to approach the red tangent line in the figure. Since the line appears to pass through the points (14, 0) and (35, 350), it has slope 350  0 = 16.7 flies>day (approximately). 35  14 On day 23 the population was increasing at a rate of about 16.7 flies>day. The instantaneous rates in Example 2 were found to be the values of the average speeds, or average rates of change, as the time interval of length h approached 0. That is, the instantaneous rate is the value the average rate approaches as the length h of the interval over which the change occurs approaches zero. The average rate of change corresponds to the slope of a secant line; the instantaneous rate corresponds to the slope of the tangent line as the independent variable approaches a fixed value. In Example 2, the independent variable t approached the values t = 1 and t = 2. In Example 3, the independent variable x approached the value x = 2. So we see that instantaneous rates and slopes of tangent lines are closely connected. We investigate this connection thoroughly in the next chapter, but to do so we need the concept of a limit.
Exercises 2.1 Average Rates of Change In Exercises 1–6, find the average rate of change of the function over the given interval or intervals. 1. ƒsxd = x 3 + 1 b. [1, 1]
a. [2, 3] 2. g sxd = x
2
a. [1, 1]
b. [2, 0]
3. hstd = cot t a. [p>4, 3p>4]
b. [p>6, p>2]
[1, 2]
Slope of a Curve at a Point In Exercises 7–14, use the method in Example 3 to find (a) the slope of the curve at the given point P, and (b) an equation of the tangent line at P. 7. y = x 2  3,
P(2, 1)
8. y = 5  x ,
P(1, 4)
9. y = x 2  2x  3, 10. y = x  4x, 2
b. [p, p]
[0, 2]
6. Psud = u3  4 u2 + 5u;
2
4. g std = 2 + cos t a. [0, p]
5. Rsud = 24u + 1;
11. y = x 3,
P(2, 3)
P(1, 3)
P(2, 8)
58
Chapter 2: Limits and Continuity b. What is the average rate of increase of the profits between 2002 and 2004?
P(1, 1)
13. y = x  12x,
P(1, 11)
3
14. y = x 3  3x 2 + 4,
c. Use your graph to estimate the rate at which the profits were changing in 2002.
P(2, 0)
Instantaneous Rates of Change 15. Speed of a car The accompanying figure shows the timetodistance graph for a sports car accelerating from a standstill.
a. Find the average rate of change of F(x) over the intervals [1, x] for each x Z 1 in your table.
s P
650 600
Distance (m)
b. Extending the table if necessary, try to determine the rate of change of F(x) at x = 1 .
Q4 Q3
500 400
T 19. Let g sxd = 2x for x Ú 0 .
Q2
a. Find the average rate of change of g(x) with respect to x over the intervals [1, 2], [1, 1.5] and [1, 1 + h] .
300 200
b. Make a table of values of the average rate of change of g with respect to x over the interval [1, 1 + h] for some values of h approaching zero, say h = 0.1, 0.01, 0.001, 0.0001, 0.00001 , and 0.000001.
Q1
100 0
5 10 15 20 Elapsed time (sec)
t
c. What does your table indicate is the rate of change of g(x) with respect to x at x = 1 ?
a. Estimate the slopes of secants PQ1 , PQ2 , PQ3 , and PQ4 , arranging them in order in a table like the one in Figure 2.6. What are the appropriate units for these slopes? b. Then estimate the car’s speed at time t = 20 sec . 16. The accompanying figure shows the plot of distance fallen versus time for an object that fell from the lunar landing module a distance 80 m to the surface of the moon. a. Estimate the slopes of the secants PQ1 , PQ2 , PQ3 , and PQ4 , arranging them in a table like the one in Figure 2.6. b. About how fast was the object going when it hit the surface?
Distance fallen (m)
80
T 20. Let ƒstd = 1>t for t Z 0 . a. Find the average rate of change of ƒ with respect to t over the intervals (i) from t = 2 to t = 3 , and (ii) from t = 2 to t = T . b. Make a table of values of the average rate of change of ƒ with respect to t over the interval [2, T], for some values of T approaching 2, say T = 2.1, 2.01, 2.001, 2.0001, 2.00001 , and 2.000001. c. What does your table indicate is the rate of change of ƒ with respect to t at t = 2 ?
21. The accompanying graph shows the total distance s traveled by a bicyclist after t hours.
Q3
60 Q2
40
0
P
Q4
d. Calculate the limit as h approaches zero of the average rate of change of g(x) with respect to x over the interval [1, 1 + h] .
d. Calculate the limit as T approaches 2 of the average rate of change of ƒ with respect to t over the interval from 2 to T. You will have to do some algebra before you can substitute T = 2 .
y
20
T 18. Make a table of values for the function Fsxd = sx + 2d>sx  2d at the points x = 1.2, x = 11>10, x = 101>100, x = 1001>1000, x = 10001>10000 , and x = 1 .
s
Q1
5 Elapsed time (sec)
10
t
T 17. The profits of a small company for each of the first five years of its operation are given in the following table: Year
Profit in $1000s
2000 2001 2002 2003 2004
6 27 62 111 174
a. Plot points representing the profit as a function of year, and join them by as smooth a curve as you can.
Distance traveled (mi)
12. y = 2  x 3,
40 30 20 10 0
1
2 3 Elapsed time (hr)
4
t
a. Estimate the bicyclist’s average speed over the time intervals [0, 1], [1, 2.5], and [2.5, 3.5]. b. Estimate the bicyclist’s instantaneous speed at the times t = 12, t = 2, and t = 3. c. Estimate the bicyclist’s maximum speed and the specific time at which it occurs.
2.2 22. The accompanying graph shows the total amount of gasoline A in the gas tank of an automobile after being driven for t days.
59
a. Estimate the average rate of gasoline consumption over the time intervals [0, 3], [0, 5], and [7, 10]. b. Estimate the instantaneous rate of gasoline consumption at the times t = 1, t = 4, and t = 8.
A Remaining amount (gal)
Limit of a Function and Limit Laws
c. Estimate the maximum rate of gasoline consumption and the specific time at which it occurs.
16 12 8 4 0
1
2
3 4 5 6 7 8 Elapsed time (days)
9 10
t
Limit of a Function and Limit Laws
2.2
In Section 2.1 we saw that limits arise when finding the instantaneous rate of change of a function or the tangent to a curve. Here we begin with an informal definition of limit and show how we can calculate the values of limits. A precise definition is presented in the next section.
Limits of Function Values
HISTORICAL ESSAY
Frequently when studying a function y = ƒ(x), we find ourselves interested in the function’s behavior near a particular point c, but not at c. This might be the case, for instance, if c is an irrational number, like p or 22, whose values can only be approximated by “close” rational numbers at which we actually evaluate the function instead. Another situation occurs when trying to evaluate a function at c leads to division by zero, which is undefined. We encountered this last circumstance when seeking the instantaneous rate of change in y by considering the quotient function ¢y>h for h closer and closer to zero. Here’s a specific example where we explore numerically how a function behaves near a particular point at which we cannot directly evaluate the function.
Limits y
2 2 y f (x) x 1 x 1
1
–1
0
1
x
EXAMPLE 1
How does the function ƒsxd =
y
x2  1 x  1
behave near x = 1? 2
The given formula defines ƒ for all real numbers x except x = 1 (we cannot divide by zero). For any x Z 1, we can simplify the formula by factoring the numerator and canceling common factors:
Solution
yx1 1
–1
0
1
FIGURE 2.7 The graph of ƒ is identical with the line y = x + 1 except at x = 1 , where ƒ is not defined (Example 1).
x
ƒsxd =
sx  1dsx + 1d = x + 1 x  1
for
x Z 1.
The graph of ƒ is the line y = x + 1 with the point (1, 2) removed. This removed point is shown as a “hole” in Figure 2.7. Even though ƒ(1) is not defined, it is clear that we can make the value of ƒ(x) as close as we want to 2 by choosing x close enough to 1 (Table 2.2).
60
Chapter 2: Limits and Continuity
TABLE 2.2 The closer x gets to 1, the closer ƒ(x) = (x 2  1)>(x  1)
seems to get to 2 x2 1 x 1, x1
Values of x below and above 1
ƒ(x)
0.9 1.1 0.99 1.01 0.999 1.001 0.999999 1.000001
1.9 2.1 1.99 2.01 1.999 2.001 1.999999 2.000001
x1
Let’s generalize the idea illustrated in Example 1. Suppose ƒ(x) is defined on an open interval about c, except possibly at c itself. If ƒ(x) is arbitrarily close to L (as close to L as we like) for all x sufficiently close to c, we say that ƒ approaches the limit L as x approaches c, and write lim ƒsxd = L,
x:c
which is read “the limit of ƒ(x) as x approaches c is L.” For instance, in Example 1 we would say that ƒ(x) approaches the limit 2 as x approaches 1, and write lim ƒsxd = 2,
x:1
or
x2  1 = 2. x:1 x  1 lim
Essentially, the definition says that the values of ƒ(x) are close to the number L whenever x is close to c (on either side of c). This definition is “informal” because phrases like arbitrarily close and sufficiently close are imprecise; their meaning depends on the context. (To a machinist manufacturing a piston, close may mean within a few thousandths of an inch. To an astronomer studying distant galaxies, close may mean within a few thousand lightyears.) Nevertheless, the definition is clear enough to enable us to recognize and evaluate limits of specific functions. We will need the precise definition of Section 2.3, however, when we set out to prove theorems about limits. Here are several more examples exploring the idea of limits.
EXAMPLE 2
The limiting value of a function does not depend on how the function is defined at the point being approached. Consider the three functions in Figure 2.8. The function ƒ has limit 2 as x : 1 even though ƒ is not defined at x = 1. The function g has y
–1
y
y
2
2
2
1
1
1
0
2 (a) f (x) x 1 x 1
1
x
–1
0
1
x
⎧ x2 1 , x 1 ⎪ (b) g(x) ⎨ x 1 ⎪ 1, x1 ⎩
–1
0
1
x
(c) h(x) x 1
FIGURE 2.8 The limits of ƒ(x), g(x), and h(x) all equal 2 as x approaches 1. However, only h(x) has the same function value as its limit at x = 1 (Example 2).
2.2
61
limit 2 as x : 1 even though 2 Z gs1d. The function h is the only one of the three functions in Figure 2.8 whose limit as x : 1 equals its value at x = 1. For h, we have limx:1 hsxd = hs1d. This equality of limit and function value is significant, and we return to it in Section 2.5.
y yx c
EXAMPLE 3
x
c
Limit of a Function and Limit Laws
(a) If ƒ is the identity function ƒsxd = x, then for any value of c (Figure 2.9a), lim ƒsxd = lim x = c.
(a) Identity function
x:c
y
(b) If ƒ is the constant function ƒsxd = k (function with the constant value k), then for any value of c (Figure 2.9b),
yk
k
x:c
lim ƒsxd = lim k = k.
x:c
c
0
x:c
For instances of each of these rules we have
x
lim x = 3
x:3
(b) Constant function
lim s4d = lim s4d = 4.
and
x: 7
x:2
We prove these rules in Example 3 in Section 2.3.
FIGURE 2.9 The functions in Example 3 have limits at all points c.
Some ways that limits can fail to exist are illustrated in Figure 2.10 and described in the next example.
y
1
y
0
y ⎧1 ⎪ , x0 y ⎨x ⎪ 0, x 0 ⎩
⎧ 0, x 0 y⎨ ⎩ 1, x 0
x
1
x
0
x
0
⎧ x0 ⎪ 0, y⎨ 1 ⎪ sin x , x 0 ⎩
–1 (a) Unit step function U(x)
(b) g(x)
(c) f (x)
FIGURE 2.10 None of these functions has a limit as x approaches 0 (Example 4).
Discuss the behavior of the following functions as x : 0.
EXAMPLE 4 (a) Usxd = e
(b) gsxd = L (c) ƒsxd = •
0, 1,
x 6 0 x Ú 0
1 x,
x Z 0
0,
x = 0
0,
x … 0
1 sin x ,
x 7 0
62
Chapter 2: Limits and Continuity Solution
(a) It jumps: The unit step function U(x) has no limit as x : 0 because its values jump at x = 0. For negative values of x arbitrarily close to zero, Usxd = 0. For positive values of x arbitrarily close to zero, Usxd = 1. There is no single value L approached by U(x) as x : 0 (Figure 2.10a). (b) It grows too “large” to have a limit: g(x) has no limit as x : 0 because the values of g grow arbitrarily large in absolute value as x : 0 and do not stay close to any fixed real number (Figure 2.10b). (c) It oscillates too much to have a limit: ƒ(x) has no limit as x : 0 because the function’s values oscillate between +1 and 1 in every open interval containing 0. The values do not stay close to any one number as x : 0 (Figure 2.10c).
The Limit Laws To calculate limits of functions that are arithmetic combinations of functions having known limits, we can use several easy rules.
THEOREM 1—Limit Laws lim ƒsxd = L
x:c
1. Sum Rule: 2. Difference Rule: 3. Constant Multiple Rule: 4. Product Rule: 5. Quotient Rule: 6. Power Rule: 7. Root Rule:
If L, M, c, and k are real numbers and lim gsxd = M,
and
x:c
then
lim sƒsxd + gsxdd = L + M
x:c
lim sƒsxd  gsxdd = L  M
x:c
lim sk # ƒsxdd = k # L
x:c
lim sƒsxd # gsxdd = L # M
x:c
lim
x:c
ƒsxd L = , M gsxd
M Z 0
lim [ƒ(x)]n = L n, n a positive integer
x:c
n
n
lim 2ƒ(x) = 2L = L 1>n, n a positive integer
x:c
(If n is even, we assume that lim ƒ(x) = L 7 0.) x:c
In words, the Sum Rule says that the limit of a sum is the sum of the limits. Similarly, the next rules say that the limit of a difference is the difference of the limits; the limit of a constant times a function is the constant times the limit of the function; the limit of a product is the product of the limits; the limit of a quotient is the quotient of the limits (provided that the limit of the denominator is not 0); the limit of a positive integer power (or root) of a function is the integer power (or root) of the limit (provided that the root of the limit is a real number). It is reasonable that the properties in Theorem 1 are true (although these intuitive arguments do not constitute proofs). If x is sufficiently close to c, then ƒ(x) is close to L and g (x) is close to M, from our informal definition of a limit. It is then reasonable that ƒsxd + gsxd is close to L + M; ƒsxd  gsxd is close to L  M; kƒ(x) is close to kL; ƒ(x)g(x) is close to LM; and ƒ(x)>g(x) is close to L>M if M is not zero. We prove the Sum Rule in Section 2.3, based on a precise definition of limit. Rules 2–5 are proved in Appendix 4. Rule 6 is obtained by applying Rule 4 repeatedly. Rule 7 is proved in more
2.2
Limit of a Function and Limit Laws
63
advanced texts. The sum, difference, and product rules can be extended to any number of functions, not just two. Use the observations limx:c k = k and limx:c x = c (Example 3) and the properties of limits to find the following limits.
EXAMPLE 5
x4 + x2  1 x:c x2 + 5
(a) lim sx 3 + 4x 2  3d
(b) lim
x:c
(c) lim 24x 2  3 x: 2
Solution
(a) lim sx 3 + 4x 2  3d = lim x 3 + lim 4x 2  lim 3 x:c
x:c
x:c
x:c
= c 3 + 4c 2  3 x4 + x2  1 = x:c x2 + 5
(b) lim
=
(c)
Power and Multiple Rules
lim sx + x  1d 4
2
x:c
Quotient Rule
lim sx 2 + 5d
x:c
lim x 4 + lim x 2  lim 1
x:c
x:c
x:c
lim x 2 + lim 5
x:c
=
Sum and Difference Rules
x:c
c + c  1 c2 + 5 4
2
Power or Product Rule
lim 24x 2  3 = 2 lim s4x 2  3d
x: 2
Sum and Difference Rules
x: 2
Root Rule with n = 2
= 2 lim 4x 2  lim 3
Difference Rule
= 24s 2d2  3
Product and Multiple Rules
x: 2
x: 2
= 216  3 = 213 Two consequences of Theorem 1 further simplify the task of calculating limits of polynomials and rational functions. To evaluate the limit of a polynomial function as x approaches c, merely substitute c for x in the formula for the function. To evaluate the limit of a rational function as x approaches a point c at which the denominator is not zero, substitute c for x in the formula for the function. (See Examples 5a and 5b.) We state these results formally as theorems.
THEOREM 2—Limits of Polynomials If Psxd = an x n + an  1 x n  1 + Á + a0 , then lim Psxd = Pscd = an c n + an  1 c n  1 + Á + a 0 .
x:c
THEOREM 3—Limits of Rational Functions If P(x) and Q(x) are polynomials and Qscd Z 0, then lim
x:c
Psxd Pscd = . Qsxd Qscd
64
Chapter 2: Limits and Continuity
EXAMPLE 6
The following calculation illustrates Theorems 2 and 3: s 1d3 + 4s 1d2  3 x 3 + 4x 2  3 0 = = = 0 2 2 6 x: 1 x + 5 s 1d + 5 lim
Identifying Common Factors It can be shown that if Q(x) is a polynomial and Qscd = 0 , then sx  cd is a factor of Q(x). Thus, if the numerator and denominator of a rational function of x are both zero at x = c , they have sx  cd as a common factor.
Eliminating Zero Denominators Algebraically Theorem 3 applies only if the denominator of the rational function is not zero at the limit point c. If the denominator is zero, canceling common factors in the numerator and denominator may reduce the fraction to one whose denominator is no longer zero at c. If this happens, we can find the limit by substitution in the simplified fraction.
EXAMPLE 7
Evaluate
y 2 x2 y x x2 x (1, 3)
3
x2 + x  2 . x:1 x2  x lim
We cannot substitute x = 1 because it makes the denominator zero. We test the numerator to see if it, too, is zero at x = 1. It is, so it has a factor of sx  1d in common with the denominator. Canceling the sx  1d’s gives a simpler fraction with the same values as the original for x Z 1: Solution
–2
0
x
1
sx  1dsx + 2d x + 2 x2 + x  2 = = x , 2 xsx 1d x  x
(a) y yx2 x (1, 3)
3
if x Z 1.
Using the simpler fraction, we find the limit of these values as x : 1 by substitution: x2 + x  2 x + 2 1 + 2 = lim x = = 3. 2 1 x:1 x:1 x  x lim
–2
0
1
x
See Figure 2.11.
(b)
FIGURE 2.11 The graph of ƒsxd = sx 2 + x  2d>sx 2  xd in part (a) is the same as the graph of g sxd = sx + 2d>x in part (b) except at x = 1, where ƒ is undefined. The functions have the same limit as x : 1 (Example 7).
Using Calculators and Computers to Estimate Limits When we cannot use the Quotient Rule in Theorem 1 because the limit of the denominator is zero, we can try using a calculator or computer to guess the limit numerically as x gets closer and closer to c. We used this approach in Example 1, but calculators and computers can sometimes give false values and misleading impressions for functions that are undefined at a point or fail to have a limit there, as we now illustrate.
EXAMPLE 8
Estimate the value of lim
x:0
2x 2 + 100  10 . x2
Solution Table 2.3 lists values of the function for several values near x = 0. As x approaches 0 through the values ;1, ;0.5, ;0.10, and ;0.01, the function seems to approach the number 0.05. As we take even smaller values of x, ;0.0005, ;0.0001, ;0.00001, and ;0.000001, the function appears to approach the value 0. Is the answer 0.05 or 0, or some other value? We resolve this question in the next example.
2.2
Limit of a Function and Limit Laws
TABLE 2.3 Computer values of ƒ(x) =
65
2x 2 + 100  10 near x = 0 x2
x
ƒ(x)
;1 ;0.5 ;0.1 ;0.01
0.049876 0.049969 t approaches 0.05? 0.049999 0.050000
;0.0005 ;0.0001 ;0.00001 ;0.000001
0.050000 0.000000 t approaches 0? 0.000000 0.000000
Using a computer or calculator may give ambiguous results, as in the last example. The calculator does not keep track of enough digits to avoid rounding errors in computing the values of f (x) when x is very small. We cannot substitute x = 0 in the problem, and the numerator and denominator have no obvious common factors (as they did in Example 7). Sometimes, however, we can create a common factor algebraically.
EXAMPLE 9
Evaluate
2x 2 + 100  10 . x:0 x2 Solution This is the limit we considered in Example 8. We can create a common factor by multiplying both numerator and denominator by the conjugate radical expression 2x 2 + 100 + 10 (obtained by changing the sign after the square root). The preliminary algebra rationalizes the numerator: lim
2x 2 + 100  10 2x 2 + 100  10 # 2x 2 + 100 + 10 = 2 x x2 2x 2 + 100 + 10 2 x + 100  100 = 2 x A 2x 2 + 100 + 10 B
Therefore, lim
x:0
x2
=
x 2 A 2x 2 + 100 + 10 B
=
1 . 2x + 100 + 10 2
Common factor x2
Cancel x2 for x Z 0.
2x 2 + 100  10 1 = lim 2 2 x:0 x 2x + 100 + 10 =
1 20 + 100 + 10
=
1 = 0.05. 20
2
Denominator not 0 at x = 0; substitute.
This calculation provides the correct answer, in contrast to the ambiguous computer results in Example 8. We cannot always algebraically resolve the problem of finding the limit of a quotient where the denominator becomes zero. In some cases the limit might then be found with the
66
Chapter 2: Limits and Continuity
aid of some geometry applied to the problem (see the proof of Theorem 7 in Section 2.4), or through methods of calculus (illustrated in Section 4.5). The next theorem is also useful.
y h f L
The Sandwich Theorem g x
c
0
The following theorem enables us to calculate a variety of limits. It is called the Sandwich Theorem because it refers to a function ƒ whose values are sandwiched between the values of two other functions g and h that have the same limit L at a point c. Being trapped between the values of two functions that approach L, the values of ƒ must also approach L (Figure 2.12). You will find a proof in Appendix 5.
FIGURE 2.12 The graph of ƒ is sandwiched between the graphs of g and h.
THEOREM 4—The Sandwich Theorem Suppose that gsxd … ƒsxd … hsxd for all x in some open interval containing c, except possibly at x = c itself. Suppose also that lim gsxd = lim hsxd = L.
y
2
The Sandwich Theorem is also called the Squeeze Theorem or the Pinching Theorem.
2 y1 x 4
0
x:c
Then limx:c ƒsxd = L.
y u(x)
1
–1
x:c
2 y1 x 2
EXAMPLE 10
Given that
x
1
FIGURE 2.13 Any function u(x) whose graph lies in the region between y = 1 + sx 2>2d and y = 1  sx 2>4d has limit 1 as x : 0 (Example 10).
1 
x2 x2 … usxd … 1 + 4 2
find limx:0 usxd, no matter how complicated u is. Solution
Since lim s1  sx 2>4dd = 1
y ⎢ ⎢
EXAMPLE 11
y sin
–
u:0
(c) For any function ƒ, lim ƒ ƒ(x) ƒ = 0 implies lim ƒ(x) = 0. x:c
x:c
(a) In Section 1.3 we established that  ƒ u ƒ … sin u … ƒ u ƒ for all u (see Figure 2.14a). Since limu:0 s  ƒ u ƒ d = limu:0 ƒ u ƒ = 0, we have
y
lim sin u = 0 .
y ⎢ ⎢
1 0
(b) lim cos u = 1
u:0
Solution
(a)
–1
The Sandwich Theorem helps us establish several important limit rules:
(a) lim sin u = 0
y – ⎢ ⎢
–1
–2
x:0
the Sandwich Theorem implies that limx:0 usxd = 1 (Figure 2.13).
1
2
lim s1 + sx 2>2dd = 1,
and
x:0
y
for all x Z 0,
u:0
y 1 cos 1
2
(b)
FIGURE 2.14 The Sandwich Theorem confirms the limits in Example 11.
(b) From Section 1.3, 0 … 1  cos u … ƒ u ƒ for all u (see Figure 2.14b), and we have limu:0 s1  cos ud = 0 or lim cos u = 1 .
u:0
(c) Since  ƒ ƒsxd ƒ … ƒsxd … ƒ ƒsxd ƒ and  ƒ ƒsxd ƒ and ƒ ƒsxd ƒ have limit 0 as x : c, it follows that lim x:c ƒ(x) = 0 .
2.2
Limit of a Function and Limit Laws
67
Another important property of limits is given by the next theorem. A proof is given in the next section.
THEOREM 5 If ƒsxd … gsxd for all x in some open interval containing c, except possibly at x = c itself, and the limits of ƒ and g both exist as x approaches c, then lim ƒsxd … lim gsxd.
x:c
x:c
The assertion resulting from replacing the less than or equal to (…) inequality by the strict less than (6) inequality in Theorem 5 is false. Figure 2.14a shows that for u Z 0,  ƒ u ƒ 6 sin u 6 ƒ u ƒ , but in the limit as u : 0, equality holds.
Exercises 2.2 Limits from Graphs 1. For the function g(x) graphed here, find the following limits or explain why they do not exist. a. lim g sxd x: 1
b. lim g sxd
c. lim g sxd
x: 2
y y f (x)
1
d. lim g sxd
x: 3
x:2.5
–1
y
1
2
x
–1 y g(x) 1
1
x
3
2
4. Which of the following statements about the function y = ƒsxd graphed here are true, and which are false? a. lim ƒsxd does not exist. x:2
2. For the function ƒ(t) graphed here, find the following limits or explain why they do not exist. a. lim ƒstd t: 2
b. lim ƒstd t: 1
c. lim ƒstd
d.
t: 0
lim ƒstd
t: 0.5
b. lim ƒsxd = 2 x:2
c. lim ƒsxd does not exist. x:1
d. lim ƒsxd exists at every point c in s 1, 1d . x:c
e. lim ƒsxd exists at every point c in (1, 3).
s
x:c
y s f (t)
1
–1
0
–2
y f (x)
1 1
t –1
–1
1
2
3
x
–1 –2
3. Which of the following statements about the function y = ƒsxd graphed here are true, and which are false? a. lim ƒsxd exists. x: 0
b. lim ƒsxd = 0 x: 0
c. lim ƒsxd = 1 x: 0
d. lim ƒsxd = 1 x: 1
e. lim ƒsxd = 0 x: 1
f. lim ƒsxd exists at every point c in s 1, 1d . x: c
g. lim ƒsxd does not exist. x: 1
Existence of Limits In Exercises 5 and 6, explain why the limits do not exist. x 1 6. lim x:1 x  1 ƒxƒ 7. Suppose that a function ƒ(x) is defined for all real values of x except x = c . Can anything be said about the existence of limx:c ƒsxd ? Give reasons for your answer. 5. lim
x:0
8. Suppose that a function ƒ(x) is defined for all x in [1, 1] . Can anything be said about the existence of limx:0 ƒsxd ? Give reasons for your answer.
68
Chapter 2: Limits and Continuity
9. If limx:1 ƒsxd = 5 , must ƒ be defined at x = 1 ? If it is, must ƒs1d = 5 ? Can we conclude anything about the values of ƒ at x = 1 ? Explain. 10. If ƒs1d = 5 , must limx:1 ƒsxd exist? If it does, then must limx:1 ƒsxd = 5 ? Can we conclude anything about limx:1 ƒsxd ? Explain.
Using Limit Rules 51. Suppose limx:0 ƒsxd = 1 and limx:0 g sxd = 5 . Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation. lim
x:0
2ƒsxd  g sxd sƒsxd + 7d2>3
Calculating Limits Find the limits in Exercises 11–22. 12. lim s x + 5x  2d
13. lim 8st  5dst  7d
14. lim sx 3  2x 2 + 4x + 8d
t:6
15. lim
x:2
x + 3 x + 6
19. lim s5  yd4>3 y: 3
h: 0
=
x: 2
3 23h + 1 + 1
=
y 2 + 5y + 6 20. lim s2z  8d1>3
x:5
x  5 x 2  25
x 2 + 3x  10 25. lim x + 5 x: 5
z: 0
22. lim
h :0
25h + 4  2 h
x: 3
lim
2x  4 29. lim 3 x: 2 x + 2x 2
30. lim
1 x
 1
x  1
x:1
u  1 u3  1
u: 1
2x  3 35. lim x:9 x  9 37. lim
x:1
x  1 2x + 3  2
2x 2 + 12  4 39. lim x  2 x:2 41. lim
x: 3
2  2x 2  5 x + 3
t: 1
y: 0
1 x  1
32. lim
+
y: 2
36. lim
x: 4
x + 2
x: 2
x: 4
2x 2 + 5  3 4  x
5  2x 2 + 9
lim sin2 x
=
7 4
(a)
lim 5hsxd 4x:1
A lim p(x) B A lim A 4  r(x) B B lim hsxd 45x:1
A lim p(x) B A lim 4  lim r (x) B x:1
x:1
2s5ds5d 5 = 2 s1ds4  2d
a. lim ƒsxdg sxd
b. lim 2ƒsxdg sxd
c. lim sƒsxd + 3g sxdd
d. lim
x:c
x:c
ƒsxd ƒsxd  g sxd
54. Suppose limx:4 ƒsxd = 0 and limx:4 g sxd = 3 . Find a. lim sg sxd + 3d
b. lim xƒsxd
c. lim sg sxdd2
d. lim
x:4
x:4
x:4
x:4
g sxd ƒsxd  1
55. Suppose limx:b ƒsxd = 7 and limx:b g sxd = 3 . Find a. lim sƒsxd + g sxdd
b. lim ƒsxd # g sxd
c. lim 4g sxd
d. lim ƒsxd>g sxd
x:b x:b
lim tan x
x: p>3
48. lim (x 2  1)(2  cos x)
x:b
x: 0
49. lim 2x + 4 cos (x + p) 50. lim 27 + sec x
a. lim spsxd + r sxd + ssxdd x: 2
b.
lim psxd # r sxd # ssxd
x: 2
c. lim s 4psxd + 5r sxdd>ssxd x: 2
(b)
x:1
x:b
56. Suppose that limx:2 psxd = 4, limx:2 r sxd = 0 , and limx:2 ssxd = 3 . Find
x: p>4
x: 0
=
x:c
2
x: p
s1 + 7d
2>3
x:1
x:c
2  2x
40. lim 42. lim
=
4x  x 2
2x 2 + 8  3 38. lim x + 1 x: 1
46.
x:0
(c)
x:0
s2ds1d  s 5d
53. Suppose limx:c ƒsxd = 5 and limx:c g sxd = 2 . Find
y  8 y4  16
34. lim
45. lim sec x 1 + x + sin x 3 cos x
x:0
x:1
1 x + 1
3
44.
47. lim
=
x
x: 0
43. lim (2 sin x  1) x:0
x:0
A lim ƒ(x) + lim 7 B 2>3
x:1
3y 4  16y 2
Limits with trigonometric functions Find the limits in Exercises 43–50. x:0
x:1
t 2 + 3t + 2 t2  t  2 5y 3 + 8y 2
4
33. lim
2 lim ƒsxd  lim g sxd
lim 25hsxd 25hsxd x:1 = psxds4  rsxdd lim spsxds4  rsxddd
x 2  7x + 10 26. lim x  2 x: 2 28. lim
31. lim
(b)
52. Let limx:1 hsxd = 5, limx:1 psxd = 1 , and limx:1 r sxd = 2 . Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation.
x + 3 x 2 + 4x + 3
24. lim
t2 + t  2 t:1 t2  1
27. lim
x:0
A lim A ƒsxd + 7 B B 2>3
y: 2
Limits of quotients Find the limits in Exercises 23–42. 23. lim
lim 2ƒsxd  lim g sxd
x:0
x:0
y + 2
18. lim
(a)
lim sƒsxd + 7d2>3
x:0
x: 2
s: 2>3
x: 1
21. lim
2
16. lim 3ss2s  1d
17. lim 3s2x  1d2
lim s2ƒsxd  g sxdd
x:0
x:0
=
11. lim s2x + 5d x: 7
=
(c)
2.2 Limits of Average Rates of Change Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form lim
h :0
ƒsx + hd  ƒsxd h
59. ƒsxd = 3x  4, 61. ƒsxd = 2x,
x = 2
x = 7
x = 2
60. ƒsxd = 1>x,
x = 2
62. ƒsxd = 23x + 1,
x = 0
Using the Sandwich Theorem 63. If 25  2x 2 … ƒsxd … 25  x 2 for limx:0 ƒsxd .
1 … x … 1, find
64. If 2  x 2 … g sxd … 2 cos x for all x, find limx:0 g sxd . 65. a. It can be shown that the inequalities x sin x x2 6 6 1 6 2  2 cos x hold for all values of x close to zero. What, if anything, does this tell you about 1 
x sin x ? 2  2 cos x Give reasons for your answer. lim
x: 0
T b. Graph y = 1  sx 2>6d, y = sx sin xd>s2  2 cos xd, and y = 1 together for 2 … x … 2 . Comment on the behavior of the graphs as x : 0 . 66. a. Suppose that the inequalities 1  cos x x2 1 1 6 6 2 24 2 x2 hold for values of x close to zero. (They do, as you will see in Section 9.9.) What, if anything, does this tell you about lim
x: 0
b. Support your conclusion in part (a) by graphing g near c = 22 and using Zoom and Trace to estimate yvalues on the graph as x : 22 . 69. Let Gsxd = sx + 6d>sx 2 + 4x  12d .
58. ƒsxd = x 2,
x = 1
69
c. Find limx:22 g sxd algebraically.
occur frequently in calculus. In Exercises 57–62, evaluate this limit for the given value of x and function ƒ. 57. ƒsxd = x 2,
Limit of a Function and Limit Laws
1  cos x ? x2
Give reasons for your answer.
T b. Graph the equations y = s1>2d  sx 2>24d, y = s1  cos xd>x 2 , and y = 1>2 together for 2 … x … 2 . Comment on the behavior of the graphs as x : 0 . Estimating Limits T You will find a graphing calculator useful for Exercises 67–76. 67. Let ƒsxd = sx 2  9d>sx + 3d . a. Make a table of the values of ƒ at the points x = 3.1, 3.01, 3.001 , and so on as far as your calculator can go. Then estimate limx:3 ƒsxd . What estimate do you arrive at if you evaluate ƒ at x = 2.9, 2.99, 2.999, Á instead? b. Support your conclusions in part (a) by graphing ƒ near c = 3 and using Zoom and Trace to estimate yvalues on the graph as x : 3 . c. Find limx:3 ƒsxd algebraically, as in Example 7. 68. Let g sxd = sx 2  2d>(x  22). a. Make a table of the values of g at the points x = 1.4, 1.41, 1.414 , and so on through successive decimal approximations of 22 . Estimate limx:22 g sxd .
a. Make a table of the values of G at x = 5.9, 5.99,  5.999, and so on. Then estimate limx:6 Gsxd . What estimate do you arrive at if you evaluate G at x =  6.1, 6.01, 6.001, Á instead? b. Support your conclusions in part (a) by graphing G and using Zoom and Trace to estimate yvalues on the graph as x : 6 . c. Find limx:6 Gsxd algebraically. 70. Let hsxd = sx 2  2x  3d>sx 2  4x + 3d . a. Make a table of the values of h at x = 2.9, 2.99, 2.999, and so on. Then estimate limx:3 hsxd . What estimate do you arrive at if you evaluate h at x = 3.1, 3.01, 3.001, Á instead? b. Support your conclusions in part (a) by graphing h near c = 3 and using Zoom and Trace to estimate yvalues on the graph as x : 3. c. Find limx:3 hsxd algebraically. 71. Let ƒsxd = sx 2  1d>s ƒ x ƒ  1d . a. Make tables of the values of ƒ at values of x that approach c = 1 from above and below. Then estimate limx:1 ƒsxd . b. Support your conclusion in part (a) by graphing ƒ near c = 1 and using Zoom and Trace to estimate yvalues on the graph as x : 1 . c. Find limx:1 ƒsxd algebraically. 72. Let Fsxd = sx 2 + 3x + 2d>s2  ƒ x ƒ d . a. Make tables of values of F at values of x that approach c = 2 from above and below. Then estimate limx:2 Fsxd . b. Support your conclusion in part (a) by graphing F near c = 2 and using Zoom and Trace to estimate yvalues on the graph as x : 2 . c. Find limx:2 Fsxd algebraically. 73. Let g sud = ssin ud>u . a. Make a table of the values of g at values of u that approach u0 = 0 from above and below. Then estimate lim u : 0 g sud . b. Support your conclusion in part (a) by graphing g near u0 = 0 . 74. Let Gstd = s1  cos td>t 2 . a. Make tables of values of G at values of t that approach t0 = 0 from above and below. Then estimate lim t:0 Gstd . b. Support your conclusion in part (a) by graphing G near t0 = 0 . 75. Let ƒsxd = x 1>s1  xd . a. Make tables of values of ƒ at values of x that approach c = 1 from above and below. Does ƒ appear to have a limit as x : 1 ? If so, what is it? If not, why not? b. Support your conclusions in part (a) by graphing ƒ near c = 1.
70
Chapter 2: Limits and Continuity
76. Let ƒsxd = s3x  1d>x . a. Make tables of values of ƒ at values of x that approach c = 0 from above and below. Does ƒ appear to have a limit as x : 0 ? If so, what is it? If not, why not?
b. If lim
x:2
82. If lim
Theory and Examples 77. If x 4 … ƒsxd … x 2 for x in [1, 1] and x 2 … ƒsxd … x 4 for x 6 1 and x 7 1 , at what points c do you automatically know limx:c ƒsxd ? What can you say about the value of the limit at these points? 78. Suppose that g sxd … ƒsxd … hsxd for all x Z 2 and suppose that
ƒsxd x2
x: 2
= 1 , find b. lim
x: 2
x:0
b. Confirm your estimate in part (a) with a proof. T 84. a. Graph hsxd = x 2 cos s1>x 3 d to estimate limx:0 hsxd , zooming in on the origin as necessary. b. Confirm your estimate in part (a) with a proof. COMPUTER EXPLORATIONS Graphical Estimates of Limits In Exercises 85–90, use a CAS to perform the following steps: a. Plot the function near the point c being approached. b. From your plot guess the value of the limit.
ƒsxd  5 = 3 , find lim ƒsxd . x: 2 x  2 x: 2
x:2
x 4  16 x  2
86. lim
x: 1
3 2 1 + x  1 87. lim x x:0
81. a. If lim
89. lim
x:0
2.3
ƒsxd x:0 x
b. lim
T 83. a. Graph g sxd = x sin s1>xd to estimate limx:0 g sxd , zooming in on the origin as necessary.
85. lim ƒsxd x: 2 x
a. lim ƒsxd
= 1 , find
x: 2
Can we conclude anything about the values of ƒ, g, and h at x = 2 ? Could ƒs2d = 0 ? Could limx:2 ƒsxd = 0 ? Give reasons for your answers. ƒsxd  5 = 1 , find lim ƒsxd . 79. If lim x:4 x  2 x: 4 80. If lim
x2
a. lim ƒsxd
lim g sxd = lim hsxd = 5 .
x:2
ƒsxd
x:0
b. Support your conclusions in part (a) by graphing ƒ near c = 0
ƒsxd  5 = 4 , find lim ƒsxd . x  2 x:2
1  cos x x sin x
88. lim
x 3  x 2  5x  3 sx + 1d2 x2  9
2x 2 + 7  4 2x 2 90. lim x:0 3  3 cos x x:3
The Precise Definition of a Limit We now turn our attention to the precise definition of a limit. We replace vague phrases like “gets arbitrarily close to” in the informal definition with specific conditions that can be applied to any particular example. With a precise definition, we can prove the limit properties given in the preceding section and establish many important limits. To show that the limit of ƒ(x) as x : c equals the number L, we need to show that the gap between ƒ(x) and L can be made “as small as we choose” if x is kept “close enough” to c. Let us see what this would require if we specified the size of the gap between ƒ(x) and L.
y
Upper bound: y9
⎧9 To satisfy ⎪ ⎨7 this ⎪ ⎩5
Lower bound: y5
3 4 5
x
⎧ ⎨ ⎩
0
Consider the function y = 2x  1 near x = 4. Intuitively it appears that y is close to 7 when x is close to 4, so limx:4 s2x  1d = 7. However, how close to x = 4 does x have to be so that y = 2x  1 differs from 7 by, say, less than 2 units?
EXAMPLE 1
y 2x 1
Restrict to this
FIGURE 2.15 Keeping x within 1 unit of x = 4 will keep y within 2 units of y = 7 (Example 1).
We are asked: For what values of x is ƒ y  7 ƒ 6 2? To find the answer we first express ƒ y  7 ƒ in terms of x: Solution
ƒ y  7 ƒ = ƒ s2x  1d  7 ƒ = ƒ 2x  8 ƒ . The question then becomes: what values of x satisfy the inequality ƒ 2x  8 ƒ 6 2? To find out, we solve the inequality: ƒ 2x  8 ƒ 2 6 3
6 6 6 6
2 2x  8 6 2 2x 6 10 x 6 5
1 6 x  4 6 1.
Solve for x. Solve for x 4.
Keeping x within 1 unit of x = 4 will keep y within 2 units of y = 7 (Figure 2.15).
2.3
1 10
f (x) f(x) lies in here
L L
1 10
Definition of Limit for all x c in here
x
c
0
c
c
x
FIGURE 2.16 How should we define d 7 0 so that keeping x within the interval sc  d, c + dd will keep ƒ(x) within the 1 1 ,L + b? interval aL 10 10
y
L L
71
In the previous example we determined how close x must be to a particular value c to ensure that the outputs ƒ(x) of some function lie within a prescribed interval about a limit value L. To show that the limit of ƒ(x) as x : c actually equals L, we must be able to show that the gap between ƒ(x) and L can be made less than any prescribed error, no matter how small, by holding x close enough to c.
y
L
The Precise Definition of a Limit
f(x)
f(x) lies in here
Suppose we are watching the values of a function ƒ(x) as x approaches c (without taking on the value of c itself ). Certainly we want to be able to say that ƒ(x) stays within onetenth of a unit from L as soon as x stays within some distance d of c (Figure 2.16). But that in itself is not enough, because as x continues on its course toward c, what is to prevent ƒ(x) from jittering about within the interval from L  (1>10) to L + (1>10) without tending toward L? We can be told that the error can be no more than 1>100 or 1>1000 or 1>100,000. Each time, we find a new dinterval about c so that keeping x within that interval satisfies the new error tolerance. And each time the possibility exists that ƒ(x) jitters away from L at some stage. The figures on the next page illustrate the problem. You can think of this as a quarrel between a skeptic and a scholar. The skeptic presents Pchallenges to prove that the limit does not exist or, more precisely, that there is room for doubt. The scholar answers every challenge with a dinterval around c that keeps the function values within P of L. How do we stop this seemingly endless series of challenges and responses? By proving that for every error tolerance P that the challenger can produce, we can find, calculate, or conjure a matching distance d that keeps x “close enough” to c to keep ƒ(x) within that tolerance of L (Figure 2.17). This leads us to the precise definition of a limit.
L
DEFINITION Let ƒ(x) be defined on an open interval about c, except possibly at c itself. We say that the limit of ƒ(x) as x approaches c is the number L, and write
for all x c in here x 0
c
lim ƒsxd = L,
x c
c
FIGURE 2.17 The relation of d and P in the definition of limit.
x:c
if, for every number P 7 0, there exists a corresponding number d 7 0 such that for all x, 0 6 ƒx  cƒ 6 d
Q
ƒ ƒsxd  L ƒ 6 P.
One way to think about the definition is to suppose we are machining a generator shaft to a close tolerance. We may try for diameter L, but since nothing is perfect, we must be satisfied with a diameter ƒ(x) somewhere between L  P and L + P. The d is the measure of how accurate our control setting for x must be to guarantee this degree of accuracy in the diameter of the shaft. Notice that as the tolerance for error becomes stricter, we may have to adjust d. That is, the value of d, how tight our control setting must be, depends on the value of P, the error tolerance.
Examples: Testing the Definition The formal definition of limit does not tell how to find the limit of a function, but it enables us to verify that a suspected limit is correct. The following examples show how the definition can be used to verify limit statements for specific functions. However, the real purpose of the definition is not to do calculations like this, but rather to prove general theorems so that the calculation of specific limits can be simplified.
72
Chapter 2: Limits and Continuity
y
L
y f(x)
1 10
L
L
c c 1/10 c 1/10 Response: x c 1/10 (a number) 0
The challenge: Make f(x) – L 1 10
1 100 L 1 L 100
1 100
L
L 1 L 100
1 10
x
c
0
x
New challenge: Make f (x) – L 1 100
y f (x)
y f (x) L
1 1000
L
L
1 1000
L
1 1000
x
c
0
x
c
0
New challenge: 1 1000
Response: x c 1/1000
y
y y f (x) 1 L 100,000
L
L
1 100,000
L
c New challenge: 1 100,000
y y f (x)
1 L 100,000
0
0
y
1 L 1000
L
x c c 1/100 c 1/100 Response: x c 1/100
x
c
0
y
L
y f(x)
y f (x)
L
1 10
y
y f (x)
1 L 10
L L
y
y
y f (x) L L L
1 100,000
x 0
c
x
Response: x c 1/100,000
EXAMPLE 2
c
0
x
New challenge: ...
Show that lim s5x  3d = 2.
x:1
Solution Set c = 1, ƒsxd = 5x  3, and L = 2 in the definition of limit. For any given P 7 0, we have to find a suitable d 7 0 so that if x Z 1 and x is within distance d of c = 1, that is, whenever
0 6 ƒ x  1 ƒ 6 d, it is true that ƒ(x) is within distance P of L = 2, so ƒ ƒsxd  2 ƒ 6 P.
2.3
y
The Precise Definition of a Limit
73
We find d by working backward from the Pinequality:
y 5x 3
ƒ s5x  3d  2 ƒ = ƒ 5x  5 ƒ 6 P
2
5ƒx  1ƒ 6 P
2
ƒ x  1 ƒ 6 P>5.
2
Thus, we can take d = P>5 (Figure 2.18). If 0 6 ƒ x  1 ƒ 6 d = P>5, then 1 1 1 5 5
0
ƒ s5x  3d  2 ƒ = ƒ 5x  5 ƒ = 5 ƒ x  1 ƒ 6 5sP>5d = P,
x
which proves that limx:1s5x  3d = 2. The value of d = P>5 is not the only value that will make 0 6 ƒ x  1 ƒ 6 d imply 5x  5 ƒ 6 P. Any smaller positive d will do as well. The definition does not ask for a ƒ “best” positive d, just one that will work.
–3
EXAMPLE 3
NOT TO SCALE
(a) lim x = c
FIGURE 2.18 If ƒsxd = 5x  3 , then 0 6 ƒ x  1 ƒ 6 P>5 guarantees that ƒ ƒsxd  2 ƒ 6 P (Example 2).
x:c
Prove the following results presented graphically in Section 2.2. (b) lim k = k
(k constant)
x:c
Solution
(a) Let P 7 0 be given. We must find d 7 0 such that for all x 0 6 ƒx  cƒ 6 d
implies
ƒ x  c ƒ 6 P.
The implication will hold if d equals P or any smaller positive number (Figure 2.19). This proves that limx:c x = c. (b) Let P 7 0 be given. We must find d 7 0 such that for all x
y yx c c c c
0 6 ƒx  cƒ 6 d
implies
ƒ k  k ƒ 6 P.
Since k  k = 0, we can use any positive number for d and the implication will hold (Figure 2.20). This proves that limx:c k = k.
c
c c c
0
x
FIGURE 2.19 For the function ƒsxd = x , we find that 0 6 ƒ x  c ƒ 6 d will guarantee ƒ ƒsxd  c ƒ 6 P whenever d … P (Example 3a).
Finding Deltas Algebraically for Given Epsilons In Examples 2 and 3, the interval of values about c for which ƒ ƒsxd  L ƒ was less than P was symmetric about c and we could take d to be half the length of that interval. When such symmetry is absent, as it usually is, we can take d to be the distance from c to the interval’s nearer endpoint.
EXAMPLE 4 For the limit limx:5 2x  1 = 2, find a d 7 0 that works for P = 1. That is, find a d 7 0 such that for all x 0 6 ƒx  5ƒ 6 d
y yk
k k k
Solution
1.
Q
ƒ 2x  1  2 ƒ 6 1.
We organize the search into two steps.
Solve the inequality ƒ 2x  1  2 ƒ 6 1 to find an interval containing x = 5 on which the inequality holds for all x Z 5. ƒ 2x  1  2 ƒ 6 1
0
c
c
c
x
FIGURE 2.20 For the function ƒsxd = k , we find that ƒ ƒsxd  k ƒ 6 P for any positive d (Example 3b).
1 6 2x  1  2 6 1 1 6 2x  1 6 3 1 6 x  1 6 9 2 6 x 6 10
74 (
Chapter 2: Limits and Continuity
3
3
2
( 10
8
5
x
2. FIGURE 2.21 An open interval of radius 3 about x = 5 will lie inside the open interval (2, 10).
The inequality holds for all x in the open interval (2, 10), so it holds for all x Z 5 in this interval as well. Find a value of d 7 0 to place the centered interval 5  d 6 x 6 5 + d (centered at x = 5) inside the interval (2, 10). The distance from 5 to the nearer endpoint of (2, 10) is 3 (Figure 2.21). If we take d = 3 or any smaller positive number, then the inequality 0 6 ƒ x  5 ƒ 6 d will automatically place x between 2 and 10 to make ƒ 2x  1  2 ƒ 6 1 (Figure 2.22): 0 6 ƒx  5ƒ 6 3
y
Q
ƒ 2x  1  2 ƒ 6 1.
y x 1 3
How to Find Algebraically a D for a Given ƒ, L, c, and P>0 The process of finding a d 7 0 such that for all x
2
0 6 ƒx  cƒ 6 d
ƒ ƒsxd  L ƒ 6 P
can be accomplished in two steps.
1 3 0
Q
1 2
1. Solve the inequality ƒ ƒsxd  L ƒ 6 P to find an open interval (a, b) containing c on which the inequality holds for all x Z c.
3 5
8
x
10
2. Find a value of d 7 0 that places the open interval sc  d, c + dd centered at c inside the interval (a, b). The inequality ƒ ƒsxd  L ƒ 6 P will hold for all x Z c in this dinterval.
NOT TO SCALE
FIGURE 2.22 The function and intervals in Example 4.
EXAMPLE 5
Prove that limx:2 ƒsxd = 4 if ƒsxd = e
Solution
1.
y x2 4
x Z 2 x = 2.
Our task is to show that given P 7 0 there exists a d 7 0 such that for all x 0 6 ƒx  2ƒ 6 d
y
x 2, 1,
Q
ƒ ƒsxd  4 ƒ 6 P.
Solve the inequality ƒ ƒsxd  4 ƒ 6 P to find an open interval containing x = 2 on which the inequality holds for all x Z 2. For x Z c = 2, we have ƒsxd = x 2 , and the inequality to solve is ƒ x 2  4 ƒ 6 P: ƒ x2  4 ƒ 6 P
(2, 4)
4
P 6 x 2  4 6 P 4
4  P 6 x2 6 4 + P 24  P 6 ƒ x ƒ 6 24 + P
(2, 1) 0
4
2
24  P 6 x 6 24 + P.
x
The inequality ƒ ƒsxd  4 ƒ 6 P holds for all x Z 2 in the open interval A 24  P, 24 + P B (Figure 2.23).
4
FIGURE 2.23 An interval containing x = 2 so that the function in Example 5 satisfies ƒ ƒsxd  4 ƒ 6 P .
Assumes P 6 4 ; see below. An open interval about x = 2 that solves the inequality
2.
Find a value of d 7 0 that places the centered interval s2  d, 2 + dd inside the interval A 24  P, 24 + P B . Take d to be the distance from x = 2 to the nearer endpoint of A 24  P, 24 + P B .
In other words, take d = min E 2  24  P, 24 + P  2 F , the minimum (the
2.3
The Precise Definition of a Limit
75
smaller) of the two numbers 2  24  P and 24 + P  2. If d has this or any smaller positive value, the inequality 0 6 ƒ x  2 ƒ 6 d will automatically place x between 24  P and 24 + P to make ƒ ƒsxd  4 ƒ 6 P. For all x, 0 6 ƒx  2ƒ 6 d
Q
ƒ ƒsxd  4 ƒ 6 P.
This completes the proof for P 6 4. If P Ú 4, then we take d to be the distance from x = 2 to the nearer endpoint of the interval A 0, 24 + P B . In other words, take d = min E 2, 24 + P  2 F . (See Figure 2.23.)
Using the Definition to Prove Theorems We do not usually rely on the formal definition of limit to verify specific limits such as those in the preceding examples. Rather, we appeal to general theorems about limits, in particular the theorems of Section 2.2. The definition is used to prove these theorems (Appendix 5). As an example, we prove part 1 of Theorem 1, the Sum Rule.
EXAMPLE 6
Given that limx:c ƒsxd = L and limx:c gsxd = M, prove that lim sƒsxd + gsxdd = L + M.
x:c
Solution
Let P 7 0 be given. We want to find a positive number d such that for all x 0 6 ƒx  cƒ 6 d
Q
ƒ ƒsxd + gsxd  sL + Md ƒ 6 P.
Regrouping terms, we get ƒ ƒsxd + gsxd  sL + Md ƒ = ƒ sƒsxd  Ld + sgsxd  Md ƒ … ƒ ƒsxd  L ƒ + ƒ gsxd  M ƒ .
Triangle Inequality: ƒa + bƒ … ƒaƒ + ƒbƒ
Since limx:c ƒsxd = L, there exists a number d1 7 0 such that for all x 0 6 ƒ x  c ƒ 6 d1
Q
ƒ ƒsxd  L ƒ 6 P>2.
Similarly, since limx:c gsxd = M, there exists a number d2 7 0 such that for all x 0 6 ƒ x  c ƒ 6 d2
Q
ƒ gsxd  M ƒ 6 P>2.
Let d = min 5d1, d26, the smaller of d1 and d2 . If 0 6 ƒ x  c ƒ 6 d then ƒ x  c ƒ 6 d1 , so ƒ ƒsxd  L ƒ 6 P>2, and ƒ x  c ƒ 6 d2 , so ƒ gsxd  M ƒ 6 P>2. Therefore P P ƒ ƒsxd + gsxd  sL + Md ƒ 6 2 + 2 = P. This shows that limx:c sƒsxd + gsxdd = L + M. Next we prove Theorem 5 of Section 2.2.
EXAMPLE 7 Given that limx:c ƒsxd = L and limx:c gsxd = M, and that ƒsxd … gsxd for all x in an open interval containing c (except possibly c itself), prove that L … M. We use the method of proof by contradiction. Suppose, on the contrary, that L 7 M. Then by the limit of a difference property in Theorem 1,
Solution
lim s gsxd  ƒsxdd = M  L.
x:c
76
Chapter 2: Limits and Continuity
Therefore, for any P 7 0, there exists d 7 0 such that ƒ sgsxd  ƒsxdd  sM  Ld ƒ 6 P
0 6 ƒ x  c ƒ 6 d.
whenever
Since L  M 7 0 by hypothesis, we take P = L  M in particular and we have a number d 7 0 such that ƒ sg sxd  ƒsxdd  sM  Ld ƒ 6 L  M
whenever
0 6 ƒ x  c ƒ 6 d.
whenever
0 6 ƒx  cƒ 6 d
Since a … ƒ a ƒ for any number a, we have sgsxd  ƒsxdd  sM  Ld 6 L  M which simplifies to gsxd 6 ƒsxd
whenever
0 6 ƒ x  c ƒ 6 d.
But this contradicts ƒsxd … gsxd. Thus the inequality L 7 M must be false. Therefore L … M.
Exercises 2.3 Centering Intervals About a Point In Exercises 1–6, sketch the interval (a, b) on the xaxis with the point c inside. Then find a value of d 7 0 such that for all x, 0 6 ƒ x  c ƒ 6 d Q a 6 x 6 b . 1. a = 1,
b = 7,
c = 5
2. a = 1,
b = 7,
c = 2
3. a = 7>2, 4. a = 7>2,
b = 1>2,
c = 3
b = 1>2,
c = 3>2
b = 4>7,
5. a = 4>9,
y 5 4 1 3 4
10. f (x) x c1 L1 1 y x 4
c = 3
0
y f (x) 2x 1 c3 L4 0.2 y 2x 1
4.2 4 3.8
c = 1>2
b = 3.2391,
6. a = 2.7591,
9.
1
9 16
x
25 16
2
–1 0
Finding Deltas Graphically In Exercises 7–14, use the graphs to find a d 7 0 such that for all x 0 6 ƒ x  c ƒ 6 d Q ƒ ƒsxd  L ƒ 6 P . 7.
12.
y 2x 4 f(x) 2x 4 c5 L6 0.2
6.2 6 5.8
0
f (x) 4 x 2 c –1 L3 0.25
f (x) x 2 c2 L4 1 y x2
y –3 x 3 2
y 4 x2
5 4
7.65 7.5 7.35
x
5 4.9
y
y
y
f (x) – 3 x 3 2 c –3 L 7.5 0.15
x
NOT TO SCALE
11.
8.
y
2.61 3 3.41
3.25 3 2.75
3 0
5.1
3
2
x 5
NOT TO SCALE NOT TO SCALE
–3.1
–3
–2.9
0
NOT TO SCALE
x –
5 –1 3 – 2 2 NOT TO SCALE
0
x
2.3 13.
y
2 –x c –1 L2 0.5
f(x)
y
x:c
f (x) 1x c1 2 L2 0.01
2.01
2 –x
that for all x 0 6 ƒx  cƒ 6 d 31. ƒsxd = 3  2 x,
2 2.5
1.99 y 1x
1.5
x
0
x
1 1 1 2 2.01 1.99
0
33. ƒsxd =
x2  4 , x  2
34. ƒsxd =
x2 + 6x + 5 , x + 5
NOT TO SCALE
36. ƒsxd = 4>x,
Each of Exercises 15–30 gives a function ƒ(x) and numbers L, c, and P 7 0 . In each case, find an open interval about c on which the inequality ƒ ƒsxd  L ƒ 6 P holds. Then give a value for d 7 0 such that for all x satisfying 0 6 ƒ x  c ƒ 6 d the inequality ƒ ƒsxd  L ƒ 6 P holds.
16. ƒsxd = 2x  2,
L = 6,
17. ƒsxd = 2x + 1, 18. ƒsxd = 2x,
L = 1, L = 1>2,
19. ƒsxd = 219  x, 20. ƒsxd = 2x  7, 21. ƒsxd = 1>x,
L = 3, L = 4,
L = 1>4,
P = 0.1
c = 1>4,
P = 0.1
c = 10,
c = 3,
c = 2,
L = 3,
c = 23,
23. ƒsxd = x2,
L = 4,
c = 2,
25. ƒsxd = x2  5, 26. ƒsxd = 120>x,
L = 11, L = 5,
c = 1, c = 4, c = 24,
P = 0.4
39. lim 2x  5 = 2
40. lim 24  x = 2
P = 1 P = 0.05
x:1
42. lim ƒsxd = 4 x: 2
x:3
x:0
ƒsxd = e
if
x Z 2 x = 2
x 2, 1,
44.
x2  9 = 6 x + 3
x Z 1 x = 1
x 2, 2,
ƒsxd = e
if
1 43. lim x = 1 x:1
lim
x: 23
46. lim
x:1
47. lim ƒsxd = 2
if
ƒsxd = e
4  2x, 6x  4,
48. lim ƒsxd = 0
if
ƒsxd = e
2x, x>2,
x:0
x 6 1 x Ú 1 x 6 0 x Ú 0
x:0
P = 0.5
y
P = 0.1 P = 1 P = 1 c = 2,
P = 0.03
28. ƒsxd = mx,
m 7 0,
L = 3m,
c = 3,
P = c 7 0
L = sm>2d + b, L = m + b,
c = 1,
– 1 2 1 –
y x sin 1x
1 2 1
1 1 = 3 x2
x2  1 = 2 x  1
1 49. lim x sin x = 0
P = 0.1
L = 2m,
m 7 0,
41. lim ƒsxd = 1
x:1
m 7 0,
30. ƒsxd = mx + b, P = 0.05
P = 0.5
38. lim s3x  7d = 2
x: 3
27. ƒsxd = mx,
m 7 0, 29. ƒsxd = mx + b, c = 1>2, P = c 7 0
P = 0.05
37. lim s9  xd = 5
45. lim
P = 1
c = 23, c = 4,
L = 1,
P = 0.02
c = 0,
22. ƒsxd = x2,
24. ƒsxd = 1>x,
P = 0.01
c = 2,
P = 0.05
c = 5,
x:9
c = 4,
P = 0.03
Prove the limit statements in Exercises 37–50.
Finding Deltas Algebraically
L = 5,
P = 0.02
c = 2,
x:4
15. ƒsxd = x + 1,
ƒ ƒsxd  L ƒ 6 P .
c = 1,
35. ƒsxd = 21  5x, –1 – 16 25
Q
c = 3,
32. ƒsxd = 3x  2,
2
–16 9
77
Using the Formal Definition Each of Exercises 31–36 gives a function ƒ(x), a point c, and a positive number P . Find L = lim ƒsxd. Then find a number d 7 0 such
14. y
The Precise Definition of a Limit
x
78
Chapter 2: Limits and Continuity When Is a Number L Not the Limit of ƒ(x) as x : c?
1 50. lim x 2 sin x = 0
Showing L is not a limit We can prove that limx:c ƒsxd Z L by providing an P 7 0 such that no possible d 7 0 satisfies the condition
x:0
y y x2
1
for all x,
0 6 ƒx  cƒ 6 d
Q
ƒ ƒsxd  L ƒ 6 P .
We accomplish this for our candidate P by showing that for each d 7 0 there exists a value of x such that y x 2 sin 1x –1
0
2 –
2
0 6 ƒx  cƒ 6 d
and
ƒ ƒsxd  L ƒ Ú P .
x
1
y y f (x) L
–1
y –x 2
L L
Theory and Examples
f (x)
51. Define what it means to say that lim g sxd = k . x: 0
52. Prove that lim ƒsxd = L if and only if lim ƒsh + cd = L . h :0
x:c
0
53. A wrong statement about limits Show by example that the following statement is wrong.
57. Let ƒsxd = e
y
The number L is the limit of ƒ(x) as x approaches c if, given any P 7 0 , there exists a value of x for which ƒ ƒsxd  L ƒ 6 P.
yx1
Explain why the function in your example does not have the given value of L as a limit as x : c .
2
T 55. Grinding engine cylinders Before contracting to grind engine cylinders to a crosssectional area of 9 in2 , you need to know how much deviation from the ideal cylinder diameter of c = 3.385 in. you can allow and still have the area come within 0.01 in2 of the required 9 in2 . To find out, you let A = psx>2d2 and look for the interval in which you must hold x to make ƒ A  9 ƒ … 0.01 . What interval do you find?
1
y f (x)
I
R
x
1 yx
a. Let P = 1>2 . Show that no possible d 7 0 satisfies the following condition: For all x,
0 6 ƒx  1ƒ 6 d
Q
ƒ ƒsxd  2 ƒ 6 1>2.
That is, for each d 7 0 show that there is a value of x such that 0 6 ƒx  1ƒ 6 d
V
x
c
x, x 6 1 x + 1, x 7 1.
54. Another wrong statement about limits Show by example that the following statement is wrong.
56. Manufacturing electrical resistors Ohm’s law for electrical circuits like the one shown in the accompanying figure states that V = RI . In this equation, V is a constant voltage, I is the current in amperes, and R is the resistance in ohms. Your firm has been asked to supply the resistors for a circuit in which V will be 120 volts and I is to be 5 ; 0.1 amp . In what interval does R have to lie for I to be within 0.1 amp of the value I0 = 5 ?
c
a value of x for which 0 x c and f (x) L
The number L is the limit of ƒ(x) as x approaches c if ƒ(x) gets closer to L as x approaches c. Explain why the function in your example does not have the given value of L as a limit as x : c .
c
and
ƒ ƒsxd  2 ƒ Ú 1>2.
This will show that limx:1 ƒsxd Z 2 . b. Show that limx:1 ƒsxd Z 1 . c. Show that limx:1 ƒsxd Z 1.5 .
2.4 x 6 2 x = 2 x 7 2.
x 2, 58. Let hsxd = • 3, 2,
OneSided Limits
79
y
2
y y g(x) 1
y h(x)
4 3
y2
2 1
y x2
0
2
–1
x
0
x
COMPUTER EXPLORATIONS In Exercises 61–66, you will further explore finding deltas graphically. Use a CAS to perform the following steps: a. Plot the function y = ƒsxd near the point c being approached.
Show that
b. Guess the value of the limit L and then evaluate the limit symbolically to see if you guessed correctly.
a. lim hsxd Z 4 x: 2
b. lim hsxd Z 3
c. Using the value P = 0.2 , graph the banding lines y1 = L  P and y2 = L + P together with the function ƒ near c.
x: 2
c. lim hsxd Z 2 x: 2
d. From your graph in part (c), estimate a d 7 0 such that for all x
59. For the function graphed here, explain why
0 6 ƒx  cƒ 6 d Q ƒ ƒsxd  L ƒ 6 P . Test your estimate by plotting ƒ, y1 , and y2 over the interval 0 6 ƒ x  c ƒ 6 d . For your viewing window use c  2d … x … c + 2d and L  2P … y … L + 2P . If any function values lie outside the interval [L  P, L + P] , your choice of d was too large. Try again with a smaller estimate.
a. lim ƒsxd Z 4 x: 3
b. lim ƒsxd Z 4.8 x: 3
c. lim ƒsxd Z 3 x: 3
y
4.8
61.
4
y f (x)
3
62. 63.
0
x
3
60. a. For the function graphed here, show that limx : 1 g sxd Z 2 . b. Does limx : 1 g sxd appear to exist? If so, what is the value of the limit? If not, why not?
2.4
64.
e. Repeat parts (c) and (d) successively for P = 0.1, 0.05 , and 0.001. x4  81 ƒsxd = , c = 3 x  3 5x3 + 9x2 ƒsxd = 5 , c = 0 2x + 3x2 sin 2x ƒsxd = , c = 0 3x xs1  cos xd ƒsxd = , c = 0 x  sin x
65. ƒsxd =
3 x  1 2 , x  1
66. ƒsxd =
3x  s7x + 1d2x + 5 , x  1
c = 1
2
c = 1
OneSided Limits In this section we extend the limit concept to onesided limits, which are limits as x approaches the number c from the lefthand side (where x 6 c) or the righthand side sx 7 cd only.
Approaching a Limit from One Side To have a limit L as x approaches c, a function ƒ must be defined on both sides of c and its values ƒ(x) must approach L as x approaches c from either side. Because of this, ordinary limits are called twosided.
80
Chapter 2: Limits and Continuity
y y x x 1
x
0
–1
If ƒ fails to have a twosided limit at c, it may still have a onesided limit, that is, a limit if the approach is only from one side. If the approach is from the right, the limit is a righthand limit. From the left, it is a lefthand limit. The function ƒsxd = x> ƒ x ƒ (Figure 2.24) has limit 1 as x approaches 0 from the right, and limit 1 as x approaches 0 from the left. Since these onesided limit values are not the same, there is no single number that ƒ(x) approaches as x approaches 0. So ƒ(x) does not have a (twosided) limit at 0. Intuitively, if ƒ(x) is defined on an interval (c, b), where c 6 b, and approaches arbitrarily close to L as x approaches c from within that interval, then ƒ has righthand limit L at c. We write lim ƒsxd = L.
x:c +
c+ ”
FIGURE 2.24 Different righthand and lefthand limits at the origin.
The symbol “x : means that we consider only values of x greater than c. Similarly, if ƒ(x) is defined on an interval (a, c), where a 6 c and approaches arbitrarily close to M as x approaches c from within that interval, then ƒ has lefthand limit M at c. We write lim ƒsxd = M.
x:c 
The symbol “x : c  ” means that we consider only xvalues less than c. These informal definitions of onesided limits are illustrated in Figure 2.25. For the function ƒsxd = x> ƒ x ƒ in Figure 2.24 we have lim ƒsxd = 1
and
x:0 +
lim ƒsxd = 1.
x:0 
y
y
f (x)
L 0
c
M
f (x) x
x
x
0
(a) lim+ f (x) L
(b)
x: c
c
x
lim _ f (x) M
x: c
FIGURE 2.25 (a) Righthand limit as x approaches c. (b) Lefthand limit as x approaches c. y
The domain of ƒsxd = 24  x 2 is [2, 2]; its graph is the semicircle in Figure 2.26. We have
EXAMPLE 1 y 4 x 2
lim 24  x 2 = 0
x: 2 +
–2
0
2
x
FIGURE 2.26 The function f (x) = 24  x2 has righthand limit 0 at x = 2 and lefthand limit 0 at x = 2 (Example 1).
and
lim 24  x 2 = 0.
x:2 
The function does not have a lefthand limit at x = 2 or a righthand limit at x = 2. It does not have ordinary twosided limits at either 2 or 2. Onesided limits have all the properties listed in Theorem 1 in Section 2.2. The righthand limit of the sum of two functions is the sum of their righthand limits, and so on. The theorems for limits of polynomials and rational functions hold with onesided limits, as do the Sandwich Theorem and Theorem 5. Onesided limits are related to limits in the following way. THEOREM 6 A function ƒ(x) has a limit as x approaches c if and only if it has lefthand and righthand limits there and these onesided limits are equal: lim ƒsxd = L
x:c
3
lim ƒsxd = L
x:c 
and
lim ƒsxd = L.
x:c +
2.4 y
EXAMPLE 2
1
limx:0+ ƒsxd = 1, limx:0 ƒsxd and limx:0 ƒsxd do not exist. The function is not defined to the left of x = 0. limx:1 ƒsxd = 0 even though ƒs1d = 1, limx:1+ ƒsxd = 1, limx:1 ƒsxd does not exist. The right and lefthand limits are not equal. limx:2 ƒsxd = 1, limx:2+ ƒsxd = 1, limx:2 ƒsxd = 1 even though ƒs2d = 2. limx:3 ƒsxd = limx:3+ ƒsxd = limx:3 ƒsxd = ƒs3d = 2. limx:4 ƒsxd = 1 even though ƒs4d Z 1, limx:4+ ƒsxd and limx:4 ƒsxd do not exist. The function is not defined to the right of x = 4.
At x = 1:
0
1
2
3
x
4
FIGURE 2.27 Graph of the function in Example 2.
At x = 2:
At x = 3: At x = 4:
y
81
For the function graphed in Figure 2.27,
At x = 0:
y f (x)
2
OneSided Limits
At every other point c in [0, 4], ƒ(x) has limit ƒ(c).
L
f(x)
Precise Definitions of OneSided Limits
f(x) lies in here
L
The formal definition of the limit in Section 2.3 is readily modified for onesided limits.
L for all x c in here
DEFINITIONS
x 0
c
c
We say that ƒ(x) has righthand limit L at c, and write lim ƒsxd = L
x
(see Figure 2.28)
x:c +
if for every number P 7 0 there exists a corresponding number d 7 0 such that for all x
FIGURE 2.28 Intervals associated with the definition of righthand limit.
c 6 x 6 c + d
Q
ƒ ƒsxd  L ƒ 6 P.
We say that ƒ has lefthand limit L at c, and write y
lim ƒsxd = L
x:c 
L L
if for every number P 7 0 there exists a corresponding number d 7 0 such that for all x c  d 6 x 6 c Q ƒ ƒsxd  L ƒ 6 P.
f(x) f(x) lies in here
EXAMPLE 3
Prove that
L
lim 2x = 0.
x:0 +
for all x c in here
Let P 7 0 be given. Here c = 0 and L = 0, so we want to find a d 7 0 such that for all x
Solution
x 0
(see Figure 2.29)
c
x
0 6 x 6 d
c
FIGURE 2.29 Intervals associated with the definition of lefthand limit.
Q
ƒ 2x  0 ƒ 6 P,
or 0 6 x 6 d
Q
2x 6 P.
82
Chapter 2: Limits and Continuity
Squaring both sides of this last inequality gives
y
x 6 P2
f(x) x
If we choose d = P2 we have
Q
0 6 x 6 d = P2
f(x)
L0
0 6 x 6 d.
if
x
FIGURE 2.30
2
x
lim 1x = 0 in Example 3.
x: 0 +
2x 6 P,
or 0 6 x 6 P2
Q
ƒ 2x  0 ƒ 6 P.
According to the definition, this shows that limx:0+ 2x = 0 (Figure 2.30). The functions examined so far have had some kind of limit at each point of interest. In general, that need not be the case. Show that y = sin s1>xd has no limit as x approaches zero from either side (Figure 2.31).
EXAMPLE 4
y 1
x
0
y sin 1x –1
FIGURE 2.31 The function y = sin s1>xd has neither a righthand nor a lefthand limit as x approaches zero (Example 4). The graph here omits values very near the yaxis. Solution As x approaches zero, its reciprocal, 1>x, grows without bound and the values of sin (1>x) cycle repeatedly from 1 to 1. There is no single number L that the function’s values stay increasingly close to as x approaches zero. This is true even if we restrict x to positive values or to negative values. The function has neither a righthand limit nor a lefthand limit at x = 0.
Limits Involving (sin U)/U A central fact about ssin ud>u is that in radian measure its limit as u : 0 is 1. We can see this in Figure 2.32 and confirm it algebraically using the Sandwich Theorem. You will see the importance of this limit in Section 3.5, where instantaneous rates of change of the trigonometric functions are studied. y 1
–3
–2
y sin (radians)
–
2
3
NOT TO SCALE
FIGURE 2.32 The graph of ƒsud = ssin ud>u suggests that the rightand lefthand limits as u approaches 0 are both 1.
2.4
83
OneSided Limits
y
THEOREM 7—Limit of the ratio sin U>U as U : 0 sin u lim = 1 su in radiansd u:0 u
T 1
(1)
P
tan
Proof The plan is to show that the righthand and lefthand limits are both 1. Then we will know that the twosided limit is 1 as well. To show that the righthand limit is 1, we begin with positive values of u less than p>2 (Figure 2.33). Notice that
1 sin
cos Q
A(1, 0)
Area ¢OAP 6 area sector OAP 6 area ¢OAT.
x
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
O 1
We can express these areas in terms of u as follows:
FIGURE 2.33 The figure for the proof of Theorem 7. By definition, TA>OA = tan u , but OA = 1 , so TA = tan u .
1 1 1 base * height = s1dssin ud = sin u 2 2 2 u 1 2 1 Area sector OAP = r u = s1d2u = 2 2 2 1 1 1 Area ¢OAT = base * height = s1dstan ud = tan u. 2 2 2 Area ¢OAP =
(2)
Thus, Equation (2) is where radian measure comes in: The area of sector OAP is u>2 only if u is measured in radians.
1 1 1 sin u 6 u 6 tan u. 2 2 2 This last inequality goes the same way if we divide all three terms by the number (1>2) sin u , which is positive since 0 6 u 6 p>2: 1 6
u 1 6 . cos u sin u
Taking reciprocals reverses the inequalities: 1 7
sin u 7 cos u. u
Since limu:0+ cos u = 1 (Example 11b, Section 2.2), the Sandwich Theorem gives lim
u:0 +
sin u = 1. u
Recall that sin u and u are both odd functions (Section 1.1). Therefore, ƒsud = ssin ud>u is an even function, with a graph symmetric about the yaxis (see Figure 2.32). This symmetry implies that the lefthand limit at 0 exists and has the same value as the righthand limit: lim
u:0 
sin u sin u = 1 = lim+ , u u u:0
so limu:0 ssin ud>u = 1 by Theorem 6.
EXAMPLE 5
Show that (a) lim
h:0
cos h  1 = 0 h
and
(b) lim
x:0
sin 2x 2 = . 5x 5
84
Chapter 2: Limits and Continuity Solution (a) Using the halfangle formula cos h = 1  2 sin2 sh>2d, we calculate
2 sin2 sh>2d cos h  1 = lim h h h:0 h:0 lim
=  lim
u:0
sin u sin u u
Let u = h>2 . Eq. (1) and Example 11a in Section 2.2
= s1ds0d = 0.
(b) Equation (1) does not apply to the original fraction. We need a 2x in the denominator, not a 5x. We produce it by multiplying numerator and denominator by 2>5: s2>5d # sin 2x sin 2x = lim x:0 5x x:0 s2>5d # 5x lim
EXAMPLE 6
Find lim
t:0
=
sin 2x 2 lim 5 x:0 2x
=
2 2 s1d = 5 5
Now, Eq. (1) applies with u = 2x.
tan t sec 2t . 3t
From the definition of tan t and sec 2t, we have
Solution
tan t sec 2t sin t 1 1 1 = lim t # cos t # 3t 3 cos 2t t:0 t:0 lim
=
1 1 (1)(1)(1) = . 3 3
Eq. (1) and Example 11b in Section 2.2
Exercises 2.4 Finding Limits Graphically 1. Which of the following statements about the function y = ƒsxd graphed here are true, and which are false?
y y f (x) 2
y y f (x)
1
1
–1
0
2
x
–1
0
lim ƒsxd = 1
1
2
3
x
b. lim ƒsxd = 0
a.
c. lim ƒsxd = 1
d. lim ƒsxd = lim+ ƒsxd
c. lim ƒsxd = 2
d. lim ƒsxd = 2
e. lim ƒsxd exists.
f. lim ƒsxd = 0
f. lim ƒsxd does not exist.
g. lim ƒsxd = 1
x: 0
e. lim+ ƒsxd = 1
h. lim ƒsxd = 1
i. lim ƒsxd = 0
j. lim ƒsxd = 2
x: 1
g. lim+ ƒsxd = lim ƒsxd
x: 2
h. lim ƒsxd exists at every c in the open interval s 1, 1d .
l. lim+ ƒsxd = 0
i. lim ƒsxd exists at every c in the open interval (1, 3).
a.
lim ƒsxd = 1
1
x: 1 + x:0 x:0 x:0 x:1
k.
lim ƒsxd does not exist .
x: 1 
x: 0 x: 0
x: 2
x: 0
2. Which of the following statements about the function y = ƒsxd graphed here are true, and which are false?
x: 1 + x:2 x:1 x:0
b. lim ƒsxd does not exist. x:2 x:1 x:1
x:0
x:c x:c
j.
lim ƒsxd = 0
x: 1 
k. lim+ ƒsxd does not exist. x:3
2.4 3  x, 3. Let ƒsxd = • x + 1, 2
x 6 2
OneSided Limits
85
6. Let g sxd = 2x sins1>xd .
x 7 2. y 1
y
y x
y3x 3
y x sin 1x
y x1 2
2
0
1 2 0
x
4
1
2
1
x
a. Find limx:2+ ƒsxd and limx:2 ƒsxd . b. Does limx:2 ƒsxd exist? If so, what is it? If not, why not? y –x
–1
c. Find limx:4 ƒsxd and limx:4+ ƒsxd . d. Does limx:4 ƒsxd exist? If so, what is it? If not, why not?
4. Let ƒsxd = d
a. Does limx:0+ g sxd exist? If so, what is it? If not, why not?
3  x, 2,
x 6 2 x = 2
x , 2
x 7 2.
b. Does limx:0 g sxd exist? If so, what is it? If not, why not? c. Does limx:0 g sxd exist? If so, what is it? If not, why not? 7. a. Graph ƒsxd = e
y
b. Find limx:1 ƒsxd and limx:1+ ƒsxd . c. Does limx:1 ƒsxd exist? If so, what is it? If not, why not?
y3x
8. a. Graph ƒsxd = e
3 y x 2 –2
0
2
x
c. Does limx:1 ƒsxd exist? If so, what is it? If not, why not? Graph the functions in Exercises 9 and 10. Then answer these questions. a. What are the domain and range of ƒ?
b. Does limx:2 ƒsxd exist? If so, what is it? If not, why not?
b. At what points c, if any, does limx:c ƒsxd exist?
c. Find limx:1 ƒsxd and limx:1+ ƒsxd .
c. At what points does only the lefthand limit exist?
d. Does limx:1 ƒsxd exist? If so, what is it? If not, why not? 0,
x … 0
1 sin x ,
x 7 0.
d. At what points does only the righthand limit exist? 21  x 2, 9. ƒsxd = • 1, 2, x, 10. ƒsxd = • 1, 0,
y 1
0
x Z 1 x = 1.
1  x 2, 2,
b. Find limx:1+ ƒsxd and limx:1 ƒsxd .
a. Find limx:2+ ƒsxd, limx:2 ƒsxd , and ƒ(2).
5. Let ƒsxd = •
x Z 1 x = 1.
x 3, 0,
x ⎧ x0 ⎪0, y⎨ 1 ⎪ sin x , x 0 ⎩
–1
a. Does limx:0+ ƒsxd exist? If so, what is it? If not, why not? b. Does limx:0 ƒsxd exist? If so, what is it? If not, why not? c. Does limx:0 ƒsxd exist? If so, what is it? If not, why not?
0 … x 6 1 1 … x 6 2 x = 2
1 … x 6 0, or 0 6 x … 1 x = 0 x 6 1 or x 7 1
Finding OneSided Limits Algebraically Find the limits in Exercises 11–18. 11. 13.
x + 2 + 1
lim
x: 0.5  A x
lim a
x: 2 +
14. lim a x:1
15. lim+ h:0
12. lim+
2x + 5 x b ba 2 x + 1 x + x
x + 6 3  x 1 b ba x ba 7 x + 1
2h 2 + 4h + 5  25 h
x:1
x  1 Ax + 2
86
Chapter 2: Limits and Continuity
26  25h 2 + 11h + 6 h ƒx + 2ƒ 17. a. lim +sx + 3d b. x + 2 x: 2
41. lim
16. lim
u :0
h: 0
18. a. lim+ x: 1
22x sx  1d ƒx  1ƒ
lim sx + 3d
x: 2
b. limx: 1
ƒx + 2ƒ x + 2
22x sx  1d ƒx  1ƒ
Use the graph of the greatest integer function y = :x; , Figure 1.10 in Section 1.1, to help you find the limits in Exercises 19 and 20. :u; :u; 19. a. lim+ b. limu u u: 3 u: 3 20. a. lim+st  : t; d
b. limst  : t; d
t:4
t: 4
sin U 1 U Find the limits in Exercises 21–42. Using lim
U:0
21. lim
sin 22u
22. lim
22u sin 3y 23. lim y:0 4y tan 2x 25. lim x x:0 u: 0
t: 0
sin kt t
sk constantd
h h :0 sin 3h 2t 26. lim t: 0 tan t 24. lim
x csc 2x x:0 cos 5x
27. lim
28. lim 6x 2scot xdscsc 2xd
x + x cos x 29. lim x:0 sin x cos x
30. lim
x: 0
x 2  x + sin x 2x x: 0
1  cos u 31. lim u: 0 sin 2u sin s1  cos td 1  cos t sin u 35. lim u: 0 sin 2u
x  x cos x 32. lim x: 0 sin2 3x sin ssin hd 34. lim sin h h :0 sin 5x 36. lim x: 0 sin 4x
37. lim u cos u
38. lim sin u cot 2u
tan 3x 39. lim x:0 sin 8x
40. lim
33. lim
t:0
u: 0
2.5
u: 0
y: 0
sin 3y cot 5y y cot 4y
tan u u2 cot 3u
42. lim
u :0
u cot 4u sin2 u cot2 2u
Theory and Examples 43. Once you know limx:a+ ƒsxd and limx:a ƒsxd at an interior point of the domain of ƒ, do you then know limx:a ƒsxd ? Give reasons for your answer. 44. If you know that limx:c ƒsxd exists, can you find its value by calculating limx:c+ ƒsxd ? Give reasons for your answer. 45. Suppose that ƒ is an odd function of x. Does knowing that limx:0+ ƒsxd = 3 tell you anything about limx:0 ƒsxd ? Give reasons for your answer. 46. Suppose that ƒ is an even function of x. Does knowing that limx:2 ƒsxd = 7 tell you anything about either limx: 2 ƒsxd or limx:2+ ƒsxd ? Give reasons for your answer. Formal Definitions of OneSided Limits 47. Given P 7 0 , find an interval I = s5, 5 + dd, d 7 0 , such that if x lies in I, then 2x  5 6 P . What limit is being verified and what is its value? 48. Given P 7 0 , find an interval I = s4  d, 4d, d 7 0 , such that if x lies in I, then 24  x 6 P . What limit is being verified and what is its value? Use the definitions of righthand and lefthand limits to prove the limit statements in Exercises 49 and 50. x x  2 49. lim50. lim+ = 1 = 1 x:0 ƒ x ƒ x:2 ƒ x  2 ƒ
51. Greatest integer function Find (a) limx:400+ :x; and (b) limx:400 :x; ; then use limit definitions to verify your findings. (c) Based on your conclusions in parts (a) and (b), can you say anything about limx:400 :x; ? Give reasons for your answer. 52. Onesided limits
Let ƒsxd = e
x 2 sin s1>xd, x 6 0 x 7 0. 2x,
Find (a) limx:0+ ƒsxd and (b) limx:0 ƒsxd ; then use limit definitions to verify your findings. (c) Based on your conclusions in parts (a) and (b), can you say anything about limx:0 ƒsxd ? Give reasons for your answer.
Continuity When we plot function values generated in a laboratory or collected in the field, we often connect the plotted points with an unbroken curve to show what the function’s values are likely to have been at the times we did not measure (Figure 2.34). In doing so, we are assuming that we are working with a continuous function, so its outputs vary continuously with the inputs and do not jump from one value to another without taking on the values in between. The limit of a continuous function as x approaches c can be found simply by calculating the value of the function at c. (We found this to be true for polynomials in Theorem 2.) Intuitively, any function y = ƒsxd whose graph can be sketched over its domain in one continuous motion without lifting the pencil is an example of a continuous function. In this section we investigate more precisely what it means for a function to be continuous.
2.5
87
We also study the properties of continuous functions, and see that many of the function types presented in Section 1.1 are continuous.
y 500 Distance fallen (m)
Continuity
Q4 Q3
375
Continuity at a Point
Q2
250
To understand continuity, it helps to consider a function like that in Figure 2.35, whose limits we investigated in Example 2 in the last section.
Q1
125 0
5 Elapsed time (sec)
10
FIGURE 2.34 Connecting plotted points by an unbroken curve from experimental data Q1 , Q2 , Q3 , Á for a falling object.
t
EXAMPLE 1 Find the points at which the function ƒ in Figure 2.35 is continuous and the points at which ƒ is not continuous. The function ƒ is continuous at every point in its domain [0, 4] except at x = 1, x = 2, and x = 4. At these points, there are breaks in the graph. Note the relationship between the limit of ƒ and the value of ƒ at each point of the function’s domain.
Solution
Points at which ƒ is continuous:
At x = 0, At x = 3, y
At 0 6 c 6 4, c Z 1, 2, y f (x)
2
At x = 1, 1
2
3
At x = 2,
x
4
lim ƒsxd = ƒs3d.
x:3
lim ƒsxd = ƒscd.
x:c
Points at which ƒ is not continuous:
1
0
lim ƒsxd = ƒs0d.
x:0 +
FIGURE 2.35 The function is continuous on [0, 4] except at x = 1, x = 2 , and x = 4 (Example 1).
At x = 4, At c 6 0, c 7 4,
lim ƒsxd does not exist.
x:1
lim ƒsxd = 1, but 1 Z ƒs2d.
x:2
lim ƒsxd = 1, but 1 Z ƒs4d.
x:4 
these points are not in the domain of ƒ.
To define continuity at a point in a function’s domain, we need to define continuity at an interior point (which involves a twosided limit) and continuity at an endpoint (which involves a onesided limit) (Figure 2.36).
Continuity from the right
Twosided continuity
Continuity from the left
DEFINITION Interior point: A function y = ƒsxd is continuous at an interior point c of its domain if lim ƒsxd = ƒscd.
x:c
y f (x) a
c
b
x
FIGURE 2.36 Continuity at points a, b, and c.
Endpoint: A function y = ƒsxd is continuous at a left endpoint a or is continuous at a right endpoint b of its domain if lim ƒsxd = ƒsad
x:a +
or
lim ƒsxd = ƒsbd,
x:b 
respectively.
If a function ƒ is not continuous at a point c, we say that ƒ is discontinuous at c and that c is a point of discontinuity of ƒ. Note that c need not be in the domain of ƒ. A function ƒ is rightcontinuous (continuous from the right) at a point x = c in its domain if limx:c+ ƒsxd = ƒscd. It is leftcontinuous (continuous from the left) at c if limx:c ƒsxd = ƒscd. Thus, a function is continuous at a left endpoint a of its domain if it
88
Chapter 2: Limits and Continuity
is rightcontinuous at a and continuous at a right endpoint b of its domain if it is leftcontinuous at b. A function is continuous at an interior point c of its domain if and only if it is both rightcontinuous and leftcontinuous at c (Figure 2.36).
y y 4
2
–2
0
x2
The function ƒsxd = 24  x 2 is continuous at every point of its domain [2, 2] (Figure 2.37), including x = 2, where ƒ is rightcontinuous, and x = 2, where ƒ is leftcontinuous.
EXAMPLE 2
x
2
FIGURE 2.37 A function that is continuous at every domain point (Example 2).
EXAMPLE 3
The unit step function U(x), graphed in Figure 2.38, is rightcontinuous at x = 0, but is neither leftcontinuous nor continuous there. It has a jump discontinuity at x = 0.
y y U(x)
1
We summarize continuity at a point in the form of a test.
x
0
Continuity Test A function ƒ(x) is continuous at an interior point x = c of its domain if and only if it meets the following three conditions.
FIGURE 2.38 A function that has a jump discontinuity at the origin (Example 3).
1. 2. 3.
(c lies in the domain of ƒ). (ƒ has a limit as x : c). (the limit equals the function value).
When we say a function is continuous at c, we are asserting that all three conditions hold. For onesided continuity and continuity at an endpoint, the limits in parts 2 and 3 of the test should be replaced by the appropriate onesided limits.
y 4 3
The function y = :x; introduced in Section 1.1 is graphed in Figure 2.39. It is discontinuous at every integer because the lefthand and righthand limits are not equal as x : n:
y ⎣x⎦
EXAMPLE 4
2 1 x –1
ƒ(c) exists limx:c ƒsxd exists limx:c ƒsxd = ƒscd
1
2
3
4
–2
FIGURE 2.39 The greatest integer function is continuous at every noninteger point. It is rightcontinuous, but not leftcontinuous, at every integer point (Example 4).
lim :x; = n  1
x:n 
and
lim :x; = n.
x:n +
Since :n ; = n, the greatest integer function is rightcontinuous at every integer n (but not leftcontinuous). The greatest integer function is continuous at every real number other than the integers. For example, lim :x; = 1 = :1.5; .
x:1.5
In general, if n  1 6 c 6 n, n an integer, then
lim :x; = n  1 = :c; .
x:c
Figure 2.40 displays several common types of discontinuities. The function in Figure 2.40a is continuous at x = 0. The function in Figure 2.40b would be continuous if it had ƒs0d = 1. The function in Figure 2.40c would be continuous if ƒ(0) were 1 instead of 2. The discontinuities in Figure 2.40b and c are removable. Each function has a limit as x : 0, and we can remove the discontinuity by setting ƒ(0) equal to this limit. The discontinuities in Figure 2.40d through f are more serious: limx:0 ƒsxd does not exist, and there is no way to improve the situation by changing ƒ at 0. The step function in Figure 2.40d has a jump discontinuity: The onesided limits exist but have different values. The function ƒsxd = 1>x 2 in Figure 2.40e has an infinite discontinuity. The function in Figure 2.40f has an oscillating discontinuity: It oscillates too much to have a limit as x : 0.
2.5
y f (x)
0
2
y f (x)
1
1 x
x
0
(b)
(c) y
y y f (x) 12 x
1
1 0
y f (x)
1
0
(a)
y
y
y
y
0
89
Continuity
1 x
y f (x)
0
x
(d) y sin 1x
x
x –1
(e)
(f)
FIGURE 2.40 The function in (a) is continuous at x = 0 ; the functions in (b) through (f ) are not.
Continuous Functions A function is continuous on an interval if and only if it is continuous at every point of the interval. For example, the semicircle function graphed in Figure 2.37 is continuous on the interval [2, 2], which is its domain. A continuous function is one that is continuous at every point of its domain. A continuous function need not be continuous on every interval. y
EXAMPLE 5 y 1x
0
x
(a) The function y = 1>x (Figure 2.41) is a continuous function because it is continuous at every point of its domain. It has a point of discontinuity at x = 0, however, because it is not defined there; that is, it is discontinuous on any interval containing x = 0. (b) The identity function ƒsxd = x and constant functions are continuous everywhere by Example 3, Section 2.3. Algebraic combinations of continuous functions are continuous wherever they are defined.
FIGURE 2.41 The function y = 1>x is continuous at every value of x except x = 0 . It has a point of discontinuity at x = 0 (Example 5).
THEOREM 8—Properties of Continuous Functions If the functions ƒ and g are continuous at x = c, then the following combinations are continuous at x = c. 1. 2. 3. 4. 5. 6. 7.
Sums: Differences: Constant multiples: Products: Quotients: Powers: Roots:
ƒ + g ƒ  g k # ƒ, for any number k ƒ#g ƒ>g, provided gscd Z 0 ƒ n, n a positive integer n 2ƒ, provided the root is defined on an open interval containing c, where n is a positive integer
90
Chapter 2: Limits and Continuity
Most of the results in Theorem 8 follow from the limit rules in Theorem 1, Section 2.2. For instance, to prove the sum property we have lim sƒ + gdsxd = lim sƒsxd + gsxdd
x:c
x:c
= lim ƒsxd + lim gsxd,
Sum Rule, Theorem 1
= ƒscd + gscd = sƒ + gdscd.
Continuity of ƒ, g at c
x:c
x:c
This shows that ƒ + g is continuous.
EXAMPLE 6 (a) Every polynomial Psxd = an x n + an  1x n  1 + Á + a0 is continuous because lim Psxd = Pscd by Theorem 2, Section 2.2. x:c
(b) If P(x) and Q(x) are polynomials, then the rational function Psxd>Qsxd is continuous wherever it is defined sQscd Z 0d by Theorem 3, Section 2.2. The function ƒsxd = ƒ x ƒ is continuous at every value of x. If x 7 0, we have ƒsxd = x, a polynomial. If x 6 0, we have ƒsxd = x, another polynomial. Finally, at the origin, limx:0 ƒ x ƒ = 0 = ƒ 0 ƒ .
EXAMPLE 7
The functions y = sin x and y = cos x are continuous at x = 0 by Example 11 of Section 2.2. Both functions are, in fact, continuous everywhere (see Exercise 70). It follows from Theorem 8 that all six trigonometric functions are then continuous wherever they are defined. For example, y = tan x is continuous on Á ´ s p>2, p>2d ´ sp>2, 3p>2d ´ Á .
Inverse Functions and Continuity The inverse function of any function continuous on an interval is continuous over its domain. This result is suggested from the observation that the graph of ƒ1, being the reflection of the graph of ƒ across the line y = x, cannot have any breaks in it when the graph of ƒ has no breaks. A rigorous proof that ƒ1 is continuous whenever ƒ is continuous on an interval is given in more advanced texts. It follows that the inverse trigonometric functions are all continuous over their domains. We defined the exponential function y = a x in Section 1.5 informally by its graph. Recall that the graph was obtained from the graph of y = a x for x a rational number by filling in the holes at the irrational points x, so the function y = a x was defined to be continuous over the entire real line. The inverse function y = loga x is also continuous. In particular, the natural exponential function y = e x and the natural logarithm function y = ln x are both continuous over their domains.
Composites All composites of continuous functions are continuous. The idea is that if ƒ(x) is continuous at x = c and g(x) is continuous at x = ƒscd, then g ⴰ ƒ is continuous at x = c (Figure 2.42). In this case, the limit as x : c is g(ƒ(c)). g˚ f Continuous at c
c
f
g
Continuous at c
Continuous at f (c) f (c)
FIGURE 2.42 Composites of continuous functions are continuous.
g( f (c))
2.5
Continuity
91
THEOREM 9—Composite of Continuous Functions If ƒ is continuous at c and g is continuous at ƒ(c), then the composite g ⴰ ƒ is continuous at c. Intuitively, Theorem 9 is reasonable because if x is close to c, then ƒ(x) is close to ƒ(c), and since g is continuous at ƒ(c), it follows that g(ƒ(x)) is close to g(ƒ(c)). The continuity of composites holds for any finite number of functions. The only requirement is that each function be continuous where it is applied. For an outline of a proof of Theorem 9, see Exercise 6 in Appendix 5.
EXAMPLE 8
Show that the following functions are continuous everywhere on their respective domains. x 2>3 1 + x4
(a) y = 2x 2  2x  5
(b) y =
(c) y = `
(d) y = `
x  2 ` x2  2
x sin x ` x2 + 2
Solution y 0.4 0.3 0.2 0.1 –2
–
0
2
FIGURE 2.43 The graph suggests that y = ƒ sx sin xd>sx 2 + 2d ƒ is continuous (Example 8d).
x
(a) The square root function is continuous on [0, q d because it is a root of the continuous identity function ƒsxd = x (Part 7, Theorem 8). The given function is then the composite of the polynomial ƒsxd = x 2  2x  5 with the square root function gstd = 2t , and is continuous on its domain. (b) The numerator is the cube root of the identity function squared; the denominator is an everywherepositive polynomial. Therefore, the quotient is continuous. (c) The quotient sx  2d>sx 2  2d is continuous for all x Z ; 22 , and the function is the composition of this quotient with the continuous absolute value function (Example 7). (d) Because the sine function is everywherecontinuous (Exercise 70), the numerator term x sin x is the product of continuous functions, and the denominator term x 2 + 2 is an everywherepositive polynomial. The given function is the composite of a quotient of continuous functions with the continuous absolute value function (Figure 2.43). Theorem 9 is actually a consequence of a more general result which we now state and prove.
THEOREM 10—Limits of Continuous Functions b and limx:c ƒ(x) = b, then
If g is continuous at the point
limx:c g(ƒ(x)) = g(b) = g(limx:c ƒ(x)). Proof
Let P 7 0 be given. Since g is continuous at b, there exists a number d1 7 0 such that
ƒ g( y)  g(b) ƒ 6 P whenever 0 6 ƒ y  b ƒ 6 d1. Since limx:c ƒ(x) = b, there exists a d 7 0 such that ƒ ƒ(x)  b ƒ 6 d1 If we let y = ƒ(x), we then have that
whenever
0 6 ƒ x  c ƒ 6 d.
ƒ y  b ƒ 6 d1 whenever 0 6 ƒ x  c ƒ 6 d, which implies from the first statement that ƒ g( y)  g(b) ƒ = ƒ g(ƒ(x))  g(b) ƒ 6 P whenever 0 6 ƒ x  c ƒ 6 d. From the definition of limit, this proves that limx:c g(ƒ(x)) = g(b).
92
Chapter 2: Limits and Continuity
EXAMPLE 9 (a)
lim cos a2x + sin a
x:p/2
(b) lim sin1 a x:1
We sometimes denote e u by exp u when u is a complicated mathematical expression.
As an application of Theorem 10, we have the following calculations. 3p 3p + xb b = cos a lim 2x + lim sin a + xb b 2 2 x:p/2 x:p/2 = cos (p + sin 2p) = cos p = 1.
1  x 1  x b = sin1 a lim b 2 x:1 1  x 2 1  x 1 = sin1 a lim b 1 + x x:1 p 1 = sin1 = 2 6
Arcsine is continuous. Cancel common factor (1  x).
(c) lim 2x + 1 e tan x = lim 2x + 1 # exp a lim tan xb x:0
x:0
Exponential is continuous.
x:0
= 1 # e0 = 1
Intermediate Value Theorem for Continuous Functions Functions that are continuous on intervals have properties that make them particularly useful in mathematics and its applications. One of these is the Intermediate Value Property. A function is said to have the Intermediate Value Property if whenever it takes on two values, it also takes on all the values in between.
THEOREM 11—The Intermediate Value Theorem for Continuous Functions If ƒ is a continuous function on a closed interval [a, b], and if y0 is any value between ƒ(a) and ƒ(b), then y0 = ƒscd for some c in [a, b]. y y f (x) f (b)
y0 f (a)
0
y
a
c
b
x
3 2 1
0
1
2
3
4
x
FIGURE 2.44 The function 2x  2, 1 … x 6 2 ƒsxd = e 3, 2 … x … 4 does not take on all values between ƒs1d = 0 and ƒs4d = 3 ; it misses all the values between 2 and 3.
Theorem 11 says that continuous functions over finite closed intervals have the Intermediate Value Property. Geometrically, the Intermediate Value Theorem says that any horizontal line y = y0 crossing the yaxis between the numbers ƒ(a) and ƒ(b) will cross the curve y = ƒsxd at least once over the interval [a, b]. The proof of the Intermediate Value Theorem depends on the completeness property of the real number system (Appendix 7) and can be found in more advanced texts. The continuity of ƒ on the interval is essential to Theorem 11. If ƒ is discontinuous at even one point of the interval, the theorem’s conclusion may fail, as it does for the function graphed in Figure 2.44 (choose y0 as any number between 2 and 3). A Consequence for Graphing: Connectedness Theorem 11 implies that the graph of a function continuous on an interval cannot have any breaks over the interval. It will be connected—a single, unbroken curve. It will not have jumps like the graph of the greatest integer function (Figure 2.39), or separate branches like the graph of 1>x (Figure 2.41).
2.5
93
Continuity
A Consequence for Root Finding We call a solution of the equation ƒsxd = 0 a root of the equation or zero of the function ƒ. The Intermediate Value Theorem tells us that if ƒ is continuous, then any interval on which ƒ changes sign contains a zero of the function. In practical terms, when we see the graph of a continuous function cross the horizontal axis on a computer screen, we know it is not stepping across. There really is a point where the function’s value is zero.
EXAMPLE 10
Show that there is a root of the equation x 3  x  1 = 0 between 1 and 2.
Let ƒ(x) = x 3  x  1. Since ƒ(1) = 1  1  1 = 1 6 0 and ƒ(2) = 2  2  1 = 5 7 0, we see that y0 = 0 is a value between ƒ(1) and ƒ(2). Since ƒ is continuous, the Intermediate Value Theorem says there is a zero of ƒ between 1 and 2. Figure 2.45 shows the result of zooming in to locate the root near x = 1.32. Solution 3
5
1
1 –1
1.6
2
–2
–1 (a)
(b)
0.02
0.003
1.320
1.330
–0.02
1.3240
1.3248
–0.003 (c)
(d)
FIGURE 2.45 Zooming in on a zero of the function ƒsxd = x 3  x  1 . The zero is near x = 1.3247 (Example 10).
EXAMPLE 11
y 4
y542 x
Use the Intermediate Value Theorem to prove that the equation 22x + 5 = 4  x 2
2
3
has a solution (Figure 2.46).
2
Solution
22x + 5 + x 2 = 4,
y 5 2x 1 5 1 0
We rewrite the equation as
c
2
x
FIGURE 2.46 The curves y = 22x + 5 and y = 4  x 2 have the same value at x = c where 22x + 5 = 4  x 2 (Example 11).
and set ƒ(x) = 22x + 5 + x 2. Now g(x) = 22x + 5 is continuous on the interval [5>2, q) since it is the composite of the square root function with the nonnegative linear function y = 2x + 5. Then ƒ is the sum of the function g and the quadratic function y = x 2, and the quadratic function is continuous for all values of x. It follows that ƒ(x) = 22x + 5 + x 2 is continuous on the interval [5>2, q). By trial and error, we find the function values ƒ(0) = 25 L 2.24 and ƒ(2) = 29 + 4 = 7, and note that ƒ is also continuous on the finite closed interval [0, 2] ( [5>2, q). Since the value y0 = 4 is between the numbers 2.24 and 7, by the Intermediate Value Theorem there is a number c H [0, 2] such that ƒ(c) = 4. That is, the number c solves the original equation.
94
Chapter 2: Limits and Continuity
Continuous Extension to a Point Sometimes the formula that describes a function f does not make sense at a point x = c. It might nevertheless be possible to extend the domain of f, to include x = c, creating a new function that is continuous at x = c. For example, the function y = ƒ(x) = ssin xd>x is continuous at every point except x = 0. In this it is like the function y = 1>x. But y = ssin xd>x is different from y = 1>x in that it has a finite limit as x : 0 (Theorem 7). It is therefore possible to extend the function’s domain to include the point x = 0 in such a way that the extended function is continuous at x = 0. We define a new function Fsxd =
L
sin x x ,
x Z 0
1,
x = 0.
The function F(x) is continuous at x = 0 because sin x = Fs0d lim x:0 x (Figure 2.47). y
y
(0, 1) ⎛– , 2⎛ ⎝ 2 ⎝ – 2
(0, 1)
f (x) ⎛ , 2⎛ ⎝ 2 ⎝
⎛– , 2⎛ ⎝ 2 ⎝ 2
0
x
– 2
⎛ , 2⎛ ⎝ 2 ⎝ 0
(a)
F(x)
2
x
(b)
FIGURE 2.47 The graph (a) of ƒsxd = ssin xd>x for p>2 … x … p>2 does not include the point (0, 1) because the function is not defined at x = 0 . (b) We can remove the discontinuity from the graph by defining the new function F(x) with Fs0d = 1 and Fsxd = ƒsxd everywhere else. Note that Fs0d = limx:0 ƒsxd .
More generally, a function (such as a rational function) may have a limit even at a point where it is not defined. If ƒ(c) is not defined, but limx:c ƒsxd = L exists, we can define a new function F(x) by the rule y 2
y
Fsxd = e
x2 x 6 x2 4
1 –1
0
1
2
3
4
x
(a)
EXAMPLE 12
y
5 4 –1
2
x 3 y x2
Show that ƒsxd =
1 0
ƒsxd, if x is in the domain of ƒ L, if x = c. The function F is continuous at x = c. It is called the continuous extension of ƒ to x = c. For rational functions ƒ, continuous extensions are usually found by canceling common factors.
x2 + x  6 , x2  4
x Z 2
has a continuous extension to x = 2, and find that extension. 1
2
3
4
x
(b)
FIGURE 2.48 (a) The graph of ƒ(x) and (b) the graph of its continuous extension F(x) (Example 12).
Solution
Although ƒ(2) is not defined, if x Z 2 we have sx  2dsx + 3d x2 + x  6 x + 3 ƒsxd = = = . 2 x + 2 sx  2dsx + 2d x  4
The new function Fsxd =
x + 3 x + 2
2.5
Continuity
95
is equal to ƒ(x) for x Z 2, but is continuous at x = 2, having there the value of 5>4. Thus F is the continuous extension of ƒ to x = 2, and x2 + x  6 5 = lim ƒsxd = . 4 x:2 x:2 x2  4 The graph of ƒ is shown in Figure 2.48. The continuous extension F has the same graph except with no hole at (2, 5>4). Effectively, F is the function ƒ with its point of discontinuity at x = 2 removed. lim
Exercises 2.5 Continuity from Graphs In Exercises 1–4, say whether the function graphed is continuous on [1, 3] . If not, where does it fail to be continuous and why? 1.
2.
y g(x)
–1
c. Does limx:1 ƒsxd = ƒs1d ? d. Is ƒ continuous at x = 1 ?
1
0
1
2
x
3
–1
3.
0
1
2
3
x
y h(x) 2
1
1 1
2
x
3
b. Is ƒ continuous at x = 2 ? 9. What value should be assigned to ƒ(2) to make the extended function continuous at x = 2 ?
y
2
0
7. a. Is ƒ defined at x = 2 ? (Look at the definition of ƒ.) 8. At what values of x is ƒ continuous?
4. y
–1
b. Does limx:1 ƒsxd exist?
2
1
c. Does limx:1+ ƒsxd = ƒs 1d ? 6. a. Does ƒ(1) exist?
y f(x) 2
b. Does limx: 1+ ƒsxd exist? d. Is ƒ continuous at x = 1 ?
y
y
5. a. Does ƒs 1d exist?
–1
y k(x)
1
0
2
Exercises 5–10 refer to the function x 2  1, 2x, ƒsxd = e 1, 2x + 4, 0,
1 0 x 1 2
… 6 = 6 6
x x 1 x x
10. To what new value should ƒ(1) be changed to remove the discontinuity?
3
x
Applying the Continuity Test At which points do the functions in Exercises 11 and 12 fail to be continuous? At which points, if any, are the discontinuities removable? Not removable? Give reasons for your answers.
6 0 6 1
11. Exercise 1, Section 2.4
6 2 6 3
13. y =
1  3x x  2
14. y =
1 + 4 sx + 2d2
15. y =
x + 1 x 2  4x + 3
16. y =
x + 3 x 2  3x  10
18. y =
x2 1 2 ƒxƒ + 1
At what points are the functions in Exercises 13–30 continuous?
graphed in the accompanying figure.
17. y = ƒ x  1 ƒ + sin x
y y f (x) 2
19. y =
(1, 2)
y 2x
y –2x 4
cos x x
21. y = csc 2x
(1, 1) –1 y x2 1
0
1
2
–1
The graph for Exercises 5–10.
3
12. Exercise 2, Section 2.4
x
23. y =
x tan x x2 + 1
x + 2 20. y = cos x px 22. y = tan 2 24. y =
2x 4 + 1 1 + sin2 x 4
25. y = 22x + 3
26. y = 23x  1
27. y = s2x  1d1>3
28. y = s2  xd1>5
96
Chapter 2: Limits and Continuity
x2  x  6 , 29. gsxd = • x  3 5, x3  8 , x2  4 30. ƒsxd = d 3, 4,
47. For what values of a and b is
x Z 3
2, ƒsxd = • ax  b, 3,
x = 3
x Z 2, x Z 2
continuous at every x?
x = 2 x = 2
48. For what values of a and b is
Limits Involving Trigonometric Functions Find the limits in Exercises 31–38. Are the functions continuous at the point being approached? p 31. lim sin sx  sin xd 32. lim sin a cos stan tdb 2 x:p t: 0 33. lim sec s y sec y  tan y  1d 2
2
y:1
34. lim tan a x:0
p cos ssin x 1>3 db 4
35. lim cos a t:0
p 219  3 sec 2t
37. lim+ sin a x:0
p 2x e b 2
x … 1 1 6 x 6 1 x Ú 1
b 36.
lim 2csc2 x + 5 13 tan x
x: p/6
38. lim cos1 (ln 2x) x: 1
Continuous Extensions 39. Define g(3) in a way that extends g sxd = sx 2  9d>sx  3d to be continuous at x = 3 . 40. Define h(2) in a way that extends hstd = st 2 + 3t  10d>st  2d to be continuous at t = 2 .
ax + 2b, gsxd = • x 2 + 3a  b, 3x  5,
x … 0 0 6 x … 2 x 7 2
continuous at every x? T In Exercises 49–52, graph the function ƒ to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function’s value at x = 0 . If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function’s value(s) should be? 10 ƒ x ƒ  1 x
49. ƒsxd =
10 x  1 x
50. ƒsxd =
51. ƒsxd =
sin x ƒxƒ
52. ƒsxd = s1 + 2xd1>x
Theory and Examples 53. A continuous function y = ƒsxd is known to be negative at x = 0 and positive at x = 1 . Why does the equation ƒsxd = 0 have at least one solution between x = 0 and x = 1? Illustrate with a sketch.
41. Define ƒ(1) in a way that extends ƒssd = ss 3  1d>ss 2  1d to be continuous at s = 1 .
54. Explain why the equation cos x = x has at least one solution.
42. Define g(4) in a way that extends
56. A function value Show that the function Fsxd = sx  ad2 # sx  bd2 + x takes on the value sa + bd>2 for some value of x.
g sxd = sx 2  16d> sx 2  3x  4d to be continuous at x = 4 . 43. For what value of a is ƒsxd = e
x 2  1, 2ax,
x 6 3 x Ú 3
continuous at every x? x 6 2 x Ú 2
continuous at every x?
58. Explain why the following five statements ask for the same information.
a 2x  2a, ƒsxd = b 12,
b. Find the xcoordinates of the points where the curve y = x 3 crosses the line y = 3x + 1 . c. Find all the values of x for which x 3  3x = 1 . d. Find the xcoordinates of the points where the cubic curve y = x 3  3x crosses the line y = 1 .
45. For what values of a is x Ú 2 x 6 2
continuous at every x? 46. For what value of b is x  b , x 6 0 gsxd = • b + 1 x 2 + b, x 7 0 continuous at every x?
57. Solving an equation If ƒsxd = x 3  8x + 10 , show that there are values c for which ƒ(c) equals (a) p ; (b)  23 ; (c) 5,000,000.
a. Find the roots of ƒsxd = x 3  3x  1 .
44. For what value of b is x, g sxd = e 2 bx ,
55. Roots of a cubic Show that the equation x 3  15x + 1 = 0 has three solutions in the interval [4, 4] .
e. Solve the equation x 3  3x  1 = 0 . 59. Removable discontinuity Give an example of a function ƒ(x) that is continuous for all values of x except x = 2 , where it has a removable discontinuity. Explain how you know that ƒ is discontinuous at x = 2 , and how you know the discontinuity is removable. 60. Nonremovable discontinuity Give an example of a function g(x) that is continuous for all values of x except x = 1 , where it has a nonremovable discontinuity. Explain how you know that g is discontinuous there and why the discontinuity is not removable.
2.6 61. A function discontinuous at every point a. Use the fact that every nonempty interval of real numbers contains both rational and irrational numbers to show that the function ƒsxd = e
1, if x is rational 0, if x is irrational
Limits Involving Infinity; Asymptotes of Graphs
97
68. The signpreserving property of continuous functions Let ƒ be defined on an interval (a, b) and suppose that ƒscd Z 0 at some c where ƒ is continuous. Show that there is an interval sc  d, c + dd about c where ƒ has the same sign as ƒ(c). 69. Prove that ƒ is continuous at c if and only if lim ƒsc + hd = ƒscd .
h:0
is discontinuous at every point. b. Is ƒ rightcontinuous or leftcontinuous at any point?
70. Use Exercise 69 together with the identities
62. If functions ƒ(x) and g(x) are continuous for 0 … x … 1 , could ƒ(x)>g (x) possibly be discontinuous at a point of [0, 1]? Give reasons for your answer.
sin sh + cd = sin h cos c + cos h sin c , cos sh + cd = cos h cos c  sin h sin c
63. If the product function hsxd = ƒsxd # g sxd is continuous at x = 0 , must ƒ(x) and g(x) be continuous at x = 0 ? Give reasons for your answer.
64. Discontinuous composite of continuous functions Give an example of functions ƒ and g, both continuous at x = 0 , for which the composite ƒ ⴰ g is discontinuous at x = 0 . Does this contradict Theorem 9? Give reasons for your answer. 65. Neverzero continuous functions Is it true that a continuous function that is never zero on an interval never changes sign on that interval? Give reasons for your answer. 66. Stretching a rubber band Is it true that if you stretch a rubber band by moving one end to the right and the other to the left, some point of the band will end up in its original position? Give reasons for your answer. 67. A fixed point theorem Suppose that a function ƒ is continuous on the closed interval [0, 1] and that 0 … ƒsxd … 1 for every x in [0, 1]. Show that there must exist a number c in [0, 1] such that ƒscd = c (c is called a fixed point of ƒ).
to prove that both ƒsxd = sin x and g sxd = cos x are continuous at every point x = c . Solving Equations Graphically T Use the Intermediate Value Theorem in Exercises 71–78 to prove that each equation has a solution. Then use a graphing calculator or computer grapher to solve the equations. 71. x 3  3x  1 = 0 72. 2x 3  2x 2  2x + 1 = 0 73. xsx  1d2 = 1
sone rootd
74. x = 2 x
75. 2x + 21 + x = 4 76. x 3  15x + 1 = 0 77. cos x = x 78. 2 sin x = x mode.
sthree rootsd
sone rootd . Make sure you are using radian mode. sthree rootsd . Make sure you are using radian
Limits Involving Infinity; Asymptotes of Graphs
2.6
In this section we investigate the behavior of a function when the magnitude of the independent variable x becomes increasingly large, or x : ; q . We further extend the concept of limit to infinite limits, which are not limits as before, but rather a new use of the term limit. Infinite limits provide useful symbols and language for describing the behavior of functions whose values become arbitrarily large in magnitude. We use these limit ideas to analyze the graphs of functions having horizontal or vertical asymptotes.
y 4 3 y 1x
2 1 –1 0 –1
1
2
3
4
FIGURE 2.49 The graph of y = 1>x approaches 0 as x : q or x :  q .
x
Finite Limits as x : —ˆ The symbol for infinity s q d does not represent a real number. We use q to describe the behavior of a function when the values in its domain or range outgrow all finite bounds. For example, the function ƒsxd = 1>x is defined for all x Z 0 (Figure 2.49). When x is positive and becomes increasingly large, 1>x becomes increasingly small. When x is negative and its magnitude becomes increasingly large, 1>x again becomes small. We summarize these observations by saying that ƒsxd = 1>x has limit 0 as x : q or x :  q , or that 0 is a limit of ƒsxd = 1>x at infinity and negative infinity. Here are precise definitions.
98
Chapter 2: Limits and Continuity
DEFINITIONS 1. We say that ƒ(x) has the limit L as x approaches infinity and write lim ƒsxd = L
x: q
if, for every number P 7 0, there exists a corresponding number M such that for all x x 7 M
Q
ƒ ƒsxd  L ƒ 6 P. 2. We say that ƒ(x) has the limit L as x approaches minus infinity and write lim ƒsxd = L
x:  q
if, for every number P 7 0, there exists a corresponding number N such that for all x x 6 N
y y 1x
M 1 –
and
lim
x: ; q
1 x = 0.
(1)
We prove the second result in Example 1, and leave the first to Exercises 87 and 88.
EXAMPLE 1 0
y –
lim k = k
x: ; q
y
N – 1
ƒ ƒsxd  L ƒ 6 P.
Intuitively, limx: q ƒsxd = L if, as x moves increasingly far from the origin in the positive direction, ƒ(x) gets arbitrarily close to L. Similarly, limx: q ƒsxd = L if, as x moves increasingly far from the origin in the negative direction, ƒ(x) gets arbitrarily close to L. The strategy for calculating limits of functions as x : ; q is similar to the one for finite limits in Section 2.2. There we first found the limits of the constant and identity functions y = k and y = x. We then extended these results to other functions by applying Theorem 1 on limits of algebraic combinations. Here we do the same thing, except that the starting functions are y = k and y = 1>x instead of y = k and y = x. The basic facts to be verified by applying the formal definition are
No matter what positive number is, the graph enters this band at x 1 and stays.
Q
x
Show that
1 (a) lim x = 0 x: q
(b)
lim
x:  q
1 x = 0.
Solution No matter what positive number is, the graph enters this band at x – 1 and stays.
(a) Let P 7 0 be given. We must find a number M such that for all x
FIGURE 2.50 The geometry behind the argument in Example 1.
The implication will hold if M = 1>P or any larger positive number (Figure 2.50). This proves limx: q s1>xd = 0. (b) Let P 7 0 be given. We must find a number N such that for all x
x 7 M
x 6 N
Q
Q
1 1 ` x  0 ` = ` x ` 6 P.
1 1 ` x  0 ` = ` x ` 6 P.
The implication will hold if N = 1>P or any number less than 1>P (Figure 2.50). This proves limx: q s1>xd = 0. Limits at infinity have properties similar to those of finite limits.
THEOREM 12 All the limit laws in Theorem 1 are true when we replace limx:c by limx: q or limx: q . That is, the variable x may approach a finite number c or ; q .
2.6
Limits Involving Infinity; Asymptotes of Graphs
99
EXAMPLE 2 The properties in Theorem 12 are used to calculate limits in the same way as when x approaches a finite number c. 1 1 (a) lim a5 + x b = lim 5 + lim x x: q x: q x: q
Sum Rule
= 5 + 0 = 5 (b)
lim
x:  q
p23 = x2 =
Known limits
1 1 lim p23 # x # x
x:  q
1 lim p23 # lim x x:  q
x:  q
#
1 lim x
x:  q
= p23 # 0 # 0 = 0
2
–5
1
Line y 5 3
0
5
10
To determine the limit of a rational function as x : ; q , we first divide the numerator and denominator by the highest power of x in the denominator. The result then depends on the degrees of the polynomials involved. x
EXAMPLE 3
These examples illustrate what happens when the degree of the numerator is less than or equal to the degree of the denominator. 5 + s8>xd  s3>x 2 d 5x 2 + 8x  3 = lim x: q x: q 3x 2 + 2 3 + s2>x 2 d
–1
(a) lim –2
NOT TO SCALE
= FIGURE 2.51 The graph of the function in Example 3a. The graph approaches the line y = 5>3 as ƒ x ƒ increases.
8 y 6
lim
x:  q
See Fig. 2.51.
s11>x 2 d + s2>x 3 d
Divide numerator and denominator by x3.
2  s1>x d 3
0 + 0 = 0 2  0
See Fig. 2.52.
Horizontal Asymptotes
2
0
11x + 2 = x:  q 2x 3  1 lim
5 + 0  0 5 = 3 + 0 3
Divide numerator and denominator by x2.
A case for which the degree of the numerator is greater than the degree of the denominator is illustrated in Example 10.
11x 2 2x 3 1
4
–2
(b)
=
y
–4
Known limits
Limits at Infinity of Rational Functions
2 y 5x 2 8x 3 3x 2
y
Product Rule
2
4
6
–2
x
If the distance between the graph of a function and some fixed line approaches zero as a point on the graph moves increasingly far from the origin, we say that the graph approaches the line asymptotically and that the line is an asymptote of the graph. Looking at ƒsxd = 1>x (see Figure 2.49), we observe that the xaxis is an asymptote of the curve on the right because 1 lim x = 0
x: q
–4 –6
and on the left because 1 lim x = 0.
x:  q
–8
FIGURE 2.52 The graph of the function in Example 3b. The graph approaches the xaxis as ƒ x ƒ increases.
We say that the xaxis is a horizontal asymptote of the graph of ƒsxd = 1>x.
DEFINITION A line y = b is a horizontal asymptote of the graph of a function y = ƒsxd if either lim ƒsxd = b
x: q
or
lim ƒsxd = b.
x:  q
100
Chapter 2: Limits and Continuity
The graph of the function 5x 2 + 8x  3 3x 2 + 2
ƒsxd =
sketched in Figure 2.51 (Example 3a) has the line y = 5>3 as a horizontal asymptote on both the right and the left because 5 3
lim ƒsxd =
x: q
EXAMPLE 4
ƒsxd =
2 y1
y –1
f(x) –2
For x Ú 0:
x3 – 2 x3 + 1
For x 6 0:
FIGURE 2.53 The graph of the function in Example 4 has two horizontal asymptotes.
5 . 3
x3  2 ƒ x ƒ 3 + 1.
We calculate the limits as x : ; q .
x
0
x:  q
Find the horizontal asymptotes of the graph of
y
Solution
lim ƒsxd =
and
lim
x: q
lim
x:  q
1  (2>x 3) x3  2 x3  2 = lim = lim = 1. x: q x 3 + 1 x: q 1 + (1>x 3) ƒxƒ3 + 1 x3  2 = ƒxƒ3 + 1
lim
x:  q
x3  2 = (x) 3 + 1
lim
x:  q
1  (2>x 3) 1 + (1>x 3)
= 1.
The horizontal asymptotes are y = 1 and y = 1. The graph is displayed in Figure 2.53. Notice that the graph crosses the horizontal asymptote y = 1 for a positive value of x.
EXAMPLE 5 y = e x because
The xaxis (the line y = 0) is a horizontal asymptote of the graph of
y
lim e x = 0.
x:  q
y ex
To see this, we use the definition of a limit as x approaches  q. So let P 7 0 be given, but arbitrary. We must find a constant N such that for all x,
1
x 6 N N ln
x
FIGURE 2.54 The graph of y = e x approaches the xaxis as x :  q (Example 5).
Q
ƒex
 0 ƒ 6 P.
Now ƒ e x  0 ƒ = e x, so the condition that needs to be satisfied whenever x 6 N is e x 6 P. Let x = N be the number where e x = P. Since e x is an increasing function, if x 6 N , then e x 6 P. We find N by taking the natural logarithm of both sides of the equation e N = P, so N = ln P (see Figure 2.54). With this value of N the condition is satisfied, and we conclude that limx: q e x = 0.
EXAMPLE 6
Find (a) lim sin s1>xd and x: q
(b)
lim x sin s1>xd.
x: ; q
Solution (a) We introduce the new variable t = 1>x. From Example 1, we know that t : 0 + as
x : q (see Figure 2.49). Therefore,
1 lim sin x = lim+ sin t = 0. t:0
x: q
2.6
Limits Involving Infinity; Asymptotes of Graphs
101
(b) We calculate the limits as x : q and x :  q :
y
sin t 1 lim x sin x = lim+ t = 1 q x: t:0
1
sin t 1 lim x sin x = lim t = 1. q x: t:0
The graph is shown in Figure 2.55, and we see that the line y = 1 is a horizontal asymptote.
1 y x sin x –1
and
x
1
Likewise, we can investigate the behavior of y = ƒ(1>x) as x : 0 by investigating y = ƒ(t) as t : ; q , where t = 1>x.
FIGURE 2.55 The line y = 1 is a horizontal asymptote of the function graphed here (Example 6b).
EXAMPLE 7
Find lime 1>x. x:0
We let t = 1>x. From Figure 2.49, we can see that t :  q as x : 0 . (We make this idea more precise further on.) Therefore, Solution
y
y
–3
–2
lim e 1>x =
1 0.8 0.6 0.4 0.2
e1 ⁄x
–1
0
x:0 
lim e t = 0
(Figure 2.56). The Sandwich Theorem also holds for limits as x : ; q . You must be sure, though, that the function whose limit you are trying to find stays between the bounding functions at very large values of x in magnitude consistent with whether x : q or x :  q .
x
FIGURE 2.56 The graph of y = e 1>x for x 6 0 shows limx:0 e 1>x = 0 (Example 7).
EXAMPLE 8
Using the Sandwich Theorem, find the horizontal asymptote of the curve y = 2 +
Solution
sin x 1 0 … ` x ` … `x`
y 2 sinx x
and limx: ; q ƒ 1>x ƒ = 0, we have limx:; q ssin xd>x = 0 by the Sandwich Theorem. Hence,
2 1 0
2
3
sin x x .
We are interested in the behavior as x : ; q. Since
y
–3 –2 –
Example 5
t:  q
lim a2 +
x: ; q
x
FIGURE 2.57 A curve may cross one of its asymptotes infinitely often (Example 8).
sin x x b = 2 + 0 = 2,
and the line y = 2 is a horizontal asymptote of the curve on both left and right (Figure 2.57). This example illustrates that a curve may cross one of its horizontal asymptotes many times.
EXAMPLE 9
Find lim A x  2x 2 + 16 B . x: q
Both of the terms x and 2x 2 + 16 approach infinity as x : q , so what happens to the difference in the limit is unclear (we cannot subtract q from q because the symbol does not represent a real number). In this situation we can multiply the numerator and the denominator by the conjugate radical expression to obtain an equivalent algebraic result: Solution
lim A x  2x 2 + 16 B = lim A x  2x 2 + 16 B
x: q
x: q
= lim
x: q
x 2  (x 2 + 16) x + 2x + 16 2
x + 2x 2 + 16 x + 2x 2 + 16
= lim
x: q
16 x + 2x 2 + 16
.
102
Chapter 2: Limits and Continuity
As x : q , the denominator in this last expression becomes arbitrarily large, so we see that the limit is 0. We can also obtain this result by a direct calculation using the Limit Laws: 16  x 0 lim = lim = = 0. x: q x + 2x 2 + 16 x: q 16 x2 1 + 21 + 0 1 + + 2 A x2 x 16
Oblique Asymptotes 2 y5 x 235x111 1 2x 2 4 2 2x 2 4
y
The vertical distance between curve and line goes to zero as x → `
6 5 4 3
y5 x 11 2
Find the oblique asymptote of the graph of
2
3
4
ƒsxd =
x2  3 2x  4
in Figure 2.58.
1 1
EXAMPLE 10
Oblique asymptote
x52
2
–1 0 –1
If the degree of the numerator of a rational function is 1 greater than the degree of the denominator, the graph has an oblique or slant line asymptote. We find an equation for the asymptote by dividing numerator by denominator to express ƒ as a linear function plus a remainder that goes to zero as x : ; q .
x
x
Solution
sx  3d: 2
We are interested in the behavior as x : ; q . We divide s2x  4d into
–2
x + 1 2
–3
2x  4 x 2  3 x 2  2x 2x  3 2x  4 1
FIGURE 2.58 The graph of the function in Example 10 has an oblique asymptote.
This tells us that ƒsxd =
x2  3 x 1 = ¢ + 1≤ + ¢ ≤. 2x  4 2 2x  4 123 linear g(x)
y
B
You can get as high as you want by taking x close enough to 0. No matter how high B is, the graph goes higher. y 1x
x 0
x
x No matter how low –B is, the graph goes lower.
You can get as low as you want by taking x close enough to 0.
14243 remainder
As x : ; q , the remainder, whose magnitude gives the vertical distance between the graphs of ƒ and g, goes to zero, making the slanted line x + 1 2 an asymptote of the graph of ƒ (Figure 2.58). The line y = g(x) is an asymptote both to the right and to the left. The next subsection will confirm that the function ƒ(x) grows arbitrarily large in absolute value as x : 2 (where the denominator is zero), as shown in the graph. g(x) =
Notice in Example 10 that if the degree of the numerator in a rational function is greater than the degree of the denominator, then the limit as ƒ x ƒ becomes large is + q or  q , depending on the signs assumed by the numerator and denominator.
–B
FIGURE 2.59 Onesided infinite limits: 1 1 lim = q and lim x =  q . x:0 + x x: 0
Infinite Limits Let us look again at the function ƒsxd = 1>x. As x : 0 + , the values of ƒ grow without bound, eventually reaching and surpassing every positive real number. That is, given any positive real number B, however large, the values of ƒ become larger still (Figure 2.59).
2.6
Limits Involving Infinity; Asymptotes of Graphs
103
Thus, ƒ has no limit as x : 0 + . It is nevertheless convenient to describe the behavior of ƒ by saying that ƒ(x) approaches q as x : 0 + . We write 1 lim ƒsxd = lim+ x = q . x:0
x:0 +
In writing this equation, we are not saying that the limit exists. Nor are we saying that there is a real number q , for there is no such number. Rather, we are saying that limx:0+ s1>xd does not exist because 1>x becomes arbitrarily large and positive as x : 0 + . As x : 0  , the values of ƒsxd = 1>x become arbitrarily large and negative. Given any negative real number B, the values of ƒ eventually lie below B. (See Figure 2.59.) We write 1 lim ƒsxd = lim x =  q . x:0
x:0 
y
Again, we are not saying that the limit exists and equals the number  q . There is no real number  q . We are describing the behavior of a function whose limit as x : 0  does not exist because its values become arbitrarily large and negative.
y 1 x1 1
EXAMPLE 11 –1
0
1
2
x
3
Find lim+ x:1
1 x  1
and
lim
x:1 
1 . x  1
Geometric Solution The graph of y = 1>sx  1d is the graph of y = 1>x shifted 1 unit to the right (Figure 2.60). Therefore, y = 1>sx  1d behaves near 1 exactly the way y = 1>x behaves near 0: FIGURE 2.60 Near x = 1 , the function y = 1>sx  1d behaves the way the function y = 1>x behaves near x = 0 . Its graph is the graph of y = 1>x shifted 1 unit to the right (Example 11).
lim
x:1 +
1 = q x  1
lim
x:1 
1 = q. x  1
Think about the number x  1 and its reciprocal. As x : 1+ , we have sx  1d : 0 and 1>sx  1d : q . As x : 1 , we have sx  1d : 0  and 1>sx  1d : q. Analytic Solution +
EXAMPLE 12
Discuss the behavior of ƒsxd =
1 x2
as
x : 0.
As x approaches zero from either side, the values of 1>x 2 are positive and become arbitrarily large (Figure 2.61). This means that
Solution
y No matter how high B is, the graph goes higher.
B
lim ƒsxd = lim
x:0
f(x) 12 x
x 0
and
x
x
FIGURE 2.61 The graph of ƒ (x) in Example 12 approaches infinity as x : 0.
x:0
1 = q. x2
The function y = 1>x shows no consistent behavior as x : 0. We have 1>x : q if x : 0 + , but 1>x :  q if x : 0  . All we can say about limx:0 s1>xd is that it does not exist. The function y = 1>x 2 is different. Its values approach infinity as x approaches zero from either side, so we can say that limx:0 s1>x 2 d = q .
EXAMPLE 13
These examples illustrate that rational functions can behave in various ways near zeros of the denominator. (a) lim
sx  2d2 sx  2d2 x  2 = lim = lim = 0 2 x:2 sx  2dsx + 2d x:2 x + 2 x  4
(b) lim
x  2 x  2 1 1 = lim = lim = 2 x + 2 4 sx 2dsx + 2d x:2 x:2 x  4
x:2
x:2
104
Chapter 2: Limits and Continuity
(c)
lim
x:2 +
(d) limx:2
x  3 x  3 = lim+ = q x:2 sx  2dsx + 2d x2  4
The values are negative for x 7 2, x near 2.
x  3 x  3 = lim= q x:2 sx  2dsx + 2d x2  4
The values are positive for x 6 2, x near 2.
x  3 x  3 = lim does not exist. 2 sx 2dsx + 2d x:2 x  4 sx  2d 2  x 1 (f) lim = lim = lim = q 3 x:2 sx  2d x:2 sx  2d3 x:2 sx  2d2 (e) lim
See parts (c) and (d).
x:2
y
In parts (a) and (b) the effect of the zero in the denominator at x = 2 is canceled because the numerator is zero there also. Thus a finite limit exists. This is not true in part (f ), where cancellation still leaves a zero factor in the denominator.
y f (x)
Precise Definitions of Infinite Limits
B
0
c
c
x c
FIGURE 2.62 For c  d 6 x 6 c + d, the graph of ƒ(x) lies above the line y = B.
y c
c
Instead of requiring ƒ(x) to lie arbitrarily close to a finite number L for all x sufficiently close to c, the definitions of infinite limits require ƒ(x) to lie arbitrarily far from zero. Except for this change, the language is very similar to what we have seen before. Figures 2.62 and 2.63 accompany these definitions.
c
DEFINITIONS 1. We say that ƒ(x) approaches infinity as x approaches c, and write lim ƒsxd = q , x:c
if for every positive real number B there exists a corresponding d 7 0 such that for all x 0 6 ƒx  cƒ 6 d Q ƒsxd 7 B. 2. We say that ƒ(x) approaches minus infinity as x approaches c, and write lim ƒsxd =  q , x:c
x
0
if for every negative real number B there exists a corresponding d 7 0 such that for all x Q ƒsxd 6 B. 0 6 ƒx  cƒ 6 d
–B
The precise definitions of onesided infinite limits at c are similar and are stated in the exercises. y f (x)
EXAMPLE 14 FIGURE 2.63 For c  d 6 x 6 c + d, the graph of ƒ(x) lies below the line y = B.
Solution
Prove that lim
x:0
1 = q. x2
Given B 7 0, we want to find d 7 0 such that 0 6 ƒx  0ƒ 6 d
implies
1 7 B. x2
Now, 1 7 B x2
if and only if x 2 6
or, equivalently, ƒxƒ 6
1 . 2B
1 B
2.6
Limits Involving Infinity; Asymptotes of Graphs
105
Thus, choosing d = 1> 2B (or any smaller positive number), we see that ƒxƒ 6 d
implies
1 1 7 2 Ú B. x2 d
Therefore, by definition, 1 = q. x2
lim
x:0
Vertical Asymptotes
y
Notice that the distance between a point on the graph of ƒsxd = 1>x and the yaxis approaches zero as the point moves vertically along the graph and away from the origin (Figure 2.64). The function ƒ(x) = 1>x is unbounded as x approaches 0 because
Vertical asymptote
Horizontal asymptote
y 1x
1 lim x = q
0
x
1
and
x:0 +
1 Horizontal asymptote, y0
Vertical asymptote, x0
We say that the line x = 0 (the yaxis) is a vertical asymptote of the graph of ƒ(x) = 1>x. Observe that the denominator is zero at x = 0 and the function is undefined there.
DEFINITION A line x = a is a vertical asymptote of the graph of a function y = ƒsxd if either
FIGURE 2.64 The coordinate axes are asymptotes of both branches of the hyperbola y = 1>x .
lim ƒsxd = ; q
or
x:a +
EXAMPLE 15
We are interested in the behavior as x : ; q and the behavior as x : 2, where the denominator is zero. The asymptotes are quickly revealed if we recast the rational function as a polynomial with a remainder, by dividing sx + 2d into sx + 3d:
6 4
Horizontal asymptote, y1
3
y
x3 x2
1
1 x2
1
3
1 x + 2 x + 3 x + 2 1
2 1
–5 –4 –3 –2 –1 0 –1
x + 3 . x + 2
Solution
y
5
lim ƒsxd = ; q .
x:a 
Find the horizontal and vertical asymptotes of the curve y =
Vertical asymptote, x –2
1 lim x =  q .
x:0 
2
–2 –3 –4
FIGURE 2.65 The lines y = 1 and x =  2 are asymptotes of the curve in Example 15.
x
This result enables us to rewrite y as: y = 1 +
1 . x + 2
As x : ; q , the curve approaches the horizontal asymptote y = 1; as x : 2, the curve approaches the vertical asymptote x = 2. We see that the curve in question is the graph of ƒ(x) = 1>x shifted 1 unit up and 2 units left (Figure 2.65). The asymptotes, instead of being the coordinate axes, are now the lines y = 1 and x = 2.
106
Chapter 2: Limits and Continuity y
EXAMPLE 16
8 7 6 5 4 3 2 1
Vertical asymptote, x –2
Find the horizontal and vertical asymptotes of the graph of
y – 28 x 4
ƒsxd = 
Vertical asymptote, x 2 Horizontal asymptote, y 0
–4 –3 –2 –1 0
1 2 3 4
x
8 . x2  4
Solution We are interested in the behavior as x : ; q and as x : ;2, where the denominator is zero. Notice that ƒ is an even function of x, so its graph is symmetric with respect to the yaxis. (a) The behavior as x : ; q . Since limx: q ƒsxd = 0, the line y = 0 is a horizontal asymptote of the graph to the right. By symmetry it is an asymptote to the left as well (Figure 2.66). Notice that the curve approaches the xaxis from only the negative side (or from below). Also, ƒs0d = 2.
(b) The behavior as x : ;2. Since lim ƒsxd =  q
x:2 +
FIGURE 2.66 Graph of the function in Example 16. Notice that the curve approaches the xaxis from only one side. Asymptotes do not have to be twosided.
y
lim ƒsxd = q ,
x:2 
the line x = 2 is a vertical asymptote both from the right and from the left. By symmetry, the line x = 2 is also a vertical asymptote. There are no other asymptotes because ƒ has a finite limit at every other point.
EXAMPLE 17 The graph of the natural logarithm function has the yaxis (the line x = 0) as a vertical asymptote. We see this from the graph sketched in Figure 2.67 (which is the reflection of the graph of the natural exponential function across the line y = x) and the fact that the xaxis is a horizontal asymptote of y = e x (Example 5). Thus,
y ex
4
lim ln x =  q .
3
x:0 +
2
y ln x
The same result is true for y = loga x whenever a 7 1.
1 –1
and
1
2
3
4
x
–1
FIGURE 2.67 The line x = 0 is a vertical asymptote of the natural logarithm function (Example 17).
EXAMPLE 18
The curves 1 y = sec x = cos x
and
sin x y = tan x = cos x
both have vertical asymptotes at oddinteger multiples of p>2, where cos x = 0 (Figure 2.68). y
y
y sec x
1 – 3 – – 2 2
0
y tan x
1 2
3 2
x
0 – 3 – – –1 2 2 2
3 2
FIGURE 2.68 The graphs of sec x and tan x have infinitely many vertical asymptotes (Example 18).
Dominant Terms In Example 10 we saw that by long division we could rewrite the function ƒ(x) =
x2  3 2x  4
x
2.6
Limits Involving Infinity; Asymptotes of Graphs
107
as a linear function plus a remainder term: ƒ(x) = a
x 1 + 1b + a b. 2 2x  4
This tells us immediately that y 20 15 10 f (x) 5 –2
–1
0
g(x) 3x 4 1
2
x
ƒsxd L
x + 1 2
For x large,
ƒsxd L
1 2x  4
For x near 2, this term is very large.
1 is near 0. 2x  4
If we want to know how ƒ behaves, this is the way to find out. It behaves like y = sx>2d + 1 when x is large and the contribution of 1>s2x  4d to the total value of ƒ is insignificant. It behaves like 1>s2x  4d when x is so close to 2 that 1>s2x  4d makes the dominant contribution. We say that sx>2d + 1 dominates when x is numerically large, and we say that 1>s2x  4d dominates when x is near 2. Dominant terms like these help us predict a function’s behavior.
–5
Let ƒsxd = 3x 4  2x 3 + 3x 2  5x + 6 and g sxd = 3x 4 . Show that although ƒ and g are quite different for numerically small values of x, they are virtually identical for ƒ x ƒ very large, in the sense that their ratios approach 1 as x : q or x :  q.
EXAMPLE 19
(a) y 500,000
The graphs of ƒ and g behave quite differently near the origin (Figure 2.69a), but appear as virtually identical on a larger scale (Figure 2.69b). We can test that the term 3x 4 in ƒ, represented graphically by g, dominates the polynomial ƒ for numerically large values of x by examining the ratio of the two functions as x : ; q . We find that
Solution 300,000
100,000 –20
–10
0
10
20
x
ƒsxd = x: ; q gsxd lim
–100,000 (b)
FIGURE 2.69 The graphs of ƒ and g are (a) distinct for ƒ x ƒ small, and (b) nearly identical for ƒ x ƒ large (Example 19).
=
3x 4  2x 3 + 3x 2  5x + 6 x: ; q 3x 4 lim
lim a1 
x: ; q
5 2 2 1 + 4b + 2 3 3x x 3x x
= 1, which means that ƒ and g appear nearly identical when ƒ x ƒ is large.
Summary In this chapter we presented several important calculus ideas that are made meaningful and precise by the concept of the limit. These include the three ideas of the exact rate of change of a function, the slope of the graph of a function at a point, and the continuity of a function. The primary methods used for calculating limits of many functions are captured in the algebraic limit laws of Theorem 1 and in the Sandwich Theorem, all of which are proved from the precise definition of the limit. We saw that these computational rules also apply to onesided limits and to limits at infinity. Moreover, we can sometimes apply these rules when calculating limits of simple transcendental functions, as illustrated by our examples or in cases like the following: lim
x:0
ex  1 ex  1 1 1 1 = lim x = = . = lim x x 2x 1 + 1 2 (e 1)(e + 1) e + 1 x:0 x:0 e  1
108
Chapter 2: Limits and Continuity
However, calculating more complicated limits involving transcendental functions such as x
x ln x 1 and lim a1 + x b , , lim x:0 e 2x  1 x:0 x x:0 requires more than simple algebraic techniques. The derivative is exactly the tool we need to calculate limits such as these (see Section 4.5), and this notion is the main subject of our next chapter. lim
Exercises 2.6 Finding Limits 1. For the function ƒ whose graph is given, determine the following limits. b.
d. lim ƒ(x)
e. lim+ ƒ(x)
f. lim ƒ(x)
g. lim ƒ(x)
h. lim ƒ(x)
i.
x: 2
x: 3
x:0
lim + ƒ(x)
c.
x: 3 x: 0
x: 3 x: 0
x: q
sin 2x x 2  t + sin t 11. lim t + cos t t:  q 9. lim
x: q
3 2
f
1 –6 –5 –4 –3 –2 –1 –1
1
2
3
4 5
6
x
–2 –3
2. For the function ƒ whose graph is given, determine the following limits. c. lim ƒ(x)
a. lim ƒ(x)
b. lim+ ƒ(x)
d. lim ƒ(x)
e.
g. lim ƒ(x)
h. lim+ ƒ(x)
i. lim ƒ(x)
j. lim ƒ(x)
k. lim ƒ(x)
l.
x: 2
x: 3 x: 0
x: 2
x: 2
lim + ƒ(x)
f.
x: 3 x: 0
2
3
2x + 3 5x + 7
14. ƒsxd =
2x 3 + 7 x  x2 + x + 7
15. ƒsxd =
x + 1 x2 + 3
16. ƒsxd =
3x + 7 x2  2
17. hsxd =
7x 3 x  3x 2 + 6x
18. g sxd =
1 x 3  4x + 1
19. g sxd =
10x 5 + x 4 + 31 x6
20. hsxd =
9x 4 + x 2x + 5x 2  x + 6
21. hsxd =
2x 3  2x + 3 3x 3 + 3x 2  5x
22. hsxd =
x 4 x 4  7x 3 + 7x 2 + 9
4 5
6
x
23. lim
x: q
–2
25.
–3
3
8x 2  3 A 2x 2 + x
lim ¢
x:  q
x: q
In Exercises 3–8, find the limit of each function (a) as x : q and (b) as x :  q . (You may wish to visualize your answer with a graphing calculator or computer.) 4. ƒsxd = p 
2 x2
1  x3 ≤ x 2 + 7x
lim
x:  q
31. lim
x: q
5
3
4
5
3 2 x + 2x
2x 5>3  x 1>3 + 7 x
8>5
lim ¢
x:  q
+ 3x + 2x
x2 + x  1 ≤ 8x 2  3
x 2  5x x: q A x + x  2
26. lim 28. lim
x: q
3
2 + 2x 2  2x
5
2x  2x 3
29.
24.
22x + x 1 3x  7
27. lim
2 3. ƒsxd = x  3
lim
u: q
Limits as x : ˆ or x : ˆ The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of x: divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits in Exercises 23–36.
1 1
cos u 3u r + sin r 12. lim r: q 2r + 7  5 sin r 10.
lim ƒ(x)
x:  q
f
2
4 + (22>x 2)
13. ƒsxd =
lim ƒ(x)
x: 3 
y 3
3  s2>xd
8. hsxd =
Limits of Rational Functions In Exercises 13–22, find the limit of each rational function (a) as x : q and (b) as x :  q .
x: 0
x: q
–6 –5 –4 –3 –2 –1 –1
3  s1>x 2 d
1 8  s5>x 2 d
6. g sxd =
Find the limits in Exercises 9–12.
lim ƒ(x)
x:  q
y
x: 4
5 + s7>xd
7. hsxd =
lim  ƒ(x)
a. lim ƒ(x)
1 2 + s1>xd
5. g sxd =
30. lim
x: q
32.
lim
x:  q
x 1 + x 4 x 2  x 3 3 2 x  5x + 3 2x + x 2>3  4
1>3
2.6
33. lim
x: q
35. lim
x: q
2x 2 + 1 x + 1
34.
x  3
36.
24x 2 + 25
2x 2 + 1 x + 1
lim
x:  q
4  3x 3
lim
x:  q
2x 6 + 9
Infinite Limits Find the limits in Exercises 37–48. 5 1 37. lim+ 38. limx: 0 3x x: 0 2x 3 1 39. lim40. lim+ x: 2 x  2 x: 3 x  3 3x 2x 41. lim + 42. lim x: 8 x + 8 x: 5 2x + 10 4 sx  7d2 2 45. a. lim+ 1>3 x: 0 3x 2 46. a. lim+ 1>5 x: 0 x 4 47. lim 2>5 x: 0 x 43. lim
44. lim
x: 7
x: 0
1 x 2sx + 1d
51.
lim
x: sp>2d
60. lim a
1 + 7b as t 3>5 a. t : 0 +
tan x 
48. lim
x: 0
lim s1 + csc ud
u :0 
50.
1 x 2>3
lim
x: sp>2d
sec x +
52. lim s2  cot ud u: 0
Find the limits in Exercises 53–58. 1 as 53. lim 2 x  4 a. x : 2+ b. c. x : 2+ d. x as 54. lim 2 x  1 a. x : 1+ b. c. x : 1+ d.
61. lim a
1 2 + b as x 2>3 sx  1d2>3 a. x : 0 + b. x : 0 c. x : 1+
x : 2
x : 2
x : 1x : 1
x 1 55. lim a  x b as 2
1 1 b as x 1>3 sx  1d4>3 a. x : 0 + b. x : 0 d. x : 1
Graphing Simple Rational Functions Graph the rational functions in Exercises 63–68. Include the graphs and equations of the asymptotes and dominant terms. 1 1 63. y = 64. y = x  1 x + 1 65. y =
1 2x + 4
66. y =
3 x  3
67. y =
x + 3 x + 2
68. y =
2x x + 1
Inventing Graphs and Functions In Exercises 69–72, sketch the graph of a function y = ƒsxd that satisfies the given conditions. No formulas are required—just label the coordinate axes and sketch an appropriate graph. (The answers are not unique, so your graphs may not be exactly like those in the answer section.) 69. ƒs0d = 0, ƒs1d = 2, ƒs 1d = 2, lim ƒsxd = 1, and x:  q lim ƒsxd = 1 x: q
70. ƒs0d = 0, lim ƒsxd = 0, lim+ ƒsxd = 2, and x: ; q
x:0
lim ƒsxd = 2
x:0
b. x : 0 d. x : 1
71. ƒs0d = 0, lim ƒsxd = 0, lim ƒsxd = lim + ƒsxd = q , x: ; q
x: 1
x:1
lim+ ƒsxd =  q , and lim  ƒsxd =  q x: 1
x:1
72. ƒs2d = 1, ƒs 1d = 0, lim ƒsxd = 0, lim+ ƒsxd = q ,
x  1 as 2x + 4 + a. x : 2 c. x : 1+
b. x : 2 d. x : 0 
x 2  3x + 2 as 57. lim x 3  2x 2 + a. x : 0 c. x : 2
b. x : 2+ d. x : 2
2
x: q
56. lim
e. What, if anything, can be said about the limit as x : 0 ?
x:  q
x:0
In Exercises 73–76, find a function that satisfies the given conditions and sketch its graph. (The answers here are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.) 73. lim ƒsxd = 0, lim ƒsxd = q , and lim+ ƒsxd = q x: ; q
74.
58. lim
b. x : 2+ d. x : 1+
x:0
lim ƒsxd =  q , and lim ƒsxd = 1

e. What, if anything, can be said about the limit as x : 0 ? x 2  3x + 2 as x 3  4x a. x : 2+ c. x : 0 
d. x : 1
62. lim a
2
a. x : 0 + 3 2 c. x : 2
b. t : 0 
c. x : 1+
2 x: 0 3x 1>3 2 b. lim 1>5 x: 0 x
109
Find the limits in Exercises 59–62. 3 59. lim a2  1>3 b as t a. t : 0 + b. t : 0 
b. lim
Find the limits in Exercises 49–52. 49.
Limits Involving Infinity; Asymptotes of Graphs
75.
x:2
x:2
lim g sxd = 0, lim g sxd =  q , and lim+ g sxd = q
x: ; q
x:3
x: 3
lim hsxd = 1, lim hsxd = 1, lim hsxd = 1, and
x:  q
x: q
x:0
lim+ hsxd = 1
x:0
76.
lim k sxd = 1, lim k sxd = q , and lim+ k sxd =  q
x: ; q
x:1
x:1
110
Chapter 2: Limits and Continuity
77. Suppose that ƒ(x) and g(x) are polynomials in x and that limx: q sƒsxd>g sxdd = 2 . Can you conclude anything about limx:  q sƒsxd>g sxdd ? Give reasons for your answer.
Modify the definition to cover the following cases. a. lim ƒsxd = q x:c
b. lim+ ƒsxd =  q
78. Suppose that ƒ(x) and g(x) are polynomials in x. Can the graph of ƒ(x)>g (x) have an asymptote if g(x) is never zero? Give reasons for your answer. 79. How many horizontal asymptotes can the graph of a given rational function have? Give reasons for your answer. Finding Limits of Differences when x : ; ˆ Find the limits in Exercises 80–86. 80. lim
x: q
81. lim
x: q
82. 83.
A 2x + 9  2x + 4 B
lim
x:  q
lim
x:  q
84. lim
x: q
85. lim
x: q
c. lim ƒsxd =  q x:c
Use the formal definitions from Exercise 93 to prove the limit statements in Exercises 94–98. 1 1 94. lim+ x = q 95. lim x =  q x:0 x:0 1 = q x  2 1 = q 98. limx:1 1  x 2 x:2
2
A 2x 2 + 3 + x B
A 2x + 24x 2 + 3x  2 B
A 29x  x  3x B
A 2x 2 + 3x  2x 2  2x B
86. lim A 2x 2 + x  2x 2  x B x: q
x2  4 x  1 x2  1 103. y = x
87. If ƒ has the constant value ƒsxd = k , then lim ƒsxd = k . x: q
88. If ƒ has the constant value ƒsxd = k , then lim ƒsxd = k . x:  q
Use formal definitions to prove the limit statements in Exercises 89–92. 1 1 = q 89. lim 2 =  q 90. lim x:0 x x: 0 ƒ x ƒ 2 1 = q = q 91. lim 92. lim x:3 sx  3d2 x: 5 sx + 5d2 93. Here is the definition of infinite righthand limit.
x2  1 2x + 4 x3 + 1 104. y = x2
Additional Graphing Exercises T Graph the curves in Exercises 105–108. Explain the relationship between the curve’s formula and what you see. 105. y =
ƒsxd 7 B.
106. y =
1
24  x 2 p 108. y = sin a 2 b x + 1
T Graph the functions in Exercises 109 and 110. Then answer the following questions. a. How does the graph behave as x : 0 +? b. How does the graph behave as x : ; q ? c. How does the graph behave near x = 1 and x = 1?
x: c
if, for every positive real number B, there exists a corresponding number d 7 0 such that for all x
x
24  x 2 1 107. y = x 2>3 + 1>3 x
We say that ƒ(x) approaches infinity as x approaches c from the right, and write lim+ ƒsxd = q ,
Q
1 = q x  2
102. y =
101. y =
Using the Formal Definitions Use the formal definitions of limits as x : ; q to establish the limits in Exercises 87 and 88.
Chapter 2
x:2
Oblique Asymptotes Graph the rational functions in Exercises 99–104. Include the graphs and equations of the asymptotes. x2 + 1 x2 99. y = 100. y = x  1 x  1
2
c 6 x 6 c + d
97. lim+
96. lim
A 2x + 25  2x  1 B 2
x:c
Give reasons for your answers. 109. y =
3 1 ax  x b 2
2>3
110. y =
3 x a b 2 x  1
Questions to Guide Your Review
1. What is the average rate of change of the function g(t) over the interval from t = a to t = b ? How is it related to a secant line? 2. What limit must be calculated to find the rate of change of a function g(t) at t = t0 ?
3. Give an informal or intuitive definition of the limit lim ƒsxd = L.
x:c
Why is the definition “informal”? Give examples.
2>3
Chapter 2
Practice Exercises
111
4. Does the existence and value of the limit of a function ƒ(x) as x approaches c ever depend on what happens at x = c ? Explain and give examples.
13. What does it mean for a function to be rightcontinuous at a point? Leftcontinuous? How are continuity and onesided continuity related?
5. What function behaviors might occur for which the limit may fail to exist? Give examples.
14. What does it mean for a function to be continuous on an interval? Give examples to illustrate the fact that a function that is not continuous on its entire domain may still be continuous on selected intervals within the domain.
6. What theorems are available for calculating limits? Give examples of how the theorems are used. 7. How are onesided limits related to limits? How can this relationship sometimes be used to calculate a limit or prove it does not exist? Give examples. 8. What is the value of lim u:0 sssin ud>ud ? Does it matter whether u is measured in degrees or radians? Explain. 9. What exactly does limx:c ƒsxd = L mean? Give an example in which you find a d 7 0 for a given ƒ, L, c , and P 7 0 in the precise definition of limit. 10. Give precise definitions of the following statements. a. limx:2 ƒsxd = 5 c. limx:2 ƒsxd = q
b. limx:2+ ƒsxd = 5 d. limx:2 ƒsxd =  q
11. What conditions must be satisfied by a function if it is to be continuous at an interior point of its domain? At an endpoint? 12. How can looking at the graph of a function help you tell where the function is continuous?
Chapter 2
1, x, ƒsxd = e 1, x, 1,
17. Under what circumstances can you extend a function ƒ(x) to be continuous at a point x = c ? Give an example. 18. What exactly do limx: q ƒsxd = L and limx: q ƒsxd = L mean? Give examples. 19. What are limx:; q k (k a constant) and limx:; q s1>xd ? How do you extend these results to other functions? Give examples. 20. How do you find the limit of a rational function as x : ; q ? Give examples. 21. What are horizontal and vertical asymptotes? Give examples.
e. cos (g(t)) g. ƒstd + g std x 1 x 0 x
… 6 = 6 Ú
1 x 6 0 0 x 6 1 1.
Then discuss, in detail, limits, onesided limits, continuity, and onesided continuity of ƒ at x = 1, 0 , and 1. Are any of the discontinuities removable? Explain. 2. Repeat the instructions of Exercise 1 for 0, 1>x, ƒsxd = d 0, 1,
x 0 x x
… 6 = 7
1 ƒxƒ 6 1 1 1.
3. Suppose that ƒ(t) and g(t) are defined for all t and that limt:t0 ƒstd = 7 and limt:t0 g std = 0 . Find the limit as t : t0 of the following functions. c. ƒstd # g std
16. What does it mean for a function to have the Intermediate Value Property? What conditions guarantee that a function has this property over an interval? What are the consequences for graphing and solving the equation ƒsxd = 0 ?
Practice Exercises
Limits and Continuity 1. Graph the function
a. 3ƒ(t)
15. What are the basic types of discontinuity? Give an example of each. What is a removable discontinuity? Give an example.
b. sƒstdd2 ƒstd d. g std  7
f. ƒ ƒstd ƒ h. 1>ƒ(t)
4. Suppose the functions ƒ(x) and g(x) are defined for all x and that limx:0 ƒsxd = 1>2 and limx:0 g sxd = 22 . Find the limits as x : 0 of the following functions. a. g sxd c. ƒsxd + g sxd e. x + ƒsxd
b. g sxd # ƒsxd
d. 1>ƒ(x) ƒsxd # cos x f. x  1
In Exercises 5 and 6, find the value that limx:0 g sxd must have if the given limit statements hold. 4  g sxd b = 1 5. lim a x x:0 6. lim ax lim g sxdb = 2 x: 4
x:0
7. On what intervals are the following functions continuous? a. ƒsxd = x 1>3
b. g sxd = x 3>4
c. hsxd = x 2>3
d. ksxd = x 1>6
8. On what intervals are the following functions continuous? a. ƒsxd = tan x
b. g sxd = csc x
cos x c. hsxd = x  p
d. ksxd =
sin x x
112
Chapter 2: Limits and Continuity
Finding Limits In Exercises 9–28, find the limit or explain why it does not exist. x 2  4x + 4 x + 5x 2  14x a. as x : 0
9. lim
3
x2 + x 10. lim 5 x + 2x 4 + x 3 a. as x : 0 11. lim
x:1
1  2x 1  x
x:0
s2 + xd3  8 x x: 0
16. lim
x 2>3  16
18. lim
2x  1 tan (2x) 19. lim x:0 tan (px)
23. lim
x  a x4  a4 sx + hd2  x 2 14. lim h x: 0 2
x: 64
2x  8
19  1b A 27
1>3
 a
19 + 1b A 27
1>3
.
Evaluate this exact answer and compare it with the value you found in part (b).
x: a
x:1
x:p
a
b. as x : 1 12. lim
x 1>3  1
21. lim sin a
c. It can be shown that the exact value of the solution in part (b) is
2
sx + hd2  x 2 13. lim h h :0 1 1 2 + x 2 15. lim x x:0 17. lim
b. as x : 2
3 T 34. Let ƒsud = u  2u + 2 . a. Use the Intermediate Value Theorem to show that ƒ has a zero between 2 and 0. b. Solve the equation ƒsud = 0 graphically with an error of magnitude at most 10 4 .
Continuous Extension 35. Can ƒsxd = xsx 2  1d> ƒ x 2  1 ƒ be extended to be continuous at x = 1 or 1 ? Give reasons for your answers. (Graph the function—you will find the graph interesting.) 36. Explain why the function ƒsxd = sin s1>xd has no continuous extension to x = 0 .
20. lim csc x x: p 
x + sin x b 2
8x 3 sin x  x
22. lim cos2 sx  tan xd x: p
24. lim
x: 0
cos 2x  1 sin x
25. lim+ ln (t  3)
26. lim t 2 ln A 2  2t B
27. lim+ 2u ecos (p>u)
28. lim+
t:3
t: 1
u: 0
z: 0
2e1>z 1>z e + 1
In Exercises 29–32, find the limit of g(x) as x approaches the indicated value. 29. lim+ s4g sxdd1>3 = 2
T In Exercises 37–40, graph the function to see whether it appears to have a continuous extension to the given point a. If it does, use Trace and Zoom to find a good candidate for the extended function’s value at a. If the function does not appear to have a continuous extension, can it be extended to be continuous from the right or left? If so, what do you think the extended function’s value should be? x  1 , a = 1 4 x  2x 5 cos u 38. g sud = , a = p>2 4u  2p 37. ƒsxd =
39. hstd = s1 + ƒ t ƒ d1>t, a = 0 x 40. k sxd = , a = 0 1  2ƒ x ƒ
x:0
30.
lim
x: 25
31. lim
x:1
32. lim
1 = 2 x + g sxd
3x 2 + 1 = q g sxd
x: 2
5  x2 2g sxd
= 0
Limits at Infinity Find the limits in Exercises 41–54. 2x + 3 41. lim 42. x: q 5x + 7 43. 45.
T Roots 33. Let ƒsxd = x 3  x  1 . a. Use the Intermediate Value Theorem to show that ƒ has a zero between 1 and 2. b. Solve the equation ƒsxd = 0 graphically with an error of magnitude at most 10 8 . c. It can be shown that the exact value of the solution in part (b) is a
269 1>3 269 1>3 1 1 + a + b b . 2 18 2 18
Evaluate this exact answer and compare it with the value you found in part (b).
lim
x:  q
x 2  4x + 8 3x 3
x 2  7x x:  q x + 1 lim
47. lim
x: q
sin x :x;
cos u  1 u u: q
48. lim
lim
x:  q
44. lim
x: q
2x2 + 3 5x 2 + 7 1 x 2  7x + 1
x4 + x3 x: q 12x 3 + 128 sIf you have a grapher, try graphing the function for 5 … x … 5.d 46. lim
sIf you have a grapher, try graphing ƒsxd = xscos s1>xd  1d near the origin to “see” the limit at infinity.d
x + sin x + 22x x + sin x x: q
x 2>3 + x 1 x 2>3 + cos2 x
49. lim
50. lim
1 51. lim e1>x cos x x: q
1 52. lim ln a1 + t b t: q
53.
lim tan1 x
x:  q
x: q
54.
1 lim e3t sin1 t
t:  q
Chapter 2 Horizontal and Vertical Asymptotes 55. Use limits to determine the equations for all vertical asymptotes. x2 + 4 a. y = x  3
x2  x  2 b. ƒ(x) = 2 x  2x + 1
x2 + x  6 c. y = 2 x + 2x  8
Chapter 2 T
113
56. Use limits to determine the equations for all horizontal asymptotes. a. y =
1  x2 x2 + 1
c. g(x) =
b. ƒ(x) =
2x + 4 x 2
d. y =
2x + 4 2x + 4
x2 + 9 A 9x 2 + 1
Additional and Advanced Exercises
1. Assigning a value to 00 The rules of exponents tell us that a 0 = 1 if a is any number different from zero. They also tell us that 0 n = 0 if n is any positive number. If we tried to extend these rules to include the case 0 0 , we would get conflicting results. The first rule would say 0 0 = 1 , whereas the second would say 0 0 = 0 . We are not dealing with a question of right or wrong here. Neither rule applies as it stands, so there is no contradiction. We could, in fact, define 0 0 to have any value we wanted as long as we could persuade others to agree. What value would you like 0 0 to have? Here is an example that might help you to decide. (See Exercise 2 below for another example.) a. Calculate x x for x = 0.1 , 0.01, 0.001, and so on as far as your calculator can go. Record the values you get. What pattern do you see? b. Graph the function y = x x for 0 6 x … 1 . Even though the function is not defined for x … 0 , the graph will approach the yaxis from the right. Toward what yvalue does it seem to be headed? Zoom in to further support your idea.
T
Additional and Advanced Exercises
2. A reason you might want 00 to be something other than 0 or 1 As the number x increases through positive values, the numbers 1>x and 1> (ln x) both approach zero. What happens to the number 1 ƒsxd = a x b
1>sln xd
as x increases? Here are two ways to find out. a. Evaluate ƒ for x = 10 , 100, 1000, and so on as far as your calculator can reasonably go. What pattern do you see? b. Graph ƒ in a variety of graphing windows, including windows that contain the origin. What do you see? Trace the yvalues along the graph. What do you find? 3. Lorentz contraction In relativity theory, the length of an object, say a rocket, appears to an observer to depend on the speed at which the object is traveling with respect to the observer. If the observer measures the rocket’s length as L0 at rest, then at speed y the length will appear to be L = L0
1 
B
y2 . c2
This equation is the Lorentz contraction formula. Here, c is the speed of light in a vacuum, about 3 * 10 8 m>sec . What happens to L as y increases? Find limy:c L . Why was the lefthand limit needed?
4. Controlling the flow from a draining tank Torricelli’s law says that if you drain a tank like the one in the figure shown, the rate y at which water runs out is a constant times the square root of the water’s depth x. The constant depends on the size and shape of the exit valve.
x Exit rate y ft 3min
Suppose that y = 2x>2 for a certain tank. You are trying to maintain a fairly constant exit rate by adding water to the tank with a hose from time to time. How deep must you keep the water if you want to maintain the exit rate a. within 0.2 ft3>min of the rate y0 = 1 ft3>min ?
b. within 0.1 ft3>min of the rate y0 = 1 ft3>min ? 5. Thermal expansion in precise equipment As you may know, most metals expand when heated and contract when cooled. The dimensions of a piece of laboratory equipment are sometimes so critical that the shop where the equipment is made must be held at the same temperature as the laboratory where the equipment is to be used. A typical aluminum bar that is 10 cm wide at 70°F will be y = 10 + st  70d * 10 4 centimeters wide at a nearby temperature t. Suppose that you are using a bar like this in a gravity wave detector, where its width must stay within 0.0005 cm of the ideal 10 cm. How close to t0 = 70°F must you maintain the temperature to ensure that this tolerance is not exceeded? 6. Stripes on a measuring cup The interior of a typical 1L measuring cup is a right circular cylinder of radius 6 cm (see accompanying figure). The volume of water we put in the cup is therefore a function of the level h to which the cup is filled, the formula being V = p62h = 36ph . How closely must we measure h to measure out 1 L of water s1000 cm3 d with an error of no more than 1% s10 cm3 d ?
114
Chapter 2: Limits and Continuity 17. A function continuous at only one point ƒsxd = e
Let
x, if x is rational 0, if x is irrational.
a. Show that ƒ is continuous at x = 0 . Stripes about 1 mm wide
b. Use the fact that every nonempty open interval of real numbers contains both rational and irrational numbers to show that ƒ is not continuous at any nonzero value of x. 18. The Dirichlet ruler function If x is a rational number, then x can be written in a unique way as a quotient of integers m>n where n 7 0 and m and n have no common factors greater than 1. (We say that such a fraction is in lowest terms. For example, 6>4 written in lowest terms is 3>2.) Let ƒ(x) be defined for all x in the interval [0, 1] by
(a) r 6 cm
ƒsxd = e Liquid volume V 36h
h
1>n, 0,
if x = m>n is a rational number in lowest terms if x is irrational.
For instance, ƒs0d = ƒs1d = 1, ƒs1>2d = 1>2, ƒs1>3d = ƒ(2>3) = 1>3, ƒs1>4d = ƒs3>4d = 1>4 , and so on. a. Show that ƒ is discontinuous at every rational number in [0, 1].
(b)
A 1L measuring cup (a), modeled as a right circular cylinder (b) of radius r = 6 cm Precise Definition of Limit In Exercises 7–10, use the formal definition of limit to prove that the function is continuous at c. 7. ƒsxd = x2  7,
8. g sxd = 1>s2xd,
c = 1
9. hsxd = 22x  3,
c = 1>4
c = 2 10. Fsxd = 29  x,
c = 5
11. Uniqueness of limits Show that a function cannot have two different limits at the same point. That is, if limx:c ƒsxd = L1 and limx:c ƒsxd = L2 , then L1 = L2 . 12. Prove the limit Constant Multiple Rule:
13. Onesided limits find
x: c
If limx:0 ƒsxd = A and limx:0 ƒsxd = B , +

a. limx:0+ ƒsx 3  xd
b. limx:0 ƒsx 3  xd
c. limx:0+ ƒsx 2  x 4 d
d. limx:0 ƒsx 2  x 4 d
a. If limx:c ƒsxd exists but limx:c g sxd does not exist, then limx:csƒsxd + g sxdd does not exist. b. If neither limx:c ƒsxd nor limx:c g sxd exists, then limx:c sƒsxd + g sxdd does not exist. c. If ƒ is continuous at x, then so is ƒ ƒ ƒ . d. If ƒ ƒ ƒ is continuous at c, then so is ƒ. In Exercises 15 and 16, use the formal definition of limit to prove that the function has a continuous extension to the given value of x. x2  1 , x + 1
x = 1
16. g sxd =
19. Antipodal points Is there any reason to believe that there is always a pair of antipodal (diametrically opposite) points on Earth’s equator where the temperatures are the same? Explain. 20. If lim sƒsxd + g sxdd = 3 and lim sƒsxd  g sxdd = 1 , find x:c
x 2  2x  3 , 2x  6
x:c
lim ƒsxdg sxd .
x:c
21. Roots of a quadratic equation that is almost linear The equation ax 2 + 2x  1 = 0 , where a is a constant, has two roots if a 7 1 and a Z 0 , one positive and one negative: 1 + 21 + a , a
rsad =
1  21 + a . a
a. What happens to r+sad as a : 0 ? As a : 1+ ? b. What happens to rsad as a : 0 ? As a : 1+ ? c. Support your conclusions by graphing r+sad and rsad as functions of a. Describe what you see.
14. Limits and continuity Which of the following statements are true, and which are false? If true, say why; if false, give a counterexample (that is, an example confirming the falsehood).
15. ƒsxd =
c. Sketch the graph of ƒ. Why do you think ƒ is called the “ruler function”?
r+sad =
lim kƒsxd = k lim ƒsxd for any constant k .
x:c
b. Show that ƒ is continuous at every irrational number in [0, 1]. (Hint: If P is a given positive number, show that there are only finitely many rational numbers r in [0, 1] such that ƒsrd Ú P .)
x = 3
d. For added support, graph ƒsxd = ax 2 + 2x  1 simultaneously for a = 1, 0.5, 0.2, 0.1, and 0.05. 22. Root of an equation Show that the equation x + 2 cos x = 0 has at least one solution. 23. Bounded functions A realvalued function ƒ is bounded from above on a set D if there exists a number N such that ƒsxd … N for all x in D. We call N, when it exists, an upper bound for ƒ on D and say that ƒ is bounded from above by N. In a similar manner, we say that ƒ is bounded from below on D if there exists a number M such that ƒsxd Ú M for all x in D. We call M, when it exists, a lower bound for ƒ on D and say that ƒ is bounded from below by M. We say that ƒ is bounded on D if it is bounded from both above and below. a. Show that ƒ is bounded on D if and only if there exists a number B such that ƒ ƒsxd ƒ … B for all x in D.
Chapter 2 b. Suppose that ƒ is bounded from above by N. Show that if limx:c ƒsxd = L , then L … N . c. Suppose that ƒ is bounded from below by M. Show that if limx:c ƒsxd = L , then L Ú M .
24. Max 5a, b6 and min 5a, b6 a. Show that the expression
ƒa  bƒ a + b + 2 2 equals a if a Ú b and equals b if b Ú a . In other words, max {a, b} gives the larger of the two numbers a and b. max 5a, b6 =
b. Find a similar expression for min 5a, b6, the smaller of a and b.
sin U U The formula limu:0ssin ud>u = 1 can be generalized. If limx:c ƒsxd = 0 and ƒ(x) is never zero in an open interval containing the point x = c, except possibly c itself, then sin ƒsxd lim = 1. x: c ƒsxd Here are several examples.
sin sx 2  x  2d sin sx 2  x  2d # = lim x + 1 x: 1 x: 1 sx 2  x  2d sx 2  x  2d sx + 1dsx  2d = 1 # lim = 3 x + 1 x + 1 x: 1 x: 1 lim
d. lim
sin A 1  2x B x  1
x:1
1 # lim
x:1
x:1
sin A 1  2x B 1  2x = x  1 1  2x = lim
sx  1d A 1 + 2x B
x:1
1  x
sx  1d A 1 + 2x B
Find the limits in Exercises 25–30. 25. lim
sin s1  cos xd x
sin ssin xd 27. lim x x:0 sin sx 2  4d x  2 x:2
29. lim
26. lim+ x:0
sin x sin 2x
sin sx 2 + xd 28. lim x x:0 30. lim
x:9
sin A 2x  3 B x  9
Oblique Asymptotes Find all possible oblique asymptotes in Exercises 31–34. 31. y =
sin x 2 sin x 2 x2 lim x = 1 # 0 = 0 = lim x 2 x: 0 x: 0 x x: 0
33. y = 2x 2 + 1
b. lim
= lim
A 1  2x B A 1 + 2x B
sin x 2 = 1 x: 0 x 2
a. lim
115
c. lim
x:0
Generalized Limits Involving
Additional and Advanced Exercises
2x 3>2 + 2x  3 2x + 1
1 32. y = x + x sin x 34. y = 2x 2 + 2x
=
1 2
3 DIFFERENTIATION OVERVIEW In the beginning of Chapter 2, we discussed how to determine the slope of a curve at a point and how to measure the rate at which a function changes. Now that we have studied limits, we can define these ideas precisely and see that both are interpretations of the derivative of a function at a point. We then extend this concept from a single point to the derivative function, and we develop rules for finding this derivative function easily, without having to calculate any limits directly. These rules are used to find derivatives of most of the common functions reviewed in Chapter 1, as well as various combinations of them. The derivative is one of the key ideas in calculus, and we use it to solve a wide range of problems involving tangents and rates of change.
Tangents and the Derivative at a Point
3.1
In this section we define the slope and tangent to a curve at a point, and the derivative of a function at a point. We will see that the derivative gives a way to find both the slope of a graph and the instantaneous rate of change of a function.
Finding a Tangent to the Graph of a Function
y y f(x) Q(x 0 h, f(x 0 h)) f (x 0 h) f (x 0)
To find a tangent to an arbitrary curve y = ƒ(x) at a point Psx0 , ƒsx0 dd, we use the procedure introduced in Section 2.1. We calculate the slope of the secant through P and a nearby point Qsx0 + h, ƒsx0 + hdd. We then investigate the limit of the slope as h : 0 (Figure 3.1). If the limit exists, we call it the slope of the curve at P and define the tangent at P to be the line through P having this slope.
P(x 0, f(x 0)) h 0
x0
x0 h
FIGURE 3.1 The slope of the tangent ƒsx0 + hd  ƒsx0 d line at P is lim . h h:0
x
DEFINITIONS number
The slope of the curve y = ƒsxd at the point Psx0 , ƒsx0 dd is the ƒsx0 + hd  ƒsx0 d h h:0
m = lim
(provided the limit exists).
The tangent line to the curve at P is the line through P with this slope.
In Section 2.1, Example 3, we applied these definitions to find the slope of the parabola ƒ(x) = x 2 at the point P(2, 4) and the tangent line to the parabola at P. Let’s look at another example.
116
3.1
117
EXAMPLE 1
y
(a) Find the slope of the curve y = 1>x at any point x = a Z 0. What is the slope at the point x = 1? (b) Where does the slope equal 1>4? (c) What happens to the tangent to the curve at the point (a, 1>a) as a changes?
y 5 1x slope is – 12 a
0
Tangents and the Derivative at a Point
x
a
Solution
(a) Here ƒsxd = 1>x. The slope at (a, 1>a) is slope is –1 at x 5 –1
FIGURE 3.2 The tangent slopes, steep near the origin, become more gradual as the point of tangency moves away (Example 1). y y 1x slope is – 1 4
1 1  a ƒsa + hd  ƒsad a + h 1 a  sa + hd lim = lim = lim h h h:0 h:0 h:0 h asa + hd = lim
h:0
Notice how we had to keep writing “limh:0” before each fraction until the stage where we could evaluate the limit by substituting h = 0. The number a may be positive or negative, but not 0. When a = 1, the slope is 1>(1) 2 = 1 (Figure 3.2). (b) The slope of y = 1>x at the point where x = a is 1>a 2. It will be 1>4 provided that
⎛2, 1 ⎛ ⎝ 2⎝
x
⎛–2, – 1 ⎛ ⎝ 2⎝
slope is – 1 4
FIGURE 3.3 The two tangent lines to y = 1>x having slope 1>4 (Example 1).
h 1 1 = lim =  2. hasa + hd h:0 asa + hd a
1 1 =  . 4 a2
This equation is equivalent to a 2 = 4, so a = 2 or a = 2. The curve has slope 1>4 at the two points (2, 1>2) and s 2, 1>2d (Figure 3.3). (c) The slope 1>a 2 is always negative if a Z 0. As a : 0 +, the slope approaches  q and the tangent becomes increasingly steep (Figure 3.2). We see this situation again as a : 0  . As a moves away from the origin in either direction, the slope approaches 0 and the tangent levels off to become horizontal.
Rates of Change: Derivative at a Point The expression ƒsx0 + hd  ƒsx0 d , h
h Z 0
is called the difference quotient of ƒ at x0 with increment h. If the difference quotient has a limit as h approaches zero, that limit is given a special name and notation.
DEFINITION
The derivative of a function ƒ at a point x0, denoted ƒ¿(x0), is ƒ¿(x0) = lim
h:0
ƒ(x0 + h)  ƒ(x0) h
provided this limit exists.
If we interpret the difference quotient as the slope of a secant line, then the derivative gives the slope of the curve y = ƒ(x) at the point P(x0, ƒ(x0)). Exercise 31 shows
118
Chapter 3: Differentiation
that the derivative of the linear function ƒ(x) = mx + b at any point x0 is simply the slope of the line, so ƒ¿(x0) = m, which is consistent with our definition of slope. If we interpret the difference quotient as an average rate of change (Section 2.1), the derivative gives the function’s instantaneous rate of change with respect to x at the point x = x0 . We study this interpretation in Section 3.4.
EXAMPLE 2
In Examples 1 and 2 in Section 2.1, we studied the speed of a rock falling freely from rest near the surface of the earth. We knew that the rock fell y = 16t 2 feet during the first t sec, and we used a sequence of average rates over increasingly short intervals to estimate the rock’s speed at the instant t = 1. What was the rock’s exact speed at this time?
We let ƒstd = 16t 2 . The average speed of the rock over the interval between t = 1 and t = 1 + h seconds, for h 7 0, was found to be Solution
ƒs1 + hd  ƒs1d 16s1 + hd2  16s1d2 16sh 2 + 2hd = = = 16sh + 2d. h h h The rock’s speed at the instant t = 1 is then lim 16sh + 2d = 16s0 + 2d = 32 ft>sec.
h:0
Our original estimate of 32 ft > sec in Section 2.1 was right.
Summary We have been discussing slopes of curves, lines tangent to a curve, the rate of change of a function, and the derivative of a function at a point. All of these ideas refer to the same limit.
The following are all interpretations for the limit of the difference quotient, ƒsx0 + hd  ƒsx0 d . h h:0 lim
1. 2. 3. 4.
The slope of the graph of y = ƒsxd at x = x0 The slope of the tangent to the curve y = ƒsxd at x = x0 The rate of change of ƒ(x) with respect to x at x = x0 The derivative ƒ¿(x0) at a point
In the next sections, we allow the point x0 to vary across the domain of the function ƒ.
3.1
Tangents and the Derivative at a Point
119
Exercises 3.1 Slopes and Tangent Lines In Exercises 1–4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in yunits per xunit) at the points P1 and P2 . 1.
2.
y
2
23. ƒsxd = x 2 + 4x  1
24. g sxd = x 3  3x
25. Find equations of all lines having slope 1 that are tangent to the curve y = 1>sx  1d .
y P2
2
Tangent Lines with Specified Slopes At what points do the graphs of the functions in Exercises 23 and 24 have horizontal tangents?
26. Find an equation of the straight line having slope 1>4 that is tangent to the curve y = 2x .
P2 1
1
–2
1
0
2
x
P1 –1
P1 0
3.
–1
x
1
–2
4.
y
28. Speed of a rocket At t sec after liftoff, the height of a rocket is 3t 2 ft. How fast is the rocket climbing 10 sec after liftoff ? 29. Circle’s changing area What is the rate of change of the area of a circle sA = pr 2 d with respect to the radius when the radius is r = 3?
y
3
30. Ball’s changing volume What is the rate of change of the volume of a ball sV = s4>3dpr 3 d with respect to the radius when the radius is r = 2 ?
2
31. Show that the line y = mx + b is its own tangent line at any point (x0, mx0 + b).
4 2
P2
P1
1
P2
P1
32. Find the slope of the tangent to the curve y = 1> 2x at the point where x = 4.
1 0
1
2
x
–2
–1
Rates of Change 27. Object dropped from a tower An object is dropped from the top of a 100mhigh tower. Its height above ground after t sec is 100  4.9t 2 m. How fast is it falling 2 sec after it is dropped?
0
1
2
x
Testing for Tangents 33. Does the graph of
In Exercises 5–10, find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. 5. y = 4  x 2, 7. y = 2 2x,
s 1, 3d s1, 2d
s 2, 8d
9. y = x 3,
6. y = sx  1d2 + 1, 1 8. y = 2 , x 10. y =
1 , x3
s1, 1d
s2, 5d
12. ƒsxd = x  2x 2,
x , x  2
s3, 3d
14. g sxd =
15. hstd = t 3,
s2, 8d
17. ƒsxd = 2x,
s4, 2d
g sxd = e
21. y =
x = 1
1 , x  1
x = 3
s8, 3d
20. y = 1  x 2,
x = 2
x  1 , x + 1
x = 0
22. y =
x Z 0 x = 0
have a tangent at the origin? Give reasons for your answer. Vertical Tangents We say that a continuous curve y = ƒsxd has a vertical tangent at the point where x = x0 if lim h:0 sƒsx0 + hd  ƒsx0 dd>h = q or  q . For example, y = x 1>3 has a vertical tangent at x = 0 (see accompanying figure):
In Exercises 19–22, find the slope of the curve at the point indicated. 19. y = 5x 2,
x sin s1>xd, 0,
s1, 1d
8 , s2, 2d x2 16. hstd = t 3 + 3t, s1, 4d 18. ƒsxd = 2x + 1,
x Z 0 x = 0
34. Does the graph of
1 a2,  b 8
11. ƒsxd = x 2 + 1,
x 2 sin s1>xd, 0,
have a tangent at the origin? Give reasons for your answer.
s 1, 1d
In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.
13. g sxd =
ƒsxd = e
lim
h:0
ƒs0 + hd  ƒs0d h 1>3  0 = lim h h h:0 1 = lim 2>3 = q . h:0 h
120
Chapter 3: Differentiation 36. Does the graph of
y y f (x)
0, x 6 0 1, x Ú 0 have a vertical tangent at the point (0, 1)? Give reasons for your answer. Usxd = e
x 13
x
0
T Graph the curves in Exercises 37–46. a. Where do the graphs appear to have vertical tangents? b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 35 and 36.
VERTICAL TANGENT AT ORIGIN
However, y = x 2>3 has no vertical tangent at x = 0 (see next figure): g s0 + hd  g s0d h 2>3  0 lim = lim h h h:0 h :0 1 = lim 1>3 h :0 h does not exist, because the limit is q from the right and  q from the left.
37. y = x 2>5
38. y = x 4>5
39. y = x 1>5
40. y = x 3>5
41. y = 4x
2>5
 2x
42. y = x 5>3  5x 2>3
45. y = e
 2ƒ x ƒ , x … 0 x 7 0 2x,
y g(x)
COMPUTER EXPLORATIONS Use a CAS to perform the following steps for the functions in Exercises 47–50: b. Holding x0 fixed, the difference quotient
x 23
qshd =
0
ƒsx0 + hd  ƒsx0 d h
at x0 becomes a function of the step size h. Enter this function into your CAS workspace.
x
c. Find the limit of q as h : 0 .
NO VERTICAL TANGENT AT ORIGIN
d. Define the secant lines y = ƒsx0 d + q # sx  x0 d for h = 3, 2 , and 1. Graph them together with ƒ and the tangent line over the interval in part (a). 5 47. ƒsxd = x 3 + 2x, x0 = 0 48. ƒsxd = x + x , x0 = 1 49. ƒsxd = x + sin s2xd, x0 = p>2
35. Does the graph of 1, x 6 0 0, x = 0 1, x 7 0
have a vertical tangent at the origin? Give reasons for your answer.
3.2
46. y = 2 ƒ 4  x ƒ
a. Plot y = ƒsxd over the interval sx0  1>2d … x … sx0 + 3d .
y
ƒsxd = •
44. y = x 1>3 + sx  1d1>3
43. y = x 2>3  sx  1d1>3
50. ƒsxd = cos x + 4 sin s2xd,
x0 = p
The Derivative as a Function In the last section we defined the derivative of y = ƒsxd at the point x = x0 to be the limit
HISTORICAL ESSAY
ƒsx0 + hd  ƒsx0 d . h h:0
ƒ¿sx0 d = lim
The Derivative
We now investigate the derivative as a function derived from ƒ by considering the limit at each point x in the domain of ƒ.
DEFINITION The derivative of the function ƒ(x) with respect to the variable x is the function ƒ¿ whose value at x is ƒ¿sxd = lim
h:0
provided the limit exists.
ƒsx + hd  ƒsxd , h
3.2 y f (x) Secant slope is f (z) f (x) zx
Q(z, f(z))
f (z) f (x)
P(x, f(x)) hzx
The Derivative as a Function
121
We use the notation ƒ(x) in the definition to emphasize the independent variable x with respect to which the derivative function ƒ¿(x) is being defined. The domain of ƒ¿ is the set of points in the domain of ƒ for which the limit exists, which means that the domain may be the same as or smaller than the domain of ƒ. If ƒ¿ exists at a particular x, we say that ƒ is differentiable (has a derivative) at x. If ƒ¿ exists at every point in the domain of ƒ, we call ƒ differentiable. If we write z = x + h, then h = z  x and h approaches 0 if and only if z approaches x. Therefore, an equivalent definition of the derivative is as follows (see Figure 3.4). This formula is sometimes more convenient to use when finding a derivative function.
zxh
x
Derivative of f at x is f(x h) f (x) f '(x) lim h h→0 lim
z→x
Alternative Formula for the Derivative ƒszd  ƒsxd z  x z:x
ƒ¿sxd = lim
f(z) f (x) zx
FIGURE 3.4 Two forms for the difference quotient.
Calculating Derivatives from the Definition The process of calculating a derivative is called differentiation. To emphasize the idea that differentiation is an operation performed on a function y = ƒsxd, we use the notation d ƒsxd dx
Derivative of the Reciprocal Function d 1 1 a b =  2, dx x x
as another way to denote the derivative ƒ¿sxd. Example 1 of Section 3.1 illustrated the differentiation process for the function y = 1>x when x = a. For x representing any point in the domain, we get the formula d 1 1 a b =  2. dx x x
x Z 0
Here are two more examples in which we allow x to be any point in the domain of ƒ.
EXAMPLE 1
Differentiate ƒsxd =
x . x  1
We use the definition of derivative, which requires us to calculate ƒ(x + h) and then subtract ƒ(x) to obtain the numerator in the difference quotient. We have
Solution
ƒsxd =
x x  1
and ƒsx + hd =
sx + hd , so sx + hd  1
ƒsx + hd  ƒsxd h h:0
ƒ¿sxd = lim
Definition
x + h x x  1 x + h  1 = lim h h:0 = lim
1 # sx + hdsx  1d  xsx + h  1d h sx + h  1dsx  1d
a c ad  cb  = b d bd
= lim
h 1 # h sx + h  1dsx  1d
Simplify.
= lim
1 1 = . sx + h  1dsx  1d sx  1d2
Cancel h Z 0.
h:0
h:0
h:0
122
Chapter 3: Differentiation
EXAMPLE 2 (a) Find the derivative of ƒsxd = 1x for x 7 0. (b) Find the tangent line to the curve y = 1x at x = 4. Solution Derivative of the Square Root Function d 1 2x = , dx 2 2x
(a) We use the alternative formula to calculate ƒ¿ : ƒ¿sxd = lim
x 7 0
z:x
= lim
z:x
ƒszd  ƒsxd z  x 1z  1x z  x
= lim
y y 1x1 4
(4, 2)
y x
ƒ¿s4d =
1 0
1z  1x
A 1z  1x B A 1z + 1x B 1 1 = lim = . z:x 1z + 1x 21x (b) The slope of the curve at x = 4 is z:x
4
x
FIGURE 3.5 The curve y = 1x and its tangent at (4, 2). The tangent’s slope is found by evaluating the derivative at x = 4 (Example 2).
1 1 = . 4 224
The tangent is the line through the point (4, 2) with slope 1>4 (Figure 3.5): 1 y = 2 + sx  4d 4 1 y = x + 1. 4
Notations There are many ways to denote the derivative of a function y = ƒsxd, where the independent variable is x and the dependent variable is y. Some common alternative notations for the derivative are ƒ¿sxd = y¿ =
dy dƒ d = = ƒsxd = Dsƒdsxd = Dx ƒsxd. dx dx dx
The symbols d>dx and D indicate the operation of differentiation. We read dy>dx as “the derivative of y with respect to x,” and dƒ>dx and (d>dx)ƒ(x) as “the derivative of ƒ with respect to x.” The “prime” notations y¿ and ƒ¿ come from notations that Newton used for derivatives. The d>dx notations are similar to those used by Leibniz. The symbol dy>dx should not be regarded as a ratio (until we introduce the idea of “differentials” in Section 3.11). To indicate the value of a derivative at a specified number x = a, we use the notation ƒ¿sad =
dy df d ` = ` = ƒsxd ` . dx x = a dx x = a dx x=a
For instance, in Example 2 ƒ¿s4d =
d 1 1 1 1x ` = ` = = . 4 dx 21x x = 4 x=4 224
Graphing the Derivative We can often make a reasonable plot of the derivative of y = ƒsxd by estimating the slopes on the graph of ƒ. That is, we plot the points sx, ƒ¿sxdd in the xyplane and connect them with a smooth curve, which represents y = ƒ¿sxd.
3.2
EXAMPLE 3
y
123
Graph the derivative of the function y = ƒsxd in Figure 3.6a.
We sketch the tangents to the graph of ƒ at frequent intervals and use their slopes to estimate the values of ƒ¿sxd at these points. We plot the corresponding sx, ƒ¿sxdd pairs and connect them with a smooth curve as sketched in Figure 3.6b.
Solution
Slope 0 A 10
The Derivative as a Function
Slope –1 B C
y f (x) ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
Slope – 4 3 E D Slope 0
⎧ ⎪ ⎨ ⎪ ⎩
5
What can we learn from the graph of y = ƒ¿sxd? At a glance we can see
8
0
4 xunits 10 15
5
x
1. 2. 3.
where the rate of change of ƒ is positive, negative, or zero; the rough size of the growth rate at any x and its size in relation to the size of ƒ(x); where the rate of change itself is increasing or decreasing.
(a) Slope
Differentiable on an Interval; OneSided Derivatives A function y = ƒsxd is differentiable on an open interval (finite or infinite) if it has a derivative at each point of the interval. It is differentiable on a closed interval [a, b] if it is differentiable on the interior (a, b) and if the limits
4 y f '(x)
3
E'
2 1 A' –1 –2
D' 10
5 C' B' Vertical coordinate –1
15
x
h:0 +
lim
ƒsa + hd  ƒsad h
Righthand derivative at a
lim
ƒsb + hd  ƒsbd h
Lefthand derivative at b
h:0
(b)
FIGURE 3.6 We made the graph of y = ƒ¿sxd in (b) by plotting slopes from the graph of y = ƒsxd in (a). The vertical coordinate of B¿ is the slope at B and so on. The slope at E is approximately 8>4 = 2. In (b) we see that the rate of change of ƒ is negative for x between A¿ and D¿; the rate of change is positive for x to the right of D¿.
exist at the endpoints (Figure 3.7). Righthand and lefthand derivatives may be defined at any point of a function’s domain. Because of Theorem 6, Section 2.4, a function has a derivative at a point if and only if it has lefthand and righthand derivatives there, and these onesided derivatives are equal.
EXAMPLE 4 Show that the function y = ƒ x ƒ is differentiable on s  q , 0d and s0, q d but has no derivative at x = 0. Solution From Section 3.1, the derivative of y = mx + b is the slope m. Thus, to the right of the origin,
d d d sxd = sxd = s1 # xd = 1. dx ƒ ƒ dx dx
d smx + bd = m, ƒ x ƒ = x dx
d d d sxd = s xd = s 1 # xd = 1 dx ƒ ƒ dx dx
ƒ x ƒ = x
To the left,
Slope f(a h) f(a) lim h h→0
(Figure 3.8). There is no derivative at the origin because the onesided derivatives differ there:
Slope f (b h) f (b) lim h h→0
Righthand derivative of ƒ x ƒ at zero = lim+ h:0
= lim+ h:0
y f (x)
ƒ0 + hƒ  ƒ0ƒ ƒhƒ = lim+ h h:0 h h h
ƒ h ƒ = h when h 7 0
= lim+1 = 1 h:0
a
ah h0
bh h0
b
x
Lefthand derivative of ƒ x ƒ at zero = limh:0
= limh:0
FIGURE 3.7 Derivatives at endpoints are onesided limits.
ƒ0 + hƒ  ƒ0ƒ ƒhƒ = limh h:0 h h h
ƒ h ƒ = h when h 6 0
= lim 1 = 1. h:0
124
Chapter 3: Differentiation
EXAMPLE 5
y
In Example 2 we found that for x 7 0,
y ⏐x⏐ y' –1
d 1 . 1x = dx 21x
y' 1
0
x
We apply the definition to examine if the derivative exists at x = 0:
y' not defined at x 0: righthand derivative lefthand derivative
FIGURE 3.8 The function y = ƒ x ƒ is not differentiable at the origin where the graph has a “corner” (Example 4).
lim
h:0 +
20 + h  20 1 = lim+ = q. h h:0 1h
Since the (righthand) limit is not finite, there is no derivative at x = 0. Since the slopes of the secant lines joining the origin to the points (h, 1h) on a graph of y = 1x approach q , the graph has a vertical tangent at the origin. (See Figure 1.17 on page 9).
When Does a Function Not Have a Derivative at a Point? A function has a derivative at a point x0 if the slopes of the secant lines through Psx0, ƒsx0 dd and a nearby point Q on the graph approach a finite limit as Q approaches P. Whenever the secants fail to take up a limiting position or become vertical as Q approaches P, the derivative does not exist. Thus differentiability is a “smoothness” condition on the graph of ƒ. A function can fail to have a derivative at a point for many reasons, including the existence of points where the graph has
P
P
Q Q
Q
Q
1. a corner, where the onesided
2. a cusp, where the slope of PQ approaches q from one side and  q from the other.
derivatives differ.
P
Q
P
Q P Q
Q
Q
3. a vertical tangent, where the slope of PQ approaches q from both sides or approaches  q from both sides (here,  q ).
4. a discontinuity (two examples shown).
Q
3.2
The Derivative as a Function
125
Another case in which the derivative may fail to exist occurs when the function’s slope is oscillating rapidly near P, as with ƒ(x) = sin (1>x) near the origin, where it is discontinuous (see Figure 2.31).
Differentiable Functions Are Continuous A function is continuous at every point where it has a derivative.
THEOREM 1—Differentiability Implies Continuity x = c, then ƒ is continuous at x = c.
If ƒ has a derivative at
Proof Given that ƒ¿scd exists, we must show that limx:c ƒsxd = ƒscd, or equivalently, that limh:0 ƒsc + hd = ƒscd. If h Z 0, then ƒsc + hd = ƒscd + sƒsc + hd  ƒscdd = ƒscd +
ƒsc + hd  ƒscd # h. h
Now take limits as h : 0. By Theorem 1 of Section 2.2, lim ƒsc + hd = lim ƒscd + lim
h:0
h:0
h:0
ƒsc + hd  ƒscd h
# lim h
= ƒscd + ƒ¿scd # 0
h: 0
= ƒscd + 0 = ƒscd. Similar arguments with onesided limits show that if ƒ has a derivative from one side (right or left) at x = c, then ƒ is continuous from that side at x = c. Theorem 1 says that if a function has a discontinuity at a point (for instance, a jump discontinuity), then it cannot be differentiable there. The greatest integer function y = :x; fails to be differentiable at every integer x = n (Example 4, Section 2.5). Caution The converse of Theorem 1 is false. A function need not have a derivative at a point where it is continuous, as we saw in Example 4.
Exercises 3.2 Finding Derivative Functions and Values Using the definition, calculate the derivatives of the functions in Exercises 1–6. Then find the values of the derivatives as specified. 1. ƒsxd = 4  x 2;
ƒ¿s  3d, ƒ¿s0d, ƒ¿s1d
2. Fsxd = sx  1d2 + 1;
F¿s 1d, F¿s0d, F¿s2d
1 ; g¿s  1d, g¿s2d, g¿ A 23 B t2 1  z ; k¿s 1d, k¿s1d, k¿ A 22 B 4. k szd = 2z 3. g std =
5. psud = 23u ;
p¿s1d, p¿s3d, p¿s2>3d
6. r ssd = 22s + 1 ;
r¿s0d, r¿s1d, r¿s1>2d
In Exercises 7–12, find the indicated derivatives. dy dr 7. if y = 2x 3 8. if r = s 3  2s 2 + 3 dx ds 9.
ds dt
if
s =
11.
dp dq
if
p =
t 2t + 1 1 2q + 1
10.
dy dt
if
1 y = t  t
12.
dz dw
if
z =
1 23w  2
126
Chapter 3: Differentiation
Slopes and Tangent Lines In Exercises 13–16, differentiate the functions and find the slope of the tangent line at the given value of the independent variable. 9 13. ƒsxd = x + x ,
27.
14. k sxd =
x = 3
8 2x  2
,
sx, yd = s6, 4d
18. w = g szd = 1 + 24  z,
y
y f 2 (x)
y f1(x)
1 , x = 2 2 + x x + 3 , x = 2 15. s = t 3  t 2, t = 1 16. y = 1  x In Exercises 17–18, differentiate the functions. Then find an equation of the tangent line at the indicated point on the graph of the function. 17. y = ƒsxd =
28.
y
x
0
29.
30.
y
sz, wd = s3, 2d
x
0
y
y f3(x)
y f4(x)
In Exercises 19–22, find the values of the derivatives. 19.
ds ` dt t = 1
20.
dy ` dx x = 23
21.
dr ` du u = 0
if
r =
dw ` 22. dz z = 4
if
w = z + 1z
if
x
0
s = 1  3t 2
x
0
1 y = 1  x
if
31. a. The graph in the accompanying figure is made of line segments joined end to end. At which points of the interval [4, 6] is ƒ¿ not defined? Give reasons for your answer.
2 24  u
y (6, 2)
(0, 2)
Using the Alternative Formula for Derivatives Use the formula
y f (x)
ƒszd  ƒsxd z  x z: x
ƒ¿sxd = lim
0
(– 4, 0)
1
x
6
to find the derivative of the functions in Exercises 23–26. (1, –2)
1 x + 2 x 25. g sxd = x  1 23. ƒsxd =
(4, –2)
24. ƒsxd = x 2  3x + 4 26. g sxd = 1 + 1x
b. Graph the derivative of ƒ. The graph should show a step function.
Graphs Match the functions graphed in Exercises 27–30 with the derivatives graphed in the accompanying figures (a)– (d). y'
32. Recovering a function from its derivative a. Use the following information to graph the function ƒ over the closed interval [2, 5] . i) The graph of ƒ is made of closed line segments joined end to end.
y'
ii) The graph starts at the point s 2, 3d . iii) The derivative of ƒ is the step function in the figure shown here.
x
0
y'
x
0 (a)
(b)
y'
y'
y' f '(x) 1 –2
0
(c)
x
0
(d)
x
0
1
3
5
x
–2
b. Repeat part (a), assuming that the graph starts at s 2, 0d instead of s  2, 3d .
3.2 33. Growth in the economy The graph in the accompanying figure shows the average annual percentage change y = ƒstd in the U.S. gross national product (GNP) for the years 1983–1988. Graph dy>dt (where defined). 7%
127
The Derivative as a Function
36. Weight loss Jared Fogle, also known as the “Subway Sandwich Guy,” weighed 425 lb in 1997 before losing more than 240 lb in 12 months (http://en.wikipedia.org/wiki/Jared_Fogle). A chart showing his possible dramatic weight loss is given in the accompanying figure.
6 W
5 4
500
3
425
1 0 1983
1984
1985
1986
1987
Weight (lbs)
2
1988
t
1 2 3 4 5 6 7 8 9 10 11 12 Time (months)
0
a. Estimate Jared’s rate of weight loss when i) t = 1
a. Use the graphical technique of Example 3 to graph the derivative of the fruit fly population. The graph of the population is reproduced here.
ii) t = 4
iii) t = 11
b. When does Jared lose weight most rapidly and what is this rate of weight loss? c. Use the graphical technique of Example 3 to graph the derivative of weight W.
p
Number of flies
200 100
34. Fruit flies (Continuation of Example 4, Section 2.1.) Populations starting out in closed environments grow slowly at first, when there are relatively few members, then more rapidly as the number of reproducing individuals increases and resources are still abundant, then slowly again as the population reaches the carrying capacity of the environment.
350 300 250 200 150 100 50 0
300
OneSided Derivatives Compute the righthand and lefthand derivatives as limits to show that the functions in Exercises 37–40 are not differentiable at the point P. 37. 10
20 30 Time (days)
40
y
38.
y
y f (x)
t
50
y 2x
y f (x)
y x2
y2 2
b. During what days does the population seem to be increasing fastest? Slowest?
P(1, 2)
yx
35. Temperature The given graph shows the temperature T in °F at Davis, CA, on April 18, 2008, between 6 A.M. and 6 P.M.
1 x
P(0, 0)
0
1
2
x
T
Temperature (°F)
80
y
39.
70
40.
y y f (x)
y f (x)
P(1, 1)
y 2x 1
60
1
50 1
40 0 6 a.m.
3
6
9
9 a.m. 12 noon 3 p.m. Time (hrs)
12
t
ii) 9 A.M.
iii) 2 P.M.
x
1
yx
1
x
6 p.m.
a. Estimate the rate of temperature change at the times i) 7 A.M.
0
P(1, 1) y x
y 1x
iv) 4 P.M.
b. At what time does the temperature increase most rapidly? Decrease most rapidly? What is the rate for each of those times? c. Use the graphical technique of Example 3 to graph the derivative of temperature T versus time t.
In Exercises 41 and 42, determine if the piecewisedefined function is differentiable at the origin. 41. ƒsxd = e
2x  1, x 2 + 2x + 7,
42. gsxd = e
x 2>3, x 1>3,
x Ú 0 x 6 0
x Ú 0 x 6 0
128
Chapter 3: Differentiation
Differentiability and Continuity on an Interval Each figure in Exercises 43–48 shows the graph of a function over a closed interval D. At what domain points does the function appear to be
55. Derivative of ƒ Does knowing that a function ƒ(x) is differentiable at x = x0 tell you anything about the differentiability of the function ƒ at x = x0 ? Give reasons for your answer.
a. differentiable? b. continuous but not differentiable? c. neither continuous nor differentiable?
56. Derivative of multiples Does knowing that a function g(t) is differentiable at t = 7 tell you anything about the differentiability of the function 3g at t = 7 ? Give reasons for your answer.
Give reasons for your answers. 43.
44. y
y
y f (x) D: –3 x 2 2
2
1
1
–3
–2
–1
0 1
2
x
–2
y f (x) D: –2 x 3
–1
0
–1
–1
–2
–2
45.
54. Tangent to y 1x Does any tangent to the curve y = 1x cross the xaxis at x = 1 ? If so, find an equation for the line and the point of tangency. If not, why not?
1
2
3
x
57. Limit of a quotient Suppose that functions g(t) and h(t) are defined for all values of t and g s0d = hs0d = 0 . Can limt:0 sg stdd>shstdd exist? If it does exist, must it equal zero? Give reasons for your answers. 58. a. Let ƒ(x) be a function satisfying ƒ ƒsxd ƒ … x 2 for 1 … x … 1 . Show that ƒ is differentiable at x = 0 and find ƒ¿s0d . b. Show that
ƒsxd = L
46. y
1 x 2 sin x ,
x Z 0
0,
x = 0
is differentiable at x = 0 and find ƒ¿s0d .
y y f (x) D: –3 x 3
T 59. Graph y = 1> A 21x B in a window that has 0 … x … 2 . Then, on the same screen, graph
y f (x) D: –2 x 3 3
y =
1 –3 –2 –1 0 –1
1
2
3
2
x
1
–2 –2
47.
–1
1
0
3
x
48. y
y y f (x) D: –1 x 2
4
y f (x) D: –3 x 3
2
1
–1
2
0
1
2
x
–3 –2 –1 0
1 2
3
x
Theory and Examples In Exercises 49–52, a. Find the derivative ƒ¿sxd of the given function y = ƒsxd . b. Graph y = ƒsxd and y = ƒ¿sxd side by side using separate sets of coordinate axes, and answer the following questions.
1x + h  1x h
for h = 1, 0.5, 0.1 . Then try h = 1, 0.5, 0.1 . Explain what is going on. T 60. Graph y = 3x 2 in a window that has 2 … x … 2, 0 … y … 3 . Then, on the same screen, graph y =
sx + hd3  x 3 h
for h = 2, 1, 0.2 . Then try h = 2, 1, 0.2 . Explain what is going on. 61. Derivative of y x Graph the derivative of ƒsxd = ƒ x ƒ . Then graph y = s ƒ x ƒ  0d>sx  0d = ƒ x ƒ >x . What can you conclude? T 62. Weierstrass’s nowhere differentiable continuous function The sum of the first eight terms of the Weierstrass function ƒ(x) = g nq= 0 s2>3dn cos s9npxd is g sxd = cos spxd + s2>3d1 cos s9pxd + s2>3d2 cos s92pxd + s2>3d3 cos s93pxd + Á + s2>3d7 cos s97pxd . Graph this sum. Zoom in several times. How wiggly and bumpy is this graph? Specify a viewing window in which the displayed portion of the graph is smooth.
c. For what values of x, if any, is ƒ¿ positive? Zero? Negative? d. Over what intervals of xvalues, if any, does the function y = ƒsxd increase as x increases? Decrease as x increases? How is this related to what you found in part (c)? (We will say more about this relationship in Section 4.3.) 49. y = x
2
51. y = x 3>3
50. y = 1>x 52. y = x 4>4
53. Tangent to a parabola Does the parabola y = 2x 2  13x + 5 have a tangent whose slope is 1 ? If so, find an equation for the line and the point of tangency. If not, why not?
COMPUTER EXPLORATIONS Use a CAS to perform the following steps for the functions in Exercises 63–68. a. Plot y = ƒsxd to see that function’s global behavior. b. Define the difference quotient q at a general point x, with general step size h. c. Take the limit as h : 0 . What formula does this give? d. Substitute the value x = x0 and plot the function y = ƒsxd together with its tangent line at that point.
3.3 e. Substitute various values for x larger and smaller than x0 into the formula obtained in part (c). Do the numbers make sense with your picture? f. Graph the formula obtained in part (c). What does it mean when its values are negative? Zero? Positive? Does this make sense with your plot from part (a)? Give reasons for your answer. 63. ƒsxd = x 3 + x 2  x,
x0 = 1
64. ƒsxd = x 1>3 + x 2>3, 65. ƒsxd =
4x , x2 + 1
Differentiation Rules
129
x0 = 1
x0 = 2
x  1 , x0 = 1 3x 2 + 1 67. ƒsxd = sin 2x, x0 = p>2 66. ƒsxd =
68. ƒsxd = x 2 cos x,
x0 = p>4
Differentiation Rules
3.3
This section introduces several rules that allow us to differentiate constant functions, power functions, polynomials, exponential functions, rational functions, and certain combinations of them, simply and directly, without having to take limits each time.
Powers, Multiples, Sums, and Differences A simple rule of differentiation is that the derivative of every constant function is zero. y c
(x h, c)
(x, c)
yc
dƒ d scd = 0. = dx dx
h 0
x
Derivative of a Constant Function If ƒ has the constant value ƒsxd = c, then
xh
FIGURE 3.9 The rule sd>dxdscd = 0 is another way to say that the values of constant functions never change and that the slope of a horizontal line is zero at every point.
x
Proof We apply the definition of the derivative to ƒsxd = c, the function whose outputs have the constant value c (Figure 3.9). At every value of x, we find that ƒsx + hd  ƒsxd c  c = lim = lim 0 = 0. h h h:0 h:0 h:0
ƒ¿sxd = lim
From Section 3.1, we know that d 1 1 a b =  2, dx x x
or
d 1 A x B = x  2. dx
From Example 2 of the last section we also know that d A 2x B = 1 , dx 22x
or
d 1>2 A x B = 12 x  1>2. dx
These two examples illustrate a general rule for differentiating a power x n. We first prove the rule when n is a positive integer.
Power Rule for Positive Integers: If n is a positive integer, then d n x = nx n  1 . dx
130
Chapter 3: Differentiation
HISTORICAL BIOGRAPHY
Proof of the Positive Integer Power Rule
The formula
z n  x n = sz  xdsz n  1 + z n  2 x + Á + zx n  2 + x n  1 d
Richard Courant (1888–1972)
can be verified by multiplying out the righthand side. Then from the alternative formula for the definition of the derivative, ƒszd  ƒsxd z n  xn = lim z  x z x z:x z:x
ƒ¿sxd = lim
= lim sz n  1 + z n  2x + Á + zx n  2 + x n  1 d z:x
n terms
= nx n  1. The Power Rule is actually valid for all real numbers n. The number n could be a negative integer or a fractional power or an irrational number. To apply the Power Rule, we subtract 1 from the original exponent n and multiply the result by n. Here we state the general version of the rule, but postpone its proof until Section 3.8.
Power Rule (General Version) If n is any real number, then d n x = nx n  1 dx for all x where the powers x n and x n  1 are defined.
EXAMPLE 1 (a) x 3
Differentiate the following powers of x.
(b) x 2/3
(c) x 22
(d)
1 x4
(e) x 4>3
(f) 2x 2 + p
Solution
(a)
d 3 (x ) = 3x 3  1 = 3x 2 dx
(b)
d 2>3 2 2 (x ) = x (2>3)  1 = x 1>3 3 3 dx
(c)
d 22 A x B = 22x 22  1 dx
(d)
d 1 d 4 4 a b = (x ) = 4x 4  1 = 4x 5 =  5 dx x 4 dx x
(e)
d 4>3 4 4 (x ) =  x (4>3)  1 =  x 7>3 3 3 dx
(f)
d d p A 2x 2 + p B = dx A x 1 + (p>2) B = a1 + 2 bx 1 + (p>2)  1 = 12 (2 + p)2x p dx
The next rule says that when a differentiable function is multiplied by a constant, its derivative is multiplied by the same constant.
Derivative Constant Multiple Rule If u is a differentiable function of x, and c is a constant, then du d scud = c . dx dx
In particular, if n is any real number, then d scx n d = cnx n  1 . dx
3.3
131
Proof
y y 3x 2
3
Differentiation Rules
cusx + hd  cusxd d cu = lim dx h h:0 usx + hd  usxd h h:0
= c lim
Slope 3(2x) 6x Slope 6(1) 6
(1, 3)
= c
du dx
Derivative definition with ƒsxd = cusxd Constant Multiple Limit Property
u is differentiable.
y x2
EXAMPLE 2
2
(a) The derivative formula 1
0
Slope 2x Slope 2(1) 2 (1, 1)
1
2
x
FIGURE 3.10 The graphs of y = x 2 and y = 3x 2 . Tripling the ycoordinate triples the slope (Example 2).
Denoting Functions by u and Y The functions we are working with when we need a differentiation formula are likely to be denoted by letters like ƒ and g. We do not want to use these same letters when stating general differentiation rules, so instead we use letters like u and y that are not likely to be already in use.
d s3x 2 d = 3 # 2x = 6x dx says that if we rescale the graph of y = x 2 by multiplying each ycoordinate by 3, then we multiply the slope at each point by 3 (Figure 3.10). (b) Negative of a function The derivative of the negative of a differentiable function u is the negative of the function’s derivative. The Constant Multiple Rule with c = 1 gives d d d du s ud = s 1 # ud = 1 # sud =  . dx dx dx dx The next rule says that the derivative of the sum of two differentiable functions is the sum of their derivatives.
Derivative Sum Rule If u and y are differentiable functions of x, then their sum u + y is differentiable at every point where u and y are both differentiable. At such points, du dy d su + yd = + . dx dx dx
For example, if y = x 4 + 12x, then y is the sum of u(x) = x 4 and y(x) = 12x. We then have dy d 4 d = (x ) + (12x) = 4x 3 + 12. dx dx dx Proof We apply the definition of the derivative to ƒsxd = usxd + ysxd: [usx + hd + ysx + hd]  [usxd + ysxd] d [usxd + ysxd] = lim dx h h:0 = lim c h:0
= lim
h:0
usx + hd  usxd ysx + hd  ysxd + d h h
usx + hd  usxd ysx + hd  ysxd du dy + lim = + . h h dx dx h:0
Combining the Sum Rule with the Constant Multiple Rule gives the Difference Rule, which says that the derivative of a difference of differentiable functions is the difference of their derivatives: d d du dy du dy su  yd = [u + s 1dy] = + s 1d = . dx dx dx dx dx dx
132
Chapter 3: Differentiation
The Sum Rule also extends to finite sums of more than two functions. If u1 , u2 , Á , un are differentiable at x, then so is u1 + u2 + Á + un , and dun du2 du1 d su + u2 + Á + un d = + + Á + . dx 1 dx dx dx For instance, to see that the rule holds for three functions we compute du3 du3 du1 du2 d d d su + u2 + u3 d = ssu1 + u2 d + u3 d = su + u2 d + = + + . dx 1 dx dx 1 dx dx dx dx
A proof by mathematical induction for any finite number of terms is given in Appendix 2.
EXAMPLE 3 Solution
Find the derivative of the polynomial y = x 3 +
dy d 4 2 d d 3 d = x + a x b s5xd + s1d dx dx dx 3 dx dx = 3x 2 +
4 2 x  5x + 1. 3
Sum and Difference Rules
8 4# 2x  5 + 0 = 3x 2 + x  5 3 3
We can differentiate any polynomial term by term, the way we differentiated the polynomial in Example 3. All polynomials are differentiable at all values of x.
EXAMPLE 4
Does the curve y = x 4  2x 2 + 2 have any horizontal tangents? If so,
where? y
Solution
y x 4 2x 2 2
dy d 4 = sx  2x 2 + 2d = 4x 3  4x. dx dx dy = 0 for x: Now solve the equation dx
(0, 2)
(–1, 1)
–1
1
0
The horizontal tangents, if any, occur where the slope dy>dx is zero. We have
(1, 1)
1
x
FIGURE 3.11 The curve in Example 4 and its horizontal tangents.
4x 3  4x = 0 4xsx 2  1d = 0 x = 0, 1, 1. The curve y = x 4  2x 2 + 2 has horizontal tangents at x = 0, 1, and 1. The corresponding points on the curve are (0, 2), (1, 1), and s 1, 1d. See Figure 3.11. We will see in Chapter 4 that finding the values of x where the derivative of a function is equal to zero is an important and useful procedure.
Derivatives of Exponential Functions We briefly reviewed exponential functions in Section 1.5. When we apply the definition of the derivative to ƒ(x) a x, we get d x a x+h  ax (a ) = lim dx h h:0
a x # ah  a x h h:0
= lim
Derivative definition
#
a xh = ax ah
= lim a x #
ah  1 h
Factoring out ax
= a x # lim
ah  1 h
a x is constant as h : 0.
h:0
h:0
= ¢ lim
h:0
ah  1 # x ≤ a. h
(++)++* a fixed numberL
(1)
3.3
y
a 3 a e a 2.5
133
Thus we see that the derivative of a x is a constant multiple L of a x. The constant L is a limit unlike any we have encountered before. Note, however, that it equals the derivative of ƒ(x) = a x at x = 0: ah  a0 ah  1 = lim = L. h h h:0 h:0
a2
L 1.0 1.1
Differentiation Rules
ƒ¿(0) = lim
0.92 h y a 1,a0 h
0.69
h
0
FIGURE 3.12 The position of the curve y = (a h  1)>h, a 7 0, varies continuously with a. The limit L of y as h : 0 changes with different values of a. The number e between a = 2 and a = 3 is the number for which the limit equals 1 as h : 0.
The limit L is therefore the slope of the graph of ƒ(x) = a x where it crosses the yaxis. In Chapter 7, where we carefully develop the logarithmic and exponential functions, we prove that the limit L exists and has the value ln a. For now we investigate values of L by graphing the function y = (a h  1)>h and studying its behavior as h approaches 0. Figure 3.12 shows the graphs of y = (a h  1)>h for four different values of a. The limit L is approximately 0.69 if a = 2, about 0.92 if a = 2.5, and about 1.1 if a = 3. It appears that the value of L is 1 at some number a chosen between 2.5 and 3. That number is given by a = e L 2.718281828. With this choice of base we obtain the natural exponential function ƒ(x) = e x as in Section 1.5, and see that it satisfies the property eh  1 = 1. h h:0
ƒ¿(0) = lim
(2)
That the limit is 1 implies an important relationship between the natural exponential function e x and its derivative: d x eh  1 # x (e ) = lim ¢ ≤ e dx h h:0 = 1 # e x = e x.
Eq. (1) with a = e Eq. (2)
Therefore the natural exponential function is its own derivative.
Derivative of the Natural Exponential Function d x (e ) = e x dx
Find an equation for a line that is tangent to the graph of y = e x and goes through the origin.
EXAMPLE 5
Since the line passes through the origin, its equation is of the form y = mx, where m is the slope. If it is tangent to the graph at the point (a, e a), the slope is m = (e a  0)>(a  0). The slope of the natural exponential at x = a is e a. Because these slopes are the same, we then have that e a = e a>a. It follows that a = 1 and m = e, so the equation of the tangent line is y = ex . See Figure 3.13. Solution
y 6
y ex
4 (a,
2 –1
a
ea) x
FIGURE 3.13 The line through the origin is tangent to the graph of y = e x when a = 1 (Example 5).
We might ask if there are functions other than the natural exponential function that are their own derivatives. The answer is that the only functions that satisfy the property that ƒ¿(x) = ƒ(x) are functions that are constant multiples of the natural exponential function, ƒ(x) = c # e x, c any constant. We prove this fact in Section 7.2. Note from the Constant Multiple Rule that indeed d d x (c # e x ) = c # (e ) = c # e x. dx dx
134
Chapter 3: Differentiation
Products and Quotients While the derivative of the sum of two functions is the sum of their derivatives, the derivative of the product of two functions is not the product of their derivatives. For instance, d 2 d sx # xd = sx d = 2x, dx dx
while
d d sxd # sxd = 1 # 1 = 1. dx dx
The derivative of a product of two functions is the sum of two products, as we now explain. Derivative Product Rule If u and y are differentiable at x, then so is their product uy, and d dy du suyd = u + y . dx dx dx The derivative of the product uy is u times the derivative of y plus y times the derivative of u. In prime notation, suyd¿ = uy¿ + yu¿ . In function notation, d [ƒsxdgsxd] = ƒsxdg¿sxd + gsxdƒ¿sxd. dx
EXAMPLE 6
1 Find the derivative of (a) y = x A x 2 + e x B ,
(b) y = e 2x.
Solution
(a) We apply the Product Rule with u = 1>x and y = x 2 + e x : d 1 2 1 1 c A x + e x B d = x A 2x + e x B + A x 2 + e x B a 2 b dx x x ex ex = 2 + x  1  2 x Picturing the Product Rule Suppose u(x) and y(x) are positive and increase when x increases, and h 7 0.
(b)
y(x h) y
= 1 + (x  1)
d dy du suyd = u + y , and dx dx dx d 1 1 a b =  2 dx x x
ex . x2
d 2x d x# x d x d x (e ) = (e e ) = e x # (e ) + e x # (e ) = 2e x # e x = 2e 2x dx dx dx dx
u(x h) y
Proof of the Derivative Product Rule
y(x) u(x)y(x)
0
usx + hdysx + hd  usxdysxd d suyd = lim dx h h:0
y(x) u
u(x)
u u(x h)
Then the change in the product uy is the difference in areas of the larger and smaller “squares,” which is the sum of the upper and righthand reddishshaded rectangles. That is, ¢(uy) = u(x + h)y(x + h)  u(x)y(x) = u(x + h)¢y + y(x)¢u.
Division by h gives ¢(uy) ¢u ¢y = u(x + h) + y(x) . h h h
The limit as h : 0 + gives the Product Rule.
To change this fraction into an equivalent one that contains difference quotients for the derivatives of u and y, we subtract and add usx + hdysxd in the numerator: usx + hdysx + hd  usx + hdysxd + usx + hdysxd  usxdysxd d suyd = lim dx h h:0 = lim cusx + hd h:0
ysx + hd  ysxd usx + hd  usxd + ysxd d h h
ysx + hd  ysxd usx + hd  usxd + ysxd # lim . h h h:0 h:0
= lim usx + hd # lim h:0
As h approaches zero, usx + hd approaches u(x) because u, being differentiable at x, is continuous at x. The two fractions approach the values of dy>dx at x and du>dx at x. In short, d dy du suyd = u + y . dx dx dx
3.3
EXAMPLE 7
Differentiation Rules
135
Find the derivative of y = sx 2 + 1dsx 3 + 3d.
Solution
(a) From the Product Rule with u = x 2 + 1 and y = x 3 + 3, we find d C sx 2 + 1dsx 3 + 3d D = sx 2 + 1ds3x 2 d + sx 3 + 3ds2xd dx
d dy du suyd = u + y dx dx dx
= 3x 4 + 3x 2 + 2x 4 + 6x = 5x 4 + 3x 2 + 6x. (b) This particular product can be differentiated as well (perhaps better) by multiplying out the original expression for y and differentiating the resulting polynomial: y = sx 2 + 1dsx 3 + 3d = x 5 + x 3 + 3x 2 + 3 dy = 5x 4 + 3x 2 + 6x. dx This is in agreement with our first calculation. The derivative of the quotient of two functions is given by the Quotient Rule.
Derivative Quotient Rule If u and y are differentiable at x and if ysxd Z 0, then the quotient u>y is differentiable at x, and d u a b = dx y
y
du dy  u dx dx . 2 y
In function notation, gsxdƒ¿sxd  ƒsxdg¿sxd d ƒsxd d = . c dx gsxd g 2sxd
EXAMPLE 8
Find the derivative of (a) y =
t2  1 , (b) y = e x. t3 + 1
Solution
(a) We apply the Quotient Rule with u = t 2  1 and y = t 3 + 1: dy st 3 + 1d # 2t  st 2  1d # 3t 2 = dt st 3 + 1d2
(b)
=
2t 4 + 2t  3t 4 + 3t 2 st 3 + 1d2
=
t 4 + 3t 2 + 2t . st 3 + 1d2
ex # 0  1 # ex d 1 d x 1 = x = e  x (e ) = a xb = dx dx e e (e x) 2
ysdu>dtd  usdy>dtd d u a b = dt y y2
136
Chapter 3: Differentiation
Proof of the Derivative Quotient Rule usxd usx + hd ysx + hd ysxd d u a b = lim dx y h h:0 ysxdusx + hd  usxdysx + hd = lim hysx + hdysxd h:0 To change the last fraction into an equivalent one that contains the difference quotients for the derivatives of u and y, we subtract and add y(x)u(x) in the numerator. We then get ysxdusx + hd  ysxdusxd + ysxdusxd  usxdysx + hd d u a b = lim dx y hysx + hdysxd h:0 usx + hd  usxd ysx + hd  ysxd ysxd  usxd h h = lim . ysx + hdysxd h:0 Taking the limits in the numerator and denominator now gives the Quotient Rule. Exercise 74 outlines another proof. The choice of which rules to use in solving a differentiation problem can make a difference in how much work you have to do. Here is an example.
EXAMPLE 9
Find the derivative of y =
sx  1dsx2  2xd . x4
Using the Quotient Rule here will result in a complicated expression with many terms. Instead, use some algebra to simplify the expression. First expand the numerator and divide by x 4 : Solution
y =
sx  1dsx 2  2xd x 3  3x 2 + 2x = = x 1  3x 2 + 2x 3 . x4 x4
Then use the Sum and Power Rules: dy = x 2  3s 2dx 3 + 2s 3dx 4 dx 6 6 1 =  2 + 3  4. x x x
Second and HigherOrder Derivatives If y = ƒsxd is a differentiable function, then its derivative ƒ¿sxd is also a function. If ƒ¿ is also differentiable, then we can differentiate ƒ¿ to get a new function of x denoted by ƒ–. So ƒ– = sƒ¿d¿ . The function ƒ– is called the second derivative of ƒ because it is the derivative of the first derivative. It is written in several ways: ƒ–sxd =
d 2y dx 2
=
dy¿ d dy a b = = y– = D 2sƒdsxd = D x2 ƒsxd. dx dx dx
The symbol D 2 means the operation of differentiation is performed twice. If y = x 6 , then y¿ = 6x 5 and we have y– = Thus D 2(x 6) = 30x 4 .
dy¿ d = (6x 5) = 30x 4 . dx dx
3.3
How to Read the Symbols for Derivatives “y prime” y¿ “y double prime” y– d 2y “d squared y dx squared” dx 2 y‡ “y triple prime” y snd “y super n” d ny “d to the n of y by dx to the n” dx n D n “D to the n”
Differentiation Rules
137
If y– is differentiable, its derivative, y‡ = dy–>dx = d 3y>dx 3, is the third derivative of y with respect to x. The names continue as you imagine, with y snd =
d ny d sn  1d y = = D ny dx dx n
denoting the nth derivative of y with respect to x for any positive integer n. We can interpret the second derivative as the rate of change of the slope of the tangent to the graph of y = ƒsxd at each point. You will see in the next chapter that the second derivative reveals whether the graph bends upward or downward from the tangent line as we move off the point of tangency. In the next section, we interpret both the second and third derivatives in terms of motion along a straight line.
EXAMPLE 10
The first four derivatives of y = x 3  3x 2 + 2 are First derivative:
y¿ = 3x 2  6x
Second derivative:
y– = 6x  6
Third derivative:
y‡ = 6
Fourth derivative:
y s4d = 0.
The function has derivatives of all orders, the fifth and later derivatives all being zero.
Exercises 3.3 Derivative Calculations In Exercises 1–12, find the first and second derivatives. 1. y =  x 2 + 3
2. y = x 2 + x + 8
3. s = 5t  3t
4. w = 3z 7  7z 3 + 21z 2
3
5
4x 3  x + 2e x 3 1 7. w = 3z 2  z 5. y =
9. y = 6x 2  10x  5x 2 5 1 11. r = 2 2s 3s
x3 x2 x + + 3 2 4 4 8. s = 2t 1 + 2 t 10. y = 4  2x  x 3 6. y =
12 1 4  3 + 4 12. r = u u u
27. y =
sx + 1dsx + 2d 1 28. y = 2 sx  1dsx  2d sx  1dsx + x + 1d 2
29. y = 2e x + e 3x 31. y = x 3e x 33. y = x 9>4 + e 2x 35. s = 2t 3>2 + 3e 2
x 2 + 3e x 2e x  x 32. w = re r 30. y =
34. y = x 3>5 + p3>2 p 1 36. w = 1.4 + z 2z
7 2 37. y = 2x  xe
3 9.6 38. y = 2 x + 2e 1.3
es 39. r = s
40. r = e u a
1 + up>2 b u2
In Exercises 13–16, find y¿ (a) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate.
Find the derivatives of all orders of the functions in Exercises 41–44.
13. y = s3  x 2 dsx 3  x + 1d 14. y = s2x + 3ds5x 2  4xd
41. y =
1 15. y = sx 2 + 1d ax + 5 + x b 16. y = s1 + x 2 dsx 3>4  x 3 d
43. y = sx  1dsx 2 + 3x  5d 44. y = s4x 3 + 3xds2  xd
Find the derivatives of the functions in Exercises 17–40. 2x + 5 17. y = 3x  2 x2  4 19. g sxd = x + 0.5 2 1
21. y = s1  tds1 + t d 1s  1 23. ƒssd = 1s + 1 1 + x  4 1x 25. y = x
4  3x 18. z = 3x 2 + x t2  1 20. ƒstd = 2 t + t  2 22. w = s2x  7d1sx + 5d 5x + 1 24. u = 2 1x 26. r = 2 a
1 2u
3 x4  x2  x 2 2
x5 120
Find the first and second derivatives of the functions in Exercises 45–52. 45. y = 47. r =
x3 + 7 x
46. s =
su  1dsu 2 + u + 1d
49. w = a + 2ub
42. y =
u3 1 + 3z bs3  zd 3z
51. w = 3z 2e 2z
48. u =
t 2 + 5t  1 t2 2 sx + xdsx 2  x + 1d x4 q + 3 2
50. p =
sq  1d3 + sq + 1d3
52. w = e zsz  1dsz 2 + 1d
138
Chapter 3: Differentiation
53. Suppose u and y are functions of x that are differentiable at x = 0 and that us0d = 5,
u¿s0d = 3,
ys0d = 1,
y¿s0d = 2 .
Find the values of the following derivatives at x = 0 . a.
d suyd dx
b.
d u a b dx y
c.
d y a b dx u
d.
d s7y  2ud dx
62. Find all points (x, y) on the graph of gsxd = 13 x 3 tangent lines parallel to the line 8x  2y = 1.
u¿s1d = 0,
ys1d = 5,
64. Find all points (x, y) on the graph of ƒsxd = x 2 with tangent lines passing through the point (3, 8). y 10
y¿s1d = 1 .
f (x) 5 x 2 (3, 8)
Find the values of the following derivatives at x = 1 . a.
d suyd dx
b.
d u a b dx y
c.
d y a b dx u
d.
x 2 + 1 with
63. Find all points (x, y) on the graph of y = x>(x  2) with tangent lines perpendicular to the line y = 2x + 3.
54. Suppose u and y are differentiable functions of x and that us1d = 2,
3 2
6
d s7y  2ud dx
(x, y) 2
Slopes and Tangents 55. a. Normal to a curve Find an equation for the line perpendicular to the tangent to the curve y = x 3  4x + 1 at the point (2, 1).
–2
2
4
x
b. Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope? c. Tangents having specified slope Find equations for the tangents to the curve at the points where the slope of the curve is 8. 56. a. Horizontal tangents Find equations for the horizontal tangents to the curve y = x 3  3x  2 . Also find equations for the lines that are perpendicular to these tangents at the points of tangency. b. Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope? Find an equation for the line that is perpendicular to the curve’s tangent at this point. 57. Find the tangents to Newton’s serpentine (graphed here) at the origin and the point (1, 2). y y 24x x 1
(1, 2)
1 2 3 4
0
8 x2 4
T b. Graph the curve and tangent together. The tangent intersects the curve at another point. Use Zoom and Trace to estimate the point’s coordinates.
x 50  1 x:1 x  1
67. lim
68.
x 2>9  1 x: 1 x + 1 lim
69. Find the value of a that makes the following function differentiable for all xvalues.
(2, 1) 1 2 3
66. a. Find an equation for the line that is tangent to the curve y = x 3  6x 2 + 5x at the origin.
Theory and Examples For Exercises 67 and 68 evaluate each limit by first converting each to a derivative at a particular xvalue.
y
2 1
T c. Confirm your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).
x
58. Find the tangent to the Witch of Agnesi (graphed here) at the point (2, 1). y
T b. Graph the curve and tangent line together. The tangent intersects the curve at another point. Use Zoom and Trace to estimate the point’s coordinates.
T c. Confirm your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).
2 1 0
65. a. Find an equation for the line that is tangent to the curve y = x 3  x at the point s 1, 0d .
x
59. Quadratic tangent to identity function The curve y = ax 2 + bx + c passes through the point (1, 2) and is tangent to the line y = x at the origin. Find a, b, and c. 60. Quadratics having a common tangent The curves y = x 2 + ax + b and y = cx  x 2 have a common tangent line at the point (1, 0). Find a, b, and c. 61. Find all points (x, y) on the graph of ƒsxd = 3x 2  4x with tangent lines parallel to the line y = 8x + 5.
gsxd = e
ax, x 2  3x,
if x 6 0 if x Ú 0
70. Find the values of a and b that make the following function differentiable for all xvalues. ƒsxd = e
ax + b, bx 2  3,
x 7 1 x … 1
71. The general polynomial of degree n has the form Psxd = an x n + an  1 x n  1 + Á + a2 x 2 + a1 x + a0 where an Z 0 . Find P¿sxd .
3.4 72. The body’s reaction to medicine The reaction of the body to a dose of medicine can sometimes be represented by an equation of the form R = M2 a
The Derivative as a Rate of Change
139
c. What is the formula for the derivative of a product u1 u2 u3 Á un of a finite number n of differentiable functions of x? 76. Power Rule for negative integers Use the Derivative Quotient Rule to prove the Power Rule for negative integers, that is,
C M  b, 2 3
where C is a positive constant and M is the amount of medicine absorbed in the blood. If the reaction is a change in blood pressure, R is measured in millimeters of mercury. If the reaction is a change in temperature, R is measured in degrees, and so on. Find dR>dM . This derivative, as a function of M, is called the sensitivity of the body to the medicine. In Section 4.5, we will see how to find the amount of medicine to which the body is most sensitive.
d m (x ) = mx m  1 dx where m is a positive integer. 77. Cylinder pressure If gas in a cylinder is maintained at a constant temperature T, the pressure P is related to the volume V by a formula of the form nRT an 2  2, V  nb V in which a, b, n, and R are constants. Find dP>dV . (See accompanying figure.) P =
73. Suppose that the function y in the Derivative Product Rule has a constant value c. What does the Derivative Product Rule then say? What does this say about the Derivative Constant Multiple Rule? 74. The Reciprocal Rule a. The Reciprocal Rule says that at any point where the function y(x) is differentiable and different from zero, d 1 1 dy a b =  2 . dx y y dx Show that the Reciprocal Rule is a special case of the Derivative Quotient Rule. b. Show that the Reciprocal Rule and the Derivative Product Rule together imply the Derivative Quotient Rule. 75. Generalizing the Product Rule The Derivative Product Rule gives the formula d dy du suyd = u + y dx dx dx
78. The best quantity to order One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is hq km Asqd = q + cm + , 2
for the derivative of the product uy of two differentiable functions of x. a. What is the analogous formula for the derivative of the product uyw of three differentiable functions of x? b. What is the formula for the derivative of the product u1 u2 u3 u4 of four differentiable functions of x?
3.4
where q is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be); k is the cost of placing an order (the same, no matter how often you order); c is the cost of one item (a constant); m is the number of items sold each week (a constant); and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). Find dA>dq and d 2A>dq 2 .
The Derivative as a Rate of Change In Section 2.1 we introduced average and instantaneous rates of change. In this section we study further applications in which derivatives model the rates at which things change. It is natural to think of a quantity changing with respect to time, but other variables can be treated in the same way. For example, an economist may want to study how the cost of producing steel varies with the number of tons produced, or an engineer may want to know how the power output of a generator varies with its temperature.
Instantaneous Rates of Change If we interpret the difference quotient sƒsx + hd  ƒsxdd>h as the average rate of change in ƒ over the interval from x to x + h, we can interpret its limit as h : 0 as the rate at which ƒ is changing at the point x.
140
Chapter 3: Differentiation
DEFINITION
The instantaneous rate of change of ƒ with respect to x at x0 is
the derivative ƒ¿sx0 d = lim
h:0
ƒsx0 + hd  ƒsx0 d , h
provided the limit exists. Thus, instantaneous rates are limits of average rates. It is conventional to use the word instantaneous even when x does not represent time. The word is, however, frequently omitted. When we say rate of change, we mean instantaneous rate of change.
EXAMPLE 1
The area A of a circle is related to its diameter by the equation A =
p 2 D . 4
How fast does the area change with respect to the diameter when the diameter is 10 m? Solution
The rate of change of the area with respect to the diameter is dA p pD = # 2D = . 4 2 dD
When D = 10 m, the area is changing with respect to the diameter at the rate of sp>2d10 = 5p m2>m L 15.71 m2>m.
Motion Along a Line: Displacement, Velocity, Speed, Acceleration, and Jerk Suppose that an object is moving along a coordinate line (an saxis), usually horizontal or vertical, so that we know its position s on that line as a function of time t: s = ƒstd. Position at time t … s f(t)
Δs
The displacement of the object over the time interval from t to t + ¢t (Figure 3.14) is
and at time t Δ t s Δs f (t Δt)
FIGURE 3.14 The positions of a body moving along a coordinate line at time t and shortly later at time t + ¢t . Here the coordinate line is horizontal.
¢s = ƒst + ¢td  ƒstd,
s
and the average velocity of the object over that time interval is yay =
ƒst + ¢td  ƒstd displacement ¢s = = . travel time ¢t ¢t
To find the body’s velocity at the exact instant t, we take the limit of the average velocity over the interval from t to t + ¢t as ¢t shrinks to zero. This limit is the derivative of ƒ with respect to t.
DEFINITION Velocity (instantaneous velocity) is the derivative of position with respect to time. If a body’s position at time t is s = ƒstd, then the body’s velocity at time t is ystd =
ƒst + ¢td  ƒstd ds = lim . dt ¢t ¢t:0
3.4
s s f (t) ds 0 dt
t
0 (a) s increasing: positive slope so moving upward
141
Besides telling how fast an object is moving along the horizontal line in Figure 3.14, its velocity tells the direction of motion. When the object is moving forward (s increasing), the velocity is positive; when the object is moving backward (s decreasing), the velocity is negative. If the coordinate line is vertical, the object moves upward for positive velocity and downward for negative velocity. The blue curves in Figure 3.15 represent position along the line over time; they do not portray the path of motion, which lies along the saxis. If we drive to a friend’s house and back at 30 mph, say, the speedometer will show 30 on the way over but it will not show 30 on the way back, even though our distance from home is decreasing. The speedometer always shows speed, which is the absolute value of velocity. Speed measures the rate of progress regardless of direction.
DEFINITION
s
The Derivative as a Rate of Change
Speed is the absolute value of velocity. ds Speed = ƒ ystd ƒ = ` ` dt
s f (t) ds 0 dt
Figure 3.16 shows the graph of the velocity y = ƒ¿std of a particle moving along a horizontal line (as opposed to showing a position function s = ƒstd such as in Figure 3.15). In the graph of the velocity function, it’s not the slope of the curve that tells us if the particle is moving forward or backward along the line (which is not shown in the figure), but rather the sign of the velocity. Looking at Figure 3.16, we see that the particle moves forward for the first 3 sec (when the velocity is positive), moves backward for the next 2 sec (the velocity is negative), stands motionless for a full second, and then moves forward again. The particle is speeding up when its positive velocity increases during the first second, moves at a steady speed during the next second, and then slows down as the velocity decreases to zero during the third second. It stops for an instant at t = 3 sec (when the velocity is zero) and reverses direction as the velocity starts to become negative. The particle is now moving backward and gaining in speed until t = 4 sec, at which time it achieves its greatest speed during its backward motion. Continuing its backward motion at time t = 4, the particle starts to slow down again until it finally stops at time t = 5 (when the velocity is once again zero). The particle now remains motionless for one full second, and then moves forward again at t = 6 sec, speeding up during the final second of the forward motion indicated in the velocity graph.
EXAMPLE 2 t
0 (b) s decreasing: negative slope so moving downward
FIGURE 3.15 For motion s = ƒstd along a straight line (the vertical axis), y = ds>dt is (a) positive when s increases and (b) negative when s decreases.
HISTORICAL BIOGRAPHY Bernard Bolzano (1781–1848)
The rate at which a body’s velocity changes is the body’s acceleration. The acceleration measures how quickly the body picks up or loses speed. In Chapter 12 we will study motion in the plane and in space, where acceleration of an object may also lead to a change in direction. A sudden change in acceleration is called a jerk. When a ride in a car or a bus is jerky, it is not that the accelerations involved are necessarily large but that the changes in acceleration are abrupt.
DEFINITIONS Acceleration is the derivative of velocity with respect to time. If a body’s position at time t is s = ƒstd, then the body’s acceleration at time t is dy d 2s astd = = 2. dt dt Jerk is the derivative of acceleration with respect to time: da d 3s jstd = = 3. dt dt Near the surface of the Earth all bodies fall with the same constant acceleration. Galileo’s experiments with free fall (see Section 2.1) lead to the equation s =
1 2 gt , 2
142
Chapter 3: Differentiation y MOVES FORWARD
FORWARD AGAIN
(y 0)
(y 0)
Velocity y f '(t) Speeds up
Steady
Slows down
(y const)
Speeds up Stands still (y 0)
0
1
2
3
4
5
6
7
t (sec)
Greatest speed
Speeds up
Slows down
MOVES BACKWARD
(y 0)
FIGURE 3.16 The velocity graph of a particle moving along a horizontal line, discussed in Example 2.
where s is the distance fallen and g is the acceleration due to Earth’s gravity. This equation holds in a vacuum, where there is no air resistance, and closely models the fall of dense, heavy objects, such as rocks or steel tools, for the first few seconds of their fall, before the effects of air resistance are significant. The value of g in the equation s = s1>2dgt 2 depends on the units used to measure t and s. With t in seconds (the usual unit), the value of g determined by measurement at sea level is approximately 32 ft>sec2 (feet per second squared) in English units, and g = 9.8 m>sec2 (meters per second squared) in metric units. (These gravitational constants depend on the distance from Earth’s center of mass, and are slightly lower on top of Mt. Everest, for example.) The jerk associated with the constant acceleration of gravity sg = 32 ft>sec2 d is zero: j = t (seconds) t0
s (meters)
t1
5
0
10 15 t2
20 25
d sgd = 0. dt
An object does not exhibit jerkiness during free fall.
EXAMPLE 3
Figure 3.17 shows the free fall of a heavy ball bearing released from rest at time t = 0 sec. (a) How many meters does the ball fall in the first 3 sec? (b) What is its velocity, speed, and acceleration when t = 3?
30 35 40 t3
45
Solution
(a) The metric freefall equation is s = 4.9t 2 . During the first 3 sec, the ball falls ss3d = 4.9s3d2 = 44.1 m. (b) At any time t, velocity is the derivative of position:
FIGURE 3.17 A ball bearing falling from rest (Example 3).
ystd = s¿std =
d s4.9t 2 d = 9.8t. dt
3.4
The Derivative as a Rate of Change
143
At t = 3, the velocity is ys3d = 29.4 m>sec in the downward (increasing s) direction. The speed at t = 3 is speed = ƒ ys3d ƒ = 29.4 m>sec. The acceleration at any time t is astd = y¿std = s–std = 9.8 m>sec2 . At t = 3, the acceleration is 9.8 m>sec2 .
EXAMPLE 4
A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft> sec (about 109 mph) (Figure 3.18a). It reaches a height of s = 160t  16t 2 ft after t sec.
s y0
Height (ft)
smax
(a) How high does the rock go? (b) What are the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down? (c) What is the acceleration of the rock at any time t during its flight (after the blast)? (d) When does the rock hit the ground again?
t?
256
Solution
(a) In the coordinate system we have chosen, s measures height from the ground up, so the velocity is positive on the way up and negative on the way down. The instant the rock is at its highest point is the one instant during the flight when the velocity is 0. To find the maximum height, all we need to do is to find when y = 0 and evaluate s at this time. At any time t during the rock’s motion, its velocity is
s0 (a) s, y 400
y =
s 160t 16t 2
ds d = s160t  16t 2 d = 160  32t ft>sec. dt dt
The velocity is zero when 160
0 –160
160  32t = 0 5
10
t
y ds 160 32t dt (b)
FIGURE 3.18 (a) The rock in Example 4. (b) The graphs of s and y as functions of time; s is largest when y = ds>dt = 0 . The graph of s is not the path of the rock: It is a plot of height versus time. The slope of the plot is the rock’s velocity, graphed here as a straight line.
or
t = 5 sec.
The rock’s height at t = 5 sec is smax = ss5d = 160s5d  16s5d2 = 800  400 = 400 ft. See Figure 3.18b. (b) To find the rock’s velocity at 256 ft on the way up and again on the way down, we first find the two values of t for which sstd = 160t  16t 2 = 256. To solve this equation, we write 16t 2  160t + 256 = 0 16st 2  10t + 16d = 0 st  2dst  8d = 0 t = 2 sec, t = 8 sec. The rock is 256 ft above the ground 2 sec after the explosion and again 8 sec after the explosion. The rock’s velocities at these times are ys2d = 160  32s2d = 160  64 = 96 ft>sec. ys8d = 160  32s8d = 160  256 = 96 ft>sec.
144
Chapter 3: Differentiation
At both instants, the rock’s speed is 96 ft> sec. Since ys2d 7 0, the rock is moving upward (s is increasing) at t = 2 sec; it is moving downward (s is decreasing) at t = 8 because ys8d 6 0. (c) At any time during its flight following the explosion, the rock’s acceleration is a constant a =
dy d = s160  32td = 32 ft>sec2 . dt dt
The acceleration is always downward. As the rock rises, it slows down; as it falls, it speeds up. (d) The rock hits the ground at the positive time t for which s = 0. The equation 160t  16t 2 = 0 factors to give 16ts10  td = 0, so it has solutions t = 0 and t = 10. At t = 0, the blast occurred and the rock was thrown upward. It returned to the ground 10 sec later.
Derivatives in Economics Cost y (dollars) Slope marginal cost
y c (x)
xh x Production (tons/week)
0
x
FIGURE 3.19 Weekly steel production: c(x) is the cost of producing x tons per week. The cost of producing an additional h tons is csx + hd  csxd .
Engineers use the terms velocity and acceleration to refer to the derivatives of functions describing motion. Economists, too, have a specialized vocabulary for rates of change and derivatives. They call them marginals. In a manufacturing operation, the cost of production c(x) is a function of x, the number of units produced. The marginal cost of production is the rate of change of cost with respect to level of production, so it is dc>dx. Suppose that c(x) represents the dollars needed to produce x tons of steel in one week. It costs more to produce x + h tons per week, and the cost difference, divided by h, is the average cost of producing each additional ton: csx + hd  csxd average cost of each of the additional = h tons of steel produced. h The limit of this ratio as h : 0 is the marginal cost of producing more steel per week when the current weekly production is x tons (Figure 3.19): csx + hd  csxd dc = lim = marginal cost of production. dx h h:0 Sometimes the marginal cost of production is loosely defined to be the extra cost of producing one additional unit:
y
csx + 1d  csxd ¢c = , 1 ¢x which is approximated by the value of dc>dx at x. This approximation is acceptable if the slope of the graph of c does not change quickly near x. Then the difference quotient will be close to its limit dc>dx, which is the rise in the tangent line if ¢x = 1 (Figure 3.20). The approximation works best for large values of x. Economists often represent a total cost function by a cubic polynomial
y c(x)
x 1
⎧ ⎪ ⎨ c ⎪ ⎩
dc dx
csxd = ax 3 + bx 2 + gx + d
0
x
x1
x
FIGURE 3.20 The marginal cost dc>dx is approximately the extra cost ¢c of producing ¢ x = 1 more unit.
where d represents fixed costs, such as rent, heat, equipment capitalization, and management costs. The other terms represent variable costs, such as the costs of raw materials, taxes, and labor. Fixed costs are independent of the number of units produced, whereas variable costs depend on the quantity produced. A cubic polynomial is usually adequate to capture the cost behavior on a realistic quantity interval.
EXAMPLE 5
Suppose that it costs csxd = x 3  6x 2 + 15x
3.4
The Derivative as a Rate of Change
145
dollars to produce x radiators when 8 to 30 radiators are produced and that rsxd = x 3  3x 2 + 12x gives the dollar revenue from selling x radiators. Your shop currently produces 10 radiators a day. About how much extra will it cost to produce one more radiator a day, and what is your estimated increase in revenue for selling 11 radiators a day? Solution
The cost of producing one more radiator a day when 10 are produced is about
c¿s10d: c¿sxd =
d 3 A x  6x 2 + 15x B = 3x 2  12x + 15 dx
c¿s10d = 3s100d  12s10d + 15 = 195. The additional cost will be about $195. The marginal revenue is d 3 (x  3x 2 + 12x) = 3x 2  6x + 12. dx The marginal revenue function estimates the increase in revenue that will result from selling one additional unit. If you currently sell 10 radiators a day, you can expect your revenue to increase by about r¿sxd =
r¿s10d = 3s100d  6s10d + 12 = $252 if you increase sales to 11 radiators a day.
EXAMPLE 6 To get some feel for the language of marginal rates, consider marginal tax rates. If your marginal income tax rate is 28% and your income increases by $1000, you can expect to pay an extra $280 in taxes. This does not mean that you pay 28% of your entire income in taxes. It just means that at your current income level I, the rate of increase of taxes T with respect to income is dT>dI = 0.28. You will pay $0.28 in taxes out of every extra dollar you earn. Of course, if you earn a lot more, you may land in a higher tax bracket and your marginal rate will increase.
y 1
Sensitivity to Change
y 2p p 2 0
1
p
(a)
When a small change in x produces a large change in the value of a function ƒ(x), we say that the function is relatively sensitive to changes in x. The derivative ƒ¿sxd is a measure of this sensitivity.
EXAMPLE 7
dy /dp
Genetic Data and Sensitivity to Change
The Austrian monk Gregor Johann Mendel (1822–1884), working with garden peas and other plants, provided the first scientific explanation of hybridization. His careful records showed that if p (a number between 0 and 1) is the frequency of the gene for smooth skin in peas (dominant) and s1  pd is the frequency of the gene for wrinkled skin in peas, then the proportion of smoothskinned peas in the next generation will be
2
dy 2 2p dp
y = 2ps1  pd + p 2 = 2p  p 2 . 0
1
p
(b)
FIGURE 3.21 (a) The graph of y = 2p  p 2 , describing the proportion of smoothskinned peas in the next generation. (b) The graph of dy>dp (Example 7).
The graph of y versus p in Figure 3.21a suggests that the value of y is more sensitive to a change in p when p is small than when p is large. Indeed, this fact is borne out by the derivative graph in Figure 3.21b, which shows that dy>dp is close to 2 when p is near 0 and close to 0 when p is near 1. The implication for genetics is that introducing a few more smooth skin genes into a population where the frequency of wrinkled skin peas is large will have a more dramatic effect on later generations than will a similar increase when the population has a large proportion of smooth skin peas.
146
Chapter 3: Differentiation
Exercises 3.4 Motion Along a Coordinate Line Exercises 1–6 give the positions s = ƒstd of a body moving on a coordinate line, with s in meters and t in seconds. a. Find the body’s displacement and average velocity for the given time interval. b. Find the body’s speed and acceleration at the endpoints of the interval. c. When, if ever, during the interval does the body change direction? 1. s = t 2  3t + 2, 2. s = 6t  t , 2
0 … t … 2
0 … t … 6
3. s = t 3 + 3t 2  3t,
0 … t … 3
4. s = st >4d  t + t , 0 … t … 3 5 25 5. s = 2  t , 1 … t … 5 t 25 6. s = , 4 … t … 0 t + 5 4
3
2
7. Particle motion At time t, the position of a body moving along the saxis is s = t 3  6t 2 + 9t m . a. Find the body’s acceleration each time the velocity is zero. b. Find the body’s speed each time the acceleration is zero. c. Find the total distance traveled by the body from t = 0 to t = 2 . 8. Particle motion At time t Ú 0 , the velocity of a body moving along the horizontal saxis is y = t 2  4t + 3 . a. Find the body’s acceleration each time the velocity is zero.
12. Speeding bullet A 45caliber bullet shot straight up from the surface of the moon would reach a height of s = 832t  2.6t 2 ft after t sec. On Earth, in the absence of air, its height would be s = 832t  16t 2 ft after t sec. How long will the bullet be aloft in each case? How high will the bullet go? 13. Free fall from the Tower of Pisa Had Galileo dropped a cannonball from the Tower of Pisa, 179 ft above the ground, the ball’s height above the ground t sec into the fall would have been s = 179  16t 2 . a. What would have been the ball’s velocity, speed, and acceleration at time t ? b. About how long would it have taken the ball to hit the ground? c. What would have been the ball’s velocity at the moment of impact? 14. Galileo’s freefall formula Galileo developed a formula for a body’s velocity during free fall by rolling balls from rest down increasingly steep inclined planks and looking for a limiting formula that would predict a ball’s behavior when the plank was vertical and the ball fell freely; see part (a) of the accompanying figure. He found that, for any given angle of the plank, the ball’s velocity t sec into motion was a constant multiple of t. That is, the velocity was given by a formula of the form y = kt . The value of the constant k depended on the inclination of the plank. In modern notation—part (b) of the figure—with distance in meters and time in seconds, what Galileo determined by experiment was that, for any given angle u , the ball’s velocity t sec into the roll was y = 9.8ssin udt m>sec .
b. When is the body moving forward? Backward?
Freefall position
c. When is the body’s velocity increasing? Decreasing? FreeFall Applications 9. Free fall on Mars and Jupiter The equations for free fall at the surfaces of Mars and Jupiter (s in meters, t in seconds) are s = 1.86t 2 on Mars and s = 11.44t 2 on Jupiter. How long does it take a rock falling from rest to reach a velocity of 27.8 m> sec (about 100 km> h) on each planet? 10. Lunar projectile motion A rock thrown vertically upward from the surface of the moon at a velocity of 24 m> sec (about 86 km> h) reaches a height of s = 24t  0.8t 2 m in t sec. a. Find the rock’s velocity and acceleration at time t. (The acceleration in this case is the acceleration of gravity on the moon.) b. How long does it take the rock to reach its highest point? c. How high does the rock go?
?
θ
a. What is the equation for the ball’s velocity during free fall? b. Building on your work in part (a), what constant acceleration does a freely falling body experience near the surface of Earth? Understanding Motion from Graphs 15. The accompanying figure shows the velocity y = ds>dt = ƒstd (m> sec) of a body moving along a coordinate line.
d. How long does it take the rock to reach half its maximum height? e. How long is the rock aloft? 11. Finding g on a small airless planet Explorers on a small airless planet used a spring gun to launch a ball bearing vertically upward from the surface at a launch velocity of 15 m> sec. Because the acceleration of gravity at the planet’s surface was gs m>sec2 , the explorers expected the ball bearing to reach a height of s = 15t  s1>2dgs t 2 m t sec later. The ball bearing reached its maximum height 20 sec after being launched. What was the value of gs ?
(b)
(a)
y (m/sec) y f (t)
3 0
2
4
6
8 10
t (sec)
–3
a. When does the body reverse direction? b. When (approximately) is the body moving at a constant speed?
3.4 c. Graph the body’s speed for 0 … t … 10 .
The Derivative as a Rate of Change
147
18. The accompanying figure shows the velocity y = ƒstd of a particle moving on a horizontal coordinate line.
d. Graph the acceleration, where defined. 16. A particle P moves on the number line shown in part (a) of the accompanying figure. Part (b) shows the position of P as a function of time t.
y
y f(t)
P
s (cm)
0
0
1 2 3 4 5 6 7 8 9
t (sec)
(a) s (cm) s f (t)
2 0
1
2
3
4
a. When does the particle move forward? Move backward? Speed up? Slow down? 5
6
t (sec)
–2
c. When does the particle move at its greatest speed?
(6, 4)
–4
b. When is the particle’s acceleration positive? Negative? Zero? d. When does the particle stand still for more than an instant?
(b)
a. When is P moving to the left? Moving to the right? Standing still? b. Graph the particle’s velocity and speed (where defined).
19. Two falling balls The multiflash photograph in the accompanying figure shows two balls falling from rest. The vertical rulers are marked in centimeters. Use the equation s = 490t 2 (the freefall equation for s in centimeters and t in seconds) to answer the following questions.
17. Launching a rocket When a model rocket is launched, the propellant burns for a few seconds, accelerating the rocket upward. After burnout, the rocket coasts upward for a while and then begins to fall. A small explosive charge pops out a parachute shortly after the rocket starts down. The parachute slows the rocket to keep it from breaking when it lands. The figure here shows velocity data from the flight of the model rocket. Use the data to answer the following.
a. How fast was the rocket climbing when the engine stopped? b. For how many seconds did the engine burn? 200
Velocity (ft /sec)
150 100 50 0 –50 –100 0
2
4 6 8 10 Time after launch (sec)
12
c. When did the rocket reach its highest point? What was its velocity then? d. When did the parachute pop out? How fast was the rocket falling then? e. How long did the rocket fall before the parachute opened? f. When was the rocket’s acceleration greatest? g. When was the acceleration constant? What was its value then (to the nearest integer)?
a. How long did it take the balls to fall the first 160 cm? What was their average velocity for the period? b. How fast were the balls falling when they reached the 160cm mark? What was their acceleration then? c. About how fast was the light flashing (flashes per second)?
148
Chapter 3: Differentiation
20. A traveling truck The accompanying graph shows the position s of a truck traveling on a highway. The truck starts at t = 0 and returns 15 h later at t = 15 . a. Use the technique described in Section 3.2, Example 3, to graph the truck’s velocity y = ds>dt for 0 … t … 15 . Then repeat the process, with the velocity curve, to graph the truck’s acceleration dy>dt. b. Suppose that s = 15t 2  t 3 . Graph ds>dt and d 2s>dt 2 and compare your graphs with those in part (a).
Position, s (km)
500
Economics 23. Marginal cost Suppose that the dollar cost of producing x washing machines is csxd = 2000 + 100x  0.1x 2 . a. Find the average cost per machine of producing the first 100 washing machines. b. Find the marginal cost when 100 washing machines are produced. c. Show that the marginal cost when 100 washing machines are produced is approximately the cost of producing one more washing machine after the first 100 have been made, by calculating the latter cost directly.
400
24. Marginal revenue Suppose that the revenue from selling x washing machines is
300
1 rsxd = 20,000 a1  x b dollars.
200
a. Find the marginal revenue when 100 machines are produced. 100 0
5 10 Elapsed time, t (hr)
15
21. The graphs in the accompanying figure show the position s, velocity y = ds>dt , and acceleration a = d 2s>dt 2 of a body moving along a coordinate line as functions of time t. Which graph is which? Give reasons for your answers. y A
B
b. Use the function r¿sxd to estimate the increase in revenue that will result from increasing production from 100 machines a week to 101 machines a week. c. Find the limit of r¿sxd as x : q . How would you interpret this number? Additional Applications 25. Bacterium population When a bactericide was added to a nutrient broth in which bacteria were growing, the bacterium population continued to grow for a while, but then stopped growing and began to decline. The size of the population at time t (hours) was b = 10 6 + 10 4t  10 3t 2 . Find the growth rates at a. t = 0 hours .
C
b. t = 5 hours . c. t = 10 hours . t
0
22. The graphs in the accompanying figure show the position s, the velocity y = ds>dt , and the acceleration a = d 2s>dt 2 of a body moving along the coordinate line as functions of time t. Which graph is which? Give reasons for your answers. y A
26. Draining a tank The number of gallons of water in a tank t minutes after the tank has started to drain is Qstd = 200s30  td2 . How fast is the water running out at the end of 10 min? What is the average rate at which the water flows out during the first 10 min? T 27. Draining a tank It takes 12 hours to drain a storage tank by opening the valve at the bottom. The depth y of fluid in the tank t hours after the valve is opened is given by the formula y = 6 a1 
2
t b m. 12
a. Find the rate dy>dt (m> h) at which the tank is draining at time t. b. When is the fluid level in the tank falling fastest? Slowest? What are the values of dy>dt at these times?
0
t B
c. Graph y and dy>dt together and discuss the behavior of y in relation to the signs and values of dy>dt. 28. Inflating a balloon The volume V = s4>3dpr 3 of a spherical balloon changes with the radius. a. At what rate sft3>ftd does the volume change with respect to the radius when r = 2 ft ?
C
b. By approximately how much does the volume increase when the radius changes from 2 to 2.2 ft?
3.5
Derivatives of Trigonometric Functions
149
29. Airplane takeoff Suppose that the distance an aircraft travels along a runway before takeoff is given by D = s10>9dt 2 , where D is measured in meters from the starting point and t is measured in seconds from the time the brakes are released. The aircraft will become airborne when its speed reaches 200 km> h. How long will it take to become airborne, and what distance will it travel in that time?
velocity function ystd = ds>dt = ƒ¿std and the acceleration function astd = d 2s>dt 2 = ƒ–std . Comment on the object’s behavior in relation to the signs and values of y and a. Include in your commentary such topics as the following:
30. Volcanic lava fountains Although the November 1959 Kilauea Iki eruption on the island of Hawaii began with a line of fountains along the wall of the crater, activity was later confined to a single vent in the crater’s floor, which at one point shot lava 1900 ft straight into the air (a Hawaiian record). What was the lava’s exit velocity in feet per second? In miles per hour? (Hint: If y0 is the exit velocity of a particle of lava, its height t sec later will be s = y0 t  16t 2 ft . Begin by finding the time at which ds>dt = 0 . Neglect air resistance.)
b. When does it move to the left (down) or to the right (up)?
Analyzing Motion Using Graphs T Exercises 31–34 give the position function s = ƒstd of an object moving along the saxis as a function of time t. Graph ƒ together with the
3.5
a. When is the object momentarily at rest? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? 31. s = 200t  16t 2, 0 … t … 12.5 (a heavy object fired straight up from Earth’s surface at 200 ft> sec) 32. s = t 2  3t + 2,
0 … t … 5
33. s = t 3  6t 2 + 7t,
0 … t … 4
34. s = 4  7t + 6t  t 3, 2
0 … t … 4
Derivatives of Trigonometric Functions Many phenomena of nature are approximately periodic (electromagnetic fields, heart rhythms, tides, weather). The derivatives of sines and cosines play a key role in describing periodic changes. This section shows how to differentiate the six basic trigonometric functions.
Derivative of the Sine Function To calculate the derivative of ƒsxd = sin x, for x measured in radians, we combine the limits in Example 5a and Theorem 7 in Section 2.4 with the angle sum identity for the sine function: sin sx + hd = sin x cos h + cos x sin h. If ƒsxd = sin x, then ƒ¿sxd = lim
h:0
= lim
h:0
ƒsx + hd  ƒsxd sin sx + hd  sin x = lim h h h:0
Derivative definition
ssin x cos h + cos x sin hd  sin x sin x scos h  1d + cos x sin h = lim h h h:0
= lim asin x # h:0
= sin x # lim
h:0
cos h  1 sin h b + lim acos x # b h h h:0
cos h  1 sin h + cos x # lim = sin x # 0 + cos x # 1 = cos x. h h:0 h
(+++)+++* limit 0
(+)+* limit 1
The derivative of the sine function is the cosine function: d ssin xd = cos x. dx
Example 5a and Theorem 7, Section 2.4
150
Chapter 3: Differentiation
EXAMPLE 1
We find derivatives of the sine function involving differences, products,
and quotients. dy d = 2x (sin x) dx dx
(a) y = x 2  sin x:
Difference Rule
= 2x  cos x dy d d x = ex (sin x) + (e ) sin x dx dx dx = e x cos x + e x sin x = e x (cos x + sin x)
x (b) y = e sin x:
d x# (sin x)  sin x # 1 dy dx = dx x2
sin x (c) y = x :
=
Product Rule
Quotient Rule
x cos x  sin x x2
Derivative of the Cosine Function With the help of the angle sum formula for the cosine function, cos sx + hd = cos x cos h  sin x sin h, we can compute the limit of the difference quotient: cossx + hd  cos x d scos xd = lim dx h h:0 = lim
h:0
scos x cos h  sin x sin hd  cos x h
Derivative definition
Cosine angle sum identity
cos xscos h  1d  sin x sin h h h:0
= lim
= lim cos x # h:0
cos h  1 sin h  lim sin x # h h h:0
= cos x # lim
cos h  1 sin h  sin x # lim h h:0 h:0 h
y y cos x
1 –
0 –1 y'
= cos x # 0  sin x # 1
Example 5a and Theorem 7, Section 2.4
x
= sin x. y' –sin x
1 –
0 –1
FIGURE 3.22 The curve y¿ = sin x as the graph of the slopes of the tangents to the curve y = cos x .
x
The derivative of the cosine function is the negative of the sine function: d scos xd = sin x. dx
3.5
Derivatives of Trigonometric Functions
151
Figure 3.22 shows a way to visualize this result in the same way we did for graphing derivatives in Section 3.2, Figure 3.6.
EXAMPLE 2
We find derivatives of the cosine function in combinations with other
functions. (a) y = 5e x + cos x: dy d d = s5e x d + (cos x) dx dx dx
Sum Rule
= 5e x  sin x (b) y = sin x cos x: dy d d = sin x (cos x) + cos x (sin x) dx dx dx
Product Rule
= sin xs sin xd + cos xscos xd = cos2 x  sin2 x (c) y =
cos x : 1  sin x d d (1  sin x) (cos x)  cos x (1  sin x) dy dx dx = dx s1  sin xd2 =
s1  sin xds sin xd  cos xs0  cos xd s1  sin xd2
=
1  sin x s1  sin xd2
=
1 1  sin x
Quotient Rule
sin2 x + cos2 x = 1
Simple Harmonic Motion The motion of an object or weight bobbing freely up and down with no resistance on the end of a spring is an example of simple harmonic motion. The motion is periodic and repeats indefinitely, so we represent it using trigonometric functions. The next example describes a case in which there are no opposing forces such as friction to slow the motion.
EXAMPLE 3
A weight hanging from a spring (Figure 3.23) is stretched down 5 units beyond its rest position and released at time t = 0 to bob up and down. Its position at any later time t is
–5
s = 5 cos t. 0
5
Rest position Position at t0
s
FIGURE 3.23 A weight hanging from a vertical spring and then displaced oscillates above and below its rest position (Example 3).
What are its velocity and acceleration at time t ? We have s = 5 cos t Position:
Solution
Velocity:
y =
ds d = s5 cos td = 5 sin t dt dt
Acceleration:
a =
dy d = s 5 sin td = 5 cos t. dt dt
152
Chapter 3: Differentiation
Notice how much we can learn from these equations:
s, y 5
y –5 sin t
1.
s 5 cos t
2. 0
2
3 2
2 5 2
t
–5
3.
FIGURE 3.24 The graphs of the position and velocity of the weight in Example 3.
4.
As time passes, the weight moves down and up between s = 5 and s = 5 on the saxis. The amplitude of the motion is 5. The period of the motion is 2p, the period of the cosine function. The velocity y = 5 sin t attains its greatest magnitude, 5, when cos t = 0, as the graphs show in Figure 3.24. Hence, the speed of the weight, ƒ y ƒ = 5 ƒ sin t ƒ , is greatest when cos t = 0, that is, when s = 0 (the rest position). The speed of the weight is zero when sin t = 0. This occurs when s = 5 cos t = ;5, at the endpoints of the interval of motion. The weight is acted on by the spring and by gravity. When the weight is below the rest position, the combined forces pull it up, and when it is above the rest position, they pull it down. The weight’s acceleration is always proportional to the negative of its displacement. This property of springs is called Hooke’s Law, and is studied further in Section 6.5. The acceleration, a = 5 cos t, is zero only at the rest position, where cos t = 0 and the force of gravity and the force from the spring balance each other. When the weight is anywhere else, the two forces are unequal and acceleration is nonzero. The acceleration is greatest in magnitude at the points farthest from the rest position, where cos t = ;1.
EXAMPLE 4
The jerk associated with the simple harmonic motion in Example 3 is j =
da d = s 5 cos td = 5 sin t. dt dt
It has its greatest magnitude when sin t = ;1, not at the extremes of the displacement but at the rest position, where the acceleration changes direction and sign.
Derivatives of the Other Basic Trigonometric Functions Because sin x and cos x are differentiable functions of x, the related functions sin x tan x = cos x ,
cot x =
cos x , sin x
1 sec x = cos x ,
and
csc x =
1 sin x
are differentiable at every value of x at which they are defined. Their derivatives, calculated from the Quotient Rule, are given by the following formulas. Notice the negative signs in the derivative formulas for the cofunctions.
The derivatives of the other trigonometric functions: d stan xd = sec2 x dx
d scot xd = csc2 x dx
d ssec xd = sec x tan x dx
d scsc xd = csc x cot x dx
To show a typical calculation, we find the derivative of the tangent function. The other derivations are left to Exercise 60.
3.5
EXAMPLE 5 Solution
Derivatives of Trigonometric Functions
Find d(tan x)> dx.
We use the Derivative Quotient Rule to calculate the derivative: d d ssin xd  sin x scos xd dx dx cos2 x cos x cos x  sin x s sin xd = cos2 x
d sin x d b = stan xd = a dx dx cos x
EXAMPLE 6 Solution
153
cos x
=
cos2 x + sin2 x cos2 x
=
1 = sec2 x. cos2 x
Quotient Rule
Find y– if y = sec x.
Finding the second derivative involves a combination of trigonometric deriva
tives. y = sec x y¿ = sec x tan x y– =
Derivative rule for secant function
d ssec x tan xd dx
= sec x
d d (tan x) + tan x (sec x) dx dx
= sec xssec2 xd + tan xssec x tan xd = sec3 x + sec x tan2 x
Derivative Product Rule Derivative rules
The differentiability of the trigonometric functions throughout their domains gives another proof of their continuity at every point in their domains (Theorem 1, Section 3.2). So we can calculate limits of algebraic combinations and composites of trigonometric functions by direct substitution.
EXAMPLE 7
We can use direct substitution in computing limits provided there is no division by zero, which is algebraically undefined. 22 + sec x 22 + sec 0 22 + 1 23 =  23 = = = 1 cossp tan xd cossp tan 0d cossp 0d x:0 lim
Exercises 3.5 Derivatives In Exercises 1–18, find dy>dx.
5. y = csc x  41x + 7
1. y = 10x + 3 cos x
3 2. y = x + 5 sin x
3. y = x cos x
4. y = 2x sec x + 3
2
7. ƒsxd = sin x tan x
1 x2 8. gsxd = csc x cot x 6. y = x 2 cot x 
9. y = ssec x + tan xdssec x  tan xd 10. y = ssin x + cos xd sec x
154 11. y =
Chapter 3: Differentiation cot x 1 + cot x
4 1 13. y = cos x + tan x
12. y =
cos x 1 + sin x
14. y =
cos x x x + cos x
44. Find all points on the curve y = cot x, 0 6 x 6 p , where the tangent line is parallel to the line y = x . Sketch the curve and tangent(s) together, labeling each with its equation. In Exercises 45 and 46, find an equation for (a) the tangent to the curve at P and (b) the horizontal tangent to the curve at Q.
15. y = x sin x + 2x cos x  2 sin x 2
16. y = x 2 cos x  2x sin x  2 cos x 17. ƒsxd = x sin x cos x 3
45.
18. gsxd = s2  xd tan x
21. s =
y Q P ⎛ , 2⎛ ⎝2 ⎝
2
In Exercises 19–22, find ds>dt. 19. s = tan t  et
46.
y
2
20. s = t 2  sec t + 5e t
1 + csc t 1  csc t
22. s =
sin t 1  cos t
1
P ⎛ , 4⎛ ⎝4 ⎝
4
In Exercises 23–26, find dr>du . 23. r = 4  u 2 sin u
24. r = u sin u + cos u
25. r = sec u csc u
26. r = s1 + sec ud sin u
0
2 2 y 4 cot x 2csc x 1
x Q
In Exercises 27–32, find dp>dq. 27. p = 5 + 29. p = 31. p =
1 cot q
28. p = s1 + csc qd cos q
sin q + cos q cos q q sin q
q  1 33. Find y– if 2
30. p =
tan q 1 + tan q
32. p =
3q + tan q q sec q
a. y = csc x . a. y = 2 sin x .
b. y = 9 cos x .
49.
3p>2 … x … 2p
36. y = tan x,
p>2 6 x 6 p>2
x = p>3, 0, p>3 37. y = sec x,
p>2 6 x 6 p>2
x = p>3, p>4 38. y = 1 + cos x,
3p>2 … x … 2p
x = p>3, 3p>2 T Do the graphs of the functions in Exercises 39–42 have any horizontal tangents in the interval 0 … x … 2p ? If so, where? If not, why not? Visualize your findings by graphing the functions with a grapher. 39. y = x + sin x 40. y = 2x + sin x 41. y = x  cot x 42. y = x + 2 cos x 43. Find all points on the curve y = tan x, p>2 6 x 6 p>2 , where the tangent line is parallel to the line y = 2x . Sketch the curve and tangent(s) together, labeling each with its equation.
x
1 1 47. lim sin a x  b 2 x:2 48.
x = p, 0, 3p>2
3
Trigonometric Limits Find the limits in Exercises 47–54.
b. y = sec x .
Tangent Lines In Exercises 35–38, graph the curves over the given intervals, together with their tangents at the given values of x. Label each curve and tangent with its equation.
2
y 1 2 csc x cot x
lim
x:  p>6
34. Find y s4d = d 4 y>dx 4 if
35. y = sin x,
1 4
0
lim
u :p>6
21 + cos sp csc xd
sin u u 
1 2
51. lim sec ce x + p tan a x:0
52. lim sin a x:0
u :p>4
tan u  1 u  p4
p b  1d 4 sec x
p + tan x b tan x  2 sec x
53. lim tan a1 t: 0
lim
50.
p 6
sin t t b
54. lim cos a u :0
pu b sin u
Theory and Examples The equations in Exercises 55 and 56 give the position s = ƒstd of a body moving on a coordinate line (s in meters, t in seconds). Find the body’s velocity, speed, acceleration, and jerk at time t = p>4 sec . 55. s = 2  2 sin t
56. s = sin t + cos t
57. Is there a value of c that will make sin2 3x , x2 ƒsxd = L c,
x Z 0 x = 0
continuous at x = 0 ? Give reasons for your answer. 58. Is there a value of b that will make g sxd = e
x + b, cos x,
x 6 0 x Ú 0
continuous at x = 0 ? Differentiable at x = 0 ? Give reasons for your answers.
3.5
Derivatives of Trigonometric Functions
59. By computing the first few derivatives and looking for a pattern, find d 999>dx 999 scos xd .
See the accompanying figure.
60. Derive the formula for the derivative with respect to x of
y
a. sec x.
b. csc x.
155
Slope f '(x)
c. cot x.
61. A weight is attached to a spring and reaches its equilibrium position sx = 0d. It is then set in motion resulting in a displacement of
Slope
C
x = 10 cos t,
B
f(x h) f(x) h
A
where x is measured in centimeters and t is measured in seconds. See the accompanying figure.
Slope
f (x h) f(x h) 2h
y f (x) h 0
xh
h xh
x
x
–10
0
Equilibrium position at x 5 0
10
a. To see how rapidly the centered difference quotient for ƒsxd = sin x converges to ƒ¿sxd = cos x , graph y = cos x together with y =
sin sx + hd  sin sx  hd 2h
x
a. Find the spring’s displacement when t = 0, t = p>3, and t = 3p>4. b. Find the spring’s velocity when t = 0, t = p>3, and t = 3p>4. 62. Assume that a particle’s position on the xaxis is given by x = 3 cos t + 4 sin t,
over the interval [p, 2p] for h = 1, 0.5 , and 0.3. Compare the results with those obtained in Exercise 63 for the same values of h. b. To see how rapidly the centered difference quotient for ƒsxd = cos x converges to ƒ¿sxd = sin x , graph y = sin x together with
where x is measured in feet and t is measured in seconds. a. Find the particle’s position when t = 0, t = p>2, and t = p.
y =
b. Find the particle’s velocity when t = 0, t = p>2, and t = p. T 63. Graph y = cos x for p … x … 2p . On the same screen, graph y =
sin sx + hd  sin x h
for h = 1, 0.5, 0.3, and 0.1. Then, in a new window, try h = 1, 0.5 , and 0.3 . What happens as h : 0 + ? As h : 0  ? What phenomenon is being illustrated here? T 64. Graph y = sin x for p … x … 2p . On the same screen, graph cos sx + hd  cos x y = h for h = 1, 0.5, 0.3, and 0.1. Then, in a new window, try h = 1, 0.5 , and 0.3 . What happens as h : 0 + ? As h : 0  ? What phenomenon is being illustrated here? 65. Centered difference quotients The centered difference quotient T
ƒsx + hd  ƒsx  hd 2h is used to approximate ƒ¿sxd in numerical work because (1) its limit as h : 0 equals ƒ¿sxd when ƒ¿sxd exists, and (2) it usually gives a better approximation of ƒ¿sxd for a given value of h than the difference quotient ƒsx + hd  ƒsxd . h
cos sx + hd  cos sx  hd 2h
over the interval [p, 2p] for h = 1, 0.5 , and 0.3. Compare the results with those obtained in Exercise 64 for the same values of h. 66. A caution about centered difference quotients (Continuation of Exercise 65.) The quotient ƒsx + hd  ƒsx  hd 2h may have a limit as h : 0 when ƒ has no derivative at x. As a case in point, take ƒsxd = ƒ x ƒ and calculate lim
h:0
ƒ0 + hƒ  ƒ0  hƒ . 2h
As you will see, the limit exists even though ƒsxd = ƒ x ƒ has no derivative at x = 0 . Moral: Before using a centered difference quotient, be sure the derivative exists. T 67. Slopes on the graph of the tangent function Graph y = tan x and its derivative together on s p>2, p>2d . Does the graph of the tangent function appear to have a smallest slope? A largest slope? Is the slope ever negative? Give reasons for your answers.
156
Chapter 3: Differentiation
T 68. Slopes on the graph of the cotangent function Graph y = cot x and its derivative together for 0 6 x 6 p . Does the graph of the cotangent function appear to have a smallest slope? A largest slope? Is the slope ever positive? Give reasons for your answers.
b. With your grapher still in degree mode, estimate lim
h:0
T 69. Exploring (sin kx) / x Graph y = ssin xd>x , y = ssin 2xd>x , and y = ssin 4xd>x together over the interval 2 … x … 2 . Where does each graph appear to cross the yaxis? Do the graphs really intersect the axis? What would you expect the graphs of y = ssin 5xd>x and y = ssin s 3xdd>x to do as x : 0 ? Why? What about the graph of y = ssin kxd>x for other values of k ? Give reasons for your answers.
c. Now go back to the derivation of the formula for the derivative of sin x in the text and carry out the steps of the derivation using degreemode limits. What formula do you obtain for the derivative? d. Work through the derivation of the formula for the derivative of cos x using degreemode limits. What formula do you obtain for the derivative?
T 70. Radians versus degrees: degree mode derivatives What happens to the derivatives of sin x and cos x if x is measured in degrees instead of radians? To find out, take the following steps.
e. The disadvantages of the degreemode formulas become apparent as you start taking derivatives of higher order. Try it. What are the second and third degreemode derivatives of sin x and cos x?
a. With your graphing calculator or computer grapher in degree mode, graph ƒshd =
cos h  1 . h
sin h h
and estimate limh:0 ƒshd . Compare your estimate with p>180 . Is there any reason to believe the limit should be p>180 ?
3.6
The Chain Rule How do we differentiate F(x) = sin (x 2  4)? This function is the composite ƒ ⴰ g of two functions y = ƒ(u) = sin u and u = gsxd = x 2  4 that we know how to differentiate. The answer, given by the Chain Rule, says that the derivative is the product of the derivatives of ƒ and g. We develop the rule in this section.
Derivative of a Composite Function The function y =
3 1 1 x = s3xd is the composite of the functions y = u and u = 3x. 2 2 2
We have dy 3 = , 2 dx 2
3 1
C: y turns B: u turns A: x turns
FIGURE 3.25 When gear A makes x turns, gear B makes u turns and gear C makes y turns. By comparing circumferences or counting teeth, we see that y = u>2 (C turns onehalf turn for each B turn) and u = 3x (B turns three times for A’s one), so y = 3x>2 . Thus, dy>dx = 3>2 = s1>2ds3d = sdy>dudsdu>dxd .
Since
dy 1 = , 2 du
and
du = 3. dx
3 1 = # 3, we see in this case that 2 2 dy dy du # . = dx du dx
If we think of the derivative as a rate of change, our intuition allows us to see that this relationship is reasonable. If y = ƒsud changes half as fast as u and u = gsxd changes three times as fast as x, then we expect y to change 3>2 times as fast as x. This effect is much like that of a multiple gear train (Figure 3.25). Let’s look at another example.
EXAMPLE 1
The function y = s3x 2 + 1d2
3.6
The Chain Rule
157
is the composite of y = ƒ(u) = u 2 and u = g(x) = 3x 2 + 1. Calculating derivatives, we see that dy du # = 2u # 6x du dx = 2s3x 2 + 1d # 6x = 36x 3 + 12x. Calculating the derivative from the expanded formula (3x 2 + 1) 2 = 9x 4 + 6x 2 + 1 gives the same result: dy d = (9x 4 + 6x 2 + 1) dx dx = 36x 3 + 12x. The derivative of the composite function ƒ(g(x)) at x is the derivative of ƒ at g(x) times the derivative of g at x. This is known as the Chain Rule (Figure 3.26). Composite f g
˚
Rate of change at x is f '(g(x)) • g'(x).
x
g
f
Rate of change at x is g'(x).
Rate of change at g(x) is f '( g(x)).
u g(x)
y f (u) f(g(x))
FIGURE 3.26 Rates of change multiply: The derivative of ƒ ⴰ g at x is the derivative of ƒ at g(x) times the derivative of g at x.
THEOREM 2—The Chain Rule If ƒ(u) is differentiable at the point u = gsxd and g(x) is differentiable at x, then the composite function sƒ ⴰ gdsxd = ƒsgsxdd is differentiable at x, and sƒ ⴰ gd¿sxd = ƒ¿sgsxdd # g¿sxd. In Leibniz’s notation, if y = ƒsud and u = gsxd, then dy du dy # , = dx du dx where dy>du is evaluated at u = gsxd.
A Proof of One Case of the Chain Rule: Let ¢u be the change in u when x changes by ¢x, so that ¢u = gsx + ¢xd  gsxd. Then the corresponding change in y is ¢y = ƒsu + ¢ud  ƒsud. If ¢u Z 0, we can write the fraction ¢y>¢x as the product ¢y ¢y ¢u # = ¢x ¢u ¢x
(1)
158
Chapter 3: Differentiation
and take the limit as ¢x : 0: dy ¢y = lim dx ¢x:0 ¢x ¢y ¢u # = lim ¢x:0 ¢u ¢x ¢y # lim ¢u = lim ¢x:0 ¢u ¢x:0 ¢x ¢y # lim ¢u = lim ¢u:0 ¢u ¢x:0 ¢x dy du # . = du dx
(Note that ¢u : 0 as ¢x : 0 since g is continuous.)
The problem with this argument is that if the function g(x) oscillates rapidly near x, then ¢u can be zero even when ¢x Z 0, so the cancellation of ¢u in Equation (1) would be invalid. A complete proof requires a different approach that avoids this problem, and we give one such proof in Section 3.11.
EXAMPLE 2 An object moves along the xaxis so that its position at any time t Ú 0 is given by xstd = cosst 2 + 1d. Find the velocity of the object as a function of t. We know that the velocity is dx>dt. In this instance, x is a composite function: x = cossud and u = t 2 + 1. We have
Solution
dx = sin sud du du = 2t. dt
x = cossud u = t2 + 1
By the Chain Rule, dx dx # du = dt du dt = sin sud # 2t = sin st 2 + 1d # 2t = 2t sinst 2 + 1d.
dx evaluated at u du
“OutsideInside” Rule A difficulty with the Leibniz notation is that it doesn’t state specifically where the derivatives in the Chain Rule are supposed to be evaluated. So it sometimes helps to think about the Chain Rule using functional notation. If y = ƒ(g(x)), then dy = ƒ¿sgsxdd # g¿sxd. dx In words, differentiate the “outside” function ƒ and evaluate it at the “inside” function g(x) left alone; then multiply by the derivative of the “inside function.”
EXAMPLE 3 Solution
Differentiate sin sx 2 + e x d with respect to x.
We apply the Chain Rule directly and find d sin (x 2 + e x ) = cos (x 2 + e x ) # (2x + e x). (+)+* (+)+* (+)+* dx inside
inside left alone
derivative of the inside
3.6
EXAMPLE 4
The Chain Rule
159
Differentiate y = e cos x.
Here the inside function is u = g (x) = cos x and the outside function is the exponential function ƒ(x) = e x. Applying the Chain Rule, we get
Solution
dy d cos x d = (e ) = e cos x (cos x) = e cos x (sin x) = e cos x sin x. dx dx dx Generalizing Example 4, we see that the Chain Rule gives the formula d u du e = eu . dx dx For example, d d kx (e ) = e kx # (kx) = ke kx, dx dx
for any constant k
and 2 2 d d x2 A e B = e x # dx (x 2) = 2xe x . dx
Repeated Use of the Chain Rule We sometimes have to use the Chain Rule two or more times to find a derivative. HISTORICAL BIOGRAPHY
EXAMPLE 5
Johann Bernoulli (1667–1748)
Solution
Find the derivative of gstd = tan s5  sin 2td.
Notice here that the tangent is a function of 5  sin 2t, whereas the sine is a function of 2t, which is itself a function of t. Therefore, by the Chain Rule, d (tan (5  sin 2t)) dt d = sec2 s5  sin 2td # (5  sin 2t) dt
Derivative of tan u with u = 5  sin 2t
= sec2 s5  sin 2td # a0  cos 2t #
Derivative of 5  sin u with u = 2t
g¿std =
= sec2 s5  sin 2td # s cos 2td # 2 = 2scos 2td sec2 s5  sin 2td.
d (2t)b dt
The Chain Rule with Powers of a Function If ƒ is a differentiable function of u and if u is a differentiable function of x, then substituting y = ƒsud into the Chain Rule formula dy dy du # = dx du dx leads to the formula d du ƒsud = ƒ¿sud . dx dx If n is any real number and ƒ is a power function, ƒsud = u n , the Power Rule tells us that ƒ¿sud = nu n  1 . If u is a differentiable function of x, then we can use the Chain Rule to extend this to the Power Chain Rule: d n du su d = nu n  1 . dx dx
d A u n B = nu n  1 du
160
Chapter 3: Differentiation
EXAMPLE 6
The Power Chain Rule simplifies computing the derivative of a power of
an expression. (a)
d d s5x 3  x 4 d7 = 7s5x 3  x 4 d6 (5x 3  x 4) dx dx
= 7s5x 3  x 4 d6s5 # 3x 2  4x 3 d
Power Chain Rule with u = 5x 3  x 4, n = 7
= 7s5x 3  x 4 d6s15x 2  4x 3 d (b)
d d 1 a b = s3x  2d1 dx 3x  2 dx d s3x  2d dx = 1s3x  2d2s3d 3 = s3x  2d2 = 1s3x  2d2
Power Chain Rule with u = 3x  2, n =  1
In part (b) we could also find the derivative with the Derivative Quotient Rule. (c)
d d (sin5 x) = 5 sin4 x # sin x dx dx 4 = 5 sin x cos x
(d)
d d 23x + 1 Ae B = e 23x + 1 # dx A 23x + 1 B dx = e 23x + 1 # =
Power Chain Rule with u = sin x, n = 5, because sinn x means ssin xdn, n Z 1 .
1 (3x + 1) 1>2 # 3 2
3 223x + 1
Power Chain Rule with u = 3x + 1, n = 1>2
e 23x + 1
In Section 3.2, we saw that the absolute value function y = ƒ x ƒ is not differentiable at x 0. However, the function is differentiable at all other real numbers, as we now show. Since ƒ x ƒ = 2x 2, we can derive the following formula:
EXAMPLE 7
d d ( x ) = 2x 2 dx ƒ ƒ dx 1 # d 2 = (x ) 22x 2 dx 1 # = 2x 2ƒxƒ x = , x Z 0. ƒxƒ
Derivative of the Absolute Value Function d x , (ƒxƒ) = dx ƒxƒ
x Z 0
EXAMPLE 8
Power Chain Rule with u = x 2, n = 1>2, x Z 0 2x 2 = ƒ x ƒ
Show that the slope of every line tangent to the curve y = 1>s1  2xd3 is
positive. We find the derivative:
Solution
dy d = s1  2xd3 dx dx = 3s1  2xd4 #
d s1  2xd dx
= 3s1  2xd4 # s 2d =
6 . s1  2xd4
Power Chain Rule with u = s1  2xd, n = 3
3.6
161
The Chain Rule
At any point (x, y) on the curve, x Z 1>2 and the slope of the tangent line is dy 6 , = dx s1  2xd4 the quotient of two positive numbers.
EXAMPLE 9
The formulas for the derivatives of both sin x and cos x were obtained under the assumption that x is measured in radians, not degrees. The Chain Rule gives us new insight into the difference between the two. Since 180° = p radians, x° = px>180 radians where x° is the size of the angle measured in degrees. By the Chain Rule, d px d px p p b = cos a b = cossx°d. sin sx°d = sin a 180 180 180 180 dx dx See Figure 3.27. Similarly, the derivative of cossx°d is sp>180d sin sx°d. The factor p>180 would compound with repeated differentiation. We see here the advantage for the use of radian measure in computations. y sin(x°) sin x 180
y 1
x
180
y sin x
FIGURE 3.27 The function sin sx°d oscillates only p>180 times as often as sin x oscillates. Its maximum slope is p>180 at x = 0 (Example 9).
Exercises 3.6 Derivative Calculations In Exercises 1–8, given y = ƒsud and u = gsxd , find dy>dx = ƒ¿sgsxddg¿sxd . 1. y = 6u  9,
u = s1>2dx 4
2. y = 2u 3,
u = 8x  1
3. y = sin u,
u = 3x + 1
4. y = cos u,
u = x>3
5. y = cos u,
u = sin x
6. y = sin u,
u = x  cos x
7. y = tan u,
u = 10x  5
8. y = sec u,
u = x 2 + 7x
In Exercises 9–22, write the function in the form y = ƒsud and u = gsxd . Then find dy>dx as a function of x. 9. y = s2x + 1d5 11. y = a1 13. y = a
x b 7
10. y = s4  3xd9
7
x2 1 + x  xb 8
12. y = a
x  1b 2
10
4
19. y = e5x
20. y = e 2x>3
22. y = e A42x + x B
21. y = e 5  7x
2
Find the derivatives of the functions in Exercises 23–50. 23. p = 23  t 25. s =
3 2r  r 2 24. q = 2
4 4 sin 3t + cos 5t 5p 3p
26. s = sin a
3pt 3pt b + cos a b 2 2
27. r = scsc u + cot ud1
28. r = 6ssec u  tan ud3>2 x 1 29. y = x 2 sin4 x + x cos2 x 30. y = x sin5 x  cos3 x 3 1 1 1 b s3x  2d7 + a4 31. y = 2 21 2x 32. y = s5  2xd3 +
1 2 a + 1b 8 x
4
14. y = 23x 2  4x + 6
33. y = s4x + 3d4sx + 1d3
34. y = s2x  5d1sx 2  5xd6
35. y = xe x + e 3x
36. y = (1 + 2x) e2x
15. y = sec stan xd
1 16. y = cot ap  x b
37. y = (x  2x + 2) e
17. y = sin3 x
18. y = 5 cos4 x
2
5x>2
39. hsxd = x tan A 21x B + 7
38. y = (9x 2  6x + 2) e x 1 40. k sxd = x 2 sec a x b
3
162
Chapter 3: Differentiation
41. ƒsxd = 27 + x sec x 43. ƒsud = a
sin u b 1 + cos u
42. g sxd =
2
44. g std = a
t 2t + 1
49. y = cos A e u
2
1 + sin 3t b 3  2t
87. Suppose that functions ƒ and g and their derivatives with respect to x have the following values at x = 2 and x = 3 . 1
1 46. r = sec2u tan a b u
45. r = sin su2 d cos s2ud 47. q = sin a
tan 3x sx + 7d4
b
sin t 48. q = cot a t b
B
50. y = u3e2u cos 5u
51. y = sin2 spt  2d
52. y = sec2 pt
53. y = s1 + cos 2td4
54. y = s1 + cot st>2dd2
55. y = st tan td10
56. y = st 3>4 sin td4>3
2
59. y = a
58. y = A e sin (t>2) B
(pt  1)
t2 b t  4t
3
60. y = a
3
t bb 12
5
t 62. y = cos a5 sin a b b 3
61. y = sin scos s2t  5dd 63. y = a1 + tan4 a
3
3t  4 b 5t + 2
3
64. y =
1 A 1 + cos2 s7td B 3 6
65. y = 21 + cos st 2 d
66. y = 4 sin A 21 + 1t B
67. y = tan2 ssin3 td
68. y = cos4 ssec2 3td
69. y = 3t s2t 2  5d4
70. y = 43t + 32 + 21  t
1
1 73. y = cot s3x  1d 9
x 74. y = 9 tan a b 3
75. y = x s2x + 1d4
76. y = x 2 sx 3  1d5
77. y = e
x2
+ 5x
78. y = sin (x 2e x )
Finding Derivative Values In Exercises 79–84, find the value of sƒ ⴰ gd¿ at the given value of x. 79. ƒsud = u 5 + 1,
u = g sxd = 1x,
x = 1
1 1 , x = 1 80. ƒsud = 1  u , u = g sxd = 1  x pu , u = g sxd = 5 1x, x = 1 81. ƒsud = cot 10 1 , u = g sxd = px, x = 1>4 82. ƒsud = u + cos2 u 2u , u = g sxd = 10x 2 + x + 1, x = 0 83. ƒsud = 2 u + 1 84. ƒsud = a
2
u  1 b , u + 1
ƒ(x)
g(x)
2 3
8 3
2 4
1>3 2p
3 5
a. 2ƒsxd,
b. ƒsxd + g sxd,
x = 2
c. ƒsxd # g sxd,
x = 3
x = 3
x = 2
d. ƒsxd>g sxd,
e. ƒsg sxdd,
x = 2
f. 2ƒsxd,
g. 1>g 2sxd,
x = 3
h. 2ƒ2sxd + g 2sxd,
x = 2 x = 2
88. Suppose that the functions ƒ and g and their derivatives with respect to x have the following values at x = 0 and x = 1 . x
ƒ(x)
g(x)
ƒ(x)
g(x)
0 1
1 3
1 4
5 1>3
1>3 8>3
Find the derivatives with respect to x of the following combinations at the given value of x. a. 5ƒsxd  g sxd, x = 1 ƒsxd c. , x = 1 g sxd + 1 e. g sƒsxdd,
x = 0
b. ƒsxdg 3sxd, d. ƒsg sxdd,
x = 0
f. sx 11 + ƒsxdd2,
x = 0
x = 1
x = 0
89. Find ds>dt when u = 3p>2 if s = cos u and d u>dt = 5 . 72. y = A 1  1x B
3
g(x)
g. ƒsx + g sxdd,
Second Derivatives Find y– in Exercises 71–78. 1 71. y = a1 + x b
ƒ(x)
Find the derivatives with respect to x of the following combinations at the given value of x.
In Exercises 51–70, find dy>dt.
57. y = ecos
x
u = g sxd =
1  1, x2
x = 1
85. Assume that ƒ¿s3d = 1, g¿s2d = 5, gs2d = 3, and y = ƒsgsxdd. What is y¿ at x = 2? 86. If r = sin sƒstdd, ƒs0d = p>3, and ƒ¿s0d = 4, then what is dr>dt at t = 0?
90. Find dy>dt when x = 1 if y = x 2 + 7x  5 and dx>dt = 1>3 . Theory and Examples What happens if you can write a function as a composite in different ways? Do you get the same derivative each time? The Chain Rule says you should. Try it with the functions in Exercises 91 and 92. 91. Find dy>dx if y = x by using the Chain Rule with y as a composite of a. y = su>5d + 7
and
u = 5x  35
b. y = 1 + s1>ud
and
u = 1>sx  1d .
92. Find dy>dx if y = x 3>2 by using the Chain Rule with y as a composite of a. y = u 3 b. y = 1u
and and
u = 1x u = x3 .
93. Find the tangent to y = ssx  1d>sx + 1dd2 at x = 0. 94. Find the tangent to y = 2x 2  x + 7 at x = 2. 95. a. Find the tangent to the curve y = 2 tan spx>4d at x = 1 . b. Slopes on a tangent curve What is the smallest value the slope of the curve can ever have on the interval 2 6 x 6 2 ? Give reasons for your answer. 96. Slopes on sine curves a. Find equations for the tangents to the curves y = sin 2x and y = sin sx>2d at the origin. Is there anything special about how the tangents are related? Give reasons for your answer.
3.6
The Chain Rule
163
b. Can anything be said about the tangents to the curves y = sin mx and y = sin sx>md at the origin sm a constant Z 0d ? Give reasons for your answer.
102. Particle acceleration A particle moves along the xaxis with velocity dx>dt = ƒsxd . Show that the particle’s acceleration is ƒsxdƒ¿sxd .
c. For a given m, what are the largest values the slopes of the curves y = sin mx and y =  sin sx>md can ever have? Give reasons for your answer.
103. Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period T and the length L of a simple pendulum with the equation
d. The function y = sin x completes one period on the interval [0, 2p] , the function y = sin 2x completes two periods, the function y = sin sx>2d completes half a period, and so on. Is there any relation between the number of periods y = sin mx completes on [0, 2p] and the slope of the curve y = sin mx at the origin? Give reasons for your answer. 97. Running machinery too fast Suppose that a piston is moving straight up and down and that its position at time t sec is s = A cos s2pbtd , with A and b positive. The value of A is the amplitude of the motion, and b is the frequency (number of times the piston moves up and down each second). What effect does doubling the frequency have on the piston’s velocity, acceleration, and jerk? (Once you find out, you will know why some machinery breaks when you run it too fast.) 98. Temperatures in Fairbanks, Alaska The graph in the accompanying figure shows the average Fahrenheit temperature in Fairbanks, Alaska, during a typical 365day year. The equation that approximates the temperature on day x is y = 37 sin c
2p sx  101d d + 25 365
L , Ag
T = 2p
where g is the constant acceleration of gravity at the pendulum’s location. If we measure g in centimeters per second squared, we measure L in centimeters and T in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to L. In symbols, with u being temperature and k the proportionality constant, dL = kL . du Assuming this to be the case, show that the rate at which the period changes with respect to temperature is kT>2. Suppose that ƒsxd = x 2 and g sxd = ƒ x ƒ . Then the
104. Chain Rule composites
sƒ ⴰ gdsxd = ƒ x ƒ 2 = x 2
T 105. The derivative of sin 2x Graph the function y = 2 cos 2x for 2 … x … 3.5 . Then, on the same screen, graph y =
a. On what day is the temperature increasing the fastest? b. About how many degrees per day is the temperature increasing when it is increasing at its fastest? y
40 20 0 ...... ..
for h = 1.0, 0.5 , and 0.2. Experiment with other values of h, including negative values. What do you see happening as h : 0 ? Explain this behavior.
y = . ... .... ............ ..... .
. .... ....
x
Ja
–20
. .. .. .. ... .... ....
sin 2sx + hd  sin 2x h
106. The derivative of cos sx 2 d Graph y = 2x sin sx 2 d for 2 … x … 3 . Then, on the same screen, graph
n Fe b M ar A pr M ay Ju n Ju l A ug Se p O ct N ov D ec Ja n Fe b M ar
Temperature (˚F)
60
sg ⴰ ƒdsxd = ƒ x 2 ƒ = x 2
are both differentiable at x = 0 even though g itself is not differentiable at x = 0 . Does this contradict the Chain Rule? Explain.
and is graphed in the accompanying figure.
....... .... . ........ .. . .... .... ... . .... . ... ... ... ... ... ... ..
and
for h = 1.0, 0.7, and 0.3 . Experiment with other values of h. What do you see happening as h : 0 ? Explain this behavior. Using the Chain Rule, show that the Power Rule sd>dxdx n = nx n  1 holds for the functions xn in Exercises 107 and 108. 107. x1>4 = 2 1x
99. Particle motion The position of a particle moving along a coordinate line is s = 21 + 4t , with s in meters and t in seconds. Find the particle’s velocity and acceleration at t = 6 sec. 100. Constant acceleration Suppose that the velocity of a falling body is y = k1s m>sec (k a constant) at the instant the body has fallen s m from its starting point. Show that the body’s acceleration is constant. 101. Falling meteorite The velocity of a heavy meteorite entering Earth’s atmosphere is inversely proportional to 1s when it is s km from Earth’s center. Show that the meteorite’s acceleration is inversely proportional to s 2 .
cos ssx + hd2 d  cos sx 2 d h
108. x3>4 = 2x1x
COMPUTER EXPLORATIONS Trigonometric Polynomials 109. As the accompanying figure shows, the trigonometric “polynomial” s = ƒstd = 0.78540  0.63662 cos 2t  0.07074 cos 6t  0.02546 cos 10t  0.01299 cos 14t gives a good approximation of the sawtooth function s = g std on the interval [p, p] . How well does the derivative of ƒ approximate the derivative of g at the points where dg>dt is defined? To find out, carry out the following steps.
164
Chapter 3: Differentiation s = hstd = 1.2732 sin 2t + 0.4244 sin 6t + 0.25465 sin 10t
a. Graph dg>dt (where defined) over [p, p] .
+ 0.18189 sin 14t + 0.14147 sin 18t
b. Find dƒ>dt. c. Graph dƒ>dt. Where does the approximation of dg>dt by dƒ>dt seem to be best? Least good? Approximations by trigonometric polynomials are important in the theories of heat and oscillation, but we must not expect too much of them, as we see in the next exercise. s
graphed in the accompanying figure approximates the step function s = kstd shown there. Yet the derivative of h is nothing like the derivative of k. s s k(t)
s g(t) s f (t)
2
– –
0
s h(t)
1
t
– 2
2
0
t
–1
110. (Continuation of Exercise 109.) In Exercise 109, the trigonometric polynomial ƒ(t) that approximated the sawtooth function g(t) on [p, p] had a derivative that approximated the derivative of the sawtooth function. It is possible, however, for a trigonometric polynomial to approximate a function in a reasonable way without its derivative approximating the function’s derivative at all well. As a case in point, the “polynomial”
a. Graph dk>dt (where defined) over [p, p] . b. Find dh>dt. c. Graph dh>dt to see how badly the graph fits the graph of dk>dt. Comment on what you see.
Implicit Differentiation
3.7
Most of the functions we have dealt with so far have been described by an equation of the form y = ƒsxd that expresses y explicitly in terms of the variable x. We have learned rules for differentiating functions defined in this way. Another situation occurs when we encounter equations like x 3 + y 3  9xy = 0,
y
5
(x 0, y 1) A
x 3 y 3 9xy 0
x 2 + y 2  25 = 0.
y f2(x)
(x 0, y 2) x0
(x 0, y3)
or
(See Figures 3.28, 3.29, and 3.30.) These equations define an implicit relation between the variables x and y. In some cases we may be able to solve such an equation for y as an explicit function (or even several functions) of x. When we cannot put an equation Fsx, yd = 0 in the form y = ƒsxd to differentiate it in the usual way, we may still be able to find dy>dx by implicit differentiation. This section describes the technique.
y f1(x)
0
y 2  x = 0,
5
x
y f3 (x)
FIGURE 3.28 The curve x 3 + y 3  9xy = 0 is not the graph of any one function of x. The curve can, however, be divided into separate arcs that are the graphs of functions of x. This particular curve, called a folium, dates to Descartes in 1638.
Implicitly Defined Functions We begin with examples involving familiar equations that we can solve for y as a function of x to calculate dy>dx in the usual way. Then we differentiate the equations implicitly, and find the derivative to compare the two methods. Following the examples, we summarize the steps involved in the new method. In the examples and exercises, it is always assumed that the given equation determines y implicitly as a differentiable function of x so that dy>dx exists.
EXAMPLE 1
Find dy>dx if y 2 = x.
The equation y 2 = x defines two differentiable functions of x that we can actually find, namely y1 = 1x and y2 =  1x (Figure 3.29). We know how to calculate the derivative of each of these for x 7 0:
Solution
dy1 1 = dx 21x
and
dy2 1 = . dx 21x
3.7
y2 x
y
Slope 1 1 2y1 2x
y1 x
P(x, x )
165
But suppose that we knew only that the equation y 2 = x defined y as one or more differentiable functions of x for x 7 0 without knowing exactly what these functions were. Could we still find dy>dx? The answer is yes. To find dy>dx, we simply differentiate both sides of the equation y 2 = x with respect to x, treating y = ƒsxd as a differentiable function of x:
x
0
Implicit Differentiation
y2 = x dy = 1 2y dx
Q(x, x ) y 2 x Slope 1 1 2y 2 2x
The Chain Rule gives
d 2 Ay B = dx
dy d 2 [ƒsxd] = 2ƒsxdƒ¿sxd = 2y . dx dx
dy 1 . = 2y dx
FIGURE 3.29 The equation y 2  x = 0 , or y 2 = x as it is usually written, defines two differentiable functions of x on the interval x 7 0. Example 1 shows how to find the derivatives of these functions without solving the equation y 2 = x for y.
This one formula gives the derivatives we calculated for both explicit solutions y1 = 1x and y2 =  1x: dy1 1 1 = = 2y1 dx 21x
y y1 25
EXAMPLE 2 x2
and
dy2 1 1 1 = = . = 2y2 dx 21x 2 A  1x B
Find the slope of the circle x 2 + y 2 = 25 at the point s3, 4d.
The circle is not the graph of a single function of x. Rather, it is the combined graphs of two differentiable functions, y1 = 225  x 2 and y2 =  225  x 2 (Figure 3.30). The point s3, 4d lies on the graph of y2 , so we can find the slope by calculating the derivative directly, using the Power Chain Rule:
Solution
–5
0
5
x
(3, – 4) y2 –25 x 2
Slope – xy 3 4
dy2 6 3 2x ` = ` = = . 2 x=3 4 dx x = 3 2225  x 2225  9
d a(25  x2)1>2 b = dx 1  (25  x 2) 1>2(2x) 2
We can solve this problem more easily by differentiating the given equation of the circle implicitly with respect to x:
FIGURE 3.30 The circle combines the graphs of two functions. The graph of y2 is the lower semicircle and passes through s3, 4d .
d 2 d 2 d (x ) + (y ) = (25) dx dx dx 2x + 2y
dy = 0 dx dy x = y. dx
x The slope at s3, 4d is  y `
s3, 4d
= 
3 3 = . 4 4
Notice that unlike the slope formula for dy2>dx, which applies only to points below the xaxis, the formula dy>dx = x>y applies everywhere the circle has a slope. Notice also that the derivative involves both variables x and y, not just the independent variable x. To calculate the derivatives of other implicitly defined functions, we proceed as in Examples 1 and 2: We treat y as a differentiable implicit function of x and apply the usual rules to differentiate both sides of the defining equation.
166
Chapter 3: Differentiation
Implicit Differentiation 1. Differentiate both sides of the equation with respect to x, treating y as a differentiable function of x. 2. Collect the terms with dy>dx on one side of the equation and solve for dy>dx.
EXAMPLE 3
y 4
y2 x2 sin xy
Solution
We differentiate the equation implicitly. y 2 = x 2 + sin xy
2
–4
–2
0
Find dy>dx if y 2 = x 2 + sin xy (Figure 3.31).
2
4
d 2 d d A y B = dx A x 2 B + dx A sin xy B dx
x
–2 –4
FIGURE 3.31 The graph of y 2 = x 2 + sin xy in Example 3.
2y
Differentiate both sides with respect to x Á
2y
dy d = 2x + scos xyd A xy B dx dx
Á treating y as a function of x and using the Chain Rule.
2y
dy dy = 2x + scos xyd ay + x b dx dx
Treat xy as a product.
dy dy  scos xyd ax b = 2x + scos xydy dx dx s2y  x cos xyd
Collect terms with dy>dx.
dy = 2x + y cos xy dx dy 2x + y cos xy = 2y  x cos xy dx
Solve for dy>dx.
Notice that the formula for dy>dx applies everywhere that the implicitly defined curve has a slope. Notice again that the derivative involves both variables x and y, not just the independent variable x.
Derivatives of Higher Order Implicit differentiation can also be used to find higher derivatives.
EXAMPLE 4
Find d 2y>dx 2 if 2x 3  3y 2 = 8.
To start, we differentiate both sides of the equation with respect to x in order to find y¿ = dy>dx.
Solution
d d (2x 3  3y 2) = s8d dx dx 6x 2  6yy¿ = 0
Treat y as a function of x. 2
x y¿ = y ,
when y Z 0
Solve for y¿.
We now apply the Quotient Rule to find y– . y– =
2xy  x 2y¿ d x2 2x x2 ay b = = y  2 # y¿ 2 dx y y
Finally, we substitute y¿ = x 2>y to express y– in terms of x and y. 2x x2 x2 2x x4 y– = y  2 a y b = y  3 , y y
when y Z 0
3.7
Curve of lens surface
Normal line
167
Lenses, Tangents, and Normal Lines
Tangent
Light ray
Implicit Differentiation
A Point of entry P
In the law that describes how light changes direction as it enters a lens, the important angles are the angles the light makes with the line perpendicular to the surface of the lens at the point of entry (angles A and B in Figure 3.32). This line is called the normal to the surface at the point of entry. In a profile view of a lens like the one in Figure 3.32, the normal is the line perpendicular to the tangent of the profile curve at the point of entry.
B
Show that the point (2, 4) lies on the curve x 3 + y 3  9xy = 0. Then find the tangent and normal to the curve there (Figure 3.33).
EXAMPLE 5
FIGURE 3.32 The profile of a lens, showing the bending (refraction) of a ray of light as it passes through the lens surface.
The point (2, 4) lies on the curve because its coordinates satisfy the equation given for the curve: 23 + 43  9s2ds4d = 8 + 64  72 = 0. To find the slope of the curve at (2, 4), we first use implicit differentiation to find a formula for dy>dx:
Solution
y
x 3 + y 3  9xy = 0
nt
ge
n Ta
d 3 d 3 d d (x ) + (y ) (9xy) = (0) dx dx dx dx
4 No al
rm
x 3 y 3 9xy 0
0
2
3x 2 + 3y 2 x
FIGURE 3.33 Example 5 shows how to find equations for the tangent and normal to the folium of Descartes at (2, 4).
dy dy dx  9 ax + y b = 0 dx dx dx
s3y 2  9xd
dy + 3x 2  9y = 0 dx 3sy 2  3xd
Differentiate both sides with respect to x. Treat xy as a product and y as a function of x.
dy = 9y  3x 2 dx dy 3y  x 2 = 2 . dx y  3x
Solve for dy>dx.
We then evaluate the derivative at sx, yd = s2, 4d: dy 3y  x 2 3s4d  22 8 4 = = . ` = 2 ` = 2 5 10 dx s2, 4d y  3x s2, 4d 4  3s2d The tangent at (2, 4) is the line through (2, 4) with slope 4>5: y = 4 + y =
4 sx  2d 5
4 12 x + . 5 5
The normal to the curve at (2, 4) is the line perpendicular to the tangent there, the line through (2, 4) with slope 5>4: y = 4 y = 
5 sx  2d 4
5 13 x + . 4 2
168
Chapter 3: Differentiation
Exercise 3.7 Differentiating Implicitly Use implicit differentiation to find dy>dx in Exercises 1–16. 1. x 2y + xy 2 = 6
2. x 3 + y 3 = 18xy
3. 2xy + y = x + y
4. x 3  xy + y 3 = 1
5. x 2sx  yd2 = x 2  y 2
6. s3xy + 7d2 = 6y 2x  y 8. x 3 = x + 3y
2
7. y 2 =
x  1 x + 1
9. x = tan y
41. Parallel tangents Find the two points where the curve x 2 + xy + y 2 = 7 crosses the xaxis, and show that the tangents to the curve at these points are parallel. What is the common slope of these tangents? 42. Normals parallel to a line Find the normals to the curve xy + 2x  y = 0 that are parallel to the line 2x + y = 0 . 43. The eight curve Find the slopes of the curve y 4 = y 2  x 2 at the two points shown here.
10. xy = cot sxyd
11. x + tan (xy) = 0
12. x 4 + sin y = x 3y 2
1 13. y sin a y b = 1  xy
14. x cos s2x + 3yd = y sin x
15. e 2x = sin (x + 3y)
16. e x
2
y
y
⎛3 , 3 ⎛ ⎝ 4 2 ⎝
1
⎛3 , 1 ⎛ ⎝ 4 2⎝
= 2x + 2y
Find dr>du in Exercises 17–20. + r
x
17. u
1>2
Second Derivatives In Exercises 21–26, use implicit differentiation to find dy>dx and then d 2y>dx 2 . 21. x + y = 1 2
2
23. y = e
x2
2
0
3 4 18. r  2 2u = u2>3 + u3>4 2 3 20. cos r + cot u = e ru
= 1 1 19. sin sr ud = 2 1>2
22. x
+ 2x
2>3
+ y
2>3
= 1
24. y  2x = 1  2y
y4 5 y2 2 x2 –1
44. The cissoid of Diocles (from about 200 B.C.) Find equations for the tangent and normal to the cissoid of Diocles y 2s2  xd = x 3 at (1, 1).
2
y
26. xy + y 2 = 1
25. 2 1y = x  y
y 2(2 2 x) 5 x 3
27. If x + y = 16 , find the value of d y>dx at the point (2, 2). 3
3
2
2
28. If xy + y 2 = 1 , find the value of d 2y>dx 2 at the point s0, 1d . 29. y 2 + x 2 = y 4  2x
s 2, 1d and s 2, 1d
at
30. sx 2 + y 2 d2 = sx  yd2
s1, 0d and s1, 1d
at
Slopes, Tangents, and Normals In Exercises 31–40, verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. 31. x 2 + xy  y 2 = 1, 32. x + y = 25, 2
2
33. x 2y 2 = 9,
s2, 3d
0
1
x
45. The devil’s curve (Gabriel Cramer, 1750) Find the slopes of the devil’s curve y 4  4y 2 = x 4  9x 2 at the four indicated points. y
s3, 4d
s 1, 3d
y 4 2 4y 2 5 x 4 2 9x 2
34. y 2  2x  4y  1 = 0,
s 2, 1d
35. 6x 2 + 3xy + 2y 2 + 17y  6 = 0, 36. x  23xy + 2y = 5, 2
(1, 1)
1
In Exercises 29 and 30, find the slope of the curve at the given points.
2
37. 2xy + p sin y = 2p,
A 23, 2 B
s1, p>2d
38. x sin 2y = y cos 2x,
sp>4, p>2d
39. y = 2 sin spx  yd,
s1, 0d
40. x 2 cos2 y  sin y = 0,
s0, pd
s 1, 0d
(–3, 2)
2
–3
(–3, –2)
–2
(3, 2) x 3 (3, –2)
3.7 46. The folium of Descartes (See Figure 3.28.) a. Find the slope of the folium of Descartes x + y  9xy = 0 at the points (4, 2) and (2, 4). 3
3
Implicit Differentiation
169
52. The graph of y 2 = x 3 is called a semicubical parabola and is shown in the accompanying figure. Determine the constant b so that the line y =  13 x + b meets this graph orthogonally.
b. At what point other than the origin does the folium have a horizontal tangent?
y
y2 5 x3
c. Find the coordinates of the point A in Figure 3.28, where the folium has a vertical tangent. Theory and Examples 47. Intersecting normal The line that is normal to the curve x 2 + 2xy  3y 2 = 0 at (1, 1) intersects the curve at what other point?
y 5 21 x 1 b 3 x
0
48. Power rule for rational exponents Let p and q be integers with q 7 0. If y = x p>q, differentiate the equivalent equation y q = x p implicitly and show that, for y Z 0, d p>q p (p>q) 1 . x = qx dx 49. Normals to a parabola Show that if it is possible to draw three normals from the point (a, 0) to the parabola x = y 2 shown in the accompanying diagram, then a must be greater than 1>2. One of the normals is the xaxis. For what value of a are the other two normals perpendicular?
T In Exercises 53 and 54, find both dy>dx (treating y as a differentiable function of x) and dx>dy (treating x as a differentiable function of y). How do dy>dx and dx>dy seem to be related? Explain the relationship geometrically in terms of the graphs. 53. xy 3 + x 2y = 6 54. x 3 + y 2 = sin2 y
y x y2
x
(a, 0)
0
COMPUTER EXPLORATIONS Use a CAS to perform the following steps in Exercises 55–62.
b. Using implicit differentiation, find a formula for the derivative dy>dx and evaluate it at the given point P.
50. Is there anything special about the tangents to the curves y 2 = x 3 and 2x 2 + 3y 2 = 5 at the points s1, ;1d ? Give reasons for your answer. y
3y 2
5 (1, 1) x
0 (1, –1)
51. Verify that the following pairs of curves meet orthogonally. a. x 2 + y 2 = 4, b. x = 1  y 2,
x 2 = 3y 2 x =
1 2 y 3
c. Use the slope found in part (b) to find an equation for the tangent line to the curve at P. Then plot the implicit curve and tangent line together on a single graph. 55. x 3  xy + y 3 = 7,
Ps2, 1d
56. x 5 + y 3x + yx 2 + y 4 = 4, 2 + x , 1  x
Ps0, 1d
58. y 3 + cos xy = x 2,
Ps1, 0d
y 59. x + tan a x b = 2,
P a1,
57. y 2 + y =
y2 x3 2x 2
a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point P satisfies the equation.
Ps1, 1d
p b 4
60. xy 3 + tan (x + yd = 1,
p P a , 0b 4
61. 2y 2 + sxyd1>3 = x 2 + 2,
Ps1, 1d
62. x21 + 2y + y = x 2,
P s1, 0d
170
Chapter 3: Differentiation
Derivatives of Inverse Functions and Logarithms
3.8
In Section 1.6 we saw how the inverse of a function undoes, or inverts, the effect of that function. We defined there the natural logarithm function ƒ 1(x) = ln x as the inverse of the natural exponential function ƒ(x) = e x. This is one of the most important functioninverse pairs in mathematics and science. We learned how to differentiate the exponential function in Section 3.3. Here we learn a rule for differentiating the inverse of a differentiable function and we apply the rule to find the derivative of the natural logarithm function. y
y 2x 2
Derivatives of Inverses of Differentiable Functions yx
y 1x1 2
We calculated the inverse of the function ƒsxd = s1>2dx + 1 as ƒ 1sxd = 2x  2 in Example 3 of Section 1.6. Figure 3.34 shows again the graphs of both functions. If we calculate their derivatives, we see that d d 1 1 ƒsxd = a x + 1b = 2 dx dx 2
1 –2
1
x
d d 1 ƒ sxd = s2x  2d = 2. dx dx
–2
FIGURE 3.34 Graphing a line and its inverse together shows the graphs’ symmetry with respect to the line y = x . The slopes are reciprocals of each other.
The derivatives are reciprocals of one another, so the slope of one line is the reciprocal of the slope of its inverse line. (See Figure 3.34.) This is not a special case. Reflecting any nonhorizontal or nonvertical line across the line y = x always inverts the line’s slope. If the original line has slope m Z 0, the reflected line has slope 1> m. y
y y = f(x)
b = f (a)
(a, b)
a = f –1(b)
0
a
x
(b, a)
0
The slopes are reciprocal: ( f –1)'(b) =
y = f –1(x) b
x
1 or ( f –1) (b) 1 ' = f '(a) f '( f –1(b))
FIGURE 3.35 The graphs of inverse functions have reciprocal slopes at corresponding points.
The reciprocal relationship between the slopes of ƒ and ƒ 1 holds for other functions as well, but we must be careful to compare slopes at corresponding points. If the slope of y = ƒsxd at the point (a, ƒ(a)) is ƒ¿sad and ƒ¿sad Z 0, then the slope of y = ƒ 1sxd at the point (ƒ(a), a) is the reciprocal 1>ƒ¿sad (Figure 3.35). If we set b = ƒsad, then sƒ 1 d¿sbd =
1 1 . = ƒ¿sad ƒ¿sƒ 1sbdd
If y = ƒsxd has a horizontal tangent line at (a, ƒ(a)), then the inverse function ƒ 1 has a vertical tangent line at (ƒ(a), a), and this infinite slope implies that ƒ 1 is not differentiable at ƒ(a). Theorem 3 gives the conditions under which ƒ 1 is differentiable in its domain (which is the same as the range of ƒ).
3.8
171
Derivatives of Inverse Functions and Logarithms
THEOREM 3—The Derivative Rule for Inverses If ƒ has an interval I as domain and ƒ¿sxd exists and is never zero on I, then ƒ 1 is differentiable at every point in its domain (the range of ƒ). The value of sƒ 1 d¿ at a point b in the domain of ƒ 1 is the reciprocal of the value of ƒ¿ at the point a = ƒ 1sbd: sƒ 1 d¿sbd =
1 ƒ¿sƒ 1sbdd
(1)
or dƒ 1 1 . ` = dx x = b dƒ ` dx x = ƒ 1sbd
Theorem 3 makes two assertions. The first of these has to do with the conditions under which ƒ 1 is differentiable; the second assertion is a formula for the derivative of ƒ 1 when it exists. While we omit the proof of the first assertion, the second one is proved in the following way: ƒsƒ 1sxdd = x
Inverse function relationship
d ƒsƒ 1sxdd = 1 dx ƒ¿sƒ 1sxdd #
Differentiating both sides
d 1 ƒ sxd = 1 dx
Chain Rule
d 1 1 ƒ sxd = . dx ƒ¿sƒ 1sxdd
Solving for the derivative
The function ƒsxd = x 2, x Ú 0 and its inverse ƒ 1sxd = 1x have derivatives ƒ¿sxd = 2x and sƒ 1 d¿sxd = 1> A 21x B . Let’s verify that Theorem 3 gives the same formula for the derivative of ƒ 1sxd:
EXAMPLE 1
1 ƒ¿sƒ 1sxdd 1 = 2sƒ 1sxdd 1 = . 2s 1xd
sƒ 1 d¿sxd = y y x 2, x 0
4
Theorem 3 gives a derivative that agrees with the known derivative of the square root function. Let’s examine Theorem 3 at a specific point. We pick x = 2 (the number a) and ƒs2d = 4 (the value b). Theorem 3 says that the derivative of ƒ at 2, ƒ¿s2d = 4, and the derivative of ƒ 1 at ƒ(2), sƒ 1 d¿s4d, are reciprocals. It states that
Slope 4 (2, 4)
3
Slope 1– 4
2
(4, 2)
y x
1 0
sƒ 1 d¿s4d = 1
2
3
4
ƒ¿sxd = 2x with x replaced by ƒ 1sxd
x
1 1 1 1 = = ` = . 2x x = 2 4 ƒ¿s2d ƒ¿sƒ 1s4dd
See Figure 3.36. FIGURE 3.36 The derivative of ƒ 1sxd = 1x at the point (4, 2) is the reciprocal of the derivative of ƒsxd = x 2 at (2, 4) (Example 1).
We will use the procedure illustrated in Example 1 to calculate formulas for the derivatives of many inverse functions throughout this chapter. Equation (1) sometimes enables us to find specific values of dƒ 1>dx without knowing a formula for ƒ 1 .
172
Chapter 3: Differentiation y 6
(2, 6)
y x3 2 Slope 3x 2 3(2)2 12
Let ƒsxd = x 3  2. Find the value of dƒ 1>dx at x = 6 = ƒs2d without finding a formula for ƒ 1sxd.
EXAMPLE 2
Solution
We apply Theorem 3 to obtain the value of the derivative of ƒ 1 at x = 6:
Reciprocal slope: 1 12 (6, 2)
–2
0
6
dƒ ` = 3x 2 ` = 12 dx x = 2 x=2 dƒ 1 1 1 ` = = . 12 dx x = ƒs2d dƒ ` dx x = 2
x
Eq. (1)
–2
See Figure 3.37. FIGURE 3.37 The derivative of ƒsxd = x 3  2 at x = 2 tells us the derivative of ƒ 1 at x = 6 (Example 2).
Derivative of the Natural Logarithm Function Since we know the exponential function ƒ(x) = e x is differentiable everywhere, we can apply Theorem 3 to find the derivative of its inverse ƒ 1(x) = ln x: (ƒ 1)¿(x) = = =
1 ƒ¿(ƒ 1(x)) 1 e
ƒ 1(x)
Theorem 3
ƒ¿(u) = e u
1 e ln x
1 = x.
Inverse function relationship
Alternate Derivation Instead of applying Theorem 3 directly, we can find the derivative of y = ln x using implicit differentiation, as follows: y = ln x ey = x d y d (e ) = (x) dx dx ey
Inverse function relationship Differentiate implicitly.
dy = 1 dx
Chain Rule
dy 1 1 = y = x. dx e
ey = x
No matter which derivation we use, the derivative of y = ln x with respect to x is d 1 (ln x) = x , dx
x 7 0.
The Chain Rule extends this formula for positive functions usxd:
d 1 du ln u = , u dx dx
u 7 0.
(2)
3.8
EXAMPLE 3 (a)
Derivatives of Inverse Functions and Logarithms
173
We use Equation (2) to find derivatives.
d 1 1 d 1 s2d = x , ln 2x = s2xd = 2x dx 2x dx
x 7 0
(b) Equation (2) with u = x 2 + 3 gives 2x d 1 # d 2 1 # 2x = 2 . ln sx 2 + 3d = 2 sx + 3d = 2 dx x + 3 dx x + 3 x + 3 (c) Equation (2) with u = ƒ x ƒ gives an important derivative: d d du ln x = ln u # dx ƒ ƒ du dx 1# x = u ƒxƒ 1 ƒxƒ x = 2 x 1 = x.
= Derivative of ln x d 1 ln ƒ x ƒ = x , x Z 0 dx
#
x ƒxƒ
u = ƒxƒ,x Z 0 x d A ƒxƒ B = dx ƒxƒ Substitute for u.
So 1>x is the derivative of ln x on the domain x 7 0, and the derivative of ln (x) on the domain x 6 0. Notice from Example 3a that the function y = ln 2x has the same derivative as the function y = ln x. This is true of y = ln bx for any constant b, provided that bx 7 0: d 1 # d 1 1 ln bx = sbxd = sbd = x . dx bx dx bx
(3)
EXAMPLE 4
A line with slope m passes through the origin and is tangent to the graph of y = ln x. What is the value of m?
Suppose the point of tangency occurs at the unknown point x = a 7 0. Then we know that the point (a, ln a) lies on the graph and that the tangent line at that point has slope m = 1>a (Figure 3.38). Since the tangent line passes through the origin, its slope is Solution
y
2 1
0
(a, ln a)
1 2 y ln x
ln a  0 ln a = a . a  0
m = 1 Slope a 3
4
Setting these two formulas for m equal to each other, we have 5
x
ln a 1 a = a ln a = 1
FIGURE 3.38 The tangent line intersects the curve at some point (a, ln a), where the slope of the curve is 1>a (Example 4).
e ln a = e 1 a = e 1 m = e.
The Derivatives of au and loga u We start with the equation a x = e ln (a
x
)
= e x ln a , which was seen in Section 1.6:
d x d x ln a d a = e = e x ln a # sx ln ad dx dx dx = a x ln a.
d u du e = eu dx dx
174
Chapter 3: Differentiation
If a 7 0, then d x a = a x ln a. dx
(4)
This equation shows why e x is the exponential function preferred in calculus. If a = e, then ln a = 1 and the derivative of a x simplifies to d x e = e x ln e = e x . dx With the Chain Rule, we get a more general form for the derivative of a general exponential function.
If a 7 0 and u is a differentiable function of x, then a u is a differentiable function of x and d u du a = a u ln a . dx dx
EXAMPLE 5
(5)
Here are some derivatives of general exponential functions.
(a)
d x 3 = 3x ln 3 dx
Eq. (5) with a = 3, u = x
(b)
d x d 3 = 3x sln 3d s xd = 3x ln 3 dx dx
Eq. (5) with a = 3, u = x
(c)
d sin x d 3 = 3sin x sln 3d ssin xd = 3sin x sln 3d cos x dx dx
Á , u = sin x
In Section 3.3 we looked at the derivative ƒ¿(0) for the exponential functions f(x) = a x at various values of the base a. The number ƒ¿(0) is the limit, limh:0 (a h  1)>h, and gives the slope of the graph of a x when it crosses the yaxis at the point (0, 1). We now see from Equation (4) that the value of this slope is lim
h:0
ah  1 = ln a. h
(6)
In particular, when a = e we obtain lim
h:0
eh  1 = ln e = 1. h
However, we have not fully justified that these limits actually exist. While all of the arguments given in deriving the derivatives of the exponential and logarithmic functions are correct, they do assume the existence of these limits. In Chapter 7 we will give another development of the theory of logarithmic and exponential functions which fully justifies that both limits do in fact exist and have the values derived above. To find the derivative of loga u for an arbitrary base (a 7 0, a Z 1), we start with the changeofbase formula for logarithms (reviewed in Section 1.6) and express loga u in terms of natural logarithms, loga x =
ln x . ln a
3.8
Derivatives of Inverse Functions and Logarithms
175
Taking derivatives, we have d d ln x loga x = a b dx dx ln a =
1 # d ln x ln a dx
=
1 #1 ln a x
=
1 . x ln a
ln a is a constant.
If u is a differentiable function of x and u 7 0, the Chain Rule gives the following formula.
For a 7 0 and a Z 1, d 1 du loga u = . dx u ln a dx
(7)
Logarithmic Differentiation The derivatives of positive functions given by formulas that involve products, quotients, and powers can often be found more quickly if we take the natural logarithm of both sides before differentiating. This enables us to use the laws of logarithms to simplify the formulas before differentiating. The process, called logarithmic differentiation, is illustrated in the next example.
EXAMPLE 6
Find dy> dx if y =
sx 2 + 1dsx + 3d1>2 , x  1
x 7 1.
We take the natural logarithm of both sides and simplify the result with the algebraic properties of logarithms from Theorem 1 in Section 1.6:
Solution
ln y = ln
sx 2 + 1dsx + 3d1>2 x  1
= ln ssx 2 + 1dsx + 3d1>2 d  ln sx  1d
Rule 2
= ln sx 2 + 1d + ln sx + 3d1>2  ln sx  1d
Rule 1
1 ln sx + 3d  ln sx  1d. 2
Rule 4
= ln sx 2 + 1d +
We then take derivatives of both sides with respect to x, using Equation (2) on the left: 1 dy 1 1 # y dx = x 2 + 1 2x + 2
#
1 1 . x + 3 x  1
Next we solve for dy> dx:
dy 2x 1 1 + b. = ya 2 2x + 6 x  1 dx x + 1
176
Chapter 3: Differentiation
Finally, we substitute for y: dy sx 2 + 1dsx + 3d1>2 2x 1 1 = a 2 b. + x  1 2x + 6 x  1 dx x + 1
Irrational Exponents and the Power Rule (General Version) The definition of the general exponential function enables us to raise any positive number to any real power n, rational or irrational. That is, we can define the power function y = x n for any exponent n.
DEFINITION
For any x 7 0 and for any real number n, x n = e n ln x.
Because the logarithm and exponential functions are inverses of each other, the definition gives ln x n = n ln x,
for all real numbers n.
That is, the rule for taking the natural logarithm of any power holds for all real exponents n, not just for rational exponents. The definition of the power function also enables us to establish the derivative Power Rule for any real power n, as stated in Section 3.3.
General Power Rule for Derivatives For x 7 0 and any real number n, d n x = nx n  1. dx If x … 0, then the formula holds whenever the derivative, x n, and x n  1 all exist.
Proof Differentiating x n with respect to x gives d n d n ln x x = e dx dx = e n ln x #
d sn ln xd dx
Definition of x n, x 7 0 Chain Rule for e u
n = xn # x
Definition and derivative of ln x
= nx n  1 .
x n # x 1 = x n  1
In short, whenever x 7 0, d n x = nx n  1 . dx For x 6 0, if y = x n, y¿ , and x n  1 all exist, then ln ƒ y ƒ = ln ƒ x ƒ n = n ln ƒ x ƒ .
3.8
Derivatives of Inverse Functions and Logarithms
177
Using implicit differentiation (which assumes the existence of the derivative y¿ ) and Example 3(c), we have y¿ n y = x. Solving for the derivative, y xn y¿ = n x = n x = nx n  1.
y = xn
It can be shown directly from the definition of the derivative that the derivative equals 0 when x = 0 and n Ú 1. This completes the proof of the general version of the Power Rule for all values of x.
EXAMPLE 7 Solution
Differentiate ƒ(x) = x x, x 7 0.
We note that ƒ(x) = x x = e x ln x, so differentiation gives ƒ¿(x) =
d x ln x (e ) dx
= e x ln x
d (x ln x) dx
d dx
e u, u = x ln x
1 = e x ln x aln x + x # x b = x x (ln x + 1).
x 7 0
The Number e Expressed as a Limit In Section 1.5 we defined the number e as the base value for which the exponential function y = a x has slope 1 when it crosses the yaxis at (0, 1). Thus e is the constant that satisfies the equation lim
h:0
eh  1 = ln e = 1. h
Slope equals ln e from Eq. (6).
We now prove that e can be calculated as a certain limit.
THEOREM 4—The Number e as a Limit limit
The number e can be calculated as the
e = lim s1 + xd1>x . x:0
y e 3 2
Proof If ƒsxd = ln x, then ƒ¿sxd = 1>x, so ƒ¿s1d = 1. But, by the definition of derivative,
y (1 x)1/x
ƒs1 + hd  ƒs1d ƒs1 + xd  ƒs1d = lim x h h:0 x:0
ƒ¿s1d = lim
1
0
x
FIGURE 3.39 The number e is the limit of the function graphed here as x : 0.
ln s1 + xd  ln 1 1 = lim x ln s1 + xd x x:0 x:0
= lim
ln 1 = 0
= lim ln s1 + xd1>x = ln c lim s1 + xd1>x d.
ln is continuous, Theorem 10 in Chapter 2.
x:0
x:0
178
Chapter 3: Differentiation
Because ƒ¿s1d = 1, we have
ln c lim s1 + xd1>x d = 1. x:0
Therefore, exponentiating both sides we get lim s1 + xd1>x = e.
x:0
See Figure 3.39 on the previous page. Approximating the limit in Theorem 4 by taking x very small gives approximations to e. Its value is e L 2.718281828459045 to 15 decimal places.
Exercises 3.8 Derivatives of Inverse Functions In Exercises 1–4: a. Find ƒ 1sxd . b. Graph ƒ and ƒ 1 together.
c. Evaluate dƒ> dx at x = a and dƒ 1>dx at x = ƒsad to show that at these points dƒ 1>dx = 1>sdƒ>dxd .
1. ƒsxd = 2x + 3,
a = 1
2. ƒsxd = s1>5dx + 7,
3. ƒsxd = 5  4x,
a = 1>2
4. ƒsxd = 2x2,
x Ú 0,
a = 1 a = 5
3 x are inverses of one an5. a. Show that ƒsxd = x3 and g sxd = 1 other.
b. Graph ƒ and g over an xinterval large enough to show the graphs intersecting at (1, 1) and s 1, 1d . Be sure the picture shows the required symmetry about the line y = x . c. Find the slopes of the tangents to the graphs of ƒ and g at (1, 1) and s  1, 1d (four tangents in all). d. What lines are tangent to the curves at the origin?
6. a. Show that hsxd = x3>4 and ksxd = s4xd1>3 are inverses of one another. b. Graph h and k over an xinterval large enough to show the graphs intersecting at (2, 2) and s 2, 2d . Be sure the picture shows the required symmetry about the line y = x . c. Find the slopes of the tangents to the graphs at h and k at (2, 2) and s 2, 2d . d. What lines are tangent to the curves at the origin?
7. Let ƒsxd = x3  3x2  1, x Ú 2 . Find the value of dƒ 1>dx at the point x = 1 = ƒs3d .
8. Let ƒsxd = x  4x  5, x 7 2 . Find the value of dƒ >dx at the point x = 0 = ƒs5d . 2
1
9. Suppose that the differentiable function y = ƒsxd has an inverse and that the graph of ƒ passes through the point (2, 4) and has a slope of 1> 3 there. Find the value of dƒ 1>dx at x = 4 .
13. y = ln st 2 d 3 15. y = ln x 17. y = ln su + 1d
14. y = ln st 3>2 d 10 16. y = ln x 18. y = ln s2u + 2d
19. y = ln x3
20. y = sln xd3
21. y = t sln td2 4
22. y = t2ln t 4
x x ln x 4 16 ln t 25. y = t ln x 27. y = 1 + ln x
24. y = (x 2 ln x) 4
29. y = ln sln xd
30. y = ln sln sln xdd
23. y =
1 + ln t t x ln x 28. y = 1 + ln x 26. y =
31. y = ussin sln ud + cos sln udd 32. y = ln ssec u + tan ud 1 33. y = ln x2x + 1 1 + ln t 35. y = 1  ln t
sx2 + 1d5 21  x
1 1 + x ln 2 1  x
36. y = 2ln 1t 38. y = ln a
37. y = ln ssec sln udd 39. y = ln a
34. y =
b
40. y = ln
2sin u cos u b 1 + 2 ln u sx + 1d5
C sx + 2d20
Logarithmic Differentiation In Exercises 41–54, use logarithmic differentiation to find the derivative of y with respect to the given independent variable. 41. y = 2xsx + 1d 43. y =
t At + 1
42. y = 2sx2 + 1dsx  1d2 44. y =
1 A t st + 1d
10. Suppose that the differentiable function y = gsxd has an inverse and that the graph of g passes through the origin with slope 2. Find the slope of the graph of g1 at the origin.
45. y = (sin u)2u + 3
46. y = stan ud22u + 1
47. y = t st + 1dst + 2d
48. y =
Derivatives of Logarithms In Exercises 11–40, find the derivative of y with respect to x, t, or u , as appropriate.
49. y =
u + 5 u cos u
50. y =
11. y = ln 3x
51. y =
x2x2 + 1 sx + 1d2>3
52. y =
12. y = ln k x, k constant
1 t st + 1dst + 2d u sin u 2sec u sx + 1d10
C s2x + 1d5
3.8
53. y =
3
xsx  2d
54. y =
Bx + 1 2
3
xsx + 1dsx  2d
Bsx + 1ds2x + 3d 2
Finding Derivatives In Exercises 55–62, find the derivative of y with respect to x, t, or u , as appropriate. 55. y = ln (cos2 u)
56. y = ln s3ueu d
57. y = ln s3tet d
58. y = ln s2et sin td
59. y = ln a
60. y = ln a
eu b 1 + eu
61. y = escos t + ln td
In Exercises 63–66, find dy> dx.
2u 1 + 2u
64. ln xy = e x + y
65. x y = y x
66. tan y = e x + ln x
70. y = 2ss
2
x 2y– + xy¿ + y = 0. 100. Using mathematical induction, show that
d
72. y = t
73. y = log 2 5u
74. y = log 3 s1 + u ln 3d
75. y = log 4 x + log 4 x 2
76. y = log 25 e x  log 5 1x
77. y = log 2 r # log 4 r
x + 1 79. y = log 3 a a b x  1
ln 3
78. y = log 3 r # log 9 r b
7x 80. y = log 5 a b 3x + 2 B
ln 5
81. y = u sin slog 7 ud
sin u cos u 82. y = log 7 a b eu 2u
83. y = log 5 e x
84. y = log 2 a
x 2e 2
2 2x + 1 86. y = 3 log 8 slog 2 td
85. y = 3
log 2 t
b
88. y = t log 3 A essin tdsln 3d B
87. y = log 2 s8t ln 2 d
Logarithmic Differentiation with Exponentials In Exercises 89–96, use logarithmic differentiation to find the derivative of y with respect to the given independent variable. sx + 1d
89. y = sx + 1d
90. y = x
91. y = s 1tdt
92. y = t2t
93. y = ssin xdx
94. y = x sin x
95. y = x
96. y = sln xdln x
x
ln x
COMPUTER EXPLORATIONS In Exercises 101–108, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS: a. Plot the function y = ƒsxd together with its derivative over the given interval. Explain why you know that ƒ is onetoone over the interval.
1e
71. y = x
p
n
x 98. Show that lim n: q a1 + n b = e x for any x 7 0.
b
In Exercises 67–88, find the derivative of y with respect to the given independent variable. 69. y = 52s
sides of this equation with respect to x, using the Chain Rule to express sg ⴰ ƒd¿sxd as a product of derivatives of g and ƒ. What do you find? (This is not a proof of Theorem 3 because we assume here the theorem’s conclusion that g = ƒ 1 is differentiable.)
(n  1)! dn ln x = (1) n  1 . dx n xn
63. ln y = e y sin x
68. y = 3x
b. Solve the equation y = ƒsxd for x as a function of y, and name the resulting inverse function g. c. Find the equation for the tangent line to ƒ at the specified point sx0 , ƒsx0 dd . d. Find the equation for the tangent line to g at the point sƒsx0 d, x0 d located symmetrically across the 45° line y = x (which is the graph of the identity function). Use Theorem 3 to find the slope of this tangent line. e. Plot the functions ƒ and g, the identity, the two tangent lines, and the line segment joining the points sx0 , ƒsx0 dd and sƒsx0 d, x0 d . Discuss the symmetries you see across the main diagonal. 101. y = 23x  2,
2 … x … 4, 3
3x + 2 , 2 … x … 2, x0 = 1>2 2x  11 4x 103. y = 2 , 1 … x … 1, x0 = 1>2 x + 1 3 x 104. y = 2 , 1 … x … 1, x0 = 1>2 x + 1 105. y = x 3  3x 2  1,
97. If we write g(x) for ƒ 1sxd , Equation (1) can be written as
106. y = 2  x  x3,
g¿sƒsadd # ƒ¿sad = 1 .
If we then write x for a, we get g¿sƒsxdd # ƒ¿sxd = 1 . The latter equation may remind you of the Chain Rule, and indeed there is a connection. Assume that ƒ and g are differentiable functions that are inverses of one another, so that sg ⴰ ƒdsxd = x . Differentiate both
x0 = 3
102. y =
Theory and Applications 1 g¿sƒsadd = , or ƒ¿sad
179
99. If y = A sin (ln x) + B cos (ln x), where A and B are constants, show that
62. y = e sin t sln t 2 + 1d
67. y = 2x
Derivatives of Inverse Functions and Logarithms
2 … x … 5, 2 … x … 2,
27 10 3 x0 = 2 x0 =
3 … x … 5, x0 = 1 p p 108. y = sin x,  … x … , x0 = 1 2 2 107. y = e , x
In Exercises 109 and 110, repeat the steps above to solve for the functions y = ƒsxd and x = ƒ 1s yd defined implicitly by the given equations over the interval. 109. y1>3  1 = sx + 2d3, 110. cos y = x , 1>5
5 … x … 5,
0 … x … 1,
x0 = 1>2
x0 = 3>2
180
Chapter 3: Differentiation
Inverse Trigonometric Functions
3.9
We introduced the six basic inverse trigonometric functions in Section 1.6, but focused there on the arcsine and arccosine functions. Here we complete the study of how all six inverse trigonometric functions are defined, graphed, and evaluated, and how their derivatives are computed.
Inverses of tan x, cot x, sec x, and csc x The graphs of these four basic inverse trigonometric functions are shown again in Figure 3.40. We obtain these graphs by reflecting the graphs of the restricted trigonometric functions (as discussed in Section 1.6) through the line y = x. Let’s take a closer look at the arctangent, arccotangent, arcsecant, and arccosecant functions. Domain: – ∞ x ∞ Range: – y 2 2 y 2 –2
–1
– 2
Domain: – ∞ x ∞ Range: 0 y y
y tan –1x x 1 2
(a)
2
2
–2
Domain: x –1 or x 1 Range: – y , y 0 2 2 y
Domain: x –1 or x 1 Range: 0 y , y 2 y
–1
y
2
y sec
–1x
–2 1
(b)
cot –1x
2
x
–2
–1
1
2
y csc–1x
–1
1
x
– 2
x
(c)
2
(d)
FIGURE 3.40 Graphs of the arctangent, arccotangent, arcsecant, and arccosecant functions.
The arctangent of x is a radian angle whose tangent is x. The arccotangent of x is an angle whose cotangent is x, and so forth. The angles belong to the restricted domains of the tangent, cotangent, secant, and cosecant functions.
DEFINITION y tan1 x is the number in s p>2, p>2d for which tan y = x. y cot1 x is the number in s0, pd for which cot y = x. y sec1 x is the number in [0, p/2) ´ (p/2, p] for which sec y = x. y csc1 x is the number in [p/2, 0) ´ (0, p/2] for which csc y = x.
We use open or halfopen intervals when describing the ranges to avoid values where the tangent, cotangent, secant, and cosecant functions are undefined. (See Figure 3.40.) The graph of y = tan1 x is symmetric about the origin because it is a branch of the graph x = tan y that is symmetric about the origin (Figure 3.40a). Algebraically this means that tan1 s xd = tan1 x; the arctangent is an odd function. The graph of y = cot1 x has no such symmetry (Figure 3.40b). Notice from Figure 3.40a that the graph of the arctangent function has two horizontal asymptotes; one at y = p>2 and the other at y = p>2.
3.9 Domain: x 1 Range: 0 y , y 2 y
B
y
2
sec –1x
–1
0
x
1
Caution There is no general agreement about how to define sec1 x for negative values of x. We chose angles in the second quadrant between p>2 and p. This choice makes sec1 x = cos1 s1>xd. It also makes sec1 x an increasing function on each interval of its domain. Some tables choose sec1 x to lie in [p, p>2d for x 6 0 and some texts choose it to lie in [p, 3p>2d (Figure 3.41). These choices simplify the formula for the derivative (our formula needs absolute value signs) but fail to satisfy the computational equation sec1 x = cos1 s1>xd. From this, we can derive the identity
– 2
p 1 1 sec1 x = cos1 a x b =  sin1 a x b 2
C – – 3 2
FIGURE 3.41 There are several logical choices for the lefthand branch of y = sec1 x . With choice A, sec1 x = cos1 s1>xd , a useful identity employed by many calculators.
EXAMPLE 1
The accompanying figures show two values of tan1 x. y
tan–1 1 tan–1 3 3 6 3 6 2 1 x 0 3
y
tan–1 ⎛⎝–3⎛⎝ – 3
1 0
x
y sin–1x Domain: –1 x 1 Range: – /2 y /2
–1
1
– 3 x –3
tan ⎛– ⎛ –3 ⎝ 3⎝
tan 1 6 3
y
(1)
by applying Equation (5) in Section 1.6.
2
2
181
The inverses of the restricted forms of sec x and csc x are chosen to be the functions graphed in Figures 3.40c and 3.40d. 3 2
A
Inverse Trigonometric Functions
tan1 x
23 1
p>3 p>4
23>3  23>3 1  23
p>6 p>6 p>4 p>3
The angles come from the first and fourth quadrants because the range of tan1 x is s p>2, p>2d.
x
– 2
FIGURE 3.42 The graph of y = sin1 x has vertical tangents at x = 1 and x = 1.
The Derivative of y = sin1 u We know that the function x = sin y is differentiable in the interval p>2 6 y 6 p>2 and that its derivative, the cosine, is positive there. Theorem 3 in Section 3.8 therefore assures us that the inverse function y = sin1 x is differentiable throughout the interval 1 6 x 6 1. We cannot expect it to be differentiable at x = 1 or x = 1 because the tangents to the graph are vertical at these points (see Figure 3.42).
182
Chapter 3: Differentiation
We find the derivative of y = sin1 x by applying Theorem 3 with ƒsxd = sin x and ƒ sxd = sin1 x: 1
1 ƒ¿sƒ 1sxdd
Theorem 3
=
1 cos ssin1 xd
ƒ¿sud = cos u
=
1 21  sin2 ssin1 xd
cos u = 21  sin2 u
=
1 . 21  x 2
sin ssin1 xd = x
sƒ 1 d¿sxd =
If u is a differentiable function of x with ƒ u ƒ 6 1, we apply the Chain Rule to get
d du 1 ssin1 ud = , 2 dx dx 21  u
EXAMPLE 2
ƒ u ƒ 6 1.
Using the Chain Rule, we calculate the derivative d 1 ssin1 x 2 d = dx 21  sx 2 d2
#
d 2 2x sx d = . dx 21  x 4
The Derivative of y = tan1 u We find the derivative of y = tan1 x by applying Theorem 3 with ƒsxd = tan x and ƒ 1sxd = tan1 x. Theorem 3 can be applied because the derivative of tan x is positive for p>2 6 x 6 p>2: 1 ƒ¿sƒ 1sxdd
Theorem 3
=
1 sec2 stan1 xd
ƒ¿sud = sec2 u
=
1 1 + tan2 stan1 xd
sec2 u = 1 + tan2 u
=
1 . 1 + x2
tan stan1 xd = x
sƒ 1 d¿sxd =
The derivative is defined for all real numbers. If u is a differentiable function of x, we get the Chain Rule form:
d du 1 . stan1 ud = dx 1 + u 2 dx
The Derivative of y = sec1 u Since the derivative of sec x is positive for 0 6 x 6 p/2 and p>2 6 x 6 p, Theorem 3 says that the inverse function y = sec1 x is differentiable. Instead of applying the formula
3.9
Inverse Trigonometric Functions
183
in Theorem 3 directly, we find the derivative of y = sec1 x, ƒ x ƒ 7 1, using implicit differentiation and the Chain Rule as follows: y = sec1 x sec y = x
Inverse function relationship
d d ssec yd = x dx dx sec y tan y
Differentiate both sides.
dy = 1 dx
Chain Rule
dy 1 = sec y tan y . dx
Since ƒ x ƒ 7 1, y lies in s0, p>2d ´ sp>2, pd and sec y tan y Z 0 .
To express the result in terms of x, we use the relationships sec y = x
and
tan y = ; 2sec2 y  1 = ; 2x 2  1
to get y
dy 1 . = ; dx x2x 2  1
y sec –1x
2 –1
0
1
Can we do anything about the ; sign? A glance at Figure 3.43 shows that the slope of the graph y = sec1 x is always positive. Thus,
x
FIGURE 3.43 The slope of the curve y = sec1 x is positive for both x 6 1 and x 7 1 .
d sec1 x = d dx
1 x2x 2  1 1 x2x 2  1 +
if x 7 1 if x 6 1.
With the absolute value symbol, we can write a single expression that eliminates the “;” ambiguity: d 1 sec1 x = . dx ƒ x ƒ 2x 2  1 If u is a differentiable function of x with ƒ u ƒ 7 1, we have the formula
d du 1 ssec1 ud = , 2 dx dx u 2u 1 ƒ ƒ
EXAMPLE 3
ƒ u ƒ 7 1.
Using the Chain Rule and derivative of the arcsecant function, we find d d 1 sec1 s5x 4 d = s5x 4 d 4 4 2 dx dx 5x 2s5x d 1 ƒ ƒ =
1 s20x 3 d 5x 225x 8  1
=
4 . x225x 8  1
4
5x 4 7 1 7 0
184
Chapter 3: Differentiation
Derivatives of the Other Three Inverse Trigonometric Functions We could use the same techniques to find the derivatives of the other three inverse trigonometric functions—arccosine, arccotangent, and arccosecant—but there is an easier way, thanks to the following identities.
Inverse Function–Inverse Cofunction Identities cos1 x = p>2  sin1 x cot1 x = p>2  tan1 x csc1 x = p>2  sec1 x
We saw the first of these identities in Equation (5) of Section 1.6. The others are derived in a similar way. It follows easily that the derivatives of the inverse cofunctions are the negatives of the derivatives of the corresponding inverse functions. For example, the derivative of cos1 x is calculated as follows: d d p (cos1 x) = a  sin1 xb dx dx 2 = 
d (sin1 x) dx
= 
1 . 21  x 2
Identity
Derivative of arcsine
The derivatives of the inverse trigonometric functions are summarized in Table 3.1.
TABLE 3.1
Derivatives of the inverse trigonometric functions
1.
dssin1 ud du 1 = , 2 dx dx 21  u
2.
dscos1 ud du 1 = , 2 dx dx 21  u
3.
dstan1 ud du 1 = dx 1 + u 2 dx
4.
dscot1 ud du 1 = 2 dx dx 1 + u
5.
dssec1 ud du 1 = , 2 dx dx u 2u 1 ƒ ƒ
6.
dscsc1 ud du 1 = , 2 dx ƒ u ƒ 2u  1 dx
ƒuƒ 6 1 ƒuƒ 6 1
ƒuƒ 7 1 ƒuƒ 7 1
3.9
Inverse Trigonometric Functions
185
Exercises 3.9 Common Values Use reference triangles in an appropriate quadrant, as in Example 1, to find the angles in Exercises 1–8. 1 1. a. tan1 1 b. tan1 A  23 B c. tan1 a b 23 1 2. a. tan1s 1d b. tan1 23 c. tan1 a b 23
4. a. sin
b. sin1 a
1 b 2
1 a b 2
1
b. sin
1
a
1 22 1
b
b 22 1 a b 22 2 a b 23 2 a b 23
1 5. a. cos1 a b 2
b. cos1
6. a. csc1 22
b. csc1
7. a. sec1 A  22 B
b. sec1
8. a. cot1 s 1d
b. cot1 A 23 B
c. sin1 a c. sin
1
 23 b 2
23 a b 2
c. cos1 a
23 b 2
22 bb 2
1 11. tan asin1 a b b 2
x: 1
15. lim tan
1
1 23
12. cot asin1 a
14.
17. lim sec1 x
18.
19. lim csc x: q
1
x
20.
lim tan
43. You are sitting in a classroom next to the wall looking at the blackboard at the front of the room. The blackboard is 12 ft long and starts 3 ft from the wall you are sitting next to. Show that your viewing angle is
3'
23 bb 2
lim csc
You
Wall x
44. Find the angle a .
65°
x
x:  q x:  q
x x  cot1 15 3
b
lim sec1 x 1
41. y = x sin1 x + 21  x 2
Theory and Examples
lim cos1 x
x:  q
x 7 1
x 42. y = ln sx 2 + 4d  x tan1 a b 2
x: 1 +
1
x,
if you are x ft from the front wall.
c. sec1s 2d c. cot1 a
38. y = 2s 2  1  sec1 s 1
1 40. y = cot1 x  tan1 x
1 10. sec acos1 b 2
16.
x: q
39. y = tan 2x  1 + csc 2
12'
x
x: q
1
a = cot1
Limits Find the limits in Exercises 13–20. (If in doubt, look at the function’s graph.) 13. lim sin1 x
37. y = s 21  s 2 + cos1 s
c. csc1 2
Evaluations Find the values in Exercises 9–12. 9. sin acos1 a
36. y = cos1 se t d
Blackboard
3. a. sin1 a
35. y = csc1 se t d
x
Finding Derivatives In Exercises 21–42, find the derivative of y with respect to the appropriate variable. 21. y = cos1 sx 2 d
22. y = cos1 s1>xd
23. y = sin1 22 t
24. y = sin1 s1  td
25. y = sec1 s2s + 1d
26. y = sec1 5s
27. y = csc1 sx 2 + 1d, x 7 0 x 28. y = csc1 2 3 1 29. y = sec1 t , 0 6 t 6 1 30. y = sin1 2 t 31. y = cot1 2t
32. y = cot1 2t  1
33. y = ln stan1 xd
34. y = tan1 sln xd
21 50
45. Here is an informal proof that tan1 1 + tan1 2 + tan1 3 = p . Explain what is going on.
186
Chapter 3: Differentiation
46. Two derivations of the identity sec1 s xd = P sec1 x
55. What is special about the functions
1
a. (Geometric) Here is a pictorial proof that sec s xd = p  sec1 x . See if you can tell what is going on. y
y sec –1x
ƒsxd = sin1
0
1
x
x
cos1 s xd = p  cos1 x sec
Eq. (4), Section 1.6
1
x = cos s1>xd
48. a. csc 49. a. sec 50. a. cot
1 1
1
b. cos1 2 b. csc1 2
(1>2)
b. sin 22 1
s 1>2d
1 2x + 1 2
and
1 g sxd = tan1 x ?
T 57. Find the values of a. sec1 1.5
b. csc1 s 1.5d
c. cot1 2
T 58. Find the values of a. sec1s 3d
b. csc1 1.7
c. cot1 s 2d
T In Exercises 59–61, find the domain and range of each composite function. Then graph the composites on separate screens. Do the graphs make sense in each case? Give reasons for your answers. Comment on any differences you see. 59. a. y = tan1 stan xd
1
0
g sxd = 2 tan1 1x ?
Eq. (1)
Which of the expressions in Exercises 47–50 are defined, and which are not? Give reasons for your answers. 47. a. tan1 2
and
Explain.
b. (Algebraic) Derive the identity sec1 s xd = p  sec1 x by combining the following two equations from the text: 1
x Ú 0,
56. What is special about the functions
2 –1
x  1 , x + 1
Explain.
–x
ƒsxd = sin1
b. cos s 5d
1
b. y = tan stan1 xd
60. a. y = sin ssin xd
b. y = sin ssin1 xd
61. a. y = cos1 scos xd
b. y = cos scos1 xd
51. Use the identity csc1 u =
p  sec1 u 2
to derive the formula for the derivative of csc1 u in Table 3.1 from the formula for the derivative of sec1 u . 52. Derive the formula dy 1 = dx 1 + x2 1
for the derivative of y = tan x by differentiating both sides of the equivalent equation tan y = x . 53. Use the Derivative Rule in Section 3.8, Theorem 3, to derive d 1 sec1 x = , ƒ x ƒ 7 1. 2 dx x 2x  1 ƒ ƒ 54. Use the identity p cot1 u =  tan1 u 2 to derive the formula for the derivative of cot1 u in Table 3.1 from the formula for the derivative of tan1 u .
3.10
T Use your graphing utility for Exercises 62–66. 62. Graph y = sec ssec1 xd = sec scos1s1>xdd . Explain what you see. 63. Newton’s serpentine Graph Newton’s serpentine, y = 4x>sx2 + 1d . Then graph y = 2 sin s2 tan1 xd in the same graphing window. What do you see? Explain. 64. Graph the rational function y = s2  x 2 d>x 2 . Then graph y = cos s2 sec1 xd in the same graphing window. What do you see? Explain. 65. Graph ƒsxd = sin1 x together with its first two derivatives. Comment on the behavior of ƒ and the shape of its graph in relation to the signs and values of ƒ¿ and ƒ–. 66. Graph ƒsxd = tan1 x together with its first two derivatives. Comment on the behavior of ƒ and the shape of its graph in relation to the signs and values of ƒ¿ and ƒ–.
Related Rates In this section we look at problems that ask for the rate at which some variable changes when it is known how the rate of some other related variable (or perhaps several variables) changes. The problem of finding a rate of change from other known rates of change is called a related rates problem.
3.10
Related Rates
187
Related Rates Equations Suppose we are pumping air into a spherical balloon. Both the volume and radius of the balloon are increasing over time. If V is the volume and r is the radius of the balloon at an instant of time, then V =
4 3 pr . 3
Using the Chain Rule, we differentiate both sides with respect to t to find an equation relating the rates of change of V and r, dV dr dV dr = = 4pr 2 . dt dr dt dt So if we know the radius r of the balloon and the rate dV>dt at which the volume is increasing at a given instant of time, then we can solve this last equation for dr>dt to find how fast the radius is increasing at that instant. Note that it is easier to directly measure the rate of increase of the volume (the rate at which air is being pumped into the balloon) than it is to measure the increase in the radius. The related rates equation allows us to calculate dr>dt from dV>dt. Very often the key to relating the variables in a related rates problem is drawing a picture that shows the geometric relations between them, as illustrated in the following example. Water runs into a conical tank at the rate of 9 ft3>min. The tank stands point down and has a height of 10 ft and a base radius of 5 ft. How fast is the water level rising when the water is 6 ft deep?
EXAMPLE 1
dV 9 ft 3/min dt 5 ft
Solution
Figure 3.44 shows a partially filled conical tank. The variables in the problem are V = volume sft3 d of the water in the tank at time t smind
x dy ? dt when y 6 ft
x = radius sftd of the surface of the water at time t
10 ft y
y = depth sftd of the water in the tank at time t. We assume that V, x, and y are differentiable functions of t. The constants are the dimensions of the tank. We are asked for dy>dt when
FIGURE 3.44 The geometry of the conical tank and the rate at which water fills the tank determine how fast the water level rises (Example 1).
y = 6 ft
dV = 9 ft3>min. dt
and
The water forms a cone with volume V =
1 2 px y. 3
This equation involves x as well as V and y. Because no information is given about x and dx>dt at the time in question, we need to eliminate x. The similar triangles in Figure 3.44 give us a way to express x in terms of y: 5 x y = 10
or
x =
y . 2
Therefore, find V =
y 2 p 3 1 pa b y = y 3 2 12
to give the derivative dV p # 2 dy p dy 3y = = y2 . 12 4 dt dt dt
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Chapter 3: Differentiation
Finally, use y = 6 and dV>dt = 9 to solve for dy>dt. dy p s6d2 4 dt dy 1 = p L 0.32 dt 9 =
At the moment in question, the water level is rising at about 0.32 ft> min.
Related Rates Problem Strategy 1. Draw a picture and name the variables and constants. Use t for time. Assume that all variables are differentiable functions of t. 2. Write down the numerical information (in terms of the symbols you have chosen). 3. Write down what you are asked to find (usually a rate, expressed as a derivative). 4. Write an equation that relates the variables. You may have to combine two or more equations to get a single equation that relates the variable whose rate you want to the variables whose rates you know. 5. Differentiate with respect to t. Then express the rate you want in terms of the rates and variables whose values you know. 6. Evaluate. Use known values to find the unknown rate.
EXAMPLE 2
A hot air balloon rising straight up from a level field is tracked by a range finder 150 m from the liftoff point. At the moment the range finder’s elevation angle is p>4, the angle is increasing at the rate of 0.14 rad> min. How fast is the balloon rising at that moment?
Balloon
d 0.14 rad/min dt when /4
Range finder
dy ? y dt when /4
150 m
FIGURE 3.45 The rate of change of the balloon’s height is related to the rate of change of the angle the range finder makes with the ground (Example 2).
We answer the question in six steps. 1. Draw a picture and name the variables and constants (Figure 3.45). The variables in the picture are u = the angle in radians the range finder makes with the ground. y = the height in feet of the balloon. We let t represent time in minutes and assume that u and y are differentiable functions of t. The one constant in the picture is the distance from the range finder to the liftoff point (150 m). There is no need to give it a special symbol. 2. Write down the additional numerical information. Solution
3. 4.
du p = 0.14 rad>min when u = 4 dt Write down what we are to find. We want dy>dt when u = p>4. Write an equation that relates the variables y and u. y = tan u or y = 150 tan u 150
5.
Differentiate with respect to t using the Chain Rule. The result tells how dy>dt (which we want) is related to du>dt (which we know). dy du = 150 ssec2 ud dt dt
6.
Evaluate with u = p>4 and du>dt = 0.14 to find dy>dt. dy = 150 A 22 B 2s0.14d = 42 dt
sec
p = 22 4
At the moment in question, the balloon is rising at the rate of 42 m > min.
3.10
Related Rates
189
EXAMPLE 3
y
A police cruiser, approaching a rightangled intersection from the north, is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.6 mi north of the intersection and the car is 0.8 mi to the east, the police determine with radar that the distance between them and the car is increasing at 20 mph. If the cruiser is moving at 60 mph at the instant of measurement, what is the speed of the car?
Situation when x 0.8, y 0.6 y dy –60 dt
ds 20 dt
We picture the car and cruiser in the coordinate plane, using the positive xaxis as the eastbound highway and the positive yaxis as the southbound highway (Figure 3.46). We let t represent time and set
Solution
0
dx ? dt
x
x
FIGURE 3.46 The speed of the car is related to the speed of the police cruiser and the rate of change of the distance between them (Example 3).
x = position of car at time t y = position of cruiser at time t s = distance between car and cruiser at time t. We assume that x, y, and s are differentiable functions of t. We want to find dx>dt when x = 0.8 mi,
dy = 60 mph, dt
y = 0.6 mi,
ds = 20 mph. dt
Note that dy>dt is negative because y is decreasing. We differentiate the distance equation s2 = x2 + y2 (we could also use s = 2x 2 + y 2), and obtain 2s
dy ds dx = 2x + 2y dt dt dt dy ds dx 1 = s ax + y b dt dt dt 1
=
2x 2 + y 2
ax
dy dx + y b. dt dt
Finally, we use x = 0.8, y = 0.6, dy>dt = 60, ds>dt = 20, and solve for dx>dt. 20 =
1 2s0.8d + s0.6d 2
2
a0.8
dx + (0.6)(60)b dt
202s0.8d2 + s0.6d2 + s0.6ds60d dx = = 70 0.8 dt At the moment in question, the car’s speed is 70 mph.
EXAMPLE 4
y P 10 u 0
Q (x, 0)
x
A particle P moves clockwise at a constant rate along a circle of radius 10 m centered at the origin. The particle’s initial position is (0, 10) on the yaxis, and its final destination is the point (10, 0) on the xaxis. Once the particle is in motion, the tangent line at P intersects the xaxis at a point Q (which moves over time). If it takes the particle 30 sec to travel from start to finish, how fast is the point Q moving along the xaxis when it is 20 m from the center of the circle? We picture the situation in the coordinate plane with the circle centered at the origin (see Figure 3.47). We let t represent time and let u denote the angle from the xaxis to the radial line joining the origin to P. Since the particle travels from start to finish in 30 sec, it is traveling along the circle at a constant rate of p>2 radians in 1>2 min, or p rad>min. In other words, du>dt = p, with t being measured in minutes. The negative sign appears because u is decreasing over time. Solution
FIGURE 3.47 The particle P travels clockwise along the circle (Example 4).
190
Chapter 3: Differentiation
Setting xstd to be the distance at time t from the point Q to the origin, we want to find dx>dt when x = 20 m
du = p rad>min. dt
and
To relate the variables x and u, we see from Figure 3.47 that x cos u = 10, or x = 10 sec u. Differentiation of this last equation gives du dx = 10 sec u tan u = 10p sec u tan u. dt dt Note that dx>dt is negative because x is decreasing (Q is moving toward the origin). When x = 20, cos u = 1>2 and sec u = 2. Also, tan u = 2sec2 u  1 = 23. It follows that dx = s 10pds2d A 23 B = 20 23p. dt At the moment in question, the point Q is moving toward the origin at the speed of 2023p L 108.8 m>min. A
12,000 u R
FIGURE 3.48 Jet airliner A traveling at constant altitude toward radar station R (Example 5).
x
EXAMPLE 5 A jet airliner is flying at a constant altitude of 12,000 ft above sea level as it approaches a Pacific island. The aircraft comes within the direct line of sight of a radar station located on the island, and the radar indicates the initial angle between sea level and its line of sight to the aircraft is 30°. How fast (in miles per hour) is the aircraft approaching the island when first detected by the radar instrument if it is turning upward (counterclockwise) at the rate of 2>3 deg>sec in order to keep the aircraft within its direct line of sight? Solution The aircraft A and radar station R are pictured in the coordinate plane, using the positive xaxis as the horizontal distance at sea level from R to A, and the positive yaxis as the vertical altitude above sea level. We let t represent time and observe that y = 12,000 is a constant. The general situation and lineofsight angle u are depicted in Figure 3.48. We want to find dx>dt when u = p>6 rad and du>dt = 2>3 deg>sec. From Figure 3.48, we see that
12,000 = tan u x
x = 12,000 cot u.
or
Using miles instead of feet for our distance units, the last equation translates to x =
12,000 cot u. 5280
Differentiation with respect to t gives du 1200 dx = csc2 u . 528 dt dt When u = p>6, sin2 u = 1>4, so csc2 u = 4. Converting du>dt = 2>3 deg>sec to radians per hour, we find du 2 p = a bs3600d rad>hr. 3 180 dt
1 hr = 3600 sec, 1 deg = p>180 rad
Substitution into the equation for dx>dt then gives 1200 dx p 2 = abs4d a b a bs3600d L 380. 528 3 180 dt The negative sign appears because the distance x is decreasing, so the aircraft is approaching the island at a speed of approximately 380 mi>hr when first detected by the radar.
3.10
Related Rates
191
EXAMPLE 6
Figure 3.49a shows a rope running through a pulley at P and bearing a weight W at one end. The other end is held 5 ft above the ground in the hand M of a worker. Suppose the pulley is 25 ft above ground, the rope is 45 ft long, and the worker is walking rapidly away from the vertical line PW at the rate of 6 ft>sec. How fast is the weight being raised when the worker’s hand is 21 ft away from PW ?
P
W M 6 ft/sec
O
We let OM be the horizontal line of length x ft from a point O directly below the pulley to the worker’s hand M at any instant of time (Figure 3.49). Let h be the height of the weight W above O, and let z denote the length of rope from the pulley P to the worker’s hand. We want to know dh>dt when x = 21 given that dx>dt = 6. Note that the height of P above O is 20 ft because O is 5 ft above the ground. We assume the angle at O is a right angle. At any instant of time t we have the following relationships (see Figure 3.49b): Solution
5 ft
x (a)
P
20  h + z = 45 20 2 + x 2 = z 2.
z
20 ft W h O
x
M
(b)
FIGURE 3.49 A worker at M walks to the right, pulling the weight W upward as the rope moves through the pulley P (Example 6).
Total length of rope is 45 ft. Angle at O is a right angle.
If we solve for z = 25 + h in the first equation, and substitute into the second equation, we have 20 2 + x 2 = s25 + hd2.
(1)
Differentiating both sides with respect to t gives 2x
dx dh = 2s25 + hd , dt dt
and solving this last equation for dh>dt we find x dx dh = . dt 25 + h dt
(2)
Since we know dx>dt, it remains only to find 25 + h at the instant when x = 21. From Equation (1), 20 2 + 212 = s25 + hd2 so that s25 + hd2 = 841,
or
25 + h = 29.
Equation (2) now gives 126 dh 21 # = 6 = L 4.3 ft>sec 29 29 dt as the rate at which the weight is being raised when x = 21 ft.
Exercises 3.10 1. Area Suppose that the radius r and area A = pr 2 of a circle are differentiable functions of t. Write an equation that relates dA>dt to dr>dt.
6. If x = y 3  y and dy>dt = 5, then what is dx>dt when y = 2?
2. Surface area Suppose that the radius r and surface area S = 4pr 2 of a sphere are differentiable functions of t. Write an equation that relates dS>dt to dr>dt.
8. If x 2y 3 = 4>27 and dy>dt = 1>2, then what is dx>dt when x = 2?
3. Assume that y = 5x and dx>dt = 2. Find dy>dt. 4. Assume that 2x + 3y = 12 and dy>dt = 2. Find dx>dt. 5. If y = x 2 and dx>dt = 3, then what is dy>dt when x = 1?
7. If x 2 + y 2 = 25 and dx>dt =  2, then what is dy>dt when x = 3 and y = 4? 9. If L = 2x 2 + y 2, dx>dt = 1, and dy>dt = 3, find dL>dt when x = 5 and y = 12. 10. If r + s 2 + y3 = 12, dr>dt = 4, and ds>dt = 3, find dy>dt when r = 3 and s = 1.
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Chapter 3: Differentiation
11. If the original 24 m edge length x of a cube decreases at the rate of 5 m>min, when x = 3 m at what rate does the cube’s
a. Assuming that x, y, and z are differentiable functions of t, how is ds>dt related to dx>dt, dy>dt, and dz>dt?
a. surface area change?
b. How is ds>dt related to dy>dt and dz>dt if x is constant?
b. volume change?
c. How are dx>dt, dy>dt, and dz>dt related if s is constant?
12. A cube’s surface area increases at the rate of 72 in2>sec. At what rate is the cube’s volume changing when the edge length is x = 3 in? 13. Volume The radius r and height h of a right circular cylinder are related to the cylinder’s volume V by the formula V = pr 2h . a. How is dV>dt related to dh>dt if r is constant? b. How is dV>dt related to dr>dt if h is constant? c. How is dV>dt related to dr>dt and dh>dt if neither r nor h is constant? 14. Volume The radius r and height h of a right circular cone are related to the cone’s volume V by the equation V = s1>3dpr 2h . a. How is dV>dt related to dh>dt if r is constant? b. How is dV>dt related to dr>dt if h is constant? c. How is dV>dt related to dr>dt and dh>dt if neither r nor h is constant? 15. Changing voltage The voltage V (volts), current I (amperes), and resistance R (ohms) of an electric circuit like the one shown here are related by the equation V = IR . Suppose that V is increasing at the rate of 1 volt> sec while I is decreasing at the rate of 1> 3 amp> sec. Let t denote time in seconds. V
19. Area The area A of a triangle with sides of lengths a and b enclosing an angle of measure u is A =
1 ab sin u . 2
a. How is dA>dt related to du>dt if a and b are constant? b. How is dA>dt related to du>dt and da>dt if only b is constant? c. How is dA>dt related to du>dt, da>dt , and db>dt if none of a, b, and u are constant? 20. Heating a plate When a circular plate of metal is heated in an oven, its radius increases at the rate of 0.01 cm> min. At what rate is the plate’s area increasing when the radius is 50 cm? 21. Changing dimensions in a rectangle The length l of a rectangle is decreasing at the rate of 2 cm> sec while the width w is increasing at the rate of 2 cm> sec. When l = 12 cm and w = 5 cm , find the rates of change of (a) the area, (b) the perimeter, and (c) the lengths of the diagonals of the rectangle. Which of these quantities are decreasing, and which are increasing? 22. Changing dimensions in a rectangular box Suppose that the edge lengths x, y, and z of a closed rectangular box are changing at the following rates: dx = 1 m>sec, dt
dy = 2 m>sec, dt
dz = 1 m>sec . dt
Find the rates at which the box’s (a) volume, (b) surface area, and (c) diagonal length s = 2x 2 + y 2 + z 2 are changing at the instant when x = 4, y = 3 , and z = 2 .
I
R
a. What is the value of dV>dt?
23. A sliding ladder A 13ft ladder is leaning against a house when its base starts to slide away (see accompanying figure). By the time the base is 12 ft from the house, the base is moving at the rate of 5 ft> sec.
b. What is the value of dI>dt?
a. How fast is the top of the ladder sliding down the wall then?
c. What equation relates dR>dt to dV>dt and dI>dt?
b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then?
d. Find the rate at which R is changing when V = 12 volts and I = 2 amp. Is R increasing, or decreasing? 16. Electrical power The power P (watts) of an electric circuit is related to the circuit’s resistance R (ohms) and current I (amperes) by the equation P = RI 2 . a. How are dP>dt, dR>dt, and dI>dt related if none of P, R, and I are constant?
c. At what rate is the angle u between the ladder and the ground changing then? y
y(t)
b. How is dR>dt related to dI>dt if P is constant?
13ft ladder
17. Distance Let x and y be differentiable functions of t and let s = 2x 2 + y 2 be the distance between the points (x, 0) and (0, y) in the xyplane. a. How is ds>dt related to dx>dt if y is constant? b. How is ds>dt related to dx>dt and dy>dt if neither x nor y is constant? c. How is dx>dt related to dy>dt if s is constant? 18. Diagonals If x, y, and z are lengths of the edges of a rectangular box, the common length of the box’s diagonals is s = 2x 2 + y 2 + z 2 .
0
x(t)
x
24. Commercial air traffic Two commercial airplanes are flying at an altitude of 40,000 ft along straightline courses that intersect at right angles. Plane A is approaching the intersection point at a speed of 442 knots (nautical miles per hour; a nautical mile is 2000 yd). Plane B is approaching the intersection at 481 knots. At what rate is the distance between the planes changing when A is 5
3.10 nautical miles from the intersection point and B is 12 nautical miles from the intersection point?
27. A growing sand pile Sand falls from a conveyor belt at the rate of 10 m3>min onto the top of a conical pile. The height of the pile is always threeeighths of the base diameter. How fast are the (a) height and (b) radius changing when the pile is 4 m high? Answer in centimeters per minute.
193
Ring at edge of dock
25. Flying a kite A girl flies a kite at a height of 300 ft, the wind carrying the kite horizontally away from her at a rate of 25 ft> sec. How fast must she let out the string when the kite is 500 ft away from her? 26. Boring a cylinder The mechanics at Lincoln Automotive are reboring a 6in.deep cylinder to fit a new piston. The machine they are using increases the cylinder’s radius one thousandth of an inch every 3 min. How rapidly is the cylinder volume increasing when the bore (diameter) is 3.800 in.?
Related Rates
6'
33. A balloon and a bicycle A balloon is rising vertically above a level, straight road at a constant rate of 1 ft> sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft> sec passes under it. How fast is the distance s(t) between the bicycle and balloon increasing 3 sec later? y
28. A draining conical reservoir Water is flowing at the rate of 50 m3>min from a shallow concrete conical reservoir (vertex down) of base radius 45 m and height 6 m. a. How fast (centimeters per minute) is the water level falling when the water is 5 m deep? b. How fast is the radius of the water’s surface changing then? Answer in centimeters per minute.
y(t)
29. A draining hemispherical reservoir Water is flowing at the rate of 6 m3>min from a reservoir shaped like a hemispherical bowl of radius 13 m, shown here in profile. Answer the following questions, given that the volume of water in a hemispherical bowl of radius R is V = sp>3dy 2s3R  yd when the water is y meters deep.
s(t)
Center of sphere 13 Water level y
a. At what rate is the water level changing when the water is 8 m deep?
x(t)
0
r
x
34. Making coffee Coffee is draining from a conical filter into a cylindrical coffeepot at the rate of 10 in3>min. a. How fast is the level in the pot rising when the coffee in the cone is 5 in. deep? b. How fast is the level in the cone falling then?
b. What is the radius r of the water’s surface when the water is y m deep? 6"
c. At what rate is the radius r changing when the water is 8 m deep? 30. A growing raindrop Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks up moisture at a rate proportional to its surface area. Show that under these circumstances the drop’s radius increases at a constant rate.
6" How fast is this level falling?
31. The radius of an inflating balloon A spherical balloon is inflated with helium at the rate of 100p ft3>min. How fast is the balloon’s radius increasing at the instant the radius is 5 ft? How fast is the surface area increasing? 32. Hauling in a dinghy A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft above the bow. The rope is hauled in at the rate of 2 ft> sec.
How fast is this level rising?
a. How fast is the boat approaching the dock when 10 ft of rope are out? b. At what rate is the angle u changing at this instant (see the figure)?
6"
194
Chapter 3: Differentiation
35. Cardiac output In the late 1860s, Adolf Fick, a professor of physiology in the Faculty of Medicine in Würzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about 7 L> min. At rest it is likely to be a bit under 6 L> min. If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30 L> min. Your cardiac output can be calculated with the formula
Light Ball at time t 0 1/2 sec later 50ft pole
Q y = , D where Q is the number of milliliters of CO2 you exhale in a minute and D is the difference between the CO2 concentration (ml> L) in the blood pumped to the lungs and the CO2 concentration in the blood returning from the lungs. With Q = 233 ml>min and D = 97  56 = 41 ml>L , y =
away from the light. (See accompanying figure.) How fast is the shadow of the ball moving along the ground 1>2 sec later? (Assume the ball falls a distance s = 16t 2 ft in t sec .)
233 ml>min L 5.68 L>min , 41 ml>L
fairly close to the 6 L> min that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D., Quillan College of Medicine, East Tennessee State University.) Suppose that when Q = 233 and D = 41 , we also know that D is decreasing at the rate of 2 units a minute but that Q remains unchanged. What is happening to the cardiac output?
Shadow 0
x(t)
30
x
NOT TO SCALE
40. A building’s shadow On a morning of a day when the sun will pass directly overhead, the shadow of an 80ft building on level ground is 60 ft long. At the moment in question, the angle u the sun makes with the ground is increasing at the rate of 0.27°> min. At what rate is the shadow decreasing? (Remember to use radians. Express your answer in inches per minute, to the nearest tenth.)
36. Moving along a parabola A particle moves along the parabola y = x 2 in the first quadrant in such a way that its xcoordinate (measured in meters) increases at a steady 10 m> sec. How fast is the angle of inclination u of the line joining the particle to the origin changing when x = 3 m ? 37. Motion in the plane The coordinates of a particle in the metric xyplane are differentiable functions of time t with dx>dt = 1 m>sec and dy>dt = 5 m>sec . How fast is the particle’s distance from the origin changing as it passes through the point (5, 12)? 38. Videotaping a moving car You are videotaping a race from a stand 132 ft from the track, following a car that is moving at 180 mi> h (264 ft> sec), as shown in the accompanying figure. How fast will your camera angle u be changing when the car is right in front of you? A half second later? Camera
132'
80'
41. A melting ice layer A spherical iron ball 8 in. in diameter is coated with a layer of ice of uniform thickness. If the ice melts at the rate of 10 in3>min , how fast is the thickness of the ice decreasing when it is 2 in. thick? How fast is the outer surface area of ice decreasing? 42. Highway patrol A highway patrol plane flies 3 mi above a level, straight road at a steady 120 mi> h. The pilot sees an oncoming car and with radar determines that at the instant the lineofsight distance from plane to car is 5 mi, the lineofsight distance is decreasing at the rate of 160 mi> h. Find the car’s speed along the highway. 43. Baseball players A baseball diamond is a square 90 ft on a side. A player runs from first base to second at a rate of 16 ft> sec.
Car
39. A moving shadow A light shines from the top of a pole 50 ft high. A ball is dropped from the same height from a point 30 ft
a. At what rate is the player’s distance from third base changing when the player is 30 ft from first base? b. At what rates are angles u1 and u2 (see the figure) changing at that time?
3.11 c. The player slides into second base at the rate of 15 ft> sec. At what rates are angles u1 and u2 changing as the player touches base?
Third base
1
2
195
44. Ships Two ships are steaming straight away from a point O along routes that make a 120° angle. Ship A moves at 14 knots (nautical miles per hour; a nautical mile is 2000 yd). Ship B moves at 21 knots. How fast are the ships moving apart when OA = 5 and OB = 3 nautical miles?
Second base 90'
Linearization and Differentials
Player 30'
First base
Home
3.11
Linearization and Differentials Sometimes we can approximate complicated functions with simpler ones that give the accuracy we want for specific applications and are easier to work with. The approximating functions discussed in this section are called linearizations, and they are based on tangent lines. Other approximating functions, such as polynomials, are discussed in Chapter 9. We introduce new variables dx and dy, called differentials, and define them in a way that makes Leibniz’s notation for the derivative dy>dx a true ratio. We use dy to estimate error in measurement, which then provides for a precise proof of the Chain Rule (Section 3.6).
Linearization As you can see in Figure 3.50, the tangent to the curve y = x 2 lies close to the curve near the point of tangency. For a brief interval to either side, the yvalues along the tangent line 4
2 y x2
y x2
y 2x 1
y 2x 1
(1, 1)
(1, 1) –1
3 0
0
2 0
y x 2 and its tangent y 2x 1 at (1, 1). 1.2
Tangent and curve very close near (1, 1). 1.003
y x2
y x2 y 2x 1
(1, 1)
(1, 1)
y 2x 1 0.8
1.2 0.8
Tangent and curve very close throughout entire xinterval shown.
0.997 0.997
1.003
Tangent and curve closer still. Computer screen cannot distinguish tangent from curve on this xinterval.
FIGURE 3.50 The more we magnify the graph of a function near a point where the function is differentiable, the flatter the graph becomes and the more it resembles its tangent.
196
Chapter 3: Differentiation
y
give good approximations to the yvalues on the curve. We observe this phenomenon by zooming in on the two graphs at the point of tangency or by looking at tables of values for the difference between ƒ(x) and its tangent line near the xcoordinate of the point of tangency. The phenomenon is true not just for parabolas; every differentiable curve behaves locally like its tangent line. In general, the tangent to y = ƒsxd at a point x = a , where ƒ is differentiable (Figure 3.51), passes through the point (a, ƒ(a)), so its pointslope equation is
y f(x) Slope f '(a) (a, f(a))
0
y L(x)
x
a
y = ƒsad + ƒ¿sadsx  ad.
FIGURE 3.51 The tangent to the curve y = ƒsxd at x = a is the line Lsxd = ƒsad + ƒ¿sadsx  ad .
Thus, this tangent line is the graph of the linear function Lsxd = ƒsad + ƒ¿sadsx  ad. For as long as this line remains close to the graph of ƒ, L(x) gives a good approximation to ƒ(x). If ƒ is differentiable at x = a, then the approximating function
DEFINITIONS
Lsxd = ƒsad + ƒ¿sadsx  ad is the linearization of ƒ at a. The approximation ƒsxd L Lsxd of ƒ by L is the standard linear approximation of ƒ at a. The point x = a is the center of the approximation.
Find the linearization of ƒsxd = 21 + x at x = 0 (Figure 3.52).
EXAMPLE 1
y
y 5 x 4 4
y1 x 2
y1 x 2
1.1
2 y 1 x
y 1 x
1.0
1
–1
0
1
2
3
x
4
FIGURE 3.52 The graph of y = 21 + x and its linearizations at x = 0 and x = 3 . Figure 3.53 shows a magnified view of the small window about 1 on the yaxis. Solution
0.9 –0.1
0
0.1
0.2
FIGURE 3.53 Magnified view of the window in Figure 3.52.
Since ƒ¿sxd =
1 s1 + xd1>2 , 2
we have ƒs0d = 1 and ƒ¿s0d = 1>2, giving the linearization Lsxd = ƒsad + ƒ¿sadsx  ad = 1 +
x 1 sx  0d = 1 + . 2 2
See Figure 3.53. The following table shows how accurate the approximation 21 + x L 1 + sx>2d from Example 1 is for some values of x near 0. As we move away from zero, we lose
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197
accuracy. For example, for x = 2, the linearization gives 2 as the approximation for 23 , which is not even accurate to one decimal place. Approximation
True value
ƒ True value approximation ƒ
21.2 L 1 +
0.2 2
= 1.10
1.095445
0.004555 6 102
21.05 L 1 +
0.05 2
= 1.025
1.024695
0.000305 6 103
1.002497
0.000003 6 105
21.005 L 1 +
0.005 = 1.00250 2
Do not be misled by the preceding calculations into thinking that whatever we do with a linearization is better done with a calculator. In practice, we would never use a linearization to find a particular square root. The utility of a linearization is its ability to replace a complicated formula by a simpler one over an entire interval of values. If we have to work with 21 + x for x close to 0 and can tolerate the small amount of error involved, we can work with 1 + sx>2d instead. Of course, we then need to know how much error there is. We further examine the estimation of error in Chapter 9. A linear approximation normally loses accuracy away from its center. As Figure 3.52 suggests, the approximation 21 + x L 1 + sx>2d will probably be too crude to be useful near x = 3. There, we need the linearization at x = 3.
EXAMPLE 2 Solution
Find the linearization of ƒsxd = 21 + x at x = 3.
We evaluate the equation defining Lsxd at a = 3. With ƒs3d = 2,
ƒ¿s3d =
1 1 = , s1 + xd1>2 ` 2 4 x=3
we have Lsxd = 2 +
5 x 1 (x  3) = + . 4 4 4
At x = 3.2, the linearization in Example 2 gives 21 + x = 21 + 3.2 L
5 3.2 + = 1.250 + 0.800 = 2.050, 4 4
which differs from the true value 24.2 L 2.04939 by less than one onethousandth. The linearization in Example 1 gives 21 + x = 21 + 3.2 L 1 +
y
3.2 = 1 + 1.6 = 2.6, 2
a result that is off by more than 25%.
EXAMPLE 3 0
2
Find the linearization of ƒsxd = cos x at x = p>2 (Figure 3.54).
x y cos x
y –x 2
FIGURE 3.54 The graph of ƒsxd = cos x and its linearization at x = p>2 . Near x = p>2, cos x L x + sp>2d (Example 3).
Since ƒsp>2d = cossp>2d = 0, ƒ¿sxd = sin x, and ƒ¿sp>2d = sin sp>2d = 1, we find the linearization at a = p>2 to be
Solution
Lsxd = ƒsad + ƒ¿sadsx  ad = 0 + s 1d ax = x +
p . 2
p b 2
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Chapter 3: Differentiation
An important linear approximation for roots and powers is s1 + xdk L 1 + kx
sx near 0; any number kd
(Exercise 15). This approximation, good for values of x sufficiently close to zero, has broad application. For example, when x is small, 21 + x L 1 +
1 x 2
k = 1>2
1 = s1  xd1 L 1 + s 1ds xd = 1 + x 1  x 3 2 1 + 5x 4 = s1 + 5x 4 d1>3 L 1 +
k = 1; replace x by x .
5 1 s5x 4 d = 1 + x 4 3 3
k = 1>3; replace x by 5x 4 .
1 1 1 = s1  x 2 d1>2 L 1 + a bs x 2 d = 1 + x 2 2 2 2 21  x
k =  1>2; replace x by x 2 .
Differentials We sometimes use the Leibniz notation dy>dx to represent the derivative of y with respect to x. Contrary to its appearance, it is not a ratio. We now introduce two new variables dx and dy with the property that when their ratio exists, it is equal to the derivative.
DEFINITION Let y = ƒsxd be a differentiable function. The differential dx is an independent variable. The differential dy is dy = ƒ¿sxd dx.
Unlike the independent variable dx, the variable dy is always a dependent variable. It depends on both x and dx. If dx is given a specific value and x is a particular number in the domain of the function ƒ, then these values determine the numerical value of dy.
EXAMPLE 4 (a) Find dy if y = x 5 + 37x. (b) Find the value of dy when x = 1 and dx = 0.2. Solution
(a) dy = s5x 4 + 37d dx (b) Substituting x = 1 and dx = 0.2 in the expression for dy, we have dy = s5 # 14 + 37d0.2 = 8.4. The geometric meaning of differentials is shown in Figure 3.55. Let x = a and set dx = ¢x. The corresponding change in y = ƒsxd is ¢y = ƒsa + dxd  ƒsad. The corresponding change in the tangent line L is ¢L = L(a + dx)  L(a) = ƒ(a) + ƒ¿(a)[(a + dx)  a]  ƒ(a) (++++++)++++++* L(a dx)
= ƒ¿(a) dx.
()* L(a)
3.11 y
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199
y f (x) (a dx, f (a dx)) y f(a dx) f (a) L f '(a)dx (a, f (a)) dx x When dx is a small change in x, the corresponding change in the linearization is precisely dy.
Tangent line 0
x
a dx
a
FIGURE 3.55 Geometrically, the differential dy is the change ¢L in the linearization of ƒ when x = a changes by an amount dx = ¢x .
That is, the change in the linearization of ƒ is precisely the value of the differential dy when x = a and dx = ¢x. Therefore, dy represents the amount the tangent line rises or falls when x changes by an amount dx = ¢x. If dx Z 0, then the quotient of the differential dy by the differential dx is equal to the derivative ƒ¿sxd because dy ƒ¿sxd dx = ƒ¿sxd = . dx dx
dy , dx = We sometimes write
dƒ = ƒ¿sxd dx in place of dy = ƒ¿sxd dx, calling dƒ the differential of ƒ. For instance, if ƒsxd = 3x 2  6, then dƒ = ds3x 2  6d = 6x dx. Every differentiation formula like dsu + yd du dy = + dx dx dx
or
dssin ud du = cos u dx dx
or
dssin ud = cos u du.
has a corresponding differential form like dsu + yd = du + dy
EXAMPLE 5 We can use the Chain Rule and other differentiation rules to find differentials of functions. (a) dstan 2xd = sec2s2xd ds2xd = 2 sec2 2x dx sx + 1d dx  x dsx + 1d x x dx + dx  x dx dx (b) d a = = b = x + 1 sx + 1d2 sx + 1d2 sx + 1d2
Estimating with Differentials Suppose we know the value of a differentiable function ƒ(x) at a point a and want to estimate how much this value will change if we move to a nearby point a + dx. If dx ¢x is small, then we can see from Figure 3.55 that ¢y is approximately equal to the differential dy. Since ƒsa + dxd = ƒsad + ¢y,
¢x = dx
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Chapter 3: Differentiation
the differential approximation gives ƒsa + dxd L ƒsad + dy when dx = ¢x. Thus the approximation ¢y L dy can be used to estimate ƒsa + dxd when ƒ(a) is known and dx is small. dr 0.1
a 10
EXAMPLE 6 The radius r of a circle increases from a = 10 m to 10.1 m (Figure 3.56). Use dA to estimate the increase in the circle’s area A. Estimate the area of the enlarged circle and compare your estimate to the true area found by direct calculation. Solution
Since A = pr 2 , the estimated increase is dA = A¿sad dr = 2pa dr = 2ps10ds0.1d = 2p m2 .
A ≈ dA 2 a dr
FIGURE 3.56 When dr is small compared with a, the differential dA gives the estimate Asa + drd = pa 2 + dA (Example 6).
Thus, since Asr + ¢rd L Asrd + dA, we have As10 + 0.1d L As10d + 2p = ps10d2 + 2p = 102p. The area of a circle of radius 10.1 m is approximately 102p m2 . The true area is As10.1d = ps10.1d2 = 102.01p m2 . The error in our estimate is 0.01p m2 , which is the difference ¢A  dA.
EXAMPLE 7
Use differentials to estimate
(a) 7.971>3 (b) sin (p>6 + 0.01). Solution
(a) The differential associated with the cube root function y = x1>3 is dy =
1 dx. 3x2>3
We set a = 8, the closest number near 7.97 where we can easily compute f (a) and f ¿(a). To arrange that a + dx = 7.97, we choose dx = 0.03. Approximating with the differential gives f (7.97) = f (a + dx) L f (a) + dy 1 (0.03) = 81>3 + 3(8)2>3 1 = 2 + (0.03) = 1.9975 12 This gives an approximation to the true value of 7.971>3, which is 1.997497 to 6 decimals. (b) The differential associated with y = sin x is dy = cos x dx.
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201
To estimate sin (p>6 + 0.01), we set a = p>6 and dx = 0.01. Then f (p>6 + 0.01) = f (a + dx) L f (a) + dy = sin =
p p + acos b (0.01) 6 6
23 1 + (0.01) L 0.5087 2 2
For comparison, the true value of sin (p>6 + 0.01) to 6 decimals is 0.508635. The method in part (b) of Example 7 is used by some calculator and computer algorithms to give values of trigonometric functions. The algorithms store a large table of sine and cosine values between 0 and p>4. Values between these stored values are computed using differentials as in Example 7b. Values outside of [0, p>4] are computed from values in this interval using trigonometric identities.
Error in Differential Approximation Let ƒ(x) be differentiable at x = a and suppose that dx = ¢x is an increment of x. We have two ways to describe the change in ƒ as x changes from a to a + ¢x: The true change:
¢ƒ = ƒsa + ¢xd  ƒsad
The differential estimate:
dƒ = ƒ¿sad ¢x.
How well does dƒ approximate ¢ƒ? We measure the approximation error by subtracting dƒ from ¢ƒ: Approximation error = ¢ƒ  dƒ = ¢ƒ  ƒ¿sad¢x = ƒ(a + ¢x)  ƒ(a)  ƒ¿(a)¢x (++++)++++* ƒ
= a
ƒ(a + ¢x)  ƒ(a)  ƒ¿(a)b # ¢x ¢x
(+++++++)+++++++* Call this part P.
= P # ¢x. As ¢x : 0, the difference quotient
ƒsa + ¢xd  ƒsad ¢x approaches ƒ¿sad (remember the definition of ƒ¿sad), so the quantity in parentheses becomes a very small number (which is why we called it P). In fact, P : 0 as ¢x : 0. When ¢x is small, the approximation error P ¢x is smaller still. ¢ƒ = ƒ¿(a)¢x + P ¢x
()* true change
(+)+* estimated change
()* error
Although we do not know the exact size of the error, it is the product P # ¢x of two small quantities that both approach zero as ¢x : 0. For many common functions, whenever ¢x is small, the error is still smaller.
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Chapter 3: Differentiation
Change in y ƒsxd near x a If y = ƒsxd is differentiable at x = a and x changes from a to a + ¢x, the change ¢y in ƒ is given by ¢y = ƒ¿sad ¢x + P ¢x
(1)
in which P : 0 as ¢x : 0.
In Example 6 we found that 2 ¢A = p(10.1) 2  p(10) 2 = (102.01  100)p = (2p + 0.01p) ()* ()* m
dA
error
so the approximation error is ¢A  dA = P¢r = 0.01p and P = 0.01p>¢r = 0.01p>0.1 = 0.1p m.
Proof of the Chain Rule Equation (1) enables us to prove the Chain Rule correctly. Our goal is to show that if ƒ(u) is a differentiable function of u and u = gsxd is a differentiable function of x, then the composite y = ƒsgsxdd is a differentiable function of x. Since a function is differentiable if and only if it has a derivative at each point in its domain, we must show that whenever g is differentiable at x0 and ƒ is differentiable at gsx0 d, then the composite is differentiable at x0 and the derivative of the composite satisfies the equation dy ` = ƒ¿s gsx0 dd # g¿sx0 d. dx x = x0 Let ¢x be an increment in x and let ¢u and ¢y be the corresponding increments in u and y. Applying Equation (1) we have ¢u = g¿sx0 d¢x + P1 ¢x = sg¿sx0 d + P1 d¢x, where P1 : 0 as ¢x : 0. Similarly, ¢y = ƒ¿su0 d¢u + P2 ¢u = sƒ¿su0 d + P2 d¢u, where P2 : 0 as ¢u : 0. Notice also that ¢u : 0 as ¢x : 0. Combining the equations for ¢u and ¢y gives ¢y = sƒ¿su0 d + P2 dsg¿sx0 d + P1 d¢x, so ¢y = ƒ¿su0 dg¿sx0 d + P2 g¿sx0 d + ƒ¿su0 dP1 + P2P1 . ¢x Since P1 and P2 go to zero as ¢x goes to zero, three of the four terms on the right vanish in the limit, leaving dy ¢y ` = ƒ¿su0 dg¿sx0 d = ƒ¿sgsx0 dd # g¿sx0 d. = lim dx x = x0 ¢x:0 ¢x
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203
Sensitivity to Change The equation df = ƒ¿sxd dx tells how sensitive the output of ƒ is to a change in input at different values of x. The larger the value of ƒ¿ at x, the greater the effect of a given change dx. As we move from a to a nearby point a + dx, we can describe the change in ƒ in three ways:
Absolute change Relative change Percentage change
True
Estimated
¢ƒ = ƒsa + dxd  ƒsad ¢ƒ ƒsad ¢ƒ * 100 ƒsad
dƒ = ƒ¿sad dx dƒ ƒsad dƒ * 100 ƒsad
You want to calculate the depth of a well from the equation s = 16t 2 by timing how long it takes a heavy stone you drop to splash into the water below. How sensitive will your calculations be to a 0.1sec error in measuring the time?
EXAMPLE 8
The size of ds in the equation ds = 32t dt depends on how big t is. If t = 2 sec, the change caused by dt = 0.1 is about ds = 32s2ds0.1d = 6.4 ft. Three seconds later at t = 5 sec, the change caused by the same dt is ds = 32s5ds0.1d = 16 ft. For a fixed error in the time measurement, the error in using ds to estimate the depth is larger when it takes a longer time before the stone splashes into the water. Solution
Exercises 3.11 Finding Linearizations In Exercises 1–5, find the linearization L(x) of ƒ(x) at x = a . 1. ƒsxd = x 3  2x + 3,
a = 2
14. ƒ(x) = sin1 x,
2. ƒsxd = 2x + 9, a = 4 1 3. ƒsxd = x + x , a = 1 2
3 x, 4. ƒsxd = 2
a = 8
5. ƒ(x) = tan x,
a = p
(b) cos x
16. Use the linear approximation s1 + xdk L 1 + kx to find an approximation for the function ƒ(x) for values of x near zero.
(c) tan x
(d) e x
(e) ln (1 + x)
Linearization for Approximation In Exercises 7–14, find a linearization at a suitably chosen integer near a at which the given function and its derivative are easy to evaluate. 7. ƒsxd = x + 2x, 2
1
8. ƒsxd = x ,
a = 0.1
a = 0.9
9. ƒsxd = 2x + 3x  3, 2
10. ƒsxd = 1 + x, 3 x, 11. ƒsxd = 2
a = 8.1 a = 8.5
a = p>12
15. Show that the linearization of ƒsxd = s1 + xdk at x = 0 is Lsxd = 1 + kx .
6. Common linear approximations at x = 0 Find the linearizations of the following functions at x = 0. (a) sin x
x , a = 1.3 x + 1 x 13. ƒ(x) = e , a = 0.1 12. ƒsxd =
a = 0.9
a. ƒsxd = s1  xd6 c. ƒsxd =
1 21 + x
e. ƒsxd = s4 + 3xd1>3
b. ƒsxd =
2 1  x
d. ƒsxd = 22 + x 2 f. ƒsxd =
3
B
a1 
1 b 2 + x
2
17. Faster than a calculator Use the approximation s1 + xdk L 1 + kx to estimate the following. a. s1.0002d50
3 1.009 b. 2
18. Find the linearization of ƒsxd = 2x + 1 + sin x at x = 0 . How is it related to the individual linearizations of 2x + 1 and sin x at x = 0 ?
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Chapter 3: Differentiation
Derivatives in Differential Form In Exercises 19–38, find dy.
49. The change in the volume V = pr 2h of a right circular cylinder when the radius changes from r0 to r0 + dr and the height does not change
19. y = x 3  3 1x
20. y = x21  x 2
2x 1 + x2 23. 2y 3>2 + xy  x = 0
22. y =
2 1x 3s1 + 1xd 24. xy 2  4x 3>2  y = 0
50. The change in the lateral surface area S = 2prh of a right circular cylinder when the height changes from h0 to h0 + dh and the radius does not change
25. y = sin s5 1xd
26. y = cos sx 2 d
Applications 51. The radius of a circle is increased from 2.00 to 2.02 m.
21. y =
27. y = 4 tan sx >3d
28. y = sec sx 2  1d
3
29. y = 3 csc s1  21xd
30. y = 2 cot a
31. y = e2x
32. y = xex
33. y = ln (1 + x2)
34. y = ln a
1 b 1x
b 2x  1 1 36. y = cot1 a 2 b + cos1 2x x
35. y = tan1 (e x ) 2
37. y = sec1 (ex )
38. y = etan
1
x + 1
a. Estimate the resulting change in area. b. Express the estimate as a percentage of the circle’s original area. 52. The diameter of a tree was 10 in. During the following year, the circumference increased 2 in. About how much did the tree’s diameter increase? The tree’s crosssection area? 53. Estimating volume Estimate the volume of material in a cylindrical shell with length 30 in., radius 6 in., and shell thickness 0.5 in.
2x 2 + 1
0.5 in. 6 in. 30 in.
Approximation Error In Exercises 39–44, each function ƒ(x) changes value when x changes from x0 to x0 + dx . Find a. the change ¢ƒ = ƒsx0 + dxd  ƒsx0 d ; b. the value of the estimate dƒ = ƒ¿sx0 d dx ; and c. the approximation error ƒ ¢ƒ  dƒ ƒ . y
54. Estimating height of a building A surveyor, standing 30 ft from the base of a building, measures the angle of elevation to the top of the building to be 75°. How accurately must the angle be measured for the percentage error in estimating the height of the building to be less than 4%? 55. Tolerance The radius r of a circle is measured with an error of at most 2%. What is the maximum corresponding percentage error in computing the circle’s
y f (x)
a. circumference? f f (x 0 dx) f (x 0) df f '(x 0 ) dx
(x 0, f(x 0 )) dx 0
x0 = 1,
40. ƒsxd = 2x + 4x  3, 42. ƒsxd = x 4, 43. ƒsxd = x 1,
x0 = 1,
x0 = 1,
dx = 0.1
dx = 0.1 dx = 0.1
x0 = 2,
b. volume?
57. Tolerance The height and radius of a right circular cylinder are equal, so the cylinder’s volume is V = ph 3 . The volume is to be calculated with an error of no more than 1% of the true value. Find approximately the greatest error that can be tolerated in the measurement of h, expressed as a percentage of h. 58. Tolerance
dx = 0.1
x0 = 0.5,
44. ƒsxd = x 3  2x + 3,
dx = 0.1
x0 = 1,
2
41. ƒsxd = x 3  x,
x
x 0 dx
x0
39. ƒsxd = x 2 + 2x,
56. Tolerance The edge x of a cube is measured with an error of at most 0.5%. What is the maximum corresponding percentage error in computing the cube’s a. surface area?
Tangent
b. area?
dx = 0.1
Differential Estimates of Change In Exercises 45–50, write a differential formula that estimates the given change in volume or surface area.
a. About how accurately must the interior diameter of a 10mhigh cylindrical storage tank be measured to calculate the tank’s volume to within 1% of its true value? b. About how accurately must the tank’s exterior diameter be measured to calculate the amount of paint it will take to paint the side of the tank to within 5% of the true amount?
45. The change in the volume V = s4>3dpr of a sphere when the radius changes from r0 to r0 + dr
59. The diameter of a sphere is measured as 100 ; 1 cm and the volume is calculated from this measurement. Estimate the percentage error in the volume calculation.
46. The change in the volume V = x 3 of a cube when the edge lengths change from x0 to x0 + dx
60. Estimate the allowable percentage error in measuring the diameter D of a sphere if the volume is to be calculated correctly to within 3%.
47. The change in the surface area S = 6x 2 of a cube when the edge lengths change from x0 to x0 + dx
61. The effect of flight maneuvers on the heart The amount of work done by the heart’s main pumping chamber, the left ventricle, is given by the equation
3
48. The change in the lateral surface area S = pr2r 2 + h 2 of a right circular cone when the radius changes from r0 to r0 + dr and the height does not change
W = PV +
Vdy2 , 2g
3.11 where W is the work per unit time, P is the average blood pressure, V is the volume of blood pumped out during the unit of time, d (“delta”) is the weight density of the blood, y is the average velocity of the exiting blood, and g is the acceleration of gravity. When P, V, d , and y remain constant, W becomes a function of g, and the equation takes the simplified form b W = a + g sa, b constantd . As a member of NASA’s medical team, you want to know how sensitive W is to apparent changes in g caused by flight maneuvers, and this depends on the initial value of g. As part of your investigation, you decide to compare the effect on W of a given change dg on the moon, where g = 5.2 ft>sec2 , with the effect the same change dg would have on Earth, where g = 32 ft>sec2 . Use the simplified equation above to find the ratio of dWmoon to dWEarth . 62. Measuring acceleration of gravity When the length L of a clock pendulum is held constant by controlling its temperature, the pendulum’s period T depends on the acceleration of gravity g. The period will therefore vary slightly as the clock is moved from place to place on the earth’s surface, depending on the change in g. By keeping track of ¢T , we can estimate the variation in g from the equation T = 2psL>gd1>2 that relates T, g, and L. a. With L held constant and g as the independent variable, calculate dT and use it to answer parts (b) and (c). b. If g increases, will T increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a 100cm pendulum is moved from a location where g = 980 cm>sec2 to a new location. This increases the period by dT = 0.001 sec . Find dg and estimate the value of g at the new location. 63. The linearization is the best linear approximation Suppose that y = ƒsxd is differentiable at x = a and that g sxd = msx  ad + c is a linear function in which m and c are constants. If the error Esxd = ƒsxd  g sxd were small enough near x = a , we might think of using g as a linear approximation of ƒ instead of the linearization Lsxd = ƒsad + ƒ¿sadsx  ad . Show that if we impose on g the conditions 1. Esad = 0 Esxd 2. lim x  a = 0 x: a
The approximation error is zero at x = a . The error is negligible when compared with x  a .
then g sxd = ƒsad + ƒ¿sadsx  ad . Thus, the linearization L(x) gives the only linear approximation whose error is both zero at x = a and negligible in comparison with x  a . The linearization, L(x): y f(a) f '(a)(x a)
Some other linear approximation, g(x): y m(x a) c y f (x)
a
x
64. Quadratic approximations a. Let Qsxd = b0 + b1sx  ad + b2sx  ad2 be a quadratic approximation to ƒ(x) at x = a with the properties: i) Qsad = ƒsad ii) Q¿sad = ƒ¿sad iii) Q–sad = ƒ–sad.
205
Determine the coefficients b0 , b1 , and b2 . b. Find the quadratic approximation to ƒsxd = 1>s1  xd at x = 0. T c. Graph ƒsxd = 1>s1  xd and its quadratic approximation at x = 0 . Then zoom in on the two graphs at the point (0, 1). Comment on what you see. T d. Find the quadratic approximation to gsxd = 1>x at x = 1 . Graph g and its quadratic approximation together. Comment on what you see. T e. Find the quadratic approximation to hsxd = 21 + x at x = 0 . Graph h and its quadratic approximation together. Comment on what you see. f. What are the linearizations of ƒ, g, and h at the respective points in parts (b), (d), and (e)? 65. The linearization of 2x a. Find the linearization of ƒsxd = 2x at x = 0 . Then round its coefficients to two decimal places. T b. Graph the linearization and function together for 3 … x … 3 and 1 … x … 1 . 66. The linearization of log3 x a. Find the linearization of ƒsxd = log 3 x at x = 3 . Then round its coefficients to two decimal places. T b. Graph the linearization and function together in the window 0 … x … 8 and 2 … x … 4 .
COMPUTER EXPLORATIONS In Exercises 67–72, use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: a. Plot the function ƒ over I. b. Find the linearization L of the function at the point a. c. Plot ƒ and L together on a single graph. d. Plot the absolute error ƒ ƒsxd  Lsxd ƒ over I and find its maximum value. e. From your graph in part (d), estimate as large a d 7 0 as you can, satisfying ƒx  aƒ 6 d
Q
ƒ ƒsxd  Lsxd ƒ 6 P
for P = 0.5, 0.1, and 0.01. Then check graphically to see if your destimate holds true. 67. ƒsxd = x 3 + x 2  2x, 68. ƒsxd =
(a, f (a))
Linearization and Differentials
x  1 , 4x 2 + 1
a = 1
[1, 2],
3 1 c , 1 d, a = 4 2
69. ƒsxd = x 2>3sx  2d,
[2, 3],
a = 2
70. ƒsxd = 1x  sin x,
[0, 2p],
a = 2
71. ƒ(x) = x2 ,
a = 1
x
[0, 2],
72. ƒ(x) = 2x sin1 x, [0, 1],
a =
1 2
206
Chapter 3: Differentiation
Chapter 3
Questions to Guide Your Review
1. What is the derivative of a function ƒ ? How is its domain related to the domain of ƒ? Give examples. 2. What role does the derivative play in defining slopes, tangents, and rates of change? 3. How can you sometimes graph the derivative of a function when all you have is a table of the function’s values? 4. What does it mean for a function to be differentiable on an open interval? On a closed interval? 5. How are derivatives and onesided derivatives related? 6. Describe geometrically when a function typically does not have a derivative at a point. 7. How is a function’s differentiability at a point related to its continuity there, if at all? 8. What rules do you know for calculating derivatives? Give some examples. 9. Explain how the three formulas d n sx d = nx n  1 a. dx d du scud = c b. dx dx du1 du2 Á dun d su + u2 + Á + un d = + + + c. dx 1 dx dx dx enable us to differentiate any polynomial.
20. What is the rule for calculating the derivative of a composite of two differentiable functions? How is such a derivative evaluated? Give examples. 21. If u is a differentiable function of x, how do you find sd>dxdsu n d if n is an integer? If n is a real number? Give examples. 22. What is implicit differentiation? When do you need it? Give examples. 23. What is the derivative of the natural logarithm function ln x? How does the domain of the derivative compare with the domain of the function? 24. What is the derivative of the exponential function ax, a 7 0 and a Z 1? What is the geometric significance of the limit of (ah  1)>h as h : 0? What is the limit when a is the number e? 26. What is logarithmic differentiation? Give an example. 27. How can you write any real power of x as a power of e? Are there any restrictions on x? How does this lead to the Power Rule for differentiating arbitrary real powers?
11. What is a second derivative? A third derivative? How many derivatives do the functions you know have? Give examples. x
12. What is the derivative of the exponential function e ? How does the domain of the derivative compare with the domain of the function? 13. What is the relationship between a function’s average and instantaneous rates of change? Give an example. 14. How do derivatives arise in the study of motion? What can you learn about a body’s motion along a line by examining the derivatives of the body’s position function? Give examples. 15. How can derivatives arise in economics? 16. Give examples of still other applications of derivatives. 17. What do the limits limh:0 sssin hd>hd and limh:0 sscos h  1d>hd have to do with the derivatives of the sine and cosine functions? What are the derivatives of these functions?
28. What is one way of expressing the special number e as a limit? What is an approximate numerical value of e correct to 7 decimal places? 29. What are the derivatives of the inverse trigonometric functions? How do the domains of the derivatives compare with the domains of the functions? 30. How do related rates problems arise? Give examples. 31. Outline a strategy for solving related rates problems. Illustrate with an example. 32. What is the linearization L(x) of a function ƒ(x) at a point x = a ? What is required of ƒ at a for the linearization to exist? How are linearizations used? Give examples. 33. If x moves from a to a nearby value a + dx , how do you estimate the corresponding change in the value of a differentiable function ƒ(x)? How do you estimate the relative change? The percentage change? Give an example.
Practice Exercises
Derivatives of Functions Find the derivatives of the functions in Exercises 1– 64. 1. y = x  0.125x + 0.25x 5
19. At what points are the six basic trigonometric functions continuous? How do you know?
25. What is the derivative of loga x? Are there any restrictions on a?
10. What formula do we need, in addition to the three listed in Question 9, to differentiate rational functions?
Chapter 3
18. Once you know the derivatives of sin x and cos x, how can you find the derivatives of tan x, cot x, sec x, and csc x? What are the derivatives of these functions?
2
3. y = x 3  3sx 2 + p2 d 5. y = sx + 1d2sx 2 + 2xd
2. y = 3  0.7x + 0.3x 1 4. y = x 7 + 27x p + 1 3
7
6. y = s2x  5ds4  xd1
7. y = su2 + sec u + 1d3 9. s =
1t 1 + 1t
11. y = 2 tan2 x  sec2 x
8. y = a1 
csc u u2  b 2 4
1 1t  1 1 2 12. y = sin x sin2 x 10. s =
2
Chapter 3
13. s = cos4 s1  2td
2 14. s = cot3 a t b
71. y 2 =
15. s = ssec t + tan td5
16. s = csc5 s1  t + 3t 2 d
73. e x + 2y = 1
74. y 2 = 2e 1>x
17. r = 22u sin u
18. r = 2u 2cos u
75. ln (x>y) = 1
76. x sin1 y = 1 + x 2
20. r = sin A u + 2u + 1 B
19. r = sin 22u 1 2 2 x csc x 2 23. y = x 1>2 sec s2xd2
22. y = 2 1x sin 1x
25. y = 5 cot x
26. y = x cot 5x
21. y =
2
2
81. r cos 2s + sin2 s = p
1 30. s = 15s15t  1d3
x + x 33. y = B x2
34. y = 4x2x + 1x
2
sin u b cos u  1
2
36. r = a
1 + sin u b 1  cos u
38. y = 20s3x  4d s3x  4d
3 s5x 2 + sin 2xd3>2 41. y = 10ex>5
40. y = s3 + cos3 3xd1>3
1>4
42. y = 22e22x
1 4x 1 4x xe e 4 16
44. y = x2e2>x 46. y = ln ssec2 ud
45. y = ln ssin2 ud
47. y = log2 sx >2d
48. y = log5 s3x  7d
2
t
50. y = 92t
51. y = 5x3.6
52. y = 22x22
53. y = sx + 2dx + 2
54. y = 2sln xdx>2
1
55. y = sin 21  u ,
0 6 u 6 1
2
56. y = sin1 a
1
2y 57. y = ln cos1 x
b,
2 b. y 2 = 1  x
a. x 3 + y 3 = 1 84. a. By differentiating dy>dx = x>y .
62. y = 22x  1 sec 63. y = csc
1
1>5
64. y = s1 + x2 detan
1
that
Numerical Values of Derivatives 85. Suppose that functions ƒ(x) and g(x) and their first derivatives have the following values at x = 0 and x = 1 . x
ƒ(x)
g (x)
ƒ(x)
g(x)
0 1
1 3
1 5
3 1>2
1>2 4
Find the first derivatives of the following combinations at the given value of x. a. 6ƒsxd  g sxd, x = 1 ƒsxd , x = 1 c. g sxd + 1 e. g sƒsxdd,
b. ƒsxdg 2sxd, d. ƒsg sxdd,
x = 0
g. ƒsx + g sxdd,
z 7 1
1x
x = 0 x = 0
f. sx + ƒsxdd3>2,
x = 1
x = 0
x
ƒ(x)
ƒ(x)
0 1
9 3
2 1>5
Find the first derivatives of the following combinations at the given value of x. a. 1x ƒsxd,
0 6 u 6 p>2
ssec ud,
show
b. Then show that d 2y>dx 2 = 1>y 3 .
58. y = z cos1 z  21  z2 1 59. y = t tan1 t  ln t 2 60. y = s1 + t 2 d cot1 2t 1
implicitly,
86. Suppose that the function ƒ(x) and its first derivative have the following values at x = 0 and x = 1 .
y 7 1
61. y = z sec1 z  2z2  1,
x2  y2 = 1
2
37. y = s2x + 1d22x + 1 39. y =
82. 2rs  r  s + s 2 = 3
83. Find d 2y>dx 2 by implicit differentiation:
2 2 1x b 32. y = a 2 1x + 1
49. y = 8
80. q = s5p 2 + 2pd3>2
In Exercises 81 and 82, find dr>ds.
28. y = x 2 sin2 sx 3 d
2
1x 2 b 31. y = a 1 + x
43. y =
78. x y = 22
2
4t b 29. s = a t + 1
35. r = a
= 2
79. p 3 + 4pq  3q 2 = 2
24. y = 1x csc sx + 1d
2
x
1 + x A1  x
72. y 2 =
In Exercises 79 and 80, find dp>dq. 3
27. y = x sin s2x d 2
77. ye tan
1
x x + 1
207
Practice Exercises
x = 1
c. ƒs 1xd, x = 1 ƒsxd , x = 0 e. 2 + cos x
x
Implicit Differentiation In Exercises 65–78, find dy>dx by implicit differentiation. 65. xy + 2x + 3y = 1
66. x 2 + xy + y 2  5x = 2
67. x 3 + 4xy  3y 4>3 = 2x
68. 5x 4>5 + 10y 6>5 = 15
69. 1xy = 1
70. x y = 1 2 2
b. 2ƒsxd,
x = 0
d. ƒs1  5 tan xd,
x = 0
px f. 10 sin a b ƒ 2sxd, 2
x=1
87. Find the value of dy>dt at t = 0 if y = 3 sin 2x and x = t 2 + p . 88. Find the value of ds>du at u = 2 if s = t 2 + 5t and t = su 2 + 2ud1>3 .
89. Find the value of dw>ds at s = 0 if w = sin A e 1r B and r = 3 sin ss + p>6d .
208
Chapter 3: Differentiation
90. Find the value of dr>dt at t = 0 if r = su2 + 7d1>3 and u2t + u = 1 . 91. If y 3 + y = 2 cos x , find the value of d 2y>dx 2 at the point (0, 1). 92. If x 1>3 + y 1>3 = 4 , find d 2y>dx 2 at the point (8, 8). Applying the Derivative Definition In Exercises 93 and 94, find the derivative using the definition. 93. ƒstd =
1 2t + 1
94. g sxd = 2x 2 + 1 95. a. Graph the function ƒsxd = e
x 2, x 2,
1 … x 6 0 0 … x … 1.
b. Is ƒ continuous at x = 0 ? c. Is ƒ differentiable at x = 0 ? Give reasons for your answers. 96. a. Graph the function x, ƒsxd = e tan x, b. Is ƒ continuous at x = 0 ?
1 … x 6 0 0 … x … p>4.
c. Is ƒ differentiable at x = 0 ? Give reasons for your answers. 97. a. Graph the function x, 0 … x … 1 ƒsxd = e 2  x, 1 6 x … 2. b. Is ƒ continuous at x = 1 ? c. Is ƒ differentiable at x = 1 ? Give reasons for your answers. 98. For what value or values of the constant m, if any, is sin 2x, x … 0 ƒsxd = e mx, x 7 0 a. continuous at x = 0 ?
105. Normals parallel to a line Find the points on the curve y = tan x, p>2 6 x 6 p>2 , where the normal is parallel to the line y = x>2 . Sketch the curve and normals together, labeling each with its equation. 106. Tangent and normal lines Find equations for the tangent and normal to the curve y = 1 + cos x at the point sp>2, 1d . Sketch the curve, tangent, and normal together, labeling each with its equation. 107. Tangent parabola The parabola y = x 2 + C is to be tangent to the line y = x . Find C. 108. Slope of tangent Show that the tangent to the curve y = x 3 at any point sa, a 3 d meets the curve again at a point where the slope is four times the slope at sa, a 3 d . 109. Tangent curve For what value of c is the curve y = c>sx + 1d tangent to the line through the points s0, 3d and s5, 2d ? 110. Normal to a circle Show that the normal line at any point of the circle x 2 + y 2 = a 2 passes through the origin. In Exercises 111–116, find equations for the lines that are tangent and normal to the curve at the given point. 111. x 2 + 2y 2 = 9, s1, 2d 112. e x + y 2 = 2, s0, 1d 113. xy + 2x  5y = 2, s3, 2d 114. s y  xd2 = 2x + 4, s6, 2d 115. x + 1xy = 6, s4, 1d 116. x 3>2 + 2y 3>2 = 17, s1, 4d 117. Find the slope of the curve x 3y 3 + y 2 = x + y at the points (1, 1) and s1, 1d . 118. The graph shown suggests that the curve y = sin sx  sin xd might have horizontal tangents at the xaxis. Does it? Give reasons for your answer. y
y sin (x sin x)
1
b. differentiable at x = 0 ?
–2
0
–
x
2
Give reasons for your answers. Slopes, Tangents, and Normals 99. Tangents with specified slope Are there any points on the curve y = sx>2d + 1>s2x  4d where the slope is 3>2 ? If so, find them. 100. Tangents with specified slope Are there any points on the curve y = x  e x where the slope is 2? If so, find them. 101. Horizontal tangents Find the points on the curve y = 2x 3  3x 2  12x + 20 where the tangent is parallel to the xaxis.
–1
Analyzing Graphs Each of the figures in Exercises 119 and 120 shows two graphs, the graph of a function y = ƒsxd together with the graph of its derivative ƒ¿sxd . Which graph is which? How do you know? y
119. A
A
2
1
B
1 –1
0
a. perpendicular to the line y = 1  sx>24d . b. parallel to the line y = 22  12x . 104. Intersecting tangents Show that the tangents to the curve y = sp sin xd>x at x = p and x = p intersect at right angles.
4
2
3
102. Tangent intercepts Find the x and yintercepts of the line that is tangent to the curve y = x 3 at the point s 2, 8d . 103. Tangents perpendicular or parallel to lines Find the points on the curve y = 2x 3  3x 2  12x + 20 where the tangent is
y
120.
–1 B –2
1
x
0
1
2
x
Chapter 3 121. Use the following information to graph the function y = ƒsxd for 1 … x … 6 . i) The graph of ƒ is made of line segments joined end to end. ii) The graph starts at the point s 1, 2d . iii) The derivative of ƒ, where defined, agrees with the step function shown here. y
–1
1
2
3
4
5
6
x
u :sp>2d
–2
122. Repeat Exercise 121, supposing that the graph starts at s 1, 0d instead of s 1, 2d . Exercises 123 and 124 are about the accompanying graphs. The graphs in part (a) show the numbers of rabbits and foxes in a small arctic population. They are plotted as functions of time for 200 days. The number of rabbits increases at first, as the rabbits reproduce. But the foxes prey on rabbits and, as the number of foxes increases, the rabbit population levels off and then drops. Part (b) shows the graph of the derivative of the rabbit population, made by plotting slopes. 123. a. What is the value of the derivative of the rabbit population when the number of rabbits is largest? Smallest? b. What is the size of the rabbit population when its derivative is largest? Smallest (negative value)? 124. In what units should the slopes of the rabbit and fox population curves be measured? Number of rabbits Initial no. rabbits 1000 Initial no. foxes 40

4 tan2 u + tan u + 1 tan2 u + 5
130. lim+
1  2 cot2 u 5 cot u  7 cot u  8
131. lim
x sin x 2  2 cos x
x:0
–1
2000
lim
u :0
1
209
Trigonometric Limits Find the limits in Exercises 125–132. sin x 3x  tan 7x 125. lim 126. lim 2x x:0 2x 2  x x:0 sin ssin ud sin r 127. lim 128. lim u tan 2r r: 0 u :0 129.
y f '(x)
Practice Exercises
2
132. lim
u :0
1  cos u u2
Show how to extend the functions in Exercises 133 and 134 to be continuous at the origin. 133. g sxd =
tan stan xd tan x
134. ƒsxd =
tan stan xd sin ssin xd
Logarithmic Differentiation In Exercises 135–140, use logarithmic differentiation to find the derivative of y with respect to the appropriate variable. 135. y =
2sx2 + 1d
2cos 2x st + 1dst  1d 5 137. y = a b , st  2dst + 3d 2u2u 138. y = 2u2 + 1 139. y = ssin ud2u
136. y =
3x + 4 A 2x  4
10
t 7 2
140. y = sln xd1>sln xd
Related Rates 141. Right circular cylinder The total surface area S of a right circular cylinder is related to the base radius r and height h by the equation S = 2pr 2 + 2prh .
(20, 1700) 1000
a. How is dS>dt related to dr>dt if h is constant? b. How is dS>dt related to dh>dt if r is constant?
Number of foxes 0
50
100 Time (days) (a)
150
200
c. How is dS>dt related to dr>dt and dh>dt if neither r nor h is constant? d. How is dr>dt related to dh>dt if S is constant? 142. Right circular cone The lateral surface area S of a right circular cone is related to the base radius r and height h by the equation S = pr2r 2 + h 2 .
100
a. How is dS>dt related to dr>dt if h is constant? 50
b. How is dS>dt related to dh>dt if r is constant? (20, 40)
c. How is dS>dt related to dr>dt and dh>dt if neither r nor h is constant?
0
143. Circle’s changing area The radius of a circle is changing at the rate of 2>p m>sec. At what rate is the circle’s area changing when r = 10 m ?
–50 –100
0
50
100 150 Time (days) Derivative of the rabbit population (b)
200
144. Cube’s changing edges The volume of a cube is increasing at the rate of 1200 cm3>min at the instant its edges are 20 cm long. At what rate are the lengths of the edges changing at that instant?
210
Chapter 3: Differentiation
145. Resistors connected in parallel If two resistors of R1 and R2 ohms are connected in parallel in an electric circuit to make an Rohm resistor, the value of R can be found from the equation 1 1 1 + . = R R1 R2 1.2'
R1
R2 R
If R1 is decreasing at the rate of 1 ohm> sec and R2 is increasing at the rate of 0.5 ohm> sec, at what rate is R changing when R1 = 75 ohms and R2 = 50 ohms ?
151. Moving searchlight beam The figure shows a boat 1 km offshore, sweeping the shore with a searchlight. The light turns at a constant rate, du>dt = 0.6 rad/sec.
146. Impedance in a series circuit The impedance Z (ohms) in a series circuit is related to the resistance R (ohms) and reactance X (ohms) by the equation Z = 2R 2 + X 2 . If R is increasing at 3 ohms> sec and X is decreasing at 2 ohms> sec, at what rate is Z changing when R = 10 ohms and X = 20 ohms ?
a. How fast is the light moving along the shore when it reaches point A?
b. How many revolutions per minute is 0.6 rad> sec?
147. Speed of moving particle The coordinates of a particle moving in the metric xyplane are differentiable functions of time t with dx>dt = 10 m>sec and dy>dt = 5 m>sec . How fast is the particle moving away from the origin as it passes through the point s3, 4d ? 148. Motion of a particle A particle moves along the curve y = x 3>2 in the first quadrant in such a way that its distance from the origin increases at the rate of 11 units per second. Find dx>dt when x = 3 . 149. Draining a tank Water drains from the conical tank shown in the accompanying figure at the rate of 5 ft3>min . a. What is the relation between the variables h and r in the figure? b. How fast is the water level dropping when h = 6 ft ?
4'
x
A 1 km
152. Points moving on coordinate axes Points A and B move along the x and yaxes, respectively, in such a way that the distance r (meters) along the perpendicular from the origin to the line AB remains constant. How fast is OA changing, and is it increasing, or decreasing, when OB = 2r and B is moving toward O at the rate of 0.3r m> sec? Linearization 153. Find the linearizations of a. tan x at x = p>4
r
b. sec x at x = p>4 .
Graph the curves and linearizations together. 10'
154. We can obtain a useful linear approximation of the function ƒsxd = 1>s1 + tan xd at x = 0 by combining the approximations
h Exit rate: 5 ft 3/min
150. Rotating spool As television cable is pulled from a large spool to be strung from the telephone poles along a street, it unwinds from the spool in layers of constant radius (see accompanying figure). If the truck pulling the cable moves at a steady 6 ft> sec (a touch over 4 mph), use the equation s = r u to find how fast (radians per second) the spool is turning when the layer of radius 1.2 ft is being unwound.
1 L 1  x 1 + x
and
tan x L x
to get 1 L 1  x. 1 + tan x Show that this result is the standard linear approximation of 1>s1 + tan xd at x = 0 . 155. Find the linearization of ƒsxd = 21 + x + sin x  0.5 at x = 0 . 156. Find the linearization of ƒsxd = 2>s1  xd + 21 + x  3.1 at x = 0 .
Chapter 3 Differential Estimates of Change 157. Surface area of a cone Write a formula that estimates the change that occurs in the lateral surface area of a right circular cone when the height changes from h0 to h0 + dh and the radius does not change.
Additional and Advanced Exercises
211
159. Compounding error The circumference of the equator of a sphere is measured as 10 cm with a possible error of 0.4 cm. This measurement is used to calculate the radius. The radius is then used to calculate the surface area and volume of the sphere. Estimate the percentage errors in the calculated values of a. the radius. b. the surface area. c. the volume.
h
160. Finding height To find the height of a lamppost (see accompanying figure), you stand a 6 ft pole 20 ft from the lamp and measure the length a of its shadow, finding it to be 15 ft, give or take an inch. Calculate the height of the lamppost using the value a = 15 and estimate the possible error in the result.
r V 5 1 pr 2h 3 S 5 pr r 2 1 h 2 (Lateral surface area)
158. Controlling error a. How accurately should you measure the edge of a cube to be reasonably sure of calculating the cube’s surface area with an error of no more than 2%?
h
b. Suppose that the edge is measured with the accuracy required in part (a). About how accurately can the cube’s volume be calculated from the edge measurement? To find out, estimate the percentage error in the volume calculation that might result from using the edge measurement.
Chapter 3
6 ft 20 ft a
Additional and Advanced Exercises
1. An equation like sin2 u + cos2 u = 1 is called an identity because it holds for all values of u . An equation like sin u = 0.5 is not an identity because it holds only for selected values of u , not all. If you differentiate both sides of a trigonometric identity in u with respect to u , the resulting new equation will also be an identity. Differentiate the following to show that the resulting equations hold for all u . a. sin 2u = 2 sin u cos u
satisfy the conditions ƒs0d = g s0d
3. a. Find values for the constants a, b, and c that will make ƒsxd = cos x
and
g sxd = a + bx + cx 2
satisfy the conditions ƒs0d = g s0d,
ƒ¿s0d = g¿s0d,
and
ƒ–s0d = g–s0d .
b. Find values for b and c that will make ƒsxd = sin sx + ad
and
g sxd = b sin x + c cos x
ƒ¿s0d = g¿s0d .
c. For the determined values of a, b, and c, what happens for the third and fourth derivatives of ƒ and g in each of parts (a) and (b)? 4. Solutions to differential equations a. Show that y = sin x, y = cos x , and y = a cos x + b sin x (a and b constants) all satisfy the equation
b. cos 2u = cos2 u  sin2 u 2. If the identity sin sx + ad = sin x cos a + cos x sin a is differentiated with respect to x, is the resulting equation also an identity? Does this principle apply to the equation x 2  2x  8 = 0 ? Explain.
and
y– + y = 0 . b. How would you modify the functions in part (a) to satisfy the equation y– + 4y = 0 ? Generalize this result. 5. An osculating circle Find the values of h, k, and a that make the circle sx  hd2 + s y  kd2 = a 2 tangent to the parabola y = x 2 + 1 at the point (1, 2) and that also make the second derivatives d 2y>dx 2 have the same value on both curves there. Circles like this one that are tangent to a curve and have the same second derivative as the curve at the point of tangency are called osculating circles (from the Latin osculari, meaning “to kiss”). We encounter them again in Chapter 12.
212
Chapter 3: Differentiation
6. Marginal revenue A bus will hold 60 people. The number x of people per trip who use the bus is related to the fare charged ( p dollars) by the law p = [3  sx>40d]2 . Write an expression for the total revenue r(x) per trip received by the bus company. What number of people per trip will make the marginal revenue dr>dx equal to zero? What is the corresponding fare? (This fare is the one that maximizes the revenue, so the bus company should probably rethink its fare policy.)
c. Find the particle’s velocity and acceleration at the points in part (b). d. When does the particle first reach the origin? What are its velocity, speed, and acceleration then? 11. Shooting a paper clip On Earth, you can easily shoot a paper clip 64 ft straight up into the air with a rubber band. In t sec after firing, the paper clip is s = 64t  16t 2 ft above your hand. a. How long does it take the paper clip to reach its maximum height? With what velocity does it leave your hand?
7. Industrial production a. Economists often use the expression “rate of growth” in relative rather than absolute terms. For example, let u = ƒstd be the number of people in the labor force at time t in a given industry. (We treat this function as though it were differentiable even though it is an integervalued step function.) Let y = g std be the average production per person in the labor force at time t. The total production is then y = uy . If the labor force is growing at the rate of 4% per year sdu>dt = 0.04ud and the production per worker is growing at the rate of 5% per year sdy>dt = 0.05yd , find the rate of growth of the total production, y. b. Suppose that the labor force in part (a) is decreasing at the rate of 2% per year while the production per person is increasing at the rate of 3% per year. Is the total production increasing, or is it decreasing, and at what rate? 8. Designing a gondola The designer of a 30ftdiameter spherical hot air balloon wants to suspend the gondola 8 ft below the bottom of the balloon with cables tangent to the surface of the balloon, as shown. Two of the cables are shown running from the top edges of the gondola to their points of tangency, s 12,  9d and s12, 9d . How wide should the gondola be? y x 2 y 2 225
b. On the moon, the same acceleration will send the paper clip to a height of s = 64t  2.6t 2 ft in t sec. About how long will it take the paper clip to reach its maximum height, and how high will it go? 12. Velocities of two particles At time t sec, the positions of two particles on a coordinate line are s1 = 3t 3  12t 2 + 18t + 5 m and s2 = t 3 + 9t 2  12t m . When do the particles have the same velocities? 13. Velocity of a particle A particle of constant mass m moves along the xaxis. Its velocity y and position x satisfy the equation 1 1 msy 2  y0 2 d = k sx0 2  x 2 d , 2 2 where k, y0 , and x0 are constants. Show that whenever y Z 0 , m
dy = kx . dt
14. Average and instantaneous velocity a. Show that if the position x of a moving point is given by a quadratic function of t, x = At 2 + Bt + C , then the average velocity over any time interval [t1, t2] is equal to the instantaneous velocity at the midpoint of the time interval. b. What is the geometric significance of the result in part (a)? 15. Find all values of the constants m and b for which the function y = e
x
0 (–12, –9)
(12, –9)
Suspension cables Gondola
sin x, x 6 p mx + b, x Ú p
15 ft
is
8 ft
b. differentiable at x = p .
a. continuous at x = p . 16. Does the function
Width NOT TO SCALE
9. Pisa by parachute On August 5, 1988, Mike McCarthy of London jumped from the top of the Tower of Pisa. He then opened his parachute in what he said was a world record lowlevel parachute jump of 179 ft. Make a rough sketch to show the shape of the graph of his speed during the jump. (Source: Boston Globe, Aug. 6, 1988.) 10. Motion of a particle The position at time t Ú 0 of a particle moving along a coordinate line is s = 10 cos st + p>4d . a. What is the particle’s starting position st = 0d ? b. What are the points farthest to the left and right of the origin reached by the particle?
ƒsxd =
L
1  cos x , x Z 0 x 0,
x = 0
have a derivative at x = 0 ? Explain. 17. a. For what values of a and b will ƒsxd = e
ax, ax 2  bx + 3,
x 6 2 x Ú 2
be differentiable for all values of x? b. Discuss the geometry of the resulting graph of ƒ.
Chapter 3 18. a. For what values of a and b will g sxd = e
ax + b, ax 3 + x + 2b,
x … 1 x 7 1
a. b.
be differentiable for all values of x? b. Discuss the geometry of the resulting graph of g. 19. Odd differentiable functions Is there anything special about the derivative of an odd differentiable function of x? Give reasons for your answer. 20. Even differentiable functions Is there anything special about the derivative of an even differentiable function of x? Give reasons for your answer. 21. Suppose that the functions ƒ and g are defined throughout an open interval containing the point x0 , that ƒ is differentiable at x0 , that ƒsx0 d = 0 , and that g is continuous at x0 . Show that the product ƒg is differentiable at x0 . This process shows, for example, that although ƒ x ƒ is not differentiable at x = 0 , the product x ƒ x ƒ is differentiable at x = 0 . 22. (Continuation of Exercise 21.) Use the result of Exercise 21 to show that the following functions are differentiable at x = 0 . a. ƒ x ƒ sin x d. hsxd = e
b. x 2>3 sin x x 2 sin s1>xd, 0,
3 c. 2 x s1  cos xd
x Z 0 x = 0 x 2 sin s1>xd, 0,
dx
2
d 3suyd dx
3
=
d 2u du dy d 2y y + 2 + u 2. 2 dx dx dx dx
=
d 3u d 2u dy d 3y du d 2y y + 3 2 + u 3. + 3 3 2 dx dx dx dx dx dx
213
d suyd d nu d n  1u dy Á = + n n y + n dx dx dx n  1 dx nsn  1d Á sn  k + 1d d n  ku d ky + k! dx n  k dx k n d y + Á + u n. dx The equations in parts (a) and (b) are special cases of the equation in part (c). Derive the equation in part (c) by mathematical induction, using n
c.
m m m! m! a b + a b = + . k!sm  kd! sk + 1d!sm  k  1d! k k + 1 27. The period of a clock pendulum The period T of a clock pendulum (time for one full swing and back) is given by the formula T 2 = 4p2L>g , where T is measured in seconds, g = 32.2 ft>sec2 , and L, the length of the pendulum, is measured in feet. Find approximately a. the length of a clock pendulum whose period is T = 1 sec . b. the change dT in T if the pendulum in part (a) is lengthened 0.01 ft.
23. Is the derivative of hsxd = e
d 2suyd
Additional and Advanced Exercises
x Z 0 x = 0
continuous at x = 0 ? How about the derivative of k sxd = xhsxd ? Give reasons for your answers. 24. Suppose that a function ƒ satisfies the following conditions for all real values of x and y: i) ƒsx + yd = ƒsxd # ƒs yd .
ii) ƒsxd = 1 + xg sxd , where limx:0 g sxd = 1 . Show that the derivative ƒ¿sxd exists at every value of x and that ƒ¿sxd = ƒsxd . 25. The generalized product rule Use mathematical induction to prove that if y = u1 u2 Á un is a finite product of differentiable functions, then y is differentiable on their common domain and dy dun du2 Á du1 Á un + u1 = u un + Á + u1 u2 Á un  1 . dx dx 2 dx dx 26. Leibniz’s rule for higherorder derivatives of products Leibniz’s rule for higherorder derivatives of products of differentiable functions says that
c. the amount the clock gains or loses in a day as a result of the period’s changing by the amount dT found in part (b). 28. The melting ice cube Assume that an ice cube retains its cubical shape as it melts. If we call its edge length s, its volume is V = s 3 and its surface area is 6s 2 . We assume that V and s are differentiable functions of time t. We assume also that the cube’s volume decreases at a rate that is proportional to its surface area. (This latter assumption seems reasonable enough when we think that the melting takes place at the surface: Changing the amount of surface changes the amount of ice exposed to melt.) In mathematical terms, dV = k s6s 2 d, dt
k 7 0.
The minus sign indicates that the volume is decreasing. We assume that the proportionality factor k is constant. (It probably depends on many things, such as the relative humidity of the surrounding air, the air temperature, and the incidence or absence of sunlight, to name only a few.) Assume a particular set of conditions in which the cube lost 1> 4 of its volume during the first hour, and that the volume is V0 when t = 0 . How long will it take the ice cube to melt?
4 APPLICATIONS OF DERIVATIVES OVERVIEW In this chapter we use derivatives to find extreme values of functions, to determine and analyze the shapes of graphs, and to solve equations numerically. We also introduce the idea of recovering a function from its derivative. The key to many of these applications is the Mean Value Theorem, which paves the way to integral calculus in Chapter 5.
Extreme Values of Functions
4.1
This section shows how to locate and identify extreme (maximum or minimum) values of a function from its derivative. Once we can do this, we can solve a variety of problems in which we find the optimal (best) way to do something in a given situation (see Section 4.6). Finding maximum and minimum values is one of the most important applications of the derivative.
DEFINITIONS Let ƒ be a function with domain D. Then ƒ has an absolute maximum value on D at a point c if ƒsxd … ƒscd
and an absolute minimum value on D at c if
y 1
ƒsxd Ú ƒscd
y sin x
y cos x
– 2
0
2
for all x in D.
x
–1
FIGURE 4.1 Absolute extrema for the sine and cosine functions on [p>2, p>2] . These values can depend on the domain of a function.
214
for all x in D
Maximum and minimum values are called extreme values of the function ƒ. Absolute maxima or minima are also referred to as global maxima or minima. For example, on the closed interval [p>2, p>2] the function ƒsxd = cos x takes on an absolute maximum value of 1 (once) and an absolute minimum value of 0 (twice). On the same interval, the function gsxd = sin x takes on a maximum value of 1 and a minimum value of 1 (Figure 4.1). Functions with the same defining rule or formula can have different extrema (maximum or minimum values), depending on the domain. We see this in the following example.
4.1
Extreme Values of Functions
215
EXAMPLE 1 The absolute extrema of the following functions on their domains can be seen in Figure 4.2. Each function has the same defining equation, y = x2, but the domains vary. Notice that a function might not have a maximum or minimum if the domain is unbounded or fails to contain an endpoint.
y x2
Function rule
Domain D
Absolute extrema on D
(a) y = x 2
s  q, q d
(b) y = x 2
[0, 2]
(c) y = x 2
(0, 2]
No absolute maximum Absolute minimum of 0 at x = 0 Absolute maximum of 4 at x = 2 Absolute minimum of 0 at x = 0 Absolute maximum of 4 at x = 2 No absolute minimum
(d) y = x 2
(0, 2)
y x2
y
D (– , )
FIGURE 4.2
HISTORICAL BIOGRAPHY Daniel Bernoulli (1700–1789)
y x2
y
x
2 (b) abs max and min
y x2
y
D (0, 2]
D [0, 2]
2 (a) abs min only
No absolute extrema
x
D (0, 2)
2 (c) abs max only
y
x
2 (d) no max or min
x
Graphs for Example 1.
Some of the functions in Example 1 did not have a maximum or a minimum value. The following theorem asserts that a function which is continuous at every point of a finite closed interval [a, b] has an absolute maximum and an absolute minimum value on the interval. We look for these extreme values when we graph a function.
THEOREM 1—The Extreme Value Theorem If ƒ is continuous on a closed interval [a, b], then ƒ attains both an absolute maximum value M and an absolute minimum value m in [a, b]. That is, there are numbers x1 and x2 in [a, b] with ƒsx1 d = m, ƒsx2 d = M, and m … ƒsxd … M for every other x in [a, b].
The proof of the Extreme Value Theorem requires a detailed knowledge of the real number system (see Appendix 7) and we will not give it here. Figure 4.3 illustrates possible locations for the absolute extrema of a continuous function on a closed interval [a, b]. As we observed for the function y = cos x, it is possible that an absolute minimum (or absolute maximum) may occur at two or more different points of the interval. The requirements in Theorem 1 that the interval be closed and finite, and that the function be continuous, are key ingredients. Without them, the conclusion of the theorem
216
Chapter 4: Applications of Derivatives
(x2, M) y f (x)
y f (x)
M
M x1
a
x2
b
m
m
x
a
b
x
Maximum and minimum at endpoints
(x1, m) Maximum and minimum at interior points
y f (x) y f (x)
M
m
m a
M
x2
b
x
Maximum at interior point, minimum at endpoint
a
x1
b
x
Minimum at interior point, maximum at endpoint
FIGURE 4.3 Some possibilities for a continuous function’s maximum and minimum on a closed interval [a, b].
need not hold. Example 1 shows that an absolute extreme value may not exist if the interval fails to be both closed and finite. Figure 4.4 shows that the continuity requirement cannot be omitted.
y No largest value 1 yx 0 x1 0
1 Smallest value
Local (Relative) Extreme Values x
FIGURE 4.4 Even a single point of discontinuity can keep a function from having either a maximum or minimum value on a closed interval. The function x, 0 … x 6 1 0, x = 1 is continuous at every point of [0, 1] except x = 1 , yet its graph over [0, 1] does not have a highest point. y = e
Figure 4.5 shows a graph with five points where a function has extreme values on its domain [a, b]. The function’s absolute minimum occurs at a even though at e the function’s value is smaller than at any other point nearby. The curve rises to the left and falls to the right around c, making ƒ(c) a maximum locally. The function attains its absolute maximum at d. We now define what we mean by local extrema.
DEFINITIONS A function ƒ has a local maximum value at a point c within its domain D if ƒsxd … ƒscd for all x H D lying in some open interval containing c. A function ƒ has a local minimum value at a point c within its domain D if ƒsxd Ú ƒscd for all x H D lying in some open interval containing c.
If the domain of ƒ is the closed interval [a, b], then ƒ has a local maximum at the endpoint x = a, if ƒ(x) … ƒ(a) for all x in some halfopen interval [a, a + d), d 7 0. Likewise, ƒ has a local maximum at an interior point x = c if ƒ(x) … ƒ(c) for all x in some open interval (c  d, c + d), d 7 0, and a local maximum at the endpoint x = b if ƒ(x) … ƒ(b) for all x in some halfopen interval (b  d, b], d 7 0. The inequalities are reversed for local minimum values. In Figure 4.5, the function ƒ has local maxima at c and d and local minima at a, e, and b. Local extrema are also called relative extrema. Some functions can have infinitely many local extrema, even over a finite interval. One example is the function ƒ(x) = sin (1>x) on the interval (0, 1]. (We graphed this function in Figure 2.40.)
4.1
217
Extreme Values of Functions
Absolute maximum No greater value of f anywhere. Also a local maximum.
Local maximum No greater value of f nearby.
Local minimum No smaller value of f nearby.
y f (x)
Absolute minimum No smaller value of f anywhere. Also a local minimum.
Local minimum No smaller value of f nearby. a
c
e
d
b
x
FIGURE 4.5 How to identify types of maxima and minima for a function with domain a … x … b.
An absolute maximum is also a local maximum. Being the largest value overall, it is also the largest value in its immediate neighborhood. Hence, a list of all local maxima will automatically include the absolute maximum if there is one. Similarly, a list of all local minima will include the absolute minimum if there is one.
Finding Extrema Local maximum value
The next theorem explains why we usually need to investigate only a few values to find a function’s extrema. y f (x)
THEOREM 2—The First Derivative Theorem for Local Extreme Values If ƒ has a local maximum or minimum value at an interior point c of its domain, and if ƒ¿ is defined at c, then ƒ¿scd = 0. Secant slopes 0 (never negative)
x
Secant slopes 0 (never positive)
c
x
x
FIGURE 4.6 A curve with a local maximum value. The slope at c, simultaneously the limit of nonpositive numbers and nonnegative numbers, is zero.
Proof To prove that ƒ¿scd is zero at a local extremum, we show first that ƒ¿scd cannot be positive and second that ƒ¿scd cannot be negative. The only number that is neither positive nor negative is zero, so that is what ƒ¿scd must be. To begin, suppose that ƒ has a local maximum value at x = c (Figure 4.6) so that ƒsxd  ƒscd … 0 for all values of x near enough to c. Since c is an interior point of ƒ’s domain, ƒ¿scd is defined by the twosided limit lim
x:c
ƒsxd  ƒscd x  c .
This means that the righthand and lefthand limits both exist at x = c and equal ƒ¿scd. When we examine these limits separately, we find that ƒ¿scd = lim+
ƒsxd  ƒscd … 0. x  c
ƒ¿scd = lim
ƒsxd  ƒscd Ú 0. x  c
x:c
Because sx  cd 7 0 and ƒsxd … ƒscd
(1)
Similarly, x:c
Because sx  cd 6 0 and ƒsxd … ƒscd
(2)
Together, Equations (1) and (2) imply ƒ¿scd = 0. This proves the theorem for local maximum values. To prove it for local minimum values, we simply use ƒsxd Ú ƒscd, which reverses the inequalities in Equations (1) and (2).
218
Chapter 4: Applications of Derivatives
Theorem 2 says that a function’s first derivative is always zero at an interior point where the function has a local extreme value and the derivative is defined. Hence the only places where a function ƒ can possibly have an extreme value (local or global) are
y y x3 1
–1
0
1
x
1. 2. 3.
interior points where ƒ¿ = 0, interior points where ƒ¿ is undefined, endpoints of the domain of ƒ.
At x = c and x = e in Fig. 4.5 At x = d in Fig. 4.5 At x = a and x = b in Fig. 4.5
The following definition helps us to summarize.
–1 (a)
DEFINITION An interior point of the domain of a function ƒ where ƒ¿ is zero or undefined is a critical point of ƒ.
y 1 y x1/3 –1
0
1
x
–1 (b)
FIGURE 4.7 Critical points without extreme values. (a) y¿ = 3x 2 is 0 at x = 0 , but y = x 3 has no extremum there. (b) y¿ = s1>3dx 2>3 is undefined at x = 0 , but y = x 1>3 has no extremum there.
Thus the only domain points where a function can assume extreme values are critical points and endpoints. However, be careful not to misinterpret what is being said here. A function may have a critical point at x = c without having a local extreme value there. For instance, both of the functions y = x 3 and y = x 1>3 have critical points at the origin, but neither function has a local extreme value at the origin. Instead, each function has a point of inflection there (see Figure 4.7). We define and explore inflection points in Section 4.4. Most problems that ask for extreme values call for finding the absolute extrema of a continuous function on a closed and finite interval. Theorem 1 assures us that such values exist; Theorem 2 tells us that they are taken on only at critical points and endpoints. Often we can simply list these points and calculate the corresponding function values to find what the largest and smallest values are, and where they are located. Of course, if the interval is not closed or not finite (such as a 6 x 6 b or a 6 x 6 q), we have seen that absolute extrema need not exist. If an absolute maximum or minimum value does exist, it must occur at a critical point or at an included right or lefthand endpoint of the interval.
How to Find the Absolute Extrema of a Continuous Function ƒ on a Finite Closed Interval 1. Evaluate ƒ at all critical points and endpoints. 2. Take the largest and smallest of these values.
EXAMPLE 2
Find the absolute maximum and minimum values of ƒsxd = x 2 on
[2, 1]. The function is differentiable over its entire domain, so the only critical point is where ƒ¿sxd = 2x = 0, namely x = 0. We need to check the function’s values at x = 0 and at the endpoints x = 2 and x = 1: Solution
Critical point value: ƒs0d = 0 Endpoint values:
ƒs 2d = 4 ƒs1d = 1.
The function has an absolute maximum value of 4 at x = 2 and an absolute minimum value of 0 at x = 0.
EXAMPLE 3 Find the absolute maximum and minimum values of ƒ(x) = 10x (2  ln x) on the interval [1, e 2].
4.1
Figure 4.8 suggests that ƒ has its absolute maximum value near x = 3 and its absolute minimum value of 0 at x = e 2. Let’s verify this observation. We evaluate the function at the critical points and endpoints and take the largest and smallest of the resulting values. The first derivative is
Solution
y 30
(e, 10e)
25 20
219
Extreme Values of Functions
(1, 20)
15
1 ƒ¿(x) = 10(2  ln x)  10x a x b = 10(1  ln x).
10 5
(e 2, 0) 1
0
2
3
4
5
6
7
x
8
The only critical point in the domain [1, e 2] is the point x = e, where ln x = 1. The values of ƒ at this one critical point and at the endpoints are
FIGURE 4.8 The extreme values of ƒ(x) = 10x(2  ln x) on [1, e 2] occur at x = e and x = e 2 (Example 3).
Critical point value:
ƒ(e) = 10e
Endpoint values:
ƒ(1) = 10(2  ln 1) = 20 ƒ(e 2) = 10e 2(2  2 ln e) = 0.
We can see from this list that the function’s absolute maximum value is 10e L 27.2; it occurs at the critical interior point x = e. The absolute minimum value is 0 and occurs at the right endpoint x = e 2. Find the absolute maximum and minimum values of ƒsxd = x 2>3 on the
EXAMPLE 4 interval [2, 3].
We evaluate the function at the critical points and endpoints and take the largest and smallest of the resulting values. The first derivative
Solution y
y x 2/3, –2 ≤ x ≤ 3
ƒ¿sxd =
Absolute maximum; also a local maximum
Local maximum 2
has no zeros but is undefined at the interior point x = 0. The values of ƒ at this one critical point and at the endpoints are
1 –2
–1
0
2 1>3 2 = x 3 3 32 x
Critical point value: ƒs0d = 0 1 2 3 Absolute minimum; also a local minimum
x
Endpoint values:
3 ƒs 2d = s 2d2>3 = 2 4 3 ƒs3d = s3d2>3 = 2 9.
FIGURE 4.9 The extreme values of ƒsxd = x 2>3 on [2, 3] occur at x = 0 and x = 3 (Example 4).
3 9 L 2.08, and it We can see from this list that the function’s absolute maximum value is 2 occurs at the right endpoint x = 3. The absolute minimum value is 0, and it occurs at the interior point x = 0 where the graph has a cusp (Figure 4.9).
Exercises 4.1 Finding Extrema from Graphs In Exercises 1–6, determine from the graph whether the function has any absolute extreme values on [a, b]. Then explain how your answer is consistent with Theorem 1. 1.
2.
y
3.
y f (x) y h(x)
y f (x) 0
a
c1
c2
b
x
y
y
y h(x)
0
4.
y
0
a
c
b
x
a
c
b
x 0
a
c
b
x
220
Chapter 4: Applications of Derivatives
5.
6.
y
In Exercises 15–20, sketch the graph of each function and determine whether the function has any absolute extreme values on its domain. Explain how your answer is consistent with Theorem 1.
y
y g(x)
y g(x)
15. ƒ(x) = ƒ x ƒ , 1 6 x 6 2 6 , 1 6 x 6 1 16. y = 2 x + 2 0
a
c
b
x
a
0
c
x
b
17. g(x) = e
x, 0 … x 6 1 x  1, 1 … x … 2
In Exercises 7–10, find the absolute extreme values and where they occur. 7.
8.
y
18. h(x) =
y 2
x
1
9.
0
–2
–1
10.
y
x
2
(1, 2)
2
–3
x
2 –1
x
ƒ (x)
a b c
0 0 5
x + 1,
L cos x,
12.
14.
x
ƒ(x)
a b c
does not exist 0 2
x
ƒ (x)
a b c
0 0 5
a b c
24. ƒsxd = 4  x 2,
3 … x … 1
1 , 0.5 … x … 2 x2 1 26. Fsxd =  x , 2 … x … 1
29. g sxd = 24  x , 30. g sxd =  25  x ,
c
(b)
b (c)
c
a (d)

36. ƒstd = ƒ t  5 ƒ ,
b
c
 25 … x … 0
5p p … u … 2 6 p p ƒsud = tan u,  … u … 3 4 p 2p g sxd = csc x, … x … 3 3 p p g sxd = sec x,  … x … 3 6 ƒstd = 2  ƒ t ƒ , 1 … t … 3
31. ƒsud = sin u,
37. g(x) = xe x,
a
2 … x … 1 2
35. (a)
1 … x … 1 2
34. b
1 … x … 8
28. hsxd = 3x 2>3,
does not exist does not exist 1.7
a
4 … x … 1
25. Fsxd = 
ƒ (x)
x
2 … x … 3
1 … x … 2
33. b c
p 2
23. ƒsxd = x 2  1,
32.
a
0 … x …
Absolute Extrema on Finite Closed Intervals In Exercises 21–40, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.
3 x, 27. hsxd = 2
13.
1 … x 6 0
2 x  5, 3 22. ƒsxd =  x  4,
In Exercises 11–14, match the table with a graph. 11.
0 6 x 6 2p
21. ƒsxd =
x
2
0 … x … 4
y
5
0
20. ƒ(x) =
1 … x 6 0
2x,
19. y = 3 sin x,
1 –1
L
1 x,
4 … t … 7
1 … x … 1
38. h(x) = ln (x + 1),
0 … x … 3
1 39. ƒ(x) = x + ln x,
0.5 … x … 4
40. g(x) = e x , 2
2 … x … 1
4.1 In Exercises 41–44, find the function’s absolute maximum and minimum values and say where they are assumed. 41. ƒsxd = x 4>3,
1 … x … 8
42. ƒsxd = x 5>3,
1 … x … 8
43. g(ud = u ,
32 … u … 1
3>5
44. hsud = 3u2>3,
Extreme Values of Functions
221
Theory and Examples 79. A minimum with no derivative The function ƒsxd = ƒ x ƒ has an absolute minimum value at x = 0 even though ƒ is not differentiable at x = 0 . Is this consistent with Theorem 2? Give reasons for your answer. 80. Even functions If an even function ƒ(x) has a local maximum value at x = c , can anything be said about the value of ƒ at x = c ? Give reasons for your answer.
27 … u … 8
Finding Critical Points In Exercises 45–52, determine all critical points for each function. 45. y = x 2  6x + 7
46. ƒ(x) = 6x 2  x 3
47. ƒ(x) = x(4  x) 3 2 49. y = x 2 + x
48. g(x) = (x  1) 2(x  3) 2 x2 50. ƒ(x) = x  2
51. y = x 2  32 2x
52. g(x) = 22x  x 2
81. Odd functions If an odd function g(x) has a local minimum value at x = c , can anything be said about the value of g at x = c ? Give reasons for your answer. 82. We know how to find the extreme values of a continuous function ƒ(x) by investigating its values at critical points and endpoints. But what if there are no critical points or endpoints? What happens then? Do such functions really exist? Give reasons for your answers. 83. The function
Finding Extreme Values In Exercises 53–68, find the extreme values (absolute and local) of the function and where they occur.
V sxd = xs10  2xds16  2xd,
0 6 x 6 5,
models the volume of a box.
54. y = x 3  2x + 4
a. Find the extreme values of V.
55. y = x 3 + x 2  8x + 5
56. y = x 3(x  5) 2
b. Interpret any values found in part (a) in terms of the volume of the box.
57. y = 2x 2  1 1 59. y = 3 2 1  x2 x 61. y = 2 x + 1 63. y = e x + e x
58. y = x  4 2x
65. y = x ln x
66. y = x 2 ln x
67. y = cos1 (x 2)
68. y = sin1 (e x )
53. y = 2x 2  8x + 9
60. y = 23 + 2x  x
84. Cubic functions Consider the cubic function 2
x + 1 x 2 + 2x + 2 64. y = e x  e x 62. y =
Local Extrema and Critical Points In Exercises 69–76, find the critical points, domain endpoints, and extreme values (absolute and local) for each function. 69. y = x
sx + 2d
70. y = x
2>3
4  2x, x + 1,
2
72. y = x 2 23  x
71. y = x24  x 2 73. y = e
sx  4d
2>3
x … 1 x 7 1
x 2  2x + 4, x … 75. y = e 2 x + 6x  4, x 7 15 1 1  x2  x + , 4 2 4 y = • 76. x 3  6x 2 + 8x,
74. y = e
3  x, x 6 0 3 + 2x  x 2, x Ú 0
1 1 x … 1 x 7 1
In Exercises 77 and 78, give reasons for your answers. 77. Let ƒsxd = sx  2d2>3 . a. Does ƒ¿s2d exist? b. Show that the only local extreme value of ƒ occurs at x = 2 . c. Does the result in part (b) contradict the Extreme Value Theorem? d. Repeat parts (a) and (b) for ƒsxd = sx  ad2>3 , replacing 2 by a. 78. Let ƒsxd = ƒ x 3  9x ƒ . a. Does ƒ¿s0d exist? c. Does ƒ¿s 3d exist?
ƒsxd = ax 3 + bx 2 + cx + d . a. Show that ƒ can have 0, 1, or 2 critical points. Give examples and graphs to support your argument. b. How many local extreme values can ƒ have? 85. Maximum height of a vertically moving body body moving vertically is given by s = 
1 2 gt + y0 t + s0, 2
The height of a
g 7 0,
with s in meters and t in seconds. Find the body’s maximum height. 86. Peak alternating current Suppose that at any given time t (in seconds) the current i (in amperes) in an alternating current circuit is i = 2 cos t + 2 sin t . What is the peak current for this circuit (largest magnitude)? T Graph the functions in Exercises 87–90. Then find the extreme values of the function on the interval and say where they occur. 87. ƒsxd = ƒ x  2 ƒ + ƒ x + 3 ƒ , 88. gsxd = ƒ x  1 ƒ  ƒ x  5 ƒ , 89. hsxd = ƒ x + 2 ƒ  ƒ x  3 ƒ , 90. ksxd = ƒ x + 1 ƒ + ƒ x  3 ƒ ,
5 … x … 5 2 … x … 7 q 6 x 6 q q 6 x 6 q
COMPUTER EXPLORATIONS In Exercises 91–98, you will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps. a. Plot the function over the interval to see its general behavior there.
b. Does ƒ¿s3d exist?
b. Find the interior points where ƒ¿ = 0 . (In some exercises, you may have to use the numerical equation solver to approximate a solution.) You may want to plot ƒ¿ as well.
d. Determine all extrema of ƒ.
c. Find the interior points where ƒ¿ does not exist.
222
Chapter 4: Applications of Derivatives
d. Evaluate the function at all points found in parts (b) and (c) and at the endpoints of the interval. e. Find the function’s absolute extreme values on the interval and identify where they occur. 91. ƒsxd = x 4  8x 2 + 4x + 2,
[20>25, 64>25]
92. ƒsxd = x + 4x  4x + 1, 4
3
93. ƒsxd = x 2>3s3  xd,
[1, 15]
We know that constant functions have zero derivatives, but could there be a more complicated function whose derivative is always zero? If two functions have identical derivatives over an interval, how are the functions related? We answer these and other questions in this chapter by applying the Mean Value Theorem. First we introduce a special case, known as Rolle’s Theorem, which is used to prove the Mean Value Theorem. f '(c) 0
Rolle’s Theorem y f (x)
a
c
As suggested by its graph, if a differentiable function crosses a horizontal line at two different points, there is at least one point between them where the tangent to the graph is horizontal and the derivative is zero (Figure 4.10). We now state and prove this result.
x
b
(a)
THEOREM 3—Rolle’s Theorem Suppose that y = ƒsxd is continuous at every point of the closed interval [a, b] and differentiable at every point of its interior (a, b). If ƒsad = ƒsbd, then there is at least one number c in (a, b) at which ƒ¿scd = 0.
y f '(c1 ) 0
a
[0, 2p] 1 96. ƒsxd = x  sin x + , [0, 2p] 2 97. ƒ(x) = px 2e  3x>2, [0, 5] 3>4
[2, 2]
y
0
95. ƒsxd = 2x + cos x,
98. ƒ(x) = ln (2x + x sin x),
[3>4, 3]
[1, 10>3]
The Mean Value Theorem
4.2
0
94. ƒsxd = 2 + 2x  3x 2>3,
c1
f '(c3 ) 0 y f (x)
f '(c2 ) 0
c2
c3
b
x
(b)
FIGURE 4.10 Rolle’s Theorem says that a differentiable curve has at least one horizontal tangent between any two points where it crosses a horizontal line. It may have just one (a), or it may have more (b).
HISTORICAL BIOGRAPHY Michel Rolle (1652–1719)
Proof Being continuous, ƒ assumes absolute maximum and minimum values on [a, b] by Theorem 1. These can occur only 1. 2. 3.
at interior points where ƒ¿ is zero, at interior points where ƒ¿ does not exist, at the endpoints of the function’s domain, in this case a and b.
By hypothesis, ƒ has a derivative at every interior point. That rules out possibility (2), leaving us with interior points where ƒ¿ = 0 and with the two endpoints a and b. If either the maximum or the minimum occurs at a point c between a and b, then ƒ¿scd = 0 by Theorem 2 in Section 4.1, and we have found a point for Rolle’s Theorem. If both the absolute maximum and the absolute minimum occur at the endpoints, then because ƒsad = ƒsbd it must be the case that ƒ is a constant function with ƒsxd = ƒsad = ƒsbd for every x H [a, b]. Therefore ƒ¿sxd = 0 and the point c can be taken anywhere in the interior (a, b). The hypotheses of Theorem 3 are essential. If they fail at even one point, the graph may not have a horizontal tangent (Figure 4.11). Rolle’s Theorem may be combined with the Intermediate Value Theorem to show when there is only one real solution of an equation ƒsxd = 0, as we illustrate in the next example.
EXAMPLE 1
Show that the equation x 3 + 3x + 1 = 0
has exactly one real solution.
4.2
y
y
y
y f (x)
a
y f (x)
b
x
a
(a) Discontinuous at an endpoint of [a, b]
y
223
The Mean Value Theorem
x0 b
y f(x)
x
(b) Discontinuous at an interior point of [a, b]
(1, 5)
a
x0
x
b
(c) Continuous on [a, b] but not differentiable at an interior point
FIGURE 4.11 There may be no horizontal tangent if the hypotheses of Rolle’s Theorem do not hold.
–1
Solution
y x 3 3x 1
1 0
1
We define the continuous function ƒsxd = x 3 + 3x + 1.
x
Since ƒ(1) = 3 and ƒ(0) = 1, the Intermediate Value Theorem tells us that the graph of ƒ crosses the xaxis somewhere in the open interval (1, 0). (See Figure 4.12.) The derivative
(–1, –3)
FIGURE 4.12 The only real zero of the polynomial y = x 3 + 3x + 1 is the one shown here where the curve crosses the xaxis between 1 and 0 (Example 1).
ƒ¿sxd = 3x 2 + 3 is never zero (because it is always positive). Now, if there were even two points x = a and x = b where ƒ(x) was zero, Rolle’s Theorem would guarantee the existence of a point x = c in between them where ƒ¿ was zero. Therefore, ƒ has no more than one zero. Our main use of Rolle’s Theorem is in proving the Mean Value Theorem.
The Mean Value Theorem Tangent parallel to chord
y Slope f'(c)
B Slope
f (b) f (a) ba
A 0
a y f(x)
The Mean Value Theorem, which was first stated by JosephLouis Lagrange, is a slanted version of Rolle’s Theorem (Figure 4.13). The Mean Value Theorem guarantees that there is a point where the tangent line is parallel to the chord AB.
c
b
x
FIGURE 4.13 Geometrically, the Mean Value Theorem says that somewhere between a and b the curve has at least one tangent parallel to chord AB.
THEOREM 4—The Mean Value Theorem Suppose y = ƒsxd is continuous on a closed interval [a, b] and differentiable on the interval’s interior (a, b). Then there is at least one point c in (a, b) at which ƒsbd  ƒsad = ƒ¿scd. b  a
(1)
Proof We picture the graph of ƒ and draw a line through the points A(a, ƒ(a)) and B(b, ƒ(b)). (See Figure 4.14.) The line is the graph of the function gsxd = ƒsad +
ƒsbd  ƒsad sx  ad b  a
(2)
(pointslope equation). The vertical difference between the graphs of ƒ and g at x is HISTORICAL BIOGRAPHY JosephLouis Lagrange (1736–1813)
hsxd = ƒsxd  gsxd = ƒsxd  ƒsad 
ƒsbd  ƒsad sx  ad. b  a
Figure 4.15 shows the graphs of ƒ, g, and h together.
(3)
224
Chapter 4: Applications of Derivatives
B(b, f (b))
y f (x)
y f (x)
B h(x)
A(a, f (a)) y g(x)
A
h(x) f (x) g(x)
a
y y 1 x 2, –1 x 1 1
–1
0
FIGURE 4.14 The graph of ƒ and the chord AB over the interval [a, b].
x
b
x
FIGURE 4.15 The chord AB is the graph of the function g(x). The function hsxd = ƒsxd  g sxd gives the vertical distance between the graphs of ƒ and g at x.
The function h satisfies the hypotheses of Rolle’s Theorem on [a, b]. It is continuous on [a, b] and differentiable on (a, b) because both ƒ and g are. Also, hsad = hsbd = 0 because the graphs of ƒ and g both pass through A and B. Therefore h¿scd = 0 at some point c H sa, bd. This is the point we want for Equation (1). To verify Equation (1), we differentiate both sides of Equation (3) with respect to x and then set x = c: h¿sxd = ƒ¿sxd 
ƒsbd  ƒsad b  a
Derivative of Eq. (3) . . .
h¿scd = ƒ¿scd 
ƒsbd  ƒsad b  a
. . . with x = c
0 = ƒ¿scd 
ƒsbd  ƒsad b  a
h¿scd = 0
y B(2, 4)
3 y x2 2
ƒ¿scd = 1
a
x
1
FIGURE 4.16 The function ƒsxd = 21  x 2 satisfies the hypotheses (and conclusion) of the Mean Value Theorem on [1, 1] even though ƒ is not differentiable at 1 and 1.
4
x
b
ƒsbd  ƒsad , b  a
Rearranged
(1, 1)
which is what we set out to prove. 1
A(0, 0)
x
2
FIGURE 4.17 As we find in Example 2, c = 1 is where the tangent is parallel to the chord. s
Distance (ft)
400
s f (t)
A Physical Interpretation
240
80 0
The function ƒsxd = x 2 (Figure 4.17) is continuous for 0 … x … 2 and differentiable for 0 6 x 6 2. Since ƒs0d = 0 and ƒs2d = 4, the Mean Value Theorem says that at some point c in the interval, the derivative ƒ¿sxd = 2x must have the value s4  0d>s2  0d = 2. In this case we can identify c by solving the equation 2c = 2 to get c = 1. However, it is not always easy to find c algebraically, even though we know it always exists.
EXAMPLE 2
(8, 352)
320
160
The hypotheses of the Mean Value Theorem do not require ƒ to be differentiable at either a or b. Continuity at a and b is enough (Figure 4.16).
At this point, the car’s speed was 30 mph. t
5 Time (sec)
FIGURE 4.18 Distance versus elapsed time for the car in Example 3.
We can think of the number sƒsbd  ƒsadd>sb  ad as the average change in ƒ over [a, b] and ƒ¿scd as an instantaneous change. Then the Mean Value Theorem says that at some interior point the instantaneous change must equal the average change over the entire interval.
EXAMPLE 3 If a car accelerating from zero takes 8 sec to go 352 ft, its average velocity for the 8sec interval is 352>8 = 44 ft>sec. The Mean Value Theorem says that at some point during the acceleration the speedometer must read exactly 30 mph (44 ft>sec) (Figure 4.18).
4.2
The Mean Value Theorem
225
Mathematical Consequences At the beginning of the section, we asked what kind of function has a zero derivative over an interval. The first corollary of the Mean Value Theorem provides the answer that only constant functions have zero derivatives.
COROLLARY 1 If ƒ¿sxd = 0 at each point x of an open interval (a, b), then ƒsxd = C for all x H sa, bd, where C is a constant.
Proof We want to show that ƒ has a constant value on the interval (a, b). We do so by showing that if x1 and x2 are any two points in (a, b) with x1 6 x2, then ƒsx1 d = ƒsx2 d. Now ƒ satisfies the hypotheses of the Mean Value Theorem on [x1 , x2]: It is differentiable at every point of [x1, x2] and hence continuous at every point as well. Therefore, ƒsx2 d  ƒsx1 d = ƒ¿scd x2  x1 at some point c between x1 and x2. Since ƒ¿ = 0 throughout (a, b), this equation implies successively that ƒsx2 d  ƒsx1 d = 0, x2  x1
ƒsx2 d  ƒsx1 d = 0,
and
ƒsx1 d = ƒsx2 d.
At the beginning of this section, we also asked about the relationship between two functions that have identical derivatives over an interval. The next corollary tells us that their values on the interval have a constant difference.
y
y 5 x2 1 C
COROLLARY 2 If ƒ¿sxd = g¿sxd at each point x in an open interval (a, b), then there exists a constant C such that ƒsxd = gsxd + C for all x H sa, bd. That is, ƒ  g is a constant function on (a, b).
C52 C51
Proof At each point x H sa, bd the derivative of the difference function h = ƒ  g is
C50
2
C 5 –1
h¿sxd = ƒ¿sxd  g¿sxd = 0.
C 5 –2
Thus, hsxd = C on (a, b) by Corollary 1. That is, ƒsxd  gsxd = C on (a, b), so ƒsxd = gsxd + C.
1 0
x
–1 –2
FIGURE 4.19 From a geometric point of view, Corollary 2 of the Mean Value Theorem says that the graphs of functions with identical derivatives on an interval can differ only by a vertical shift there. The graphs of the functions with derivative 2x are the parabolas y = x 2 + C , shown here for selected values of C.
Corollaries 1 and 2 are also true if the open interval (a, b) fails to be finite. That is, they remain true if the interval is sa, q d, s  q , bd, or s  q , q d. Corollary 2 plays an important role when we discuss antiderivatives in Section 4.8. It tells us, for instance, that since the derivative of ƒsxd = x 2 on s  q , q d is 2x, any other function with derivative 2x on s  q , q d must have the formula x 2 + C for some value of C (Figure 4.19).
EXAMPLE 4 Find the function ƒ(x) whose derivative is sin x and whose graph passes through the point (0, 2). Since the derivative of gsxd = cos x is g¿(x) = sin x, we see that ƒ and g have the same derivative. Corollary 2 then says that ƒsxd = cos x + C for some
Solution
226
Chapter 4: Applications of Derivatives
constant C. Since the graph of ƒ passes through the point (0, 2), the value of C is determined from the condition that ƒs0d = 2: ƒs0d = cos s0d + C = 2,
so
C = 3.
The function is ƒsxd = cos x + 3.
Finding Velocity and Position from Acceleration We can use Corollary 2 to find the velocity and position functions of an object moving along a vertical line. Assume the object or body is falling freely from rest with acceleration 9.8 m>sec2. We assume the position s(t) of the body is measured positive downward from the rest position (so the vertical coordinate line points downward, in the direction of the motion, with the rest position at 0). We know that the velocity y(t) is some function whose derivative is 9.8. We also know that the derivative of gstd = 9.8t is 9.8. By Corollary 2, ystd = 9.8t + C for some constant C. Since the body falls from rest, ys0d = 0. Thus 9.8s0d + C = 0,
and
C = 0.
The velocity function must be ystd = 9.8t. What about the position function s(t)? We know that s(t) is some function whose derivative is 9.8t. We also know that the derivative of ƒstd = 4.9t 2 is 9.8t. By Corollary 2, sstd = 4.9t 2 + C for some constant C. Since ss0d = 0, 4.9s0d2 + C = 0,
and
C = 0.
The position function is sstd = 4.9t 2 until the body hits the ground. The ability to find functions from their rates of change is one of the very powerful tools of calculus. As we will see, it lies at the heart of the mathematical developments in Chapter 5.
Proofs of the Laws of Logarithms The algebraic properties of logarithms were stated in Section 1.6. We can prove those properties by applying Corollary 2 of the Mean Value Theorem to each of them. The steps in the proofs are similar to those used in solving problems involving logarithms. Proof that ln bx ln b ln x have the same derivative:
The argument starts by observing that ln bx and ln x
d b d 1 ln (bx) = = x = ln x. dx bx dx According to Corollary 2 of the Mean Value Theorem, then, the functions must differ by a constant, which means that ln bx = ln x + C for some C. Since this last equation holds for all positive values of x, it must hold for x = 1. Hence, ln (b # 1) = ln 1 + C ln b = 0 + C C = ln b.
ln 1 = 0
4.2
The Mean Value Theorem
227
By substituting, we conclude ln bx = ln b + ln x. Proof that ln x r r ln x values of x,
We use the samederivative argument again. For all positive d 1 d r ln x r = r (x ) dx x dx 1 = r rx r  1 x d 1 = r# x = (r ln x). dx
Chain Rule Derivative Power Rule
Since ln x r and r ln x have the same derivative, ln x r = r ln x + C for some constant C. Taking x to be 1 identifies C as zero, and we’re done. You are asked to prove the Quotient Rule for logarithms, b ln a x b = ln b  ln x, in Exercise 75. The Reciprocal Rule, ln (1>x) = ln x, is a special case of the Quotient Rule, obtained by taking b = 1 and noting that ln 1 = 0.
Laws of Exponents The laws of exponents for the natural exponential e x are consequences of the algebraic properties of ln x. They follow from the inverse relationship between these functions.
Laws of Exponents for e x For all numbers x, x1, and x2, the natural exponential e x obeys the following laws: 1 1. e x1 # e x2 = e x1 + x2 2. e x = x e e x1 3. x2 = e x1  x2 4. (e x1) x2 = e x1x2 = (e x2) x1 e
Proof of Law 1 Let y1 = e x1
and
x1 = ln y1
and
y2 = e x2.
(4)
Then x2 = ln y2
Take logs of both sides of Eqs. (4).
x1 + x2 = ln y1 + ln y2 e
x1 + x2
= ln y1 y2
Product Rule for logarithms
= e = y1 y2 = e x1e x2.
Exponentiate. e ln u = u
ln y1 y2
The proof of Law 4 is similar. Laws 2 and 3 follow from Law 1 (Exercises 77 and 78).
228
Chapter 4: Applications of Derivatives
Exercises 4.2 Checking the Mean Value Theorem Find the value or values of c that satisfy the equation ƒsbd  ƒsad = ƒ¿scd b  a in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises 1–8. 1. ƒsxd = x 2 + 2x  1, 2. ƒsxd = x 2>3,
1 c , 2d 2
4. ƒsxd = 2x  1, 5. ƒsxd = sin1 x, 7. ƒsxd = x  x , 2
x 3, 8. g(x) = e 2 x ,
[1, 2]
20. Show that a cubic polynomial can have at most three real zeros.
2 … x … 0 0 6 x … 2
Show that the functions in Exercises 21–28 have exactly one zero in the given interval.
Which of the functions in Exercises 9–14 satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers. ,
[1, 8]
10. ƒsxd = x
4>5
,
[0, 1]
11. ƒsxd = 2xs1  xd, 12. ƒsxd =
L
sin x x , 0,
iv) y = x 3  33x 2 + 216x = xsx  9dsx  24d
19. Show that if ƒ– 7 0 throughout an interval [a, b], then ƒ¿ has at most one zero in [a, b]. What if ƒ– 6 0 throughout [a, b] instead?
[2, 4]
9. ƒsxd = x
iii) y = x 3  3x 2 + 4 = sx + 1dsx  2d2
18. Suppose that ƒ– is continuous on [a, b] and that ƒ has three zeros in the interval. Show that ƒ– has at least one zero in (a, b). Generalize this result.
[1, 1]
2>3
ii) y = x 2 + 8x + 15
nx n  1 + sn  1dan  1x n  2 + Á + a1.
[1, 3]
6. ƒsxd = ln (x  1),
i) y = x 2  4
b. Use Rolle’s Theorem to prove that between every two zeros of x n + an  1x n  1 + Á + a1 x + a0 there lies a zero of
[0, 1]
1 3. ƒsxd = x + x ,
3
[0, 1]
Roots (Zeros) 17. a. Plot the zeros of each polynomial on a line together with the zeros of its first derivative.
21. ƒsxd = x 4 + 3x + 1,
[2, 1]
4 + 7, x2
s  q , 0d
22. ƒsxd = x 3 +
s0, q d
23. g std = 2t + 21 + t  4, 24. g std =
[0, 1]
u 25. r sud = u + sin2 a b  8, 3
p … x 6 0 x = 0
13. ƒ(x) = e
x  x, 2x 2  3x  3,
14. ƒ(x) = e
2x  3, 6x  x 2  7,
15. The function x, 0 … x 6 1 0, x = 1
is zero at x = 0 and x = 1 and differentiable on (0, 1), but its derivative on (0, 1) is never zero. How can this be? Doesn’t Rolle’s Theorem say the derivative has to be zero somewhere in (0, 1)? Give reasons for your answer. 16. For what values of a, m, and b does the function 3, ƒsxd = • x 2 + 3x + a, mx + b,
1 + 5, u3
28. r sud = tan u  cot u  u,
0 … x … 2 2 6 x … 3
ƒsxd = e
27. r sud = sec u 
x = 0 0 6 x 6 1 1 … x … 2
satisfy the hypotheses of the Mean Value Theorem on the interval [0, 2]?
s 1, 1d
s  q, q d s  q, q d
26. r sud = 2u  cos2 u + 22,
2 … x …  1 1 6 x … 0
2
1 + 21 + t  3.1, 1  t
s0, p>2d s0, p>2d
Finding Functions from Derivatives 29. Suppose that ƒs 1d = 3 and that ƒ¿sxd = 0 for all x. Must ƒsxd = 3 for all x? Give reasons for your answer. 30. Suppose that ƒs0d = 5 and that ƒ¿sxd = 2 for all x. Must ƒsxd = 2x + 5 for all x? Give reasons for your answer. 31. Suppose that ƒ¿sxd = 2x for all x. Find ƒ(2) if a. ƒs0d = 0
b. ƒs1d = 0
c. ƒs 2d = 3.
32. What can be said about functions whose derivatives are constant? Give reasons for your answer. In Exercises 33–38, find all possible functions with the given derivative. 33. a. y¿ = x
b. y¿ = x 2
c. y¿ = x 3
34. a. y¿ = 2x
b. y¿ = 2x  1
c. y¿ = 3x 2 + 2x  1
35. a. y¿ = 
1 x2
b. y¿ = 1 
1 x2
c. y¿ = 5 +
1 x2
4.2
36. a. y¿ =
1
b. y¿ =
2 2x
1 2x t 2
37. a. y¿ = sin 2t
b. y¿ = cos
38. a. y¿ = sec2 u
b. y¿ = 2u
c. y¿ = 4x 
1 2x
c. y¿ = sin 2t + cos
t 2
c. y¿ = 2u  sec2 u
In Exercises 39–42, find the function with the given derivative whose graph passes through the point P. 39. ƒ¿sxd = 2x  1, 40. g¿(x) =
Ps0, 0d
1 + 2x, x2
41. ƒ¿(x) = e ,
Ps0, 0d
43. y = 9.8t + 5,
ss0d = 10
44. y = 32t  2,
ss0.5d = 4 ss0d = 0 s(p2) = 1
48. a = 9.8,
y(0) = 20, ys0d = 3,
s(0) = 5 ss0d = 0
49. a = 4 sin 2t,
ys0d = 2,
ss0d = 3
9 3t cos p , p2
ys0d = 0,
ss0d = 1
50. a =
What does the graph do? Why does the function behave this way? Give reasons for your answers. 60. Rolle’s Theorem a. Construct a polynomial ƒ(x) that has zeros at x = 2, 1, 0, 1, and 2 . b. Graph ƒ and its derivative ƒ¿ together. How is what you see related to Rolle’s Theorem? c. Do gsxd = sin x and its derivative g¿ illustrate the same phenomenon as ƒ and ƒ¿?
Exercises 47–50 give the acceleration a = d 2s>dt 2 , initial velocity, and initial position of a body moving on a coordinate line. Find the body’s position at time t. 47. a = e t,
58. The arithmetic mean of a and b The arithmetic mean of two numbers a and b is the number sa + bd>2 . Show that the value of c in the conclusion of the Mean Value Theorem for ƒsxd = x 2 on any interval [a, b] is c = sa + bd>2 .
ƒsxd = sin x sin sx + 2d  sin2 sx + 1d.
Finding Position from Velocity or Acceleration Exercises 43–46 give the velocity y = ds>dt and initial position of a body moving along a coordinate line. Find the body’s position at time t.
2t 2 46. y = p cos p ,
Theory and Examples 57. The geometric mean of a and b The geometric mean of two positive numbers a and b is the number 2ab . Show that the value of c in the conclusion of the Mean Value Theorem for ƒsxd = 1>x on an interval of positive numbers [a, b] is c = 2ab .
T 59. Graph the function
42. r¿std = sec t tan t  1,
45. y = sin pt,
229
P( 1, 1)
3 P a0, b 2
2x
The Mean Value Theorem
Applications 51. Temperature change It took 14 sec for a mercury thermometer to rise from 19°C to 100 C when it was taken from a freezer and placed in boiling water. Show that somewhere along the way the mercury was rising at the rate of 8.5 C>sec. 52. A trucker handed in a ticket at a toll booth showing that in 2 hours she had covered 159 mi on a toll road with speed limit 65 mph. The trucker was cited for speeding. Why?
61. Unique solution Assume that ƒ is continuous on [a, b] and differentiable on (a, b). Also assume that ƒ(a) and ƒ(b) have opposite signs and that ƒ¿ Z 0 between a and b. Show that ƒsxd = 0 exactly once between a and b. 62. Parallel tangents Assume that ƒ and g are differentiable on [a, b] and that ƒsad = g sad and ƒsbd = g sbd . Show that there is at least one point between a and b where the tangents to the graphs of ƒ and g are parallel or the same line. Illustrate with a sketch. 63. Suppose that ƒ¿(x) … 1 for 1 … x … 4. Show that ƒ(4) ƒ(1) … 3. 64. Suppose that 0 6 ƒ¿(x) 6 1>2 for all xvalues. Show that ƒ(1) 6 ƒ(1) 6 2 + ƒ(1). 65. Show that ƒ cos x  1 ƒ … ƒ x ƒ for all xvalues. (Hint: Consider ƒ(t) = cos t on [0, x].) 66. Show that for any numbers a and b, the sine inequality ƒ sin b  sin a ƒ … ƒ b  a ƒ is true. 67. If the graphs of two differentiable functions ƒ(x) and g(x) start at the same point in the plane and the functions have the same rate of change at every point, do the graphs have to be identical? Give reasons for your answer.
53. Classical accounts tell us that a 170oar trireme (ancient Greek or Roman warship) once covered 184 sea miles in 24 hours. Explain why at some point during this feat the trireme’s speed exceeded 7.5 knots (sea miles per hour).
68. If ƒ ƒ(w)  ƒ(x) ƒ … ƒ w  x ƒ for all values w and x and ƒ is a differentiable function, show that 1 … ƒ¿(x) … 1 for all xvalues.
54. A marathoner ran the 26.2mi New York City Marathon in 2.2 hours. Show that at least twice the marathoner was running at exactly 11 mph, assuming the initial and final speeds are zero.
70. Let ƒ be a function defined on an interval [a, b]. What conditions could you place on ƒ to guarantee that
55. Show that at some instant during a 2hour automobile trip the car’s speedometer reading will equal the average speed for the trip. 56. Free fall on the moon On our moon, the acceleration of gravity is 1.6 m>sec2 . If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 30 sec later?
69. Assume that ƒ is differentiable on a … x … b and that ƒsbd 6 ƒsad. Show that ƒ¿ is negative at some point between a and b.
min ƒ¿ …
ƒsbd  ƒsad … max ƒ¿, b  a
where min ƒ¿ and max ƒ¿ refer to the minimum and maximum values of ƒ¿ on [a, b]? Give reasons for your answers.
230
Chapter 4: Applications of Derivatives
T 71. Use the inequalities in Exercise 70 to estimate ƒs0.1d if ƒ¿sxd = 1>s1 + x 4 cos xd for 0 … x … 0.1 and ƒs0d = 1 . T 72. Use the inequalities in Exercise 70 to estimate ƒs0.1d if ƒ¿sxd = 1>s1  x 4 d for 0 … x … 0.1 and ƒs0d = 2 . 73. Let ƒ be differentiable at every value of x and suppose that ƒs1d = 1 , that ƒ¿ 6 0 on s  q , 1d , and that ƒ¿ 7 0 on s1, q d. a. Show that ƒsxd Ú 1 for all x. b. Must ƒ¿s1d = 0 ? Explain. 74. Let ƒsxd = px 2 + qx + r be a quadratic function defined on a closed interval [a, b]. Show that there is exactly one point c in (a, b) at which ƒ satisfies the conclusion of the Mean Value Theorem.
4.3
75. Use the samederivative argument, as was done to prove the Product and Power Rules for logarithms, to prove the Quotient Rule property. 76. Use the samederivative argument to prove the identities a. tan1 x + cot1 x =
p 2
b. sec1 x + csc1 x =
p 2
77. Starting with the equation e x1e x2 = e x1 + x2, derived in the text, show that e x = 1>e x for any real number x. Then show that e x1>e x2 = e x1  x2 for any numbers x1 and x2. 78. Show that (e x1) x2 = e x1 x2 = (e x2) x1 for any numbers x1 and x2.
Monotonic Functions and the First Derivative Test In sketching the graph of a differentiable function, it is useful to know where it increases (rises from left to right) and where it decreases (falls from left to right) over an interval. This section gives a test to determine where it increases and where it decreases. We also show how to test the critical points of a function to identify whether local extreme values are present.
Increasing Functions and Decreasing Functions As another corollary to the Mean Value Theorem, we show that functions with positive derivatives are increasing functions and functions with negative derivatives are decreasing functions. A function that is increasing or decreasing on an interval is said to be monotonic on the interval.
COROLLARY 3
Suppose that ƒ is continuous on [a, b] and differentiable on
(a, b). If ƒ¿sxd 7 0 at each point x H sa, bd, then ƒ is increasing on [a, b]. If ƒ¿sxd 6 0 at each point x H sa, bd, then ƒ is decreasing on [a, b]. Proof Let x1 and x2 be any two points in [a, b] with x1 6 x2 . The Mean Value Theorem applied to ƒ on [x1, x2] says that ƒsx2 d  ƒsx1 d = ƒ¿scdsx2  x1 d for some c between x1 and x2 . The sign of the righthand side of this equation is the same as the sign of ƒ ¿scd because x2  x1 is positive. Therefore, ƒsx2 d 7 ƒsx1 d if ƒ¿ is positive on (a, b) and ƒsx2 d 6 ƒsx1 d if ƒ¿ is negative on (a, b). Corollary 3 tells us that f (x) = 1x is increasing on the interval [0, b] for any b 7 0 because f ¿(x) = 1> 1x is positive on (0, b). The derivative does not exist at x = 0, but Corollary 3 still applies. The corollary is valid for infinite as well as finite intervals, so f (x) = 1x is increasing on [0, q ). To find the intervals where a function ƒ is increasing or decreasing, we first find all of the critical points of ƒ. If a 6 b are two critical points for ƒ, and if the derivative ƒ¿ is continuous but never zero on the interval (a, b), then by the Intermediate Value Theorem applied to ƒ¿ , the derivative must be everywhere positive on (a, b), or everywhere negative there. One way we can determine the sign of ƒ¿ on (a, b) is simply by evaluating the derivative at a single point c in (a, b). If ƒ¿(c) 7 0, then ƒ¿(x) 7 0 for all x in (a, b) so ƒ is increasing on [a, b] by Corollary 3; if ƒ¿(c) 6 0, then ƒ is decreasing on [a, b]. The next example illustrates how we use this procedure.
4.3 y y x3 – 12x – 5
Monotonic Functions and the First Derivative Test
231
20
Find the critical points of ƒsxd = x 3  12x  5 and identify the intervals on which ƒ is increasing and on which ƒ is decreasing.
10
Solution
EXAMPLE 1
(–2, 11)
–4 –3 –2 –1 0
1
2
3
The function ƒ is everywhere continuous and differentiable. The first derivative ƒ¿sxd = 3x 2  12 = 3sx 2  4d = 3sx + 2dsx  2d
x
4
is zero at x = 2 and x = 2. These critical points subdivide the domain of ƒ to create nonoverlapping open intervals s  q , 2d, s 2, 2d, and s2, q d on which ƒ¿ is either positive or negative. We determine the sign of ƒ¿ by evaluating ƒ¿ at a convenient point in each subinterval. The behavior of ƒ is determined by then applying Corollary 3 to each subinterval. The results are summarized in the following table, and the graph of ƒ is given in Figure 4.20.
–10
–20 (2, –21)
FIGURE 4.20 The function ƒsxd = x 3  12x  5 is monotonic on three separate intervals (Example 1).
 q 6 x 6 2 ƒ¿s 3d = 15 + increasing
Interval ƒ œ evaluated Sign of ƒ œ Behavior of ƒ
2 6 x 6 q ƒ¿s3d = 15 + increasing
2 6 x 6 2 ƒ¿s0d = 12 decreasing
x –3
–2
–1
0
1
2
3
We used “strict” lessthan inequalities to specify the intervals in the summary table for Example 1. Corollary 3 says that we could use … inequalities as well. That is, the function ƒ in the example is increasing on  q 6 x … 2, decreasing on 2 … x … 2, and increasing on 2 … x 6 q . We do not talk about whether a function is increasing or decreasing at a single point.
HISTORICAL BIOGRAPHY
First Derivative Test for Local Extrema
Edmund Halley (1656–1742)
In Figure 4.21, at the points where ƒ has a minimum value, ƒ¿ 6 0 immediately to the left and ƒ¿ 7 0 immediately to the right. (If the point is an endpoint, there is only one side to consider.) Thus, the function is decreasing on the left of the minimum value and it is increasing on its right. Similarly, at the points where ƒ has a maximum value, ƒ¿ 7 0 immediately to the left and ƒ¿ 6 0 immediately to the right. Thus, the function is increasing on the left of the maximum value and decreasing on its right. In summary, at a local extreme point, the sign of ƒ¿sxd changes. Absolute max f ' undefined Local max f'0 No extremum f'0 f'0
y f(x)
No extremum f'0
f'0
f'0
f'0
f'0 Local min
Local min f'0
f'0 Absolute min a
c1
c2
c3
c4
c5
b
x
FIGURE 4.21 The critical points of a function locate where it is increasing and where it is decreasing. The first derivative changes sign at a critical point where a local extremum occurs.
These observations lead to a test for the presence and nature of local extreme values of differentiable functions.
232
Chapter 4: Applications of Derivatives
First Derivative Test for Local Extrema Suppose that c is a critical point of a continuous function ƒ, and that ƒ is differentiable at every point in some interval containing c except possibly at c itself. Moving across this interval from left to right, 1. if ƒ¿ changes from negative to positive at c, then ƒ has a local minimum at c; 2. if ƒ¿ changes from positive to negative at c, then ƒ has a local maximum at c; 3. if ƒ¿ does not change sign at c (that is, ƒ¿ is positive on both sides of c or negative on both sides), then ƒ has no local extremum at c.
The test for local extrema at endpoints is similar, but there is only one side to consider. Proof of the First Derivative Test Part (1). Since the sign of ƒ¿ changes from negative to positive at c, there are numbers a and b such that a 6 c 6 b, ƒ¿ 6 0 on (a, c), and ƒ¿ 7 0 on (c, b). If x H sa, cd, then ƒscd 6 ƒsxd because ƒ¿ 6 0 implies that ƒ is decreasing on [a, c]. If x H sc, bd, then ƒscd 6 ƒsxd because ƒ¿ 7 0 implies that ƒ is increasing on [c, b]. Therefore, ƒsxd Ú ƒscd for every x H sa, bd. By definition, ƒ has a local minimum at c. Parts (2) and (3) are proved similarly.
EXAMPLE 2
Find the critical points of ƒsxd = x 1>3sx  4d = x 4>3  4x 1>3.
Identify the intervals on which ƒ is increasing and decreasing. Find the function’s local and absolute extreme values. Solution The function ƒ is continuous at all x since it is the product of two continuous functions, x 1>3 and sx  4d. The first derivative
d 4>3 4 4  4x 1>3R = x 1>3  x 2>3 x 3 3 dx Q 4sx  1d 4 = x 2>3 Qx  1R = 3 3x 2>3
ƒ¿sxd =
is zero at x = 1 and undefined at x = 0. There are no endpoints in the domain, so the critical points x = 0 and x = 1 are the only places where ƒ might have an extreme value. The critical points partition the xaxis into intervals on which ƒ¿ is either positive or negative. The sign pattern of ƒ¿ reveals the behavior of ƒ between and at the critical points, as summarized in the following table. Interval Sign of ƒ œ
y 4
Behavior of ƒ y x1/3(x 4)
2 1 –1 0 –1
1
2
3
4
x
–2 –3
(1, 3)
FIGURE 4.22 The function ƒsxd = x 1>3 sx  4d decreases when x 6 1 and increases when x 7 1 (Example 2).
x 6 0 decreasing
0 6 x 6 1 decreasing
x 7 1 + increasing x
–1
0
1
2
Corollary 3 to the Mean Value Theorem tells us that ƒ decreases on s  q , 0], decreases on [0, 1], and increases on [1, q d. The First Derivative Test for Local Extrema tells us that ƒ does not have an extreme value at x = 0 (ƒ¿ does not change sign) and that ƒ has a local minimum at x = 1 (ƒ¿ changes from negative to positive). The value of the local minimum is ƒs1d = 11>3s1  4d = 3. This is also an absolute minimum since ƒ is decreasing on s  q , 1] and increasing on [1, q d. Figure 4.22 shows this value in relation to the function’s graph. Note that limx:0 ƒ¿sxd =  q , so the graph of ƒ has a vertical tangent at the origin.
4.3
233
Monotonic Functions and the First Derivative Test
EXAMPLE 3
Find the critical points of ƒ(x) = (x 2  3)e x. Identify the intervals on which ƒ is increasing and decreasing. Find the function’s local and absolute extreme values. The function ƒ is continuous and differentiable for all real numbers, so the critical points occur only at the zeros of ƒ¿.
Solution
Using the Derivative Product Rule, we find the derivative d x d 2 e + (x  3) # e x ƒ¿(x) = (x 2  3) # dx dx = (x 2  3) # e x + (2x) # e x = (x 2 + 2x  3)e x. Since e x is never zero, the first derivative is zero if and only if x 2 + 2x  3 = 0 (x + 3)(x  1) = 0. y
y (x 2 3)e x
The zeros x = 3 and x = 1 partition the xaxis into intervals as follows.
4 3
Interval Sign of ƒ œ
2 1 –5 –4 –3 –2 –1
–1
1
2
3
x
Behavior of ƒ
x 6 3 + increasing
1 6 x + increasing x
–4
–2
3 6 x 6 1 decreasing
–3
–2
–1
0
1
2
3
–3 –4
We can see from the table that there is a local maximum (about 0.299) at x = 3 and a local minimum (about 5.437) at x = 1. The local minimum value is also an absolute minimum because ƒ(x) 7 0 for ƒ x ƒ 7 23. There is no absolute maximum. The function increases on ( q , 3) and (1, q ) and decreases on (3, 1). Figure 4.23 shows the graph.
–5 –6
FIGURE 4.23 The graph of ƒ(x) = (x 2  3) e x (Example 3).
Exercises 4.3 Analyzing Functions from Derivatives Answer the following questions about the functions whose derivatives are given in Exercises 1–14: a. What are the critical points of ƒ? b. On what intervals is ƒ increasing or decreasing? c. At what points, if any, does ƒ assume local maximum and minimum values? 1. ƒ¿sxd = xsx  1d
2. ƒ¿sxd = sx  1dsx + 2d
3. ƒ¿sxd = sx  1d2sx + 2d
4. ƒ¿sxd = sx  1d2sx + 2d2
5. ƒ¿(x) = (x  1)e x 6. ƒ¿sxd = sx  7dsx + 1dsx + 5d x 2(x  1) , x Z 2 7. ƒ¿(x) = x + 2 (x  2)(x + 4) , x Z 1, 3 8. ƒ¿(x) = (x + 1)(x  3) 6 4 , 9. ƒ¿(x) = 1  2 , x Z 0 10. ƒ¿(x) = 3 x 2x
11. ƒ¿sxd = x 1>3sx + 2d
13. ƒ¿(x) = (sin x  1)(2 cos x + 1), 0 … x … 2p 14. ƒ¿(x) = (sin x + cos x)(sin x  cos x), 0 … x … 2p Identifying Extrema In Exercises 15–44: a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function’s local and absolute extreme values, if any, saying where they occur. 15.
16.
y 2 1
–3 –2 –1 –1
x Z 0
12. ƒ¿sxd = x 1>2sx  3d
–2
2
y 5 f (x) 1
2
y
1 3
x
–3 –2 –1 –1 –2
y 5 f (x) 1
2
3
x
234
Chapter 4: Applications of Derivatives
17.
18.
y
57. ƒsxd = sin 2x,
y
0 … x … p
58. ƒsxd = sin x  cos x, 2
y 5 f (x)
1 –3 –2 –1 –1
1
2
2
y 5 f (x) x
3
1
–3 –2 –1 –1
1
2
3
–2
–2
19. g std = t 2  3t + 3
20. g std = 3t 2 + 9t + 5
x
0 … x … 2p
59. ƒsxd = 23 cos x + sin x, p 60. ƒsxd = 2x + tan x, 2 x x 61. ƒsxd =  2 sin , 0 … 2 2 62. ƒsxd = 2 cos x  cos2 x, 63. ƒsxd = csc x  2 cot x, 2
0 … x … 2p p 6 x 6 2 x … 2p p … x … p
0 6 x 6 p p p 6 x 6 2 2
21. hsxd = x 3 + 2x 2
22. hsxd = 2x 3  18x
64. ƒsxd = sec x  2 tan x,
23. ƒsud = 3u2  4u3
24. ƒsud = 6u  u3
25. ƒsrd = 3r + 16r
26. hsrd = sr + 7d3
27. ƒsxd = x 4  8x 2 + 16 3 29. Hstd = t 4  t 6 2
28. g sxd = x 4  4x 3 + 4x 2
Theory and Examples Show that the functions in Exercises 65 and 66 have local extreme values at the given values of u , and say which kind of local extreme the function has.
31. ƒsxd = x  6 2x  1
32. g sxd = 4 2x  x + 3
33. g sxd = x28  x
2
34. g sxd = x 2 25  x
x Z 2
36. ƒsxd =
3
x2  3 , x  2
30. Kstd = 15t 3  t 5
37. ƒsxd = x 1>3sx + 8d 39. hsxd = x
40. k sxd = x
41. ƒ(x) = e
1>3
2x
sx  4d 2
+ e
x
42. ƒ(x) = e
2
sx  4d
2>3
2
2x
44. ƒ(x) = x ln x
43. ƒ(x) = x ln x
2
a. ƒ¿sxd 7 0 for x 6 1 and ƒ¿sxd 6 0 for x 7 1; b. ƒ¿sxd 6 0 for x 6 1 and ƒ¿sxd 7 0 for x 7 1; c. ƒ¿sxd 7 0 for x Z 1; d. ƒ¿sxd 6 0 for x Z 1. 68. Sketch the graph of a differentiable function y = ƒsxd that has
In Exercises 45–56: a. Identify the function’s local extreme values in the given domain, and say where they occur. b. Which of the extreme values, if any, are absolute? T c. Support your findings with a graphing calculator or computer grapher. 45. ƒsxd = 2x  x 2,  q 6 x … 2 q 6 x … 0
46. ƒsxd = sx + 1d , 2
1 … x 6 q
47. g sxd = x 2  4x + 4,
4 … x 6 q 3 … t 6 q q 6 t … 3
48. g sxd = x  6x  9, 2
49. ƒstd = 12t  t , 3
50. ƒstd = t 3  3t 2, 51. hsxd =
u , 0 … u … 2p, at u = 0 and u = 2p 2 u 66. hsud = 5 sin , 0 … u … p, at u = 0 and u = p 2 67. Sketch the graph of a differentiable function y = ƒsxd through the point (1, 1) if ƒ¿s1d = 0 and 65. hsud = 3 cos
2
x3 3x + 1 38. g sxd = x 2>3sx + 5d
35. ƒsxd =
2
x3  2x 2 + 4x, 3
0 … x 6 q
52. k sxd = x 3 + 3x 2 + 3x + 1, 53. ƒsxd = 225  x 2,
q 6 x … 0
5 … x … 5
54. ƒsxd = 2x 2  2x  3, 3 … x 6 q x  2 , 0 … x 6 1 55. g sxd = 2 x  1 x2 , 2 6 x … 1 56. g sxd = 4  x2 In Exercises 57–64: a. Find the local extrema of each function on the given interval, and say where they occur. T b. Graph the function and its derivative together. Comment on the behavior of ƒ in relation to the signs and values of ƒ¿.
a. a local minimum at (1, 1) and a local maximum at (3, 3); b. a local maximum at (1, 1) and a local minimum at (3, 3); c. local maxima at (1, 1) and (3, 3); d. local minima at (1, 1) and (3, 3). 69. Sketch the graph of a continuous function y = g sxd such that a. g s2d = 2, 0 6 g¿ 6 1 for x 6 2, g¿sxd : 1 as x : 2, 1 6 g¿ 6 0 for x 7 2, and g¿sxd : 1+ as x : 2+ ; b. g s2d = 2, g¿ 6 0 for x 6 2, g¿sxd :  q as x : 2, g¿ 7 0 for x 7 2, and g¿sxd : q as x : 2+ . 70. Sketch the graph of a continuous function y = hsxd such that a. hs0d = 0, 2 … hsxd … 2 for all x, h¿sxd : q as x : 0 , and h¿sxd : q as x : 0 + ; b. hs0d = 0, 2 … hsxd … 0 for all x, h¿sxd : q as x : 0 , and h¿sxd :  q as x : 0 + . 71. Discuss the extremevalue behavior of the function ƒ(x) = x sin (1>x), x Z 0. How many critical points does this function have? Where are they located on the xaxis? Does ƒ have an absolute minimum? An absolute maximum? (See Exercise 49 in Section 2.3.) 72. Find the intervals on which the function ƒsxd = ax 2 + bx + c, a Z 0 , is increasing and decreasing. Describe the reasoning behind your answer. 73. Determine the values of constants a and b so that ƒ(x) = ax 2 + bx has an absolute maximum at the point (1, 2). 74. Determine the values of constants a, b, c, and d so that ƒ(x) = ax 3 + bx 2 + cx + d has a local maximum at the point (0, 0) and a local minimum at the point (1, 1).
4.4
Concavity and Curve Sketching
235
a. ln (cos x) on [p>4, p>3],
79. Find the absolute maximum value of ƒsxd = x 2 ln s1>xd and say where it is assumed.
b. cos (ln x) on [1>2, 2].
80. a. Prove that e x Ú 1 + x if x Ú 0.
75. Locate and identify the absolute extreme values of
76. a. Prove that ƒ(x) = x  ln x is increasing for x 7 1.
b. Use the result in part (a) to show that
b. Using part (a), show that ln x 6 x if x 7 1.
ex Ú 1 + x +
77. Find the absolute maximum and minimum values of ƒsxd = e x  2x on [0, 1].
81. Show that increasing functions and decreasing functions are onetoone. That is, show that for any x1 and x2 in I, x2 Z x1 implies ƒsx2 d Z ƒsx1 d.
78. Where does the periodic function ƒsxd = 2e sin sx>2d take on its extreme values and what are these values?
Use the results of Exercise 81 to show that the functions in Exercises 82–86 have inverses over their domains. Find a formula for dƒ 1>dx using Theorem 3, Section 3.8.
y y 2e sin (x/2)
x
0
y
CA
VE
UP
y x3
CO
DO
N
f ' increases x
0
CO NC AV E
N W
82. ƒsxd = s1>3dx + s5>6d
83. ƒsxd = 27x 3
84. ƒsxd = 1  8x 3
85. ƒsxd = s1  xd3
86. ƒsxd = x 5>3
Concavity and Curve Sketching
4.4
f ' decreases
1 2 x . 2
We have seen how the first derivative tells us where a function is increasing, where it is decreasing, and whether a local maximum or local minimum occurs at a critical point. In this section we see that the second derivative gives us information about how the graph of a differentiable function bends or turns. With this knowledge about the first and second derivatives, coupled with our previous understanding of symmetry and asymptotic behavior studied in Sections 1.1 and 2.6, we can now draw an accurate graph of a function. By organizing all of these ideas into a coherent procedure, we give a method for sketching graphs and revealing visually the key features of functions. Identifying and knowing the locations of these features is of major importance in mathematics and its applications to science and engineering, especially in the graphical analysis and interpretation of data.
Concavity FIGURE 4.24 The graph of ƒsxd = x is concave down on s  q , 0d and concave up on s0, q d (Example 1a). 3
As you can see in Figure 4.24, the curve y = x 3 rises as x increases, but the portions defined on the intervals s  q , 0d and s0, q d turn in different ways. As we approach the origin from the left along the curve, the curve turns to our right and falls below its tangents. The slopes of the tangents are decreasing on the interval s  q , 0d. As we move away from the origin along the curve to the right, the curve turns to our left and rises above its tangents. The slopes of the tangents are increasing on the interval s0, q d. This turning or bending behavior defines the concavity of the curve.
DEFINITION
The graph of a differentiable function y = ƒsxd is
(a) concave up on an open interval I if ƒ¿ is increasing on I; (b) concave down on an open interval I if ƒ¿ is decreasing on I.
If y = ƒsxd has a second derivative, we can apply Corollary 3 of the Mean Value Theorem to the first derivative function. We conclude that ƒ¿ increases if ƒ– 7 0 on I, and decreases if ƒ– 6 0.
236
Chapter 4: Applications of Derivatives
The Second Derivative Test for Concavity Let y = ƒsxd be twicedifferentiable on an interval I.
y 4
1. If ƒ– 7 0 on I, the graph of ƒ over I is concave up. 2. If ƒ– 6 0 on I, the graph of ƒ over I is concave down.
y x2
3 EU
P
CAV CON
NC
P
–2
–1
CO
EU
y'' 0
If y = ƒsxd is twicedifferentiable, we will use the notations ƒ– and y– interchangeably when denoting the second derivative.
AV
2 1
y'' 0
0
1
EXAMPLE 1
x
2
(a) The curve y = x 3 (Figure 4.24) is concave down on s  q , 0d where y– = 6x 6 0
and concave up on s0, q d where y– = 6x 7 0. (b) The curve y = x 2 (Figure 4.25) is concave up on s  q , q d because its second derivative y– = 2 is always positive.
FIGURE 4.25 The graph of ƒsxd = x 2 is concave up on every interval (Example 1b).
EXAMPLE 2
Determine the concavity of y = 3 + sin x on [0, 2p].
The first derivative of y = 3 + sin x is y¿ = cos x, and the second derivative is y– = sin x. The graph of y = 3 + sin x is concave down on s0, pd, where y– = sin x is negative. It is concave up on sp, 2pd, where y– = sin x is positive (Figure 4.26). Solution
Points of Inflection The curve y = 3 + sin x in Example 2 changes concavity at the point sp, 3d. Since the first derivative y¿ = cos x exists for all x, we see that the curve has a tangent line of slope 1 at the point sp, 3d. This point is called a point of inflection of the curve. Notice from Figure 4.26 that the graph crosses its tangent line at this point and that the second derivative y– = sin x has value 0 when x = p. In general, we have the following definition.
y 4 3
y 5 3 1 sin x (p, 3)
2 1 0 –1
p
2p
x
y'' 5 – sin x
DEFINITION A point where the graph of a function has a tangent line and where the concavity changes is a point of inflection.
FIGURE 4.26 Using the sign of y– to determine the concavity of y (Example 2).
We observed that the second derivative of ƒ(x) = 3 + sin x is equal to zero at the inflection point sp, 3d. Generally, if the second derivative exists at a point of inflection (c, ƒ(c)), then ƒ–(c) = 0. This follows immediately from the Intermediate Value Theorem whenever ƒ– is continuous over an interval containing x = c because the second derivative changes sign moving across this interval. Even if the continuity assumption is dropped, it is still true that ƒ–(c) = 0, provided the second derivative exists (although a more advanced agrument is required in this noncontinuous case). Since a tangent line must exist at the point of inflection, either the first derivative ƒ¿(c) exists (is finite) or a vertical tangent exists at the point. At a vertical tangent neither the first nor second derivative exists. In summary, we conclude the following result.
At a point of inflection (c, ƒ(c)), either ƒ–(c) = 0 or ƒ–(c) fails to exist.
The next example illustrates a function having a point of inflection where the first derivative exists, but the second derivative fails to exist.
4.4
The graph of ƒ(x) = x 5>3 has a horizontal tangent at the origin because ƒ¿(x) = (5>3)x = 0 when x = 0. However, the second derivative 2>3
y x5/3
2
ƒ–(x) =
1 1 Point of inflection
0 –1
237
EXAMPLE 3
y
–1
Concavity and Curve Sketching
10 1>3 d 5 2>3 a x b = x 9 dx 3
x
fails to exist at x = 0. Nevertheless, ƒ–(x) 6 0 for x 6 0 and ƒ–(x) 7 0 for x 7 0, so the second derivative changes sign at x = 0 and there is a point of inflection at the origin. The graph is shown in Figure 4.27.
–2
FIGURE 4.27 The graph of ƒ(x) = x 5>3 has a horizontal tangent at the origin where the concavity changes, although ƒ– does not exist at x = 0 (Example 3).
Here is an example showing that an inflection point need not occur even though both derivatives exist and ƒ– = 0. The curve y = x 4 has no inflection point at x = 0 (Figure 4.28). Even though the second derivative y– = 12x 2 is zero there, it does not change sign.
EXAMPLE 4
As our final illustration, we show a situation in which a point of inflection occurs at a vertical tangent to the curve where neither the first nor the second derivative exists.
y y x4 2
The graph of y = x 1>3 has a point of inflection at the origin because the second derivative is positive for x 6 0 and negative for x 7 0:
EXAMPLE 5
1 y'' 0 –1
1
0
y– =
x
FIGURE 4.28 The graph of y = x 4 has no inflection point at the origin, even though y– = 0 there (Example 4).
However, both y¿ = x 2>3>3 and y– fail to exist at x = 0, and there is a vertical tangent there. See Figure 4.29. To study the motion of an object moving along a line as a function of time, we often are interested in knowing when the object’s acceleration, given by the second derivative, is positive or negative. The points of inflection on the graph of the object’s position function reveal where the acceleration changes sign.
y Point of inflection
d2 d 1 2>3 2 ax 1>3 b = a x b =  x 5>3 . 9 dx 3 dx 2
y 5 x1/3 0
x
EXAMPLE 6 A particle is moving along a horizontal coordinate line (positive to the right) with position function sstd = 2t 3  14t 2 + 22t  5,
t Ú 0.
Find the velocity and acceleration, and describe the motion of the particle. FIGURE 4.29 A point of inflection where y¿ and y– fail to exist (Example 5).
Solution
The velocity is ystd = s¿std = 6t 2  28t + 22 = 2st  1ds3t  11d,
and the acceleration is astd = y¿std = s–std = 12t  28 = 4s3t  7d. When the function s(t) is increasing, the particle is moving to the right; when s(t) is decreasing, the particle is moving to the left. Notice that the first derivative sy = s¿d is zero at the critical points t = 1 and t = 11>3. Interval Sign of Y s œ Behavior of s Particle motion
0 6 t 6 1 + increasing right
1 6 t 6 11>3 decreasing left
11>3 6 t + increasing right
238
Chapter 4: Applications of Derivatives
The particle is moving to the right in the time intervals [0, 1) and s11>3, q d, and moving to the left in (1, 11>3). It is momentarily stationary (at rest) at t = 1 and t = 11>3. The acceleration astd = s–std = 4s3t  7d is zero when t = 7>3. Interval Sign of a s ﬂ Graph of s
0 6 t 6 7>3 concave down
7>3 6 t + concave up
The particle starts out moving to the right while slowing down, and then reverses and begins moving to the left at t = 1 under the influence of the leftward acceleration over the time interval [0, 7>3). The acceleration then changes direction at t = 7>3 but the particle continues moving leftward, while slowing down under the rightward acceleration. At t = 11>3 the particle reverses direction again: moving to the right in the same direction as the acceleration.
Second Derivative Test for Local Extrema Instead of looking for sign changes in ƒ¿ at critical points, we can sometimes use the following test to determine the presence and nature of local extrema.
THEOREM 5—Second Derivative Test for Local Extrema
Suppose ƒ– is continuous
on an open interval that contains x = c. 1. If ƒ¿scd = 0 and ƒ–scd 6 0, then ƒ has a local maximum at x = c. 2. If ƒ¿scd = 0 and ƒ–scd 7 0, then ƒ has a local minimum at x = c. 3. If ƒ¿scd = 0 and ƒ–scd = 0, then the test fails. The function ƒ may have a local maximum, a local minimum, or neither.
f ' 5 0, f '' , 0 ⇒ local max
f ' 5 0, f '' . 0 ⇒ local min
Proof Part (1). If ƒ–scd 6 0, then ƒ–sxd 6 0 on some open interval I containing the point c, since ƒ– is continuous. Therefore, ƒ¿ is decreasing on I. Since ƒ¿scd = 0, the sign of ƒ¿ changes from positive to negative at c so ƒ has a local maximum at c by the First Derivative Test. The proof of Part (2) is similar. For Part (3), consider the three functions y = x 4, y = x 4 , and y = x 3 . For each function, the first and second derivatives are zero at x = 0. Yet the function y = x 4 has a local minimum there, y = x 4 has a local maximum, and y = x 3 is increasing in any open interval containing x = 0 (having neither a maximum nor a minimum there). Thus the test fails. This test requires us to know ƒ– only at c itself and not in an interval about c. This makes the test easy to apply. That’s the good news. The bad news is that the test is inconclusive if ƒ– = 0 or if ƒ– does not exist at x = c. When this happens, use the First Derivative Test for local extreme values. Together ƒ¿ and ƒ– tell us the shape of the function’s graph—that is, where the critical points are located and what happens at a critical point, where the function is increasing and where it is decreasing, and how the curve is turning or bending as defined by its concavity. We use this information to sketch a graph of the function that captures its key features.
EXAMPLE 7
Sketch a graph of the function ƒsxd = x 4  4x 3 + 10
using the following steps. (a) Identify where the extrema of ƒ occur. (b) Find the intervals on which ƒ is increasing and the intervals on which ƒ is decreasing.
4.4
Concavity and Curve Sketching
239
(c) Find where the graph of ƒ is concave up and where it is concave down. (d) Sketch the general shape of the graph for ƒ. (e) Plot some specific points, such as local maximum and minimum points, points of in
flection, and intercepts. Then sketch the curve. The function ƒ is continuous since ƒ¿sxd = 4x 3  12x 2 exists. The domain of ƒ is s  q , q d, and the domain of ƒ¿ is also s  q , q d. Thus, the critical points of ƒ occur only at the zeros of ƒ¿ . Since
Solution
ƒ¿sxd = 4x 3  12x 2 = 4x 2sx  3d, the first derivative is zero at x = 0 and x = 3. We use these critical points to define intervals where ƒ is increasing or decreasing. x 6 0 decreasing
Interval Sign of ƒ œ Behavior of ƒ
0 6 x 6 3 decreasing
3 6 x + increasing
(a) Using the First Derivative Test for local extrema and the table above, we see that there is no extremum at x = 0 and a local minimum at x = 3. (b) Using the table above, we see that ƒ is decreasing on s  q , 0] and [0, 3], and increasing on [3, q d. (c) ƒ–sxd = 12x 2  24x = 12xsx  2d is zero at x = 0 and x = 2. We use these points to define intervals where ƒ is concave up or concave down. x 6 0 + concave up
Interval Sign of ƒ ﬂ Behavior of ƒ
0 6 x 6 2 concave down
2 6 x + concave up
We see that ƒ is concave up on the intervals s  q , 0d and s2, q d, and concave down on (0, 2). (d) Summarizing the information in the last two tables, we obtain the following.
y y x 4 4x 3 10
x<0
0<x<2
2<x<3
3<x
decreasing concave up
decreasing concave down
decreasing concave up
increasing concave up
20 15
The general shape of the curve is shown in the accompanying figure.
(0, 10) Inflection 10 point 5 –1
0 –5 –10
1 Inflection point
2
3 (2, – 6)
–15 –20
(3, –17) Local minimum
FIGURE 4.30 The graph of ƒsxd = x 4  4x 3 + 10 (Example 7).
4
x
decr
decr
decr
incr
conc up
conc down
conc up
conc up
0
2
3
infl point
infl point
local min
General shape
(e) Plot the curve’s intercepts (if possible) and the points where y¿ and y– are zero. Indicate any local extreme values and inflection points. Use the general shape as a guide to sketch the curve. (Plot additional points as needed.) Figure 4.30 shows the graph of ƒ.
240
Chapter 4: Applications of Derivatives
The steps in Example 7 give a procedure for graphing the key features of a function.
Procedure for Graphing y ƒ(x) 1. Identify the domain of ƒ and any symmetries the curve may have. 2. Find the derivatives y¿ and y– . 3. Find the critical points of ƒ, if any, and identify the function’s behavior at each one. 4. Find where the curve is increasing and where it is decreasing. 5. Find the points of inflection, if any occur, and determine the concavity of the curve. 6. Identify any asymptotes that may exist (see Section 2.6). 7. Plot key points, such as the intercepts and the points found in Steps 3–5, and sketch the curve together with any asymptotes that exist.
EXAMPLE 8
Sketch the graph of ƒsxd =
sx + 1d2 . 1 + x2
Solution
1. 2.
The domain of ƒ is s  q , q d and there are no symmetries about either axis or the origin (Section 1.1). Find ƒ¿ and ƒ– . ƒsxd =
ƒ¿sxd =
=
ƒ–sxd =
=
3.
4.
xintercept at x = 1, yintercept sy = 1d at x = 0
sx + 1d2 1 + x2 s1 + x 2 d # 2sx + 1d  sx + 1d2 # 2x s1 + x 2 d2 2s1  x 2 d s1 + x 2 d2
Critical points: x =  1, x = 1
s1 + x 2 d2 # 2s 2xd  2s1  x 2 d[2s1 + x 2 d # 2x] s1 + x 2 d4 4xsx 2  3d s1 + x 2 d3
After some algebra
Behavior at critical points. The critical points occur only at x = ;1 where ƒ¿sxd = 0 (Step 2) since ƒ¿ exists everywhere over the domain of ƒ. At x = 1, ƒ–(1) = 1 7 0, yielding a relative minimum by the Second Derivative Test. At x = 1, f –(1) = 1 6 0, yielding a relative maximum by the Second Derivative test. Increasing and decreasing. We see that on the interval s  q , 1d the derivative ƒ¿sxd 6 0, and the curve is decreasing. On the interval s 1, 1d, ƒ¿sxd 7 0 and the curve is increasing; it is decreasing on s1, q d where ƒ¿sxd 6 0 again.
4.4
Inflection points. Notice that the denominator of the second derivative (Step 2) is always positive. The second derivative ƒ– is zero when x =  23, 0, and 23. The second derivative changes sign at each of these points: negative on A  q ,  23 B , positive on A  23, 0 B , negative on A 0, 23 B , and positive again on A 23, q B . Thus each point is a point of inflection. The curve is concave down on the interval A  q ,  23 B , concave up on A  23, 0 B , concave down on A 0, 23 B , and concave up again on A 23, q B .
6.
Asymptotes. Expanding the numerator of ƒ(x) and then dividing both numerator and denominator by x 2 gives
=
(1, 2)
Point of inflection where x 3
2 y1
1 Horizontal asymptote –1 Point of inflection where x 3
1
FIGURE 4.31 The graph of y =
x
sx + 1d2
241
5.
ƒsxd =
y
Concavity and Curve Sketching
7.
1 + x2
(Example 8).
sx + 1d2 x 2 + 2x + 1 = 1 + x2 1 + x2 1 + s2>xd + s1>x 2 d s1>x 2 d + 1
.
Expanding numerator
Dividing by x 2
We see that ƒsxd : 1+ as x : q and that ƒsxd : 1 as x :  q . Thus, the line y = 1 is a horizontal asymptote. Since ƒ decreases on s  q , 1d and then increases on s 1, 1d, we know that ƒs 1d = 0 is a local minimum. Although ƒ decreases on s1, q d, it never crosses the horizontal asymptote y = 1 on that interval (it approaches the asymptote from above). So the graph never becomes negative, and ƒs 1d = 0 is an absolute minimum as well. Likewise, ƒs1d = 2 is an absolute maximum because the graph never crosses the asymptote y = 1 on the interval s  q , 1d, approaching it from below. Therefore, there are no vertical asymptotes (the range of ƒ is 0 … y … 2). The graph of ƒ is sketched in Figure 4.31. Notice how the graph is concave down as it approaches the horizontal asymptote y = 1 as x :  q , and concave up in its approach to y = 1 as x : q .
EXAMPLE 9
Sketch the graph of ƒ(x) =
x2 + 4 . 2x
Solution
1.
2.
3.
The domain of ƒ is all nonzero real numbers. There are no intercepts because neither x nor ƒ(x) can be zero. Since ƒ(x) = ƒ(x), we note that ƒ is an odd function, so the graph of ƒ is symmetric about the origin. We calculate the derivatives of the function, but first rewrite it in order to simplify our computations: ƒ(x) =
x2 + 4 x 2 = + x 2x 2
Function simplified for differentiation
ƒ¿(x) =
x2  4 1 2  2 = 2 x 2x 2
Combine fractions to solve easily ƒ¿(x) = 0.
ƒ–(x) =
4 x3
Exists throughout the entire domain of ƒ
The critical points occur at x = ;2 where ƒ¿(x) = 0. Since ƒ–(2) 6 0 and ƒ–(2) 7 0, we see from the Second Derivative Test that a relative maximum occurs at x = 2 with ƒ(2) = 2, and a relative minimum occurs at x = 2 with ƒ(2) = 2.
242
Chapter 4: Applications of Derivatives
4.
5. y 4
2 y5 x 14 2x
(2, 2) 2 –4
–2 (–2, –2)
6.
y5 x 2 2
0
4
On the interval ( q , 2) the derivative ƒ¿ is positive because x 2  4 7 0 so the graph is increasing; on the interval (2, 0) the derivative is negative and the graph is decreasing. Similarly, the graph is decreasing on the interval (0, 2) and increasing on (2, q ). There are no points of inflection because ƒ–(x) 6 0 whenever x 6 0, ƒ–(x) 7 0 whenever x 7 0, and ƒ– exists everywhere and is never zero throughout the domain of ƒ. The graph is concave down on the interval ( q , 0) and concave up on the interval (0, q ). From the rewritten formula for ƒ(x), we see that
x
lim+ a
x:0
–2 –4
FIGURE 4.32 The graph of y =
x2 + 4 2x
7.
x 2 + xb = +q 2
and
lim a
x:0
x 2 + x b =  q, 2
so the yaxis is a vertical asymptote. Also, as x : q or as x :  q , the graph of ƒ(x) approaches the line y = x>2. Thus y = x>2 is an oblique asymptote. The graph of ƒ is sketched in Figure 4.32.
(Example 9).
EXAMPLE 10
Sketch the graph of ƒ(x) = e 2>x.
The domain of ƒ is ( q , 0) h (0, q ) and there are no symmetries about either axis or the origin. The derivatives of ƒ are Solution
ƒ¿(x) = e 2>x a
2e 2>x 2 b =  2 2 x x
and ƒ–(x) =
x 2(2e 2>x)(2>x 2)  2e 2>x(2x) x4
=
4e 2>x(1 + x) x4
.
y y e 2x
5 4 3 Inflection 2 point 1 –2
–1
y1 0
1
2
3
x
FIGURE 4.33 The graph of y = e 2>x has a point of inflection at (1, e 2). The line y = 1 is a horizontal asymptote and x = 0 is a vertical asymptote (Example 10).
Both derivatives exist everywhere over the domain of ƒ. Moreover, since e 2>x and x 2 are both positive for all x Z 0, we see that ƒ¿ 6 0 everywhere over the domain and the graph is everywhere decreasing. Examining the second derivative, we see that ƒ–(x) = 0 at x = 1. Since e 2>x 7 0 and x 4 7 0, we have ƒ– 6 0 for x 6 1 and ƒ– 7 0 for x 7 1, x Z 0. Therefore, the point (1, e 2) is a point of inflection. The curve is concave down on the interval ( q , 1) and concave up over (1, 0) h (0, q ). From Example 7, Section 2.6, we see that limx:0 ƒ(x) = 0. As x : 0 +, we see that 2>x : q , so limx:0+ ƒ(x) = q and the yaxis is a vertical asymptote. Also, as x :  q , 2>x : 0  and so limx: q ƒ(x) = e 0 = 1. Therefore, y = 1 is a horizontal asymptote. There are no absolute extrema since ƒ never takes on the value 0. The graph of ƒ is sketched in Figure 4.33.
Graphical Behavior of Functions from Derivatives As we saw in Examples 7–10, we can learn much about a twicedifferentiable function y = ƒsxd by examining its first derivative. We can find where the function’s graph rises and falls and where any local extrema are located. We can differentiate y¿ to learn how the graph bends as it passes over the intervals of rise and fall. We can determine the shape of the function’s graph. Information we cannot get from the derivative is how to place the graph in the xyplane. But, as we discovered in Section 4.2, the only additional information we need to position the graph is the value of ƒ at one point. Information about the asymptotes is found using limits (Section 2.6). The following
4.4
243
Concavity and Curve Sketching
figure summarizes how the derivative and second derivative affect the shape of a graph.
y f (x)
y f (x)
Differentiable ⇒ smooth, connected; graph may rise and fall
y' 0 ⇒ rises from left to right; may be wavy
or
y f (x)
y' 0 ⇒ falls from left to right; may be wavy
or
y'' 0 ⇒ concave up throughout; no waves; graph may rise or fall
y'' 0 ⇒ concave down throughout; no waves; graph may rise or fall
y'' changes sign at an inflection point
y' 0 and y'' 0 at a point; graph has local maximum
y' 0 and y'' 0 at a point; graph has local minimum
or y' changes sign ⇒ graph has local maximum or local minimum
Exercises 4.4 Analyzing Functions from Graphs Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the intervals on which the functions are concave up and concave down. 1.
2.
3 2 y x x 2x 1 3 2 3 y
4 y x 2x2 4 4 y
0
2 3
7. y sin x , – 2 x 2 y
0
4.
y 3 (x 2 1)2/3 4 y
0
x
y
x
2
2
x
0
8.
3 y 2 cos x 2 x, – x 2 y
x
0
3.
6. y tan x 4x, – x
y x sin 2x, – 2 x 2 3 3 y
– 2 3
x
0
5.
y 9 x1/3(x 2 7) 14 y
0
x
–
0
3 2
x
NOT TO SCALE
x
Graphing Equations Use the steps of the graphing procedure on page 240 to graph the equations in Exercises 9–58. Include the coordinates of any local and absolute extreme points and inflection points. 9. y = x 2  4x + 3 11. y = x  3x + 3 3
10. y = 6  2x  x 2 12. y = xs6  2xd2
244
Chapter 4: Applications of Derivatives 14. y = 1  9x  6x 2  x 3
13. y = 2x 3 + 6x 2  3 15. y = sx  2d3 + 1 16. y = 1  sx + 1d3 17. y = x 4  2x 2 = x 2sx 2  2d
18. y = x 4 + 6x 2  4 = x 2s6  x 2 d  4 19. y = 4x 3  x 4 = x 3s4  xd 20. y = x 4 + 2x 3 = x 3sx + 2d 21. y = x 5  5x 4 = x 4sx  5d 22. y = x a
x  5b 2
4
23. y = x + sin x,
0 … x … 2p
24. y = x  sin x,
0 … x … 2p
29. y = x 1>5 31. y =
0 … x … 2p
2x 2 + 1 33. y = 2x  3x 2>3 35. y = x 2>3 a
5  xb 2
37. y = x28  x
2
39. y = 216  x 2 x2  3 41. y = x  2 8x 43. y = 2 x + 4 45. y = ƒ x 2  1 ƒ 2 x, 47. y = 2 ƒ x ƒ = e 2x,
32. y =
0 … t … 2p
74. y¿ = sin t,
0 … t … 2p
21  x 2 2x + 1
34. y = 5x 2>5  2x 36. y = x 2>3(x  5)
78. y¿ = x 4>5sx + 1d
Sketching y from Graphs of y œ and y ﬂ Each of Exercises 81–84 shows the graphs of the first and second derivatives of a function y = ƒsxd . Copy the picture and add to it a sketch of the approximate graph of ƒ, given that the graph passes through the point P. 81.
82.
y
y
y f '(x)
y f'(x)
P
x
x
42. y = 2x + 1 3
P
y f ''(x)
83. P
y f '(x)
x
0
ex 50. y = x
51. y = ln (3  x 2)
52. y = x (ln x) 2
53. y = e x  2e x  3x
54. y = xe x
55. y = ln (cos x)
56. y =
y f ''(x)
y
x 6 0 x Ú 0
49. y = xe 1>x
1 1 + e x
2x, x … 0 2x, x 7 0
x 2, x … 0 x 7 0 x 2,
38. y = (2  x ) 2 40. y = x 2 + x
5 44. y = 4 x + 5 46. y = ƒ x 2  2 x ƒ
0 6 u 6 2p
77. y¿ = x 2>3sx  1d
48. y = 2ƒ x  4 ƒ
57. y =
u 70. y¿ = csc2 , 2 p 6 2 p
76. y¿ = sx  2d1>3
2 3>2
3
66. y¿ = sx 2  2xdsx  5d2
75. y¿ = sx + 1d2>3
80. y¿ = e
30. y = x 2>5 x
73. y¿ = cos t,
79. y¿ = 2 ƒ x ƒ = e
25. y = 23x  2 cos x, 0 … x … 2p p p 4 26. y = x  tan x, 6 x 6 3 2 2 27. y = sin x cos x, 0 … x … p 28. y = cos x + 23 sin x,
65. y¿ = s8x  5x 2 d(4  x) 2 p p 67. y¿ = sec2 x,  6 x 6 2 2 p p 68. y¿ = tan x,  6 x 6 2 2 u 69. y¿ = cot , 0 6 u 6 2p 2 p 71. y¿ = tan2 u  1,  6 u 2 72. y¿ = 1  cot2 u, 0 6 u 6
y f ''(x)
84.
y y f '(x)
ln x
2x ex 58. y = 1 + ex
Sketching the General Shape, Knowing y œ Each of Exercises 59–80 gives the first derivative of a continuous function y = ƒsxd . Find y– and then use steps 2–4 of the graphing procedure on page 240 to sketch the general shape of the graph of ƒ. 59. y¿ = 2 + x  x 2
60. y¿ = x 2  x  6
61. y¿ = xsx  3d2
62. y¿ = x 2s2  xd
63. y¿ = xsx 2  12d
64. y¿ = sx  1d2s2x + 3d
x
0 y f ''(x) P
Graphing Rational Functions Graph the rational functions in Exercises 85–102. 2x 2 + x  1 x2  1 4 x + 1 87. y = x2 1 89. y = 2 x  1 85. y =
x 2  49 x + 5x  14 x2  4 88. y = 2x x2 90. y = 2 x  1 86. y =
2
4.4
95. y = 97. y = 99. y = 101. y = 102. y =
y S
y f (x) R
P
107.
s
s f(t)
0
Theory and Examples 103. The accompanying figure shows a portion of the graph of a twicedifferentiable function y = ƒsxd . At each of the five labeled points, classify y¿ and y– as positive, negative, or zero.
108.
5
5
ƒ¿s2d = ƒ¿s 2d = 0,
ƒs0d = 4,
ƒ¿sxd 6 0
for
ƒ x ƒ 6 2,
ƒs2d = 0,
ƒ–sxd 6 0
for
x 6 0,
ƒ–sxd 7 0
for
x 7 0.
ƒ x ƒ 7 2,
x 6 2 2 6 x 4 4 6 x 6 x 7
y
Derivatives
2
y¿ y¿ y¿ y¿ y¿ y¿ y¿
1 6 4 4 6 6 7 6
6 = 7 7 7 = 6
0, 0, 0, 0, 0, 0, 0,
y– y– y– y– y– y– y–
7 7 7 = 6 6 6
0 0 0 0 0 0 0
106. Sketch the graph of a twicedifferentiable function y = ƒsxd that passes through the points s 2, 2d, s 1, 1d, s0, 0d, s1, 1d, and (2, 2) and whose first two derivatives have the following sign patterns. y¿:
+
y–:

2 1

+ 0
+
2 
1

15
t
c c f (x)
105. Sketch the graph of a twicedifferentiable function y = ƒsxd with the following properties. Label coordinates where possible.
x
10 Time (sec)
Cost
ƒs  2d = 8,
for
t
109. Marginal cost The accompanying graph shows the hypothetical cost c = ƒsxd of manufacturing x items. At approximately what production level does the marginal cost change from decreasing to increasing?
x
104. Sketch a smooth connected curve y = ƒsxd with
ƒ¿sxd 7 0
15
s f (t)
Q
0
10 Time (sec)
s
0
T
245
Motion Along a Line The graphs in Exercises 107 and 108 show the position s = ƒstd of an object moving up and down on a coordinate line. (a) When is the object moving away from the origin? Toward the origin? At approximately what times is the (b) velocity equal to zero? (c) Acceleration equal to zero? (d) When is the acceleration positive? Negative?
Displacement
93. y =
x2  4 x2  2 x2  4 x + 1 x2  x + 1 x  1 x3 + x  2 x  x2 x  1 x 2(x  2)
Displacement
x2  2 92. y = x2  1 x2 94. y = x + 1 x2  x + 1 96. y = x  1 3 2 x  3x + 3x  1 98. y = x2 + x  2 x 100. y = x2  1 8 (Agnesi's witch) x2 + 4 4x (Newton's serpentine) x2 + 4
91. y = 
Concavity and Curve Sketching
20 40 60 80 100 120 Thousands of units produced
x
110. The accompanying graph shows the monthly revenue of the Widget Corporation for the last 12 years. During approximately what time intervals was the marginal revenue increasing? Decreasing? y y r(t)
0
5
10
t
111. Suppose the derivative of the function y = ƒsxd is y¿ = sx  1d2sx  2d . At what points, if any, does the graph of ƒ have a local minimum, local maximum, or point of inflection? (Hint: Draw the sign pattern for y¿ .)
246
Chapter 4: Applications of Derivatives
112. Suppose the derivative of the function y = ƒsxd is
120. Suppose that the second derivative of the function y = ƒsxd is y– = x 2(x  2) 3(x + 3).
y¿ = sx  1d2sx  2dsx  4d .
For what xvalues does the graph of ƒ have an inflection point? At what points, if any, does the graph of ƒ have a local minimum, local maximum, or point of inflection? 113. For x 7 0 , sketch a curve y = ƒsxd that has ƒs1d = 0 and ƒ¿sxd = 1>x . Can anything be said about the concavity of such a curve? Give reasons for your answer. 114. Can anything be said about the graph of a function y = ƒsxd that has a continuous second derivative that is never zero? Give reasons for your answer. 115. If b, c, and d are constants, for what value of b will the curve y = x 3 + bx 2 + cx + d have a point of inflection at x = 1 ? Give reasons for your answer. 116. Parabolas a. Find the coordinates of the vertex of the parabola y = ax 2 + bx + c, a Z 0 . b. When is the parabola concave up? Concave down? Give reasons for your answers. 117. Quadratic curves What can you say about the inflection points of a quadratic curve y = ax 2 + bx + c, a Z 0 ? Give reasons for your answer. 118. Cubic curves What can you say about the inflection points of a cubic curve y = ax 3 + bx 2 + cx + d, a Z 0 ? Give reasons for your answer. 119. Suppose that the second derivative of the function y = ƒsxd is y– = (x + 1)(x  2). For what xvalues does the graph of ƒ have an inflection point?
4.5
121. Find the values of constants a, b, and c so that the graph of y = ax 3 + bx 2 + cx has a local maximum at x = 3, local minimum at x = 1, and inflection point at (1, 11). 122. Find the values of constants a, b, and c so that the graph of y = (x 2 + a)>(bx + c) has a local minimum at x = 3 and a local maximum at (1, 2). COMPUTER EXPLORATIONS In Exercises 123–126, find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simultaneously. Add to your picture the graphs of the function’s first and second derivatives. How are the values at which these graphs intersect the xaxis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function? 123. y = x 5  5x 4  240
124. y = x 3  12x 2
4 5 x + 16x 2  25 5 x4 x3 126. y =  4x 2 + 12x + 20 4 3 125. y =
127. Graph ƒsxd = 2x 4  4x 2 + 1 and its first two derivatives together. Comment on the behavior of ƒ in relation to the signs and values of ƒ¿ and ƒ– . 128. Graph ƒsxd = x cos x and its second derivative together for 0 … x … 2p . Comment on the behavior of the graph of ƒ in relation to the signs and values of ƒ– .
Indeterminate Forms and L’Hôpital’s Rule
HISTORICAL BIOGRAPHY Guillaume François Antoine de l’Hôpital (1661–1704) Johann Bernoulli (1667–1748)
John (Johann) Bernoulli discovered a rule using derivatives to calculate limits of fractions whose numerators and denominators both approach zero or + q . The rule is known today as l’Hôpital’s Rule, after Guillaume de l’Hôpital. He was a French nobleman who wrote the first introductory differential calculus text, where the rule first appeared in print. Limits involving transcendental functions often require some use of the rule for their calculation.
Indeterminate Form 0/0 If we want to know how the function Fsxd =
x  sin x x3
behaves near x = 0 (where it is undefined), we can examine the limit of Fsxd as x : 0. We cannot apply the Quotient Rule for limits (Theorem 1 of Chapter 2) because the limit of the denominator is 0. Moreover, in this case, both the numerator and denominator approach 0, and 0>0 is undefined. Such limits may or may not exist in general, but the limit does exist for the function Fsxd under discussion by applying l’Hôpital’s Rule, as we will see in Example 1d.
4.5
Indeterminate Forms and L’Hôpital’s Rule
247
If the continuous functions ƒ(x) and g(x) are both zero at x = a, then lim
x:a
ƒsxd gsxd
cannot be found by substituting x = a. The substitution produces 0>0, a meaningless expression, which we cannot evaluate. We use 0>0 as a notation for an expression known as an indeterminate form. Other meaningless expressions often occur, such as q > q , q # 0, q  q , 0 0, and 1q, which cannot be evaluated in a consistent way; these are called indeterminate forms as well. Sometimes, but not always, limits that lead to indeterminate forms may be found by cancellation, rearrangement of terms, or other algebraic manipulations. This was our experience in Chapter 2. It took considerable analysis in Section 2.4 to find limx:0 ssin xd>x. But we have had success with the limit ƒsxd  ƒsad , x  a x:a
ƒ¿sad = lim
from which we calculate derivatives and which produces the indeterminant form 0>0 when we substitute x = a. L’Hôpital’s Rule enables us to draw on our success with derivatives to evaluate limits that otherwise lead to indeterminate forms.
THEOREM 6— L’Hôpital’s Rule Suppose that ƒsad = gsad = 0, that ƒ and g are differentiable on an open interval I containing a, and that g¿sxd Z 0 on I if x Z a. Then ƒsxd ƒ¿sxd = lim , lim x:a gsxd x:a g¿sxd assuming that the limit on the right side of this equation exists.
We give a proof of Theorem 6 at the end of this section.
Caution To apply l’Hôpital’s Rule to ƒ>g, divide the derivative of ƒ by the derivative of g. Do not fall into the trap of taking the derivative of ƒ>g. The quotient to use is ƒ¿>g¿, not sƒ>gd¿ .
EXAMPLE 1 The following limits involve 0>0 indeterminate forms, so we apply l’Hôpital’s Rule. In some cases, it must be applied repeatedly. (a) lim
x:0
3x  sin x 3  cos x 3  cos x = lim = ` = 2 x 1 1 x:0 x=0
1 21 + x  1 221 + x 1 = lim = (b) lim x 2 x:0 x:0 1 (c) lim
x:0
21 + x  1  x>2 x
0 ; apply l’Hôpital’s Rule. 0
2
s1>2ds1 + xd1>2  1>2 2x x:0
= lim = lim
x:0
s1>4ds1 + xd3>2 1 = 2 8
0 Still ; apply l’Hôpital’s Rule again. 0 Not
0 ; limit is found. 0
248
Chapter 4: Applications of Derivatives
(d) lim
x:0
x  sin x x3
0 ; apply l’Hôpital’s Rule. 0
= lim
1  cos x 3x 2
0 Still ; apply l’Hôpital’s Rule again. 0
= lim
sin x 6x
0 Still ; apply l’Hôpital’s Rule again. 0
= lim
cos x 1 = 6 6
0 Not ; limit is found. 0
x:0
x:0
x:0
Here is a summary of the procedure we followed in Example 1.
Using L’Hôpital’s Rule To find lim
x:a
ƒsxd gsxd
by l’Hôpital’s Rule, continue to differentiate ƒ and g, so long as we still get the form 0>0 at x = a. But as soon as one or the other of these derivatives is different from zero at x = a we stop differentiating. L’Hôpital’s Rule does not apply when either the numerator or denominator has a finite nonzero limit.
EXAMPLE 2
Be careful to apply l’Hôpital’s Rule correctly: lim
x:0
1  cos x x + x2
= lim
x:0
sin x 1 + 2x
0 0 Not
0 0
It is tempting to try to apply l’Hôpital’s Rule again, which would result in lim
x:0
cos x 1 = , 2 2
but this is not the correct limit. l’Hôpital’s Rule can be applied only to limits that give indeterminate forms, and limx:0 (sin x)>(1 + 2x) does not give an indeterminate form. Instead, this limit is 0>1 = 0, and the correct answer for the original limit is 0. L’Hôpital’s Rule applies to onesided limits as well.
EXAMPLE 3 (a) lim+ x:0
Recall that q and + q mean the same thing.
In this example the onesided limits are different.
sin x x2 = lim+ x:0
(b) limx:0
0 0
cos x = q 2x
sin x x2 = limx:0
Positive for x 7 0 0 0
cos x = q 2x
Negative for x 6 0
Indeterminate Forms ˆ / ˆ , ˆ # 0, ˆ ˆ Sometimes when we try to evaluate a limit as x : a by substituting x = a we get an indeterminant form like q > q , q # 0, or q  q , instead of 0>0. We first consider the form q> q.
4.5
Indeterminate Forms and L’Hôpital’s Rule
249
In more advanced treatments of calculus, it is proved that l’Hôpital’s Rule applies to the indeterminate form q > q as well as to 0>0. If ƒsxd : ; q and gsxd : ; q as x : a, then lim
x:a
ƒsxd ƒ¿sxd = lim gsxd x:a g¿sxd
provided the limit on the right exists. In the notation x : a, a may be either finite or infinite. Moreover, x : a may be replaced by the onesided limits x : a + or x : a . Find the limits of these q > q forms:
EXAMPLE 4 (a)
lim
x:p>2
sec x 1 + tan x
ln x
(b) lim
x: q
(c) lim
x: q
22x
ex . x2
Solution
(a) The numerator and denominator are discontinuous at x = p>2, so we investigate the onesided limits there. To apply l’Hôpital’s Rule, we can choose I to be any open interval with x = p>2 as an endpoint. lim
x:sp>2d 
q q from the left so we apply l’Hôpital’s Rule.
sec x 1 + tan x =
lim
x:sp>2d 
sec x tan x = sec2 x
lim
x:sp>2d 
sin x = 1
The righthand limit is 1 also, with s  q d>s  q d as the indeterminate form. Therefore, the twosided limit is equal to 1. (b) lim
x: q
ln x 22x
x: q
x
(c)
lim
x: q
1>x
= lim
1> 2x
= lim
x: q
1 = 0 2x
1>x 1> 2x
=
2x x =
1 2x
x
e e ex = lim = lim = q 2 x: q 2x x: q 2 x
Next we turn our attention to the indeterminate forms q # 0 and q  q . Sometimes these forms can be handled by using algebra to convert them to a 0>0 or q > q form. Here again we do not mean to suggest that q # 0 or q  q is a number. They are only notations for functional behaviors when considering limits. Here are examples of how we might work with these indeterminate forms.
EXAMPLE 5
Find the limits of these q # 0 forms:
1 (a) lim ax sin x b x: q
(b) lim+ 2x ln x x:0
Solution
(a)
sin h 1 1 lim ax sin x b = lim+ a sin hb = lim+ = 1 h h:0 h h:0
x: q
(b) lim+ 2x ln x = lim+ x:0
x:0
= lim+ x:0
ln x 1> 2x
q # 0 converted to q > q
1>x 1>2x 3>2
= lim+ A 22x B = 0 x:0
q # 0; let h = 1>x.
l’Hôpital’s Rule applied
250
Chapter 4: Applications of Derivatives
EXAMPLE 6
Find the limit of this q  q form: lim a
x:0
Solution
1 1  xb. sin x
If x : 0 + , then sin x : 0 + and 1 1  x : q  q. sin x
Similarly, if x : 0  , then sin x : 0  and 1 1  x :  q  s qd =  q + q . sin x Neither form reveals what happens in the limit. To find out, we first combine the fractions: x  sin x 1 1  x = sin x x sin x
Common denominator is x sin x.
Then we apply l’Hôpital’s Rule to the result: lim a
x:0
x  sin x 1 1  x b = lim sin x x:0 x sin x = lim
1  cos x sin x + x cos x
= lim
sin x 0 = = 0. 2 2 cos x  x sin x
x:0
x:0
0 0 Still
0 0
Indeterminate Powers Limits that lead to the indeterminate forms 1q, 0 0, and q 0 can sometimes be handled by first taking the logarithm of the function. We use l’Hôpital’s Rule to find the limit of the logarithm expression and then exponentiate the result to find the original function limit. This procedure is justified by the continuity of the exponential function and Theorem 10 in Section 2.5, and it is formulated as follows. (The formula is also valid for onesided limits.)
If limx:a ln ƒ(x) = L, then lim ƒ(x) = lim e ln ƒ(x) = e L.
x:a
x:a
Here a may be either finite or infinite.
EXAMPLE 7
Apply l’Hôpital’s Rule to show that limx:0+ (1 + x) 1>x = e.
The limit leads to the indeterminate form 1q. We let ƒ(x) = (1 + x) 1>x and find limx:0+ ln ƒ(x). Since Solution
1 ln ƒ(x) = ln (1 + x) 1>x = x ln (1 + x),
4.5
Indeterminate Forms and L’Hôpital’s Rule
251
l’Hôpital’s Rule now applies to give lim+ ln ƒ(x) = lim+
x:0
x:0
ln (1 + x) x
1 1 + x = lim+ 1 x:0 =
0 0
l’Hôpital’s Rule applied
1 = 1. 1
Therefore, lim+ (1 + x) 1>x = lim+ ƒ(x) = lim+ e ln ƒ(x) = e 1 = e. x:0
EXAMPLE 8
x:0
x:0
Find limx: q x 1>x.
The limit leads to the indeterminate form q 0. We let ƒ(x) = x 1>x and find limx: q ln ƒ(x). Since Solution
ln ƒ(x) = ln x 1>x =
ln x x ,
l’Hôpital’s Rule gives lim ln ƒ(x) = lim
ln x x
q q
= lim
1>x 1
l’Hôpital’s Rule applied
x: q
x: q
x: q
=
0 = 0. 1
Therefore lim x 1>x = lim ƒ(x) = lim e ln ƒ(x) = e 0 = 1. x: q
x: q
x: q
Proof of L’Hôpital’s Rule HISTORICAL BIOGRAPHY AugustinLouis Cauchy (1789–1857)
When gsxd = x, Theorem 7 is the Mean Value Theorem.
The proof of l’Hôpital’s Rule is based on Cauchy’s Mean Value Theorem, an extension of the Mean Value Theorem that involves two functions instead of one. We prove Cauchy’s Theorem first and then show how it leads to l’Hôpital’s Rule.
THEOREM 7—Cauchy’s Mean Value Theorem Suppose functions ƒ and g are continuous on [a, b] and differentiable throughout (a, b) and also suppose g¿sxd Z 0 throughout (a, b). Then there exists a number c in (a, b) at which ƒ¿scd ƒsbd  ƒsad = . g¿scd gsbd  gsad
Proof We apply the Mean Value Theorem of Section 4.2 twice. First we use it to show that gsad Z gsbd. For if gsbd did equal gsad, then the Mean Value Theorem would give g¿scd =
gsbd  gsad = 0 b  a
for some c between a and b, which cannot happen because g¿sxd Z 0 in (a, b).
252
Chapter 4: Applications of Derivatives
We next apply the Mean Value Theorem to the function Fsxd = ƒsxd  ƒsad 
ƒsbd  ƒsad [ gsxd  gsad]. gsbd  gsad
This function is continuous and differentiable where ƒ and g are, and Fsbd = Fsad = 0. Therefore, there is a number c between a and b for which F¿scd = 0. When expressed in terms of ƒ and g, this equation becomes F¿scd = ƒ¿scd 
ƒsbd  ƒsad [ g¿scd] = 0 gsbd  gsad
so that ƒ¿scd ƒsbd  ƒsad = . g¿scd gsbd  gsad
y
slope 5
f '(c) g'(c) B (g(b), f (b))
P
slope 5
A 0
f (b) 2 f (a) g(b) 2 g(a)
Cauchy’s Mean Value Theorem has a geometric interpretation for a general winding curve C in the plane joining the two points A = sgsad, ƒsadd and B = sgsbd, ƒsbdd. In Chapter 11 you will learn how the curve C can be formulated so that there is at least one point P on the curve for which the tangent to the curve at P is parallel to the secant line joining the points A and B. The slope of that tangent line turns out to be the quotient ƒ¿>g¿ evaluated at the number c in the interval sa, bd, which is the lefthand side of the equation in Theorem 7. Because the slope of the secant line joining A and B is
(g(a), f (a))
ƒsbd  ƒsad , gsbd  gsad
x
FIGURE 4.34 There is at least one point P on the curve C for which the slope of the tangent to the curve at P is the same as the slope of the secant line joining the points A(g(a), ƒ(a)) and B(g(b), ƒ(b)).
the equation in Cauchy’s Mean Value Theorem says that the slope of the tangent line equals the slope of the secant line. This geometric interpretation is shown in Figure 4.34. Notice from the figure that it is possible for more than one point on the curve C to have a tangent line that is parallel to the secant line joining A and B. Proof of l’Hôpital’s Rule We first establish the limit equation for the case x : a + . The method needs almost no change to apply to x : a , and the combination of these two cases establishes the result. Suppose that x lies to the right of a. Then g¿sxd Z 0, and we can apply Cauchy’s Mean Value Theorem to the closed interval from a to x. This step produces a number c between a and x such that ƒ¿scd ƒsxd  ƒsad = . g¿scd gsxd  gsad But ƒsad = gsad = 0, so ƒ¿scd ƒsxd = . g¿scd gsxd As x approaches a, c approaches a because it always lies between a and x. Therefore, lim+
x:a
ƒsxd ƒ¿scd ƒ¿sxd = lim+ = lim+ , gsxd c:a g¿scd x:a g¿sxd
which establishes l’Hôpital’s Rule for the case where x approaches a from above. The case where x approaches a from below is proved by applying Cauchy’s Mean Value Theorem to the closed interval [x, a], x 6 a.
4.5
Indeterminate Forms and L’Hôpital’s Rule
253
Exercises 4.5 Finding Limits in Two Ways In Exercises 1–6, use l’Hôpital’s Rule to evaluate the limit. Then evaluate the limit using a method studied in Chapter 2.
cos u  1 u :0 e u  u  1
43. lim
x + 2 1. lim 2 x: 2 x  4
sin 5x 2. lim x x: 0
45. lim
5x 2  3x 3. lim x: q 7x 2 + 1
x3  1 4. lim 3 x: 1 4x  x  3
47. lim
1  cos x 5. lim x: 0 x2
2x 2 + 3x 6. lim 3 q x: x + x + 1
49. lim
t: q
t  4t + 15 t: 3 t 2  t  12
9. lim
5x 3  2x 11. lim x: q 7x 3 + 3 sin t t t:0
8x x: 0 cos x  1
3t + 3 t: 1 4t3  t + 3
17. 19.
lim
u :p>2
2u  p cos (2p  u)
1  sin u u :p>2 1 + cos 2u lim
21. lim
x: 0
x2 ln (sec x)
t (1  cos t) 23. lim t:0 t  sin t 25.
lim
x: (p>2) 
sin x  x x: 0 x3
61. lim a
ax 
p b sec x 2
26.
lim
30. lim
a
32. lim
log2 x log3 (x + 3)
33. lim+
ln (x 2 + 2x) ln x
34. lim+
ln (e x  1) ln x
25y + 25  5 y
36. lim
x: 0
35. lim
y: 0
37. lim (ln 2x  ln (x + 1)) x: q
39. lim+ x: 0
sln x) 2 ln ssin x)
41. lim+ a x: 1
1 1 b x  1 ln x
x: q
x: 0
y: 0
2ay + a 2  a , y
x + 2 b x  1
x
lim x 2 ln x
64.
x:0 +
lim x tan a
x:0 +
62.
p  xb 2
66.
lim a
x: q
x
x2 + 1 b x + 2
1>x
lim x sln xd2
x:0 +
lim sin x # ln x
x:0 +
29x + 1
2x + 1 sec x 69. lim x:sp>2d  tan x 71.
2x  3x x x x: q 3 + 4
73.
ex x q xe x:
lim
68. lim+
lim
2x
2sin x cot x 70. lim+ csc x x:0 x:0
72.
2x + 4x x x x:  q 5  2 lim
74.
lim
x:0 +
x e 1>x
75. Which one is correct, and which one is wrong? Give reasons for your answers. 0 x  3 x  3 1 1 b. lim 2 = lim = = 0 = 6 6 x:3 2x x:3 x  3 x2  3 76. Which one is correct, and which one is wrong? Give reasons for your answers. a. lim
x:3
a 7 0
x 2  2x 2x  2 = lim x:0 x 2  sin x x:0 2x  cos x 2 2 = lim = = 1 2 + 0 x:0 2 + sin x
a. lim
3x + 1 1 b x sin x
42. lim+ (csc x  cot x + cos x) x: 0
x:0
2
x: 0
x: 0
58. lim (e x + x) 1>x 1 60. lim+ a1 + x b x:0
x: q
38. lim+ (ln x  ln sin x) 40. lim+ a
x: q
x:0
67. lim
3x  1 x x: 0 2  1
ln (x + 1) log2 x
56. lim x 1>ln x
Theory and Applications L’Hôpital’s Rule does not help with the limits in Exercises 67–74. Try it—you just keep on cycling. Find the limits some other way.
2
p  xb tan x 2
31. lim
x: q
65.
(1>2) u  1 u u: 0
x2x x: 0 2  1 x
(x  (p>2))
x: (p>2) 
28. lim
29. lim
63.
t sin t 24. lim t: 0 1  cos t
3sin u  1 u u :0
27. lim
3u + p sin (u + (p>3))
ln (csc x)
lim
x:e
lim+ x x
x: q
x  1 x: 1 ln x  sin px x: p>2
54. lim+ (ln x) 1>(x  e)
x: q
59.
lim
x:1
1>x
1>ln x
sin 5t t: 0 2t
u: p>3
sin 3x  3x + x 2 sin x sin 2x x:0
52. lim+ x 1>(x  1)
57. lim (1 + 2x) 1>(2 ln x)
20. lim 22.
50. lim
x:0
x  8x 2 12. lim x: q 12x 2 + 5x
18.
u  sin u cos u tan u  u u :0
55. lim+ x
16. lim
15. lim
48. lim
x: q
10. lim
2
se x  1d2 x:0 x sin x
x  sin x x:0 x tan x
53. lim (ln x)
14. lim
13. lim
x: q
x:1
3
2
h2
h:0
46. lim x 2e x
51. lim+ x 1>(1  x)
x 2  25 8. lim x:  5 x + 5
3
e h  (1 + h)
Indeterminate Powers and Products Find the limits in Exercise 51–66.
Applying l’Hôpital’s Rule Use l’Hôpital’s rule to find the limits in Exercises 7– 50. x  2 7. lim 2 x: 2 x  4
et + t2 et  t
44. lim
b.
x 2  2x 2x  2 2 = = 2 = lim 0  1 x:0 x 2  sin x x:0 2x  cos x lim
254
Chapter 4: Applications of Derivatives
77. Only one of these calculations is correct. Which one? Why are the others wrong? Give reasons for your answers. a. lim+ x ln x = 0 # ( q ) = 0
a. Use l’Hôpital’s Rule to show that x
1 lim a1 + x b = e. x: q
x: 0
b. lim+ x ln x = 0 # ( q ) =  q x: 0
ln x q = q = 1 (1>x) ln x d. lim+ x ln x = lim+ x: 0 x:0 (1>x) (1>x) = lim+ (x) = 0 = lim+ x: 0 (1>x 2) x: 0 78. Find all values of c that satisfy the conclusion of Cauchy’s Mean Value Theorem for the given functions and interval. c. lim+ x ln x = lim+ x: 0
T b. Graph
x:0
a. ƒsxd = x,
g sxd = x ,
sa, bd = s 2, 0d
b. ƒsxd = x,
g sxd = x ,
sa, bd arbitrary
2 2
c. ƒsxd = x 3>3  4x,
g sxd = x 2,
ƒ(x) = a1 +
9x  3 sin 3x , 5x 3 c,
x = 0
continuous at x = 0 . Explain why your value of c works.
T 81. ˆ ˆ Form
85. Show that lim a1 +
k: q
lim A x  2x 2 + x B
x: q
by graphing ƒsxd = x  2x 2 + x over a suitably large interval of xvalues. b. Now confirm your estimate by finding the limit with l’Hôpital’s Rule. As the first step, multiply ƒ(x) by the fraction sx + 2x 2 + xd>sx + 2x 2 + xd and simplify the new numerator.
A 2x 2 + 1  2x B .
Estimate the value of 2x 2  s3x + 1d2x + 2 x  1 x:1 lim
by graphing. Then confirm your estimate with l’Hôpital’s Rule. 84. This exercise explores the difference between the limit lim a1 +
x: q
k
r b = e r. k
86. Given that x 7 0, find the maximum value, if any, of a. x 1>x 2
b. x 1>x
n
c. x 1>x (n a positive integer) 87. Use limits to find horizontal asymptotes for each function. 1 a. y = x tan a x b 88. Find ƒ¿s0d for ƒsxd = e
a. Estimate the value of
T 83. 0/0 Form
x
n
tan 2x sin bx a lim a 3 + 2 + x b = 0? x:0 x x
x: q
1 g(x) = a1 + x b
d. Show that limx: q x 1>x = 1 for every positive integer n.
80. For what values of a and b is
82. Find lim
and
c. Confirm your estimate of limx: q ƒ(x) by calculating it with l’Hôpital’s Rule.
sa, bd = s0, 3d
x Z 0
x
together for x Ú 0. How does the behavior of ƒ compare with that of g? Estimate the value of limx: q ƒ(x).
79. Continuous extension Find a value of c that makes the function ƒsxd = •
1 b x2
1 b x2
x
and the limit x
1 lim a1 + x b = e. x: q
b. y = e 1/x , 0, 2
3x + e 2x 2x + e 3x
x Z 0 x = 0.
T 89. The continuous extension of (sin x)x to [0, p] a. Graph ƒ(x) = (sin x) x on the interval 0 … x … p. What value would you assign to ƒ to make it continuous at x = 0? b. Verify your conclusion in part (a) by finding limx:0+ ƒ(x) with l’Hôpital’s Rule. c. Returning to the graph, estimate the maximum value of ƒ on [0, p]. About where is max ƒ taken on? d. Sharpen your estimate in part (c) by graphing ƒ¿ in the same window to see where its graph crosses the xaxis. To simplify your work, you might want to delete the exponential factor from the expression for ƒ¿ and graph just the factor that has a zero. T 90. The function (sin x)tan x (Continuation of Exercise 89.) a. Graph ƒ(x) = (sin x) tan x on the interval 7 … x … 7. How do you account for the gaps in the graph? How wide are the gaps? b. Now graph ƒ on the interval 0 … x … p. The function is not defined at x = p>2, but the graph has no break at this point. What is going on? What value does the graph appear to give for ƒ at x = p>2? (Hint: Use l’Hôpital’s Rule to find lim ƒ as x : (p>2)  and x : (p>2) +.) c. Continuing with the graphs in part (b), find max ƒ and min ƒ as accurately as you can and estimate the values of x at which they are taken on.
4.6
Applied Optimization
255
Applied Optimization
4.6
What are the dimensions of a rectangle with fixed perimeter having maximum area? What are the dimensions for the least expensive cylindrical can of a given volume? How many items should be produced for the most profitable production run? Each of these questions asks for the best, or optimal, value of a given function. In this section we use derivatives to solve a variety of optimization problems in business, mathematics, physics, and economics.
x
Solving Applied Optimization Problems 1. Read the problem. Read the problem until you understand it. What is given? What is the unknown quantity to be optimized? 2. Draw a picture. Label any part that may be important to the problem. 3. Introduce variables. List every relation in the picture and in the problem as an equation or algebraic expression, and identify the unknown variable. 4. Write an equation for the unknown quantity. If you can, express the unknown as a function of a single variable or in two equations in two unknowns. This may require considerable manipulation. 5. Test the critical points and endpoints in the domain of the unknown. Use what you know about the shape of the function’s graph. Use the first and second derivatives to identify and classify the function’s critical points.
12
x x
x
12 (a)
x
12 2x 12 12 2x
x
EXAMPLE 1
An opentop box is to be made by cutting small congruent squares from the corners of a 12in.by12in. sheet of tin and bending up the sides. How large should the squares cut from the corners be to make the box hold as much as possible?
x
(b)
FIGURE 4.35 An open box made by cutting the corners from a square sheet of tin. What size corners maximize the box’s volume (Example 1)?
We start with a picture (Figure 4.35). In the figure, the corner squares are x in. on a side. The volume of the box is a function of this variable: Solution
Vsxd = xs12  2xd2 = 144x  48x 2 + 4x 3.
Since the sides of the sheet of tin are only 12 in. long, x … 6 and the domain of V is the interval 0 … x … 6. A graph of V (Figure 4.36) suggests a minimum value of 0 at x = 0 and x = 6 and a maximum near x = 2. To learn more, we examine the first derivative of V with respect to x:
Maximum y
Volume
y x(12 – 2x)2, 0x6
0
dV = 144  96x + 12x 2 = 12s12  8x + x 2 d = 12s2  xds6  xd. dx
min
min 2
6
V = hlw
x
Of the two zeros, x = 2 and x = 6, only x = 2 lies in the interior of the function’s domain and makes the criticalpoint list. The values of V at this one critical point and two endpoints are
NOT TO SCALE
FIGURE 4.36 The volume of the box in Figure 4.35 graphed as a function of x.
Criticalpoint value:
Vs2d = 128
Endpoint values:
Vs0d = 0,
Vs6d = 0.
The maximum volume is 128 in3 . The cutout squares should be 2 in. on a side.
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Chapter 4: Applications of Derivatives
EXAMPLE 2
You have been asked to design a oneliter can shaped like a right circular cylinder (Figure 4.37). What dimensions will use the least material?
2r
h
Solution Volume of can: If r and h are measured in centimeters, then the volume of the can in cubic centimeters is
pr 2h = 1000. Surface area of can: FIGURE 4.37 This oneliter can uses the least material when h = 2r (Example 2).
1 liter = 1000 cm3
A = ()* 2pr 2 + 2prh ()*
circular cylindrical ends wall
How can we interpret the phrase “least material”? For a first approximation we can ignore the thickness of the material and the waste in manufacturing. Then we ask for dimensions r and h that make the total surface area as small as possible while satisfying the constraint pr 2h = 1000. To express the surface area as a function of one variable, we solve for one of the variables in pr 2h = 1000 and substitute that expression into the surface area formula. Solving for h is easier: h =
1000 . pr 2
Thus, A = 2pr 2 + 2prh = 2pr 2 + 2pr a = 2pr 2 +
1000 b pr 2
2000 r .
Our goal is to find a value of r 7 0 that minimizes the value of A. Figure 4.38 suggests that such a value exists.
A Tall and thin can Short and wide can —— , r . 0 A 5 2pr 2 1 2000 r
Tall and thin
min
0 Short and wide
r 3
500 p
FIGURE 4.38 The graph of A = 2pr 2 + 2000>r is concave up.
Notice from the graph that for small r (a tall, thin cylindrical container), the term 2000>r dominates (see Section 2.6) and A is large. For large r (a short, wide cylindrical container), the term 2pr 2 dominates and A again is large.
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257
Since A is differentiable on r 7 0, an interval with no endpoints, it can have a minimum value only where its first derivative is zero. 2000 dA = 4pr dr r2 2000 0 = 4pr r2 4pr 3 = 2000 r =
Set dA>dr = 0 . Multiply by r 2.
500 L 5.42 A p 3
Solve for r.
3 What happens at r = 2 500>p? The second derivative
4000 d 2A = 4p + dr 2 r3 is positive throughout the domain of A. The graph is therefore everywhere concave up and 3 500>p is an absolute minimum. the value of A at r = 2 The corresponding value of h (after a little algebra) is h =
1000 500 = 2 3 p = 2r. A pr 2
The oneliter can that uses the least material has height equal to twice the radius, here with r L 5.42 cm and h L 10.84 cm.
Examples from Mathematics and Physics EXAMPLE 3 A rectangle is to be inscribed in a semicircle of radius 2. What is the largest area the rectangle can have, and what are its dimensions?
y x 2 1 y2 5 4
Let sx, 24  x 2 d be the coordinates of the corner of the rectangle obtained by placing the circle and rectangle in the coordinate plane (Figure 4.39). The length, height, and area of the rectangle can then be expressed in terms of the position x of the lower righthand corner:
⎛x, 4 2 x 2⎛ Solution ⎝ ⎝ 2 –2 –x
0
x 2
FIGURE 4.39 The rectangle inscribed in the semicircle in Example 3.
x
Length: 2x,
Height: 24  x 2,
Area: 2x24  x 2 .
Notice that the values of x are to be found in the interval 0 … x … 2, where the selected corner of the rectangle lies. Our goal is to find the absolute maximum value of the function Asxd = 2x24  x 2 on the domain [0, 2]. The derivative dA 2x 2 = + 224  x 2 2 dx 24  x is not defined when x = 2 and is equal to zero when 2x 2
+ 224  x 2 = 0 24  x 2 2x 2 + 2s4  x 2 d = 0 8  4x 2 = 0 x 2 = 2 or x = ; 22.
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Chapter 4: Applications of Derivatives
Of the two zeros, x = 22 and x =  22, only x = 22 lies in the interior of A’s domain and makes the criticalpoint list. The values of A at the endpoints and at this one critical point are Criticalpoint value: Endpoint values:
A A 22 B = 22224  2 = 4 As0d = 0,
As2d = 0.
The area has a maximum value of 4 when the rectangle is 24  x 2 = 22 units high and 2x = 222 units long.
EXAMPLE 4
The speed of light depends on the medium through which it travels, and is generally slower in denser media. Fermat’s principle in optics states that light travels from one point to another along a path for which the time of travel is a minimum. Describe the path that a ray of light will follow in going from a point A in a medium where the speed of light is c1 to a point B in a second medium where its speed is c2 .
HISTORICAL BIOGRAPHY Willebrord Snell van Royen (1580–1626)
y A a
u1
Angle of incidence u1 P
0
x Medium 2
Since light traveling from A to B follows the quickest route, we look for a path that will minimize the travel time. We assume that A and B lie in the xyplane and that the line separating the two media is the xaxis (Figure 4.40). In a uniform medium, where the speed of light remains constant, “shortest time” means “shortest path,” and the ray of light will follow a straight line. Thus the path from A to B will consist of a line segment from A to a boundary point P, followed by another line segment from P to B. Distance traveled equals rate times time, so Solution
Medium 1
b
u2 d2x
d Angle of refraction
x
B
Time =
FIGURE 4.40 A light ray refracted (deflected from its path) as it passes from one medium to a denser medium (Example 4).
distance rate .
From Figure 4.40, the time required for light to travel from A to P is 2a 2 + x 2 AP . t1 = c1 = c1 From P to B, the time is 2b 2 + sd  xd2 PB t2 = c2 = . c2 The time from A to B is the sum of these: t = t1 + t2 =
2b 2 + sd  xd2 2a 2 + x 2 + . c1 c2
This equation expresses t as a differentiable function of x whose domain is [0, d]. We want to find the absolute minimum value of t on this closed interval. We find the derivative dt x d  x = 2 2 2 dx c1 2a + x c2 2b + sd  xd2 dt/dx negative
0
dt/dx zero
x0
and observe that it is continuous. In terms of the angles u1 and u2 in Figure 4.40,
dt/dx positive x d
FIGURE 4.41 The sign pattern of dt>dx in Example 4.
sin u1 sin u2 dt = c1  c2 . dx The function t has a negative derivative at x = 0 and a positive derivative at x = d. Since dt>dx is continuous over the interval [0, d], by the Intermediate Value Theorem for continuous functions (Section 2.5), there is a point x0 H [0, d ] where dt>dx = 0 (Figure 4.41).
4.6
Applied Optimization
259
There is only one such point because dt>dx is an increasing function of x (Exercise 62). At this unique point we then have sin u1 sin u2 c1 = c2 . This equation is Snell’s Law or the Law of Refraction, and is an important principle in the theory of optics. It describes the path the ray of light follows.
Examples from Economics Suppose that rsxd = the revenue from selling x items csxd = the cost of producing the x items psxd = rsxd  csxd = the profit from producing and selling x items. Although x is usually an integer in many applications, we can learn about the behavior of these functions by defining them for all nonzero real numbers and by assuming they are differentiable functions. Economists use the terms marginal revenue, marginal cost, and marginal profit to name the derivatives r¿(x), c¿(x), and p¿(x) of the revenue, cost, and profit functions. Let’s consider the relationship of the profit p to these derivatives. If r(x) and c(x) are differentiable for x in some interval of production possibilities, and if psxd = rsxd  csxd has a maximum value there, it occurs at a critical point of p(x) or at an endpoint of the interval. If it occurs at a critical point, then p¿sxd = r¿sxd c¿sxd = 0 and we see that r¿(x) = c¿(x). In economic terms, this last equation means that
At a production level yielding maximum profit, marginal revenue equals marginal cost (Figure 4.42).
y
Dollars
Cost c(x)
Revenue r(x) Breakeven point B
0
Maximum profit, c'(x) r'(x)
Local maximum for loss (minimum profit), c'(x) r'(x) x Items produced
FIGURE 4.42 The graph of a typical cost function starts concave down and later turns concave up. It crosses the revenue curve at the breakeven point B. To the left of B, the company operates at a loss. To the right, the company operates at a profit, with the maximum profit occurring where c¿sxd = r¿sxd . Farther to the right, cost exceeds revenue (perhaps because of a combination of rising labor and material costs and market saturation) and production levels become unprofitable again.
260
Chapter 4: Applications of Derivatives
Suppose that rsxd = 9x and csxd = x 3  6x 2 + 15x, where x represents millions of MP3 players produced. Is there a production level that maximizes profit? If so, what is it?
EXAMPLE 5
y
Notice that r¿sxd = 9 and c¿sxd = 3x 2  12x + 15.
Solution
3x 2  12x + 15 = 9 3x 2  12x + 6 = 0
c(x) x 3 6x2 15x
Set c¿sxd = r¿sxd.
The two solutions of the quadratic equation are r(x) 9x Maximum for profit
Local maximum for loss 0 2 2
x
2 2
2
NOT TO SCALE
FIGURE 4.43 The cost and revenue curves for Example 5.
x1 =
12  272 = 2  22 L 0.586 6
x2 =
12 + 272 = 2 + 22 L 3.414. 6
and
The possible production levels for maximum profit are x L 0.586 million MP3 players or x L 3.414 million. The second derivative of psxd = rsxd  csxd is p–sxd = c–sxd since r–sxd is everywhere zero. Thus, p–(x) = 6(2  x), which is negative at x = 2 + 22 and positive at x = 2  22. By the Second Derivative Test, a maximum profit occurs at about x = 3.414 (where revenue exceeds costs) and maximum loss occurs at about x = 0.586. The graphs of r(x) and c(x) are shown in Figure 4.43.
Exercises 4.6 Mathematical Applications Whenever you are maximizing or minimizing a function of a single variable, we urge you to graph it over the domain that is appropriate to the problem you are solving. The graph will provide insight before you calculate and will furnish a visual context for understanding your answer. 1. Minimizing perimeter What is the smallest perimeter possible for a rectangle whose area is 16 in2 , and what are its dimensions? 2. Show that among all rectangles with an 8m perimeter, the one with largest area is a square. 3. The figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units long. a. Express the ycoordinate of P in terms of x. (Hint: Write an equation for the line AB.) b. Express the area of the rectangle in terms of x. c. What is the largest area the rectangle can have, and what are its dimensions?
B
P(x, ?)
A 0
5. You are planning to make an open rectangular box from an 8in.by15in. piece of cardboard by cutting congruent squares from the corners and folding up the sides. What are the dimensions of the box of largest volume you can make this way, and what is its volume? 6. You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a, 0) to (0, b). Show that the area of the triangle enclosed by the segment is largest when a = b. 7. The best fencing plan A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a singlestrand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? 8. The shortest fence A 216 m2 rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed?
y
–1
4. A rectangle has its base on the xaxis and its upper two vertices on the parabola y = 12  x 2 . What is the largest area the rectangle can have, and what are its dimensions?
x
1
x
9. Designing a tank Your iron works has contracted to design and build a 500 ft3 , squarebased, opentop, rectangular steel holding tank for a paper company. The tank is to be made by welding thin stainless steel plates together along their edges. As the production engineer, your job is to find dimensions for the base and height that will make the tank weigh as little as possible.
4.6 a. What dimensions do you tell the shop to use?
x NOT TO SCALE
b. Briefly describe how you took weight into account. 3
10. Catching rainwater A 1125 ft opentop rectangular tank with a square base x ft on a side and y ft deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy.
261
Applied Optimization
x
x
x
10"
Base
Lid
x
x x
x 15"
a. If the total cost is c = 5sx 2 + 4xyd + 10xy, what values of x and y will minimize it? b. Give a possible scenario for the cost function in part (a). 11. Designing a poster You are designing a rectangular poster to contain 50 in2 of printing with a 4in. margin at the top and bottom and a 2in. margin at each side. What overall dimensions will minimize the amount of paper used? 12. Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3.
a. Write a formula V(x) for the volume of the box. b. Find the domain of V for the problem situation and graph V over this domain. c. Use a graphical method to find the maximum volume and the value of x that gives it. d. Confirm your result in part (c) analytically. T 17. Designing a suitcase A 24in.by36in. sheet of cardboard is folded in half to form a 24in.by18in. rectangle as shown in the accompanying figure. Then four congruent squares of side length x are cut from the corners of the folded rectangle. The sheet is unfolded, and the six tabs are folded up to form a box with sides and a lid. a. Write a formula V(x) for the volume of the box. b. Find the domain of V for the problem situation and graph V over this domain.
3
3
y
c. Use a graphical method to find the maximum volume and the value of x that gives it. d. Confirm your result in part (c) analytically.
x
e. Find a value of x that yields a volume of 1120 in3 . f. Write a paragraph describing the issues that arise in part (b).
13. Two sides of a triangle have lengths a and b, and the angle between them is u . What value of u will maximize the triangle’s area? (Hint: A = s1>2dab sin u .) 14. Designing a can What are the dimensions of the lightest opentop right circular cylindrical can that will hold a volume of 1000 cm3 ? Compare the result here with the result in Example 2. 15. Designing a can You are designing a 1000 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up by the can will therefore be A = 8r 2 + 2prh rather than the A = 2pr 2 + 2prh in Example 2. In Example 2, the ratio of h to r for the most economical can was 2 to 1. What is the ratio now? T 16. Designing a box with a lid A piece of cardboard measures 10 in. by 15 in. Two equal squares are removed from the corners of a 10in. side as shown in the figure. Two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular box with lid.
x
x
x
x
24"
24" x
x x
36"
x 18"
The sheet is then unfolded.
24"
Base
36"
18. A rectangle is to be inscribed under the arch of the curve y = 4 cos s0.5xd from x = p to x = p . What are the dimensions of the rectangle with largest area, and what is the largest area?
262
Chapter 4: Applications of Derivatives
19. Find the dimensions of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius 10 cm. What is the maximum volume?
cylindrical sidewall. Determine the dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction.
20. a. The U.S. Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 108 in. What dimensions will give a box with a square end the largest possible volume?
24. The trough in the figure is to be made to the dimensions shown. Only the angle u can be varied. What value of u will maximize the trough’s volume?
Girth distance around here
1'
1' 1'
Length
Square end
T b. Graph the volume of a 108in. box (length plus girth equals 108 in.) as a function of its length and compare what you see with your answer in part (a). 21. (Continuation of Exercise 20.)
20'
25. Paper folding A rectangular sheet of 8.5in.by11in. paper is placed on a flat surface. One of the corners is placed on the opposite longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L. Try it with paper. a. Show that L 2 = 2x 3>s2x  8.5d .
b. What value of x minimizes L 2 ?
a. Suppose that instead of having a box with square ends you have a box with square sides so that its dimensions are h by h by w and the girth is 2h + 2w . What dimensions will give the box its largest volume now?
c. What is the minimum value of L? D
C
R
Girth
L2 x 2
L
Crease
Q (originally at A)
h
x x A
w h
T b. Graph the volume as a function of h and compare what you see with your answer in part (a). 22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame.
P
B
26. Constructing cylinders Compare the answers to the following two construction problems. a. A rectangular sheet of perimeter 36 cm and dimensions x cm by y cm is to be rolled into a cylinder as shown in part (a) of the figure. What values of x and y give the largest volume? b. The same sheet is to be revolved about one of the sides of length y to sweep out the cylinder as shown in part (b) of the figure. What values of x and y give the largest volume?
y x Circumference 5 x
x y
y
23. A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit of surface area is twice as great for the hemisphere as it is for the
(a)
(b)
4.6 27. Constructing cones A right triangle whose hypotenuse is 23 m long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made this way.
h
3
Applied Optimization
263
39. Shortest beam The 8ft wall shown here stands 27 ft from the building. Find the length of the shortest straight beam that will reach to the side of the building from the ground outside the wall.
Beam
Building
r
y x 28. Find the point on the line a + = 1 that is closest to the origin. b 29. Find a positive number for which the sum of it and its reciprocal is the smallest (least) possible. 30. Find a postitive number for which the sum of its reciprocal and four times its square is the smallest possible. 31. A wire b m long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle. If the sum of the areas enclosed by each part is a minimum, what is the length of each part? 32. Answer Exercise 31 if one piece is bent into a square and the other into a circle. 33. Determine the dimensions of the rectangle of largest area that can be inscribed in the right triangle shown in the accompanying figure.
w
5
4
h 3
34. Determine the dimensions of the rectangle of largest area that can be inscribed in a semicircle of radius 3. (See accompanying figure.)
w
8' wall 27'
40. Motion on a line The positions of two particles on the saxis are s1 = sin t and s2 = sin st + p>3d , with s1 and s2 in meters and t in seconds. a. At what time(s) in the interval 0 … t … 2p do the particles meet? b. What is the farthest apart that the particles ever get? c. When in the interval 0 … t … 2p is the distance between the particles changing the fastest? 41. The intensity of illumination at any point from a light source is proportional to the square of the reciprocal of the distance between the point and the light source. Two lights, one having an intensity eight times that of the other, are 6 m apart. How far from the stronger light is the total illumination least? 42. Projectile motion The range R of a projectile fired from the origin over horizontal ground is the distance from the origin to the point of impact. If the projectile is fired with an initial velocity y0 at an angle a with the horizontal, then in Chapter 12 we find that y02 R = g sin 2a,
h r5 3
35. What value of a makes ƒsxd = x 2 + sa>xd have a. a local minimum at x = 2 ? b. a point of inflection at x = 1 ? 36. What values of a and b make ƒsxd = x 3 + ax 2 + bx have a. a local maximum at x = 1 and a local minimum at x = 3 ? b. a local minimum at x = 4 and a point of inflection at x = 1 ? Physical Applications 37. Vertical motion The height above ground of an object moving vertically is given by s = 16t 2 + 96t + 112 ,
where g is the downward acceleration due to gravity. Find the angle a for which the range R is the largest possible. T 43. Strength of a beam The strength S of a rectangular wooden beam is proportional to its width times the square of its depth. (See the accompanying figure.) a. Find the dimensions of the strongest beam that can be cut from a 12in.diameter cylindrical log. b. Graph S as a function of the beam’s width w, assuming the proportionality constant to be k = 1. Reconcile what you see with your answer in part (a). c. On the same screen, graph S as a function of the beam’s depth d, again taking k = 1 . Compare the graphs with one another and with your answer in part (a). What would be the effect of changing to some other value of k? Try it.
with s in feet and t in seconds. Find a. the object’s velocity when t = 0; b. its maximum height and when it occurs; c. its velocity when s = 0 . 38. Quickest route Jane is 2 mi offshore in a boat and wishes to reach a coastal village 6 mi down a straight shoreline from the point nearest the boat. She can row 2 mph and can walk 5 mph. Where should she land her boat to reach the village in the least amount of time?
12" d
w
264
Chapter 4: Applications of Derivatives
T 44. Stiffness of a beam The stiffness S of a rectangular beam is proportional to its width times the cube of its depth. a. Find the dimensions of the stiffest beam that can be cut from a 12in.diameter cylindrical log. b. Graph S as a function of the beam’s width w, assuming the proportionality constant to be k = 1 . Reconcile what you see with your answer in part (a). c. On the same screen, graph S as a function of the beam’s depth d, again taking k = 1 . Compare the graphs with one another and with your answer in part (a). What would be the effect of changing to some other value of k? Try it. 45. Frictionless cart A small frictionless cart, attached to the wall by a spring, is pulled 10 cm from its rest position and released at time t = 0 to roll back and forth for 4 sec. Its position at time t is s = 10 cos pt . a. What is the cart’s maximum speed? When is the cart moving that fast? Where is it then? What is the magnitude of the acceleration then?
c. The visibility that day was 5 nautical miles. Did the ships ever sight each other? T d. Graph s and ds>dt together as functions of t for 1 … t … 3 , using different colors if possible. Compare the graphs and reconcile what you see with your answers in parts (b) and (c). e. The graph of ds>dt looks as if it might have a horizontal asymptote in the first quadrant. This in turn suggests that ds>dt approaches a limiting value as t : q . What is this value? What is its relation to the ships’ individual speeds? 48. Fermat’s principle in optics Light from a source A is reflected by a plane mirror to a receiver at point B, as shown in the accompanying figure. Show that for the light to obey Fermat’s principle, the angle of incidence must equal the angle of reflection, both measured from the line normal to the reflecting surface. (This result can also be derived without calculus. There is a purely geometric argument, which you may prefer.) Normal Light receiver
b. Where is the cart when the magnitude of the acceleration is greatest? What is the cart’s speed then?
0
10
Light source A
a. At what times in the interval 0 6 t do the masses pass each other? (Hint: sin 2t = 2 sin t cos t .) b. When in the interval 0 … t … 2p is the vertical distance between the masses the greatest? What is this distance? (Hint: cos 2t = 2 cos2 t  1.)
s1 0 s2
m2
1
Angle of reflection 2
B
Plane mirror
s
46. Two masses hanging side by side from springs have positions s1 = 2 sin t and s2 = sin 2t , respectively.
m1
Angle of incidence
49. Tin pest When metallic tin is kept below 13.2°C, it slowly becomes brittle and crumbles to a gray powder. Tin objects eventually crumble to this gray powder spontaneously if kept in a cold climate for years. The Europeans who saw tin organ pipes in their churches crumble away years ago called the change tin pest because it seemed to be contagious, and indeed it was, for the gray powder is a catalyst for its own formation. A catalyst for a chemical reaction is a substance that controls the rate of reaction without undergoing any permanent change in itself. An autocatalytic reaction is one whose product is a catalyst for its own formation. Such a reaction may proceed slowly at first if the amount of catalyst present is small and slowly again at the end, when most of the original substance is used up. But in between, when both the substance and its catalyst product are abundant, the reaction proceeds at a faster pace. In some cases, it is reasonable to assume that the rate y = dx>dt of the reaction is proportional both to the amount of the original substance present and to the amount of product. That is, y may be considered to be a function of x alone, and y = k xsa  xd = kax  k x 2,
s
47. Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship B was sailing east at 8 knots and continued to do so all day. a. Start counting time with t = 0 at noon and express the distance s between the ships as a function of t. b. How rapidly was the distance between the ships changing at noon? One hour later?
where x = the amount of product a = the amount of substance at the beginning k = a positive constant . At what value of x does the rate y have a maximum? What is the maximum value of y? 50. Airplane landing path An airplane is flying at altitude H when it begins its descent to an airport runway that is at horizontal ground distance L from the airplane, as shown in the figure. Assume that the
4.6
Applied Optimization
265
landing path of the airplane is the graph of a cubic polynomial function y = ax 3 + bx 2 + cx + d, where y s Ld = H and y s0d = 0 .
54. Production level Prove that the production level (if any) at which average cost is smallest is a level at which the average cost equals marginal cost.
a. What is dy>dx at x = 0 ?
55. Show that if rsxd = 6x and csxd = x 3  6x 2 + 15x are your revenue and cost functions, then the best you can do is break even (have revenue equal cost).
b. What is dy>dx at x = L ? c. Use the values for dy>dx at x = 0 and x = L together with y s0d = 0 and y s Ld = H to show that 3
2
x x y sxd = H c2 a b + 3 a b d. L L
57. You are to construct an open rectangular box with a square base and a volume of 48 ft3. If material for the bottom costs $6>ft2 and material for the sides costs $4>ft2, what dimensions will result in the least expensive box? What is the minimum cost?
y
Landing path
56. Production level Suppose that csxd = x 3  20x 2 + 20,000x is the cost of manufacturing x items. Find a production level that will minimize the average cost of making x items.
58. The 800room Mega Motel chain is filled to capacity when the room charge is $50 per night. For each $10 increase in room charge, 40 fewer rooms are filled each night. What charge per room will result in the maximum revenue per night?
H = Cruising altitude Airport x L
Business and Economics 51. It costs you c dollars each to manufacture and distribute backpacks. If the backpacks sell at x dollars each, the number sold is given by
Biology 59. Sensitivity to medicine (Continuation of Exercise 72, Section 3.3.) Find the amount of medicine to which the body is most sensitive by finding the value of M that maximizes the derivative dR>dM , where R = M2 a
a n = x  c + bs100  xd , where a and b are positive constants. What selling price will bring a maximum profit? 52. You operate a tour service that offers the following rates: $200 per person if 50 people (the minimum number to book the tour) go on the tour. For each additional person, up to a maximum of 80 people total, the rate per person is reduced by $2. It costs $6000 (a fixed cost) plus $32 per person to conduct the tour. How many people does it take to maximize your profit? 53. Wilson lot size formula One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is hq km Asqd = q + cm + , 2 where q is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be), k is the cost of placing an order (the same, no matter how often you order), c is the cost of one item (a constant), m is the number of items sold each week (a constant), and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). a. Your job, as the inventory manager for your store, is to find the quantity that will minimize A(q). What is it? (The formula you get for the answer is called the Wilson lot size formula.) b. Shipping costs sometimes depend on order size. When they do, it is more realistic to replace k by k + bq , the sum of k and a constant multiple of q. What is the most economical quantity to order now?
C M  b 2 3
and C is a constant. 60. How we cough a. When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity y can be modeled by the equation y = csr0  rdr 2 cm>sec,
r0 … r … r0 , 2
where r0 is the rest radius of the trachea in centimeters and c is a positive constant whose value depends in part on the length of the trachea. Show that y is greatest when r = s2>3dr0; that is, when the trachea is about 33% contracted. The remarkable fact is that Xray photographs confirm that the trachea contracts about this much during a cough. T b. Take r0 to be 0.5 and c to be 1 and graph y over the interval 0 … r … 0.5 . Compare what you see with the claim that y is at a maximum when r = s2>3dr0 . Theory and Examples 61. An inequality for positive integers Show that if a, b, c, and d are positive integers, then sa 2 + 1dsb 2 + 1dsc 2 + 1dsd 2 + 1d Ú 16 . abcd
266
Chapter 4: Applications of Derivatives a. Explain why you need to consider values of x only in the interval [0, 2p] .
62. The derivative dt>dx in Example 4 a. Show that ƒsxd =
b. Is ƒ ever negative? Explain.
x
65. a. The function y = cot x  22 csc x has an absolute maximum value on the interval 0 6 x 6 p . Find it.
2a 2 + x 2
is an increasing function of x.
T b. Graph the function and compare what you see with your answer in part (a).
b. Show that g sxd =
66. a. The function y = tan x + 3 cot x has an absolute minimum value on the interval 0 6 x 6 p>2 . Find it.
d  x 2b + sd  xd 2
2
T b. Graph the function and compare what you see with your answer in part (a).
is a decreasing function of x. c. Show that dt d  x x = 2 2 2 dx c1 2a + x c2 2b + sd  xd2 is an increasing function of x.
67. a. How close does the curve y = 2x come to the point (3>2, 0)? (Hint: If you minimize the square of the distance, you can avoid square roots.) T b. Graph the distance function D(x) and y = 2x together and reconcile what you see with your answer in part (a).
63. Let ƒ(x) and g(x) be the differentiable functions graphed here. Point c is the point where the vertical distance between the curves is the greatest. Is there anything special about the tangents to the two curves at c? Give reasons for your answer.
y
(x, x)
y x
y f (x) y g(x)
a
c
b
0
x
64. You have been asked to determine whether the function ƒsxd = 3 + 4 cos x + cos 2x is ever negative.
4.7
⎛ 3 , 0⎛ ⎝2 ⎝
x
68. a. How close does the semicircle y = 216  x 2 come to the point A 1, 23 B ? T b. Graph the distance function and y = 216  x 2 together and reconcile what you see with your answer in part (a).
Newton’s Method In this section we study a numerical method, called Newton’s method or the Newton–Raphson method, which is a technique to approximate the solution to an equation ƒsxd = 0. Essentially it uses tangent lines in place of the graph of y = ƒsxd near the points where ƒ is zero. (A value of x where ƒ is zero is a root of the function ƒ and a solution of the equation ƒsxd = 0.)
Procedure for Newton’s Method The goal of Newton’s method for estimating a solution of an equation ƒsxd = 0 is to produce a sequence of approximations that approach the solution. We pick the first number x0 of the sequence. Then, under favorable circumstances, the method does the rest by moving step by step toward a point where the graph of ƒ crosses the xaxis (Figure 4.44). At each step the method approximates a zero of ƒ with a zero of one of its linearizations. Here is how it works. The initial estimate, x0 , may be found by graphing or just plain guessing. The method then uses the tangent to the curve y = ƒsxd at sx0, ƒsx0 dd to approximate the curve, calling
4.7
y f (x) (x0, f (x0))
y = ƒsxn d + ƒ¿sxn dsx  xn d.
(x1, f (x1))
We can find where it crosses the xaxis by setting y = 0 (Figure 4.45):
(x2, f(x2 ))
0 = ƒsxn d + ƒ¿sxn dsx  xn d
Root sought

x x3 Fourth
x2 Third
x1 Second
x0 First
ƒsxn d = x  xn ƒ¿sxn d
APPROXIMATIONS
x = xn 
FIGURE 4.44 Newton’s method starts with an initial guess x0 and (under favorable circumstances) improves the guess one step at a time.
If ƒ¿sxn d Z 0
Newton’s Method 1. Guess a first approximation to a solution of the equation ƒsxd = 0. A graph of y = ƒsxd may help. 2. Use the first approximation to get a second, the second to get a third, and so on, using the formula
y f (x) Point: (xn, f (xn )) Slope: f'(xn ) Tangent line equation: y f (xn ) f '(xn )(x xn ) (xn, f (xn))
ƒsxn d ƒ¿sxn d
This value of x is the next approximation xn + 1 . Here is a summary of Newton’s method.
y
xn + 1 = xn 
Tangent line (graph of linearization of f at xn )
Root sought 0
267
the point x1 where the tangent meets the xaxis (Figure 4.44). The number x1 is usually a better approximation to the solution than is x0 . The point x2 where the tangent to the curve at sx1, ƒsx1 dd crosses the xaxis is the next approximation in the sequence. We continue on, using each approximation to generate the next, until we are close enough to the root to stop. We can derive a formula for generating the successive approximations in the following way. Given the approximation xn , the pointslope equation for the tangent to the curve at sxn, ƒsxn dd is
y
0
Newton’s Method
ƒsxn d , ƒ¿sxn d
if ƒ¿sxn d Z 0.
(1)
Applying Newton’s Method xn
f (xn ) xn1 xn f'(xn )
FIGURE 4.45 The geometry of the successive steps of Newton’s method. From xn we go up to the curve and follow the tangent line down to find xn + 1 .
x
Applications of Newton’s method generally involve many numerical computations, making them well suited for computers or calculators. Nevertheless, even when the calculations are done by hand (which may be very tedious), they give a powerful way to find solutions of equations. In our first example, we find decimal approximations to 22 by estimating the positive root of the equation ƒsxd = x 2  2 = 0.
EXAMPLE 1
Find the positive root of the equation ƒsxd = x 2  2 = 0.
Solution
With ƒsxd = x 2  2 and ƒ¿sxd = 2x, Equation (1) becomes xn + 1 = xn 
xn 2  2 2xn
= xn 
xn 1 + xn 2
=
xn 1 + xn . 2
268
Chapter 4: Applications of Derivatives y
The equation
20
xn + 1 =
y x3 x 1
xn 1 + xn 2
enables us to go from each approximation to the next with just a few keystrokes. With the starting value x0 = 1, we get the results in the first column of the following table. (To five decimal places, 22 = 1.41421.)
15 10
Error
Number of correct digits
0.41421 0.08579 0.00246 0.00001
1 1 3 5
5
–1
0
2
1
x
3
x0 x1 x2 x3
FIGURE 4.46 The graph of ƒsxd = x 3  x  1 crosses the xaxis once; this is the root we want to find (Example 2).
= = = =
1 1.5 1.41667 1.41422
Newton’s method is the method used by most calculators to calculate roots because it converges so fast (more about this later). If the arithmetic in the table in Example 1 had been carried to 13 decimal places instead of 5, then going one step further would have given 22 correctly to more than 10 decimal places.
y x3 x 1 (1.5, 0.875)
Find the xcoordinate of the point where the curve y = x 3  x crosses the horizontal line y = 1.
EXAMPLE 2
Root sought x1
x2
x0
x 1
1.5
The curve crosses the line when x 3  x = 1 or x 3  x  1 = 0. When does ƒsxd = x  x  1 equal zero? Since ƒs1d = 1 and ƒs2d = 5, we know by the Intermediate Value Theorem there is a root in the interval (1, 2) (Figure 4.46). We apply Newton’s method to ƒ with the starting value x0 = 1. The results are displayed in Table 4.1 and Figure 4.47. At n = 5, we come to the result x6 = x5 = 1.3247 17957. When xn + 1 = xn , Equation (1) shows that ƒsxn d = 0. We have found a solution of ƒsxd = 0 to nine decimals. Solution
3
1.3478 (1, –1)
FIGURE 4.47 The first three xvalues in Table 4.1 (four decimal places).
TABLE 4.1 The result of applying Newton’s method to ƒsxd = x3  x  1
with x0 = 1
y 25
n
xn
ƒ(xn)
ƒ(xn)
xn1 xn
0 1 2 3 4 5
1 1.5 1.3478 26087 1.3252 00399 1.3247 18174 1.3247 17957
1 0.875 0.1006 82173 0.0020 58362 0.0000 00924 1.8672E13
2 5.75 4.4499 05482 4.2684 68292 4.2646 34722 4.2646 32999
1.5 1.3478 26087 1.3252 00399 1.3247 18174 1.3247 17957 1.3247 17957
B0(3, 23) 20 y x3 x 1 15
10 B1(2.12, 6.35) 5 –13 –1
0
ƒsxn d ƒ¿sxn d
Root sought 13
x2 x1 1
1.6 2.12
x0
x
3
FIGURE 4.48 Any starting value x0 to the right of x = 1> 23 will lead to the root.
In Figure 4.48 we have indicated that the process in Example 2 might have started at the point B0s3, 23d on the curve, with x0 = 3. Point B0 is quite far from the xaxis, but the tangent at B0 crosses the xaxis at about (2.12, 0), so x1 is still an improvement over x0 . If we use Equation (1) repeatedly as before, with ƒsxd = x 3  x  1 and ƒ¿sxd = 3x 2  1, we obtain the nineplace solution x7 = x6 = 1.3247 17957 in seven steps.
4.7
Newton’s Method
269
Convergence of the Approximations
y y f (x)
r 0
x0
x1
x
In Chapter 9 we define precisely the idea of convergence for the approximations xn in Newton’s method. Intuitively, we mean that as the number n of approximations increases without bound, the values xn get arbitrarily close to the desired root r. (This notion is similar to the idea of the limit of a function g(t) as t approaches infinity, as defined in Section 2.6.) In practice, Newton’s method usually gives convergence with impressive speed, but this is not guaranteed. One way to test convergence is to begin by graphing the function to estimate a good starting value for x0 . You can test that you are getting closer to a zero of the function by evaluating ƒ ƒsxn d ƒ , and check that the approximations are converging by evaluating ƒ xn  xn + 1 ƒ . Newton’s method does not always converge. For instance, if ƒsxd = e
FIGURE 4.49 Newton’s method fails to converge. You go from x0 to x1 and back to x0 , never getting any closer to r.
 2r  x, 2x  r,
x 6 r x Ú r,
the graph will be like the one in Figure 4.49. If we begin with x0 = r  h, we get x1 = r + h, and successive approximations go back and forth between these two values. No amount of iteration brings us closer to the root than our first guess. If Newton’s method does converge, it converges to a root. Be careful, however. There are situations in which the method appears to converge but there is no root there. Fortunately, such situations are rare. When Newton’s method converges to a root, it may not be the root you have in mind. Figure 4.50 shows two ways this can happen. y f (x) Starting point Root sought
FIGURE 4.50
x0
Root found
y f (x) x1
x
x2
x1
Root sought
Root found
x
x0
Starting point
If you start too far away, Newton’s method may miss the root you want.
Exercises 4.7 Root Finding 1. Use Newton’s method to estimate the solutions of the equation x 2 + x  1 = 0 . Start with x0 = 1 for the lefthand solution and with x0 = 1 for the solution on the right. Then, in each case, find x2 .
6. Use Newton’s method to find the negative fourth root of 2 by solving the equation x 4  2 = 0 . Start with x0 = 1 and find x2 .
2. Use Newton’s method to estimate the one real solution of x 3 + 3x + 1 = 0 . Start with x0 = 0 and then find x2 .
8. Estimating pi You plan to estimate p>2 to five decimal places by using Newton’s method to solve the equation cos x = 0 . Does it matter what your starting value is? Give reasons for your answer.
3. Use Newton’s method to estimate the two zeros of the function ƒsxd = x 4 + x  3 . Start with x0 = 1 for the lefthand zero and with x0 = 1 for the zero on the right. Then, in each case, find x2 . 4. Use Newton’s method to estimate the two zeros of the function ƒsxd = 2x  x 2 + 1 . Start with x0 = 0 for the lefthand zero and with x0 = 2 for the zero on the right. Then, in each case, find x2 . 5. Use Newton’s method to find the positive fourth root of 2 by solving the equation x 4  2 = 0 . Start with x0 = 1 and find x2 .
7. Guessing a root Suppose that your first guess is lucky, in the sense that x0 is a root of ƒsxd = 0 . Assuming that ƒ¿sx0 d is defined and not 0, what happens to x1 and later approximations?
Theory and Examples 9. Oscillation Show that if h 7 0 , applying Newton’s method to ƒsxd = •
2x,
x Ú 0
2 x, x 6 0
leads to x1 = h if x0 = h and to x1 = h if x0 = h . Draw a picture that shows what is going on.
270
Chapter 4: Applications of Derivatives
10. Approximations that get worse and worse Apply Newton’s method to ƒsxd = x 1>3 with x0 = 1 and calculate x1 , x2 , x3 , and x4 . Find a formula for ƒ xn ƒ . What happens to ƒ xn ƒ as n : q ? Draw a picture that shows what is going on.
23. Intersection of curves At 2 e x = x 2  x + 1?
what
value(s)
of
x
does
24. Intersection of curves At ln (1  x 2) = x  1?
what
value(s)
of
x
does
11. Explain why the following four statements ask for the same information: iii) Find the roots of ƒsxd = x 3  3x  1.
25. Use the Intermediate Value Theorem from Section 2.5 to show that ƒsxd = x 3 + 2x  4 has a root between x = 1 and x = 2. Then find the root to five decimal places.
iii) Find the xcoordinates of the intersections of the curve y = x 3 with the line y = 3x + 1.
26. Factoring a quartic Find the approximate values of r1 through r4 in the factorization
iii) Find the xcoordinates of the points where the curve y = x 3  3x crosses the horizontal line y = 1 .
8x 4  14x 3  9x 2 + 11x  1 = 8sx  r1 dsx  r2 dsx  r3 dsx  r4 d.
iv) Find the values of x where the derivative of g sxd = s1>4dx 4  s3>2dx 2  x + 5 equals zero. 12. Locating a planet To calculate a planet’s space coordinates, we have to solve equations like x = 1 + 0.5 sin x . Graphing the function ƒsxd = x  1  0.5 sin x suggests that the function has a root near x = 1.5 . Use one application of Newton’s method to improve this estimate. That is, start with x0 = 1.5 and find x1 . (The value of the root is 1.49870 to five decimal places.) Remember to use radians. T 13. Intersecting curves The curve y = tan x crosses the line y = 2x between x = 0 and x = p>2 . Use Newton’s method to find where. T 14. Real solutions of a quartic Use Newton’s method to find the two real solutions of the equation x 4  2x 3  x 2  2x + 2 = 0 . T 15. a. How many solutions does the equation sin 3x = 0.99  x 2 have? b. Use Newton’s method to find them.
y
y 8x 4 14x 3 9x 2 11x 1
2 –1
1
–2 –4 –6 –8 –10 –12
T 27. Converging to different zeros Use Newton’s method to find the zeros of ƒsxd = 4x 4  4x 2 using the given starting values. a. x0 = 2 and x0 = 0.8 , lying in A  q ,  22>2 B b. x0 = 0.5 and x0 = 0.25 , lying in A  221>7, 221>7 B c. x0 = 0.8 and x0 = 2 , lying in A 22>2, q B
d. x0 =  221>7 and x0 = 221>7
16. Intersection of curves a. Does cos 3x ever equal x? Give reasons for your answer. b. Use Newton’s method to find where. 17. Find the four real zeros of the function ƒsxd = 2x 4  4x 2 + 1. T 18. Estimating pi Estimate p to as many decimal places as your calculator will display by using Newton’s method to solve the equation tan x = 0 with x0 = 3. 19. Intersection of curves At what value(s) of x does cos x = 2x? 20. Intersection of curves At what value(s) of x does cos x = x?
28. The sonobuoy problem In submarine location problems, it is often necessary to find a submarine’s closest point of approach (CPA) to a sonobuoy (sound detector) in the water. Suppose that the submarine travels on the parabolic path y = x 2 and that the buoy is located at the point s2, 1>2d . a. Show that the value of x that minimizes the distance between the submarine and the buoy is a solution of the equation x = 1>sx 2 + 1d . b. Solve the equation x = 1>sx 2 + 1d with Newton’s method. y
21. The graphs of y = x 2(x + 1) and y = 1>x (x 7 0) intersect at one point x = r. Use Newton’s method to estimate the value of r to four decimal places. y
x
2
y 5 x 2(x 1 1)
y x2 Submarine track in two dimensions
1 CPA
3
1 –1
0
0
⎛r, 1 ⎛ ⎝ r⎝ y 5 1x
2
1
2
1
2
x
1 Sonobuoy ⎛⎝2, – ⎛⎝ 2 x
22. The graphs of y = 2x and y = 3  x 2 intersect at one point x = r. Use Newton’s method to estimate the value of r to four decimal places.
T 29. Curves that are nearly flat at the root Some curves are so flat that, in practice, Newton’s method stops too far from the root to give a useful estimate. Try Newton’s method on ƒsxd = sx  1d40 with a starting value of x0 = 2 to see how close your machine comes to the root x = 1 . See the accompanying graph.
4.8
Antiderivatives
30. The accompanying figure shows a circle of radius r with a chord of length 2 and an arc s of length 3. Use Newton’s method to solve for r and u (radians) to four decimal places. Assume 0 6 u 6 p.
y
s5 3
r u y (x 1) 40 Slope 40
1
(2, 1)
Nearly flat
4.8
2
r
Slope –40
0
271
1
x
2
Antiderivatives We have studied how to find the derivative of a function. However, many problems require that we recover a function from its known derivative (from its known rate of change). For instance, the laws of physics tell us the acceleration of an object falling from an initial height and we can use this to compute its velocity and its height at any time. More generally, starting with a function ƒ, we want to find a function F whose derivative is ƒ. If such a function F exists, it is called an antiderivative of ƒ. We will see in the next chapter that antiderivatives are the link connecting the two major elements of calculus: derivatives and definite integrals.
Finding Antiderivatives DEFINITION A function F is an antiderivative of ƒ on an interval I if F¿sxd = ƒsxd for all x in I.
The process of recovering a function F(x) from its derivative ƒ(x) is called antidifferentiation. We use capital letters such as F to represent an antiderivative of a function ƒ, G to represent an antiderivative of g, and so forth.
EXAMPLE 1
Find an antiderivative for each of the following functions. 1 (a) ƒsxd = 2x (b) gsxd = cos x (c) h(x) = x + 2e 2x We need to think backward here: What function do we know has a derivative equal to the given function?
Solution
(a) Fsxd = x 2
(b) Gsxd = sin x
(c) H(x) = ln ƒ x ƒ + e 2x
Each answer can be checked by differentiating. The derivative of Fsxd = x 2 is 2x. The derivative of Gsxd = sin x is cos x, and the derivative of H(x) = ln ƒ x ƒ + e 2x is (1>x) + 2e 2x.
272
Chapter 4: Applications of Derivatives
The function Fsxd = x 2 is not the only function whose derivative is 2x. The function x + 1 has the same derivative. So does x 2 + C for any constant C. Are there others? Corollary 2 of the Mean Value Theorem in Section 4.2 gives the answer: Any two antiderivatives of a function differ by a constant. So the functions x 2 + C, where C is an arbitrary constant, form all the antiderivatives of ƒsxd = 2x. More generally, we have the following result. 2
THEOREM 8 If F is an antiderivative of ƒ on an interval I, then the most general antiderivative of ƒ on I is Fsxd + C where C is an arbitrary constant.
Thus the most general antiderivative of ƒ on I is a family of functions Fsxd + C whose graphs are vertical translations of one another. We can select a particular antiderivative from this family by assigning a specific value to C. Here is an example showing how such an assignment might be made.
EXAMPLE 2
y
y
x3
C 2 1
C2 C1 C0 C –1 C –2 x
0 –1
Solution
Find an antiderivative of ƒsxd = 3x 2 that satisfies Fs1d = 1.
Since the derivative of x 3 is 3x 2, the general antiderivative Fsxd = x 3 + C
gives all the antiderivatives of ƒ(x). The condition Fs1d = 1 determines a specific value for C. Substituting x = 1 into Fsxd = x 3 + C gives Fs1d = (1) 3 + C = 1 + C.
(1, –1)
–2
Since Fs1d = 1, solving 1 + C = 1 for C gives C = 2. So Fsxd = x 3  2
FIGURE 4.51 The curves y = x 3 + C fill the coordinate plane without overlapping. In Example 2, we identify the curve y = x 3  2 as the one that passes through the given point s1,  1d .
is the antiderivative satisfying Fs1d = 1. Notice that this assignment for C selects the particular curve from the family of curves y = x 3 + C that passes through the point (1, 1) in the plane (Figure 4.51). By working backward from assorted differentiation rules, we can derive formulas and rules for antiderivatives. In each case there is an arbitrary constant C in the general expression representing all antiderivatives of a given function. Table 4.2 gives antiderivative formulas for a number of important functions. The rules in Table 4.2 are easily verified by differentiating the general antiderivative formula to obtain the function to its left. For example, the derivative of (tan kx)>k + C is sec2 kx, whatever the value of the constants C or k Z 0, and this establishes Formula 4 for the most general antiderivative of sec2 kx.
EXAMPLE 3
Find the general antiderivative of each of the following functions.
(a) ƒsxd = x 5 (d) isxd = cos
1 2x (e) j(x) = e 3x
(b) gsxd =
x 2
(c) hsxd = sin 2x (f ) k(x) = 2x
4.8
Antiderivatives
273
TABLE 4.2 Antiderivative formulas, k a nonzero constant
Function
General antiderivative
1.
xn
1 x n + 1 + C, n + 1
2.
sin kx
3.
cos kx
4.
sec2 kx
5.
csc2 kx
1  cos kx + C k 1 sin kx + C k 1 tan kx + C k 1  cot kx + C k
6.
sec kx tan kx
7.
csc kx cot kx
Function
n Z 1
8.
e kx
1 kx e + C k
9.
1 x
ln ƒ x ƒ + C,
1 21  k 2x 2
1 1 sin kx + C k
1 1 + k 2x 2 1 x2k 2x 2  1
1 tan1 kx + C k
10. 11. 12.
1 sec kx + C k 1  csc kx + C k
General antiderivative
13.
sec1 kx + C, kx 7 1 a
a kx
x Z 0
1 b a kx + C, a 7 0, a Z 1 k ln a
In each case, we can use one of the formulas listed in Table 4.2.
Solution (a) Fsxd =
x6 + C 6
Formula 1 with n = 5
(b) gsxd = x 1>2 , so
Gsxd = (c) Hsxd = (d) Isxd =
x 1>2 + C = 22x + C 1>2
Formula 1 with n = 1>2
cos 2x + C 2
Formula 2 with k = 2
sin sx>2d x + C = 2 sin + C 2 1>2
(e) J(x) = 
1 3x + C e 3
(f) K(x) = a
Formula 3 with k = 1>2 Formula 8 with k =  3
1 b 2x + C ln 2
Formula 13 with a = 2, k = 1
Other derivative rules also lead to corresponding antiderivative rules. We can add and subtract antiderivatives and multiply them by constants. TABLE 4.3 Antiderivative linearity rules
1. 2. 3.
Constant Multiple Rule: Negative Rule: Sum or Difference Rule:
Function
General antiderivative
kƒ(x) ƒsxd ƒsxd ; gsxd
kFsxd + C, k a constant Fsxd + C Fsxd ; Gsxd + C
The formulas in Table 4.3 are easily proved by differentiating the antiderivatives and verifying that the result agrees with the original function. Formula 2 is the special case k = 1 in Formula 1.
274
Chapter 4: Applications of Derivatives
EXAMPLE 4
Find the general antiderivative of ƒsxd =
3 2x
+ sin 2x.
Solution We have that ƒsxd = 3gsxd + hsxd for the functions g and h in Example 3. Since Gsxd = 22x is an antiderivative of g(x) from Example 3b, it follows from the Constant Multiple Rule for antiderivatives that 3Gsxd = 3 # 22x = 62x is an antiderivative of 3gsxd = 3> 2x. Likewise, from Example 3c we know that Hsxd = s 1>2d cos 2x is an antiderivative of hsxd = sin 2x. From the Sum Rule for antiderivatives, we then get that
Fsxd = 3Gsxd + Hsxd + C = 62x 
1 cos 2x + C 2
is the general antiderivative formula for ƒ(x), where C is an arbitrary constant.
Initial Value Problems and Differential Equations Antiderivatives play several important roles in mathematics and its applications. Methods and techniques for finding them are a major part of calculus, and we take up that study in Chapter 8. Finding an antiderivative for a function ƒ(x) is the same problem as finding a function y(x) that satisfies the equation dy = ƒsxd. dx This is called a differential equation, since it is an equation involving an unknown function y that is being differentiated. To solve it, we need a function y(x) that satisfies the equation. This function is found by taking the antiderivative of ƒ(x). We fix the arbitrary constant arising in the antidifferentiation process by specifying an initial condition ysx0 d = y0 . This condition means the function y(x) has the value y0 when x = x0 . The combination of a differential equation and an initial condition is called an initial value problem. Such problems play important roles in all branches of science. The most general antiderivative Fsxd + C (such as x 3 + C in Example 2) of the function ƒ(x) gives the general solution y = Fsxd + C of the differential equation dy>dx = ƒsxd. The general solution gives all the solutions of the equation (there are infinitely many, one for each value of C). We solve the differential equation by finding its general solution. We then solve the initial value problem by finding the particular solution that satisfies the initial condition ysx0 d = y0 . In Example 2, the function y = x 3  2 is the particular solution of the differential equation dy>dx = 3x 2 satisfying the initial condition y(1) = 1.
Antiderivatives and Motion We have seen that the derivative of the position function of an object gives its velocity, and the derivative of its velocity function gives its acceleration. If we know an object’s acceleration, then by finding an antiderivative we can recover the velocity, and from an antiderivative of the velocity we can recover its position function. This procedure was used as an application of Corollary 2 in Section 4.2. Now that we have a terminology and conceptual framework in terms of antiderivatives, we revisit the problem from the point of view of differential equations.
EXAMPLE 5
A hotair balloon ascending at the rate of 12 ft>sec is at a height 80 ft above the ground when a package is dropped. How long does it take the package to reach the ground?
4.8
Antiderivatives
275
Solution Let y(t) denote the velocity of the package at time t, and let s(t) denote its height above the ground. The acceleration of gravity near the surface of the earth is 32 ft>sec2. Assuming no other forces act on the dropped package, we have
dy = 32. dt
Negative because gravity acts in the direction of decreasing s
This leads to the following initial value problem (Figure 4.52):
s v(0) 5 12
dy = 32 dt ys0d = 12.
Differential equation: Initial condition:
Balloon initially rising
This is our mathematical model for the package’s motion. We solve the initial value problem to obtain the velocity of the package. 1.
Solve the differential equation: The general formula for an antiderivative of 32 is y = 32t + C.
dv –32 5 dt s(t)
2.
Having found the general solution of the differential equation, we use the initial condition to find the particular solution that solves our problem. Evaluate C: 12 = 32s0d + C
0
ground
Initial condition ys0d = 12
C = 12.
FIGURE 4.52 A package dropped from a rising hotair balloon (Example 5).
The solution of the initial value problem is y = 32t + 12. Since velocity is the derivative of height, and the height of the package is 80 ft at time t = 0 when it is dropped, we now have a second initial value problem. Differential equation: Initial condition:
ds = 32t + 12 dt ss0d = 80
Set y = ds>dt in the previous equation.
We solve this initial value problem to find the height as a function of t. 1.
Solve the differential equation: Finding the general antiderivative of 32t + 12 gives s = 16t 2 + 12t + C.
2.
Evaluate C: 80 = 16s0d2 + 12s0d + C C = 80.
Initial condition ss0d = 80
The package’s height above ground at time t is s = 16t 2 + 12t + 80. Use the solution: To find how long it takes the package to reach the ground, we set s equal to 0 and solve for t: 16t 2 + 12t + 80 = 0 4t 2 + 3t + 20 = 0 3 ; 2329 8 t L 1.89, t L 2.64. t =
Quadratic formula
The package hits the ground about 2.64 sec after it is dropped from the balloon. (The negative root has no physical meaning.)
Chapter 4: Applications of Derivatives
Indefinite Integrals A special symbol is used to denote the collection of all antiderivatives of a function ƒ.
DEFINITION The collection of all antiderivatives of ƒ is called the indefinite integral of ƒ with respect to x, and is denoted by L
ƒsxd dx.
The symbol 1 is an integral sign. The function ƒ is the integrand of the integral, and x is the variable of integration.
After the integral sign in the notation we just defined, the integrand function is always followed by a differential to indicate the variable of integration. We will have more to say about why this is important in Chapter 5. Using this notation, we restate the solutions of Example 1, as follows: 2x dx = x 2 + C,
L L
cos x dx = sin x + C, 1 a x + 2e 2x b dx = ln ƒ x ƒ + e 2x + C.
L
This notation is related to the main application of antiderivatives, which will be explored in Chapter 5. Antiderivatives play a key role in computing limits of certain infinite sums, an unexpected and wonderfully useful role that is described in a central result of Chapter 5, called the Fundamental Theorem of Calculus.
EXAMPLE 6
Evaluate L
sx 2  2x + 5d dx.
If we recognize that sx 3>3d  x 2 + 5x is an antiderivative of x 2  2x + 5, we can evaluate the integral as Solution
antiderivative $++%++&
L
(x 2  2x + 5) dx =
x3  x 2 + 5x + C. 3 #
276
arbitrary constant
If we do not recognize the antiderivative right away, we can generate it termbyterm with the Sum, Difference, and Constant Multiple Rules: L
sx 2  2x + 5d dx = =
L L
= a =
x 2 dx 
L
x 2 dx  2
2x dx +
L
L
x dx + 5
5 dx
L
1 dx
x2 x3 + C1 b  2 a + C2 b + 5sx + C3 d 3 2
x3 + C1  x 2  2C2 + 5x + 5C3 . 3
4.8
Antiderivatives
277
This formula is more complicated than it needs to be. If we combine C1, 2C2 , and 5C3 into a single arbitrary constant C = C1  2C2 + 5C3 , the formula simplifies to x3  x 2 + 5x + C 3 and still gives all the possible antiderivatives there are. For this reason, we recommend that you go right to the final form even if you elect to integrate termbyterm. Write L
sx 2  2x + 5d dx =
x 2 dx 2x dx + 5 dx L L L x3 =  x 2 + 5x + C. 3
Find the simplest antiderivative you can for each part and add the arbitrary constant of integration at the end.
Exercises 4.8 Finding Antiderivatives In Exercises 1–24, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.
18. a. sec x tan x
b. 4 sec 3x tan 3x
c. sec
19. a. e 3x
b. e x
c. e x>2
2x
1. a. 2x
b. x
c. x  2x + 1
20. a. e
2. a. 6x
b. x 7
c. x 7  6x + 8
21. a. 3x
b. 2x
3. a. 3x 4
b. x 4
c. x 4 + 2x + 3
22. a. x 23
b. x p
2
3
b.
x + x2 2
c. x 3 + x  1
b.
5 x2
c. 2 
2 x3
b.
1 2x 3
c. x 3 
7. a.
3 2x 2
b.
8. a.
4 3 2x 3
b.
9. a.
2 1>3 x 3
10. a.
1 1>2 x 2
4. a. 2x 3 5. a.
1 x2
6. a. 
1 2 2x 1
3 32 x 1 2>3 b. x 3
b. 
1 3>2 x 2
7 b. x
1 11. a. x 1 3x
5 x2 1 x3
c. 2x + 3 x + c. 2
1 2x 1
2 5x
27.
c. 
3 5>2 x 2
29. 31.
b. 3 sin x
14. a. p cos px
px p cos b. 2 2
px + p cos x c. cos 2
15. a. sec x
x 2 b. sec2 3 3
3x c. sec 2
16. a. csc2 x
b. 
17. a. csc x cot x
b.  csc 5x cot 5x
2
3x 3 csc2 2 2
25.
1 4>3 x 3
13. a. p sin px
b.
c. 1 +
2
c. 1  8 csc2 2x c. p csc
px px cot 2 2
b.
c. x 22  1 1 c. 1 + 4x 2
1 2(x 2 + 1)
c. p x  x 1
b. x 2 + 2x
Finding Indefinite Integrals In Exercises 25–70, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.
c. 
5 c. 1  x
2
21  x 2 x 1 24. a. x  a b 2
3 2 x
4 1  2 3x x c. sin px  3 sin 3x
12. a.
23. a.
b. e
c. e x>5 x 5 c. a b 3
4x>3
2
px px tan 2 2
33. 35. 37. 39. 41.
L L
sx + 1d dx
26.
a3t 2 +
28.
t b dt 2
s2x 3  5x + 7d dx
30.
1 1 a 2  x 2  b dx 3 L x
32.
L
L L L L L
x 1>3 dx
34.
3 x B dx A 2x + 2
36.
a8y 
38.
2 b dy y 1>4
2xs1  x 3 d dx
40.
t2t + 2t dt t2
42.
L
s5  6xd dx a
t2 + 4t 3 b dt L 2 L
s1  x 2  3x 5 d dx
2 1 a  3 + 2xb dx x L 5 L
x 5>4 dx
2x 2 b dx + 2 L 2x a
a
1 1  5>4 b dy 7 y L L L
x 3sx + 1d dx 4 + 2t dt t3
278 43. 45. 47. 49. 51. 53. 55. 56. 57. 59. 61. 63. 65.
67.
69.
L L L
Chapter 4: Applications of Derivatives
s 2 cos td dt
44.
u du 3
46.
s 3 csc2 xd dx
48.
7 sin
csc u cot u du 2 L L L L
50.
se 3x + 5e x d dx
52.
(e x + 4x ) dx
54.
L L L
s 5 sin td dt
78.
3 cos 5u du
79.
a
80.
sec2 x b dx 3
2 sec u tan u du L5 L L
s2e x  3e 2x d dx s1.3dx dx
s4 sec x tan x  2 sec2 xd dx
1 scsc2 x  csc x cot xd dx L2 58.
L
1 + cos 4t dt 2
60.
L
1 ax 
L
3x
23
5 b dx x2 + 1 dx
62. 64.
L L L L
s2 cos 2x  3 sin 3xd dx 1  cos 6t dt 2 a x
2 21  y 2
22  1

dx
cot2 x dx 68. s1  cot2 xd dx L L 2 2 (Hint: 1 + cot x = csc x) cos u stan u + sec ud du 70.
csc u du csc u  sin u L
L
s7x  2d3 dx =
s7x  2d4 + C 28
s3x + 5d1 s3x + 5d2 dx = + C 72. 3 L 73. 74.
L L
sec2 s5x  1d dx = csc2 a
1 tan s5x  1d + C 5
x  1 x  1 b dx = 3 cot a b + C 3 3
dx L 2a  x 2
1 dx = ln (x + 1) + C, Lx + 1
2
x = sin1 a a b + C
tan1 x tan1 x 1 dx = ln x  ln s1 + x 2 d + C x 2 2 L x
ssin1 xd2 dx = xssin1 xd2  2x + 221  x 2 sin1 x + C L 83. Right, or wrong? Say which for each formula and give a brief reason for each answer. 82.
L L
x sin x dx =
x2 sin x + C 2
x sin x dx = x cos x + C
x sin x dx = x cos x + sin x + C L 84. Right, or wrong? Say which for each formula and give a brief reason for each answer. c.
a. b.
L L
c.
tan u sec2 u du =
sec3 u + C 3
tan u sec2 u du =
1 2 tan u + C 2
tan u sec2 u du =
1 sec2 u + C 2
L 85. Right, or wrong? Say which for each formula and give a brief reason for each answer. s2x + 1d3 s2x + 1d2 dx = + C a. 3 L b.
L
3s2x + 1d2 dx = s2x + 1d3 + C
6s2x + 1d2 dx = s2x + 1d3 + C L 86. Right, or wrong? Say which for each formula and give a brief reason for each answer. c.
a. b.
L L
22x + 1 dx = 2x 2 + x + C 22x + 1 dx = 2x 2 + x + C
1 22x + 1 dx = A 22x + 1 B 3 + C 3 L 87. Right, or wrong? Give a brief reason why. c.
15(x + 3) 2
1 1 dx = + C 2 x + 1 sx + 1d L x 1 dx = + C 76. 2 x + 1 L sx + 1d 75.
77.
1 b dy y 1>4
s2 + tan2 ud du s1 + tan2 ud du 66. L L (Hint: 1 + tan2 u = sec2 u)
L
x dx 1 = a tan1 a a b + C 2 2 a + x L
b.
ssin 2x  csc2 xd dx
L
xe x dx = xe x  e x + C
a.
Checking Antiderivative Formulas Verify the formulas in Exercises 71–82 by differentiation. 71.
81.
L
L
(x  2) 4
dx = a
3
x + 3 b + C x  2
88. Right, or wrong? Give a brief reason why. x 7 1
x cos (x 2)  sin (x 2) L
x2
dx =
sin (x 2) + C x
4.8 Initial Value Problems 89. Which of the following graphs shows the solution of the initial value problem dy = 2x, dx
y = 4 when x = 1 ?
y
y
y
100.
dr = cos pu, du
101.
dy 1 = sec t tan t, 2 dt
y s0d = 1
102.
dy = 8t + csc2 t, dt
p y a b = 7 2
279
Antiderivatives
r s0d = 1
3 dy , t 7 1, y(2) = 0 = 2 dt t2t  1 8 dy + sec2 t, y(0) = 1 = 104. dt 1 + t2 d 2y = 2  6x; y¿s0d = 4, y s0d = 1 105. dx 2 d 2y = 0; y¿s0d = 2, y s0d = 0 106. dx 2 103.
4
4
(1, 4)
3
4
(1, 4)
3
2
3 2
1
1
1
–1 0
x
1
–1 0
(a)
(1, 4)
2
1
x
–1 0
(b)
1
x
107.
d 2r 2 = 3; dt 2 t
dr = 1, ` dt t = 1
r s1d = 1
108.
3t d 2s = ; 8 dt 2
ds = 3, ` dt t = 4
s s4d = 4
(c)
Give reasons for your answer. 90. Which of the following graphs shows the solution of the initial value problem dy = x, dx
y = 1 when x = 1 ?
y
111. y s4d = sin t + cos t ;
y
y
d 3y
= 6; y–s0d = 8, y¿s0d = 0, y s0d = 5 dx 3 d3 u 1 = 0; u–s0d = 2, u¿s0d =  , us0d = 22 110. 2 dt 3 109.
y‡s0d = 7, (–1, 1) (–1, 1) 0
x
x
(b)
(a)
Give reasons for your answer.
0 (c)
y‡s0d = 0,
x
dy = 10  x, dx
dy 1 = 2 + x, x 7 0; y s2d = 1 dx x dy = 9x 2  4x + 5, y s 1d = 0 94. dx dy = 3x 2>3, 95. dx
i)
dy 1 , y s4d = 0 = dx 2 2x ds = 1 + cos t, s s0d = 4 97. dt 98.
ds = cos t + sin t, dt
99.
dr = p sin pu, du
d 2y dx 2
= 6x
ii) Its graph passes through the point (0, 1), and has a horizontal tangent there. b. How many curves like this are there? How do you know?
Solution (Integral) Curves Exercises 115–118 show solution curves of differential equations. In each exercise, find an equation for the curve through the labeled point. 115.
116.
y s 1d = 5
96.
y s0d = 3
114. a. Find a curve y = ƒsxd with the following properties:
y s0d = 1
93.
y–s0d = y¿s0d = 1,
113. Find the curve y = ƒsxd in the xyplane that passes through the point (9, 4) and whose slope at each point is 32x .
Solve the initial value problems in Exercises 91–112. dy = 2x  7, y s2d = 0 91. dx 92.
y s0d = 0
112. y s4d = cos x + 8 sin 2x ;
(–1, 1) 0
y–s0d = y¿s0d = 1,
y dy x 1 dx
dy 1 4 x1/3 3 dx
y
2 1 (1, 0.5)
0
s spd = 1
1
2
x
(–1, 1) –1
1
0
–1
r s0d = 0
–1
1
2
x
280
Chapter 4: Applications of Derivatives
117.
123. Motion along a coordinate line A particle moves on a coordinate line with acceleration a = d 2s>dt 2 = 152t  A 3> 2t B , subject to the conditions that ds>dt = 4 and s = 0 when t = 1 . Find
118. dy sin x cos x dx y
y
dy 1 sin x dx 2x
6
a. the velocity y = ds>dt in terms of t
1 0
2
x
(–, –1)
b. the position s in terms of t.
4 (1, 2)
2
0
1
2
3
x
–2
Applications 119. Finding displacement from an antiderivative of velocity
T 124. The hammer and the feather When Apollo 15 astronaut David Scott dropped a hammer and a feather on the moon to demonstrate that in a vacuum all bodies fall with the same (constant) acceleration, he dropped them from about 4 ft above the ground. The television footage of the event shows the hammer and the feather falling more slowly than on Earth, where, in a vacuum, they would have taken only half a second to fall the 4 ft. How long did it take the hammer and feather to fall 4 ft on the moon? To find out, solve the following initial value problem for s as a function of t. Then find the value of t that makes s equal to 0. Differential equation:
a. Suppose that the velocity of a body moving along the saxis is ds = y = 9.8t  3 . dt iii) Find the body’s displacement over the time interval from t = 1 to t = 3 given that s = 5 when t = 0 . iii) Find the body’s displacement from t = 1 to t = 3 given that s = 2 when t = 0 . iii) Now find the body’s displacement from t = 1 to t = 3 given that s = s0 when t = 0 . b. Suppose that the position s of a body moving along a coordinate line is a differentiable function of time t. Is it true that once you know an antiderivative of the velocity function ds>dt you can find the body’s displacement from t = a to t = b even if you do not know the body’s exact position at either of those times? Give reasons for your answer. 120. Liftoff from Earth A rocket lifts off the surface of Earth with a constant acceleration of 20 m>sec2 . How fast will the rocket be going 1 min later? 121. Stopping a car in time You are driving along a highway at a steady 60 mph (88 ft>sec) when you see an accident ahead and slam on the brakes. What constant deceleration is required to stop your car in 242 ft? To find out, carry out the following steps.
Initial conditions:
125. Motion with constant acceleration The standard equation for the position s of a body moving with a constant acceleration a along a coordinate line is s =
Initial conditions:
d 2s = k sk constantd dt 2 ds = 88 and s = 0 when t = 0 . dt Measuring time and distance from when the brakes are applied
2. Find the value of t that makes ds>dt = 0 . (The answer will involve k.) 3. Find the value of k that makes s = 242 for the value of t you found in Step 2. 122. Stopping a motorcycle The State of Illinois Cycle Rider Safety Program requires motorcycle riders to be able to brake from 30 mph (44 ft>sec) to 0 in 45 ft. What constant deceleration does it take to do that?
a 2 t + y0 t + s0 , 2
(1)
where y0 and s0 are the body’s velocity and position at time t = 0 . Derive this equation by solving the initial value problem Differential equation: Initial conditions:
d 2s = a dt 2 ds = y0 and s = s0 when t = 0 . dt
126. Free fall near the surface of a planet For free fall near the surface of a planet where the acceleration due to gravity has a constant magnitude of g lengthunits>sec2 , Equation (1) in Exercise 125 takes the form s = 
1. Solve the initial value problem Differential equation:
d 2s = 5.2 ft>sec2 dt 2 ds = 0 and s = 4 when t = 0 dt
1 2 gt + y0 t + s0 , 2
(2)
where s is the body’s height above the surface. The equation has a minus sign because the acceleration acts downward, in the direction of decreasing s. The velocity y0 is positive if the object is rising at time t = 0 and negative if the object is falling. Instead of using the result of Exercise 125, you can derive Equation (2) directly by solving an appropriate initial value problem. What initial value problem? Solve it to be sure you have the right one, explaining the solution steps as you go along. 127. Suppose that ƒsxd =
d A 1  2x B dx
and
g sxd =
d sx + 2d . dx
Find: a.
L
ƒsxd dx
b.
L
g sxd dx
Chapter 4
c.
L
[ƒsxd] dx
d.
L
[g sxd] dx
[ƒsxd + g sxd] dx [ƒsxd  g sxd] dx f. L L 128. Uniqueness of solutions If differentiable functions y = Fsxd and y = Gsxd both solve the initial value problem e.
dy = ƒsxd, dx
y sx0 d = y0 ,
on an interval I, must Fsxd = Gsxd for every x in I? Give reasons for your answer.
Chapter 4
Practice Exercises
281
COMPUTER EXPLORATIONS Use a CAS to solve the initial value problems in Exercises 129–132. Plot the solution curves. 129. y¿ = cos2 x + sin x,
y spd = 1
1 130. y¿ = x + x, y s1d = 1 1 , y s0d = 2 131. y¿ = 24  x 2 2 132. y– = x + 2x, y s1d = 0, y¿s1d = 0
Questions to Guide Your Review
1. What can be said about the extreme values of a function that is continuous on a closed interval? 2. What does it mean for a function to have a local extreme value on its domain? An absolute extreme value? How are local and absolute extreme values related, if at all? Give examples.
14. What is a cusp? Give examples. 15. List the steps you would take to graph a rational function. Illustrate with an example. 16. Outline a general strategy for solving maxmin problems. Give examples.
3. How do you find the absolute extrema of a continuous function on a closed interval? Give examples.
17. Describe l’Hôpital’s Rule. How do you know when to use the rule and when to stop? Give an example.
4. What are the hypotheses and conclusion of Rolle’s Theorem? Are the hypotheses really necessary? Explain.
18. How can you sometimes handle limits that lead to indeterminate forms q > q , q # 0, and q  q ? Give examples.
5. What are the hypotheses and conclusion of the Mean Value Theorem? What physical interpretations might the theorem have?
19. How can you sometimes handle limits that lead to indeterminate q q forms 1 , 00, and q ? Give examples.
6. State the Mean Value Theorem’s three corollaries. 7. How can you sometimes identify a function ƒ(x) by knowing ƒ¿ and knowing the value of ƒ at a point x = x0 ? Give an example.
20. Describe Newton’s method for solving equations. Give an example. What is the theory behind the method? What are some of the things to watch out for when you use the method?
8. What is the First Derivative Test for Local Extreme Values? Give examples of how it is applied.
21. Can a function have more than one antiderivative? If so, how are the antiderivatives related? Explain.
9. How do you test a twicedifferentiable function to determine where its graph is concave up or concave down? Give examples.
22. What is an indefinite integral? How do you evaluate one? What general formulas do you know for finding indefinite integrals?
10. What is an inflection point? Give an example. What physical significance do inflection points sometimes have? 11. What is the Second Derivative Test for Local Extreme Values? Give examples of how it is applied. 12. What do the derivatives of a function tell you about the shape of its graph? 13. List the steps you would take to graph a polynomial function. Illustrate with an example.
Chapter 4
23. How can you sometimes solve a differential equation of the form dy>dx = ƒsxd ? 24. What is an initial value problem? How do you solve one? Give an example. 25. If you know the acceleration of a body moving along a coordinate line as a function of time, what more do you need to know to find the body’s position function? Give an example.
Practice Exercises
Extreme Values 1. Does ƒsxd = x3 + 2x + tan x have any local maximum or minimum values? Give reasons for your answer. 2. Does g sxd = csc x + 2 cot x have any local maximum values? Give reasons for your answer.
3. Does ƒsxd = s7 + xds11  3xd1>3 have an absolute minimum value? An absolute maximum? If so, find them or give reasons why they fail to exist. List all critical points of ƒ.
282
Chapter 4: Applications of Derivatives
4. Find values of a and b such that the function ƒsxd =
7
b. Show that ƒ has a local maximum value at x = 25 L 1.2585 3 and a local minimum value at x = 2 2 L 1.2599 .
ax + b x2  1
has a local extreme value of 1 at x = 3 . Is this extreme value a local maximum, or a local minimum? Give reasons for your answer. 5. Does g(x) = e x  x have an absolute minimum value? An absolute maximum? If so, find them or give reasons why they fail to exist. List all critical points of g.
6. Does ƒ(x) = 2e x>(1 + x 2) have an absolute minimum value? An absolute maximum? If so, find them or give reasons why they fail to exist. List all critical points of ƒ.
In Exercises 7 and 8, find the absolute maximum and absolute minimum values of ƒ over the interval. 7. ƒ(x) = x  2 ln x,
1 … x … 3
8. ƒ(x) = (4>x) + ln x2,
1 … x … 4
9. The greatest integer function ƒsxd = :x; , defined for all values of x, assumes a local maximum value of 0 at each point of [0, 1). Could any of these local maximum values also be local minimum values of ƒ? Give reasons for your answer. 10. a. Give an example of a differentiable function ƒ whose first derivative is zero at some point c even though ƒ has neither a local maximum nor a local minimum at c. b. How is this consistent with Theorem 2 in Section 4.1? Give reasons for your answer. 11. The function y = 1>x does not take on either a maximum or a minimum on the interval 0 6 x 6 1 even though the function is continuous on this interval. Does this contradict the Extreme Value Theorem for continuous functions? Why? 12. What are the maximum and minimum values of the function y = ƒ x ƒ on the interval 1 … x 6 1 ? Notice that the interval is not closed. Is this consistent with the Extreme Value Theorem for continuous functions? Why? T 13. A graph that is large enough to show a function’s global behavior may fail to reveal important local features. The graph of ƒsxd = sx8>8d  sx6>2d  x5 + 5x3 is a case in point. a. Graph ƒ over the interval 2.5 … x … 2.5 . Where does the graph appear to have local extreme values or points of inflection? b. Now factor ƒ¿sxd and show that ƒ has a local maximum at x = 3 2 5 L 1.70998 and local minima at x = ; 23 L ;1.73205 . c. Zoom in on the graph to find a viewing window that shows 3 5 and x = 23 . the presence of the extreme values at x = 2 The moral here is that without calculus the existence of two of the three extreme values would probably have gone unnoticed. On any normal graph of the function, the values would lie close enough together to fall within the dimensions of a single pixel on the screen. (Source: Uses of Technology in the Mathematics Curriculum, by Benny Evans and Jerry Johnson, Oklahoma State University, published in 1990 under National Science Foundation Grant USE8950044.)
c. Zoom in to find a viewing window that shows the presence of 7 3 the extreme values at x = 25 and x = 2 2. The Mean Value Theorem 15. a. Show that g std = sin2 t  3t decreases on every interval in its domain. b. How many solutions does the equation sin2 t  3t = 5 have? Give reasons for your answer. 16. a. Show that y = tan u increases on every interval in its domain. b. If the conclusion in part (a) is really correct, how do you explain the fact that tan p = 0 is less than tan sp>4d = 1 ? 17. a. Show that the equation x4 + 2x2  2 = 0 has exactly one solution on [0, 1]. T b. Find the solution to as many decimal places as you can. 18. a. Show that ƒsxd = x>sx + 1d increases on every interval in its domain. b. Show that ƒsxd = x3 + 2x has no local maximum or minimum values. 19. Water in a reservoir As a result of a heavy rain, the volume of water in a reservoir increased by 1400 acreft in 24 hours. Show that at some instant during that period the reservoir’s volume was increasing at a rate in excess of 225,000 gal>min. (An acrefoot is 43,560 ft3 , the volume that would cover 1 acre to the depth of 1 ft. A cubic foot holds 7.48 gal.) 20. The formula Fsxd = 3x + C gives a different function for each value of C. All of these functions, however, have the same derivative with respect to x, namely F¿sxd = 3 . Are these the only differentiable functions whose derivative is 3? Could there be any others? Give reasons for your answers. 21. Show that x d d 1 a ab = b x + 1 dx x + 1 dx even though x 1 Z . x + 1 x + 1 Doesn’t this contradict Corollary 2 of the Mean Value Theorem? Give reasons for your answer.
22. Calculate the first derivatives of ƒsxd = x2>sx2 + 1d and g sxd = 1>sx2 + 1d . What can you conclude about the graphs of these functions? Analyzing Graphs In Exercises 23 and 24, use the graph to answer the questions.
23. Identify any global extreme values of ƒ and the values of x at which they occur. y y f (x) (1, 1)
T 14. (Continuation of Exercise 13.)
a. Graph ƒsxd = sx >8d  s2>5dx  5x  s5>x d + 11 over the interval 2 … x … 2 . Where does the graph appear to have local extreme values or points of inflection? 8
5
2
0
⎛2, 1⎛ ⎝ 2⎝ x
Chapter 4 24. Estimate the intervals on which the function y = ƒsxd is a. increasing.
Practice Exercises
In Exercises 49–52, graph each function. Then use the function’s first derivative to explain what you see.
b. decreasing.
49. y = x2>3 + sx  1d1>3
50. y = x2>3 + sx  1d2>3
c. Use the given graph of ƒ¿ to indicate where any local extreme values of the function occur, and whether each extreme is a relative maximum or minimum.
51. y = x1>3 + sx  1d1>3
52. y = x2>3  sx  1d1>3
Sketch the graphs of the rational functions in Exercises 53–60.
y
53. y =
x + 1 x  3
54. y =
2x x + 5
55. y =
x2 + 1 x
56. y =
x2  x + 1 x
57. y =
x3 + 2 2x
58. y =
x4  1 x2
59. y =
x2  4 x2  3
60. y =
x2 x2  4
(2, 3) y f ' (x) (–3, 1) x –1
283
–2
Using L’Hôpital’s Rule Use l’Hôpital’s Rule to find the limits in Exercises 61–72. Each of the graphs in Exercises 25 and 26 is the graph of the position function s = ƒstd of an object moving on a coordinate line (t represents time). At approximately what times (if any) is each object’s (a) velocity equal to zero? (b) Acceleration equal to zero? During approximately what time intervals does the object move (c) forward? (d) Backward? 25.
tan x x
63. lim
x:p
65. lim s f (t) 3
6
9
12 14
t
x:p>2
69. lim scsc x  cot xd x:0
26.
s
4
6
8
t
tan x x + sin x
66. lim
sin mx sin nx
x:0
68. lim+ 2x sec x x:0
70. lim a x:0
1 1  2b x4 x
x3 x3  2 b x  1 x + 1 2
Find the limits in Exercises 73–84.
Graphs and Graphing Graph the curves in Exercises 27–42. 27. y = x2  sx3>6d
64. lim
x:1
x: q x: q
2
xa  1 xb  1
71. lim A 2x2 + x + 1  2x2  x B
s f (t)
72. lim a 0
62. lim
x:0
sin2 x x:0 tan sx2 d 67. lim  sec 7x cos 3x
s
0
x2 + 3x  4 x  1 x:1
61. lim
73. lim
x:0
28. y = x3  3x2 + 3
29. y = x3 + 6x2  9x + 3
10x  1 x
74. lim
u :0
2sin x  1 x x:0 e  1
2sin x  1 x x:0 e  1
76. lim
5  5 cos x x x:0 e  x  1
78. lim
75. lim
77. lim
30. y = s1>8dsx3 + 3x2  9x  27d 31. y = x3s8  xd
32. y = x2s2x2  9d
33. y = x  3x2>3
34. y = x1>3sx  4d
35. y = x23  x
36. y = x24  x2
37. y = (x  3)2 ex
38. y = xex
39. y = ln (x2  4x + 3)
40. y = ln (sin x)
1 41. y = sin1 a x b
1 42. y = tan1 a x b
2
Each of Exercises 43–48 gives the first derivative of a function y = ƒsxd . (a) At what points, if any, does the graph of ƒ have a local maximum, local minimum, or inflection point? (b) Sketch the general shape of the graph. 43. y¿ = 16  x2
44. y¿ = x2  x  6
45. y¿ = 6xsx + 1dsx  2d
46. y¿ = x2s6  4xd
47. y¿ = x4  2x2
48. y¿ = 4x2  x4
79. lim+
t  ln s1 + 2td
e 1 81. lim+ a t  t b t: 0 t
b 83. lim a1 + x b x: q
4  4e x xe x x:0
80. lim
sin2 spxd
x:4 ex  4
t2
t: 0
3u  1 u
kx
+ 3  x
82. lim+ e1>y ln y y:0
7 2 84. lim a1 + x + 2 b x: q x
Optimization 85. The sum of two nonnegative numbers is 36. Find the numbers if a. the difference of their square roots is to be as large as possible. b. the sum of their square roots is to be as large as possible. 86. The sum of two nonnegative numbers is 20. Find the numbers a. if the product of one number and the square root of the other is to be as large as possible. b. if one number plus the square root of the other is to be as large as possible.
284
Chapter 4: Applications of Derivatives
87. An isosceles triangle has its vertex at the origin and its base parallel to the xaxis with the vertices above the axis on the curve y = 27  x2 . Find the largest area the triangle can have.
Newton’s Method 95. Let ƒsxd = 3x  x3 . Show that the equation ƒsxd = 4 has a solution in the interval [2, 3] and use Newton’s method to find it.
88. A customer has asked you to design an opentop rectangular stainless steel vat. It is to have a square base and a volume of 32 ft3 , to be welded from quarterinch plate, and to weigh no more than necessary. What dimensions do you recommend?
96. Let ƒsxd = x4  x3 . Show that the equation ƒsxd = 75 has a solution in the interval [3, 4] and use Newton’s method to find it.
89. Find the height and radius of the largest right circular cylinder that can be put in a sphere of radius 23 . 90. The figure here shows two right circular cones, one upside down inside the other. The two bases are parallel, and the vertex of the smaller cone lies at the center of the larger cone’s base. What values of r and h will give the smaller cone the largest possible volume?
Finding Indefinite Integrals Find the indefinite integrals (most general antiderivatives) in Exercises 97–120. You may need to try a solution and then adjust your guess. Check your answers by differentiation. 97.
L
sx3 + 5x  7d dx
4 99. a3 2t + 2 b dt t L dr 101. 2 L sr + 5d 103.
L
12'
105.
r
107.
h
L L
3u2u2 + 1 du x3s1 + x4 d1>4 dx sec2
s ds 10
98.
L
100.
a8t3 
t2 + tb dt 2
a

1
L 22t 6 dr 102. L A r  22 B 3 u 104. du L 27 + u2 106. 108.
L L
s2  xd3>5 dx csc2 ps ds
6'
109. 91. Manufacturing tires Your company can manufacture x hundred grade A tires and y hundred grade B tires a day, where 0 … x … 4 and 40  10x y = . 5  x
111. 112. 113.
Your profit on a grade A tire is twice your profit on a grade B tire. What is the most profitable number of each kind to make?
115.
92. Particle motion The positions of two particles on the saxis are s1 = cos t and s2 = cos st + p>4d .
117.
a. What is the farthest apart the particles ever get? 119.
b. When do the particles collide? T 93. Opentop box An opentop rectangular box is constructed from a 10in.by16in. piece of cardboard by cutting squares of equal side length from the corners and folding up the sides. Find analytically the dimensions of the box of largest volume and the maximum volume. Support your answers graphically. 94. The ladder problem What is the approximate length (in feet) of the longest ladder you can carry horizontally around the corner of the corridor shown here? Round your answer down to the nearest foot. y
(8, 6)
6 0
8
x
L
csc 22u cot 22u du
3 b dt t4
sec
110.
u u tan du 3 3
L 1  cos 2u x 2 b sin dx aHint: sin u = 4 2 L x cos2 dx 2 L 3 5 2 114. a x  xb dx a 2 + 2 b dx x + 1 L L x 1 116. a et  et b dt (5s + s5) ds L 2 L 2
L
u1  p du
3 dx L 2x2x2  1
118. 120.
L
2p + r dr
du L 216  u2
Initial Value Problems Solve the initial value problems in Exercises 121–124. dy x2 + 1 121. , y s1d = 1 = dx x2 2 dy 1 122. = ax + x b , y s1d = 1 dx 3 d 2r = 152t + ; r¿s1d = 8, r s1d = 0 dt 2 2t d3r 124. 3 = cos t; r–s0d = r¿s0d = 0, r s0d = 1 dt 123.
Applications and Examples 125. Can the integrations in (a) and (b) both be correct? Explain. dx a. = sin1 x + C L 21  x2 dx dx b. =  = cos1 x + C 2 L 21  x L 21  x2
Chapter 4 126. Can the integrations in (a) and (b) both be correct? Explain. a. b.
dx
= 
L 21  x2
L

dx 21  x2
= cos1 x + C
dx du = L 21  x2 L 21  s  ud2 =
x = u dx = du
L 21  u
2
130. y = 10xs2  ln xd,
y
u = x
132. g(x) = e23  2x  x
T 133. Graph the following functions and use what you see to locate and estimate the extreme values, identify the coordinates of the inflection points, and identify the intervals on which the graphs are concave up and concave down. Then confirm your estimates by working with the functions’ derivatives.
128. The rectangle shown here has one side on the positive yaxis, one side on the positive xaxis, and its upper righthand vertex on the curve y = sln xd>x2 . What dimensions give the rectangle its largest area, and what is that area?
1
c. y = s1 + x) ex
136. A round underwater transmission cable consists of a core of copper wires surrounded by nonconducting insulation. If x denotes the ratio of the radius of the core to the thickness of the insulation, it is known that the speed of the transmission signal is given by the equation y = x2 ln s1>xd . If the radius of the core is 1 cm, what insulation thickness h will allow the greatest transmission speed? Insulation x r h
y ln2x x
0.2 0.1
2
T 135. Graph ƒsxd = ssin xdsin x over [0, 3p] . Explain what you see.
x
y
b. y = ex
T 134. Graph ƒsxd = x ln x . Does the function appear to have an absolute minimum value? Confirm your answer with calculus.
2 e –x
0
s0, e2]
+1
a. y = sln xd> 1x
1
1 e , d 2e 2
2
y
Chapter 4
c
129. y = x ln 2x  x,
4
127. The rectangle shown here has one side on the positive yaxis, one side on the positive xaxis, and its upper righthand vertex 2 on the curve y = ex . What dimensions give the rectangle its largest area, and what is that area?
0
In Exercises 129 and 130, find the absolute maximum and minimum values of each function on the given interval.
131. ƒ(x) = ex>2x
= cos1 u + C = cos s xd + C
285
In Exercises 131 and 132, find the absolute maxima and minima of the functions and say where they are assumed.
du
1
Additional and Advanced Exercises
Core
h r
x
Additional and Advanced Exercises
Functions and Derivatives 1. What can you say about a function whose maximum and minimum values on an interval are equal? Give reasons for your answer. 2. Is it true that a discontinuous function cannot have both an absolute maximum and an absolute minimum value on a closed interval? Give reasons for your answer. 3. Can you conclude anything about the extreme values of a continuous function on an open interval? On a halfopen interval? Give reasons for your answer. 4. Local extrema Use the sign pattern for the derivative dƒ = 6sx  1dsx  2d2sx  3d3sx  4d4 dx to identify the points where ƒ has local maximum and minimum values.
5. Local extrema a. Suppose that the first derivative of y = ƒsxd is y¿ = 6sx + 1dsx  2d2 . At what points, if any, does the graph of ƒ have a local maximum, local minimum, or point of inflection? b. Suppose that the first derivative of y = ƒsxd is y¿ = 6x sx + 1dsx  2d . At what points, if any, does the graph of ƒ have a local maximum, local minimum, or point of inflection? 6. If ƒ¿sxd … 2 for all x, what is the most the values of ƒ can increase on [0, 6]? Give reasons for your answer. 7. Bounding a function Suppose that ƒ is continuous on [a, b] and that c is an interior point of the interval. Show that if ƒ¿sxd … 0 on [a, c) and ƒ¿sxd Ú 0 on (c, b], then ƒ(x) is never less than ƒ(c) on [a, b].
286
Chapter 4: Applications of Derivatives
8. An inequality a. Show that 1>2 … x>s1 + x2 d … 1>2 for every value of x. b. Suppose that ƒ is a function whose derivative is ƒ¿sxd = x>s1 + x2 d . Use the result in part (a) to show that 1 ƒ ƒsbd  ƒsad ƒ … 2 ƒ b  a ƒ for any a and b.
drill the hole near the bottom, the water will exit at a higher velocity but have only a short time to fall. Where is the best place, if any, for the hole? (Hint: How long will it take an exiting particle of water to fall from height y to the ground?) Tank kept full, top open
y
9. The derivative of ƒsxd = x2 is zero at x = 0 , but ƒ is not a constant function. Doesn’t this contradict the corollary of the Mean Value Theorem that says that functions with zero derivatives are constant? Give reasons for your answer.
h Exit velocity 64(h y) y
10. Extrema and inflection points Let h = ƒg be the product of two differentiable functions of x.
b. If the graphs of ƒ and g have inflection points at x = a , does the graph of h have an inflection point at a? In either case, if the answer is yes, give a proof. If the answer is no, give a counterexample. 11. Finding a function Use the following information to find the values of a, b, and c in the formula ƒsxd = sx + ad> sbx2 + cx + 2d .
Ground
0
a. If ƒ and g are positive, with local maxima at x = a , and if ƒ¿ and g¿ change sign at a, does h have a local maximum at a?
x
Range
16. Kicking a field goal An American football player wants to kick a field goal with the ball being on a right hash mark. Assume that the goal posts are b feet apart and that the hash mark line is a distance a 7 0 feet from the right goal post. (See the accompanying figure.) Find the distance h from the goal post line that gives the kicker his largest angle b . Assume that the football field is flat. Goal posts b
i) The values of a, b, and c are either 0 or 1.
Goal post line
a
ii) The graph of ƒ passes through the point s 1, 0d . iii) The line y = 1 is an asymptote of the graph of ƒ. 12. Horizontal tangent For what value or values of the constant k will the curve y = x3 + kx2 + 3x  4 have exactly one horizontal tangent?
h
Optimization 13. Largest inscribed triangle Points A and B lie at the ends of a diameter of a unit circle and point C lies on the circumference. Is it true that the area of triangle ABC is largest when the triangle is isosceles? How do you know? 14. Proving the second derivative test The Second Derivative Test for Local Maxima and Minima (Section 4.4) says: a. ƒ has a local maximum value at x = c if ƒ¿scd = 0 and ƒ–scd 6 0 b. ƒ has a local minimum value at x = c if ƒ¿scd = 0 and ƒ–scd 7 0 . To prove statement (a), let P = s1>2d ƒ ƒ–scd ƒ . Then use the fact that ƒ–scd = lim
h:0
Football
17. A maxmin problem with a variable answer Sometimes the solution of a maxmin problem depends on the proportions of the shapes involved. As a case in point, suppose that a right circular cylinder of radius r and height h is inscribed in a right circular cone of radius R and height H, as shown here. Find the value of r (in terms of R and H) that maximizes the total surface area of the cylinder (including top and bottom). As you will see, the solution depends on whether H … 2R or H 7 2R .
ƒ¿sc + hd  ƒ¿scd ƒ¿sc + hd = lim h h h :0
to conclude that for some d 7 0 , 0 6 ƒhƒ 6 d
Q
r
ƒ¿sc + hd 6 ƒ–scd + P 6 0 . h
Thus, ƒ¿sc + hd is positive for d 6 h 6 0 and negative for 0 6 h 6 d . Prove statement (b) in a similar way. 15. Hole in a water tank You want to bore a hole in the side of the tank shown here at a height that will make the stream of water coming out hit the ground as far from the tank as possible. If you drill the hole near the top, where the pressure is low, the water will exit slowly but spend a relatively long time in the air. If you
H
h
R
Chapter 4 18. Minimizing a parameter Find the smallest value of the positive constant m that will make mx  1 + s1>xd greater than or equal to zero for all positive values of x. Limits 19. Evaluate the following limits. a. lim
x: 0
2 sin 5x 3x
b. lim sin 5x cot 3x x: 0
c. lim x csc 22x 2
x: 0
d.
x  sin x x  tan x
e. lim
x: 0
lim ssec x  tan xd
x: p>2
f. lim
x: 0
sin x2 x sin x
sec x  1 x3  8 g. lim h. lim 2 x: 0 x: 2 x  4 x2 20. L’Hôpital’s Rule does not help with the following limits. Find them some other way. a. lim
2x + 5
x: q
b. lim
x: q
2x + 5
2x x + 7 2x
Theory and Examples 21. Suppose that it costs a company y = a + bx dollars to produce x units per week. It can sell x units per week at a price of P = c  ex dollars per unit. Each of a, b, c, and e represents a positive constant. (a) What production level maximizes the profit? (b) What is the corresponding price? (c) What is the weekly profit at this level of production? (d) At what price should each item be sold to maximize profits if the government imposes a tax of t dollars per item sold? Comment on the difference between this price and the price before the tax. 22. Estimating reciprocals without division You can estimate the value of the reciprocal of a number a without ever dividing by a if you apply Newton’s method to the function ƒsxd = s1>xd  a . For example, if a = 3 , the function involved is ƒsxd = s1>xd  3 . a. Graph y = s1>xd  3 . Where does the graph cross the xaxis? b. Show that the recursion formula in this case is xn + 1 = xns2  3xn d , so there is no need for division. q
a q1 b, x0
287
25. Free fall in the fourteenth century In the middle of the fourteenth century, Albert of Saxony (1316–1390) proposed a model of free fall that assumed that the velocity of a falling body was proportional to the distance fallen. It seemed reasonable to think that a body that had fallen 20 ft might be moving twice as fast as a body that had fallen 10 ft. And besides, none of the instruments in use at the time were accurate enough to prove otherwise. Today we can see just how far off Albert of Saxony’s model was by solving the initial value problem implicit in his model. Solve the problem and compare your solution graphically with the equation s = 16t2 . You will see that it describes a motion that starts too slowly at first and then becomes too fast too soon to be realistic. T 26. Group blood testing During World War II it was necessary to administer blood tests to large numbers of recruits. There are two standard ways to administer a blood test to N people. In method 1, each person is tested separately. In method 2, the blood samples of x people are pooled and tested as one large sample. If the test is negative, this one test is enough for all x people. If the test is positive, then each of the x people is tested separately, requiring a total of x + 1 tests. Using the second method and some probability theory it can be shown that, on the average, the total number of tests y will be 1 y = N a1  qx + x b . With q = 0.99 and N = 1000 , find the integer value of x that minimizes y. Also find the integer value of x that maximizes y. (This second result is not important to the reallife situation.) The group testing method was used in World War II with a savings of 80% over the individual testing method, but not with the given value of q. 27. Assume that the brakes of an automobile produce a constant deceleration of k ft>sec2 . (a) Determine what k must be to bring an automobile traveling 60 mi>hr (88 ft>sec) to rest in a distance of 100 ft from the point where the brakes are applied. (b) With the same k, how far would a car traveling 30 mi>hr travel before being brought to a stop? 28. Let ƒ(x), g(x) be two continuously differentiable functions satisfying the relationships ƒ¿sxd = g sxd and ƒ–sxd = ƒsxd . Let hsxd = ƒ 2sxd + g 2sxd . If hs0d = 5 , find h(10).
23. To find x = 2a , we apply Newton’s method to ƒsxd = xq  a . Here we assume that a is a positive real number and q is a positive integer. Show that x1 is a “weighted average” of x0 and a>x0q  1 , and find the coefficients m0, m1 such that x1 = m0 x0 + m1 a
Additional and Advanced Exercises
m0 7 0, m1 7 0, m0 + m1 = 1. q1
What conclusion would you reach if x0 and a>x0 What would be the value of x1 in that case?
were equal?
24. The family of straight lines y = ax + b (a, b arbitrary constants) can be characterized by the relation y– = 0 . Find a similar relation satisfied by the family of all circles sx  hd2 + sy  hd2 = r2 , where h and r are arbitrary constants. (Hint: Eliminate h and r from the set of three equations including the given one and two obtained by successive differentiation.)
29. Can there be a curve satisfying the following conditions? d 2y>dx 2 is everywhere equal to zero and, when x = 0, y = 0 and dy>dx = 1 . Give a reason for your answer. 30. Find the equation for the curve in the xyplane that passes through the point s1, 1d if its slope at x is always 3x2 + 2 . 31. A particle moves along the xaxis. Its acceleration is a = t 2 . At t = 0 , the particle is at the origin. In the course of its motion, it reaches the point x = b , where b 7 0 , but no point beyond b. Determine its velocity at t = 0 . 32. A particle moves with acceleration a = 2t  (1> 2t) . Assuming that the velocity y = 4>3 and the position s =  4>15 when t = 0 , find a. the velocity y in terms of t. b. the position s in terms of t. 33. Given ƒsxd = ax2 + 2bx + c with a 7 0 . By considering the minimum, prove that ƒsxd Ú 0 for all real x if and only if b2  ac … 0 .
288
Chapter 4: Applications of Derivatives
34. Schwarz’s inequality B
a. In Exercise 33, let ƒsxd = sa1 x + b1 d2 + sa2 x + b2 d2 + Á + san x + bn d2,
d2
and deduce Schwarz’s inequality:
b. Show that equality holds in Schwarz’s inequality only if there exists a real number x that makes ai x equal bi for every value of i from 1 to n. 35. The best branching angles for blood vessels and pipes When a smaller pipe branches off from a larger one in a flow system, we may want it to run off at an angle that is best from some energysaving point of view. We might require, for instance, that energy loss due to friction be minimized along the section AOB shown in the accompanying figure. In this diagram, B is a given point to be reached by the smaller pipe, A is a point in the larger pipe upstream from B, and O is the point where the branching occurs. A law due to Poiseuille states that the loss of energy due to friction in nonturbulent flow is proportional to the length of the path and inversely proportional to the fourth power of the radius. Thus, the loss along AO is skd1 d>R4 and along OB is skd2 d>r4 , where k is a constant, d1 is the length of AO, d2 is the length of OB, R is the radius of the larger pipe, and r is the radius of the smaller pipe. The angle u is to be chosen to minimize the sum of these two losses: L = k
d1 4
R
+ k
A
sa1 b1 + a2 b2 + Á + an bn d2 … A a1 2 + a2 2 + Á + an 2 B A b1 2 + b 2 2 + Á + bn 2 B .
d2 4
r
.
b d 2 sin
O
d1
d 2 cos
C
a
In our model, we assume that AC = a and BC = b are fixed. Thus we have the relations d1 + d2 cos u = a
d2 sin u = b ,
so that d2 = b csc u , d1 = a  d2 cos u = a  b cot u . We can express the total loss L as a function of u : L = ka
a  b cot u b csc u + b. R4 r4
a. Show that the critical value of u for which dL>du equals zero is uc = cos1
r4 . R4
b. If the ratio of the pipe radii is r>R = 5>6 , estimate to the nearest degree the optimal branching angle given in part (a).
5 INTEGRATION OVERVIEW A great achievement of classical geometry was obtaining formulas for the areas and volumes of triangles, spheres, and cones. In this chapter we develop a method to calculate the areas and volumes of very general shapes. This method, called integration, is a tool for calculating much more than areas and volumes. The integral is of fundamental importance in statistics, the sciences, and engineering. As with the derivative, the integral also arises as a limit, this time of increasingly fine approximations. We use it to calculate quantities ranging from probabilities and averages to energy consumption and the forces against a dam’s floodgates. We study a variety of these applications in the next chapter, but in this chapter we focus on the integral concept and its use in computing areas of various regions with curved boundaries.
Area and Estimating with Finite Sums
5.1
The definite integral is the key tool in calculus for defining and calculating quantities important to mathematics and science, such as areas, volumes, lengths of curved paths, probabilities, and the weights of various objects, just to mention a few. The idea behind the integral is that we can effectively compute such quantities by breaking them into small pieces and then summing the contributions from each piece. We then consider what happens when more and more, smaller and smaller pieces are taken in the summation process. Finally, if the number of terms contributing to the sum approaches infinity and we take the limit of these sums in the way described in Section 5.3, the result is a definite integral. We prove in Section 5.4 that integrals are connected to antiderivatives, a connection that is one of the most important relationships in calculus. The basis for formulating definite integrals is the construction of appropriate finite sums. Although we need to define precisely what we mean by the area of a general region in the plane, or the average value of a function over a closed interval, we do have intuitive ideas of what these notions mean. So in this section we begin our approach to integration by approximating these quantities with finite sums. We also consider what happens when we take more and more terms in the summation process. In subsequent sections we look at taking the limit of these sums as the number of terms goes to infinity, which then leads to precise definitions of the quantities being approximated here.
y
1 y 1 x2
0.5 R
0
0.5
1
FIGURE 5.1 The area of the region R cannot be found by a simple formula.
x
Area Suppose we want to find the area of the shaded region R that lies above the xaxis, below the graph of y = 1  x 2 , and between the vertical lines x = 0 and x = 1 (Figure 5.1). Unfortunately, there is no simple geometric formula for calculating the areas of general shapes having curved boundaries like the region R. How, then, can we find the area of R? While we do not yet have a method for determining the exact area of R, we can approximate it in a simple way. Figure 5.2a shows two rectangles that together contain the region R. Each rectangle has width 1>2 and they have heights 1 and 3>4, moving from left to right. The height of each rectangle is the maximum value of the function ƒ,
289
290
Chapter 5: Integration
y
1
y y 1 x2
(0, 1)
1
(0, 1)
y 1 x2
⎛ 1 , 15 ⎛ ⎝ 4 16 ⎝
⎛1 , 3⎛ ⎝2 4⎝
⎛1 , 3⎛ ⎝2 4⎝
0.5
⎛3 , 7 ⎛ ⎝ 4 16 ⎝
0.5 R
0
R
0.5
x
1
0
0.25
0.5
(a)
0.75
x
1
(b)
FIGURE 5.2 (a) We get an upper estimate of the area of R by using two rectangles containing R. (b) Four rectangles give a better upper estimate. Both estimates overshoot the true value for the area by the amount shaded in light red.
obtained by evaluating ƒ at the left endpoint of the subinterval of [0, 1] forming the base of the rectangle. The total area of the two rectangles approximates the area A of the region R, A L 1#
3 1 + 2 4
#
7 1 = = 0.875. 2 8
This estimate is larger than the true area A since the two rectangles contain R. We say that 0.875 is an upper sum because it is obtained by taking the height of each rectangle as the maximum (uppermost) value of ƒ(x) for a point x in the base interval of the rectangle. In Figure 5.2b, we improve our estimate by using four thinner rectangles, each of width 1>4, which taken together contain the region R. These four rectangles give the approximation A L 1#
15 1 + 4 16
#
#
3 1 + 4 4
7 1 + 4 16
#
25 1 = = 0.78125, 4 32
which is still greater than A since the four rectangles contain R. Suppose instead we use four rectangles contained inside the region R to estimate the area, as in Figure 5.3a. Each rectangle has width 1>4 as before, but the rectangles are y 1
y y 1 x2
⎛ 1 , 15 ⎛ ⎝ 4 16 ⎝
1
⎛ 1 , 63 ⎛ ⎝ 8 64 ⎝ ⎛ 3 , 55 ⎛ ⎝ 8 64 ⎝
y 1 x2
⎛1 , 3⎛ ⎝2 4 ⎝ ⎛ 5 , 39 ⎛ ⎝ 8 64 ⎝ ⎛3 , 7 ⎛ ⎝ 4 16 ⎝
0.5
0.5 ⎛ 7 , 15 ⎛ ⎝ 8 64 ⎝
0
0.25
0.5 (a)
0.75
1
x
0
0.25 0.5 0.75 1 0.125 0.375 0.625 0.875
x
(b)
FIGURE 5.3 (a) Rectangles contained in R give an estimate for the area that undershoots the true value by the amount shaded in light blue. (b) The midpoint rule uses rectangles whose height is the value of y = ƒsxd at the midpoints of their bases. The estimate appears closer to the true value of the area because the light red overshoot areas roughly balance the light blue undershoot areas.
5.1
Area and Estimating with Finite Sums
291
shorter and lie entirely beneath the graph of ƒ. The function ƒsxd = 1  x 2 is decreasing on [0, 1], so the height of each of these rectangles is given by the value of ƒ at the right endpoint of the subinterval forming its base. The fourth rectangle has zero height and therefore contributes no area. Summing these rectangles with heights equal to the minimum value of ƒ(x) for a point x in each base subinterval gives a lower sum approximation to the area, 15 16
A L y
#
3 1 + 4 4
#
7 1 + 4 16
#
17 1 1 + 0# = = 0.53125. 4 4 32
This estimate is smaller than the area A since the rectangles all lie inside of the region R. The true value of A lies somewhere between these lower and upper sums:
1
0.53125 6 A 6 0.78125.
y 1 x2
1
0
x
(a)
By considering both lower and upper sum approximations, we get not only estimates for the area, but also a bound on the size of the possible error in these estimates since the true value of the area lies somewhere between them. Here the error cannot be greater than the difference 0.78125  0.53125 = 0.25. Yet another estimate can be obtained by using rectangles whose heights are the values of ƒ at the midpoints of their bases (Figure 5.3b). This method of estimation is called the midpoint rule for approximating the area. The midpoint rule gives an estimate that is between a lower sum and an upper sum, but it is not quite so clear whether it overestimates or underestimates the true area. With four rectangles of width 1>4 as before, the midpoint rule estimates the area of R to be A L
y 1
63 64
#
55 1 + 4 64
#
39 1 + 4 64
#
15 1 + 4 64
#
172 1 = 4 64
#
1 = 0.671875. 4
In each of our computed sums, the interval [a, b] over which the function ƒ is defined was subdivided into n subintervals of equal width (also called length) ¢x = sb  ad>n, and ƒ was evaluated at a point in each subinterval: c1 in the first subinterval, c2 in the second subinterval, and so on. The finite sums then all take the form
y 1 x2
ƒsc1 d ¢x + ƒsc2 d ¢x + ƒsc3 d ¢x + Á + ƒscn d ¢x.
1
0
x
(b)
FIGURE 5.4 (a) A lower sum using 16 rectangles of equal width ¢x = 1>16. (b) An upper sum using 16 rectangles.
By taking more and more rectangles, with each rectangle thinner than before, it appears that these finite sums give better and better approximations to the true area of the region R. Figure 5.4a shows a lower sum approximation for the area of R using 16 rectangles of equal width. The sum of their areas is 0.634765625, which appears close to the true area, but is still smaller since the rectangles lie inside R. Figure 5.4b shows an upper sum approximation using 16 rectangles of equal width. The sum of their areas is 0.697265625, which is somewhat larger than the true area because the rectangles taken together contain R. The midpoint rule for 16 rectangles gives a total area approximation of 0.6669921875, but it is not immediately clear whether this estimate is larger or smaller than the true area.
EXAMPLE 1
Table 5.1 shows the values of upper and lower sum approximations to the area of R, using up to 1000 rectangles. In Section 5.2 we will see how to get an exact value of the areas of regions such as R by taking a limit as the base width of each rectangle goes to zero and the number of rectangles goes to infinity. With the techniques developed there, we will be able to show that the area of R is exactly 2>3.
Distance Traveled Suppose we know the velocity function y(t) of a car moving down a highway, without changing direction, and want to know how far it traveled between times t = a and t = b . The position function s(t) of the car has derivative y(t). If we can find an antiderivative F(t) of y(t) then
292
Chapter 5: Integration
TABLE 5.1 Finite approximations for the area of R
Number of subintervals
Lower sum
Midpoint rule
Upper sum
2 4 16 50 100 1000
0.375 0.53125 0.634765625 0.6566 0.66165 0.6661665
0.6875 0.671875 0.6669921875 0.6667 0.666675 0.66666675
0.875 0.78125 0.697265625 0.6766 0.67165 0.6671665
we can find the car’s position function s(t) by setting sstd = Fstd + C. The distance traveled can then be found by calculating the change in position, ssbd  ssad = F(b)  F(a). If the velocity function is known only by the readings at various times of a speedometer on the car, then we have no formula from which to obtain an antiderivative function for velocity. So what do we do in this situation? When we don’t know an antiderivative for the velocity function y(t), we can approximate the distance traveled with finite sums in a way similar to our estimates for area discussed before. We subdivide the interval [a, b] into short time intervals on each of which the velocity is considered to be fairly constant. Then we approximate the distance traveled on each time subinterval with the usual distance formula distance = velocity * time and add the results across [a, b]. Suppose the subdivided interval looks like t a
t1
t
t
t2
t3
b
t (sec)
with the subintervals all of equal length ¢t. Pick a number t1 in the first interval. If ¢t is so small that the velocity barely changes over a short time interval of duration ¢t, then the distance traveled in the first time interval is about yst1 d ¢t. If t2 is a number in the second interval, the distance traveled in the second time interval is about yst2 d ¢t. The sum of the distances traveled over all the time intervals is D L yst1 d ¢t + yst2 d ¢t + Á + ystn d ¢t, where n is the total number of subintervals.
EXAMPLE 2
The velocity function of a projectile fired straight into the air is ƒstd = 160  9.8t m>sec. Use the summation technique just described to estimate how far the projectile rises during the first 3 sec. How close do the sums come to the exact value of 435.9 m? We explore the results for different numbers of intervals and different choices of evaluation points. Notice that ƒ(t) is decreasing, so choosing left endpoints gives an upper sum estimate; choosing right endpoints gives a lower sum estimate. Solution
(a) Three subintervals of length 1, with ƒ evaluated at left endpoints giving an upper sum: t1
t2
t3
0
1
2
t
3
t
5.1
Area and Estimating with Finite Sums
293
With ƒ evaluated at t = 0, 1, and 2, we have D L ƒst1 d ¢t + ƒst2 d ¢t + ƒst3 d ¢t = [160  9.8s0d]s1d + [160  9.8s1d]s1d + [160  9.8s2d]s1d = 450.6. (b) Three subintervals of length 1, with ƒ evaluated at right endpoints giving a lower sum:
0
t1
t2
t3
1
2
3
t
t
With ƒ evaluated at t = 1, 2, and 3, we have D L ƒst1 d ¢t + ƒst2 d ¢t + ƒst3 d ¢t = [160  9.8s1d]s1d + [160  9.8s2d]s1d + [160  9.8s3d]s1d = 421.2. (c) With six subintervals of length 1>2, we get t1 t 2 t 3 t 4 t 5 t 6 0
1
2
t
t1 t 2 t 3 t 4 t 5 t 6 3
t
0
1
2
3
t
t
These estimates give an upper sum using left endpoints: D L 443.25; and a lower sum using right endpoints: D L 428.55. These sixinterval estimates are somewhat closer than the threeinterval estimates. The results improve as the subintervals get shorter. As we can see in Table 5.2, the leftendpoint upper sums approach the true value 435.9 from above, whereas the rightendpoint lower sums approach it from below. The true value lies between these upper and lower sums. The magnitude of the error in the closest entries is 0.23, a small percentage of the true value. Error magnitude = ƒ true value  calculated value ƒ = ƒ 435.9  435.67 ƒ = 0.23. Error percentage =
0.23 L 0.05%. 435.9
It would be reasonable to conclude from the table’s last entries that the projectile rose about 436 m during its first 3 sec of flight. TABLE 5.2 Traveldistance estimates
Number of subintervals
Length of each subinterval
Upper sum
Lower sum
3 6 12 24 48 96 192
1 1>2 1>4 1>8 1>16 1>32 1>64
450.6 443.25 439.58 437.74 436.82 436.36 436.13
421.2 428.55 432.23 434.06 434.98 435.44 435.67
294
Chapter 5: Integration
Displacement Versus Distance Traveled If an object with position function s(t) moves along a coordinate line without changing direction, we can calculate the total distance it travels from t = a to t = b by summing the distance traveled over small intervals, as in Example 2. If the object reverses direction one or more times during the trip, then we need to use the object’s speed ƒ ystd ƒ , which is the absolute value of its velocity function, y(t), to find the total distance traveled. Using the velocity itself, as in Example 2, gives instead an estimate to the object’s displacement, ssbd  ssad, the difference between its initial and final positions. To see why using the velocity function in the summation process gives an estimate to the displacement, partition the time interval [a, b] into small enough equal subintervals ¢t so that the object’s velocity does not change very much from time tk  1 to tk . Then ystk d gives a good approximation of the velocity throughout the interval. Accordingly, the change in the object’s position coordinate during the time interval is about
s s(5)
Height (ft)
400
256
()
()
s(2)
s(8)
144
ystk d ¢t. The change is positive if ystk d is positive and negative if ystk d is negative. In either case, the distance traveled by the object during the subinterval is about ƒ ystk d ƒ ¢t.
s0
The total distance traveled is approximately the sum ƒ yst1 d ƒ ¢t + ƒ yst2 d ƒ ¢t + Á + ƒ ystn d ƒ ¢t.
s(0)
FIGURE 5.5 The rock in Example 3. The height s = 256 ft is reached at t = 2 and t = 8 sec. The rock falls 144 ft from its maximum height when t = 8.
TABLE 5.3 Velocity Function
t
Y(t)
t
Y(t)
0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
160 144 128 112 96 80 64 48 32
4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0
16 0 16 32 48 64 80 96
We revisit these ideas in Section 5.4.
EXAMPLE 3
In Example 4 in Section 3.4, we analyzed the motion of a heavy rock blown straight up by a dynamite blast. In that example, we found the velocity of the rock at any time during its motion to be ystd = 160  32t ft>sec. The rock was 256 ft above the ground 2 sec after the explosion, continued upward to reach a maximum height of 400 ft at 5 sec after the explosion, and then fell back down to reach the height of 256 ft again at t = 8 sec after the explosion. (See Figure 5.5.) If we follow a procedure like that presented in Example 2, and use the velocity function ystd in the summation process over the time interval [0, 8], we will obtain an estimate to 256 ft, the rock’s height above the ground at t = 8. The positive upward motion (which yields a positive distance change of 144 ft from the height of 256 ft to the maximum height) is canceled by the negative downward motion (giving a negative change of 144 ft from the maximum height down to 256 ft again), so the displacement or height above the ground is being estimated from the velocity function. On the other hand, if the absolute value ƒ ystd ƒ is used in the summation process, we will obtain an estimate to the total distance the rock has traveled: the maximum height reached of 400 ft plus the additional distance of 144 ft it has fallen back down from that maximum when it again reaches the height of 256 ft at t = 8 sec. That is, using the absolute value of the velocity function in the summation process over the time interval [0, 8], we obtain an estimate to 544 ft, the total distance up and down that the rock has traveled in 8 sec. There is no cancelation of distance changes due to sign changes in the velocity function, so we estimate distance traveled rather than displacement when we use the absolute value of the velocity function (that is, the speed of the rock). As an illustration of our discussion, we subdivide the interval [0, 8] into sixteen subintervals of length ¢t = 1>2 and take the right endpoint of each subinterval in our calculations. Table 5.3 shows the values of the velocity function at these endpoints. Using ystd in the summation process, we estimate the displacement at t = 8: s144 + 128 + 112 + 96 + 80 + 64 + 48 + 32 + 16 1 + 0  16  32  48  64  80  96d # = 192 2 Error magnitude = 256  192 = 64
5.1
Area and Estimating with Finite Sums
295
Using ƒ ystd ƒ in the summation process, we estimate the total distance traveled over the time interval [0, 8]: s144 + 128 + 112 + 96 + 80 + 64 + 48 + 32 + 16 1 + 0 + 16 + 32 + 48 + 64 + 80 + 96d # = 528 2 Error magnitude = 544  528 = 16 If we take more and more subintervals of [0, 8] in our calculations, the estimates to 256 ft and 544 ft improve, approaching their true values.
Average Value of a Nonnegative Continuous Function The average value of a collection of n numbers x1, x2 , Á , xn is obtained by adding them together and dividing by n. But what is the average value of a continuous function ƒ on an interval [a, b]? Such a function can assume infinitely many values. For example, the temperature at a certain location in a town is a continuous function that goes up and down each day. What does it mean to say that the average temperature in the town over the course of a day is 73 degrees? When a function is constant, this question is easy to answer. A function with constant value c on an interval [a, b] has average value c. When c is positive, its graph over [a, b] gives a rectangle of height c. The average value of the function can then be interpreted geometrically as the area of this rectangle divided by its width b  a (Figure 5.6a). y
y
0
y g(x)
yc
c
a
(a)
b
c
x
0
a
(b)
b
x
FIGURE 5.6 (a) The average value of ƒsxd = c on [a, b] is the area of the rectangle divided by b  a . (b) The average value of g(x) on [a, b] is the area beneath its graph divided by b  a .
What if we want to find the average value of a nonconstant function, such as the function g in Figure 5.6b? We can think of this graph as a snapshot of the height of some water that is sloshing around in a tank between enclosing walls at x = a and x = b. As the water moves, its height over each point changes, but its average height remains the same. To get the average height of the water, we let it settle down until it is level and its height is constant. The resulting height c equals the area under the graph of g divided by b  a. We are led to define the average value of a nonnegative function on an interval [a, b] to be the area under its graph divided by b  a. For this definition to be valid, we need a precise understanding of what is meant by the area under a graph. This will be obtained in Section 5.3, but for now we look at an example.
y f(x) sin x 1
0
2
FIGURE 5.7 Approximating the area under ƒsxd = sin x between 0 and p to compute the average value of sin x over [0, p] , using eight rectangles (Example 4).
x
EXAMPLE 4
Estimate the average value of the function ƒsxd = sin x on the interval
[0, p]. Looking at the graph of sin x between 0 and p in Figure 5.7, we can see that its average height is somewhere between 0 and 1. To find the average we need to calculate the area A under the graph and then divide this area by the length of the interval, p  0 = p. We do not have a simple way to determine the area, so we approximate it with finite sums. To get an upper sum approximation, we add the areas of eight rectangles of equal
Solution
296
Chapter 5: Integration
width p>8 that together contain the region beneath the graph of y = sin x and above the xaxis on [0, p]. We choose the heights of the rectangles to be the largest value of sin x on each subinterval. Over a particular subinterval, this largest value may occur at the left endpoint, the right endpoint, or somewhere between them. We evaluate sin x at this point to get the height of the rectangle for an upper sum. The sum of the rectangle areas then estimates the total area (Figure 5.7): A L asin
3p 5p 3p 7p # p p p p p + sin + sin + sin + sin + sin + sin + sin b 8 4 8 2 2 8 4 8 8
L s.38 + .71 + .92 + 1 + 1 + .92 + .71 + .38d #
p p = s6.02d # L 2.365. 8 8
To estimate the average value of sin x we divide the estimated area by p and obtain the approximation 2.365>p L 0.753. Since we used an upper sum to approximate the area, this estimate is greater than the actual average value of sin x over [0, p]. If we use more and more rectangles, with each rectangle getting thinner and thinner, we get closer and closer to the true average value. Using the techniques covered in Section 5.3, we will show that the true average value is 2>p L 0.64. As before, we could just as well have used rectangles lying under the graph of y = sin x and calculated a lower sum approximation, or we could have used the midpoint rule. In Section 5.3 we will see that in each case, the approximations are close to the true area if all the rectangles are sufficiently thin.
Summary The area under the graph of a positive function, the distance traveled by a moving object that doesn’t change direction, and the average value of a nonnegative function over an interval can all be approximated by finite sums. First we subdivide the interval into subintervals, treating the appropriate function ƒ as if it were constant over each particular subinterval. Then we multiply the width of each subinterval by the value of ƒ at some point within it, and add these products together. If the interval [a, b] is subdivided into n subintervals of equal widths ¢x = sb  ad>n, and if ƒsck d is the value of ƒ at the chosen point ck in the kth subinterval, this process gives a finite sum of the form ƒsc1 d ¢x + ƒsc2 d ¢x + ƒsc3 d ¢x + Á + ƒscn d ¢x. The choices for the ck could maximize or minimize the value of ƒ in the kth subinterval, or give some value in between. The true value lies somewhere between the approximations given by upper sums and lower sums. The finite sum approximations we looked at improved as we took more subintervals of thinner width.
Exercises 5.1 Area In Exercises 1–4, use finite approximations to estimate the area under the graph of the function using a. a lower sum with two rectangles of equal width. b. a lower sum with four rectangles of equal width. c. an upper sum with two rectangles of equal width. d. an upper sum with four rectangles of equal width.
3. ƒsxd = 1>x between x = 1 and x = 5. 4. ƒsxd = 4  x 2 between x = 2 and x = 2. Using rectangles whose height is given by the value of the function at the midpoint of the rectangle’s base (the midpoint rule), estimate the area under the graphs of the following functions, using first two and then four rectangles. 5. ƒsxd = x 2 between x = 0 and x = 1. 6. ƒsxd = x 3 between x = 0 and x = 1.
1. ƒsxd = x 2 between x = 0 and x = 1.
7. ƒsxd = 1>x between x = 1 and x = 5.
2. ƒsxd = x 3 between x = 0 and x = 1.
8. ƒsxd = 4  x 2 between x = 2 and x = 2.
5.1 Distance 9. Distance traveled The accompanying table shows the velocity of a model train engine moving along a track for 10 sec. Estimate the distance traveled by the engine using 10 subintervals of length 1 with b. rightendpoint values. Velocity (in. / sec)
Time (sec)
Velocity (in. / sec)
0 1 2 3 4 5
0 12 22 10 5 13
6 7 8 9 10
11 6 2 6 0
297
12. Distance from velocity data The accompanying table gives data for the velocity of a vintage sports car accelerating from 0 to 142 mi> h in 36 sec (10 thousandths of an hour).
a. leftendpoint values.
Time (sec)
Area and Estimating with Finite Sums
Time (h)
Velocity (mi / h)
Time (h)
Velocity (mi / h)
0.0 0.001 0.002 0.003 0.004 0.005
0 40 62 82 96 108
0.006 0.007 0.008 0.009 0.010
116 125 132 137 142
mi/hr 160
10. Distance traveled upstream You are sitting on the bank of a tidal river watching the incoming tide carry a bottle upstream. You record the velocity of the flow every 5 minutes for an hour, with the results shown in the accompanying table. About how far upstream did the bottle travel during that hour? Find an estimate using 12 subintervals of length 5 with
140 120 100
a. leftendpoint values.
80
b. rightendpoint values.
60
Time (min)
Velocity (m / sec)
Time (min)
Velocity (m / sec)
0 5 10 15 20 25 30
1 1.2 1.7 2.0 1.8 1.6 1.4
35 40 45 50 55 60
1.2 1.0 1.8 1.5 1.2 0
11. Length of a road You and a companion are about to drive a twisty stretch of dirt road in a car whose speedometer works but whose odometer (mileage counter) is broken. To find out how long this particular stretch of road is, you record the car’s velocity at 10sec intervals, with the results shown in the accompanying table. Estimate the length of the road using a. leftendpoint values.
40 20 0
0.002 0.004 0.006 0.008 0.01
hours
a. Use rectangles to estimate how far the car traveled during the 36 sec it took to reach 142 mi> h. b. Roughly how many seconds did it take the car to reach the halfway point? About how fast was the car going then? 13. Free fall with air resistance An object is dropped straight down from a helicopter. The object falls faster and faster but its acceleration (rate of change of its velocity) decreases over time because of air resistance. The acceleration is measured in ft>sec2 and recorded every second after the drop for 5 sec, as shown: t
0
1
2
3
4
5
a
32.00
19.41
11.77
7.14
4.33
2.63
b. rightendpoint values.
Time (sec)
Velocity (converted to ft / sec) (30 mi / h 44 ft / sec)
0 10 20 30 40 50 60
0 44 15 35 30 44 35
Time (sec)
Velocity (converted to ft / sec) (30 mi / h 44 ft / sec)
70 80 90 100 110 120
15 22 35 44 30 35
a. Find an upper estimate for the speed when t = 5. b. Find a lower estimate for the speed when t = 5. c. Find an upper estimate for the distance fallen when t = 3. 14. Distance traveled by a projectile An object is shot straight upward from sea level with an initial velocity of 400 ft> sec. a. Assuming that gravity is the only force acting on the object, give an upper estimate for its velocity after 5 sec have elapsed. Use g = 32 ft>sec2 for the gravitational acceleration. b. Find a lower estimate for the height attained after 5 sec.
298
Chapter 5: Integration
Average Value of a Function In Exercises 15–18, use a finite sum to estimate the average value of ƒ on the given interval by partitioning the interval into four subintervals of equal length and evaluating ƒ at the subinterval midpoints. 15. ƒsxd = x 3 on
16. ƒsxd = 1>x on
[0, 2]
17. ƒstd = s1>2d + sin2 pt
on
[1, 9]
Month
Jan
Feb
Mar
Apr
May
Jun
Pollutant release rate (tons> day)
0.20
0.25
0.27
0.34
0.45
0.52
Month
Jul
Aug
Sep
Oct
Nov
Dec
Pollutant release rate (tons> day)
0.63
0.70
0.81
0.85
0.89
0.95
[0, 2]
y y 1 sin 2 t 2
1.5
Measurements are taken at the end of each month determining the rate at which pollutants are released into the atmosphere, recorded as follows.
1 0.5 0
18. ƒstd = 1  acos
1
pt b 4
t
2
4
on
y
[0, 4] a. Assuming a 30day month and that new scrubbers allow only 0.05 ton> day to be released, give an upper estimate of the total tonnage of pollutants released by the end of June. What is a lower estimate?
4 y 1 ⎛cos t ⎛ ⎝ 4⎝
1
b. In the best case, approximately when will a total of 125 tons of pollutants have been released into the atmosphere? 0
1
2
3
4
t
21. Inscribe a regular nsided polygon inside a circle of radius 1 and compute the area of the polygon for the following values of n: a. 4 (square)
Examples of Estimations 19. Water pollution Oil is leaking out of a tanker damaged at sea. The damage to the tanker is worsening as evidenced by the increased leakage each hour, recorded in the following table. Time (h)
0
1
2
3
4
Leakage (gal / h)
50
70
97
136
190
Time (h)
5
6
7
8
265
369
516
720
Leakage (gal / h)
a. Give an upper and a lower estimate of the total quantity of oil that has escaped after 5 hours. b. Repeat part (a) for the quantity of oil that has escaped after 8 hours.
c. The tanker continues to leak 720 gal> h after the first 8 hours. If the tanker originally contained 25,000 gal of oil, approximately how many more hours will elapse in the worst case before all the oil has spilled? In the best case?
20. Air pollution A power plant generates electricity by burning oil. Pollutants produced as a result of the burning process are removed by scrubbers in the smokestacks. Over time, the scrubbers become less efficient and eventually they must be replaced when the amount of pollution released exceeds government standards.
b. 8 (octagon)
c. 16
d. Compare the areas in parts (a), (b), and (c) with the area of the circle. 22. (Continuation of Exercise 21. ) a. Inscribe a regular nsided polygon inside a circle of radius 1 and compute the area of one of the n congruent triangles formed by drawing radii to the vertices of the polygon. b. Compute the limit of the area of the inscribed polygon as n: q. c. Repeat the computations in parts (a) and (b) for a circle of radius r. COMPUTER EXPLORATIONS In Exercises 23–26, use a CAS to perform the following steps. a. Plot the functions over the given interval. b. Subdivide the interval into n = 100 , 200, and 1000 subintervals of equal length and evaluate the function at the midpoint of each subinterval. c. Compute the average value of the function values generated in part (b). d. Solve the equation ƒsxd = saverage valued for x using the average value calculated in part (c) for the n = 1000 partitioning. 23. ƒsxd = sin x on [0, p]
24. ƒsxd = sin2 x on [0, p]
p 1 25. ƒsxd = x sin x on c , p d 4
p 1 26. ƒsxd = x sin2 x on c , p d 4
5.2
5.2
Sigma Notation and Limits of Finite Sums
299
Sigma Notation and Limits of Finite Sums In estimating with finite sums in Section 5.1, we encountered sums with many terms (up to 1000 in Table 5.1, for instance). In this section we introduce a more convenient notation for sums with a large number of terms. After describing the notation and stating several of its properties, we look at what happens to a finite sum approximation as the number of terms approaches infinity.
Finite Sums and Sigma Notation Sigma notation enables us to write a sum with many terms in the compact form n
Á + an  1 + an . a ak = a1 + a2 + a3 +
k=1
The Greek letter © (capital sigma, corresponding to our letter S), stands for “sum.” The index of summation k tells us where the sum begins (at the number below the © symbol) and where it ends (at the number above © ). Any letter can be used to denote the index, but the letters i, j, and k are customary. The index k ends at k 5 n.
n The summation symbol (Greek letter sigma)
ak a k is a formula for the kth term.
k51
The index k starts at k 5 1.
Thus we can write 11
12 + 22 + 32 + 42 + 52 + 62 + 72 + 82 + 92 + 10 2 + 112 = a k 2, k=1
and 100
ƒs1d + ƒs2d + ƒs3d + Á + ƒs100d = a ƒsid. i=1
The lower limit of summation does not have to be 1; it can be any integer.
EXAMPLE 1 A sum in sigma notation
The sum written out, one term for each value of k
The value of the sum
ak
1 + 2 + 3 + 4 + 5
15
k a s 1d k
s 1d1s1d + s 1d2s2d + s 1d3s3d
1 + 2  3 = 2
k a k + 1
1 2 + 1 + 1 2 + 1
7 1 2 + = 2 3 6
k2 a k  1 k=4
52 42 + 4  1 5  1
16 25 139 + = 3 4 12
5 k=1 3 k=1 2 k=1 5
300
Chapter 5: Integration
EXAMPLE 2
Express the sum 1 + 3 + 5 + 7 + 9 in sigma notation.
The formula generating the terms changes with the lower limit of summation, but the terms generated remain the same. It is often simplest to start with k = 0 or k = 1, but we can start with any integer. Solution
4
Starting with k = 0:
1 + 3 + 5 + 7 + 9 = a s2k + 1d
Starting with k = 1:
1 + 3 + 5 + 7 + 9 = a s2k  1d
Starting with k = 2:
1 + 3 + 5 + 7 + 9 = a s2k  3d
Starting with k = 3:
1 + 3 + 5 + 7 + 9 = a s2k + 7d
k=0 5 k=1 6 k=2 1
k = 3
When we have a sum such as 3 2 a sk + k d
k=1
we can rearrange its terms, 3 2 2 2 2 a sk + k d = s1 + 1 d + s2 + 2 d + s3 + 3 d
k=1
= s1 + 2 + 3d + s12 + 22 + 32 d 3
Regroup terms.
3
= a k + a k 2. k=1
k=1
This illustrates a general rule for finite sums: n
n
n
a sak + bk d = a ak + a bk .
k=1
k=1
k=1
Four such rules are given below. A proof that they are valid can be obtained using mathematical induction (see Appendix 2).
Algebra Rules for Finite Sums n
1.
2.
n
n
a (ak + bk) = a ak + a bk
Sum Rule:
k=1
k=1
k=1
n
n
n
a (ak  bk) = a ak  a bk
Difference Rule:
k=1
3.
Constant Multiple Rule:
4.
Constant Value Rule:
k=1
k=1
# a cak = c a ak
(Any number c)
# ac = n c
(c is any constant value.)
n
n
k=1
k=1
n
EXAMPLE 3
k=1
We demonstrate the use of the algebra rules.
n
n
n
Difference Rule and Constant Multiple Rule
(a) a s3k  k 2 d = 3 a k  a k 2 k=1 n
k=1
k=1
n
n
n
k=1
k=1
k=1
k=1
(b) a s ak d = a s 1d # ak = 1 # a ak =  a ak
Constant Multiple Rule
5.2 3
3
Sigma Notation and Limits of Finite Sums
3
(c) a sk + 4d = a k + a 4 k=1
k=1
Sum Rule
k=1
= s1 + 2 + 3d + s3 # 4d = 6 + 12 = 18
Constant Value Rule
1 1 (d) a n = n # n = 1 n
Constant Value Rule (1>n is constant)
k=1
HISTORICAL BIOGRAPHY Carl Friedrich Gauss (1777–1855)
301
Over the years people have discovered a variety of formulas for the values of finite sums. The most famous of these are the formula for the sum of the first n integers (Gauss is said to have discovered it at age 8) and the formulas for the sums of the squares and cubes of the first n integers.
EXAMPLE 4
Show that the sum of the first n integers is n
ak =
k=1
Solution
nsn + 1d . 2
The formula tells us that the sum of the first 4 integers is s4ds5d = 10. 2
Addition verifies this prediction: 1 + 2 + 3 + 4 = 10. To prove the formula in general, we write out the terms in the sum twice, once forward and once backward. 1 n
+ +
2 sn  1d
+ +
+ +
3 sn  2d
Á Á
+ +
n 1
If we add the two terms in the first column we get 1 + n = n + 1. Similarly, if we add the two terms in the second column we get 2 + sn  1d = n + 1. The two terms in any column sum to n + 1. When we add the n columns together we get n terms, each equal to n + 1, for a total of nsn + 1d. Since this is twice the desired quantity, the sum of the first n integers is sndsn + 1d>2. Formulas for the sums of the squares and cubes of the first n integers are proved using mathematical induction (see Appendix 2). We state them here. n
The first n squares: The first n cubes:
2 ak =
k=1 n
nsn + 1ds2n + 1d 6
3 ak = a
k=1
nsn + 1d 2 b 2
Limits of Finite Sums The finite sum approximations we considered in Section 5.1 became more accurate as the number of terms increased and the subinterval widths (lengths) narrowed. The next example shows how to calculate a limiting value as the widths of the subintervals go to zero and their number grows to infinity.
EXAMPLE 5 Find the limiting value of lower sum approximations to the area of the region R below the graph of y = 1  x 2 and above the interval [0, 1] on the xaxis using equalwidth rectangles whose widths approach zero and whose number approaches infinity. (See Figure 5.4a.)
302
Chapter 5: Integration
We compute a lower sum approximation using n rectangles of equal width ¢x = s1  0d>n, and then we see what happens as n : q . We start by subdividing [0, 1] into n equal width subintervals
Solution
n  1 n 1 2 1 c0, n d, c n , n d, Á , c n , n d . Each subinterval has width 1>n. The function 1  x 2 is decreasing on [0, 1], and its smallest value in a subinterval occurs at the subinterval’s right endpoint. So a lower sum is constructed with rectangles whose height over the subinterval [sk  1d>n, k>n] is ƒsk>nd = 1  sk>nd2 , giving the sum k n 1 2 1 1 1 1 cƒ a n b d a n b + cƒ a n b d a n b + Á + cƒ a n b d a n b + Á + cƒ a n b d a n b . We write this in sigma notation and simplify, n
n
k=1
k=1 n
2
k k 1 1 a ƒ a n b a n b = a a1  a n b b a n b k2 1 = a an  3 b n k=1 n
n
k2 1 = a n  a 3 k=1 k=1 n 1 1 = n # n  3 ak 2 n k=1 1 sndsn + 1ds2n + 1d = 1  a 3b 6 n
Difference Rule
n
= 1 
2n 3 + 3n 2 + n . 6n 3
Constant Value and Constant Multiple Rules Sum of the First n Squares
Numerator expanded
We have obtained an expression for the lower sum that holds for any n. Taking the limit of this expression as n : q , we see that the lower sums converge as the number of subintervals increases and the subinterval widths approach zero: lim a1 
n: q
2n 3 + 3n 2 + n 2 2 b = 1  = . 3 6 3 6n
The lower sum approximations converge to 2>3. A similar calculation shows that the upper sum approximations also converge to 2>3. Any finite sum approximation g nk = 1 ƒsck ds1>nd also converges to the same value, 2>3. This is because it is possible to show that any finite sum approximation is trapped between the lower and upper sum approximations. For this reason we are led to define the area of the region R as this limiting value. In Section 5.3 we study the limits of such finite approximations in a general setting.
Riemann Sums HISTORICAL BIOGRAPHY Georg Friedrich Bernhard Riemann (1826–1866)
The theory of limits of finite approximations was made precise by the German mathematician Bernhard Riemann. We now introduce the notion of a Riemann sum, which underlies the theory of the definite integral studied in the next section. We begin with an arbitrary bounded function ƒ defined on a closed interval [a, b]. Like the function pictured in Figure 5.8, ƒ may have negative as well as positive values. We subdivide the interval [a, b] into subintervals, not necessarily of equal widths (or lengths), and form sums in the same way as for the finite approximations in Section 5.1. To do so, we choose n  1 points 5x1, x2 , x3 , Á , xn  16 between a and b and satisfying a 6 x1 6 x2 6 Á 6 xn  1 6 b.
5.2
To make the notation consistent, we denote a by x0 and b by xn , so that a = x0 6 x1 6 x2 6 Á 6 xn  1 6 xn = b.
y
y 5 f(x)
0 a
303
Sigma Notation and Limits of Finite Sums
The set
b
x
P = 5x0 , x1, x2 , Á , xn  1, xn6
is called a partition of [a, b]. The partition P divides [a, b] into n closed subintervals [x0 , x1], [x1, x2], Á , [xn  1, xn].
FIGURE 5.8 A typical continuous function y = ƒsxd over a closed interval [a, b].
The first of these subintervals is [x0 , x1], the second is [x1, x2], and the kth subinterval of P is [xk  1, xk], for k an integer between 1 and n. kth subinterval x0 a
x1
•••
x2
x k1
x xk
•••
xn b
x n1
The width of the first subinterval [x0 , x1] is denoted ¢x1 , the width of the second [x1, x2] is denoted ¢x2 , and the width of the kth subinterval is ¢xk = xk  xk  1 . If all n subintervals have equal width, then the common width ¢x is equal to sb  ad>n . D x1
D x2
D xk
D xn x
x0 5 a
{{{
x2
x1
x k21
xk
{{{
xn21
xn 5 b
In each subinterval we select some point. The point chosen in the kth subinterval [xk  1, xk] is called ck . Then on each subinterval we stand a vertical rectangle that stretches from the xaxis to touch the curve at sck , ƒsck dd. These rectangles can be above or below the xaxis, depending on whether ƒsck d is positive or negative, or on the xaxis if ƒsck d = 0 (Figure 5.9). On each subinterval we form the product ƒsck d # ¢xk . This product is positive, negative, or zero, depending on the sign of ƒsck d. When ƒsck d 7 0, the product ƒsck d # ¢xk is the area of a rectangle with height ƒsck d and width ¢xk . When ƒsck d 6 0, the product ƒsck d # ¢xk is a negative number, the negative of the area of a rectangle of width ¢xk that drops from the xaxis to the negative number ƒsck d. y
y f (x)
(cn, f (cn ))
(ck, f (ck )) kth rectangle
c1 0 x0 a
c2 x1
x2
x k1
ck xk
cn x n1
xn b
x
(c1, f (c1))
(c 2, f (c 2))
FIGURE 5.9 The rectangles approximate the region between the graph of the function y = ƒsxd and the xaxis. Figure 5.8 has been enlarged to enhance the partition of [a, b] and selection of points ck that produce the rectangles.
304
Chapter 5: Integration
Finally we sum all these products to get n
SP = a ƒsck d ¢xk . k=1
The sum SP is called a Riemann sum for ƒ on the interval [a, b]. There are many such sums, depending on the partition P we choose, and the choices of the points ck in the subintervals. For instance, we could choose n subintervals all having equal width ¢x = sb  ad>n to partition [a, b], and then choose the point ck to be the righthand endpoint of each subinterval when forming the Riemann sum (as we did in Example 5). This choice leads to the Riemann sum formula n
Sn = a ƒ aa + k
y
k=1
y f(x)
0 a
b
x
b  a # b  a n b a n b.
Similar formulas can be obtained if instead we choose ck to be the lefthand endpoint, or the midpoint, of each subinterval. In the cases in which the subintervals all have equal width ¢x = sb  ad>n, we can make them thinner by simply increasing their number n. When a partition has subintervals of varying widths, we can ensure they are all thin by controlling the width of a widest (longest) subinterval. We define the norm of a partition P, written 7P7 , to be the largest of all the subinterval widths. If 7P7 is a small number, then all of the subintervals in the partition P have a small width. Let’s look at an example of these ideas.
(a)
EXAMPLE 6 The set P = {0, 0.2, 0.6, 1, 1.5, 2} is a partition of [0, 2]. There are five subintervals of P: [0, 0.2], [0.2, 0.6], [0.6, 1], [1, 1.5], and [1.5, 2]:
y
y f(x)
x1 0
0 a
b
x 2 0.2
x3 0.6
x5
x4 1
1.5
2
x
x
(b)
FIGURE 5.10 The curve of Figure 5.9 with rectangles from finer partitions of [a, b]. Finer partitions create collections of rectangles with thinner bases that approximate the region between the graph of ƒ and the xaxis with increasing accuracy.
The lengths of the subintervals are ¢x1 = 0.2, ¢x2 = 0.4, ¢x3 = 0.4, ¢x4 = 0.5, and ¢x5 = 0.5. The longest subinterval length is 0.5, so the norm of the partition is 7P7 = 0.5. In this example, there are two subintervals of this length. Any Riemann sum associated with a partition of a closed interval [a, b] defines rectangles that approximate the region between the graph of a continuous function ƒ and the xaxis. Partitions with norm approaching zero lead to collections of rectangles that approximate this region with increasing accuracy, as suggested by Figure 5.10. We will see in the next section that if the function ƒ is continuous over the closed interval [a, b], then no matter how we choose the partition P and the points ck in its subintervals to construct a Riemann sum, a single limiting value is approached as the subinterval widths, controlled by the norm of the partition, approach zero.
Exercises 5.2 Sigma Notation Write the sums in Exercises 1–6 without sigma notation. Then evaluate them. 2
6k 1. a k + 1 k=1 4
3
k  1 2. a k k=1 5
3. a cos kp
4. a sin kp
p 5. a s 1dk + 1 sin k k=1
6. a s 1dk cos kp
k=1 3
k=1 4 k=1
7. Which of the following express 1 + 2 + 4 + 8 + 16 + 32 in sigma notation? 6
a. a 2k  1 k=1
5
b. a 2k k=0
4
c. a 2k + 1 k = 1
8. Which of the following express 1  2 + 4  8 + 16  32 in sigma notation? 6
a. a s 2dk  1 k=1
5
b. a s 1dk 2k k=0
3
c. a s 1dk + 1 2k + 2 k = 2
5.3 9. Which formula is not equivalent to the other two? 4 s 1dk  1 a. a k=2 k  1
2 s 1dk b. a k=0 k + 1
1 s 1dk c. a k = 1 k + 2
10. Which formula is not equivalent to the other two? 4
1
3
a. a sk  1d2 k=1
b. a sk + 1d2 k = 1
c. a k 2 k = 3
Express the sums in Exercises 11–16 in sigma notation. The form of your answer will depend on your choice of the lower limit of summation. 11. 1 + 2 + 3 + 4 + 5 + 6 13.
12. 1 + 4 + 9 + 16
1 1 1 1 + + + 2 4 8 16
15. 1 
14. 2 + 4 + 6 + 8 + 10
1 1 1 1 +  + 5 2 3 4
3 5 2 4 1 +  + 5 5 5 5 5
16. 
Values of Finite Sums n
n
17. Suppose that a ak = 5 and a bk = 6. Find the values of k=1
n
k=1
n
n
bk b. a k=1 6
a. a 3ak k=1 n
c. a sak + bk d k=1
n
d. a sak  bk d k=1
n
18. Suppose that a ak = 0 and a bk = 1. Find the values of k=1
k=1
n
n
a. a 8ak
b. a 250bk
c. a sak + 1d
d. a sbk  1d
k=1 n
k=1 n
k=1
k=1
Evaluate the sums in Exercises 19–32. 10
19. a. a k k=1 13
20. a. a k k=1
7
10
10
b. a k 2
c. a k 3
k=1 13
k=1 13
b. a k 2
c. a k 3
k=1
k=1
5
21. a s 2kd
pk 22. a k = 1 15
23. a s3  k 2 d
24. a sk 2  5d
k=1 6 k=1
5.3
7
k=1 5
26. a ks2k + 1d 5
k3 + a a kb 27. a k = 1 225 k=1
k=1 7
3
2
7
k3 28. a a kb  a k=1 k=1 4
7
500
264
29. a. a 3
b. a 7
c. a 10
30. a. a k
b. a k 2
c. a ksk  1d
31. a. a 4
b. a c
c. a sk  1d
1 32. a. a a n + 2nb
c b. a n
k c. a 2 k=1 n
k=1 36
k=1 17
k=9 n
k=3 n
k=1 n k=1
k=1 n k=1
k=3 71
k = 18 n k=1 n
Riemann Sums In Exercises 33–36, graph each function ƒ(x) over the given interval. Partition the interval into four subintervals of equal length. Then add to your sketch the rectangles associated with the Riemann sum © 4k = 1ƒsck d ¢xk , given that ck is the (a) lefthand endpoint, (b) righthand endpoint, (c) midpoint of the kth subinterval. (Make a separate sketch for each set of rectangles.) 33. ƒsxd = x 2  1,
[0, 2]
[p, p]
34. ƒsxd = x 2,
[0, 1]
36. ƒsxd = sin x + 1,
[p, p]
37. Find the norm of the partition P = 50, 1.2, 1.5, 2.3, 2.6, 36.
k=1
n
5
25. a ks3k + 5d
35. ƒsxd = sin x,
e. a sbk  2ak d
305
The Definite Integral
38. Find the norm of the partition P = 52, 1.6, 0.5, 0, 0.8, 16. Limits of Riemann Sums For the functions in Exercises 39–46, find a formula for the Riemann sum obtained by dividing the interval [a, b] into n equal subintervals and using the righthand endpoint for each ck. Then take a limit of these sums as n : q to calculate the area under the curve over [a, b]. 39. ƒsxd = 1  x 2 over the interval [0, 1]. 40. ƒsxd = 2x over the interval [0, 3]. 41. ƒsxd = x 2 + 1 over the interval [0, 3]. 42. ƒsxd = 3x 2 over the interval [0, 1]. 43. ƒsxd = x + x 2 over the interval [0, 1]. 44. ƒsxd = 3x + 2x 2 over the interval [0, 1].
6 k=1
45. ƒsxd = 2x 3 over the interval [0, 1]. 46. ƒsxd = x 2  x 3 over the interval [1, 0].
The Definite Integral In Section 5.2 we investigated the limit of a finite sum for a function defined over a closed interval [a, b] using n subintervals of equal width (or length), sb  ad>n. In this section we consider the limit of more general Riemann sums as the norm of the partitions of [a, b] approaches zero. For general Riemann sums the subintervals of the partitions need not have equal widths. The limiting process then leads to the definition of the definite integral of a function over a closed interval [a, b].
Definition of the Definite Integral The definition of the definite integral is based on the idea that for certain functions, as the norm of the partitions of [a, b] approaches zero, the values of the corresponding Riemann
Chapter 5: Integration
sums approach a limiting value J. What we mean by this limit is that a Riemann sum will be close to the number J provided that the norm of its partition is sufficiently small (so that all of its subintervals have thin enough widths). We introduce the symbol P as a small positive number that specifies how close to J the Riemann sum must be, and the symbol d as a second small positive number that specifies how small the norm of a partition must be in order for convergence to happen. We now define this limit precisely.
DEFINITION Let ƒ(x) be a function defined on a closed interval [a, b]. We say that a number J is the definite integral of ƒ over [a, b] and that J is the limit of the Riemann sums g nk = 1 ƒsck d ¢xk if the following condition is satisfied: Given any number P 7 0 there is a corresponding number d 7 0 such that for every partition P = 5x0 , x1, Á , xn