1.Alternate EditionCalculuswith analytk: geometryEarl W SwokowskiMarquette University~Prindle, Weber & SchmidtBoston, Massachusetts

2. Dedicated to the memory of my mother and father; Sophia and john SwokowskiF=WS F=UBLISHEF=tSPnndle Weber & Schmrdt II. Wollard Granl Press ooc: Duxbury Press Statler Off1ce Buld.ng 20 Provdence Street Boston Massachusetts 02116Copyright 1983 by PWS PublishersAll rights reserved. No part of this book may be reproduced ortransmitted in any form or by any means, electronic or mechanical,including photocopying, recording, or by any information storageand retrieval system, without permission, in writing, from thepublisher.PWS Publishers is a division of Wadsworth, Inc.Portions of this book previously appeared in Calculus with Ana-lytic Geometry, Second Edition by Earl W. Swokowski. Copyright 1979 by Prindle, Weber & Schmidt.8786 85 8483 -1098 7 65 43 2ISBN 0-87150-341-7Library of Congress Cataloging in Publication DataSwokowski, Earl W.Calculus with analytic geometry.Includes index.I. Calculus. 2. Geometry, Analytic.I. TitleQA303.S94 1983515.1582-21481ISBN 0-87150-341-7Cover image courtesy of General Motors Research Laboratories.The computer graphic image depicts the location of valence elec-trons trapped near the surface of rhodium. Quantum mechanicalcalculations using the Schrodingerequation were employed to gen-erate the image. The two sets of peaks in the foreground reveal apreferential accumulation of electrons around the surface atoms.Production and design: Kathi TownesText composition: Composition House LimitedTechnical artwork: Vantage Art, Inc.Cover printing: Federated Lithographers-Printers, Inc.Text printing/binding: Von Hoffmann Press, Inc.Printed in the United States of America 3. Preface Most students study calculus for its use as a tool in areas functions leads to some nontrivial applications of the Chain other than mathematics. They desire information about why Rule and enlarges the scope of applications of the derivative. calculus is important, and where and how it can be applied. IIn Chapter4testvalues are used to determine intervals in kept these facts in mind as I wrote this text. In particular, which derivatives are positive or negative. This pedagogical when introducing new concepts I often refer to problems device is also employed to help obtain graphs of rational that are familiar to students and that require methods of functions. calculus for solutions. Numerous examples and exercisesChapters 5 and 6, on properties and applications of defi- have been designed to further motivate student interest, notnite integrals, include exercises on numerical integration only in the mathematical or physical sciences, but in other that require reference to graphs to approximate areas, vol- disciplines as well. Figures are frequently used to bridge theumes, work, and force exerted by a liquid. gap between the statement of a problem and its solution. Inverse functions are discussed in the first section ofIn addition to achieving a good balance between theory Chapter 7 and are used in Section 7. 3 to define the natural-and applications, my primary objective was to write a bookexponential function as the inverse of the natural log- that can be read and understood by college freshmen. In eacharithmic function. section I have striven for accuracy and clarity of exposition, Chapters 8- 10 contain material on transcendental func- together with a presentation that makes the transition from tions, techniques of integration, and improper integrals. precalculus mathematics to calculus as smooth as possible.Infinite series are presented in a precise manner in Chapter The comments that follow highlight some of the featuresII. Chapter 12 consists of a detailed study of conic sections. of this text.Chapters 13-15 deal with curves, vectors and vector-A review of the trigonometric functions is contained invalued functions. There are many examples and exercises the last section of Chapter I. It was placed there, instead ofpertaining to parametric and polar equations, and a strong in an appendix, to alert students to the fact that trigonometry emphasis is placed on geometric and applied aspects of is, indeed, a prerequisite for calculus, as indicated by thevectors. title of the chapter. Tests for symmetry are also introduced Functions of several variables are discussed at length in early, so that they can be used throughout the text.Chapter 16. The relevance of level curves and surfaces toIn Chapter 2 limits involving the sine and cosine func-practical situations is illustrated in examples and exercises. tions are considered after limits of algebraic functions, and The approach to increments and differentials is motivated by thus are readily available for use in obtaining derivative for- analogous single variable concepts. The definition of direc- mulas in Chapter 3. The early introduction of trigonometric tional derivative does not require the use of direction angles ;;; 4. ivPrefaceof a line, and considerable stress is given to the gradient of versity of Georgia; Stanley M. Lukawecki, Clemson Uni-a function. The study of maxima and minima includes an versity; Louise E. Moser, California State University,examination of boundary extrema. The final section, on Hayward; Norman K. Nystrom, American River College;Lagrange multipliers, includes a proof that indicatesDavid A. Petrie, Cypress College; William Robinson, Ven-the geometric nature of why the method is valid. tura College; JohnT. Scheick, Ohio State University; Jon W. Properties and applications of multiple integrals are con-Scott, Montgomery College; Monty J. Strauss, Texas Techsidered in Chapter 17. University; Richard G. Vinson, University of South Ala- Vector fields are discussed in Chapter 18, and specialbama; Loyd Wilcox, Golden West College; and T. J.attention is given to conservative fields. The physical sig- Worosz, Metropolitan State College, Denver.nificance of divergence and curl is brought out by using theI also wish to express my gratitude to Christian C.theorems of Gauss and Stokes. The last two sections con- Braunschweiger of Marquette University, who provided an-tain results on Jacobians and change of variables in multipleswers for exercises; Thomas A. Bronikowski of Marquetteintegrals. University, who authored the student supplement containing Chapter 19, on differential equations, includes two sepa- detailed solutions for one-third of the exercises; Stephen B.rate sections on applications. Rodi of Austin Community College, who developed a com- There is a review section at the end of each chapter con- plete solutions manual; Michael B. Gregory of the U niver-sisting of a list of important topics and pertinent exercises. sity of North Dakota, who supplied a number of challengingThe review exercises are similar to those that appearexercises; and Christopher L. Morgan, California State U ni-throughout the text and may be used by students to prepare versity at Hayward, and Howard Pyron, University of Mis-for examinations. Answers to odd-numbered exercises aresouri at Rolla, who prepared the computer graphics. Specialgiven at the end of the text. Instructors may obtain an answer thanks are due to Stephen J. Merrill of Marquette Universitybooklet for the even-numbered exercises from the publisher.for suggesting several interesting examples, including one Portions of this text are based on material that appears in that indicates how infinite sequences and series may be em-my book Calculus withAnalyticGeometry, Second Edition. ployed to study the time course of an epidemic, and anotherThis second edition is available for courses where a later that illustrates the use of exponential functions in the field ofintroduction of the trigonometric functions is desired.radiation therapy. I wish to thank the following individuals, who receivedI am grateful for the valuable assistance of the staff ofall, or parts of, the manuscript and offered many helpfulPWS Publishers. In particular, Mary LeQuesne and Joesuggestions: Alfred Andrew, Georgia Institute of Technol-Power were very helpful with exercise sets; Kathi Townesogy; Jan F. Andrus, University of New Orleans; Robert M. did a superlative job as copy editor; and David Pallai, whoBrooks, University of Utah; Dennis R. Dunniger, Michigan supervised the production of this large project, was a con-State University; Daniel Drucker, Wayne State University;stant source of information and advice.Joseph M. Egar, Cleveland State University; Ronald D. In addition to all of the persons named here, I express myFerguson, San Antonio State College; Stuart Goldenberg,sincere appreciation to the many unnamed students andCalifornia Polytechnic State University; Theodore Guinn, teachers who have helped shape my views on how calculusUniversity of New Mexico; Joe A. Guthrie, University ofshould be presented in the classroom.Texas, El Paso; David Hoff, Indiana University; AdamHulin, University of New Orleans; W. D. Lichtenstein, Uni- Earl W. Swokowski 5. Table of ContentsIntroduction: What Is Calculus?ix 3. 7 Implicit Differentiation 1263.8Derivatives Involving Powers of Functions 1313.9Higher Order Derivatives 1353.10 NewtonsMethod 1381 Prerequisites for Calculus13.11 Review 1411.1Real Numbers 11.2Coordinate Systems in Two Dimensions91.3Lines 18 4 Applications of the Derivative 1441.4Functions 241.5Combinations of Functions 33 4.1Local Extrema of Functions 1441.6The Trigonometric Functions 37 4.2Rolles Theorem and the Mean Value Theorem /521.7Review 474.3The First Derivative Test 1564.4Concavity and the Second Derivative Test 1624.5Horizontal and Vertical Asymptotes 1704.6Applications of Extrema 1822 Limits and Continuity of Functions494. 7 The Derivative as a Rate of Change 1922.1Introduction 494.8Related Rates 2012.2Definition of Limit 54 4. 9 Antiderivatives 2062.3Theorems on Limits 604.10 Applications to Economics 2142.4One-Sided Limits 684.11 Review 22/2.5Limits of Trigonometric Functions722.6Continuous Functions 762.7Review 865 The Definite Integral2235.1Area 2235.2Definition of Definite Integral 2323 The Derivative875.3Properties of the Definite Integral 2393 .I Introduction 875.4The Mean Value Theorem for Definite3.2The Derivative of a Function 92 Integrals 2443.3Rules for Finding Derivatives 98 5. 5 The Fundamental Theorem of Calculus 2463.4Derivatives of the Sine and Cosine Functions 106 5.6Indefinite Integrals and Change of Variables 2543.5Increments and Differentials 111 5.7Numerical Integration 2633.6The Chain Rule 119 5.8Review 271v 6. viTable of Contents6 Applications of the Definite Integral274 10 Indeterminate Forms, Improper Integrals, and 6.1Area 274 Taylors Formula 457 6.2Solids of Revolution 283 6.3Volumes Using Cylindrical Shells29110.1The Indeterminate Forms 0/0 and xfx 457 6.4Volumes by Slicing 296 10.2Other Indeterminate Forms 464 6.5Work 299 I 0.3 Integrals with Infinite Limits of Integration 468 6.6Force Exerted by a Liquid 30610.4Integrals with Discontinuous Integrands 473 6.7Arc Length 31/ 10.5Taylors Formula 479 6.8Other Applications 317 10.6Review 488 6.9Review 323 11 Infinite Series 490 11.1Infinite Sequences 4907 Exponential and Logarithmic Functions325 11.2Convergent or Divergent Infinite Series 500 7.1Inverse Functions 32511.3Positive Term Series 5// 7.2The Natural Logarithmic Function 329 11.4Alternating Series 519 7.3The Natural Exponential Function 337 11.5Absolute Convergence 523 7.4Differentiation and Integration 34511.6Power Series 530 7.5General Exponential and LogarithmicII. 7 Power Series Representations of Functions 536Functions 35211.8Taylor and Maclaurin Series 541 7.6Laws of Growth and Decay 359 11.9The Binomial Series 550 7.7Derivatives of Inverse Functions 366 11.10 Review 553 7.8Review 370 12 Topics in Analytic Geometry 556 12.1Conic Sections 5568 Other Transcendental Functions 372 12.2Parabolas 557 12.3Ellipses 565 8.1Derivatives of the Trigonometric Functions 372 12.4Hyperbolas 571 8.2Integrals of Trigonometric Functions 378 12.5Rotation of Axes 577 8.3The Inverse Trigonometric Functions 382 12.6Review 581 8.4Derivatives and Integrals Involving InverseTrigonometric Functions 387 8.5The Hyperbolic Functions 393 8.6The Inverse Hyperbolic Functions 399 13 Plane Curves and Polar Coordinates583 8.7Review 403 13. I Plane Curves 583 13.2Tangent Lines to Curves 59/ 13.3Polar Coordinate Systems 594 13.4Polar Equations of Conics 6039 Additional Techniques and Applications 13.5Areas in Polar Coordinates 608 13.6Lengths of Curves 611 of Integration405 13.7Surfaces of Revolution 615 9.1Integration by Parts 406 13.8Review 6/9 9.2Trigonometric Integrals 412 9.3Trigonometric Substitutions 418 9.4Partial Fractions 423 9.5Quadratic Expressions 430 14 Vectors and Solid Analytic Geometry 621 9.6Miscellaneous Substitutions 43314.1Vectors in Two Dimensions 621 9.7Tables of Integrals 43714.2Rectangular Coordinate Systems in Three 9.8Moments and Centroids of Plane Regions 440 Dimensions 631 9.9Centroids of Solids of Revolution 44714.3Vectors in Three Dimensions 635 9.10 Review 453 14.4The Vector Product 645 7. Table of Contentsvii14.5Lines in Space 65218 Topics In Vector Calculus82914.6Planes 65414.7Cylinders and Surfaces of Revolution661 18.1 Vector Fields 82914.8Quadric Surfaces 66518.2 Line Integrals 83614.9Cylindrical and Spherical Coordinate18.3 Independence of Path 847Systems 670 18.4 Greens Theorem 85414.10 Review 67318.5 Surface Integrals 86218.6 The Divergence Theorem 86918.7 Stokes Theorem 87518.8 Transformations of Coordinates 88215 Vector-Valued Functions 67618.9 Change of Variables in Multiple Integrals 88515.1Definitions and Graphs 67618.10Review 89115.2Limits, Derivatives, and Integrals 68015.3Motion 68815.4Curvature 69215.5Tangential and Normal Components of 19 Differential Equations893Acceleration 70019. IIntroduction 89315.6Keplers Laws 70519.2 Exact Differential Equations 89815.7Review 71019.3 Homogeneous Differential Equations 90219.4 First-Order Linear Differential Equations 90619.5 Applications 90916 Partial Differentiation713 19.6 Second-Order Linear Differential Equations 91416.1Functions of Several Variables 71319.7 Nonhomogeneous Linear Differential16.2Limits and Continuity 72/ Equations 92016.3Partial Derivatives 727 19.8 Vibrations 92616.4Increments and Differentials 73319.9 Series Solutions of Differential EquatiDns 93116.5The Chain Rule 74219.10Review 93416.6Directional Derivatives 75016.7Tangent Planes and Normal Lines to Surfaces 75816.8Extrema of Functions of Several Variables 76416.9Lagrange Multipliers 770 Appendices16.10 Review 778Mathematical Induction AlIITheorems on Limits and Definite IntegralsA817 Multiple Integrals 780 III TablesA Trigonometric Functions A 1817. I Double Integrals 780B Exponential Functions Al917.2Evaluation of Double Integrals 785C Natural Logarithms Al917.3AreasandVolumes 794IVFormulas from GeometryA2017.4Moments and Center of Mass 79817.5Double Integrals in Polar Coordinates 80417.6Triple Integrals 80917.7Applications of Triple Integrals 81617.8Triple Integrals in Cylindrical and Spherical Answers to Odd-Numbered ExercisesA21Coordinates 82017.9Surface Area 82417.10 Review 827 Index A57 8. Introduction:What is Calculus? Calculus was invented in the seventeenth century to provide a tool for solving problems involving motion. The subject matter of geometry, algebra, and trigonometry is applicable to objects which move at constant speeds; how- ever, methods introduced in calculus are required to study the orbits of planets, to calculate the flight of a rocket, to predict the path of a charged particle through an electromagnetic field and, for that matter, to deal with all aspects of motion.In order to discuss objects in motion it is essential first to define what is meant by velocity and acceleration. Roughly speaking, the velocity of an object is a measure of the rate at which the distance traveled changes with respect to time. Acceleration is a measure of the rate at which velocity changes. Velocity may vary considerably, as is evident from the motion of a drag-strip racer or the descent of a space capsule as it reenters the Earths atmosphere. In order to give precise meanings to the notions of velocity and acceleration it is necessary to use one of the fundamental concepts of calculus, the derivative.Although calculus was introduced to help solve problems in physics, it has been applied to many different fields. One of the reasons for its versatility is the fact that the derivative is useful in the study of rates of change of many entities other than objects in motion. For example, a chemist may use derivatives to forecast the outcome of various chemical reactions. A biologist may employ it in the investigation of the rate of growth of bacteria in a culture. An electrical engineer uses the derivative to describe the change in current in an electrical circuit. Economists have applied it to problems involving corporate profits and losses.The derivative is also used to find tangent lines to curves. Although this has some independent geometric interest, the significance of tangent lines is of major importance in physical problems. For example, if a particle moves along a curve, then the tangent line indicates the direction of motion. If we restrict our attention to a sufficiently small portion of the curve, then in aix 9. X Introduction: What is Calculus?certain sense the tangent line may be used to approximate the position of theparticle.Many problems involving maximum and minimum values may beattacked with the aid of the derivative. Some typical questions that can beanswered are: At what angle of elevation should a projectile be fired in orderto achieve its maximum range? If a tin can is to hold one gallon of a liquid,what dimensions require the least amount of tin? At what point between twolight sources will the illumination be greatest? How can certain corporationsmaximize their revenue? How can a manufacturer minimize the cost ofproducing a given article?Another fundamental concept of calculus is known as the definiteintegral. It, too, has many applications in the sciences. A physicist uses it tofind the work required to stretch or compress a spring. An engineer may useit to find the center of mass or moment of inertia of a solid. The definiteintegral can be used by a biologist to calculate the flow of blood through anarteriole. An economist may employ it to estimate depreciation of equipmentin a manufacturing plant. Mathematicians use definite integrals to investigatesuch concepts as areas of surfaces, volumes of geometric solids, and lengthsof curves.All the examples we have listed, and many more, will be discussed indetail as we progress through this book. There is literally no end to theapplications of calculus. Indeed, in the future perhaps you, the reader, willdiscover new uses for this important branch of mathematics.The derivative and the definite integral are defined in terms of certainlimiting processes. The notion of limit is the initial idea which separatescalculus from the more elementary branches of mathematics. Sir IsaacNewton (1642-1727) and Gottfried Wilhelm Leibniz (1646-1716) discoveredthe connection between derivatives and integrals. Because of this, and theirother contributions to the subject, they are credited with the invention ofcalculus. Many other mathematicians have added a great deal to its develop-ment.The preceding discussion has not answered the question "What iscalculus?" Actually, there is no simple answer. Calculus could be called thestudy of limits, derivatives, and integrals; however, this statement is meaning-less if definitions of the terms are unknown. Although we have given a fewexamples to illustrate what can be accomplished with derivatives and inte-grals, neither of these concepts has been given any meaning. Defining themwill be one of the principal objectives of our early work in this text. 10. Prerequisites for CalculusThis chapter contains topics necessary for the study ofattention to one of the most important concepts incalculus. After a brief review of real numbers, coordinate mathematics-the notion of function.systems, and graphs in two dimensions, we turn our J.J Real Numbers Real numbers are used considerably in precalculus mathematics, and we will assume familiarity with the fundamental properties of addition, subtraction, multiplication, division, exponents and radicals. Throughout this chapter, unless otherwise specified, lower-case letters a, b, c, ... denote real numbers. The positive integers 1, 2, 3, 4, ... may be obtained by adding the real number 1 successively to itself. The integers consist of all positive and negative integers together with the real number 0. A rational number is a real number that can be expressed as a quotient ajb, where a and bare integers and b -1= 0. Real numbers that are not rational are called irrational. The ratio of the circumference of a circle to its diameter is irrational. This real number is denoted by n and the notation n ~ 3.1416 is used to indicate that n is ap- proximately equal to 3.1416. Another example of an irrational number is .)2. Real numbers may be represented by nonterminating decimals. For example, the decimal representation for the rational number 7434/2310 is found by long division to be 3.2181818 ... , where the digits I and 8 repeat indefinitely. Rational numbers may always be represented by repeating decimals. Decimal representations for irrational numbers may also be obtained; however, they are non terminating and nonrepeating. It is possible to associate real numbers with points on a line lin such a way that to each real number a there corresponds one and only one point, and I 11. 2 1 Prerequisites for Calculusconversely, to each point Pthere corresponds precisely one real number. Suchan association between two sets is referred to as a one-to-one correspondence.We first choose an arbitrary point 0, called the origin, and associate with itthe real number 0. Points associated with the integers are then determined byconsidering successive line segments of equal length on either side of 0 asillustrated in Figure 1.1. The points corresponding to rational numbers suchas 253 and -tare obtained by subdividing the equal line segments. Pointsassociated with certain irrational numbers, such as fi, can be found bygeometric construction. For other irrational numbers such as rc, no con-struction is possible. However, the point corresponding to n can be approxi-mated to any degree of accuracy by locating successively the points corres-ponding to 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, .... It can be shown that toevery irrational number there corresponds a unique point on I and, con-versely, every point that is not associated with a rational number correspondsto an irrational number.0B A -3 -2~~r0 17; .,fi; 1f4 ; b a1323-2 25FIGURE 1.1The number a that is associated with a point A on I is called the coordinateof A. An assignment of coordinates to points on I is called a coordinatesystem for /, and I is called a coordinate line, or a real line. A direction can beassigned to I by taking the positive direction to the right and the negativedirection to the left. The positive direction is noted by placing an arrowheadon I as shown in Figure 1.1.The real numbers which correspond to points to the right of 0 in Figure1.1 are called positive real numbers, whereas those which correspond topoints to the left of 0 are negative real numbers. The real number 0 is neitherpositive nor negative. The collection of positive real numbers is closedrelative to addition and multiplication; that is, if a and b are positive, thenso is the sum a + band the product ab.If a and bare real numbers, and a - b is positive, we say that a is greaterthan h and write a > b. An equivalent statement is h is less than a, writtenb < a. The symbols > or < are called inequality signs and expressions suchas a > b or b < a are called inequalities. From the manner in which we con-structed the coordinate line I in Figure 1.1, we see that if A and B are pointswith coordinates a and b, respectively, then a > b (orb < a) if and only if Alies to the right of B. Since a - 0 = a, it follows that a > 0 if and only if a ispositive. Similarly, a < 0 means that a is negative. The following propertiesof inequalities can be proved. If a > b and b > c, then a > c. If a > b, then a + c > b + c.(J.J) If a > b and c > 0, then ac > be.If a > b and c < 0, then ac < be.Analogous properties for" less than" can also be established. 12. Real Numbers 1.1 3The symbol a ~ b, which is read a is greater than or equal to h, means thateither a > bora = b. The symbol a < b < c means that a < band b < c,inwhichcasewesaythathishetweenaandc. Thenotationsa :s; b,a < b :s; c,a :s; b < c, a :s; b :s; c, and so on, have similar meanings.Another property, called completeness, is needed to characterize the realnumbers. This property will be discussed in Chapter 11. If a is a real number, then it is the coordinate of some point A on a co-ordinate line /, and the symbol Ia I is used to denote the number of units (or 1-41=4 141=4 distance) between A and the origin, without regard to direction. Referring,------"-...~to Figure 1.2 we see that for the point with coordinate -4 we have I -41 = 4. + I I I+ I I I + ISimilarly, 141 = 4. In general, if a is negative we change its sign to find Ia I,-5 -4 -3 -2 -1 01 2 3 4 5 Iwhereas if a is nonnegative then Ia I = a. The nonnegative number Ia I isFIGURE 1.2called the absolute value of a. The following definition of absolute value summarizes our remarks. if a~ 0Definition (1.2) lal = { a-a if a< 0Example 1 Find 131, 1-31,101, IJ2- 21, and 12- fll.Solution Since 3, 2 - J2, and 0 are nonnegative, we have131 = 3,12 - J21 = 2 - J2, and I0 I = 0.Since -3 andJ2 - 2 are negative, we use the formula Ia I = -a of Definition(1.2) to obtain 1-31 = -( -3) = 3 and IJ2- 21 =- b if and only if a > b or a < - bIa I = b if and only if a = b or a = -b. It follows from the first and third properties stated in (1.4) thatIa I :s; b if and only if-b :s; a :s; b. The Triangle Inequality (1 .5) Ia + bl ;S; lal + lbl 13. 4 1 Prerequisites for CalculusProof From (1.3), -lal ~a~ lal and -lbl~ b ~ lbl. Addingcorresponding sides we obtain-(lal + lbi) ~a+ b~ lal + lbl.Using the remark preceding this theorem gives us the desired conclusion. D 5= 17-21= 12-71We shall use the concept of absolute value to define the distance between~ any two points on a coordinate line. Let us begin by noting that the distance I I I I+ I I I I+ I between the points with coordinates 2 and 7 shown in Figure 1.3 equals-2 -1 0 1 2 3 4 5 6 7 8 1 5 units on /. This distance is the difference, 7 - 2, obtained by subtractingFIGURE 1.3the smaller coordinate from the larger. If we employ absolute values, then,since 17 - 21 = 12 - 71, it is unnecessary to be concerned about the order ofsubtraction. We shall use this as our motivation for the next definition. Definition (1.6) Let a and b be the coordinates of two points A and B, respectively, on acoordinate line 1. The distance between A and B, denoted by d(A,B), isdefined byd(A, B) = lb- al. The number d(A, B) is also called the length of the line segment AB. Observe that, since d(B, A)= Ia- bl and lb- al = Ia- bl, we maywrite d(A, B) = d(B, A).Also note that the distance between the origin 0 and the point A is d(O, A)= Ia- Ol = Ia I,which agrees with the geometric interpretation of absolute value illustratedin Figure 1.2.Example 2 If A, B, C, and D have coordinates -5, -3, 1, and 6, respectively,find d(A, B), d(C, B), d(O, A), and d(C, D).ABOC DSolution The points are indicated in Figure 1.4.lltllttlllltl.By Definition (1.6),-5 -30 1 61FIGURE 1.4 d(A, B)= I - 3 - (- 5) I = I - 3 + 51 = 121 = 2. d(C, B)= 1-3- 11 = 1-41 = 4. d(O, A)= l-5- 01 =I-51= 5. d( C, D) = 16 - 11 = 151 = 5. 14. Real Numbers 1.1 5 The concept of absolute value has uses other than that of finding distances between points. Generally, it is employed whenever one is interested in the magnitude or numerical value of a real number without regard to its sign.In order to shorten explanations it is sometimes convenient to use the notation and terminology of sets. A set may be thought of as a collection of objects of some type. The objects are called elements of the set. Throughout our work ~will denote the set of real numbers. If Sis a set, then a E S means that a is an element of S, whereas a S signifies that a is not an element of S. If every element of a set Sis also an element of a set T, then Sis called a subset ofT. Two sets Sand Tare said to be equal, written S = T, if Sand Tcontain precisely the same elements. The notation S #- T means that Sand Tare not equal. If Sand Tare sets, their unionS u T consists of the elements which are either in S, in T, or in both S and T. The intersection S n T consists of the elements which the sets have in common. If the elements of a set Shave a certain property, tpen we write S = {x : ... } where the property describing the arbitrary element x is stated in the space after the colon. For example, {x: x > 3} may be used to represent the set of all real numbers greater than 3.Of major importance in calculus are certain subsets of~ called intervals. If a < b, tpe symbol (a, b) is sometimes used for all re!!l numbers between a and b. This set is called an open interval. Thus we have:(1.7)(a, b) = { x: a< x < b}. The numbers a and b are called the endpoints of the interval. The graphofa set Sofreal numbers is defined as the points on a coordinate ( ) line that correspond to the numbers inS. In particular, the graph of the open a b interval (a, b) consists of all points between the points corresponding to a and b. In Figure 1.5 we have sketched the graphs of a general open interval (a, b) ( I ) and the special open intervals (- 1, 3) and (2, 4). The parentheses in the figure -1 03 indicate that the endpoints of the intervals are not to be included. For convenience, we shall use the terms interval and graph of an interval inter- ( ) changeably.02 4 If we wish to include an endpoint of an interval, a bracket is used insteadFIGURE 1.5 Open intervals (a, b), (-I, 3), of a parenthesis. If a < b, then closed intervals, denoted by [a, b], and half-and (2, 4) open intervals, denoted by [a, b) or (a, b], are defined as follows. [a, b] = { x: a ~ x ~ b}(1.8)[a, b) = {x: a ~ x< b} (a, b] = {x: a< x ~ b} Typical graphs are sketched in Figure 1.6, where a bracket indicates that the corresponding endpoint is part of the graph.[ 3 [ ) ( 3a b aba bFIGURE 1.6 15. 6 1 Prerequisites for Calculus In future discussions of intervals, whenever the magnitudes of a and b are not stated explicitly it will always be assumed that a < b. If an interval is a subset of another interval I it is called a subinterval of/. For example, the closed interval [2, 3] is a subinterval of [0, 5]. We shall sometimes employ the following infinite intervals. (a, oo) = {x: x >a} [a,oo) = {x:x ~a} (1.9)(-oo,a) = {x:x 2x - 5.Solution The following inequalities are equivalent: 4x +3>2x- 54x > 2x- 82x > -8 X> -4Hence the solutions consist of all real numbers greater than -4, that is, thenumbers in the infinite interval (- 4, oo ).E xamp le 4 So Ive the mequa 1 -5 < - - -3x < 1. .. 1ty4-2Solution We may proceed as follows:4- 3x -5 3 or 2x- 7 < -3.The first of these two inequalities is equivalent to 2x > 10, or x > 5. Thesecond is equivalent to 2x < 4, or x < 2. Hence the solutions of 12x - 71 > 3are the numbers in the union (- oo, 2) u (5, oo ).1.1 ExercisesIn Exercises 1 and 2 replace the comma between each pair of5 If A, B, and C are points on a coordinate line with co-real numbers with the appropriate symbol , or =. ordinates -5, -I, and 7, respectively, find the following distances. 1 (a) -2, -5 (b) -2, 5(c) 6- I, 2+3 (d) !. 0.66(a) d(A, B) (b) d(B, C)(e) 2, J4 (f) 1!:, (c) d(C, B) (d) d(A, C) 2 (a) -3,0 (b) -8, -3 6 Rework Exercise 5 if A, B, and C have coordinates 2, -8, and -3, respectively. (c) 8, -3(d)i-i,-fs(e) Ji, 1.4 (f) tm. 3.6513 Solve the inequalities in Exercises 7-34 and express the solutions in terms of intervals.Rewrite the expressions in Exercises 3 and 4 without using 7 5x- 6 >II 8 3x- 5 < 10symbols for absolute values.9 2- 7x~ 1610 7- 2x~ -3 3 (a) 12-51(b) I-51+ 1-2111 12x + 11 > 512 lx+21 -7 16 5 > 2- 9x > -4(i) 15 - X I if X > 5 (j) Ia - b I if a < b 3- 7x17 -1 < -4 0 20-2--9 > 0(e) 1-W (f) 12- fil7- 2x X +(g) l-0.671 (h) -l-3121 lx- 101 < 0.3 22 12x + 31 0, and hence d(A 1 , A 2 ) = x 2 - x 1 Since M 1 is halfway from A 1 to A 2 , the x-coordinate of M 1 isx1 + !(x 2- x 1) = x 1 + !x 2 - !x 1X = !x 1 + !x 2 x 1 + x2 FIGURE 1.10 2 21. 12 1 Prerequisites for CalculusIt follows that the x-coordinate of M is also (x 1 + x 2 )/2. It can be shown insimilar fashion that they-coordinate of M is (y 1 + Y2)/2. Moreover, theseformulas hold for all positions of P 1 and P 2 . This gives us the followingresult. Midpoint Formula (1.11)Example 2 Find the midpoint M of the line segment from P 1( - 2, 3) toP 2 (4, -2). Plot the points P 1 , P 2 , M and verify that d(P" M) = d(P 2 , M).Solution Applying the Midpoint Formula (1.11), the coordinates of Mare( -2 2+ 4 , 3 + ( -2))or ( 1, ~) .2 2 yThe three points P 1 , P 2 , and Mare plotted in Figure 1.11. Using the DistanceFormula we obtain P 1 (-2, 3) d(P 1 , M) =j(-2- V + (3- !) 2 = J9 + (1}-) = j61;2 d(P2, M) =j(4- 1? + ( -2- !) 2 = J9 + eJ) = fo/2XHence d(P 1 , M) = d(P 2 , M). If W is a set of ordered pairs, then we may consider the point P(x, y) ina coordinate plane which corresponds to the ordered pair (x, y) in W. Thegraph of W is the set of all points that correspond to the ordered pairs in W. FIGURE 1.11The phrase" sketch the graph W" means to illustrate the significant featuresof the graph geometrically on a coordinate plane. yExample 3Sketch the graph of W= {(x, y): lxl s 2, IYI s 1}. 3Solution The indicated inequalities are equivalent to -2 s x s 2 and 2- 1 s y s 1. Hence the graph of W consists of all points within and on theboundary of the rectangular region shown in Figure 1.12.Example 4 Sketch the graph of W = {(x, y): y = 2x- 1}. {(x, y): lx I ,;;; 2, IY I ,;;; 1}-2SolutionWe begin by finding points with coordinates of the form (x, y)where the ordered pair (x, y) is in W. It is convenient to list these coordinates FIGURE 1.12in the following tabular form, where for each real number x the correspondingvalue for y is 2x - 1. 22. Coordinate Systems in Two Dimensions 1.2 13 y(3, 5)X ~-2-10 2 3y -5 -3 -1 3 5 After plotting, it appears that the points with these coordinates all lie on a line and we sketch the graph (see Figure 1.13). Ordinarily the few points we have plotted would not be enough to illustrate the graph; however, in this elementary case we can be reasonably sure that the graph is a line. In the nextX section we will prove that our conjecture is correct. {(x, y): y = 2x -1} The x-coordinates of points at which a graph intersects the x-axis are called the x-intercepts of the graph. Similarly, they-coordinates of points at which a graph intercepts they-axis are called they-intercepts. In Figure 1.13, there is one x-intercept 1/2 and one y-intercept - 1.FIGURE /.13It is impossible to sketch the entire graph in Example 4 since x may be assigned values which are numerically as large as desired. Nevertheless, we often call a drawing of the type given in Figure 1.13 the graph of W or a sketch of the graph where it is understood that the drawing is only a device for visualizing the actual graph and the line does not terminate as shown in the figure. In general, the sketch of a graph should illustrate enough of the graph so that the remaining parts are evident. The graph in Example 4 is determined by the equation y = 2x - 1 in the sense that for every real number x, the equation can be used to find a number y such that (x, y) is in W. Given an equation in x andy, we say that an ordered pair (a, b) is a solution of the equation if equality is obtained when a is sub- stituted for x and b for y. For example, (2, 3) is a solution of y = 2x - 1 since substitution of 2 for x and 3 for y leads to 3 = 4 - 1, or 3 = 3. Two equations in x andy are said to be equivalent if they have exactly the same solutions. The solutions of an equation in x and y determine a set S of ordered pairs, and we define the graph of the equation as the graph of S. Notice that the solutions of the equation y = 2x - 1 are the pairs (a, b) such that b = 2a - 1, and hence the solutions are identical with the set W given in Example 4. Consequently the graph of the equation y = 2x - 1 is the same as the graph of W (see Figure I.i3). For some of the equations we shall encounter in this chapter the technique used for sketching the graph will consist of plotting a sufficient number of points until some pattern emerges, and then sketching the graph accordingly. This is obviously a crude (and often inaccurate) way to arrive at the graph; however, it is a method often employed in el~mentary courses. As we progress through this text, techniques will be introduced that will enable us to sketch accurate graphs without plotting many points. Example 5 Sketch the graph of the equation y = x 2 Solution To obtain the graph, it is necessary to plot more points than in the previous example. Increasing successive x-coordinates by t, we obtain the following table.X -3! -2-~ -1t 0 t ~ 2 ! 3y9 ll4 4 *0 * 23. 141 Prerequisites for CalculusLarger numerical values of x produce even larger values of y. For example, thepoints (4, 16), (5, 25), and (6, 36) are on the graph, as are ( -4, 16), (- 5, 25),and (- 6, 36). Plotting the points given by the table and drawing a smoothcurve through these points gives us the sketch in Figure 1.14, where we havelabeled several points.The graph in Example 5 is called a parabola. The lowest point (0, 0) iscalled the vertex of the parabola and we say that the parabola opens upward.If the graph were inverted, as would be the case for y = - x 2 , then the para-bola opens downward. They-axis is called the axis of the parabola. Parabolasand their properties will be discussed in detail in Chapter 12, where it willbe shown that the graph of every equation of the form y = ax 2 + bx + c,FIGURE 1.14 with a =f. 0, is a parabola whose axis is parallel to they-axis. Parabolas mayalso open to the right or to the left (cf. Example 6).If the coordinate plane in Figure 1.14 is folded along they-axis, then thegraph which lies in the left half of the plane coincides with that in the right half.We say that the graph is symmetric with respect to they-axis. As in (i) ofFigure 1.15, a graph is symmetric with respect to the y-axis provided thatthe point ( -x, y) is on the graph whenever (x, y) is on the graph. Similarly,as in (ii) of Figure 1.15, a graph is symmetric with respect to the x-axis if,whenever a point (x, y) is on the graph, then (x, - y) is also on the graph.In this case if we fold the coordinate plane along the x-axis, the part of thegraph which lies above the x-axis will coincide with the part which lies below.Another type of symmetry which certain graphs possess is called symmetrywith respect to the origin. In this situation, whenever a point (x, y) is on thegraph, then (- x, - y) is also on the graph, as illustrated in (iii) of Figure 1.15.yy~ XX (i) y-axis(ii) x-axis(iii) originFIGURE 1.15Symmetries The following tests are useful for investigating these three types ofsymmetry for graphs of equations in x and y. Tests for Symmetry (1.12) (i) The graph of an equation is symmetric with respect to they-axis if substitution of- x for x leads to an equivalent equation.(ii) The graph of an equation is symmetric with respect to the x-axis if substitution of - y for y leads to an equivalent equation. (iii) The graph of an equation is symmetric with respect to the origin if the simultaneous substitution of -x for x and - y for y leads to an equivalent equation. 24. Coordinate Systems in Two Dimensions 1.2 15 If, in the equation of Example 5, we substitute - x for x, we obtain y = ( -x) 2 , which is equivalent toy= x 2 Hence, by Test (i), the graph is sym- metric with respect to the y-axis. If symmetry with respect to an axis exists, then it is sufficient to determine the graph in half of the coordinate plane, since the remainder of the graph is a mirror image, or reflection, of that half. Example 6 Sketch the graph ofi = x. SolutionSince substitution of - y for y does not change the equation, the graph is symmetric with respect to the x-axis. (See Symmetry Test (ii).) It is sufficient, therefore, to plot points with nonnegative y-coordinates and then reflect through the x-axis. Since i = x, they-coordinates of points x above the x-axis are given by y =Jx. Coordinates of some points on the graph are tabulated below. A portion of the graph is sketched in Figure 1.16. The graph is a parabola that opens to the right, with its vertex at the origin. In this case the x-axis is the axis of the parabola.FIGURE 1.16X0 23 4 9 y0 j2j323 Example 7 Sketch the graph of the equation 4y = x 3 . SolutionIf we substitute - x for x and - y for y, then 4(- y) = ( -x) 3 or -4y = -x 3 Multiplying both sides by -1, we see that the last equation has the same solutions as the given equation 4y = x 3 . Hence, from Symmetry Test (iii), the graph is symmetric with respect to the origin. The following table lists some points on the graph.X 0 I2t 2 5 2 By symmetry (or substitution) we see that the points ( -1, -!), ( -2, -2),FIGURE 1.17etc., are on the graph. Plotting points leads to the graph in Figure 1.17. If C(h, k) is a point in a coordinate plane, then a circle with center C and radius r > 0 may be defined as the collection of all points in the plane that are r units from C. As shown in (i) of Figure 1.18, a point P(x, y) is on the circle if and only if d( C, P) = r or, by the Distance Formula, if and only ifj(x - h) 2+ (y - k) 2 = r. The equivalent equation(1.13)(x - h)z+ (y- k)z= ,z 25. /6 1 Prerequisites for Calculusy is an equation of a circle of radius rand center C(h, k). If h = 0 and k = 0,this equation reduces to x 2 + y 2 = r 2 , which is an equation of a circle ofradius r with center at the origin (see (ii) of Figure 1.18). If r = l, the graph of(1.13) is called a unit circle.XExample 8 Find an equation of the circle with center C(- 2, 3) which passesthrough the point D( 4, 5). (i) (x- h)2 + (y- k)2 = ,2SolutionSince Dis on the circle, the radius r is d(C, D). By the DistanceFormula, yr = j(-2- 4) 2 + (3- 5) 2 = )36+4 =flO. Using ( 1.13) with h = - 2 and k = 3, we obtain (x + 2) 2 + (y- 3? = 40, (-r, 0) (r, 0) x orx2+ y 2 + 4x - 6y - 27= 0. Squaring terms in (1.13) and simplifying, we obtain an equation of the (ii) x2 + y2 = ,2form FIGURE 1.18x2 +y2 + ax + by + c =0where a, b, and c are real numbers. Conversely, if we begin with such anequation, it is always possible, by completing the squares in x andy, to obtainan equation of the form(x - h) 2 + (y - k) 2 =d.The method will be illustrated in Example 9. If d > 0, the graph is a circlewith center (h, k) and radius r =Jd. If d = 0, then, since (x - h) 2 :::::: 0and (y - k)2 :::::: 0, the only solution of the equation is (h, k), and hencethe graph consists of only one point. Finally, if d < 0, the equation has noreal solutions and there is no graph.Example 9 Find the center and radius of the circle with equation x2+ y2-4x + 6y- 3= 0.Solution We begin by arranging the equation as follows:(x 2 -4x) + (y 2 + 6y) =3.Next we complete the squares by adding appropriate numbers within theparentheses. Of course, to obtain equivalent equations we must add thenumbers to both sides of the equation. In order to complete the square foran expression of the form x 2 + ax, we add the square of half the coefficient ofx, that is, (a/2) 2 , to both sides of the equation. Similarly, for y 2 + by, we add 26. Coordinate Systems in Two Dimensions1.217(b/2) 2 to both sides. In this example a = -4, b = 6, (a/2) 2 = ( - 2) 2 = 4,and (b/2) 2 = 3 2 = 9. This leads to(x 2 - 4x + 4) + (y 2 + 6y + 9) = 3 + 4 + 9or (x - 2) 2 + (y + 3) 2 = 16.Hence, by (1.13) the center is (2, - 3) and the radius is 4. 1.2 ExercisesIn Exercises 1-6, find (a) the distance d(A, B) between the points In Exercises 21-38 sketch the graph of the equation and useA and B, and (b) the midpoint of the segment AB. ( 1.12) to test for symmetry. A(6, -2), B(2, I)2 A( -4, -I), B(2, 3)21 y = 3x + I 22 y = 4x- 3 3 A(O, -7),8(-1, -2) 4 A(4, 5), B(4, -4)23 y = -2x + 324y = 2- 3x 5 A( -3, -2), B( -8, -2) 6 A(! I, -7), B( -9, 0)25 y = 2x2 - 126 y = -x 2 + 2In Exercises 7 and 8 prove that the triangle with vertices A, B, 27 4y = x 2 283y + x2 = 0and Cis a right triangle and find its area.29 y = -tx3 30 y = tx 3 7 A( -3, 4), B(2, -I), C(9, 6) 31 y = x 3 - 232 y = 2- x 3 8 A(7, 2), B( -4, 0), C(4, 6) 33 y=Jx 34y=Jx-1 9 Prove that the following points are vertices of a parallelo- gram: A(- 4, - I), B(O, - 2), C( 6, I), D(2, 2).35 y=Fx 36 y=~10 Given A( -4, -3) and B(6, 1), find a formula which37 x 2 + l = 16 38 4x 2 + 4y 2 = 25 expresses the fact that P(x, y) is on the perpendicular In Exercises 39-46 find an equation of a circle satisfying the bisector of AB. stated conditions.11 For what values of a is the distance between (a, 3) and 39 Center C(3, - 2), radius 4 (5, 2a) greater than J26? 40 Center C(- 5, 2), radius 512 Given the points A(- 2, 0) and B(2, 0), find a formula not containing radicals that expresses the fact that the sum41 Center at the origin, passing through P(- 3, 5) of the distances from P(x, y) to A and to B, respectively,42 Center C( -4, 6), passing through P(l, 2) is 5. 43 Center C( -4, 2), tangent to the x-axis13 Prove that the midpoint of the hypotenuse of any right triangle is equidistant from the vertices. (Hint: Label 44 Center C(3, - 5), tangent to the y-axis the vertices of the triangle 0(0, 0), A(a, 0), and B(O, b).)45 Endpoints of a diameter A(4, - 3) and B(- 2, 7)14 Prove that the diagonals of any parallelogram bisect46 Tangent to both axes, center in the first quadrant, radius 2 each other. (Hint: Label three of the vertices of the paral- lelogram 0(0, 0), A(a, b), and C(c, 0).)In Exercises 47-52 find the center and radius of the circle with the given equation.In Exercises 15-20 sketch the graph of the set W. 47 x2 + l + 4x - 6y +4= 015 W = {(x, y): x = 4} 48 x2 +l - !Ox + 2y + 22 =016 W= {(x,y):y= -3}17 W = {(x, y): xy < 0} 49 x2 + y2 + 6x= 018 W = {(x, y): xy = 0}50 x 2 + y 2 + x + y - 1 = 019 W = {(x, y): lxl < 2, IYI > 1}51 2x 2 + 2y 2 - x +y- 3=020 W = {{x, y): lxl > 1, IYI::;:;; 2}52 9x 2 + 9y 2 - 6x + 12y- 31 = 0 27. 18 1 Prerequisites for Calculus ].3 Lines The following concept is fundamental to the study of lines. All lines referred to are considered to be in some fixed coordinate plane. Definition (1.14) Let I be a line that is not parallel to they-axis, and let P 1(x 1 , y 1) and P 2 (x 2 , Y2) be distinct points on l. The slope m of lis given by y If I is parallel to the y-axis, then the slope is not defined.Typical points P 1 and P 2 on a line I are shown in Figure 1.19.The numerator Y2 - y 1 in the formula for m measures the vertical change in direction in proceeding from P 1 toP 2 and may be positive, negative, or zero. The denominator x 2 - x 1 measures the amount of horizontal change in going from P 1 to P 2 , and it may be positive or negative, but never zero, because I is not parallel to they-axis. ( i) Positive slopeIn finding the slope of a line it is immaterial which point is labeled P 1 and which is labeled P 2 , since y Consequently, we may as well assume that the points are labeled so that x 1 < x 2 , as in Figure 1.19. In this event x 2 - x 1 > 0, and hence the slope is positive, negative, or zero, depending on whether Y2 > y 1 , y 2 < y 1 , or X y 2 = y 1 The slope of the line shown in (i) of Figure 1.19 is positive, whereas the slope of the line shown in (ii) of the figure is negative. A horizontal line is a line that is parallel to the x-axis. Note that a line is (ii) Negative slope horizontal if and only if its slope is 0. A vertical line is a line that is parallel to they-axis. The slope of a vertical line is undefined.FIGURE 1.19 It is important to note that the definition of slope is independent of the two points that are chosen on I, for if other points P1(xt- y1) and P~(x~, y~) y are used, then as in Figure 1.20, the triangle with vertices P~, P;, and P3(x;, y~) is similar to the triangle with vertices Pt. P 2 , and P 3 (x 2 , y 1 ). Since the ratios of corresponding sides are equal it follows that Y2 - YtY~ - it x2 - x1x~ - x1 Example 1 Sketch the lines through the following pairs of points and find X their slopes. (a) A( -1, 4) and B(3, 2)(b) A(2, 5) and B( -2, -1)FIGURE 1.20(c) A(4, 3) and B( -2, 3)(d) A(4, -1) and B(4, 4). 28. Lines 1.3 19SolutionThe lines are sketched in Figure 1.21. Using Definition (1.14)gives us2- 4 -21(a) m= =- =--3-(-1)425- ( -1) 6 3(b) m==- =-2- ( -2) 4 2 3- 3 0(c) m==- = 0-2-4 -6(d) The slope is undefined since the line is vertical. This is also seen bynoting that, if the formula for m is used, the denominator is zero. y A(-1, 4) (i) m =-!(ii) m = ! y yA (4, 3)B(4, 4)X X A(4,-1) (iii) m = 0 (iv) m undefined FIGURE 1.21Theorem (1 .15)(i) The graph of the equation x =a is a vertical line with x-intercept a.(ii) The graph of the equation y = b is a horizontal line with y- intercept b.ProofTheequationx =a, whereaisarealnumber,maybeconsideredas an equation in two variables x andy, since we can write it in the form x + 0 y =a. 29. 20 1 Prerequisites for Calculus y Some typical solutions of the equation are (a, - 2), (a, 1), and (a, 3). Evidently,y=ball solutions of the equation consist of pairs of the form (a, y), where y may (0, b)have any value and a is fixed. It follows that the graph of x = a is a line parallel to the y-axis with x-intercept a, as illustrated in Figure 1.22. This proves (i). Part (ii) is proved in similar fashion.D(a, 0) X Let us next find an equation of a line I through a point P 1 (x 1 , y 1 ) with slope m (only one such line exists). If P(x, y) is any point with x =F x 1 , then Px=ais on I if and only if the slope of the line through P 1 and Pis m; that is,Y- Yt---=m.FIGURE 1.22 x- x 1 This equation may be written in the form Note that (xl> yd is also a solution of the last equation and hence the points on I are precisely the points which correspond to the solutions. This equation for I is referred to as the Point-Slope Form. Our discussion may be summarized as follows:Point-Slope Form (1.16) An equation for the line through the point P(x 1 , y 1 ) with slope m is Example 2 Find an equation of the line through the points A(1, 7) and B( -3, 2). Solution By Definition (1.14) the slope m of the line is7- 2 5 m= = -.1 - ( -3)4 Using the coordinates of A in the Point-Slope Form (1.16) gives usy- 7 = i(x- 1) which is equivalent to 4y- 28= 5x- 5 or5x- 4y+ 23 = 0. The same equation would have been obtained if the coordinates of point B had been substituted in (1.16). The Point-Slope Form may be rewritten as y= mx - mx 1 + y 1 , which is of the formy=mx+b 30. Lines 1.321 with b = - mx 1 + y 1 The real number b is they-intercept of the graph, as may be seen by setting x = 0. Since the equation y = mx + b displays the slope m andy-intercept b of l, it is called the Slope-Intercept Form "for the equation of a line. Conversely, if we start withy = mx + b, we may write y - b = m(x - 0). Comparing with the Point-Slope Form, we see that the graph is a line with slope m and passing through the point (0, b). This gives us the next result.Slope-Intercept Form (1.17) The graph of the equation y= mx + b is a line having slope m andy-intercept b. The work we have done shows that every line is the graph of an equation of the form ax+ by+ c = 0 where a, b, and care real numbers, and a and bare not both zero. We call such an equation a linear equation in x andy. Let us show, conversely, that the graph of ax + by + c = 0, where a and b are not both zero is always a line. On the one hand, if b # 0, we may solve for y, obtaining which, by the Slope-Intercept Form, is an equation of a line with slope- ajb and y-intercept - cjb. On the other hand, if b = 0 but a # 0, then we may solve for x, obtaining x = - cja, which is the equation of a vertical line with x-intercept - cja. This establishes the following important theorem.Theorem (1.18)The graph of a linear equation ax + by + c = 0 is a line and, conversely,every line is the graph of a linear equation. y For simplicity, we shall use the terminology the line ax + by + c = 0 instead of the more accurate phrase the line with equation ax + by + c = 0. Example 3 Sketch the graph of 2x - 5y= 8. Solution From Theorem (1.18) the graph is a line, and hence it is sufficient to find two points on the graph. Let us find the x- andy-intercepts. Substitut- ing y = 0 in the given equation, we obtain the x-intercept 4. Substituting FIGURE 1.23 x = 0, we see that they-intercept is -!.This leads to the graph in Figure 1.23. 31. 22 1 Prerequisites for Calculus Another method of solution is to express the given equation in Slope- Intercept Form. To do this we begin by isolating the term involving y on one side of the equals sign, obtaining5y=2x- 8. Next, dividing both sides by 5 gives usy=sx+ 2 (-8) 5 which is in the form y = mx + b. Hence, the slope ism = ~and they-intercept is b = -l We may then sketch a line through the point (0, -!) with slope~. The following theorem can be proved geometrically. Theorem (1 .19)Two non vertical lines are parallel if and only if they have the same slope. We shall use this fact in the next example. Example 4 Find an equation of a line through the point (5, - 7) that is parallel to the line 6x + 3y - 4 = 0. SolutionLet us express the given equation in Slope-Intercept Form. We begin by writing3y = -6x+4 and then divide both sides by 3, obtainingy =-2x+ 4. The last equation is in Slope-Intercept Form with m = -2, and hence the slope is -2. Since parallel lines have the same slope, the required line also has slope -2. Applying the Point-Slope Form gives usy +7=-2(x - 5). This is equivalent toy +7= -2x + 10 or 2x +y - 3 = 0. The next result specifies conditions for perpendicular lines. Theorem (1.20) Two lines with slopes m1 and m 2 are perpendicular if and only ifmlm2 = -1. 32. Lines 1.323 yProofFor simplicity, let us consider the special case where the lines intersect at the origin 0, as illustrated in Figure 1.24. In this case equations of the lines are y = m1 x and y = m 2 x. If, as in the figure, we choose points A(x 1, m 1x 1) and B(x 2 , m 2 x 2 ) different from 0 on the lines, then the lines are perpendicular if and only if angle AOB is a right angle. Applying the Pythagorean Theorem to triangle AOB, this is equivalent to the condition [d(A, B)] 2 = [d(O, B)] 2+ [d(O, A)] 2 or, by the Distance Formula,FIGURE 1.24 (m2x2 - m1xd 2+ (x 2 - x 1) 2 = (m 2 x 2) 2 + x~ + (m 1x 1) 2 + xf, Squaring the indicated terms and simplifying gives us Dividing both sides by - 2x 1 x 2 we see that the lines are perpendicular if and only if m 1m2 + 1 = 0, or m1m2 = -1. The same type of proof may be given if the lines intersect at any point (a, b)(see Exercise 42). D A convenient way to remember the conditions for perpendicularity is to note that m1 and m 2 must be negative reciprocals of one another, that is, m 1 = -1/m 2 and m 2 = -1/m 1. Example 5 Find an equation for the perpendicular bisector of the line segment from A(l, 7) to B(- 3, 2). Solution By the Midpoint Formula (1.11), the midpoint M of the segment AB is ( -1, 1). Since the slope of AB is i (see Example 2), it follows from Theorem ( 1.20) that the slope of the perpendicular bisector is - ~. Applying the Point-Slope Form,9 4y-- = -- (x + 1).2 5 Multiplying both sides by 10 and simplifying leads to 8x + lOy - 37 = 0. 1.3 ExercisesIn Exercises 1-4 plot the points A and B and find the slope of6 Show that A(2, 3), B(5, -1), C(O, -6), and D( -6, 2) arethe line through A and B. vertices of a trapezoid. 1 A( -4, 6), B( -1, 18)2 A(6, -2), B( -3, 5) 7 Prove that the points A(6, 15), B(11, 12), C( -1, -8), and3 A(-1, -3),B(-1,2) 4 A( -3, 4), B(2, 4)D(- 6, - 5) are vertices of a rectangle.5 Show that A(- 3, 1), B(5, 3), C(3, 0), and D(- 5, - 2) are8 Prove that the points A(l, 4), B(6. -4), and C( -15, -6)vertices of a parallelogram.are vertices of a right triangle. 33. 24 I Prerequisites for Calculus 9 If three consecutive vertices of a parallelogram areand sketch the graph of each line. A( -1, -3), 8(4, 2), and C( -7, 5), find the fourth vertex. 25 3x- 4y+ 8= 026 2y- 5x =I10 Let A(x~> y 1 ), 8(x 2 , y 2 ), C(x 3 , y 3 ), and D(x 4 , y 4 ) denote 27 X+ 2y = 0 28 8x = I - 4y the vertices of an arbitrary quadrilateral. Prove that the line segments joining midpoints of adjacent sides form a29 y=4 30 x+2=!y parallelogram. 31 5x + 4y = 2032 y=OIn Exercises 11-24 find an equation for the line satisfying the33 X= 3y + 7 34 x-y=Ogiven conditions. 35 Find a real number k such that the point P(- I, 2) is onII Through A(2, -6 ), slope ! the line kx + 2y - 7 = 0.12 Slope - 3, y-intercept 536 Find a real number k such that the line 5x+ ky - 3 =0has y-intcrcept - 5.13 Through A(- 5, -7), 8(3, - 4) 37 If a line I has nonzero x- andy-intercepts a and b, respec-14 x-intercept -4, y-intercept 8tively, prove that an equation for I is (xja) + (yjb) = I.15 Through A(8, - 2), y-intercept - 3 (This is called the intercept form for the equation of aline.) Express the equation 4x - 2y = 6 in intercept16 Slope 6, x-intercept -2form.17 Through A( I 0, - 6), parallel to (a) they-axis; (b) the x-axis. 38 Prove that an equation of the line through P 1(x 1,; 1 )18 Through A( -5, 1), perpendicular to (a) the y-axis; (b)and P2 (x 2 , y 2 ) is the x-axis.19 Through A(7, - 3), perpendicular to the line with equation 2x- 5y = 8.(This is called the two-point form for the equation of aline.) Usc the two-point form to find an equation of the20 Through ( -i, -!), parallel to the line with equationline through A(7, -I) and 8(4, 6). X+ 3y = J. 39 Find all values of r such that the slope of the line through21 Given A(3, -I) and 8(- 2, 6), find an equation for the the points (r, 4) and (I, 3 - 2r) is less than 5. perpendicular bisector of the line segment A8. 40 Find all values of 1 such that the slope of the line through22 Find an equation for the line which bisects the second and (1, 31 + 1) and (I - 21, 1) is greater than 4. fourth quadrants. 41 Six years ago a house was purchased for $59,000. This23 Find equations for the altitudes of the triangle withyear it is appraised at $95,000. Assuming that the value vertices A(- 3, 2), 8(5, 4), C(3, - 8), and find the point increased by the same amount each year, find an equation at which they intersect. that specifics the value at any time after the purchase date.24 Find equations for the medians of the triangle in Exercise When was the house worth $73,000? 23, and find their point of intersection. 42 Complete the proof of Theorem (I .20) by considering thecase where the lines intersect at any point (a, b).In Exercises 25-34 use the Slope-Intercept Form (1.17) tofind the slope andy-intercept of the line with the given equation ]. 4FunctionsThe notion of correspondence is encountered frequently in everyday life. Forexample, to each book in a library there corresponds the number of pages inthe book. As another example, to each human being there corresponds abirth date. To cite a third example, if the temperature of the air is recorded 34. Functions1.4 25 throughout a day, then at each instant of time there is a corresponding temperature.These examples of correspondences involve two sets X and Y. In our first example, X denotes the set of books in a library and Y the set of positive integers. For each book x in X there corresponds a positive integer y, namely the number of pages in the book.Our examples also indicate that to each x in X there corresponds one and only one yin Y; that is, y is unique for a given x. However, the same element of Y may correspond to different elements of X. For example, two different books may have the same number of pages, two different people may have the same birthday, and so on. In most of our work X and Y will be sets of real numbers. To illustrate, let X andY both denote the set IR of real numbers, and to each real number x let us assign its square x 2 Thus, to 3 we assign 9, to - 5 we assign 25, and to y2 the number 2. This gives us a correspondence from IR to RAll the examples of correspondences we have given are functions, as defined below. Definition (1.21)A function/from a set X to a set Y is a correspondence that assigns toeach element x of X a unique element y of Y. The element y is called theimage of x under f and is denoted by f (x ). The set X is called the domainof the function. The range of the function consists of all images of ele-ments of X.The symbol f(x) used for the element associated with xis read "f of x." Sometimes f(x) is called the value off at x.y Functions may be represented pictorially by diagrams of the type shown in Figure 1.25. The curved arrows indicate that the elements f(x), f(w), f(z), andf(a) of Yare associated with the elements x, w, z, and a, respectively, of X. We might imagine a whole family of arrows of this type, where each arrow connects an element of X to some specific element of Y. Although the sets X and Y have been pictured as having no elements in common, this is not required by Definition (1.21). As a matter of fact, we often take X = Y. It is important to note that with each x in X there is associated precisely one image f(x); however, different elements such as wand z in Figure 1.25 may have the same image in Y.FIGURE 1.25If the sets X and Y of Definition (1.21) are intervals, or other sets of real numbers, then instead of using points within regions to represent elements, as in Figure 1.25, we may use two coordinate lines land l. This technique is illustrated in Figure 1.26, where two images for a function fare represented graphically. Xa X f(x) f(a) Y l FIGURE 1.26 35. 26 1 Prerequisites for Calculus Beginning students are sometimes confused by the symbols f and f(x) inDefinition (1.21). Remember that f is used to represent the function. It is neither in X nor in Y. However, f(x) is an element of Y, namely the element which f assigns to x. Two functions f and g from X to Y are said to be equal, and we write f = g, provided f(x) = g(x) for every x in X. For example, if g(x) = (!}(2x 2 - 6) + 3 and f(x) = x 2 for all x in IR, then g =f. Example I Let f be the function with domain IR such that f(x) = x 2 for every x in R Find f( -6), f(j3), and f(a +b), where a, bare real numbers. What is the range off? SolutionValues of f(or images under f) may be found by substituting for x in the equation f(x) = x 2 . Thusf( -6) = ( -6) 2 = 36, andf(a +b)= (a+ b) 2 = a 2 + 2ab + VIf T denotes the range off, then by Definition (1.21) T consists of all numbers of the form f(a) where a is in R Hence T is the set of all squares a2 , where a is a real number. Since the square of any real number is non- negative, Tis contained in the set of all nonnegative real numbers. Moreover, every nonnegative real number cis an image under f, since f(.jC) = (.jC) 2 = c. Hence the range off is the set of all nonnegative real numbers. To describe a function f it is necessary to specify the image f(x) of each element x of the domain. A common method for doing this is to use an equation, as in Example 1. In this case, the symbol used for the variable is immaterial. Thus, expressions such as f(x) = x 2 , f(s) = s 2 , and f(t) = t 2 all define the same function f. This is true because if a is any number in the domain off, then the image a 2 is obtained no matter which expression is employed. Occasionally one of the notations X~ Y,f: X ~ Y or f: x ~ f(x) is used to signify thatfis a function from X to Y. It is not unusual in this event to say fmaps X into Yor fmaps x intof(x). Iffis the function in Example 1, thenfmaps x into x 2 and we may writef: x ~ x 2 . Many formulas that occur in mathematics and the sciences determine functions. As an illustration, the formula A = nr 2 for the area A of a circle of radius r associates with each positive real number r a unique value of A and hence determines a functionfwherej(r) = nr 2 The letter r, which represents an arbitrary number from the domain off, is often called an independent variable. The letter A, which represents a number from the range off, is called a dependent variable, since its value depends on the number assigned tor. When two variables rand A are related in this manner, it is customary 36. Functions 1.4 27to use the phrase A is a function of r. To cite another example, if an auto-mobile travels at a uniform rate of 50 miles per hour, then the distance d(miles) traveled in time t (hours) is given by d = 50t "and we say that dis afunction oft.We have seen that different elements in the domain of a function may havethe same image. If images are always different, then the function is calledone-to-one.Definition (1.22) A function/from X to Y is a one-to-one function if, whenever a - bin X,thenf(a) # j(b) in Y.Iff is one-to-one, then each f(x) in the range is the image of precisely one xin X. The function illustrated in Figure 1.25 is not one-to-one since twodifferent elements wand z of X have the same image in Y. If the range ofjis Yand f is one-to-one, then sets X and Y are said to be in one-to-one corres-pondence. In this case each element of Yis the image of precisely one elementof X. The association between real numbers and points on a coordinate lineis an example of a one-to-one correspondence.Example 2(a) If f(x) = 3x + 2, where x is real, prove that f is one-to-one.(b) If g(x) = x2 + 5, where xis real, prove that g is not one-to-one.Solution(a) If a # b, then 3a - 3b and hence 3a + 2 # 3b + 2, or f(a) # f(b).Hence f is one-to-one by Definition (1.22).(b) The function g is not one-to-one since different numbers in the domainmay have the same image. For example, although -I # 1, both g( -1) andg(l) are equal to 6. ~ Iff is a function from X to X and if f(x) = x for every x, that is, everyelement x maps into itself, thenfis called the identity function on X. A functionjis a constant function if there is some (fixed) element c such thatj(x) = c forevery x in the domain. If a constant function is represented by a diagram ofthe type shown in Figure 1.25, then every arrow from X terminates at thesame point in Y. In the remainder of our work, unless specified otherwise, the phrase f is afunction will mean that the domain and range are sets of real numbers. If afunction is defined by means of some expression as in Examples 1 or 2, andthe domain X is not stated explicitly, then X is considered to be the totality of real numbers for which the given expression is meaningful. To illustrate, iff(x) = JX!(x - I), then the domain is assumed to be the set of nonnegativereal numbers different from 1. (Why?) If xis the domain we sometimes say that f is defined at x, or thatf(x) exists. If a set Sis contained in the domainwe often say that/is defined on S. The terminology fis undefined at x meansthat x is not in the domain off 37. 281 Prerequisites for CalculusThe concept of ordered pair can be used to obtain an alternate approachto functions. We first observe that a function f from X to Y determines thefollowing set W of ordered pairs: W = {(x, f(x)): xis in X}.Thus W is the totality of ordered pairs for which the first number is in X andthe second number is the image of the first. In Example 1, where f(x) =x 2 , W consists of all pairs of the form (x, x 2 ) where x is any real number.It is important to note that for each x there is exactly one ordered pair (x, y)in W having x in the first position.Conversely, if we begin with a set W of ordered pairs such that each xin X appears exactly once in the first position of an ordered pair, and numbersfrom Y appear in the second position, then W determines a function fromX to Y. Specifically, for any x in X there is a unique pair (x, y) in W, and byletting y correspond to x, we obtain a function from X to Y.It follows from the preceding discussion that the following statementcould also be used as a definition of function. We prefer, however, to thinkof it as an alternative approach to this concept. Alternative Definition (1.23) A function with domain X is a set W of ordered pairs such that, for each x in X, there is exactly one ordered pair (x, y) in W having x in the first position. Y = f(x)If/is a function, we may use a graph to exhibit the variation of/(x) as xvaries through the domain off By definition, the graph of a function/ is theset of all points (x,j(x)) in a coordinate plane, where xis in the domain offThus, the graph of/is the same as the graph of the equation y = f(x), and if P(a, b) is on the graph, then they-coordinate b is the functional valuef(a), as illustrated in Figure 1.27. It is important to note that, since there is a unique.f(a) for each a in the domain, there is only one point on the graph with x-a Xcoordinate a. Thus, every rerticalline intersects the graph of a function in atmost one point. Consequently, for graphs of functions it is impossible to obtain a sketch such as that shown in Figure 1.16 where some vertical lines FIGURE 1.27 Graph of a function f intersect the graph in more than one point. The x-intercepts of the graph of a function fare the solutions of the equationf(x) = 0. These numbers are called the zeros of the function. They-intercept of the graph is ./(0), if it exists. yExample 3 Sketch the graph off if f(x) =~.Solution The domain off consists of all real numbers x such that x 2: 1.f(x) = t vx-=--1(Why?) The following table lists some points (x,f(x)) on the graph.X 2 3 4 56 X(1, 0)f(x)0Plotting points leads to the sketch shown in Figure 1.28. Note that the x- FIGURE 1.28intercept is 1, and there is no y-intercept. 38. Functions 1.4 29 yExample 4Sketch the graph off if f(x) = 3 - x 2 .Solution We list coordinates (x,J(x)) of some points on the graph offin tabular form, as follows:X -3-2 -1 023 X.f(x) -6-1 2 3 2-1 -6f(x) = 3 -x 2The x-intercepts are the solutions of the equation f(x) = 0, that is, of3 - x 2 = 0. These are J3.The y-intercept is f(O) = 3. Plotting thepoints given by the table and using the x-intercepts leads to the sketch inFigure 1.29. FIGURE 1.29The solution to Example 4 could have been shortened by observing that,since 3 - ( -x) 2 = 3 - x 2 , the graph of y = 3 - x 2 is symmetric withrespect to the y-axis. yExample 5 Sketch the graph off if f(x) =Ix 1.SolutionIf x ;?: 0, then f(x) = x and hence the part of the graph to theright of the y-axis is identical to the graph of y = x, which is a line throughthe origin with slope 1. If x < 0, then by Definition (1.2),J(x) = lxl = -x,and hence the part ofthe graph to the left of they-axis is the same as the graphof y = - x. The graph is sketched in Figure 1.30.XExample 6Sketch the graph off if f(x) = 1/x.SolutionThe domain off is the set of all nonzero real numbers. If x isFIGURE 1.30 positive, so is f(x), and hence no part of the graph lies in quadrant IV.Quadrant II is also excluded, since if x < 0 then f(x) < 0. If x is close tozero, they-coordinate 1/x is very large numerically. As x increases throughpositive values, 1/x decreases and is close to zero when x is large. Similarly,if we let x take on numerically large negative values, they-coordinate 1/x isy close to zero. Using these remarks and plotting several points gives us thesketch in Figure 1.31.The graph of J, or equivalently, of the equation y = 1/x, is symmetricwith respect to the origin. This may be verified by using (iii) of (1.12). Example 7 Describe the graph of a constant function.XSolution If f(x) = c, where cis a real number, then the graph off is thesame as the graph of the equation y = c and hence is a horizontal line withy-intercept c.Sometimes functions are described in terms of more than one expression,FIGURE 1.31 as in the next examples. 39. 301 Prerequisites for CalculusyExample 8Sketch the graph of the function f that is defined as follows: 2x +3 ifx < 0f(x)= {x2ifO~x-->- 25 2(b) find the coordinates of the midpoint of AB. (c) find the slope of the line through Band C. 3 12x - 71 :s;; O.Ql4 16x- 71 > I 2x 2 - 3x- 20Sketch the graphs of the equations in Exercises 10-13 and 5 2x 2 < 5x- 36 x}34 f(x) = -1/(x + I)35 f(x)=lx+5116 W= {(x,y):x 2 + y2 0, there exists a b > 0, such thatif 0 < lx- al < b, then l.f(x)- Ll 0, there exists a b > 0, such that if x is in the open interval (a - b, a + b), and x # a, then .f(x) is in the open interval (L- e, L +e).To get a better understanding of the relationship between the positivenumbers e and b in Definitions (2.3) and (2.41; let us consider a geometricinterpretation similar to that in Figure 1.26, where the domain of f isrepresented by certain points ori a coordinate line /, and the range by otherpoints on a coordinate line/. The limit process may be outlined as follows. 65. 56 2 Limits and Continuity of FunctionsTo prove that lim f(x) x~a = L.Step 1. For anyB > 0 consider the open interval (L -B, L + ~>)(see Figure2.5). a L- E LL +EIFIGURE 2.5Step 2. Show that there exists an open interval (a - b, a + b) in the domain offsuch that (2.4) is true (see Figure 2.6). a-o( I~ ~Ix a a+oI L-E f(x)LL+ ) . IFIGURE 2.6It is extremely important to remember that first we consider the interval(L - s, L + s) and then, second, we show that an interval (a - b, a + b)of the required type exists in the domain off One scheme for rememberingthe proper sequence of events is to think of the function f as a cannon thatshoots a cannonball from the point on I with coordinate x to the point on Iwith coordinate f(x), as illustrated by the curved arrow in Figure 2.6. Step1 may then be regarded as setting up a target of radius B with hulls eye at L.To apply Step 2 we must find an open interval containing a in which to placethe cannon such that the cannonball hits the target. Incidentally, there is noguarantee that it will hit the hulls eye; however, if limx-a f(x) = L we canmake the cannonball land as close as we please to the hulls eye.It should be clear that the number b in the limit definition is not unique,for if a specific b can be found, then any smaller positive number b will alsosatisfy the requirements.Since a function may be described geometrically by means of a graph on arectangular coordinate system, it is of interest to interpret Definitions (2.3)and (2.4) graphically. Figure 2.7 illustrates the graph of a function f where,for any x in the domain off, the number f(x) is they-coordinate of the pointon the graph with x-coordinate x. Given any s > 0, we consider the openinterval (L - B, L + s) on the y-axis, and the horizontal lines y = L sshown in the figure. If there exists an open interval (a - b, a + b) such thatfor every x in (a - b, a + b), with the possible exception of x = a, the pointP(x, f(x)) lies between the horizontal lines, that is, within the shaded rectangleshown in Figure 2.7, thenL- s < f(x) < L + t:and hence limx-a f(x) = L. 66. Definition of Limit 2.257 y P(x,f(x)) L+E"-- - - - - - y=L + EL i-------- y =L- EL- E /I I! II II IleJX FIGURE 2.7 lim f(x) = L The next example illustrates how the geometric process pictured mFigure 2.7 may be applied to a specific function.yf(x)=x 2 Example 1 Prove that limx~a x 2= a2 Solution Let us consider the case a > 0. We shall apply (2.4) with.f(x) = x 2 and L = a 2 The graph of .f is sketched in Figure 2.8, together withtypical points on the x- andy-axes corresponding to a and a 2 , respectively.For any positive number , consider the horizontal lines y = a 2 + andy = a 2 - . These lines intersect the graph off at points with x-coordinates~ and~ as illustrated in the figure. If x is in the open interval(~,~),then ~ smaller than both ~ - a and a -~as illustrated in Figure 2.8. It follows that if x is in the interval(a - f>, a+ f>), then x is also in (~, ~), and, therefore, .f(x)is in (a 2 - , a 2 + ). Hence, by Definition (2.4), limx~a x 2 = a 2 Althoughwe have considered only a > 0, a similar argument applies if a :s; 0. To shorten explanations, whenever the notation limx~a .f(x) = L isused we shall often assume that all the conditions given in Definition (2.3) aresatisfied. Thus, it may not always be pointed out that f is defined on anopen interval containing a. Moreover, we shall not always specify L butmerely write "limx~a .f(x) exists," or "f(x) has a limit as x approaches a."The phrase "find limx~a f(x)" means "find a number L such thatIimx~a f(x) = L." If no such L exists we write "Iimx~a f(x) does not exist." 67. 582Limits and Continuity of Functions It can be proved that if f(x) has a limit as x approaches a, then that limit is unique (see Appendix II). In the following example we return to the function considered at the beginning of this section and prove that the limit exists by means of Definition (2.3). Example 2 Prove that limx~ 4 !(3x - 1) = z.1 SolutionLet .f(x) = !(3x - 1), a = 4, and L = l. According to Defini- tion (2.3) we must show that for every & > 0, there exists a {J > 0 such that if 0 < lx - 41 < b,then1!(3x - 1) - ll < &. A clue to the choice of {J can be found by examining the last inequality involving t:. The following is a list of equivalent inequalities.I!(3x -1) - z 1 0. lx- 31. X+ 211 hm-- 12hm-- x~3 X- 3x~-2IX + 21 21 lim~=faI I13lim-- 14 lim2 x~ -5 (x + 5) x~l(x-1) 69. 602 Limits and Continuity of Functions 2.3 Theorems on Limits It would be an excruciating task to solve each problem on limits by means of Definition (2.3). The purpose of this section is to introduce theorems that may be used to simplify the process. To prove the theorems it is necessary to employ the definition of limit; however, once they are established it will be possible to determine many limits without referring to an a or a b. Several theorems are proved in this section; the remaining proofs may be found in Appendix II. The simplest limit to consider involves the constant function defined by f(x) = c, where cis a real number. If.fis represented geometrically by means of coordinate Jines I and /, then every arrow from I terminates at the same point on /, namely, the point with coordinate c, as indicated in Figure 2.11.x a zI~Ic - c C ) + E [ FIGURE 2.11 It is easy to prove that for every real number a, Jimx-af(x) = c. Thus, if a > 0, consider the open interval (c - a, c + a) on / as illustrated in the figure. Since .f(x) = cis always in this interval, any number [J will satisfy the conditions of Definition (2.3); that is, for every [J > 0,if 0 < lx- al < b,then l.f(x)- cl 0 there existsa (j > 0 such thatif 0 < jx- al < b,thenj(mx +b)- (ma + b)i 0, if 0 < jx- al < b, where () = e/lmlthen the last inequality in the list is true, and hence, so is the first inequality,which is what we wished to prove. D As special cases of Theorem (2.6), we have lim x =a lim (3x - 5) = 3 4 - 5 = 7lim (l3x + x---+/2.j2) =13j2 + J2 = 14j2.The next theorem states that if a function f has a positive limit as xapproaches a, thenf(x) is positive throughout some open interval containinga, with the possible exception of a.Theorem (2.7)If limx-+a f(x) = L and L > 0, then there exists an open interval (a-(), a+ b) containing a such thatf(x) > Oforallxin(a- b, a+ b), except possibly x = a.Prooflfe = L/2,thentheinterval(L- e, L + e)containsonlypositivenumbers. By Definition (2.4) there exists a (j > 0 such that whenever xis inthe open interval (a - b, a + {j) and x =1= a, thenf(x) is in (L - e, L + e), andhence f(x) > 0. D In like manner, it can be shown that iffhas a negative limit as x approachesa, then there is an open interval I containing a such that f(x) < 0 for everyx in /, with the possible exception of x = a. 71. 62 2 Limits and Continuity of Functions Many functions may be expressed as sums, differences, products, andquotients of other functions. In particular, suppose a functions is a sum of twofunctions f and g, so that s(x) = f(x) + g(x) for every x in the domain of s.If f(x) and g(x) have limits L and M, respectively, as x approaches a, it isnatural to conclude that s(x) has the limit L + Mas x approaches a. The factthat this and analogous statements hold for products and quotients areconsequences of the next theorem.Theorem (2.8) If lim f(x) = Land lim g(x) = M, then (i) lim [f(x)+ g(x)] = L + M. (ii)lim [f(x) g(x)]= L M.(iii) . I1m [f(x)J-) L = M prov1.d edM =1- 0. x~a g(x(iv) lim [cf(x)] = cL. (v) lim [f(x) - g(x)] = L - M.Although the conclusions of Theorem (2.8) appear to be intuitivelyevident, the proofs are rather technical and require some ingenuity. Proofsfor (i)-(iii) may be found in Appendix II. Part (iv) of the theorem followsreadily from part (ii) and Theorem (2.5) as follows: lim [cf(x)] = [tim c] [limf(x)] = cL. x-ax-a x-aFinally, to prove (v) we may writef(x)- g(x) = f(x) + ( -l)g(x)and then use (i) and (iv) (with c = -I). The conclusions of Theorem (2.8) are often written as follows:(i) lim [f(x) + g(x)] = lim f(x) + lim g(x)(ii)lim [f(x) g(x)] = lim f(x) lim g(x)x~a /( )]limf(x) (iii)lim [ (x)x~ag X = ;.~a 1m g( ), if lim g(x)X x~a=1- 0 (iv) ~i~ [cf(x)] c[~i~ f(x)]=(v) lim [f(x)- g(x)] = limf(x)- lim g(x). 72. Theorems on Limits 2.3 633x + 4F m d 1. - - - ..Example I1mx-+ 2 5x + 7SolutionThe numerator and denominator of the indicated quotientdefine linear functions whose limits exist by Theorem (2.6). Moreover, thelimit of the denominator is not 0. Hence by (iii) of Theorem (2.8) and (2.6),lim (3x + 4) . 3X + 4 x-+2 3(2)+410I1m - - - = "-:--::.....,-::-----,=c-x-+2 5x + 7 lim (5x + 7) x-+2 5(2)+717. Notice the simple manner in which the limit in Example 1 was found. Itwould be a lengthy task to verify the limit by means of Definition (2.3).Theorem (2.8) may be extended to limits of sums, differences, products,and quotients that involve any number of functions. An application of (ii)to three (equal) functions is given in the next example.Example 2 Prove that for every real number a, limx .... a x 3 = a 3 SolutionSince limx .... a x = a we may writelim x 3 = lim (x x x)x-+a x-+a=(lim x) (lim x) (lim x)x~ax-+a x-+a The technique used in Example 2 can be extended to xn, where n is anypositive integer. We merely write xn as a product x x x of n factors andthen take the limit of each factor. This gives us (i) of the next theorem.Part (ii) may be proved in similar fashion by using (ii) of Theorem (2.8).Theorem (2.9)If n is a positive integer, then(i)lim xn = an. x-+a (ii)~i~ [f(x)]n = [~i~ f(x)r provided~i~ f(x) exists.Example 3 Find limx .... 2 (3x + 4) 5 .SolutionApplying (ii) of (2.9) and Theorem (2.6),lim (3x+ 4) 5 =[lim (3x+ 4)] 5x-+2x-+2=[3(2)+ 4] 5=10 5 =100,000. 73. 64 2Limits and Continuity of FunctionsExample 4Find limx~ _ 2 (5x 3 + 3x 2 - 6).SolutionWe may proceed as follows (supply reasons): lim (5x 3 + 3x 2 - 6) = lim (5x 3 ) + lim (3x 2 )- lim (6)x- -2x- -2 x--2x-+-2x--2 x-+-2 = 5(- 2? + 3(- 2) 2 -6 = 5(-8) + 3(4)- 6 =-34. Note that the limit in Example 4 is the number obtained by substituting-2 for x in 5x 3 + 3x 2 - 6. The next theorem states that the same is true forthe limit of every polynomial. Theorem (2.10)Iff is a polynomial function and a is a real number, then limf(x) = f(a).ProofWe may write .f(x) in the formf(x) = bnx" + bn_ 1 xn-! + + b 0where the bi are real numbers. As in Example 4, lim f(x) =lim (bnx") + lim (bn_ 1 x"- 1 ) + + lim b 0 .---+(1 x -a= bn lim (x") + bn-! lim (x"- 1) ++lim b 0x~ax~ax~aD Corollary (2.11)If q is a rational function and a is in the domain of q, then lim q(x)= q(a).Proof We may write q(x) = f(x)jh(x) where f and h are polynomialfunctions. If a is in the domain of q, then h(a) =1- 0. Using (iii) of Theorem(2.8) and (2.1 0), . ~i~ f(x) f(a)~~~ q(x) = lim h(x) = h(a) = q(a).D 74. Theorems on Limits 2.3 65 E xamp.e 5 F"md xi~ 5x2 - _ + 1 , I"2x6x7 Solution Applying Corollary (2.11),. 5x 2 - 2x + 1 I1m ---=----=-- 5(3) 2- 2(3) + 1 x-+3 6x -7 6(3)- 7 45- 6 + 1 4018 - 7The following theorem states that for positive integral roots of x, we may determine a limit by substitution. The proof makes use of the definition of limit and may be found in Appendix II.Theorem (2.12) If a > 0 and n is a positive integer, or if a ::s;; 0 and n is an odd positive integer, then limylx = .ja. x-+aIf m and n are positive integers and a > 0, then using (ii) of (2.9) and Theorem (2.12),lim(yfx r = (lim v~x)m = (.jar.x-+ax-+a In terms of rational exponents, x-+a This limit formula may be extended to negative exponents by writing x- = 1/x and then using (iii) of (2.8).. .x 213 + 3Jx Example 6 Fmd !~ 4 _ ( 16/x) . Solution The reader should supply reasons for each of the following steps. 2/3 3r: lim (x2i3 + 3jX) .I1mX + yX=x-+8 ~~~----x-+8 4 - (16/x) lim (4 - (16/x))x-+8lim 4 - lim (16/x)x-+8 x-+8-g2/3+ 3JS- 4- (16/8)=4+4-2 6J2 = 2 + 3J2 75. 66 2 Limits and Continuity of FunctionsTheorem (2.13) If a functionf has a limit as x approaches a, then lim j]Tx) =y/limf(x)x->a provided either n is an odd positive integer or n is an even positive integer and limx_,a f(x) > 0.The preceding theorem will be proved in Section 2.6. In the meantime we shall use it whenever applicable to gain experience in finding limits that involve roots of algebraic expressions. Example 7 Find Iimx~s J3x 2 - 4x + 9. SolutionUsing Theorems (2.13) and (2.10), lim ~hx 2 - 4x +9=Jlim (3x 2- 4x + 9) . ___,. 5 x-->5 = J75 - 20 + 9 = 164 = 4. The beginning student should not be misled by the preceding examples. It is not always possible to find limits merely by substitution. Sometimes other devices must be employed. The next theorem concerns three functions f, h, and g, where h(x) is always "sandwiched" between f(x) and g(x). If f and g have a common limit L as x approaches a, then as stated below, h must have the same limit. The Sandwich Theorem (2.14) If f(x) ~ h(x) ~ g(x) for all x in an open interval containing a, except possibly at a, and iflim f(x)= L = lim g(x),thenlim h(x) =L.x->a yY = f(x) A proof of the Sandwich Theorem based on the definition of limit may be found in Appendix II. The result is also clear from geometric considerations. Specifically, if f(x) ~ h(x) ~ g(x) for all x in an open interval containing x,aX then the graph of h lies "between" the graphs off and gin that interval, as illustrated in Figure 2.12. Iff and g have the same limit Las x approaches a,FIGURE 2.12then evidently, h also has the limit L. 76. Theorems on Limits 2.3 67 Example 8 Prove that limx-+o x sin(1/x) = 0. SolutionThe limit cannot be found by substituting 0 for x, or by using an algebraic manipulation. However, since all values of the sine function are between -1 and 1 it follows that if x =f. 0, Isin (1/x)l" :$;; 1 and, therefore, Ix sin~ I= Ix II sin~ I :$;; Ix 1. Consequently, 0 :$;; I x sin~ I:$;; lxl. It is not difficult to show that limx-+o IxI = 0. Hence, by the Sandwich Theorem (2.14), with f(x) = 0 and g(x) = IxI, we see thatlim I x sin .!_ I = 0.x-+0 X It now follows from the Definition of Limit (2.3) that.. 10. 1 xsm-1m x-+0 X=2.3 ExercisesIn Exercises 1-36 find the limits, if they exist.21 . 2Jx + hm _.!...-=-- x31222 lim 16x 213 413 1lim (3x 3-2x + 7)2 lim (5x 2 - 9x - 8) x-+16fx + 5x-+-8 4- X x-+ -2x-+4 3lim (x 2 + 3Xx - 4) 4lim (3t + 4)(7t - 9)23 r2 + 5x - 3x 3Im Jx2- 1 x-+,/ir-+-3 x-+3 5 lim .jx 2 - 5x - 4 6 lim jx 4 -4x+1 Ji6+h x-+4x-+ -225. 4- hm h-+0h 26limh-+0 1 (~)(- -1) h Ji+h 4x 2 - 6x + 3 7 limO8lim --..,.--.-- - -3 x-112 16x + 8x - 728lim (x + 4) 3(x -6) 2 9lim 1510limj2 x-+6 x-+/2 x-+ 1530 lim j3k 2 + 4.j3k + 2 . 2x + 5x- 3212limx +3v-+3 k-+211 hm --::---:2.------::---=- x-+112 6x - 7x + 2x-+,- 3 (1/x) + (1/3)31IIm (4t + 5t-4 . 2w 32lim~x-2 x2 - x- 2t-+-1 (6t + 5)(t - 5)r-+713 lim--14 lim2 x-+2 x3 - 8 x-+2(x -2)81 .X- 8X- 16 x3 + 833 lim x 2 -34II m - - -15lim--- x-+16Jx- 416lim--- x--2 x4- 16 x-+93-Jx x-+8 fx- 26s- 1 . (2 + h)- 2 - 2- 2. (9+h)- 1 -9- 117 lim--18 lim (x - 3.1416) 35 hm - - -- - - - - 36hm -----.:.....,.---- s-+4 2s - 9 h-oh h-+0h37 If r is any rational number and a > 0, prove that19 lim(~- -X -1-) X- 1 x-+1 120 lim x-+1(Jx + y ~) 6X limx-+ax =a. Under what conditions will this be true if a < 0? 77. 682Limits and Continuity of Functions38 If limx-af(x) = L =f. 0 and limx-.g(x) = 0, prove that 41 If c is a nonnegative real number and 0 ::;:; ((x) ::;:; c for limx-aLf(x)/g(x)] does not exist. (Hint: Assume there is aevery x, prove that limx-o x 2 ((x) = 0. number M such that limx-aLf(x)/g(x)] = M and consider42 Provethatlimx_ 0 x 4 sin(l/fx) = 0. (Hint: Sec Example 8.) lim f(x) = lim [g(x) f(x)].43 Prove that iff has a negative limit as x approaches a, x-ax-a g(x) then there is some open interval I containing a such39 Use the Sandwich Theorem and the fact thatthatf(x) is negative for every x in I except possibly x = a. limx-o(lxl + I)= I to prove that limx-o (x 2 + I)= I.40 Use the Sandwich Theorem withf(x) = 0 and g(x) = Ixi to prove that limlxl = 0. x-oJx 4 + 4x 2 + 7 2. 4One-Sided LimitsIf f(x) = ~and a > 2, then .f is defined throughout an open intervalcontaining a and, by Theorem (2.13),lim~ = Jlim (x - 2) = J a - 2.The case a = 2 is not covered by Definition (2.3) since there is no open intervalcontaining 2 throughout which/ is defined (note that v x - 2 is not real ifx < 2). A natural way to extend the definition of limit to include this excep-tional case is to restrict x to values greater than 2. Thus, we replace the con-dit