Solution Manual for Calculus Early Transcendentals, 3rd Edition

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SOLUTIONSMANUALMARKWOODARDFurman UniversityCALCULUSEARLYTRANSCENDENTALSTHIRDEDITIONWilliam BriggsUniversity of Colorado at DenverLyle CochranWhitworth UniversityBernhard GillettUniversity of Colorado, BoulderEric SchulzWalla Walla Community College

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Contents1Functions31.1Review of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31.2Representing Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .121.3Inverse, Exponential and Logarithmic Functions. . . . . . . . . . . . . . . . . . . . . . . . .301.4Trigonometric Functions and Their Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . .39Chapter One Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .522Limits652.1The Idea of Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .652.2Definition of a Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .702.3Techniques for Computing Limits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .852.4Infinite Limits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .952.5Limits at Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1042.6Continuity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1182.7Precise Definitions of Limits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .131Chapter Two Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1403Derivatives1533.1Introducing the Derivative. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1533.2The Derivative as a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1623.3Rules of Di↵erentiation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1803.4The Product and Quotient Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1883.5Derivatives of Trigonometric Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2003.6Derivatives as Rates of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2093.7The Chain Rule. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2243.8Implicit Di↵erentiation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2383.9Derivatives of Logarithmic and Exponential Functions . . . . . . . . . . . . . . . . . . . . . .2563.10Derivatives of Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . .2683.11Related Rates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .277Chapter Three Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2894Applications of the Derivative3054.1Maxima and Minima. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3054.2Mean Value Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3204.3What Derivatives Tell Us . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3284.4Graphing Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3464.5Optimization Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3804.6Linear Approximation and Di↵erentials. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4034.7L’Hˆopital’s Rule. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4134.8Newton’s Method. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4274.9Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .443Chapter Four Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4541

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2Contents5Integration4775.1Approximating Areas under Curves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4775.2Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4975.3Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5175.4Working with Integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5345.5Substitution Rule. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .544Chapter Five Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5556Applications of Integration5716.1Velocity and Net Change. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5716.2Regions Between Curves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5856.3Volume by Slicing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6006.4Volume by Shells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6086.5Length of Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6186.6Surface Area. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6246.7Physical Applications. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .632Chapter Six Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6427Logarithmic, Exponential, and Hyperbolic Functions6597.1Logarithmic and Exponential Functions Revisited. . . . . . . . . . . . . . . . . . . . . . . .6597.2Exponential Models. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6677.3Hyperbolic Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .673Chapter Seven Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6858Integration Techniques6918.1Basic Approaches. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6918.2Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7018.3Trigonometric Integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7208.4Trigonometric Substitutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7298.5Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7468.6Integration Strategies. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7618.7Other Methods of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7968.8Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .8058.9Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .816Chapter Eight Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .8319Di↵erential Equations8539.1Basic Ideas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .8539.2Direction Fields and Euler’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .8599.3Separable Di↵erential Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .8719.4Special First-Order Linear Di↵erential Equations . . . . . . . . . . . . . . . . . . . . . . . . .8849.5Modeling with Di↵erential Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .892Chapter Nine Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .90110 Sequences and Infinite Series90910.1An Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .90910.2Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .91710.3Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .93110.4The Divergence and Integral Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .94210.5Comparison Tests. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .95310.6Alternating Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .96110.7The Ratio and Root Tests. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .96810.8Choosing a Convergence Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .974Chapter Ten Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .991

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Contents311 Power Series100311.1Approximating Functions With Polynomials. . . . . . . . . . . . . . . . . . . . . . . . . . . 100311.2Properties of Power Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102111.3Taylor Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103011.4Working with Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1045Chapter Eleven Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105912 Parametric and Polar Curves106912.1Parametric Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106912.2Polar Coordinates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108612.3Calculus in Polar Coordinates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110612.4Conic Sections. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1121Chapter Twelve Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114113 Vectors and the Geometry of Space115913.1Vectors in the Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115913.2Vectors in Three Dimensions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116713.3Dot Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117713.4Cross Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118513.5Lines and Planes in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119413.6Cylinders and Quadric Surfaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1202Chapter Thirteen Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121714 Vector-Valued Functions123114.1Vector-Valued Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123114.2Calculus of Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123914.3Motion in Space. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124514.4Length of Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126414.5Curvature and Normal Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1270Chapter Fourteen Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128115 Functions of Several Variables129715.1Graphs and Level Curves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129715.2Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130915.3Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131515.4The Chain Rule. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132715.5Directional Derivatives and the Gradient. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133815.6Tangent Planes and Linear Approximation. . . . . . . . . . . . . . . . . . . . . . . . . . . . 135215.7Maximum/Minimum Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136015.8Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1371Chapter Fifteen Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138016 Multiple Integration139116.1Double Integrals over Rectangular Regions. . . . . . . . . . . . . . . . . . . . . . . . . . . . 139116.2Double Integrals over General Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139816.3Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141416.4Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142816.5Triple Integrals in Cylindrical and Spherical Coordinates. . . . . . . . . . . . . . . . . . . . 144016.6Integrals for Mass Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145216.7Change of Variables in Multiple Integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1463Chapter Sixteen Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1475

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4Contents17 Vector Calculus148917.1Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148917.2Line Integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150217.3Conservative Vector Fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151217.4Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151817.5Divergence and Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153217.6Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154317.7Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155417.8Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1562Chapter Seventeen Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1571D2 Second-Order Di↵erential Equations1583D2.1 Basic Ideas. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1583D2.2 Linear Homogeneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1602D2.3 Linear Nonhomogeneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1612D2.4 Applications. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1626D2.5 Complex Forcing Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1664Chapter D2 Review. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1683

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Chapter 1Functions1.1Review of Functions1.1.1A function is a rule that assigns each to each value of the independent variable in the domain a uniquevalue of the dependent variable in the range.1.1.2The independent variable belongs to the domain, while the dependent variable belongs to the range.1.1.3GraphAdoes not represent a function, while graphBdoes. Note that graphAfails the vertical linetest, while graphBpasses it.1.1.4The domain offis [1,4), while the range offis (1,5]. Note that the domain is the “shadow” of thegraph on thex-axis, while the range is the “shadow” of the graph on they-axis.1.1.5Item i. is true while item ii. isn’t necessarily true. In the definition of function, item i. is stipulated.However, item ii. need not be true – for example, the functionf(x) =x2has two di↵erent domain valuesassociated with the one range value 4, becausef(2) =f(2) = 4.1.1.6g(x) =x2+1x1=(x+1)(x1)x1=x+ 1,x6= 1.Thedomain is{x:x6= 1}and the range is{x:x6= 2}.-��=�(�)5

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6Chapter 1.Functions1.1.7The domain of this function is the set of a realnumbers. The range is [10,1).-��-��-��-��-�-�-�-��1.1.8The independent variabletis elapsed time and the dependent variabledis distance above the ground.The domain in context is [0,8]1.1.9The independent variablehis the height of the water in the tank and the dependent variableVis thevolume of water in the tank. The domain in context is [0,50]1.1.10f(2) =123+ 1 = 19 .f(y2) =1(y2)3+1=1y6+1.1.1.11f(g(1/2)) =f(2) =3;g(f(4)) =g(9) =18;g(f(x)) =g(2x+ 1) =1(2x+ 1)1 =12x.1.1.12One possible answer isg(x) =x2+ 1 andf(x) =x5, because thenf(g(x)) =f(x2+ 1) = (x2+ 1)5.Another possible answer isg(x) =x2andf(x) = (x+ 1)5, because thenf(g(x)) =f(x2) = (x2+ 1)5.1.1.13The domain offgconsists of allxin the domain ofgsuch thatg(x) is in the domain off.1.1.14(fg)(3) =f(g(3)) =f(25) =p25 = 5.(ff)(64) =f(p64) =f(8) =p8 = 2p2.(gf)(x) =g(f(x)) =g(px) =x3/22.(fg)(x) =f(g(x)) =f(x32) =px321.1.15a. (fg)(2) =f(g(2)) =f(2) = 4.b.g(f(2)) =g(4) = 1.c.f(g(4)) =f(1) = 3.d.g(f(5)) =g(6) = 3.e.f(f(8)) =f(8) = 8.f.g(f(g(5))) =g(f(2)) =g(4) = 1.1.1.16a.h(g(0)) =h(0) =1.b.g(f(4)) =g(1) =1.c.h(h(0)) =h(1) = 0.d.g(h(f(4))) =g(h(1)) =g(0) = 0.e.f(f(f(1))) =f(f(0)) =f(1) = 0.f.h(h(h(0))) =h(h(1)) =h(0) =1.g.f(h(g(2))) =f(h(3)) =f(0) = 1.h.g(f(h(4))) =g(f(4)) =g(1) =1.i.g(g(g(1))) =g(g(2)) =g(3) = 4.j.f(f(h(3))) =f(f(0)) =f(1) = 0.1.1.17f(5)f(0)50= 8365= 15.4; the radiosonde rises at an average rate of 15.4 ft/s during the first 5seconds after it is released.

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1.1.Review of Functions71.1.18f(0) = 0.f(34) = 127852.4109731 = 18121.4.f(64) = 127852.475330.4 = 52522.f(64)f(34)6434= 5252218121.430⇡1146.69 ft/s.1.1.19f(2) =f(2) = 2;g(2) =g(2) =(2) = 2;f(g(2)) =f(2) =f(2) = 2;g(f(2)) =g(f(2)) =g(2) =2.1.1.20The graph would be the result of leaving the portion of the graph in the first quadrant, and thenalso obtaining a portion in the third quadrant which would be the result of reflecting the portion in the firstquadrant around they-axis and then thex-axis.1.1.21FunctionAis symmetric about they-axis, so is even. FunctionBis symmetric about the origin sois odd. FunctionCis symmetric about they-axis, so is even.1.1.22FunctionAis symmetric about they-axis, so is even. FunctionBis symmetric about the origin, sois odd. FunctionCis also symmetric about the origin, so is odd.1.1.23f(x) =x25x+ 6x2= (x2)(x3)x2=x3,x6= 2. The domain offis{x:x6= 2}. The range is{y:y6=1}.1.1.24f(x) =x22x=x2(x2)=1,x6= 2. The domain is{x:x6= 2}. The range is{1}.1.1.25The domain of the function is the set of numbersxwhich satisfy 7x20.This is the interval[p7,p7]. Note thatf(p7) = 0 andf(0) =p7. The range is [0,p7].1.1.26The domain of the function is the set of numbersxwhich satisfy 25x20. This is the interval[5,5]. Note thatf(0) =5 andf(5) = 0. The range is [5,0].1.1.27Because the cube root function is defined for all real numbrs, the domain isR, the set of all realnumbers.1.1.28The domain consists of the set of numberswfor which 2w0, so the interval (1,2].1.1.29The domain consists of the set of numbersxfor which 9x20, so the interval [3,3].1.1.30Because 1 +t2is never zero for any real numbered value oft, the domain of this function isR, theset of all real numbers.1.1.31a. The formula for the height of the rocket isvalid fromt= 0 until the rocket hits theground, which is the positive solution to16t2+ 96t+ 80 = 0, which the quadraticformula reveals ist= 3 +p14. Thus, thedomain is [0,3 +p14].b.��(�)The maximum appears to occur att= 3.The height at that time would be 224.1.1.32a.d(0) = (10(2.2)·0)2= 100.b. The tank is first empty whend(t) = 0, which is when 10(2.2)t= 0, ort= 50/11.c. An appropriate domain would [0,50/11].

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Solution Manual for Calculus Early Transcendentals, 3rd Edition - Page 10 preview image

8Chapter 1.Functions1.1.33g(1/z) = (1/z)3=1z31.1.34F(y4) =1y431.1.35F(g(y)) =F(y3) =1y331.1.36f(g(w)) =f(w3) = (w3)24 =w641.1.37g(f(u)) =g(u24) = (u24)31.1.38f(2 +h)f(2)h= (2 +h)240h= 4 + 4h+h24h= 4h+h2h= 4 +h1.1.39F(F(x)) =F✓1x3◆=11x33 =11x33(x3)x3=1103xx3=x3103x1.1.40g(F(f(x))) =g(F(x24)) =g✓1x243◆=✓1x27◆31.1.41f(px+ 4) = (px+ 4)24 =x+ 44 =x.1.1.42F((3x+ 1)/x) =13x+1x3 =13x+13xx=x3x+ 13x=x.1.1.43g(x) =x35 andf(x) =x10.1.1.44g(x) =x6+x2+ 1 andf(x) =2x2.1.1.45g(x) =x4+ 2 andf(x) =px.1.1.46g(x) =x31 andf(x) =1px.1.1.47(fg)(x) =f(g(x)) =f(x24) =|x24|. The domain of this function is the set of all real numbers.1.1.48(gf)(x) =g(f(x)) =g(|x|) =|x|24 =x24. The domain of this function is the set of all realnumbers.1.1.49(fG)(x) =f(G(x)) =f⇣1x2⌘=1x2=1|x2|. The domain of this function is the set of all realnumbers except for the number 2.1.1.50(fgG)(x) =f(g(G(x))) =f⇣g⇣1x2⌘⌘=f✓⇣1x2⌘24◆=⇣1x2⌘24. The domain of thisfunction is the set of all real numbers except for the number 2.1.1.51(Ggf)(x) =G(g(f(x))) =G(g(|x|)) =G(x24) =1x242=1x26. The domain of this functionis the set of all real numbers except for the numbers±p6.1.1.52(gFF)(x) =g(F(F(x))) =g(F(px)) =g(ppx) =px4. The domain is [0,1).1.1.53(gg)(x) =g(g(x)) =g(x24) = (x24)24 =x48x2+ 164 =x48x2+ 12. The domain isthe set of all real numbers.1.1.54(GG)(x) =G(G(x)) =G(1/(x2)) =11x22 =112(x2)x2=x212x+ 4 =x252x. ThenGGisdefined except where the denominator vanishes, so its domain is the set of all real numbers except forx=52.1.1.55Because (x2+ 3)3 =x2, we may choosef(x) =x3.1.1.56Because the reciprocal ofx2+ 3 is1x2+3, we may choosef(x) =1x.1.1.57Because (x2+ 3)2=x4+ 6x2+ 9, we may choosef(x) =x2.

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Solution Manual for Calculus Early Transcendentals, 3rd Edition - Page 11 preview image

1.1.Review of Functions91.1.58Because (x2+ 3)2=x4+ 6x2+ 9, and the given expression is 11 more than this, we may choosef(x) =x2+ 11.1.1.59Because (x2)2+ 3 =x4+ 3, this expression results from squaringx2and adding 3 to it. Thus wemay choosef(x) =x2.1.1.60Becausex2/3+ 3 = (3px)2+ 3, we may choosef(x) =3px.1.1.61a. True. A real numberzcorresponds to the domain elementz/2 + 19, becausef(z/2 + 19) = 2(z/2 +19)38 =z+ 3838 =z.b. False. The definition of function does not require that each range element comes from a unique domainelement, rather that each domain element is paired with a unique range element.c. True.f(1/x) =11/x=x, and1f(x)=11/x=x.d. False. For example, suppose thatfis the straight line through the origin with slope 1, so thatf(x) =x.Thenf(f(x)) =f(x) =x, while (f(x))2=x2.e. False. For example, letf(x) =x+2 andg(x) = 2x1. Thenf(g(x)) =f(2x1) = 2x1+2 = 2x+1,whileg(f(x)) =g(x+ 2) = 2(x+ 2)1 = 2x+ 3.f. True. This is the definition offg.g. True.Iffis even, thenf(z) =f(z) for allz, so this is true in particular forz=ax.So ifg(x) =cf(ax), theng(x) =cf(ax) =cf(ax) =g(x), sogis even.h. False.For example,f(x) =xis an odd function, buth(x) =x+ 1 isn’t, becauseh(2) = 3, whileh(2) =1 which isn’th(2).i. True. Iff(x) =f(x) =f(x), then in particularf(x) =f(x), so 0 = 2f(x), sof(x) = 0 for allx.1.1.62f(x+h)f(x)h= 1010h= 0h= 0.1.1.63f(x+h)f(x)h= 3(x+h)3xh= 3x+ 3h3xh= 3hh= 3.1.1.64f(x+h)f(x)h= 4(x+h)3(4x3)h= 4x+ 4h34x+ 3h= 4hh= 4.1.1.65f(x+h)f(x)h= (x+h)2x2h= (x2+ 2hx+h2)x2h=h(2x+h)h= 2x+h.1.1.66f(x+h)f(x)h= 2(x+h)23(x+h) + 1(2x23x+ 1)h=2x2+ 4xh+ 2h23x3h+ 12x2+ 3x1h= 4xh+ 2h23hh=h(4x+ 2h3)h= 4x+ 2h3.1.1.67f(x+h)f(x)h=2x+h2xh=2x2(x+h)x(x+h)h= 2x2x2hhx(x+h)=2hhx(x+h) =2x(x+h).1.1.68f(x+h)f(x)h=x+hx+h+1xx+1h=(x+h)(x+1)x(x+h+1)(x+1)(x+h+1)h=x2+x+hx+hx2xhxh(x+ 1)(x+h+ 1)=hh(x+ 1)(x+h+ 1) =1(x+ 1)(x+h+ 1)

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Solution Manual for Calculus Early Transcendentals, 3rd Edition - Page 12 preview image

10Chapter 1.Functions1.1.69f(x)f(a)xa=x2+x(a2+a)xa= (x2a2) + (xa)xa= (xa)(x+a) + (xa)xa=(xa)(x+a+ 1)xa=x+a+ 1.1.1.70f(x)f(a)xa= 44xx2(44aa2)xa=4(xa)(x2a2)xa=4(xa)(xa)(x+a)xa= (xa)(4(x+a))xa=4xa.1.1.71f(x)f(a)xa=x32x(a32a)xa= (x3a3)2(xa)xa= (xa)(x2+ax+a2)2(xa)xa=(xa)(x2+ax+a22)xa=x2+ax+a22.1.1.72f(x)f(a)xa=x4a4xa= (x2a2)(x2+a2)xa= (xa)(x+a)(x2+a2)xa= (x+a)(x2+a2).1.1.73f(x)f(a)xa=4x24a2xa=4a2+4x2a2x2xa=4(x2a2)(xa)a2x2= 4(xa)(x+a)(xa)a2x2= 4(x+a)a2x2.1.1.74f(x)f(a)xa=1xx2(1aa2)xa=1x1axax2a2xa=axaxxa(xa)(x+a)xa=1ax(x+a).1.1.75a. The slope is122271049931= 864 ft/h. The hiker’s elevation increases at an average rate of 874 feet perhour.b. The slope is121441263154=487 ft/h. The hiker’s elevation decreases at an average rate of 487 feetper hour.c. The hiker might have stopped to rest during this interval of time or the trail is level on this portion ofthe hike.1.1.76a. The slope is11302995431= 674 ft/m. The elevation of the trail increases by an average of 674 feet permile for 1d3.b. The slope is122371235765=120 ft/m. The elevation of the trail decreases by an average of 120 feetper mile for 5d6.c. The elevation of the trail doesn’t change much for 4.5d5.1.1.77a.��(�� ))(�� ))b. Theslopeofthesecantlineisgivenby4006452=3363= 112 feet per second.Theobject falls at an average rate of 112 feet persecond over the interval 2t5.

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Solution Manual for Calculus Early Transcendentals, 3rd Edition - Page 13 preview image

1.1.Review of Functions111.1.78a.��(�/�� ))(�� ))b. Theslopeofthesecantlineisgivenby41.52=31.5=2 cubic centimeters per at-mosphere. The volume decreases by an av-erage of 2 cubic centimeters per atmosphereover the interval 0.5p2.1.1.79This function is symmetric about they-axis, becausef(x) = (x)4+ 5(x)212 =x4+ 5x212 =f(x).1.1.80This function is symmetric about the origin, becausef(x) = 3(x)5+ 2(x)3(x) =3x52x3+x=(3x5+ 2x3x) =f(x).1.1.81This function has none of the indicated symmetries.For example, note thatf(2) =26, whilef(2) = 22, sofis not symmetric about either the origin or about they-axis, and is not symmetric aboutthex-axis because it is a function.1.1.82This function is symmetric about they-axis. Note thatf(x) = 2|x|= 2|x|=f(x).1.1.83This curve (which is not a function) is symmetric about thex-axis, they-axis, and the origin. Notethat replacing eitherxbyxorybyy(or both) yields the same equation. This is due to the fact that(x)2/3= ((x)2)1/3= (x2)1/3=x2/3, and a similar fact holds for the term involvingy.1.1.84This function is symmetric about the origin. Writing the function asy=f(x) =x3/5, we see thatf(x) = (x)3/5=(x)3/5=f(x).1.1.85This function is symmetric about the origin. Note thatf(x) = (x)|(x)|=x|x|=f(x).1.1.86This curve (which is not a function) is symmetric about thex-axis, they-axis, and the origin. Notethat replacing eitherxbyxorybyy(or both) yields the same equation. This is due to the fact that|x|=|x|and|y|=|y|.1.1.87a.f(g(2)) =f(g(2)) =f(2) = 4b.g(f(2)) =g(f(2)) =g(4) = 1c.f(g(4)) =f(g(4)) =f(1) = 3d.g(f(5)8) =g(2) =g(2) =2e.g(g(7)) =g(g(7)) =g(4) =1f.f(1f(8)) =f(7) = 71.1.88a.f(g(1)) =f(g(1)) =f(3) = 3b.g(f(4)) =g(f(4)) =g(4) =g(4) = 2c.f(g(3)) =f(g(3)) =f(4) =4d.f(g(2)) =f(g(2)) =f(1) = 2e.g(g(1)) =g(g(1)) =g(3) =4f.f(g(0)1) =f(1) =f(1) = 2g.f(g(g(2))) =f(g(g(2))) =f(g(1)) =f(3) = 3h.g(f(f(4))) =g(f(4)) =g(4) = 2i.g(g(g(1))) =g(g(g(1))) =g(g(3)) =g(4) = 2

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Solution Manual for Calculus Early Transcendentals, 3rd Edition - Page 14 preview image

12Chapter 1.Functions1.1.89We will make heavy use of the fact that|x|isxifx >0, and isxifx <0.In the first quadrantwherexandyare both positive, this equationbecomesxy= 1 which is a straight line withslope 1 andy-intercept1.In the second quad-rant wherexis negative andyis positive, thisequation becomesxy= 1, which is a straightline with slope1 andy-intercept1. In the thirdquadrant where bothxandyare negative, we ob-tain the equationx(y) = 1, ory=x+ 1,and in the fourth quadrant, we obtainx+y= 1.Graphing these lines and restricting them to theappropriate quadrants yields the following curve:-4-224x-4-224y1.1.90We havey= 10 +px2+ 10x9, so by subtracting 10 from both sides and squaring we have(y10)2=x2+ 10x9, which can be written asx210x+ (y10)2=9.To complete the square inx, we add 25 to both sides, yieldingx210x+ 25 + (y10)2=9 + 25,or(x5)2+ (y10)2= 16.This is the equation of a circle of radius 4 centered at (5,10). Becausey10, we see that the graph offisthe upper half of this circle. The domain of the function is [1,9] and the range is [10,14].1.1.91We havey= 2px2+ 6x+ 16, so by subtracting 2 from both sides and squaring we have(y2)2=x2+ 6x+ 16, which can be written asx26x+ (y2)2= 16.To complete the square inx, we add 9 to both sides, yieldingx26x+ 9 + (y2)2= 16 + 9,or(x3)2+ (y2)2= 25.This is the equation of a circle of radius 5 centered at (3,2). Becausey2, we see that the graph offisthe lower half of this circle. The domain of the function is [2,8] and the range is [3,2].1.1.92a. No. For examplef(x) =x2+ 3 is an even function, butf(0) is not 0.b. Yes.becausef(x) =f(x), and because0 = 0, we must havef(0) =f(0) =f(0), sof(0) =f(0), and the only number which is its own additive inverse is 0, sof(0) = 0.1.1.93Because the composition offwith itself has first degree,fhas first degree as well, so letf(x) =ax+b.Then (ff)(x) =f(ax+b) =a(ax+b) +b=a2x+ (ab+b). Equating coecients, we see thata2= 9 andab+b=8. Ifa= 3, we get thatb=2, while ifa=3 we haveb= 4. So the two possible answers aref(x) = 3x2 andf(x) =3x+ 4.

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Solution Manual for Calculus Early Transcendentals, 3rd Edition - Page 15 preview image

1.1.Review of Functions131.1.94Since the square of a linear function is a quadratic, we letf(x) =ax+b. Thenf(x)2=a2x2+2abx+b2.Equating coecients yields thata=±3 andb=±2. However, a quick check shows that the middle termis correct only when one of these is positive and one is negative. So the two possible such functionsfaref(x) = 3x2 andf(x) =3x+ 2.1.1.95Letf(x) =ax2+bx+c. Then (ff)(x) =f(ax2+bx+c) =a(ax2+bx+c)2+b(ax2+bx+c) +c.Expanding this expression yieldsa3x4+ 2a2bx3+ 2a2cx2+ab2x2+ 2abcx+ac2+abx2+b2x+bc+c, whichsimplifies toa3x4+ 2a2bx3+ (2a2c+ab2+ab)x2+ (2abc+b2)x+ (ac2+bc+c). Equating coecients yieldsa3= 1, soa= 1.Then 2a2b= 0, sob= 0.It then follows thatc=6, so the original function wasf(x) =x26.1.1.96Because the square of a quadratic is a quartic, we letf(x) =ax2+bx+c.Then the square offisc2+ 2bcx+b2x2+ 2acx2+ 2abx3+a2x4.By equating coecients, we see thata2= 1 and soa=±1.Because the coecient onx3must be 0, we have thatb= 0. And the constant term reveals thatc=±6. Aquick check shows that the only possible solutions are thusf(x) =x26 andf(x) =x2+ 6.1.1.97f(x+h)f(x)h=px+hpxh=px+hpxh·px+h+pxpx+h+px=(x+h)xh(px+h+px) =1px+h+px.f(x)f(a)xa=pxpaxa=pxpaxa·px+papx+pa=xa(xa)(px+pa) =1px+pa.1.1.98f(x+h)f(x)h=p12(x+h)p12xh=p12(x+h)p12xh·p12(x+h) +p12xp12(x+h) +p12x=12(x+h)(12x)h(p12(x+h) +p12x) =2p12(x+h) +p12x.f(x)f(a)xa=p12xp12axa=p12xp12axa·p12x+p12ap12x+p12a=(12x)(12a)(xa)(p12x+p12a) =(2)(xa)(xa)(p12x+p12a) =2(p12x+p12a) .1.1.99f(x+h)f(x)h=3px+h3pxh=3(pxpx+h)hpxpx+h=3(pxpx+h)hpxpx+h·px+px+hpx+px+h=3(x(x+h))hpxpx+h(px+px+h) =3pxpx+h(px+px+h) .f(x)f(a)xa=3px3paxa=3⇣papxpapx⌘xa= (3)(papx)(xa)papx·pa+pxpa+px=(3)(xa)(xa)(papx)(pa+px) =3pax(pa+px) .1.1.100f(x+h)f(x)h=p(x+h)2+ 1px2+ 1h=p(x+h)2+ 1px2+ 1h·p(x+h)2+ 1 +px2+ 1p(x+h)2+ 1 +px2+ 1 =(x+h)2+ 1(x2+ 1)h(p(x+h)2+ 1 +px2+ 1) =x2+ 2hx+h2x2h(p(x+h)2+ 1 +px2+ 1) =2x+hp(x+h)2+ 1 +px2+ 1 .

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Solution Manual for Calculus Early Transcendentals, 3rd Edition - Page 16 preview image

14Chapter 1.Functionsf(x)f(a)xa=px2+ 1pa2+ 1xa=px2+ 1pa2+ 1xa·px2+ 1 +pa2+ 1px2+ 1 +pa2+ 1 =x2+ 1(a2+ 1)(xa)(px2+ 1 +pa2+ 1) =(xa)(x+a)(xa)(px2+ 1 +pa2+ 1) =x+apx2+ 1 +pa2+ 1 .1.1.101This would not necessarily have either kind of symmetry.For example,f(x) =x2is an evenfunction andg(x) =x3is odd, but the sum of these two is neither even nor odd.1.1.102This would be an odd function, so it would be symmetric about the origin. Supposefis even andgis odd. Then (f·g)(x) =f(x)g(x) =f(x)·(g(x)) =(f·g)(x).1.1.103This would be an even function, so it would be symmetric about they-axis. Supposefis even andgis odd. Theng(f(x)) =g(f(x)), becausef(x) =f(x).1.1.104This would be an even function, so it would be symmetric about they-axis. Supposefis even andgis odd. Thenf(g(x)) =f(g(x)) =f(g(x)).1.2Representing Functions1.2.1Functions can be defined and represented by a formula, through a graph, via a table, and by usingwords.1.2.2The domain of every polynomial is the set of all real numbers.1.2.3The slope of the line shown ism=3(1)30=2/3. They-intercept isb=1. Thus the function isgiven byf(x) =23x1.1.2.4Because it is to be parallel to a line with slope 2, it must also have slope 2. Using the point-slope formof the equation of the line, we havey0 = 2(x5), ory= 2x10.1.2.5The domain of a rational functionp(x)q(x)is the set of all real numbers for whichq(x)6= 0.1.2.6A piecewise linear function is one which is linear over intervals in the domain.1.2.7Forx <0, the graph is a line with slope 1 andy- intercept 3, while forx >0, it is a line with slope1/2 andy-intercept 3. Note that both of these lines contain the point (0,3). The function shown can thusbe writtenf(x) =8<:x+ 3ifx <0;12x+ 3ifx0.1.2.8The transformed graph would have equationy=px2 + 3.1.2.9Compared to the graph off(x), the graph off(x+ 2) will be shifted 2 units to the left.1.2.10Compared to the graph off(x), the graph of3f(x) will be scaled vertically by a factor of 3 andflipped about thexaxis.1.2.11Compared to the graph off(x), the graph off(3x) will be compressed horizontally by a factor of13.1.2.12To produce the graph ofy= 4(x+ 3)2+ 6 from the graph ofx2, one must1. shift the graph horizontally by 3 units to left2. scale the graph vertically by a factor of 43. shift the graph vertically up 6 units.

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