This page intentionally left blank Students Solutions Manualto accompany Jon RogawskisSingle VariableCALCULUSSECOND EDITIONBRIAN BRADIEChristopher Newport UniversityROGER LIPSETTW. H. FREEMAN AND COMPANYNEW YORK 2012 by W. H. Freeman and CompanyISBN-13: 978-1-4292-4290-5ISBN-10: 1-4292-4290-6All rights reservedPrinted in the United States of AmericaFirst PrintingW. H. Freeman and Company, 41 Madison Avenue, New York, NY 10010Houndmills, Basingstoke RG21 6XS, Englandwww.whfreeman.comCONTENTSChapter 1 PRECALCULUS REVIEW 11.1 Real Numbers, Functions, and Graphs 11.2 Linear and Quadratic Functions 81.3 The Basic Classes of Functions 131.4 Trigonometric Functions 161.5 Technology: Calculators and Computers 23Chapter Review Exercises 27Chapter 2 LIMITS 312.1 Limits, Rates of Change, and Tangent Lines 312.2 Limits: A Numerical and Graphical Approach 372.3 Basic Limit Laws 462.4 Limits and Continuity 492.5 Evaluating Limits Algebraically 572.6 Trigonometric Limits 612.7 Limits at Innity 662.8 Intermediate Value Theorem 732.9 The Formal Denition of a Limit 76Chapter Review Exercises 82Chapter 3 DIFFERENTIATION 913.1 Definition of the Derivative 913.2 The Derivative as a Function 1013.3 Product and Quotient Rules 1123.4 Rates of Change 1193.5 Higher Derivatives 1263.6 Trigonometric Functions 1323.7 The Chain Rule 1383.8 Implicit Differentiation 1473.9 Related Rates 157Chapter Review Exercises 165Chapter 4 APPLICATIONS OF THE DERIVATIVE 1744.1 Linear Approximation and Applications 1744.2 Extreme Values 1814.3 The Mean Value Theorem and Monotonicity 1914.4 The Shape of a Graph 1984.5 Graph Sketching and Asymptotes 2064.6 Applied Optimization 2204.7 Newtons Method 2364.8 Antiderivatives 242Chapter Review Exercises 250Chapter 5 THE INTEGRAL 2605.1 Approximating and Computing Area 2605.2 The Denite Integral 2745.3 The Fundamental Theorem of Calculus, Part I 2845.4 The Fundamental Theorem of Calculus, Part II 2905.5 Net Change as the Integral of a Rate 2965.6 Substitution Method 300Chapter Review Exercises 307Chapter 6 APPLICATIONS OF THE INTEGRAL 3176.1 Area Between Two Curves 3176.2 Setting Up Integrals: Volume, Density, Average Value 3286.3 Volumes of Revolution 3366.4 The Method of Cylindrical Shells 3466.5 Work and Energy 355Chapter Review Exercises 362Chapter 7 EXPONENTIAL FUNCTIONS 3707.1 Derivative of f (x) = bxand the Number e 3707.2 Inverse Functions 3787.3 Logarithms and Their Derivatives 3837.4 Exponential Growth and Decay 3937.5 Compound Interest and Present Value 3987.6 Models Involving y

= k ( y b) 4017.7 LHpitals Rule 4077.8 Inverse Trigonometric Functions 4157.9 Hyperbolic Functions 424Chapter Review Exercises 431Chapter 8 TECHNIQUES OF INTEGRATION 4468.1 Integration by Parts 4468.2 Trigonometric Integrals 4578.3 Trigonometric Substitution 4678.4 Integrals Involving Hyperbolic and Inverse HyperbolicFunctions 4818.5 The Method of Partial Fractions 4858.6 Improper Integrals 5038.7 Probability and Integration 5208.8 Numerical Integration 525Chapter Review Exercises 537Chapter 9 FURTHER APPLICATIONS OF THEINTEGRAL AND TAYLORPOLYNOMIALS 5559.1 Arc Length and Surface Area 5559.2 Fluid Pressure and Force 5649.3 Center of Mass 5699.4 Taylor Polynomials 577Chapter Review Exercises 593iiiiv C A L C U L U S CONTENTSChapter 10 INTRODUCTION TO DIFFERENTIALEQUATIONS 60110.1 Solving Differential Equations 60110.2 Graphical and Numerical Methods 61410.3 The Logistic Equation 62110.4 First-Order Linear Equations 626Chapter Review Exercises 637Chapter 11 INFINITE SERIES 64611.1 Sequences 64611.2 Summing an Innite Series 65811.3 Convergence of Series with Positive Terms 66911.4 Absolute and Conditional Convergence 68311.5 The Ratio and Root Tests 69011.6 Power Series 69711.7 Taylor Series 710Chapter Review Exercises 727Chapter 12 PARAMETRIC EQUATIONS, POLARCOORDINATES, AND CONICSECTIONS 74212.1 Parametric Equations 74212.2 Arc Length and Speed 75912.3 Polar Coordinates 76612.4 Area and Arc Length in Polar Coordinates 78012.5 Conic Sections 789Chapter Review Exercises 801June 7, 2011 LTSV SSM Second Pass1 PRECALCULUS REVIEW1.1 Real Numbers, Functions, and GraphsPreliminary Questions1. Giveanexampleof numbersa andb suchthat a < b and|a| > |b|.solution Takea = 3andb = 1. Thena < b but |a| = 3> 1= |b|.2. Whichnumberssatisfy|a| = a?Whichsatisfy|a| = a?What about |a| = a?solution Thenumbersa 0satisfy|a| = a and| a| = a. Thenumbersa 0satisfy|a| = a.3. Giveanexampleof numbersa andb suchthat |a +b| < |a| +|b|.solution Takea = 3andb = 1. Then|a +b| = | 3+1| = | 2| = 2, but |a| +|b| = | 3| +|1| = 3+1= 4.Thus, |a +b| < |a| +|b|.4. What arethecoordinatesof thepoint lyingat theintersectionof thelinesx = 9andy = 4?solution Thepoint (9, 4) liesat theintersectionof thelinesx = 9andy = 4.5. Inwhichquadrant dothefollowingpointslie?(a) (1, 4) (b) (3, 2) (c) (4, 3) (d) (4, 1)solution(a) Becauseboththex- andy-coordinatesof thepoint (1, 4) arepositive, thepoint (1, 4) liesintherst quadrant.(b) Becausethex-coordinateof thepoint (3, 2) is negativebut they-coordinateis positive, thepoint (3, 2) lies inthesecondquadrant.(c) Becausethex-coordinateof thepoint (4, 3) is positivebut they-coordinateis negative, thepoint (4, 3) lies inthefourthquadrant.(d) Becauseboththex- andy-coordinatesof thepoint(4, 1) arenegative, thepoint(4, 1) liesinthethirdquadrant.6. What istheradiusof thecirclewithequation(x 9)2+(y 9)2= 9?solution Thecirclewithequation(x 9)2+(y 9)2= 9hasradius3.7. Theequationf (x) = 5hasasolutionif (chooseone):(a) 5belongstothedomainof f .(b) 5belongstotherangeof f .solution Thecorrect responseis(b): theequationf (x) = 5hasasolutionif 5belongstotherangeof f .8. What kindof symmetrydoesthegraphhaveif f (x) = f (x)?solution If f (x) = f (x), thenthegraphof f issymmetricwithrespect totheorigin.Exercises1. Useacalculator tondarational number r suchthat |r 2| < 104.solution r must satisfy 2 104 < r < 2+ 104, or 9.869504 < r < 9.869705. r = 9.8696 = 123371250 wouldbeonesuchnumber.Whichof (a)(f) aretruefor a = 3andb = 2?(a) a < b (b) |a| < |b| (c) ab > 0(d) 3a < 3b (e) 4a < 4b (f) 1a 2}solution x 4> 2or x 4< 2 x > 6or x < 2 (, 2) (6, ){x : |2x +4| > 3}21. {x : |x21| > 2}solution x2 1> 2or x2 1< 2 x2 > 3or x2 < 1(thiswill never happen) x >3or x < 3(, 3) (3, ).{x : |x2+2x| > 2}23. Match(a)(f) with(i)(vi).(a) a > 3 (b) |a 5| 5(e) |a 4| < 3 (f) 1 a 5(i) a liestotheright of 3.(ii) a liesbetween1and7.(iii) Thedistancefroma to5islessthan13.(iv) Thedistancefroma to3isat most 2.(v) a islessthan5unitsfrom13.(vi) a lieseither totheleft of 5or totheright of 5.solution(a) Onthenumber line, numbersgreater than3appear totheright; hence, a > 3isequivalenttothenumberstotherightof 3: (i).(b) |a 5| measuresthedistancefroma to5; hence, |a 5| < 13issatisedbythosenumberslessthan13of aunitfrom5: (iii).(c) |a 13| measuresthedistancefroma to 13; hence, |a 13| < 5issatisedbythosenumberslessthan5unitsfrom13: (v).(d) Theinequality|a| > 5isequivalenttoa > 5or a < 5; thatis, either a liestotherightof 5or totheleftof 5: (vi).(e) Theinterval describedbytheinequality|a 4| < 3hasacenter at 4andaradiusof 3; that is, theinterval consistsof thosenumbersbetween1and7: (ii).(f) Theinterval describedbytheinequality1< x < 5hasacenter at 3andaradiusof 2; that is, theinterval consistsofthosenumberslessthan2unitsfrom3: (iv).Describe_x : xx +1 < 0_asaninterval.25. Describe{x : x2+2x < 3} asaninterval. Hint: Plot y = x2+2x 3.solution Theinequalityx2+2x < 3isequivalent tox2+2x 3< 0. Inthegurebelow, weseethat thegraphofy = x2+ 2x 3falls belowthex-axis for 3 < x < 1. Thus, theset {x : x2+ 2x < 3} corresponds totheinterval3< x < 1.June 7, 2011 LTSV SSM Second PassS E C T I ON 1.1 Real Numbers, Functions, and Graphs 34 3 2 2246810yx1 2y = x2 + 2x 3Describetheset of real numberssatisfying|x 3| = |x 2| +1asahalf-inniteinterval.27. Showthat if a > b, thenb1 > a1, providedthat a andb havethesamesign. What happensif a > 0andb < 0?solution Case1a: If a andb arebothpositive, thena > b 1> ba 1b > 1a.Case1b: If a andb arebothnegative, thena > b 1< ba (sincea isnegative) 1b > 1a (again, sinceb isnegative).Case2: If a > 0andb < 0, then 1a > 0and 1b < 0so 1b < 1a. (SeeExercise2f for anexampleof this).Whichx satisfyboth|x 3| < 2and|x 5| < 1?29. Showthat if |a 5| < 12 and|b 8| < 12, then|(a +b) 13| < 1. Hint: Usethetriangleinequality.solution|a +b 13| = |(a 5) +(b 8)| |a 5| +|b 8| (bythetriangleinequality) 0}g(t ) = cos1tIn Exercises 4952, determine where f (x) is increasing.49. f (x) = |x +1|solution A graphof thefunctiony = |x +1| isshownbelow. Fromthegraph, weseethat thefunctionisincreasingontheinterval (1, ).x3 2 1121yf (x) = x351. f (x) = x4solution A graphof thefunctiony = x4isshownbelow. Fromthegraph, weseethat thefunctionisincreasingontheinterval (0, ).x2 1 1 21248yf (x) = 1x4+x2+1In Exercises 5358, nd the zeros of f (x) and sketch its graph by plotting points. Use symmetry and increase/decreaseinformation where appropriate.53. f (x) = x24solution Zeros: 2Increasing: x > 0Decreasing: x < 0Symmetry: f (x) = f (x) (evenfunction). So, y-axissymmetry.2244yx2 1 1 2f (x) = 2x2455. f (x) = x34xsolution Zeros: 0, 2; Symmetry: f (x) = f (x) (oddfunction). Sooriginsymmetry.551010yx2 1 1 2June 7, 2011 LTSV SSM Second PassS E C T I ON 1.1 Real Numbers, Functions, and Graphs 5f (x) = x357. f (x) = 2x3solution Thisisanx-axisreectionof x3translatedup2units. Thereisonezeroat x = 32.10102020yx2 1 1 2f (x) = 1(x 1)2+159. Whichof thecurvesinFigure26isthegraphof afunction?(A)xy(B)xy(C)xy(D)xyFIGURE 26solution (B) isthegraphof afunction. (A), (C), and(D) all fail thevertical linetest.Determinewhether thefunctioniseven, odd, or neither.(a) f (x) = x5 (b) g(t ) = t3t2(c) F(t ) = 1t4+t261. Determinewhether thefunctioniseven, odd, or neither.(a) f (t ) = 1t4+t +1 1t4t +1 (b) g(t ) = 2t2t(c) G() = sin +cos (d) H() = sin(2)solution(a) Thisfunctionisoddbecausef (t ) = 1(t )4+(t ) +1 1(t )4(t ) +1= 1t4t +1 1t4+t +1 = f (t ).(b) g(t ) = 2t2(t )= 2t2t= g(t ), sothisfunctionisodd.(c) G() = sin() + cos() = sin + cos which is equal to neither G() nor G(), so this function isneither oddnor even.(d) H() = sin(()2) = sin(2) = H(), sothisfunctioniseven.Writef (x) = 2x45x3+12x23x +4asthesumof anevenandanoddfunction.63. Determinetheinterval onwhichf (x) = 1x 4isincreasingor decreasing.solution Agraphof thefunctionisshownbelow. Fromthisgraphwecanseethatf (x) isdecreasingon(, 4) andalsodecreasingon(4, ).220462462 4 6 8 10 xyJune 7, 2011 LTSV SSM Second Pass6 C HA P T E R 1 PRECALCULUS REVIEWStatewhether thefunctionisincreasing, decreasing, or neither.(a) Surfaceareaof asphereasafunctionof itsradius(b) Temperatureat apoint ontheequator asafunctionof time(c) Priceof anairlineticket asafunctionof thepriceof oil(d) Pressureof thegasinapistonasafunctionof volumeIn Exercises 6570, let f (x) be the function shown in Figure 27.1 2 3 401234xyFIGURE 2765. Findthedomainandrangeof f (x)?solution D : [0, 4]; R : [0, 4]Sketchthegraphsof f (x +2) andf (x) +2.67. Sketchthegraphsof f (2x), f_12x_, and2f (x).solution Thegraphof y = f (2x) isobtainedbycompressingthegraphof y = f (x) horizontallybyafactor of 2(seethegraphbelowontheleft). Thegraphof y = f (12x) isobtainedbystretchingthegraphof y = f (x) horizontallybyafactor of 2(seethegraphbelowinthemiddle). Thegraphof y = 2f (x) isobtainedbystretchingthegraphof y = f (x)verticallybyafactor of 2(seethegraphbelowontheright).yx12341 2 3 4f (2x)yx12342 4 6 8f (x/2)yx24681 2 3 42f (x)Sketchthegraphsof f (x) andf (x).69. Extendthegraphof f (x) to[4, 4] sothat it isanevenfunction.solution Tocontinuethegraphof f (x) totheinterval [4, 4] asanevenfunction, reect thegraphof f (x) acrossthey-axis(seethegraphbelow).2 4 x2 4y1234Extendthegraphof f (x) to[4, 4] sothat it isanoddfunction.71. Supposethat f (x) hasdomain[4, 8] andrange[2, 6]. Findthedomainandrangeof:(a) f (x) +3 (b) f (x +3)(c) f (3x) (d) 3f (x)solution(a) f (x) + 3 is obtained by shifting f (x) upward threeunits. Therefore, thedomain remains [4, 8], whiletherangebecomes[5, 9].(b) f (x +3) isobtainedbyshiftingf (x) leftthreeunits. Therefore, thedomainbecomes[1, 5], whiletherangeremains[2, 6].(c) f (3x) is obtained by compressingf (x) horizontally by afactor of three. Therefore, thedomain becomes [43, 83],whiletherangeremains[2, 6].(d) 3f (x) isobtainedbystretchingf (x) verticallybyafactor of three. Therefore, thedomainremains[4, 8], whiletherangebecomes[6, 18].Let f (x) = x2. Sketchthegraphover [2, 2] of:(a) f (x +1) (b) f (x) +1(c) f (5x) (d) 5f (x)73. Supposethat thegraphof f (x) = sinx is compressedhorizontally by afactor of 2andthenshifted5units to theright.(a) What istheequationfor thenewgraph?(b) What istheequationif yourst shift by5andthencompressby2?(c) Verifyyour answersbyplottingyour equations.June 7, 2011 LTSV SSM Second PassS E C T I ON 1.1 Real Numbers, Functions, and Graphs 7solution(a) Let f (x) = sinx. After compressingthegraphof f horizontally by afactor of 2, weobtainthefunctiong(x) =f (2x) = sin2x. Shiftingthegraph5unitstotheright thenyieldsh(x) = g(x 5) = sin2(x 5) = sin(2x 10).(b) Letf (x) = sinx. After shiftingthegraph5unitstotheright, weobtainthefunctiong(x) = f (x 5) = sin(x 5).Compressingthegraphhorizontallybyafactor of 2thenyieldsh(x) = g(2x) = sin(2x 5).(c) The gure below at the top left shows the graphs of y = sinx (the dashed curve), the sine graph compressedhorizontallybyafactor of 2(thedash, doubledotcurve) andthenshiftedright5units(thesolidcurve). Comparethislastgraphwiththegraphof y = sin(2x 10) shownat thebottomleft.Thegurebelowat thetopright showsthegraphsof y = sinx (thedashedcurve), thesinegraphshiftedtotheright5units(thedash, doubledotcurve) andthencompressedhorizontallybyafactor of 2(thesolidcurve). Comparethislastgraphwiththegraphof y = sin(2x 5) shownat thebottomright.11yx6 4 2 6 4 211yx6 4 2 6 4 211yx6 4 2 6 4 211yx6 4 2 6 4 2Figure28showsthegraphof f (x) = |x| +1. Matchthefunctions(a)(e) withtheir graphs(i)(v).(a) f (x 1) (b) f (x) (c) f (x) +2(d) f (x 1) 2 (e) f (x +1)75. Sketchthegraphof f (2x) andf_12x_, wheref (x) = |x| +1(Figure28).solution Thegraphof y = f (2x) is obtainedby compressingthegraphof y = f (x) horizontally by afactor of 2(seethegraphbelowontheleft). Thegraphof y = f (12x) isobtainedbystretchingthegraphof y = f (x) horizontallybyafactor of 2(seethegraphbelowontheright).x12462 3 1 2 3yf (2x)x12462 3 1 2 3yf (x/2)Findthefunctionf (x) whosegraphisobtainedbyshiftingtheparabolay = x2threeunitstotheright andfourunitsdown, asinFigure29.77. Denef (x) tobethelarger of x and2x. Sketchthegraphof f (x). Whatareitsdomainandrange?Expressf (x)intermsof theabsolutevaluefunction.solutionx1121 2 3yThegraphof y = f (x) isshownabove. Clearly, thedomainof f isthesetof all real numberswhiletherangeis{y | y 1}.Noticethegraphhas thestandardV-shapeassociatedwiththeabsolutevaluefunction, but thebaseof theV has beentranslatedtothepoint (1, 1). Thus, f (x) = |x 1| +1.For eachcurveinFigure30, statewhether it issymmetricwithrespect tothey-axis, theorigin, both, or neither.79. Showthat thesumof twoevenfunctionsisevenandthesumof twooddfunctionsisodd.solution Even: (f +g)(x) = f (x) +g(x) even= f (x) +g(x) = (f +g)(x)Odd: (f +g)(x) = f (x) +g(x) odd= f (x) +g(x) = (f +g)(x)June 7, 2011 LTSV SSM Second Pass8 C HA P T E R 1 PRECALCULUS REVIEWSupposethat f (x) andg(x) arebothodd. Whichof thefollowingfunctionsareeven?Whichareodd?(a) f (x)g(x) (b) f (x)3(c) f (x) g(x) (d) f (x)g(x)81. Provethat theonlyfunctionwhosegraphissymmetricwithrespect toboththey-axisandtheoriginisthefunctionf (x) = 0.solution Supposef issymmetricwithrespecttothey-axis. Thenf (x) = f (x). If f isalsosymmetricwithrespecttotheorigin, thenf (x) = f (x). Thusf (x) = f (x) or 2f (x) = 0. Finally, f (x) = 0.Further Insights and ChallengesProvethetriangleinequalitybyaddingthetwoinequalities|a| a |a|, |b| b |b|83. Showthat afractionr = a/b inlowest termshasanite decimal expansionif andonlyifb = 2n5mfor somen, m 0.Hint: Observethat r hasanitedecimal expansionwhen10Nr isaninteger for someN 0(andhenceb divides10N).solution Supposer hasanitedecimal expansion. Thenthereexistsaninteger N 0suchthat 10Nr isaninteger,call it k. Thus, r = k/10N. Becausetheonlyprimefactorsof 10are2and5, it followsthat whenr iswritteninlowestterms, itsdenominator must beof theform2n5mfor someintegersn, m 0.Conversely, suppose r = a/b in lowest with b = 2n5mfor some integers n, m 0. Then r = ab= a2n5m or2n5mr = a. If m n, then2m5mr = a2mnor r = a2mn10m andthus r has anitedecimal expansion(less thanorequal tom terms, tobeprecise). Ontheother hand, if n > m, then2n5nr = a5nmor r = a5nm10n andonceagainr hasanitedecimal expansion.Let p = p1. . . ps beaninteger withdigitsp1, . . . , ps. Showthatp10s1 = 0.p1. . . psUsethistondthedecimal expansionof r = 211. Notethatr = 211 = 18102185. A functionf (x) issymmetricwithrespect tothevertical linex = a if f (a x) = f (a +x).(a) Drawthegraphof afunctionthat issymmetricwithrespect tox = 2.(b) Showthat if f (x) issymmetricwithrespect tox = a, theng(x) = f (x +a) iseven.solution(a) Therearemanypossibilities, oneof whichisx1125 4 3 2 1yy = |x 2|(b) Let g(x) = f (x +a). Theng(x) = f (x +a) = f (a x)= f (a +x) symmetrywithrespect tox = a= g(x)Thus, g(x) iseven.Formulateaconditionfor f (x) tobesymmetricwithrespect tothepoint (a, 0) onthex-axis.1.2 Linear and Quadratic FunctionsPreliminary Questions1. What istheslopeof theliney = 4x 9?solution Theslopeof theliney = 4x 9is4, givenbythecoefcient of x.2. Arethelinesy = 2x +1andy = 2x 4perpendicular?solution Theslopesof perpendicular linesarenegativereciprocalsof oneanother. Becausetheslopeof y = 2x +1is2andtheslopeof y = 2x 4is2, thesetwolinesarenot perpendicular.3. Whenisthelineax +by = c parallel tothey-axis?Tothex-axis?solution Thelineax +by = c will beparallel tothey-axiswhenb = 0andparallel tothex-axiswhena = 0.4. Supposey = 3x +2. What isy if x increasesby3?solution Becausey = 3x +2isalinear functionwithslope3, increasingx by3will leadtoy = 3(3) = 9.June 7, 2011 LTSV SSM Second PassS E C T I ON 1.2 Linear and Quadratic Functions 95. What istheminimumof f (x) = (x +3)24?solution Because(x +3)2 0, it followsthat(x +3)24 4. Thus, theminimumvalueof (x +3)24is4.6. What istheresult of completingthesquarefor f (x) = x2+1?solution Becausethereisnox terminx2+1, completingthesquareonthisexpressionleadsto(x 0)2+1.ExercisesIn Exercises 14, nd the slope, the y-intercept, and the x-intercept of the line with the given equation.1. y = 3x +12solution Becausetheequationof thelineis giveninslope-intercept form, theslopeis thecoefcient of x andthey-intercept istheconstant term: that is, m = 3andthey-intercept is12. Todeterminethex-intercept, substitutey = 0andthensolvefor x: 0= 3x +12or x = 4.y = 4x3. 4x +9y = 3solution Todeterminetheslopeandy-intercept, werst solvetheequationfor y toobtaintheslope-intercept form.This yields y = 49x + 13. Fromhere, weseethat theslopeis m = 49 and they-intercept is 13. To determinethex-intercept, substitutey = 0andsolvefor x: 4x = 3or x = 34.y 3= 12(x 6)In Exercises 58, nd the slope of the line.5. y = 3x +2solution m = 3y = 3(x 9) +27. 3x +4y = 12solution First solvetheequationfor y toobtaintheslope-intercept form. Thisyieldsy = 34x +3. Theslopeof thelineisthereforem = 34.3x +4y = 8In Exercises 920, nd the equation of the line with the given description.9. Slope3, y-intercept 8solution Usingtheslope-intercept formfor theequationof aline, wehavey = 3x +8.Slope2, y-intercept 311. Slope3, passesthrough(7, 9)solution Usingthepoint-slopeformfor theequationof aline, wehavey 9= 3(x 7) or y = 3x 12.Slope5, passesthrough(0, 0)13. Horizontal, passesthrough(0, 2)solution Ahorizontal linehasaslopeof 0. Usingthepoint-slopeformfor theequationof aline, wehavey (2) =0(x 0) or y = 2.Passesthrough(1, 4) and(2, 7)15. Parallel toy = 3x 4, passesthrough(1, 1)solution Becausetheequationy = 3x 4isinslope-intercept form, wecanreadilyidentifythat it hasaslopeof 3.Parallel lineshavethesameslope, sotheslopeof therequestedlineisalso3. Usingthepoint-slopeformfor theequationof aline, wehavey 1= 3(x 1) or y = 3x 2.Passesthrough(1, 4) and(12, 3)17. Perpendicular to3x +5y = 9, passesthrough(2, 3)solution Westart bysolvingtheequation3x +5y = 9for y toobtaintheslope-intercept formfor theequationof aline. Thisyieldsy = 35x + 95,fromwhichweidentifytheslopeas35. Perpendicular lineshaveslopesthat arenegativereciprocalsof oneanother, sotheslopeof thedesiredlineism = 53. Usingthepoint-slopeformfor theequationof aline, wehavey 3= 53(x 2)or y = 53x 13.Vertical, passesthrough(4, 9)19. Horizontal, passesthrough(8, 4)solution Ahorizontal linehasslope0. Usingthepointslopeformfortheequationof aline, wehavey 4= 0(x 8)or y = 4.Slope3, x-intercept 6June 7, 2011 LTSV SSM Second Pass10 C HA P T E R 1 PRECALCULUS REVIEW21. Findtheequationof theperpendicularbisectorof thesegmentjoining(1, 2) and(5, 4) (Figure11). Hint: ThemidpointQof thesegment joining(a, b) and(c, d) is_a +c2 ,b +d2_.Q(1, 2)(5, 4)PerpendicularbisectorxyFIGURE 11solution Theslopeof thesegment joining(1, 2) and(5, 4) ism = 4251 = 12andthemidpoint of thesegment (Figure11) ismidpoint =_1+52 ,2+42_ = (3, 3)Theperpendicular bisector hasslope1/m = 2andpassesthrough(3, 3), soitsequationis: y 3 = 2(x 3) ory = 2x +9.Intercept-Intercept Form Showthat if a, b = 0, thenthelinewithx-intercept x = a andy-intercept y = bhasequation(Figure12)xa+ yb= 123. Findanequationof thelinewithx-intercept x = 4andy-intercept y = 3.solution FromExercise22, x4+ y3 = 1or 3x +4y = 12.Findy suchthat (3, y) liesonthelineof slopem = 2through(1, 4).25. Determinewhether thereexistsaconstant c suchthat thelinex +cy = 1:(a) Hasslope4 (b) Passesthrough(3, 1)(c) Ishorizontal (d) Isverticalsolution(a) Rewritingtheequationof thelineinslope-intercept formgivesy = xc + 1c. Tohaveslope4requires1c = 4orc = 14.(b) Substitutingx = 3andy = 1intotheequationof thelinegives3+c = 1or c = 2.(c) From(a), weknowtheslopeof thelineis1c. Thereisnovaluefor c that will makethisslopeequal to0.(d) Withc = 0, theequationbecomesx = 1. Thisistheequationof avertical line.Assumethat thenumber N of concert ticketsthat canbesoldat apriceof P dollarsper ticket isalinear functionN(P) for 10 P 40. DetermineN(P) (calledthedemandfunction) if N(10) = 500andN(40) = 0. What isthedecreaseN inthenumber of ticketssoldif thepriceisincreasedbyP = 5dollars?27. Materialsexpandwhenheated. Consider ametal rodof lengthL0at temperatureT0. If thetemperatureischangedby anamount T , thentherodslengthchangesby L = L0T , where isthethermal expansioncoefcient. Forsteel, = 1.24105C1.(a) A steel rodhaslengthL0= 40cmat T0= 40C. Finditslengthat T = 90C.(b) Finditslengthat T = 50C if itslengthat T0= 100C is65cm.(c) ExpresslengthL asafunctionof T if L0= 65cmat T0= 100C.solution(a) WithT = 90C andT0= 40C, T = 50C. Therefore,L = L0T = (1.24105)(40)(50) = 0.0248 and L = L0+L = 40.0248cm.(b) WithT = 50C andT0= 100C, T = 50C. Therefore,L = L0T = (1.24105)(65)(50) = 0.0403 and L = L0+L = 64.9597cm.(c) L = L0+L = L0+L0T = L0(1+T ) = 65(1+(T 100))Dothepoints(0.5, 1), (1, 1.2), (2, 2) lieonaline?29. Findb suchthat (2, 1), (3, 2), and(b, 5) lieonaline.solution Theslopeof thelinedeterminedbythepoints(2, 1) and(3, 2) is2(1)32 = 3.Tolieonthesameline, theslopebetween(3, 2) and(b, 5) must alsobe3. Thus, werequire52b 3 = 3b 3 = 3,or b = 4.June 7, 2011 LTSV SSM Second PassS E C T I ON 1.2 Linear and Quadratic Functions 11Findanexpressionfor thevelocityv asalinear functionof t that matchesthefollowingdata.t (s) 0 2 4 6v (m/s) 39.2 58.6 78 97.431. TheperiodT of apendulumismeasuredfor pendulumsof several different lengthsL. Basedonthefollowingdata,doesT appear tobealinear functionof L?L (cm) 20 30 40 50T (s) 0.9 1.1 1.27 1.42solution Examinetheslopebetweenconsecutivedatapoints. Therst pair of datapointsyieldsaslopeof1.10.93020 = 0.02,whilethesecondpair of datapointsyieldsaslopeof1.271.14030 = 0.017,andthelast pair of datapointsyieldsaslopeof1.421.275040 = 0.015Becausethethreeslopesarenot equal, T doesnot appear tobealinear functionof L.Showthat f (x) islinear of slopem if andonlyiff (x +h) f (x) = mh (for all x andh)33. Findtherootsof thequadraticpolynomials:(a) 4x23x 1 (b) x22x 1solution(a) x = 394(4)(1)2(4)= 3258 = 1or 14(b) x = 24(4)(1)(1)2 = 282 = 12In Exercises 3441, complete the square and nd the minimum or maximum value of the quadratic function.y = x2+2x +535. y = x26x +9solution y = (x 3)2; therefore, theminimumvalueof thequadraticpolynomial is0, andthisoccursat x = 3.y = 9x2+x37. y = x2+6x +2solution y = x2+ 6x + 9 9+ 2 = (x + 3)2 7; therefore, theminimumvalueof thequadratic polynomial is7, andthisoccursat x = 3.y = 2x24x 739. y = 4x2+3x +8solution y = 4x2+3x +8= 4(x2 34x + 964) +8+ 916 = 4(x 38)2+ 13716; therefore, themaximumvalueof thequadraticpolynomial is13716, andthisoccursat x = 38.y = 3x2+12x 541. y = 4x 12x2solution y = 12(x2 x3) = 12(x2 x3 + 136) + 13 = 12(x 16)2+ 13; therefore, themaximumvalueof thequadraticpolynomial is13, andthisoccursat x = 16.Sketchthegraphof y = x26x +8byplottingtherootsandtheminimumpoint.43. Sketchthegraphof y = x2+4x +6byplottingtheminimumpoint, they-intercept, andoneother point.solution y = x2+ 4x + 4 4+ 6= (x + 2)2+ 2sotheminimumoccursat (2, 2). If x = 0, theny = 6andifx = 4, y = 6. Thisisthegraphof x2movedleft 2unitsandup2units.4 3 2 1246810yxIf thealleles A andB of thecystic brosis geneoccur inapopulationwithfrequencies p and1 p (wherepisafractionbetween0and1), thenthefrequencyof heterozygouscarriers(carrierswithbothalleles) is2p(1p).Whichvalueof p givesthelargest frequencyof heterozygouscarriers?June 7, 2011 LTSV SSM Second Pass12 C HA P T E R 1 PRECALCULUS REVIEW45. For whichvaluesof c doesf (x) = x2+cx +1haveadoubleroot? Noreal roots?solution A doubleroot occurswhenc24(1)(1) = 0or c2= 4. Thus, c = 2.Therearenoreal rootswhenc24(1)(1) < 0or c2 < 4. Thus, 2< c < 2.Let f (x) be a quadratic function and c a constant. Which of the following statements is correct? Explaingraphically.(a) Thereisauniquevalueof c suchthat y = f (x) c hasadoubleroot.(b) Thereisauniquevalueof c suchthat y = f (x c) hasadoubleroot.47. Provethat x + 1x 2for all x > 0. Hint: Consider (x1/2x1/2)2.solution Let x > 0. Then_x1/2x1/2_2= x 2+ 1x.Because(x1/2x1/2)2 0, it followsthatx 2+ 1x 0 or x + 1x 2.Let a, b > 0. Showthat thegeometric mean ab isnot larger thanthearithmetic mean (a + b)/2. Hint: Useavariationof thehint giveninExercise47.49. If objects of weights x and w1 are suspended fromthe balance in Figure 13(A), the cross-beamis horizontal ifbx = aw1. If thelengthsa andb areknown, wemayusethisequationtodetermineanunknownweightx byselectingw1suchthat thecross-beamishorizontal. If a andb arenot knownprecisely, wemight proceedasfollows. First balancexbyw1ontheleft asin(A). Thenswitchplacesandbalancex byw2ontheright asin(B). Theaverage x = 12(w1+w2)givesanestimatefor x. Showthat x isgreater thanor equal tothetrueweight x.w1(A)axb(B)w2xa bFIGURE 13solution First notebx = aw1andax = bw2. Thus, x = 12(w1+w2)= 12_bxa+ axb_= x2_ba+ ab_ x2(2) byExercise47= xFindnumbersx andy withsum10andproduct 24. Hint: Findaquadraticpolynomial satisedbyx.51. Findapair of numberswhosesumandproduct arebothequal to8.solution Let x andy benumberswhosesumandproduct arebothequal to8. Thenx +y = 8andxy = 8. Fromthesecondequation, y = 8x. Substitutingthisexpressionfor y intherst equationgivesx + 8x = 8or x28x +8= 0. Bythequadraticformula,x = 864322 = 422.If x = 4+22, theny = 84+22 = 84+22 422422 = 422.Ontheother hand, if x = 422, theny = 8422 = 8422 4+224+22 = 4+22.Thus, thetwonumbersare4+22and422.Showthat theparabolay = x2consists of all points P suchthat d1 = d2, whered1is thedistancefromP to_0, 14_andd2isthedistancefromP totheliney = 14 (Figure14).June 7, 2011 LTSV SSM Second PassS E C T I ON 1.3 The Basic Classes of Functions 13Further Insights and Challenges53. Showthat if f (x) andg(x) arelinear, thensoisf (x) +g(x). Isthesametrueof f (x)g(x)?solution If f (x) = mx +b andg(x) = nx +d, thenf (x) +g(x) = mx +b +nx +d = (m+n)x +(b +d),whichislinear. f (x)g(x) isnot generallylinear. Take, for example, f (x) = g(x) = x. Thenf (x)g(x) = x2.Showthat if f (x) andg(x) arelinear functionssuchthat f (0) = g(0) andf (1) = g(1), thenf (x) = g(x).55. Showthaty/x for thefunctionf (x) = x2over theinterval [x1, x2] isnotaconstant, butdependsontheinterval.Determinetheexact dependenceof y/x onx1andx2.solution For x2, yx= x22x21x2x1 = x2+x1.UseEq. (2) toderivethequadraticformulafor therootsof ax2+bx +c = 0.57. Let a, c = 0. Showthat therootsofax2+bx +c = 0 and cx2+bx +a = 0arereciprocalsof eachother.solution Let r1 andr2 betheroots of ax2+ bx + c andr3 andr4 betheroots of cx2+ bx + a. Without loss ofgenerality, letr1= b +_b24ac2a 1r1 = 2ab +_b24ac b _b24acb _b24ac= 2a(b _b24ac)b2b2+4ac= b _b24ac2c= r4.Similarly, youcanshow 1r2 = r3.Show, bycompletingthesquare, that theparabolay = ax2+bx +ciscongruent toy = ax2byavertical andhorizontal translation.59. ProveVites Formulas: Thequadratic polynomial with and asrootsisx2+ bx + c, whereb = andc = .solution If aquadraticpolynomial hasroots and, thenthepolynomial is(x )(x ) = x2x x + = x2+( )x +.Thus, b = andc = .1.3 The Basic Classes of FunctionsPreliminary Questions1. Giveanexampleof arational function.solution Oneexampleis 3x227x3+x 1.2. Is|x| apolynomial function?What about |x2+1|?solution |x| isnot apolynomial; however, becausex2+1> 0for all x, it followsthat |x2+1| = x2+1, whichisapolynomial.3. Whatisunusual aboutthedomainof thecompositefunctionf g forthefunctionsf (x) = x1/2andg(x) = 1|x|?solution Recall that (f g)(x) = f (g(x)). Now, for any real number x, g(x) = 1 |x| 1< 0. Becausewecannot takethesquareroot of anegativenumber, it follows that f (g(x)) is not denedfor any real number. Inotherwords, thedomainof f (g(x)) istheemptyset.4. Isf (x) = _12_xincreasingor decreasing?solution Thefunctionf (x) = (12)xis anexponential functionwithbaseb = 12 < 1. Therefore, f is adecreasingfunction.5. Giveanexampleof atranscendental function.solution Onepossibilityisf (x) = exsinx.June 7, 2011 LTSV SSM Second Pass14 C HA P T E R 1 PRECALCULUS REVIEWExercisesIn Exercises 112, determine the domain of the function.1. f (x) = x1/4solution x 0g(t ) = t2/33. f (x) = x3+3x 4solution All realsh(z) = z3+z35. g(t ) = 1t +2solution t = 2f (x) = 1x2+47. G(u) = 1u24solution u = 2f (x) =xx299. f (x) = x4+(x 1)3solution x = 0, 1F(s) = sin_ ss +1_11. g(y) = 10y+y1solution y > 0f (x) = x +x1(x 3)(x +4)In Exercises 1324, identify each of the following functions as polynomial, rational, algebraic, or transcendental.13. f (x) = 4x3+9x28solution Polynomialf (x) = x415. f (x) = xsolution Algebraicf (x) =_1x2 17. f (x) = x2x +sinxsolution Transcendentalf (x) = 2x19. f (x) = 2x3+3x97x2solution Rationalf (x) = 3x 9x1/297x221. f (x) = sin(x2)solution Transcendentalf (x) = xx +123. f (x) = x2+3x1solution Rationalf (x) = sin(3x)25. Isf (x) = 2x2atranscendental function?solution Yes.Showthat f (x) = x2+3x1andg(x) = 3x39x +x2arerational functionsthat is, quotientsof polyno-mials.In Exercises 2734, calculate the composite functions f g and g f , and determine their domains.27. f (x) = x, g(x) = x +1solution f (g(x)) = x +1; D: x 1, g(f (x)) = x +1; D: x 0f (x) = 1x, g(x) = x429. f (x) = 2x, g(x) = x2solution f (g(x)) = 2x2; D: R, g(f (x)) = (2x)2= 22x; D: Rf (x) = |x|, g() = sin31. f () = cos, g(x) = x3+x2solution f (g(x)) = cos(x3+x2); D: R, g(f ()) = cos3 +cos2; D: Rf (x) = 1x2+1, g(x) = x233. f (t ) = 1t, g(t ) = t2solution f (g(t )) = 1_t2; D: Not validfor anyt , g(f (t )) = _ 1t_2= 1t ; D: t > 0f (t ) = t , g(t ) = 1t3June 7, 2011 LTSV SSM Second PassS E C T I ON 1.3 The Basic Classes of Functions 1535. Thepopulation(inmillions) of acountryasafunctionof timet (years) isP(t ) = 30.20.1t. Showthatthepopulationdoublesevery 10years. Showmoregenerally that for any positiveconstantsa andk, thefunctiong(t ) = a2ktdoublesafter 1/k years.solution Let P(t ) = 30 20.1t. ThenP(t +10) = 30 20.1(t +10)= 30 20.1t +1= 2(30 20.1t) = 2P(t ).Hence, thepopulationdoublesinsizeevery10years. Inthemoregeneral case, let g(t ) = a2kt. Theng_t + 1k_ = a2k(t +1/k)= a2kt +1= 2a2kt= 2g(t ).Hence, thefunctiong doublesafter 1/k years.Findall valuesof c suchthat f (x) = x +1x2+2cx +4hasdomainR. Further Insights and ChallengesIn Exercises 3743, we dene the rst differencef of a function f (x) by f (x) = f (x +1) f (x).37. Showthat if f (x) = x2, thenf (x) = 2x +1. Calculatef for f (x) = x andf (x) = x3.solution f (x) = x2: f (x) = f (x +1) f (x) = (x +1)2x2= 2x +1f (x) = x: f (x) = x +1x = 1f (x) = x3: f (x) = (x +1)3x3= 3x2+3x +1Showthat (10x) = 9 10xand, moregenerally, that (bx) = (b 1)bx.39. Showthat for anytwofunctionsf andg, (f +g) = f +g and(cf ) = c(f ), wherec isanyconstant.solution (f +g) = (f (x +1) +g(x +1)) (f (x) g(x))= (f (x +1) f (x)) +(g(x +1) g(x)) = f (x) +g(x)(cf ) = cf (x +1) cf (x) = c(f (x +1) f (x)) = cf (x).SupposewecanndafunctionP(x) suchthat P = (x + 1)kandP(0) = 0. Provethat P(1) = 1k, P(2) =1k+2k, and, moregenerally, for everywholenumber n,P(n) = 1k+2k+ +nk41. First showthatP(x) = x(x +1)2satisesP = (x +1). ThenapplyExercise40toconcludethat1+2+3+ +n = n(n +1)2solution Let P(x) = x(x +1)/2. ThenP(x) = P(x +1) P(x) = (x +1)(x +2)2 x(x +1)2 = (x +1)(x +2x)2 = x +1.Also, notethat P(0) = 0. Thus, byExercise40, withk = 1, it followsthatP(n) = n(n +1)2 = 1+2+3+ +n.Calculate(x3), (x2), and(x).Thenndapolynomial P(x) of degree3suchthatP = (x +1)2andP(0) = 0.Concludethat P(n) = 12+22+ +n2.43. ThisexercisecombinedwithExercise40showsthatforall wholenumbersk, thereexistsapolynomial P(x) satisfyingEq. (1). ThesolutionrequirestheBinomial Theoremandproof byinduction(seeAppendixC).(a) Showthat (xk+1) = (k +1) xk+ , wherethedotsindicatetermsinvolvingsmaller powersof x.(b) Showbyinductionthat thereexistsapolynomial of degreek +1withleadingcoefcient 1/(k +1):P(x) = 1k +1xk+1+ suchthat P = (x +1)kandP(0) = 0.solution(a) BytheBinomial Theorem:(xn+1) = (x +1)n+1xn+1=_xn+1+_ n +11_xn+_ n +12_xn1+ +1_xn+1=_ n +11_xn+_ n +12_xn1+ +1Thus,(xn+1) = (n +1) xn+ wherethedotsindicatetermsinvolvingsmaller powersof x.June 7, 2011 LTSV SSM Second Pass16 C HA P T E R 1 PRECALCULUS REVIEW(b) For k = 0, notethat P(x) = x satisesP = (x +1)0= 1andP(0) = 0.NowsupposethepolynomialP(x) = 1kxk+pk1xk1+ +p1xwhichclearlysatisesP(0) = 0alsosatisesP = (x +1)k1. WetrytoprovetheexistenceofQ(x) = 1k +1xk+1+qkxk+ +q1xsuchthat Q = (x +1)k. Observethat Q(0) = 0.If Q = (x +1)kandP = (x +1)k1, thenQ = (x +1)k= (x +1)P = xP(x) +PBythelinearityof (Exercise39), wendQP = xP or (QP) = xP. Bydenition,QP = 1k +1xk+1+_qk 1k_xk+ +(q1p1)x,so, bythelinearityof ,(QP) = 1k +1(xk+1) +_qk 1k_(xk) + +(q1p1) = x(x +1)k1Bypart (a),(xk+1) = (k +1)xk+Lk1,k1xk1+. . . +Lk1,1x +1(xk) = kxk1+Lk2,k2xk2+. . . +Lk2,1x +1...(x2) = 2x +1wheretheLi,j arereal numbersfor eachi, j.Toconstruct Q, wehavetogrouplikepowersof x onbothsidesof Eq. (43b). Thisyieldsthesystemof equations1k +1_(k +1)xk_ = xk1k +1Lk1,k1xk1+_qk 1k_kxk1= (k 1)xk1...1k +1+_qk 1k_+(qk1pk1) + +(q1p1) = 0.Therst equationis identically true, andthesecondequationcanbesolvedimmediately for qk. Substitutingthevalueof qk intothethirdequationof thesystem, wecanthensolvefor qk1. Wecontinuethisprocessuntil wesubstitutethevaluesof qk, qk1, . . . q2intothelast equation, andthensolvefor q1.1.4 Trigonometric FunctionsPreliminary Questions1. Howisit possiblefor twodifferent rotationstodenethesameangle?solution Workingfromthesameinitial radius, two rotations that differ by awholenumber of full revolutions willhavethesameendingradius; consequently, thetworotationswill denethesameangleeventhoughthemeasuresof therotationswill bedifferent.2. Givetwodifferent positiverotationsthat denetheangle/4.solution Theangle/4is denedby any rotationof theform4 + 2k wherek is aninteger. Thus, two differentpositiverotationsthat denetheangle/4are4 +2(1) = 94 and 4 +2(5) = 414 .June 7, 2011 LTSV SSM Second PassS E C T I ON 1.4 Trigonometric Functions 173. Giveanegativerotationthat denestheangle/3.solution Theangle/3isdenedbyanyrotationof theform3 +2k wherek isaninteger. Thus, anegativerotationthat denestheangle/3is3 +2(1) = 53 .4. Thedenitionof cos usingright trianglesapplieswhen(choosethecorrect answer):(a) 0< 2, werepeatedlysubtract 2 until wearriveat aradianmeasurethat isbetween0and2. After onesubtraction, wehave13/42 = 5/4. Because0< 5/4< 2, 5/4istheanglemeasurebetween0and2 that isequivalent to13/4.Describe = /6byanangleof negativeradianmeasure.3. Convert fromradianstodegrees:(a) 1 (b) 3 (c) 512 (d) 34solution(a) 1_180_ = 180 57.3 (b) 3_180_ = 60(c) 512_180_ = 75 23.87 (d) 34_180_ = 135Convert fromdegreestoradians:(a) 1 (b) 30 (c) 25 (d) 1205. Findthelengthsof thearcssubtendedbytheangles and radiansinFigure20.4q = 0.9f = 2FIGURE 20 Circleof radius4.solution s = r = 4(.9) = 3.6; s = r = 4(2) = 8Calculatethevaluesof thesixstandardtrigonometricfunctionsfor theangle inFigure21.7. Fill intheremainingvaluesof (cos, sin) for thepointsinFigure22.p20 (0, 0) p5p67p6 11p63p45p4 7p4 4p3 5p33p22p3( , )p6 23 12( , )23 12p3( , )p4 22 22FIGURE 22June 7, 2011 LTSV SSM Second Pass18 C HA P T E R 1 PRECALCULUS REVIEWsolution 2 23 34 56 76(cos, sin) (0, 1)_12 ,32 _ _22 ,22 _ _32 , 12_ (1, 0)_32 , 12 _ 54 43 32 53 74 116(cos, sin)_22 , 22 _ _12 , 32 _ (0, 1)_12, 32 _ _22 , 22 _ _32 , 12 _Findthevaluesof thesixstandardtrigonometricfunctionsat = 11/6.In Exercises 914, use Figure 22 to nd all angles between 0and 2 satisfying the given condition.9. cos = 12solution = 3, 53tan = 111. tan = 1solution = 34 , 74csc = 213. sinx =32solution x = 3, 23sect = 215. Fill inthefollowingtableof values:6 4 3 2 23 34 56tansecsolution6 4 3 2 23 34 56tan13 1 3 und 3 1 13sec23 2 2 und 2 2 23Completethefollowingtableof signs: sin cos tan cot sec csc0< 1, thehorizontal liney = c andthegraphof y = sinxnever intersect. If c = +1, theny = c andy = sinx intersect at thepeak of thesinecurve; that is, they intersect atx = 2. Ontheother hand, if c = 1, theny = c andy = sinx intersect at thebottomof thesinecurve; that is, theyintersect at x = 32 . Finally, if |c| < 1, thegraphsof y = c andy = sinx intersect twice.Howmanypointslieontheintersectionof thehorizontal liney = c andthegraphof y = tanx for 0 x < 2?In Exercises 4144, solve for 0 < 2 (see Example 4).41. sin2 +sin3 = 0solution sin = sin when = + 2k or = + + 2k. Substituting = 2 and = 3, wehaveeither2 = 3 +2k or2 = +3 +2k. Solvingeachof theseequationsfor yields = 25k or = 2k.Thesolutionsontheinterval 0 < 2 arethen = 0,25 ,45 , ,65 ,85 .sin = sin2June 7, 2011 LTSV SSM Second Pass22 C HA P T E R 1 PRECALCULUS REVIEW43. cos4 +cos2 = 0solution cos = cos when + = +2k or = + +2k. Substituting = 4 and = 2, wehaveeither 6 = +2k or 4 = 2 + +2k. Solvingeachof theseequationsfor yields = 6 + 3k or = 2 +k.Thesolutionsontheinterval 0 < 2 arethen = 6,2,56 ,76 ,32 ,116 .sin = cos2In Exercises 4554, derive the identity using the identities listed in this section.45. cos2 = 2cos2 1solution Startingfromthedoubleangleformulafor cosine, cos2 = 12(1+cos2), wesolvefor cos2. Thisgives2cos2 = 1+cos2 andthencos2 = 2cos2 1.cos2 2 = 1+cos247. sin2 =_1cos2solution Substitutex = /2intothedoubleangleformulafor sine, sin2x = 12(1cos2x) toobtainsin2_2_ =1cos2 . Takingthesquareroot of bothsidesyieldssin_2_ =_1cos2 .sin( +) = sin49. cos( +) = cossolution Fromtheadditionformulafor thecosinefunction, wehavecos( +) = cos cos sin sin = cos(1) = costanx = cot_2 x_ 51. tan( ) = tansolution UsingExercises48and49,tan( ) = sin( )cos( )= sin( +())cos( +())= sin()cos()= sincos= tan.Thesecondtolast equalityoccursbecausesinx isanoddfunctionandcosx isanevenfunction.tan2x = 2tanx1tan2x53. tanx = sin2x1+cos2xsolution Usingtheadditionformulafor thesinefunction, wendsin2x = sin(x +x) = sinx cosx +cosx sinx = 2sinx cosx.ByExercise45, weknowthat cos2x = 2cos2x 1. Therefore,sin2x1+cos2x= 2sinx cosx1+2cos2x 1 = 2sinx cosx2cos2x= sinxcosx= tanx.sin2x cos2x = 1cos4x855. UseExercises48and49toshowthat tan andcot areperiodicwithperiod.solution ByExercises48and49,tan( +) = sin( +)cos( +)= sincos= tan,andcot( +) = cos( +)sin( +)= cossin= cot.Thus, bothtan andcot areperiodicwithperiod.Usetheidentityof Exercise45toshowthat cos8 isequal to_12+24 .57. UsetheLawof CosinestondthedistancefromP toQinFigure26.PQ8107/9FIGURE 26solution BytheLawof Cosines, thedistancefromP toQis_102+822(10)(8) cos79 = 16.928.June 7, 2011 LTSV SSM Second PassS E C T I ON 1.5 Technology: Calculators and Computers 23Further Insights and ChallengesUseFigure27toderivetheLawof CosinesfromthePythagoreanTheorem.59. Usetheadditionformulatoprovecos3 = 4cos3 3cossolutioncos3 = cos(2 +) = cos2 cos sin2 sin = (2cos2 1) cos (2sin cos) sin= cos(2cos2 12sin2) = cos(2cos2 12(1cos2))= cos(2cos2 12+2cos2) = 4cos3 3cosUsetheadditionformulasfor sineandcosinetoprovetan(a +b) = tana +tanb1tana tanbcot(a b) = cota cotb +1cotb cota61. Let betheanglebetweentheliney = mx +b andthex-axis[Figure28(A)]. Provethat m = tan.y = mx + bqxrs(A)yqx(B)y L2L1FIGURE 28solution UsingthedistanceslabeledinFigure28(A), weseethat theslopeof thelineisgivenbytheratior/s. Thetangent of theangle isgivenbythesameratio. Therefore, m = tan.LetL1andL2bethelinesof slopem1andm2[Figure28(B)]. Showthattheangle betweenL1andL2satisescot = m2m1+1m2m1 .63. Perpendicular Lines UseExercise62toprovethat twolineswithnonzeroslopesm1andm2areperpendicular ifandonlyif m2= 1/m1.solution If linesareperpendicular, thentheanglebetweenthemis = /2cot(/2) = 1+m1m2m1m20= 1+m1m2m1m2 m1m2= 1 m1= 1m2Applythedouble-angleformulatoprove:(a) cos8 = 12_2+2(b) cos 16 = 12_2+_2+2Guessthevaluesof cos 32andof cos 2n for all n.1.5 Technology: Calculators and ComputersPreliminary Questions1. Isthereadenitewayof choosingtheoptimal viewingrectangle, or isit best toexperiment until youndaviewingrectangleappropriatetotheproblemat hand?solution It isbest toexperiment withthewindowsizeuntil oneisfoundthat isappropriatefor theproblemat hand.2. Describethecalculator screenproducedwhenthefunctiony = 3+x2isplottedwithviewingrectangle:(a) [1, 1] [0, 2] (b) [0, 1] [0, 4]solution(a) Usingtheviewingrectangle[1, 1] by[0, 2], thescreenwill displaynothingastheminimumvalueof y = 3+x2isy = 3.(b) Usingtheviewingrectangle[0, 1] by [0, 4], thescreenwill display theportionof theparabolabetweenthepoints(0, 3) and(1, 4).3. AccordingtotheevidenceinExample4, it appearsthat f (n) = (1+1/n)nnever takesonavaluegreater than3forn > 0. Doesthisevidenceprove that f (n) 3for n > 0?solution No, thisevidencedoesnot constituteaproof that f (n) 3for n 0.4. Howcanagraphingcalculator beusedtondtheminimumvalueof afunction?solution Experiment with theviewing window to zoomin on thelowest point on thegraph of thefunction. They-coordinateof thelowest point onthegraphistheminimumvalueof thefunction.June 7, 2011 LTSV SSM Second Pass24 C HA P T E R 1 PRECALCULUS REVIEWExercisesThe exercises in this section should be done using a graphing calculator or computer algebra system.1. Plot f (x) = 2x4+3x314x29x +18intheappropriateviewingrectanglesanddetermineitsroots.solution Usingaviewingrectangleof [4, 3] by[20, 20], weobtaintheplot below.10202010yx4 2 3 1 1 2 3Now, therootsof f (x) arethex-interceptsof thegraphof y = f (x). Fromtheplot, wecanidentifythex-interceptsas3, 1.5, 1, and2. Therootsof f (x) arethereforex = 3, x = 1.5, x = 1, andx = 2.Howmanysolutionsdoesx34x +8= 0have?3. Howmanypositive solutionsdoesx312x +8= 0have?solution Thegraphof y = x3 12x + 8shownbelowhastwox-interceptstotheright of theorigin; thereforetheequationx312x +8= 0hastwopositivesolutions.204060604020yx4 2 4 2Doescosx +x = 0haveasolution?A positivesolution?5. Findall thesolutionsof sinx = x for x > 0.solution Solutionstotheequationsinx = x correspondtopointsof intersectionbetweenthegraphsof y = sinxandy = x. Thetwographsareshownbelow; theonlypointof intersectionisatx = 0. Therefore, therearenosolutionsof sinx = x for x > 0.x125 4 3 2 1y1Howmanysolutionsdoescosx = x2have?7. Let f (x) = (x 100)2+ 1000. What will thedisplay showif yougraphf (x) intheviewingrectangle[10, 10]by[10, 10]? Findanappropriateviewingrectangle.solution Because(x 100)2 0for all x, it followsthat f (x) = (x 100)2+1000 1000for all x. Thus, usingaviewingrectangleof [10, 10] by [10, 10] will display nothing. Theminimumvalueof thefunctionoccurs whenx = 100, soanappropriateviewingrectanglewouldbe[50, 150] by[1000, 2000].Plot f (x) = 8x +18x 4inanappropriateviewingrectangle. What arethevertical andhorizontal asymptotes?9. Plotthegraphof f (x) = x/(4x) inaviewingrectanglethatclearlydisplaysthevertical andhorizontal asymptotes.solution Fromthegraphof y = x4xshownbelow, weseethat thevertical asymptoteisx = 4andthehorizontalasymptoteisy = 1.22yx8 44 8 12 16Illustratelocal linearityfor f (x) = x2byzoominginonthegraphat x = 0.5(seeExample6).11. Plotf (x) = cos(x2) sinx for 0 x 2. Thenillustratelocal linearityatx = 3.8bychoosingappropriateviewingrectangles.solution The following three graphs display f (x) = cos(x2) sinx over the intervals [0, 2], [3.5, 4.1] and[3.75, 3.85]. Thenal graphlookslikeastraight line.June 7, 2011 LTSV SSM Second PassS E C T I ON 1.5 Technology: Calculators and Computers 25x111 2 3 4 5 6yx113.5 3.6 3.7 3.8 3.9 4yx0.20.40.23.76 3.8 3.78 3.82 3.84yIf P0dollarsaredepositedinabankaccountpaying5%interestcompoundedmonthly, thentheaccounthasvalueP0_1+ 0.0512 _Nafter N months. Find, tothenearestinteger N, thenumber of monthsafter whichtheaccountvaluedoubles.In Exercises 1318, investigate the behavior of the function as n or x grows large by making a table of function valuesand plotting a graph (see Example 4). Describe the behavior in words.13. f (n) = n1/nsolution Thetableandgraphsbelowsuggest that asn getslarge, n1/napproaches1.n n1/n10 1.258925412102 1.047128548103 1.006931669104 1.000921458105 1.000115136106 1.000013816xy10 2 4 6 8 10 xy10 200 400 600 800 1000f (n) = 4n +16n 515. f (n) =_1+ 1n_n2solution Thetableandgraphsbelowsuggest that asn getslarge, f (n) tendstoward.n_1+ 1n_n210 13780.61234102 1.6358287111043103 1.19530660310434104 5.341783312104342105 1.7023330541043429106 1.83973874910434294xy10,0000 2 4 6 8 10 xy1 10430 20 40 60 80 100f (x) =_x +6x 4_x17. f (x) =_x tan1x_xsolution Thetableandgraphsbelowsuggest that asx getslarge, f (x) approaches1.x_x tan1x_x10 1.033975759102 1.003338973103 1.000333389104 1.000033334105 1.000003333106 1.000000333June 7, 2011 LTSV SSM Second Pass26 C HA P T E R 1 PRECALCULUS REVIEWxy11.11.21.31.41.55 10 15 20 x20 40 60 80 100y11.11.21.31.41.5f (x) =_x tan1x_x219. Thegraphof f () = Acos + B sin isasinusoidal wavefor any constantsA andB. Conrmthisfor (A, B) =(1, 1), (1, 2), and(3, 4) byplottingf ().solution Thegraphsof f () = cos +sin, f () = cos +2sin andf () = 3cos +4sin areshownbelow.yx2 2(A, B) = (1, 1)4 6 811yx2 2(A, B) = (1, 2)4 6 82121yx2 2(A, B) = (3, 4)4 6 84242Find themaximumvalueof f () for thegraphs produced in Exercise19. Can you guess theformulafor themaximumvalueintermsof A andB?21. Findtheintervalsonwhichf (x) = x(x +2)(x 3) ispositivebyplottingagraph.solution Thefunctionf (x) = x(x + 2)(x 3) is positivewhenthegraphof y = x(x + 2)(x 3) lies abovethex-axis. Thegraphof y = x(x + 2)(x 3) isshownbelow. Clearly, thegraphliesabovethex-axisandthefunctionispositivefor x (2, 0) (3, ).204020yx4 2 2 4Findtheset of solutionstotheinequality(x24)(x21) < 0byplottingagraph.Further Insights and Challenges23. Let f1(x) = x and dene a sequence of functions by fn+1(x) = 12(fn(x) + x/fn(x)). For example,f2(x) = 12(x +1). Useacomputer algebrasystemtocomputefn(x) for n = 3, 4, 5andplotfn(x) together withx forx 0. What doyounotice?solution Withf1(x) = x andf2(x) = 12(x +1), wecalculatef3(x) = 12_12(x +1) + x12(x +1)_ = x2+6x +14(x +1)f4(x) = 12x2+6x +14(x +1)+ xx2+6x+14(x+1) = x4+28x3+70x2+28x +18(1+x)(1+6x +x2)andf5(x) = 1+120x +1820x2+8008x3+12870x4+8008x5+1820x6+120x7+x816(1+x)(1+6x +x2)(1+28x +70x2+28x3+x4).A plot of f1(x), f2(x), f3(x), f4(x), f5(x) andx isshownbelow, withthegraphofx shownasadashedcurve. Itseemsasif thefn areasymptotictox.yx481240 20 60 80 100Set P0(x) = 1 and P1(x) = x. The Chebyshev polynomials (useful in approximation theory) are denedinductivelybytheformulaPn+1(x) = 2xPn(x) Pn1(x).(a) Showthat P2(x) = 2x21.(b) ComputePn(x) for 3 n 6usingacomputer algebrasystemor byhand andplot Pn(x) over [1 1]June 7, 2011 LTSV SSM Second PassChapter Review Exercises 27CHAPTER REVIEW EXERCISES1. Express(4, 10) asaset {x : |x a| < c} for suitablea andc.solution Thecenter of theinterval (4, 10) is4+102 = 7andtheradiusis1042 = 3. Therefore, theinterval (4, 10) isequivalent totheset {x : |x 7| < 3}.Expressasaninterval:(a) {x : |x 5| < 4} (b) {x : |5x +3| 2}3. Express{x : 2 |x 1| 6} asaunionof twointervals.solution Theset {x : 2 |x 1| 6} consistsof thosenumbersthat areat least 2but at most 6unitsfrom1. Thenumbers larger than1that satisfy theseconditions are3 x 7, whilethenumbers smaller than1that satisfy theseconditionsare5 x 1. Therefore{x : 2 |x 1| 6} = [5, 1] [3, 7].Giveanexampleof numbersx, y suchthat |x| +|y| = x y.5. Describethepairsof numbersx, y suchthat |x +y| = x y.solution First consider thecasewhenx +y 0. Then|x +y| = x +y andweobtaintheequationx +y = x y.Thesolutionof thisequationisy = 0. Thus, thepairs(x, 0) withx 0satisfy|x +y| = x y. Next, consider thecasewhenx +y < 0. Then|x +y| = (x +y) = x y andweobtaintheequationx y = x y. Thesolutionof thisequationisx = 0. Thus, thepairs(0, y) withy < 0alsosatisfy|x +y| = x y.Sketchthegraphof y = f (x +2) 1, wheref (x) = x2for 2 x 2.In Exercises 710, let f (x) be the function shown in Figure 1.1 2 3 41203xyFIGURE 17. Sketchthegraphsof y = f (x) +2andy = f (x +2).solution Thegraphof y = f (x) + 2isobtainedby shiftingthegraphof y = f (x) up2units(seethegraphbelowat theleft). Thegraphof y = f (x + 2) isobtainedby shiftingthegraphof y = f (x) totheleft 2units(seethegraphbelowat theright).xyyx123451 2 3 4f (x) + 21 2 xyyx123451 2 3 4f (x + 2)1 2Sketchthegraphsof y = 12f (x) andy = f_12x_.9. Continuethegraphof f (x) totheinterval [4, 4] asanevenfunction.solution Tocontinuethegraphof f (x) totheinterval [4, 4] asanevenfunction, reect thegraphof f (x) acrossthey-axis(seethegraphbelow).1 4 2 3 x1 2 3 4y123Continuethegraphof f (x) totheinterval [4, 4] asanoddfunction.In Exercises 1114, nd the domain and range of the function.11. f (x) = x +1solution Thedomainof thefunctionf (x) = x +1is{x : x 1} andtherangeis{y : y 0}.f (x) = 4x4+113. f (x) = 23xsolution Thedomainof thefunctionf (x) = 23xis{x : x = 3} andtherangeis{y : y = 0}.f (x) =_x2x +5June 7, 2011 LTSV SSM Second Pass28 C HA P T E R 1 PRECALCULUS REVIEW15. Determinewhether thefunctionisincreasing, decreasing, or neither:(a) f (x) = 3x(b) f (x) = 1x2+1(c) g(t ) = t2+t (d) g(t ) = t3+tsolution(a) Thefunctionf (x) = 3xcanberewrittenasf (x) = (13)x. Thisisanexponential functionwithabaselessthan1;therefore, thisisadecreasingfunction.(b) Fromthegraphof y = 1/(x2+1) shownbelow, weseethatthisfunctionisneither increasingnor decreasingfor allx (thoughit isincreasingfor x < 0anddecreasingfor x > 0).x3 2 1 1 2 3y0.20.40.60.81(c) Thegraphof y = t2+t isanupwardopeningparabola; therefore, thisfunctionisneither increasingnor decreasingfor all t . By completingthesquarewendy = (t + 12)2 14. Thevertex of this parabolais thenat t = 12, so thefunctionisdecreasingfor t < 12 andincreasingfor t > 12.(d) Fromthegraphof y = t3+t shownbelow, weseethat thisisanincreasingfunction.2020yx1 1 2 3 2 3Determinewhether thefunctioniseven, odd, or neither:(a) f (x) = x43x2(b) g(x) = sin(x +1)(c) f (x) = 2x2In Exercises 1722, nd the equation of the line.17. Linepassingthrough(1, 4) and(2, 6)solution Theslopeof thelinepassingthrough(1, 4) and(2, 6) ism = 642(1)= 23.Theequationof thelinepassingthrough(1, 4) and(2, 6) isthereforey 4= 23(x +1) or 2x 3y = 14.Linepassingthrough(1, 4) and(1, 6)19. Lineof slope6through(9, 1)solution Usingthepoint-slopeformfor theequationof aline, theequationof thelineof slope6andpassingthrough(9, 1) isy 1= 6(x 9) or 6x y = 53.Lineof slope32 through(4, 12)21. Linethrough(2, 3) parallel toy = 4xsolution Theequation y = 4 x is in slope-intercept form; it follows that theslopeof this lineis 1. Any lineparallel toy = 4 x will havethesameslope, sowearelookingfor theequationof thelineof slope1andpassingthrough(2, 3). Theequationof thislineisy 3= (x 2) or x +y = 5.Horizontal linethrough(3, 5)23. Doesthefollowingtableof marketdatasuggestalinear relationshipbetweenpriceandnumber of homessoldduringaone-year period? Explain.Price(thousandsof $) 180 195 220 240No. of homessold 127 118 103 91solution Examinetheslopebetweenconsecutivedatapoints. Therst pair of datapointsyieldsaslopeof118127195180 = 915 = 35,whilethesecondpair of datapointsyieldsaslopeof103118220195 = 1525 = 35June 7, 2011 LTSV SSM Second PassChapter Review Exercises 29andthelast pair of datapointsyieldsaslopeof91103240220 = 1220 = 35.Becauseall threeslopes areequal, thedatadoes suggest alinear relationshipbetweenpriceandthenumber of homessold.Doesthefollowingtableof revenuedatafor acomputer manufacturer suggest alinear relationbetweenrevenueandtime? Explain.Year 2001 2005 2007 2010Revenue(billionsof $) 13 18 15 1125. Findtherootsof f (x) = x44x2andsketchitsgraph. Onwhichintervalsisf (x) decreasing?solution Theroots of f (x) = x4 4x2areobtainedby solvingtheequationx4 4x2 = x2(x 2)(x + 2) = 0,whichyieldsx = 2, x = 0andx = 2. Thegraphof y = f (x) isshownbelow. Fromthisgraphweseethat f (x) isdecreasingfor x lessthanapproximately1.4andfor x between0andapproximately1.4.1020yx1 12 3 2 3Let h(z) = 2z2+12z +3. Completethesquareandndtheminimumvalueof h(z).27. Let f (x) bethesquareof thedistancefromthepoint (2, 1) toapoint (x, 3x +2) ontheliney = 3x +2. Showthatf (x) isaquadraticfunction, andnditsminimumvaluebycompletingthesquare.solution Letf (x) denotethesquareof thedistancefromthepoint(2, 1) toapoint(x, 3x +2) ontheliney = 3x +2.Thenf (x) = (x 2)2+(3x +21)2= x24x +4+9x2+6x +1= 10x2+2x +5,whichisaquadraticfunction. Completingthesquare, wendf (x) = 10_x2+ 15x + 1100_+5 110 = 10_x + 110_2+ 4910.Because(x + 110)2 0for all x, it followsthat f (x) 4910 for all x. Hence, theminimumvalueof f (x) is4910.Provethat x2+3x +3 0for all x.In Exercises 2934, sketch the graph by hand.29. y = t4solutionx1 0.5 1 0.5y0.20.40.60.81y = t5 31. y = sin2solution yx5 5 100.510.51y = 10x33. y = x1/3solutionx1 2 3 4 1 2 3 4y1212y = 1x2June 7, 2011 LTSV SSM Second Pass30 C HA P T E R 1 PRECALCULUS REVIEW35. Showthatthegraphof y = f_13x b_isobtainedbyshiftingthegraphof y = f_13x_totheright3b units. Usethisobservationtosketchthegraphof y = 13x 4.solution Let g(x) = f (13x). Theng(x 3b) = f_13(x 3b)_ = f_13x b_.Thus, thegraphof y = f (13x b) isobtainedbyshiftingthegraphof y = f (13x) totheright 3b units.Thegraphof y = |13x 4| isthegraphof y = |13x| shiftedright 12units(seethegraphbelow).yx12340 5 10 15 20Leth(x) = cosx andg(x) = x1. Computethecompositefunctionsh(g(x)) andg(h(x)), andndtheirdomains.37. Findfunctionsf andg suchthat thefunctionf (g(t )) = (12t +9)4solution Onepossiblechoiceisf (t ) = t4andg(t ) = 12t +9. Thenf (g(t )) = f (12t +9) = (12t +9)4asdesired.Sketchthepoints ontheunit circlecorrespondingto thefollowingthreeangles, andndthevalues of thesixstandardtrigonometricfunctionsat eachangle:(a) 23 (b) 74 (c) 7639. What istheperiodof thefunctiong() = sin2 +sin2?solution Thefunctionsin2 hasaperiodof , andthefunctionsin(/2) hasaperiodof 4. Because4 isamultipleof , theperiodof thefunctiong() = sin2 +sin/2is4.Assumethat sin = 45, where/2< < . Find:(a) tan (b) sin2 (c) csc241. Giveanexampleof valuesa, b suchthat(a) cos(a +b) = cosa +cosb (b) cosa2 = cosa2solution(a) Takea = b = /2. Thencos(a +b) = cos = 1butcosa +cosb = cos2 +cos2 = 0+0= 0.(b) Takea = . Thencos_a2_ = cos_2_ = 0butcosa2 = cos2 = 12 = 12.Let f (x) = cosx. Sketchthegraphof y = 2f_13x 4_for 0 x 6.43. Solvesin2x +cosx = 0for 0 x < 2.solution Using thedoubleangleformulafor thesinefunction, werewritetheequation as 2sinx cosx + cosx =cosx(2sinx + 1) = 0. Thus, either cosx = 0 or sinx = 1/2. Fromhereweseethat thesolutions arex = /2,x = 7/6, x = 3/2andx = 11/6.Howdoesh(n) = n2/2nbehavefor largewhole-number valuesof n? Doesh(n) tendtoinnity?45. Useagraphingcalculator todeterminewhether theequationcosx = 5x28x4hasanysolutions.solution Thegraphsof y = cosx andy = 5x2 8x4areshownbelow. Becausethegraphsdonot intersect, therearenosolutionstotheequationcosx = 5x28x4.x1 1yy = cos xy = 5x2 8x411Usingagraphingcalculator, ndthenumber of real rootsandestimatethelargest root totwodecimal places:(a) f (x) = 1.8x4x5x(b) g(x) = 1.7x4x5xJune 7, 2011 LTSV SSM Second Pass2 LIMITS2.1 Limits, Rates of Change, and Tangent LinesPreliminary Questions1. Averagevelocityisequal totheslopeof asecant linethroughtwopointsonagraph. Whichgraph?solution Averagevelocity is theslopeof asecant linethroughtwopoints onthegraphof positionas afunctionoftime.2. Caninstantaneousvelocitybedenedasaratio? If not, howisinstantaneousvelocitycomputed?solution Instantaneous velocity cannot bedened as aratio. It is dened as thelimit of averagevelocity as timeelapsedshrinkstozero.3. What isthegraphical interpretationof instantaneousvelocityat amoment t = t0?solution Instantaneousvelocityat timet = t0istheslopeof thelinetangent tothegraphof positionasafunctionoftimeat t = t0.4. What isthegraphical interpretationof thefollowingstatement?Theaveragerateof changeapproachestheinstanta-neousrateof changeastheinterval [x0, x1] shrinkstox0.solution Theslopeof thesecant lineover theinterval [x0, x1] approachestheslopeof thetangent lineat x = x0.5. Therateof changeof atmospheric temperaturewithrespect toaltitudeisequal totheslopeof thetangent linetoagraph. Whichgraph?What arepossibleunitsfor thisrate?solution Therateof changeof atmospherictemperaturewithrespect toaltitudeistheslopeof thelinetangent tothegraphof atmospherictemperatureasafunctionof altitude. Possibleunitsfor thisrateof changeareF/ft orC/m.Exercises1. A ball droppedfromastateof rest at timet = 0travelsadistances(t ) = 4.9t2mint seconds.(a) Howfar doestheball travel duringthetimeinterval [2, 2.5]?(b) Computetheaveragevelocityover [2, 2.5].(c) Computetheaveragevelocityforthetimeintervalsinthetableandestimatetheballsinstantaneousvelocityatt = 2.Interval [2, 2.01] [2, 2.005] [2, 2.001] [2, 2.00001]Averagevelocitysolution(a) Duringthetimeinterval [2, 2.5], theball travelss = s(2.5) s(2) = 4.9(2.5)24.9(2)2= 11.025m.(b) Theaveragevelocityover [2, 2.5] isst= s(2.5) s(2)2.52 = 11.0250.5 = 22.05m/s.(c)timeinterval [2, 2.01] [2, 2.005] [2, 2.001] [2, 2.00001]averagevelocity 19.649 19.6245 19.6049 19.600049Theinstantaneousvelocityat t = 2is19.6m/s.A wrenchreleasedfromastateof rest at timet = 0travelsadistances(t ) = 4.9t2mint seconds. Estimatetheinstantaneousvelocityat t = 3.3. Let v = 20T asinExample2. Estimatetheinstantaneousrateof changeof v withrespect toT whenT = 300K.solutionT interval [300, 300.01] [300, 300.005]averagerateof change 0.577345 0.577348T interval [300, 300.001] [300, 300.00001]averagerateof change 0.57735 0.57735Theinstantaneousrateof changeisapproximately0.57735m/(s K).31June 7, 2011 LTSV SSM Second Pass32 C HA P T E R 2 LIMITSComputey/x for theinterval [2, 5], wherey = 4x 9. What istheinstantaneousrateof changeof y withrespect tox at x = 2?InExercises5and6, astoneistossedverticallyintotheair fromgroundlevel withaninitial velocityof 15m/s. Itsheightat timet ish(t ) = 15t 4.9t2m.5. Computethestonesaveragevelocityover thetimeinterval [0.5, 2.5] andindicatethecorrespondingsecant lineonasketchof thegraphof h(t ).solution Theaveragevelocityisequal toh(2.5) h(0.5)2 = 0.3.Thesecant lineisplottedwithh(t ) below.20.5 1 1.5 2 2.5 346810thCompute the stones average velocity over the time intervals [1, 1.01], [1, 1.001], [1, 1.0001] and [0.99, 1],[0.999, 1], [0.9999, 1], andthenestimatetheinstantaneousvelocityat t = 1.7. Withaninitial deposit of $100, thebalanceinabankaccount after t yearsisf (t ) = 100(1.08)tdollars.(a) What aretheunitsof therateof changeof f (t )?(b) Findtheaveragerateof changeover [0, 0.5] and[0, 1].(c) Estimatetheinstantaneousrateof changeat t = 0.5by computingtheaveragerateof changeover intervalstotheleft andright of t = 0.5.solution(a) Theunitsof therateof changeof f (t ) aredollars/year or $/yr.(b) Theaveragerateof changeof f (t ) = 100(1.08)tover thetimeinterval [t1, t2] isgivenbyft= f (t2) f (t1)t2t1 .timeinterval [0, .5] [0, 1]averagerateof change 7.8461 8(c)timeinterval [0.5, 0.51] [0.5, 0.501] [0.5, 0.5001]averagerateof change 8.0011 7.9983 7.9981timeinterval [0.49, 0.5] [0.499, 0.5] [0.4999, 0.5]averagerateof change 7.9949 7.9977 7.998Therateof changeat t = 0.5isapproximately$8/yr.Thepositionof aparticleat timet is s(t ) = t3+ t . Computetheaveragevelocity over thetimeinterval [1, 4]andestimatetheinstantaneousvelocityat t = 1.9. Figure8shows theestimatednumber N of Internet users inChile, basedondatafromtheUnitedNationsStatisticsDivision.(a) Estimatetherateof changeof N at t = 2003.5.(b) Doestherateof changeincreaseor decreaseast increases? Explaingraphically.(c) Let R betheaveragerateof changeover [2001, 2005]. ComputeR.(d) Istherateof changeat t = 2002greater thanor lessthantheaveragerateR? Explaingraphically.2001 2002 2003 2004 20053.54.04.5N (Internet users in Chilein millions)t (years)FIGURE 8June 7, 2011 LTSV SSM Second PassS E C T I ON 2.1 Limits, Rates of Change, and Tangent Lines 33solution(a) Thetangent lineshowninFigure8appearstopassthroughthepoints(2002, 3.75) and(2005, 4.6). Thus, therateofchangeof N at t = 2003.5isapproximately4.63.7520052002 = 0.283millionInternet usersper year.(b) Ast increases, wemovefromlefttorightalongthegraphinFigure8. Moreover, aswemovefromlefttorightalongthegraph, theslopeof thetangent linedecreases. Thus, therateof changedecreasesast increases.(c) Thegraphof N(t ) appear topass throughthepoints (2001, 3.1) and(2005, 4.5). Thus, theaveragerateof changeover [2001, 2005] isapproximatelyR = 4.53.120052001 = 0.35millionInternet usersper year.(d) For thegurebelow, weseethat theslopeof thetangent lineat t = 2002islarger thantheslopeof thesecant linethrough theendpoints of thegraph of N(t ). Thus, therateof changeat t = 2002 is greater than theaveragerateofchangeR.3.02001 2002 2003 2004 2005 xy3.54.04.5TheatmospherictemperatureT (inC) ataltitudeh metersaboveacertainpointonearthisT = 150.0065hfor h 12,000m. What aretheaverageandinstantaneousratesof changeof T withrespect toh?Whyaretheythesame? Sketchthegraphof T for h 12,000.InExercises1118, estimatetheinstantaneousrateof changeat thepoint indicated.11. P(x) = 3x25; x = 2solutionx interval [2, 2.01] [2, 2.001] [2, 2.0001] [1.99, 2] [1.999, 2] [1.9999, 2]averagerateof change 12.03 12.003 12.0003 11.97 11.997 11.9997Therateof changeat x = 2isapproximately12.f (t ) = 12t 7; t = 4 13. y(x) = 1x +2; x = 2solutionx interval [2, 2.01] [2, 2.001] [2, 2.0001] [1.99, 2] [1.999, 2] [1.9999, 2]averagerateof change 0.0623 0.0625 0.0625 0.0627 0.0625 0.0625Therateof changeat x = 2isapproximately0.06.y(t ) = 3t +1; t = 115. f (x) = 3x; x = 0solutionx interval [0.01, 0] [0.001, 0] [0.0001, 0] [0, 0.01] [0, 0.001] [0, 0.0001]averagerateof change 1.0926 1.098 1.0986 1.1047 1.0992 1.0987Therateof changeisbetwenn1.0986and1.0987.f (x) = 3x; x = 317. f (x) = sinx; x = 6solutionx interval _6 0.01, 6_ _6 0.001, 6_ _6 0.0001, 6_ _6, 6 +0.01_ _6, 6 +0.001_ _6, 6 +0.001_averagerateof change 0.8685 0.8663 0.8660 0.8635 0.8658 0.8660Therateof changeat x = 6 isapproximately0.866.June 7, 2011 LTSV SSM Second Pass34 C HA P T E R 2 LIMITSf (x) = tanx; x = 419. Theheight(incentimeters)attimet (inseconds)of asmall massoscillatingattheendof aspringish(t ) = 8cos(12t ).(a) Calculatethemasssaveragevelocityover thetimeintervals[0, 0.1] and[3, 3.5].(b) Estimateitsinstantaneousvelocityat t = 3.solution(a) Theaveragevelocityover thetimeinterval [t1, t2] isgivenby ht= h(t2) h(t1)t2t1 .timeinterval [0, 0.1] [3, 3.5]averagevelocity 144.721cm/s 0cm/s(b)timeinterval [3, 3.0001] [3, 3.00001] [3, 3.000001] [2.9999, 3] [2.99999, 3] [2.999999, 3]averagevelocity 0.5685 0.05685 0.005685 0.5685 0.05685 0.005685Theinstantaneousvelocityat t = 3secondsisapproximately0cm/s.Thenumber P(t ) of E. coli cellsat timet (hours) inapetri dishisplottedinFigure9.(a) Calculatetheaveragerateof changeof P(t ) over thetimeinterval [1, 3] anddrawthecorrespondingsecantline.(b) Estimatetheslopem of thelineinFigure9. What doesm represent?21. AssumethattheperiodT (inseconds) of apendulum(thetimerequiredfor acompleteback-and-forthcycle)isT = 32L, whereL isthependulumslength(inmeters).(a) What aretheunitsfor therateof changeof T withrespect toL? Explainwhat thisratemeasures.(b) Whichquantitiesarerepresentedbytheslopesof linesA andB inFigure10?(c) Estimatetheinstantaneousrateof changeof T withrespect toL whenL = 3m.Period (s)Length (m)1 3AB2FIGURE 10 TheperiodT isthetimerequiredfor apendulumtoswingbackandforth.solution(a) Theunitsfor therateof changeof T withrespect toL aresecondsper meter. Thisratemeasuresthesensitivity oftheperiodof thependulumtoachangeinthelengthof thependulum.(b) Theslopeof thelineB representstheaveragerateof changeinT fromL = 1mtoL = 3m. Theslopeof thelineA representstheinstantaneousrateof changeof T at L = 3m.(c)timeinterval [3, 3.01] [3, 3.001] [3, 3.0001] [2.99, 3] [2.999, 3] [2.9999, 3]averagevelocity 0.4327 0.4330 0.4330 0.4334 0.4330 0.4330Theinstantaneousrateof changeat L = 1misapproximately0.4330s/m.ThegraphsinFigure11represent thepositionsof movingparticlesasfunctionsof time.(a) Dotheinstantaneousvelocitiesat timest1, t2, t3in(A) formanincreasingor adecreasingsequence?(b) Istheparticlespeedingupor slowingdownin(A)?(c) Istheparticlespeedingupor slowingdownin(B)?23. Anadvertisingcampaignboostedsalesof CrunchyCrustfrozenpizzatoapeaklevel of S0dollarsper month.A marketingstudyshowedthat after t months, monthlysalesdeclinedtoS(t ) = S0g(t ), whereg(t ) = 11+t.Do sales declinemoreslowly or morerapidly as timeincreases? Answer by referringto asketchof thegraphof g(t )together withseveral tangent lines.solution Wenoticefromthegurebelowthat, astimeincreases, theslopesof thetangent linestothegraphof g(t )becomelessnegative. Thus, salesdeclinemoreslowlyastimeincreases.20.20.40.60.81.0yx4 6 8 10 12June 7, 2011 LTSV SSM Second PassS E C T I ON 2.1 Limits, Rates of Change, and Tangent Lines 35Thefractionof acityspopulationinfectedbyauvirusisplottedasafunctionof time(inweeks) inFigure12.(a) Whichquantitiesarerepresentedbytheslopesof linesA andB? Estimatetheseslopes.(b) Istheuspreadingmorerapidlyat t = 1, 2, or 3?(c) Istheuspreadingmorerapidlyat t = 4, 5, or 6?25. ThegraphsinFigure13represent thepositionss of movingparticlesasfunctionsof timet . Matcheachgraphwithadescription:(a) Speedingup(b) Speedingupandthenslowingdown(c) Slowingdown(d) Slowingdownandthenspeedingup(B) (A) (D) (C)tstststsFIGURE 13solution When aparticleis speeding up over atimeinterval, its graph is bent upward over that interval. When aparticleisslowingdown, itsgraphisbent downwardover that interval. Accordingly, Ingraph(A), theparticleis(c) slowingdown. Ingraph(B), theparticleis(b) speedingupandthenslowingdown. Ingraph(C), theparticleis(d) slowingdownandthenspeedingup. Ingraph(D), theparticleis(a) speedingup.Anepidemiologist ndsthat thepercentageN(t ) of susceptiblechildrenwhowereinfectedonday t duringtherst threeweeksof ameaslesoutbreakisgiven, toareasonableapproximation, bytheformula(Figure14)N(t ) = 100t2t3+5t2100t +380(a) Drawthesecant linewhoseslopeistheaveragerateof changeininfectedchildrenover theintervals[4, 6] and[12, 14]. Thencomputetheseaveragerates(inunitsof percent per day).(b) Istherateof declinegreater at t = 8or t = 16?(c) Estimatetherateof changeof N(t ) onday12.27. ThefungusFusariumexosporiuminfectsaeldof axplantsthroughtherootsandcausestheplantstowilt.Eventually,theentireeldisinfected. Thepercentagef (t ) of infectedplantsasafunctionof timet (indays) sinceplantingisshowninFigure15.(a) What aretheunitsof therateof changeof f (t ) withrespect tot ?What doesthisratemeasure?(b) Usethegraphto rank (fromsmallest to largest) theaverageinfectionrates over theintervals [0, 12], [20, 32], and[40, 52].(c) Usethefollowingtabletocomputetheaverageratesof infectionover theintervals[30, 40], [40, 50], [30, 50].Days 0 10 20 30 40 50 60Percent infected 0 18 56 82 91 96 98(d) Drawthetangent lineat t = 40andestimateitsslope.Percent infectedDays after planting10 20 30 40 50 6010080604020FIGURE 15solution(a) Theunitsof therateof changeof f (t ) withrespecttot arepercent/dayor %/d. Thisratemeasureshowquicklythepopulationof axplantsisbecominginfected.(b) Fromsmallest tolargest, theaverageratesof infectionarethoseover theintervals[40, 52], [0, 12], [20, 32]. Thisisbecausetheslopesof thesecant linesover theseintervalsarearrangedfromsmallest tolargest.(c) Theaverageratesof infectionover theintervals[30, 40], [40, 50], [30, 50] are0.9, 0.5, 0.7%/d, respectively.(d) Thetangent linesketchedinthegraphbelowappearstopassthroughthepoints(20, 80) and(40, 91). Theestimateof theinstantaneousrateof infectionat t = 40daysistherefore91804020 = 1120 = 0.55%/d.June 7, 2011 LTSV SSM Second Pass36 C HA P T E R 2 LIMITS10 20 30 40 50 6010080604020Letv = 20T asinExample2. Istherateof changeof v withrespecttoT greater atlowtemperaturesor hightemperatures? Explainintermsof thegraph.29. If anobject inlinear motion(but withchangingvelocity) coverss metersint seconds, thenitsaveragevelocity is v0 = s/t m/s. Showthat it wouldcover thesamedistanceif it traveledat constant velocity v0over thesametimeinterval. Thisjustiesour callings/t theaveragevelocity.solution At constant velocity, thedistancetraveledisequal tovelocity timestime, soanobject movingat constantvelocityv0for t secondstravelsv0t meters. Sincev0= s/t , wenddistancetraveled= v0t =_st_t = sSotheobject coversthesamedistances bytravelingat constant velocityv0.Sketchthegraphof f (x) = x(1 x) over [0, 1]. Refer tothegraphand, without makingany computations,nd:(a) Theaveragerateof changeover [0, 1](b) The(instantaneous) rateof changeat x = 12(c) Thevaluesof x at whichtherateof changeispositive31. WhichgraphinFigure16has thefollowingproperty: For all x, theaveragerateof changeover [0, x] isgreater thantheinstantaneousrateof changeat x. Explain.(B)xy(A)xyFIGURE 16solution(a) Theaveragerateof changeover [0, x] isgreater thantheinstantaneousrateof changeat x: (B).(b) Theaveragerateof changeover [0, x] islessthantheinstantaneousrateof changeat x: (A)Thegraphin(B) bends downward, so theslopeof thesecant linethrough(0, 0) and(x, f (x)) is larger thantheslopeof thetangent lineat (x, f (x)). Ontheother hand, thegraphin(A) bends upward, so theslopeof thetangent lineat(x, f (x)) islarger thantheslopeof thesecant linethrough(0, 0) and(x, f (x)).Further Insights and ChallengesTheheight of aprojectileredintheair verticallywithinitial velocity25m/sish(t ) = 25t 4.9t2m.(a) Computeh(1). Showthat h(t ) h(1) canbefactoredwith(t 1) asafactor.(b) Usingpart (a), showthat theaveragevelocityover theinterval [1, t ] is20.14.9t .(c) Usethis formulato ndtheaveragevelocity over several intervals [1, t ] witht closeto 1. Thenestimatetheinstantaneousvelocityat timet = 1.33. Let Q(t ) = t2. Asinthepreviousexercise, ndaformulafor theaveragerateof changeof Qover theinterval [1, t ]anduseit to estimatetheinstantaneous rateof changeat t = 1. Repeat for theinterval [2, t ] andestimatetherateofchangeat t = 2.solution Theaveragerateof changeisQ(t ) Q(1)t 1 = t21t 1.Applyingthedifferenceof squaresformulagivesthattheaveragerateof changeis((t +1)(t 1))/(t 1) = (t +1) fort = 1. Ast getscloser to1, thisgetscloser to1+1= 2. Theinstantaneousrateof changeis2.For t0= 2, theaveragerateof changeisQ(t ) Q(2)t 2 = t24t 2,whichsimpliestot + 2for t = 2. Ast approaches2, theaveragerateof changeapproaches4. Theinstantaneousrateof changeistherefore4.Showthat theaveragerateof changeof f (x) = x3over [1, x] isequal tox2+x +1.Usethistoestimatetheinstantaneousrateof changeof f (x) at x = 1.June 7, 2011 LTSV SSM Second PassS E C T I ON 2.2 Limits: A Numerical and Graphical Approach 3735. Findaformulafor theaveragerateof changeof f (x) = x3over [2, x] anduseit toestimatetheinstantaneousrateof changeat x = 2.solution Theaveragerateof changeisf (x) f (2)x 2 = x38x 2.Applyingthedifferenceof cubesformulatothenumerator, wendthat theaveragerateof changeis(x2+2x +4)(x 2)x 2 = x2+2x +4for x = 2. Thecloser x getsto2, thecloser theaveragerateof changegetsto22+2(2) +4= 12.LetT = 32L asinExercise21. Thenumbersinthesecondcolumnof Table4areincreasing, andthoseinthelastcolumnaredecreasing. Explainwhyintermsof thegraphof T asafunctionof L. Also, explaingraphicallywhytheinstantaneousrateof changeat L = 3liesbetween0.4329and0.4331.2.2 Limits: A Numerical and Graphical ApproachPreliminary Questions1. What isthelimit of f (x) = 1asx ?solution limx 1= 1.2. What isthelimit of g(t ) = t ast ?solution limt t = .3. Is limx1020equal to10or 20?solution limx1020= 20.4. Canf (x) approachalimit asx c if f (c) isundened? If so, giveanexample.solution Yes. Thelimitof afunctionf asx c doesnotdependonwhathappensatx = c, onlyonthebehavior off asx c. Asanexample, consider thefunctionf (x) = x21x 1.Thefunctionisclearlynot denedat x = 1butlimx1f (x) = limx1x21x 1 = limx1(x +1) = 2.5. What doesthefollowingtablesuggest about limx1f (x) and limx1+f (x)?x 0.9 0.99 0.999 1.1 1.01 1.001f (x) 7 25 4317 3.0126 3.0047 3.00011solution Thevaluesinthetablesuggest that limx1 f (x) = andlimx1+ f (x) = 3.6. Canyoutell whether limx5f (x) existsfromaplot of f (x) for x > 5? Explain.solution No. Byexaminingvaluesof f (x) for x closetobut greater than5, wecandeterminewhether theone-sidedlimit limx5+ f (x) exists. Todeterminewhether limx5f (x) exists, wemust examinevalueof f (x) onbothsidesofx = 5.7. If youknowinadvancethat limx5f (x) exists, canyoudetermineitsvaluefromaplot of f (x) for all x > 5?solution Yes. If limx5f (x) exists, thenbothone-sidedlimitsmust exist andbeequal.June 7, 2011 LTSV SSM Second Pass38 C HA P T E R 2 LIMITSExercisesInExercises14, ll inthetablesandguessthevalueof thelimit.1. limx1f (x), wheref (x) = x31x21.x f (x) x f (x)1.002 0.9981.001 0.9991.0005 0.99951.00001 0.99999solutionx 0.998 0.999 0.9995 0.99999 1.00001 1.0005 1.001 1.002f (x) 1.498501 1.499250 1.499625 1.499993 1.500008 1.500375 1.500750 1.501500Thelimit asx 1is32.limt 0h(t ), whereh(t ) = cost 1t2 . Notethat h(t ) iseven; that is, h(t ) = h(t ).t 0.002 0.0001 0.00005 0.00001h(t )3. limy2f (y), wheref (y) = y2y 2y2+y 6.y f (y) y f (y)2.002 1.9982.001 1.9992.0001 1.9999solutiony 1.998 1.999 1.9999 2.0001 2.001 2.02f (y) 0.59984 0.59992 0.599992 0.600008 0.60008 0.601594Thelimit asy 2is35.lim0f (), wheref () = sin 3 . 0.002 0.0001 0.00005 0.00001f ()5. Determine limx0.5f (x) for f (x) asinFigure9.0.51.5xy1 f(x)FIGURE 9solution Thegraphsuggeststhat f (x) 1.5asx 0.5.Determine limx0.5g(x) for g(x) asinFigure10.InExercises7and8, evaluatethelimit.7. limx21xsolution Asx 21, f (x) = x 21. Youcanseethis, for example, onthegraphof f (x) = x.limx4.23InExercises 916, verifyeachlimit usingthelimit denition. For example, inExercise9, showthat |3x 12| canbemadeassmall asdesiredbytakingx closeto4.9. limx43x = 12solution |3x 12| = 3|x 4|. |3x 12| canbemadearbitrarilysmall bymakingx closeenoughto4, thusmaking|x 4| small.June 7, 2011 LTSV SSM Second PassS E C T I ON 2.2 Limits: A Numerical and Graphical Approach 39limx53= 311. limx3(5x +2) = 17solution |(5x + 2) 17| = |5x 15| = 5|x 3|. Therefore, if you make|x 3| small enough, you can make|(5x +2) 17| assmall asdesired.limx2(7x 4) = 1013. limx0x2= 0solution Asx 0, wehave|x20| = |x +0||x 0|. Tosimplifythings, supposethat |x| < 1, sothat |x +0||x 0| = |x||x| < |x|. Bymaking|x| sufcientlysmall, sothat |x +0||x 0| = x2isevensmaller, youcanmake|x20|assmall asdesired.limx0(3x29) = 915. limx0(4x2+2x +5) = 5solution Asx 0, wehave|4x2+2x +55| = |4x2+2x| = |x||4x +2|. If |x| < 1, |4x +2| canbenobiggerthan6, so|x||4x +2| < 6|x|. Therefore, bymaking|x 0| = |x| sufcientlysmall, youcanmake|4x2+2x +55| =|x||4x +2| assmall asdesired.limx0(x3+12) = 12In Exercises 1736, estimate the limit numerically or state that the limit does not exist. If innite, state whether theone-sidedlimitsareor .17. limx1x 1x 1solutionx 0.9995 0.99999 1.00001 1.0005f (x) 0.500063 0.500001 0.49999 0.499938Thelimit asx 1is12.limx42x232x +419. limx2x2+x 6x2x 2solutionx 1.999 1.99999 2.00001 2.001f (x) 1.666889 1.666669 1.666664 1.666445Thelimit asx 2is53.limx3x32x29x22x 321. limx0sin2xxsolutionx 0.01 0.005 0.005 0.01f (x) 1.999867 1.999967 1.999967 1.999867Thelimit asx 0is2.limx0sin5xx23. lim0cos 1solutionx 0.05 0.001 0.001 0.05f (x) 0.0249948 0.0005 0.0005 0.0249948Thelimit asx 0is0.limx0sinxx225. limx4 1(x 4)3solutionx 3.99 3.999 3.9999 4.0001 4.001 4.01f (x) 106 109 1012 1012 109 106Thelimit doesnot exist. Asx 4, f (x) ; similarly, asx 4+, f (x) .June 7, 2011 LTSV SSM Second Pass40 C HA P T E R 2 LIMITSlimx13xx 127. limx3+x 4x29solutionx 3.01 3.001 3.0001 3.00001f (x) 16.473 166.473 1666.473 16666.473Asx 3+, f (x) .limh03h1h29. limh0sinhcos1hsolutionh 0.01 0.001 0.0001 0.0001 0.001 0.01f (h) 0.008623 0.000562 0.000095 0.000095 0.000562 0.008623Thelimit asx 0is0.limh0cos1h31. limx0|x|xsolutionx 0.05 0.001 0.00001 0.00001 0.001 0.05f (x) 1.161586 1.006932 1.000115 0.999885 0.993116 0.860892Thelimit asx 0is1.limx02x3xx33. lim4tan 2sin cos 4solution 4 0.01 4 0.001 4 0.0001 4 +0.0001 4 +0.001 4 +0.01f () 1.96026 1.99600 1.99960 2.00040 2.00400 2.04027Thelimit asx 4 isapproximately2.limr0(1+r)1/r 35. lim01cos2solution 0.01 0.001 0.0001 0.0001 0.001 0.01f () 0.499996 0.500000 0.500000 0.500000 0.500000 0.499996Thelimit as 0appearstobe0.5.lim01cos337. Thegreatestinteger functionisdenedby[x] = n, wheren istheuniqueinteger suchthat n x < n +1. Sketchthegraphof y = [x]. Calculate, for c aninteger:(a) limxc[x] (b) limxc+[x]solution Hereisagraphof thegreatest integer function:211 2 3 1 xy(a) Fromthegraph, weseethat, for c aninteger,limxc[x] = c 1.June 7, 2011 LTSV SSM Second PassS E C T I ON 2.2 Limits: A Numerical and Graphical Approach 41(b) Fromthegraph, weseethat, for c aninteger,limxc+[x] = c.Determinetheone-sidedlimitsat c = 1, 2, and4of thefunctiong(x) showninFigure11, andstatewhether thelimit existsat thesepoints.InExercises3946, determinetheone-sidedlimitsnumericallyor graphically. Ifinnite, statewhether theone-sidedlimitsare or , anddescribethecorrespondingvertical asymptote. InExercise46, [x] is thegreatest integer functiondenedinExercise37.39. limx0sinx|x|solutionx 0.2 0.02 0.02 0.2f (x) 0.993347 0.999933 0.999933 0.993347Theleft-handlimit is limx0f (x) = 1, whereastheright-handlimit is limx0+f (x) = 1.limx0|x|1/x 41. limx0x sin|x|x3solutionx 0.1 0.01 0.01 0.1f (x) 199.853 19999.8 0.166666 0.166583Theleft-handlimitis limx0f (x) = , whereastheright-handlimitis limx0+f (x) = 16. Thus, thelinex = 0isaverticalasymptotefromtheleft for thegraphof y = xsin|x|x3 .limx4x +1x 443. limx24x2+7x3+8solution Thegraphof y = 4x2+7x3+8 for x near 2isshownbelow. Fromthisgraph, weseethatlimx24x2+7x3+8 = while limx2+4x2+7x3+8 = .Thus, thelinex = 2isavertical asymptotefor thegraphof y = 4x2+7x3+8.3.0 2.5 2.0 1.5 1.0 xlimx3x2x2945. limx1x5+x 2x2+x 2solution Thegraphof y = x5+x2x2+x2 for x near 1isshownbelow. Fromthisgraph, weseethatlimx1x5+x 2x2+x 2 = 2.20.5 1.0 1.5 xy46June 7, 2011 LTSV SSM Second Pass42 C HA P T E R 2 LIMITSlimx2cos_2(x [x])_ 47. Determinetheone-sidedlimits at c = 2, 4of thefunctionf (x) inFigure12. What arethevertical asymptotes off (x)?5 4 215510xyFIGURE 12solution For c = 2, wehave limx2f (x) = and limx2+f (x) = . For c = 4, wehave limx4f (x) = and limx4+f (x) = 10.Thevertical asymptotesarethevertical linesx = 2andx = 4.Determinetheinniteone- andtwo-sidedlimitsinFigure13.InExercises4952, sketchthegraphof afunctionwiththegivenlimits.49. limx1f (x) = 2, limx3f (x) = 0, limx3+f (x) = 4solution2461 2 3 4yxlimx1f (x) = , limx3f (x) = 0, limx3+f (x) = 51. limx2+f (x) = f (2) = 3, limx2f (x) = 1, limx4f (x) = 2= f (4)solution11231 2 3 4 5yxlimx1+f (x) = , limx1f (x) = 3, limx4f (x) = 53. Determinetheone-sidedlimitsof thefunctionf (x) inFigure14, at thepointsc = 1, 3, 5, 6.123412345yx1 2 3 4 5 6 7 8FIGURE 14 Graphof f (x)solution limx1f (x) = limx1+f (x) = 3 limx3f (x) = June 7, 2011 LTSV SSM Second PassS E C T I ON 2.2 Limits: A Numerical and Graphical Approach 43 limx3+f (x) = 4 limx5f (x) = 2 limx5+f (x) = 3 limx6f (x) = limx6+f (x) = Doeseither of thetwooscillatingfunctionsinFigure15appear toapproachalimit asx 0?InExercises5560, plot thefunctionandusethegraphtoestimatethevalueof thelimit.55. lim0sin5sin2solution2.422.442.462.482.50yFromthegraphof y = sin5sin2shownabove, weseethat thelimit as 0is52.limx012x14x157. limx02xcosxxsolution0.69350.69400.69300.69250.6920yy = 2x cosxxThelimit asx 0isapproximately0.693. (Theexact answer isln2.)lim0 sin24cos 159. lim0cos7 cos52solution12.011.811.611.4yFromthegraphof y = cos7 cos52 shownabove, weseethat thelimit as 0is12.lim0sin22 sin4461. Let n beapositiveinteger. For whichn arethetwoinniteone-sidedlimits limx01/xnequal?solution First, supposethat n iseven. Thenxn 0for all x, and 1xn > 0for all x = 0. Hence,limx01xn = limx0+1xn = .Next, supposethat n isodd. Then 1xn > 0for all x > 0but 1xn < 0for all x < 0. Thus,limx01xn = but limx0+1xn = .Finally, thetwoinniteone-sidedlimitsareequal whenever n iseven.Let L(n) = limx1_ n1xn 11x_for n apositiveinteger. InvestigateL(n) numericallyfor several valuesofn, andthenguessthevalueof of L(n) ingeneral.June 7, 2011 LTSV SSM Second Pass44 C HA P T E R 2 LIMITS63. Insomecases, numerical investigationscanbemisleading. Plot f (x) = cosx.(a) Does limx0f (x) exist?(b) Show, byevaluatingf (x) at x = 12, 14, 16, . . . , that youmight beabletotrick your friendsintobelievingthatthelimit existsandisequal toL = 1.(c) Whichsequenceof evaluationsmight trickthemintobelievingthat thelimit isL = 1.solution Hereisthegraphof f (x).0.050.50.5yx0.05(a) Fromthegraphof f (x), whichshowsthat thevalueof f (x) oscillatesmoreandmorerapidlyasx 0, it followsthat limx0f (x) doesnot exist.(b) Noticethatf_12_ = cos 1/2 = cos2 = 1;f_14_ = cos 1/4 = cos4 = 1;f_16_ = cos 1/6 = cos6 = 1;and, ingeneral, f ( 12n) = 1for all integersn.(c) At x = 1, 13, 15, . . ., thevalueof f (x) isalways1.Further Insights and ChallengesLight wavesof frequency passingthroughaslit of widtha produceaFraunhofer diffractionpatternof lightanddarkfringes(Figure16). Theintensityasafunctionof theangle isI () = Im_sin(Rsin)Rsin_2whereR = a/ andIm isaconstant. Showthattheintensityfunctionisnotdenedat = 0. Thenchooseanytwovaluesfor R andchecknumericallythat I () approachesIm as 0.65. Investigate lim0sinnnumericallyfor several valuesof n. Thenguessthevalueingeneral.solution For n = 3, wehave 0.1 0.01 0.001 0.001 0.01 0.1sinn2.955202 2.999550 2.999996 2.999996 2.999550 2.955202Thelimit as 0is3. For n = 5, wehave 0.1 0.01 0.001 0.001 0.01 0.1sinn4.794255 4.997917 4.999979 4.999979 4.997917 4.794255Thelimit as 0is5. Wesurmisethat, ingeneral, lim0sinn= n.Shownumerically that limx0bx1xfor b = 3, 5appearstoequal ln3, ln5, wherelnx isthenatural logarithm.Thenmakeaconjecture(guess) for thevalueingeneral andtest your conjecturefor twoadditional valuesof b.67. Investigate limx1 xn1xm1 for (m, n) equal to (2, 1), (1, 2), (2, 3), and (3, 2). Then guess thevalueof thelimit ingeneral andcheckyour guessfor twoadditional pairs.solutionx 0.99 0.9999 1.0001 1.01x 1x21 0.502513 0.500025 0.499975 0.497512June 7, 2011 LTSV SSM Second PassS E C T I ON 2.2 Limits: A Numerical and Graphical Approach 45Thelimit asx 1is12.x 0.99 0.9999 1.0001 1.01x21x 1 1.99 1.9999 2.0001 2.01Thelimit asx 1is2.x 0.99 0.9999 1.0001 1.01x21x31 0.670011 0.666700 0.666633 0.663344Thelimit asx 1is23.x 0.99 0.9999 1.0001 1.01x31x21 1.492513 1.499925 1.500075 1.507512Thelimit asx 1is32. For general m andn, wehave limx1 xn1xm1 = nm.x 0.99 0.9999 1.0001 1.01x 1x31 0.336689 0.333367 0.333300 0.330022Thelimit asx 1is13.x 0.99 0.9999 1.0001 1.01x31x 1 2.9701 2.9997 3.0003 3.0301Thelimit asx 1is3.x 0.99 0.9999 1.0001 1.01x31x71 0.437200 0.428657 0.428486 0.420058Thelimit asx 1is37 0.428571.Findbynumerical experimentationthepositiveintegersk suchthat limx0sin(sin2x)xk exists.69. Plot thegraphof f (x) = 2x8x 3.(a) ZoominonthegraphtoestimateL = limx3f (x).(b) Explainwhyf (2.99999) L f (3.00001)UsethistodetermineL tothreedecimal places.solution(a)5.5555.5655.5455.5355.525yx = 3y = 2x 8x 3June 7, 2011 LTSV SSM Second Pass46 C HA P T E R 2 LIMITS(b) It is clear that thegraphof f rises as wemoveto theright. Mathematically, wemay express this observationas:whenever u < v, f (u) < f (v). Because2.99999< 3= limx3f (x) < 3.00001,it followsthatf (2.99999) < L = limx3f (x) < f (3.00001).Withf (2.99999) 5.54516andf (3.00001) 5.545195, theaboveinequality becomes 5.54516 < L < 5.545195;hence, tothreedecimal places, L = 5.545.Thefunctionf (x) = 21/x21/x21/x+21/x isdenedfor x = 0.(a) Investigate limx0+f (x) and limx0f (x) numerically.(b) Plot thegraphof f anddescribeitsbehavior near x = 0.2.3 Basic Limit LawsPreliminary Questions1. StatetheSumLawandQuotient Law.solution Supposelimxc f (x) andlimxc g(x) bothexist. TheSumLawstatesthatlimxc(f (x) +g(x)) = limxcf (x) + limxcg(x).Providedlimxc g(x) = 0, theQuotient Lawstatesthatlimxcf (x)g(x)= limxc f (x)limxc g(x).2. Whichof thefollowingisaverbal versionof theProduct Law(assumingthelimitsexist)?(a) Theproduct of twofunctionshasalimit.(b) Thelimit of theproduct istheproduct of thelimits.(c) Theproduct of alimit isaproduct of functions.(d) A limit producesaproduct of functions.solution Theverbal versionof theProduct Lawis(b): Thelimit of theproduct istheproduct of thelimits.3. Whichstatement iscorrect?TheQuotient Lawdoesnot holdif:(a) Thelimit of thedenominator iszero.(b) Thelimit of thenumerator iszero.solution Statements(a)iscorrect. TheQuotient Lawdoesnot holdif thelimit of thedenominator iszero.ExercisesInExercises124, evaluatethelimit usingtheBasicLimit Lawsandthelimits limxcxp/q= cp/qand limxck = k.1. limx9xsolution limx9x = 9.limx3143. limx12x4solution limx12x4=_12_4= 116.limz27z2/35. limt 2t1solution limt 2t1= 21= 12.limx5x27. limx0.2(3x +4)solution UsingtheSumLawandtheConstant MultipleLaw:limx0.2(3x +4) = limx0.23x + limx0.24= 3 limx0.2x + limx0.24= 3(0.2) +4= 4.6.June 7, 2011 LTSV SSM Second PassS E C T I ON 2.3 Basic Limit Laws 47limx13(3x3+2x2)9. limx1(3x42x3+4x)solution UsingtheSumLaw, theConstant MultipleLawandthePowers