Solution Manual for A Graphical Approach to Precalculus with Limits, 7th Edition

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SOLUTIONSMANUALDAVIDATWOODRochester Community and Technical CollegeAGRAPHICALAPPROACH TOPRECALCULUS WITHLIMITSAUNITCIRCLEAPPROACHSEVENTHEDITIONJohn HornsbyUniversity of New OrleansMargaret L. LialAmerican River CollegeGary RockswoldMinnesota State University, Mankato

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ContentsChapter 1Linear Functions, Equations, and Inequalities1Chapter 2Analysis of Graphs of Functions69Chapter 3Quadratic Functions153Chapter 4Polynomial Functions of Higher Degree209Chapter 5Rational, Power, and Root Functions279Chapter 6Inverse, Exponential, and Logarithmic Functions371Chapter 7Systems and Matrices451Chapter 8Conic Sections, Nonlinear Systems, and Parametric Equations615Chapter 9The Unit Circle and the Functions of Trigonometry683Chapter 10Trigonometric Identities and Equations807Chapter 11Applications of Trigonometry and Vectors893Chapter 12Further Topics in Algebra977Chapter 13Limits, Derivatives, and Definite Integrals1043Chapter RReview: Basic Algebraic Concepts1087Appendices1113

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Section 1.11Chapter 1: Linear Functions, Equations, and Inequalities1.1: Real Numbers and the Rectangular Coordinate System1.(a)The only natural number is 10.(b)The whole numbers are 0 and 10.(c)The integers are126,(or3), 0,410.(d)The rational numbers are1256,(or3),, 0, .31, .3,48and 10.(e)The irrational numbers are3, 2and17.(f)All of the numbers listed are real numbers.2.(a)The natural numbers are6 (or3),8, and81(or 9).2(b)The whole numbers are60,(or 3),8, and81(or 9).2(c)The integers are1468,(or2), 0,(or 3),8, and81(or 9).72(d)The rational numbers are1468,(or2),.245,(or 3),8, and81(or 9).72(e)The only irrational number is12.(f)All of the numbers listed are real numbers.3.(a)There are no natural numbers listed.(b)There are no whole numbers listed.(c)The integers are100(or10) and1.(d)The rational numbers are13100 (or10),,1,5.23,9.14,3.14,6and22 .7(e)There are no irrational numbers listed.(f)All of the numbers listed are real numbers.4.(a)The natural numbers are 3, 18, and 56.(b)The whole numbers are 3, 18, and 56.(c)The integers are49(or7),3,18, and 56.(d)The rational numbers are49(or7),.405,. 3,.1,3,18, and 56.(e)The only irrational number is6.(f)All of the numbers listed are real numbers.5.The number 19,900,037,000,000 is a natural number, integer, rational number, and real number.6.The number 700,000,000,000 is a natural number, integer, rational number, and real number.7.The number24is an integer, rational, and real number.

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2Chapter 1 Linear Functions, Equations, and Inequalities8.The number17is an integer, rational number, and real number9.The number71, 060is an integer, rational number and real number.10.The number12.5is a rational number and real number.11.The number72is a real number.12.The numberis a real number.13.Natural numbers would be appropriate because population is only measured in positive wholenumbers.14.Natural numbers would be appropriate because distance on road signs is only given in positivewhole numbers.15.Rational numbers would be appropriate because shoes come in fraction sizes.16.Rational numbers would be appropriate because gas is paid for in dollars and cents, a decimal partof a dollar.17.Integers would be appropriate because temperature is given in positive and negative whole numbers.18.Integers would be appropriate because golf scores are given in positive and negative wholenumbers.19.20.21.22.23.A rational number can be written as a fraction,,0,pqqwherepandqare integers. An irrationalnumber cannot be written in this way.24.She should write21.414213562.Calculators give only approximations of irrational numbers.25.The point52, 7is in Quadrant I. See Figure 25-34.26.The point(1, 2)is in Quadrant I. See Figure 25-34.27.The point( 3, 2)is in Quadrant II. See Figure 25-34.28.The point( 4,3)is in Quadrant II. See Figure 25-34.29.The point( 5,2)is in Quadrant III. See Figure 25-34.30.The point( 2,4)is in Quadrant III. See Figure 25-34.31.The point(2,2)is in Quadrant IV. See Figure 25-34.

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Section 1.1332.The point(3,3)is in Quadrant IV. See Figure 25-34.33.The point(3, 0)is located on thex-axis, therefore is not in a quadrant. See Figure 25-34.34.The point( 2, 0)is located on thex-axis, therefore is not in a quadrant. See Figure 25-34.Figure 25-3435.If0, then either0 and0Quadrant I, or0 and0Quadrant III.xyxyxy36.If0, then either0 and< 0Quadrant IV, or0 and0Quadrant II.xyxyxy37.If0, then either0 and< 0Quadrant IV, or0 and0Quadrant II.xxyxyy38.If0, then either0 and> 0Quadrant I, or0 and0Quadrant III.xxyxyy39.Any point of the form(0,)bis located on they-axis.40.Any point of the form( , 0)ais located on thex-axis.41.[ 5,5]by[25,25]42.[ 25, 25]by[5,5]43.[ 60, 60]by[100,100]44.[ 100,100]by[60,60]45.[ 500,300]by[300,500]46.[ 300,300]by[375,150]47.See Figure 47.48.See Figure 48.49.See Figure 49.50.See Figure 50.

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4Chapter 1 Linear Functions, Equations, and Inequalities[-10,10] by [-10,10][-40,40] by [-30,30][-5,10] by [-5,10][-3.5,3.5] by [-4,10]Xscl = 1Yscl = 1Xscl = 5Yscl = 5Xscl= 3Yscl = 3Xscl = 1Yscl= 1Figure 47Figure 48Figure 49Figure 5051.See Figure 51.52.See Figure 52.[-100,100] by [-50,50][-4.7,4.7] by [-3.1,3.1]Xscl = 20Yscl = 25Xscl = 1Yscl = 1Figure 51Figure 5253.There are no tick marks, which is a result of setting Xscl and Yscl to 0.54.The axes appear thicker because the tick marks are so close together. The problem can be fixed byusing larger values for Xscl and Yscl such asXscl = Yscl =10.55.587.6157731067.61656.979.8488578029.84957.3333.207534333.20858.3914.4979414454.49859.4863.0452616463.04560.41233.3302457133.33061.1/ 2194.358898444.35962.1/3293.0723168263.07263.1.546311.9871792311.98764.2.75235555.8632685555.86365.(5.63.1) / (8.91.3).2566.(3425) / 232.5767.(^ 31)5.6668.23( 2.16 )3.24 69.23(5.9)2(5.9)698.63

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Section 1.1570.2^ 3539.6671.22(46)(71) )8.2572.22( 13 )( 53)8.25   73.(1) /(1).7274.3( 4.3E53.7E2)76.6575.32 / (15)2.82 76.14.5 / (32)1.84 77.222222228156422528917abccccc78.222222227244957662525abccccc79.2222222213851697225705684abcbbbb80.2222222214501962500230448abcbbbb81.222222225825648989abccccc82.2222222291081100181181abccccc83.22222222( 13)(29)1329164abcaaaa84.22222222(7)( 11)71142abcaaaa85.(a)2222(2( 4))(53)(6)(2)364402 10d (b)42 3528,,( 1, 4)2222M 86.(a)2222(2( 3))(14)(5)( 5)25255052d  (b)324( 1)1 313,,,222222M 87.(a)2222(6( 7))( 24)(13)( 6)16936205d   (b)764( 2)1 21,,,122222M 88.(a)2222(1( 3))(4( 3))(4)(7)164965d  (b)3134211,,1,22222M89.(a)2222(25)(117)( 3)(4)916255d(b)527117 187,,,922222M90.(a)2222(4( 2))( 35)(6)( 8)361610010d   

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6Chapter 1 Linear Functions, Equations, and Inequalities(b)24 5( 3)22,,(1,1)2222M 91.(a)2222( 3( 8))(( 5)( 2))(5)( 3)25934d   (b)8( 3)2( 5)117117,,,222222M  92.(a)2222(6( 6))(5( 10))(12)(15)144225369341d  (b)66105055,,0,22222M93.(a)2222(6.29.2)(7.43.4)( 3)(4)916255d(b)9.26.2 3.47.415.4 10.8,,(7.7,5.4)2222M94.(a)2222(3.98.9)(13.61.6)( 5)(12)2514416913d(b)8.93.9 1.613.612.8 15.2,,(6.4, 7.6)2222M95.(a)2222222(613 )(( 23 ))( 7 )(24 )4957662525dxxxxxxxxxx (b)13623192219,,,1122222xxxxxxMxx96.(a)2222222(2012 )(12( 3 ))(8 )(15 )6422528917dyyyyyyyyyy (b)12203123299,,16 ,22222yyyyyyMyy97. Using the midpoint formula we get:22222747,(8,5)87169222xyxxxand2224541014.2yyyTherefore the coordinates are:(19,14).Q98.Using the midpoint formula we get:222213513,( 2,4)2134222xyxx   2222517 and45813.2yxyy    Therefore the coordinates are:( 17,13).Q99.Using the midpoint formula we get:2225.648.215.64,( 4.04,1.60)4.04222xyx  25.648.08x 22228.2113.72 and1.608.213.205.01.2yxyy  Therefore thecoordinates are:( 13.72,5.01).Q

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Section 1.17100.Using the midpoint formula we get:222210.328.5510.32,(1.55,2.75)1.5510.323.10222xyxx22228.5513.42.2.758.555.5014.05.2yxyy   Therefore the coordinatesare:( 13.42,13.05).Q101.20112015 36.5367.394026 103.92,,(2013,51.96);2222Mthe revenue was about $51.96 billion.102.20062012750533354018 10840,,(2009,5420);2222Mthe revenue was about $5420 million.The result is quite a bit higher than the actual figure.103.In 2012 ,20112013 22,35023,550402445,900,,(2012, 22,950);2222Mpoverty levelwas approximately $22,950. In 2014,2013201523,35024, 250402847,800,,2222M(2014, 23,900);poverty level was approximately $23,900.104.For 2017,20162018 719475004034 14, 694,,(2017 , 7347);2222Menrollmentwas 7347 thousand. For 2019,20182020750077064038 15, 206,,(2019, 7603);2222MEnrollment was about 7603 thousand.105.(a)From (0, 0) to (3, 4):22221(30)(40)(3)(4)916255.dFrom (3,4) to (7, 1):22222(73)(14)(4)( 3)169255.d From (0, 0) to(7, 1):22223(70)(10)(7)(1)4915052.dSinced1=d2, the triangle isisosceles.(b)From (−1, −1) to (2, 3):22221(2( 1))(3( 1))(3)(4)916255.d  From (2, 3) to (−4, 3):22222( 42)(33)( 6)(0)360366.dFrom (−4, 3) to (−1, −1):22223( 1( 4))( 13)(3)( 4)916255.d     Since12,ddthe triangle is not equilateral.(c)From (−1, 0) to (1, 0):22221(1( 1))(00)(2)(0)4042.d From (-1, 0) to0,3:22222( 10)(03)( 1)(3)1342.d  

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8Chapter 1 Linear Functions, Equations, and InequalitiesFrom (1, 0) to0,3:22223(10)(03)(1)(3)1342.dSince123,dddthe triangle is equilateral and isosceles.(d)From (−3, 3) to (-1, 3):22221( 3( 1))(33)( 2)(0)4042.d From (-3, 3) to (−2, 5):22222( 3( 2))(35)( 1)( 2)145.d  From (−1, 3) to (−2, 5):22223( 1( 2))(35)(1)( 2)145.d   Since23,ddthe triangle is not isosceles.106.Let1drepresent the distance betweenPandMand let2drepresent the distance betweenMandQ.22221212112112111222222xxyyxxxyyydxy2222121211212()()1()()442xxyydxxyy22221212212212222222222xxyyxxxyyydxy2222212122121()()1()()442xxyydxxyySince221221()()xxxxand221221()()yyyy, the distances are the same.Since12,ddthe sum222212221212121122()()()() .2dddxxyyxxyyThat is, the sum is equal to the distance betweenPand Q.1.2: Introduction to Relations and Functions1.The interval is( 1, 4).2.The interval is[ 3,).3.The interval is(, 0).4.The interval is(3, 8).5.The interval is1, 2 .( 5,4].6.The interval is7.( 4,3){|43}xx8.[2, 7){| 27}xx9.(,1]{|1}xx  

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Section 1.2910.(3,){|3}xx11.26xx12.08xx13.4x x 14.3x x15.A parenthesis is used if the symbol is,,, or  or .A square bracket is used if the symbol isor.16.No real number is both greater than7and less than10.Part (d) should be written107.x 17.See Figure 1718.See Figure 1819.See Figure 19Figure 17Figure 18Figure 1920.See Figure 2021.See Figure 2122.See Figure 22Figure 20Figure 21Figure 2223.See Figure 2324.See Figure 24

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10Chapter 1 Linear Functions, Equations, and InequalitiesFigure 23Figure 2425.The relation is a function. Domain:5,3, 4, 7Range:1, 2,9, 6 .26.The relation is a function. Domain:8,5,9,3 ,Range:0, 4,3,8 .27.The relation is a function. Domain:1, 2,3 , Range:6 .28.The relation is a function. Domain:10,20,30 ,Range:5 .29.The relation is not a function. Domain:4,3,2 ,Range:1,5,3, 7 .30.The relation is not a function. Domain:0,1 ,Range:5,3,4 .31.The relation is a function. Domain:11,12,13,14 , Range:6,7 .32.The relation is not a function. Domain:1 ,Range:12,13,14,15 .33.The relation is a function. Domain:0,1, 2,3, 4 ,Range:2,3,5,6,7 .34.The relation is a function. Domain:11 111,,,,,24 8 16Range:0,1,2,3,4 .35.The relation is a function. Domain:,, Range:,. 36.The relation is a function. Domain:,, Range:, 4 .37.The relation is not a function. Domain:4, 4 ,Range:3,3 .38.The relation is a function. Domain:2, 2 ,Range:0, 4 .39.The relation is a function. Domain:2,,Range:0,.40.The relation is a function. Domain:,, Range:1,.41.The relation is not a function. Domain:9,,Range:,. 42.The relation is a function. Domain:,, Range:,. 43.The relation is a function. Domain:5,2,1,.5, 0,1.75,3.5 ,Range:1, 2,3,3.5, 4,5.75, 7.5 .44.The relation is a function. Domain:2,1, 0,5,9,10,13 ,Range:5, 0,3,12, 60, 77,140 .45.The relation is a function. Domain:{2,3,5,11,17} Range:1, 7, 20 .

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Solution Manual for A Graphical Approach to Precalculus with Limits, 7th Edition - Page 14 preview image

Section 1.21146.The relation is not a function. Domain:1, 2,3,5 ,Range: {10,15,19,27}47.From the diagram,( 2)2.f48.From the diagram,(5)12.f49.From the diagram,(11)7.f50.From the diagram,(5)1.f51.f(1) is undefined since 1 is not in the domain of the function.52.f(10) is undefined since 10 is not in the domain of the function.53.( 2)3( 2)46410f  54.( 5)5( 5)625619f  55.2(1)2(1)(1)32134f56.2(2)3(2)2(2)5124511f57.2(4)(4)(4)2164210f   58.2(3)(3)(3)693618f   59.(9)5f60.(12)4f 61.3( 2)( 2)1281242f62.2333(2)(2)(2)642682f63.323(8)8644f64.233( 8)( 8)644f65.Given that( )5 ,fxxthen( )5 ,(1)5(1)55, and(3 )5(3 )15faaf bbbfxxx66.Given that( )5,fxxthen( )5,(1)154, and(3 )35f aaf bbbfxx67.Given that( )25,fxxthen( )25,(1)2(1)522523, andf aaf bbbb(3 )2(3 )565fxxx68.Given that2( ),fxxthen222( ),(1)(1)(1)(1)21, andf aaf bbbbbb22(3 )(3 )9fxxx69.Given that2( )1,fxxthen2222( )1,(1)1(1)1(21)2 , andf aaf bbbbbb 22(3 )1(3 )19fxxx70.(a)Given that2( )24,fxxthen2( )24f aa(b)Given that2( )24,fxxthen22(1)2(1)42(21)4f bbbb222224226bbbb

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Solution Manual for A Graphical Approach to Precalculus with Limits, 7th Edition - Page 15 preview image

12Chapter 1 Linear Functions, Equations, and Inequalities(c)Given that2( )24,fxxthen222(3 )2(3 )42(9)4184fxxxx71.Since( 2)3,fthe point( 2,3)lies on the graph of ƒ.72.Since(3)9.7,f the point(3,9.7)lies on the graph of ƒ.73.Since the point(7,8)lies on the graph of ƒ,(7)8.f74.Since the point( 3, 2)lies on the graph of ƒ,( 3)2.f75.From the graph: (a)( 2)0,f(b)(0)4,f(c)(1)2,fand(d)(4)4.f76.From the graph: (a)( 2)5,f(b)(0)0,f(c)(1)2,fand(d)(4)4.f77.From the graph: (a)( 2)fis undefined , (b)(0)2,f (c)(1)0,fand(d)(4)2.f78.From the graph: (a)( 2)3,f(b)(0)3,f(c)(1)3,fand(d)(4)fis undefined.79.(a)Apple, 216 , Alphabet, 90 , Google, 89 , Microsoft, 85A, The U.S. total revenue in 2011 forApple was $216,000,000 dollars.(b)See Figure 81.(c)Apple, Alphabet, Google, MicrosoftD,216, 90, 89, 85R80.(a)0,1.0 , 2, 2.0 , 7,5.5 , 12,11.0T(b)See Figure 80(c)0, 2, 7, 12 ;1.0, 2.0, 5.5, 11.0DRFigure 79Figure 8081.(a)See Figure 81.(b)(2000)12.8fIn 2000 there were 12,800 radio stations on the air.(c)Domain:1990, 2000, 2005, 2012, 2015 ,Range:10.8,12.8,13.5,15.1,16.4 .82.(a)See Figure 82.(b)(20012)2366fIn 2012, there were $2366 billion spent on personal health care.(c):2010, 2011, 2012, 2013, 2014, 2015DR :2195, 2273, 2366, 2436, 2563, 2717

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Solution Manual for A Graphical Approach to Precalculus with Limits, 7th Edition - Page 16 preview image

Section 1.313Figure 81Figure 82Reviewing Basic Concepts (Sections 1.1 and 1.2)1.See Figure 1.2.The distance is22(6( 4))( 25)10049149.d  The midpoint is46 523,1,.222M3.351.168(31)4.22(12( 4))( 327)256900115634d  5.Using Pythagorean Theorem,222222211616111360060bbbbinches.6.The set25xxis the interval( 2,5].The set4x xis the interval[4,).7.The relation is not a function because it does not pass the vertical line test. Domain:2, 2 ,Range:3,3 .8.See Figure 8.9.Given( )34fxxthen( 5)34( 5)23 and(4)34(4)3416413ff aaaa 10.From the graph,(2)3fand( 1)3.f Figure 1Figure 81.3: Linear Functions1.The graph is shown in Figure 1.(a)x-intercept: 4(b)y-intercept:4(c) Domain:(,) (d) Range:(,) 

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