Solutions Manuals for Calculus Early Transcendentals PDF
Preview Solutions Manuals for Calculus Early Transcendentals
1 FUNCTIONS AND MODELS 1.1 FourWaystoRepresentaFunction 1. Thefunctions()=+√2 and()=+√2 giveexactlythesameoutputvaluesforeveryinputvalue,so − − andareequal. 2 ( 1) 2. ()= − = − =for 1=0,so and[where()=]arenotequalbecause(1)isundefinedand 1 1 − 6 − − (1)=1. 3. (a)Thepoint( 22)liesonthegraphof,so( 2)=2.Similarly,(0)= 2,(2)=1,and(3)25. − − − (b)Onlythepoint( 43)onthegraphhasavalueof3,sotheonlyvalueofforwhich()=3is 4. − − (c)Thefunctionoutputs()arenevergreaterthan3,so() 3fortheentiredomainofthefunction.Thus,() 3for ≤ ≤ 4 4(or,equivalently,ontheinterval[ 44]). − ≤ ≤ − (d)Thedomainconsistsofallvaluesonthegraphof: 4 4 =[ 44].Therangeofconsistsofallthe { |− ≤ ≤ } − valuesonthegraphof: 2 3 =[ 23]. { |− ≤ ≤ } − (e)Forany intheinterval[02],wehave( )( ).[Thegraphrisesfrom(0 2)to(21).]Thus,()is 1 2 1 2 − increasingon[02]. 4. (a)Fromthegraph,wehave( 4)= 2and(3)=4. − − (b)Since( 3)= 1and( 3)=2,orbyobservingthatthegraphofisabovethegraphof at= 3,( 3)islarger − − − − − than( 3). − (c)Thegraphsof andintersectat= 2and=2,so()=()atthesetwovaluesof. − (d)Thegraphof liesbeloworonthegraphoffor 4 2andfor2 3.Thus,theintervalsonwhich − ≤ ≤− ≤ ≤ () ()are[ 4 2]and[23]. ≤ − − (e)()= 1isequivalentto= 1,andthepointsonthegraphof withvaluesof 1are( 3 1)and(4 1),so − − − − − − thesolutionoftheequation()= 1is= 3or=4. − − (f)Forany intheinterval[ 40],wehave( )( ).Thus,()isdecreasingon[ 40]. 1 2 1 2 − − (g)Thedomainof is 4 4 =[ 44].Therangeof is 2 3 =[ 23]. { |− ≤ ≤ } − { |− ≤ ≤ } − (h)Thedomainofis 4 3 =[ 43].Estimatingthelowestpointofthegraphofashavingcoordinates { |− ≤ ≤ } − (005),therangeofisapproximately 05 4 =[054]. { | ≤ ≤ } 5. FromFigure1inthetext,thelowestpointoccursatabout()=(12 85).Thehighestpointoccursatabout(17115). − Thus,therangeoftheverticalgroundaccelerationis 85 115.Writteninintervalnotation,therangeis[ 85115]. − ≤ ≤ − c 2021CengageLearning.AllRightsReserved.Maynotbescanned,copied,orduplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart. 9 ° 10 ¤ CHAPTER1 FUNCTIONSANDMODELS 6. Example1:Acarisdrivenat60mihfor2hours.Thedistance traveledbythecarisafunctionofthetime.Thedomainofthe functionis 0 2 ,whereismeasuredinhours.Therange { | ≤ ≤ } ofthefunctionis 0 120 ,whereismeasuredinmiles. { | ≤ ≤ } Example2:Atacertainuniversity,thenumberofstudents on campusatanytimeonaparticulardayisafunctionofthetimeafter midnight.Thedomainofthefunctionis 0 24 ,whereis { | ≤ ≤ } measuredinhours.Therangeofthefunctionis 0 , { | ≤ ≤ } where isanintegerandisthelargestnumberofstudentson campusatonce. Example3:Acertainemployeeispaid$800perhourandworksa pay maximumof30hoursperweek.Thenumberofhoursworkedis 240 238 roundeddowntothenearestquarterofanhour.Thisemployee’s 236 grossweeklypay isafunctionofthenumberofhoursworked. 4 Thedomainofthefunctionis[030]andtherangeofthefunctionis 2 02004002380024000 . 0 0.25 0.50 0.75 29.5029.75 30 hours { } 7. Wesolve3 5=7for:3 5=7 5= 3+7 = 3 7.Sincetheequationdeterminesexactly − − ⇔ − − ⇔ 5 −5 onevalueofforeachvalueof,theequationdefinesasafunctionof. 8. Wesolve32 2=5for: 32 2=5 2= 32+5 = 32 5.Sincetheequationdetermines − − ⇔ − − ⇔ 2 − 2 exactlyonevalueofforeachvalueof,theequationdefinesasafunctionof. 9. Wesolve2+( 3)2 =5for: 2+( 3)2 =5 ( 3)2 =5 2 3= √5 2 − − ⇔ − − ⇔ − ± − ⇔ =3 √5 2.Someinputvaluescorrespondtomorethanoneoutput.(Forinstance,=1correspondsto=1and ± − to=5.)Thus,theequationdoesnotdefineasafunctionof. 10. Wesolve2+52 =4for: 2+52 =4 52+(2) 4=0 ⇔ − ⇔ 2 (2)2 4(5)( 4) 2 √42+80 √2+20 = − ± − − = − ± = − ± (usingthequadraticformula).Someinput 2(5) 10 5 valuescorrespondtomorethanoneoutput.(Forinstance,=4correspondsto= 2andto=25.)Thus,the − equationdoesnotdefineasafunctionof. 11. Wesolve(+3)3+1=2for: (+3)3+1=2 (+3)3 =2 1 +3= √32 1 ⇔ − ⇔ − ⇔ = 3+√32 1.Sincetheequationdeterminesexactlyonevalueofforeachvalueof,theequationdefinesasa − − functionof. c 2021CengageLearning.AllRightsReserved.Maynotbescanned,copied,orduplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart. ° SECTION1.1 FOURWAYSTOREPRESENTAFUNCTION ¤ 11 12. Wesolve2 =0for: 2 =0 =2 = 2.Someinputvaluescorrespondtomorethan −| | −| | ⇔ | | ⇔ ± oneoutput.(Forinstance,=1correspondsto= 2andto=2.)Thus,theequationdoesnotdefineasafunction − of. 13. Theheight60in(=60)correspondstoshoesizes7and8(=7and=8).Sinceaninputvaluecorrespondstomore thanoutputvalue,thetabledoesnotdefineasafunctionof. 14. Eachyearcorrespondstoexactlyonetuitioncost.Thus,thetabledefinesasafunctionof. 15. No,thecurveisnotthegraphofafunctionbecauseaverticallineintersectsthecurvemorethanonce.Hence,thecurvefails theVerticalLineTest. 16. Yes,thecurveisthegraphofafunctionbecauseitpassestheVerticalLineTest.Thedomainis[ 22]andtherange − is[ 12]. − 17. Yes,thecurveisthegraphofafunctionbecauseitpassestheVerticalLineTest.Thedomainis[ 32]andtherange − is [ 3 2) [ 13]. − − ∪ − 18. No,thecurveisnotthegraphofafunctionsincefor=0, 1,and 2,thereareinfinitelymanypointsonthecurve. ± ± 19. (a)When=1950, 138◦C,sotheglobalaveragetemperaturein1950wasabout138◦C. ≈ (b)When =142 C, 1990. ◦ ≈ (c)Theglobalaveragetemperaturewassmallestin1910(theyearcorrespondingtothelowestpointonthegraph)andlargest in2000(theyearcorrespondingtothehighestpointonthegraph). (d)When=1910, 135 C,andwhen=2000, 144 C.Thus,therangeof isabout[135,144]. ◦ ◦ ≈ ≈ 20. (a)Theringwidthvariesfromnear0mmtoabout16mm,sotherangeoftheringwidthfunctionisapproximately[016]. (b)Accordingtothegraph,theearthgraduallycooledfrom1550to1700,warmedintothelate1700s,cooledagainintothe late1800s,andhasbeensteadilywarmingsincethen.Inthemid19thcentury,therewasvariationthatcouldhavebeen associatedwithvolcaniceruptions. 21. Thewaterwillcooldownalmosttofreezingastheicemelts.Then,when theicehasmelted,thewaterwillslowlywarmuptoroomtemperature. 22. Thetemperatureofthepiewouldincreaserapidly,levelofftooven temperature,decreaserapidly,andthenlevelofftoroomtemperature. c 2021CengageLearning.AllRightsReserved.Maynotbescanned,copied,orduplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart. ° 12 ¤ CHAPTER1 FUNCTIONSANDMODELS 23. (a)Thepowerconsumptionat6AMis500 MWwhichisobtainedbyreadingthevalueofpower when=6fromthe graph.At6PMwereadthevalueof when=18obtainingapproximately730 MW (b)Theminimumpowerconsumptionisdeterminedbyfindingthetimeforthelowestpointonthegraph,=4or4AM.The maximumpowerconsumptioncorrespondstothehighestpointonthegraph,whichoccursjustbefore=12orright beforenoon.Thesetimesarereasonable,consideringthepowerconsumptionschedulesofmostindividualsand businesses. 24. RunnerAwontherace,reachingthefinishlineat100metersinabout15seconds,followedbyrunnerBwithatimeofabout 19seconds,andthenbyrunnerCwhofinishedinaround23seconds.Binitiallyledtherace,followedbyC,andthenA. CthenpassedBtoleadforawhile.ThenApassedfirstB,andthenpassedCtotaketheleadandfinishfirst.Finally, BpassedCtofinishinsecondplace.Allthreerunnerscompletedtherace. 25. Ofcourse,thisgraphdependsstronglyonthe 26.Thesummersolstice(thelongestdayoftheyear)is geographicallocation! aroundJune21,andthewintersolstice(theshortestday) isaroundDecember22.(Exchangethedatesforthe southernhemisphere.) 27. Asthepriceincreases,theamountsolddecreases. 28.Thevalueofthecardecreasesfairlyrapidlyinitially,then somewhatlessrapidly. 29. c 2021CengageLearning.AllRightsReserved.Maynotbescanned,copied,orduplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart. ° SECTION1.1 FOURWAYSTOREPRESENTAFUNCTION ¤ 13 30. (a) (b) (c) (d) 31. (a) (b)9:00AMcorrespondsto=9.When=9,the temperature isabout74 F. ◦ 32. (a) (b)Thebloodalcoholconcentrationrisesrapidly,thenslowly decreasestonearzero. 33. ()=32 +2. − (2)=3(2)2 2+2=12 2+2=12. − − ( 2)=3( 2)2 ( 2)+2=12+2+2=16. − − − − ()=32 +2. − ( )=3( )2 ( )+2=32++2. − − − − (+1)=3(+1)2 (+1)+2=3(2+2+1) 1+2=32+6+3 +1=32+5+4. − − − − 2()=2 ()=2(32 +2)=62 2+4. · − − (2)=3(2)2 (2)+2=3(42) 2+2=122 2+2. − − − [continued] c 2021CengageLearning.AllRightsReserved.Maynotbescanned,copied,orduplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart. ° 14 ¤ CHAPTER1 FUNCTIONSANDMODELS (2)=3(2)2 (2)+2=3(4) 2+2=34 2+2. − − − [()]2= 32 +2 2 = 32 +2 32 +2 − − − =94 33+62 33+2 2+62 2+4=94 63+132 4+4 − − − − − − (+)=3(+)2 (+)+2=3(2+2+2) +2=32+6+32 +2. − − − − − 34. ()= . √+1 0 (0)= =0. √0+1 3 3 (3)= = . √3+1 2 5 5()=5 = . · √+1 √+1 1 1 1 4 2 (4)= (4)= = . 2 2 · 2 · √4+1 √4+1 2 2 2 (2)= ;[()]2 = = . √2+1 √+1 +1 (+) + (+)= = . (+)+1 √++1 ( ) ( )= − = − . − ( )+1 √ +1 − − 35. ()=4+3 2,so(3+)=4+3(3+) (3+)2 =4+9+3 (9+6+2)=4 3 2, − − − − − (3+) (3) (4 3 2) 4 ( 3 ) and − = − − − = − − = 3 . − − 36. () = 3,so(+) = (+)3 = 3+32+32+3, (+) () (3+32+32+3) 3 (32+3+2) and − = − = =32+3+2. 1 1 37. ()= 1,so ()−() = − = − = − = −1(−) = 1 . ( ) ( ) − − − − − − () (1) √+2 √3 38. ()=√+2,so − = − .Dependinguponthecontext,thismaybeconsideredsimplified. 1 1 − − Note:Wemayalsorationalizethenumerator: √+2 √3 √+2 √3 √+2+√3 (+2) 3 − = − = − 1 1 · √+2+√3 ( 1) √ 2+√3 − − − − 1 1 = − = ( 1) √ 2+√3 √+2+√3 − − 39. ()=(+4)(2 9)isdefinedforallexceptwhen0=2 9 0=(+3)( 3) = 3or3,sothe − − ⇔ − ⇔ − domainis = 33 =( 3) ( 33) (3 ). { ∈ | 6 − } −∞ − ∪ − ∪ ∞ c 2021CengageLearning.AllRightsReserved.Maynotbescanned,copied,orduplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart. ° SECTION1.1 FOURWAYSTOREPRESENTAFUNCTION ¤ 15 2+1 40. Thefunction()= isdefinedforallvaluesofexceptthoseforwhich2+4 21=0 2+4 21 − ⇔ − (+7)( 3)=0 = 7or=3.Thus,thedomainis = 73 =( 7) ( 73) (3 ). − ⇔ − { ∈ | 6 − } −∞ − ∪ − ∪ ∞ 41. ()= √32 1isdefinedforallrealnumbers.Infact 3 (),where()isapolynomial,isdefinedforallrealnumbers. − Thus,thedomainisor( ). −∞ ∞ 42. ()=√3 √2+isdefinedwhen3 0 3and2+ 0 2.Thus,thedomainis − − − ≥ ⇔ ≤ ≥ ⇔ ≥− 2 3,or[ 23]. − ≤ ≤ − 43. ()=1 √42 5 isdefinedwhen2 50 ( 5)0.Notethat2 5=0sincethatwouldresultin − − ⇔ − − 6 divisionbyzero.Theexpression( 5)ispositiveif0or5.(SeeAppendixAformethodsforsolving − inequalities.)Thus,thedomainis( 0) (5 ). −∞ ∪ ∞ +1 1 1 44. ()= isdefinedwhen+1=0 [= 1] and1+ =0.Since1+ =0 1 6 6 − +1 6 +1 ⇔ 1+ +1 1 = 1 1= 1 = 2,thedomainis = 2,= 1 =( 2) ( 2 1) ( 1 ). +1 − ⇔ − − ⇔ − { | 6 − 6 − } −∞ − ∪ − − ∪ − ∞ 45. ()= 2 √isdefinedwhen 0and2 √ 0.Since2 √ 0 2 √ √ 2 − ≥ − ≥ − ≥ ⇔ ≥ ⇔ ≤ ⇔ 0 4,thedomainis[04]. ≤ ≤ 46. Thefunction()=√2 4 5isdefinedwhen2 4 5 0 (+1)( 5) 0.Thepolynomial − − − − ≥ ⇔ − ≥ ()=2 4 5maychangesignsonlyatitszeros,sowetestvaluesofontheintervalsseparatedby= 1and − − − =5: ( 2)=70,(0)= 50,and(6)=70.Thus,thedomainof,equivalenttothesolutionintervals − − of() 0,is 1or 5 =( 1] [5 ). ≥ { | ≤− ≥ } −∞ − ∪ ∞ 47. ()=√4 2.Now=√4 2 2 =4 2 2+2=4,so − − ⇒ − ⇔ thegraphisthetophalfofacircleofradius2withcenterattheorigin.Thedomain is 4 2 0 = 4 2 = 2 =[ 22].Fromthegraph, | − ≥ | ≥ { | ≥| |} − therangeis0 2,or[02]. ≤ ≤ 2 4 48. Thefunction()= − isdefinedwhen 2=0 =2,sothe 2 − 6 ⇔ 6 − domainis =2 =( 2) (2 ).Onitsdomain, { | 6 } −∞ ∪ ∞ 2 4 ( 2)(+2) ()= − = − =+2.Thus,thegraphof isthe 2 2 − − line=+2withaholeat(24). c 2021CengageLearning.AllRightsReserved.Maynotbescanned,copied,orduplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart. ° 16 ¤ CHAPTER1 FUNCTIONSANDMODELS 2+2 if 0 49. ()= if 0 ≥ ( 3)=( 3)2+2=11,(0)=0,and(2)=2. − − 5 if 2 50. ()= 1 3 if 2 2 − ≥ ( 3)=5,(0)=5,and(2)= 1(2) 3= 2. − 2 − − +1 if 1 51. ()= ≤− 2 if 1 − ( 3)= 3+1= 2,(0)=02=0,and(2)=22 =4. − − − 1 if 1 52. ()= − ≤ 7 2 if 1 − ( 3)= 1,(0)= 1,and(2)=7 2(2)=3. − − − − if 0 53. = ≥ | | if 0 − 2 if 0 so ()=+ = ≥ | | 0 if 0 Graphtheline=2for 0andgraph=0(theaxis)for0 ≥ c 2021CengageLearning.AllRightsReserved.Maynotbescanned,copied,orduplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart. ° SECTION1.1 FOURWAYSTOREPRESENTAFUNCTION ¤ 17 +2 if +2 0 54. ()= +2 = ≥ | | (+2) if +20 − +2 if 2 = ≥− 2 if 2 − − − 1 3 if 1 3 0 55. ()= 1 3 = − − ≥ | − | (1 3) if 1 30 − − − 1 3 if 1 = − ≤ 3 3 1 if 1 − 3 56. ()= | | Thedomainof is =0 and =if0, = if0. { | 6 } | | | | − Sowecanwrite − = 1 if 0 − ()= =1 if 0 if 1 57. Tograph()= | | | |≤ ,graph= [Figure16] 1 if 1 | | | | for 1 1andgraph=1for1andfor 1. − ≤ ≤ − 1 if 1 − if 1 0 Wecouldrewritef as()= − − ≤ . if 0≤≤1 1 if 1 1 if 1 0 58. ()= 1 = | |− | |− ≥ | |− ( 1) if 10 − | |− | |− 1 if 1 = | |− | |≥ +1 if 1 −| | | | 1 if 1and 0 1 if 1 − | |≥ ≥ − ≥ 1 if 1and0 1 if 1 = − − | |≥ = − − ≤− −+1 if ||1and≥0 −+1 if 0≤1 ( )+1 if 1and0 +1 if 10 − − | | − c 2021CengageLearning.AllRightsReserved.Maynotbescanned,copied,orduplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart. ° 18 ¤ CHAPTER1 FUNCTIONSANDMODELS 59. Recallthattheslopeofalinebetweenthetwopoints(11)and(22)is= 2−1 andanequationoftheline 2 1 − connectingthosetwopointsis =( ).Theslopeofthelinesegmentjoiningthepoints(1 3)and(57)is 1 1 − − − 7 ( 3) 5 − − = ,soanequationis ( 3)= 5( 1).Thefunctionis()= 5 11,1 5. 5 1 2 − − 2 − 2 − 2 ≤ ≤ − 10 10 5 60. Theslopeofthelinesegmentjoiningthepoints( 510)and(7 10)is − − = ,soanequationis − − 7 ( 5) −3 − − 10= 5[ ( 5)].Thefunctionis()= 5+ 5, 5 7. − −3 − − −3 3 − ≤ ≤ 61. Weneedtosolvethegivenequationfor. +( 1)2 =0 ( 1)2 = 1= √ − ⇔ − − ⇔ − ± − ⇔ =1 √ .Theexpressionwiththepositiveradicalrepresentsthetophalfoftheparabola,andtheonewiththenegative ± − radicalrepresentsthebottomhalf.Hence,wewant()=1 √ .Notethatthedomainis 0. − − ≤ 62. 2+( 2)2 =4 ( 2)2 =4 2 2= √4 2 =2 √4 2.Thetophalfisgivenby − ⇔ − − ⇔ − ± − ⇔ ± − thefunction()=2+√4 2, 2 2. − − ≤ ≤ 63. For0 3,thegraphisthelinewithslope 1andintercept3,thatis,= +3.For3 5,thegraphistheline ≤ ≤ − − ≤ withslope2passingthrough(30);thatis, 0=2( 3),or=2 6.Sothefunctionis − − − +3 if 0 3 ()= − ≤ ≤ 2 6 if 3 5 − ≤ 64. For 4 2,thegraphisthelinewithslope 3 passingthrough( 20);thatis, 0= 3[ ( 2)],or − ≤ ≤− −2 − − −2 − − = 3 3.For 22,thegraphisthetophalfofthecirclewithcenter(00)andradius2.Anequationofthecircle −2 − − is2+2 =4,soanequationofthetophalfis=√4 2.For2 4,thegraphisthelinewithslope 3 passing − ≤ ≤ 2 through(20);thatis, 0= 3( 2),or= 3 3.Sothefunctionis − 2 − 2 − 3 3 if 4 2 −2 − − ≤ ≤− ()=√4 2 if 22 3−3 if −2 4 2 − ≤ ≤ 65. Letthelengthandwidthoftherectanglebeand.Thentheperimeteris2+2 =20andtheareais=. 20 2 Solvingthefirstequationfor intermsofgives = − =10 .Thus,()=(10 )=10 2.Since 2 − − − lengthsarepositive,thedomainofis010.Ifwefurtherrestricttobelargerthan,then510wouldbe thedomain. 66. Letthelengthandwidthoftherectanglebeand.Thentheareais =16,sothat =16.Theperimeteris =2+2,so()=2+2(16)=2+32,andthedomainof is0,sincelengthsmustbepositive quantities.Ifwefurtherrestricttobelargerthan,then4wouldbethedomain. c 2021CengageLearning.AllRightsReserved.Maynotbescanned,copied,orduplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart. °
The list of books you might like
