FUNCTIONSANDMODELS
1.1FourWaystoRepresentaFunction
1. Thefunctions ()= + √2 and ()= + √2 giveexactlythesameoutputvaluesforeveryinputvalue,so and areequal.
2. ()= 2 1 = ( 1) 1 = for 1 =0,so and [where ()= ]arenotequalbecause (1) isundefinedand (1)=1.
3. (a)Thepoint ( 2 2) liesonthegraphof ,so ( 2)=2.Similarly, (0)= 2, (2)=1,and (3) 2 5.
(b)Onlythepoint ( 4 3) onthegraphhasa valueof 3,sotheonlyvalueof forwhich ()=3 is 4
(c)Thefunctionoutputs () arenevergreaterthan3,so () ≤ 3 fortheentiredomainofthefunction.Thus, () ≤ 3 for 4 ≤ ≤ 4 (or,equivalently,ontheinterval [ 4 4]).
(d)Thedomainconsistsofall valuesonthegraphof : { | 4 ≤ ≤ 4} =[ 4 4].Therangeof consistsofallthe valuesonthegraphof : { | 2 ≤ ≤ 3} =[ 2 3]
(e)Forany 1 2 intheinterval [0 2],wehave (1 ) (2 ).[Thegraphrisesfrom (0 2) to (2 1).]Thus, () is increasingon [0 2]
4. (a)Fromthegraph,wehave ( 4)= 2 and (3)=4
(b)Since ( 3)= 1 and ( 3)=2,orbyobservingthatthegraphof isabovethegraphof at = 3, ( 3) islarger than ( 3)
(c)Thegraphsof and intersectat = 2 and =2,so ()= () atthesetwovaluesof .
(d)Thegraphof liesbeloworonthegraphof for 4 ≤ ≤−2 andfor 2 ≤ ≤ 3.Thus,theintervalsonwhich () ≤ () are [ 4 2] and [2 3].
(e) ()= 1 isequivalentto = 1,andthepointsonthegraphof with valuesof 1 are ( 3 1) and (4 1),so thesolutionoftheequation ()= 1 is = 3 or =4.
(f)Forany 1 2 intheinterval [ 4 0],wehave (1 ) (2 ).Thus, () isdecreasingon [ 4 0]
(g)Thedomainof is { | 4 ≤ ≤ 4} =[ 4 4].Therangeof is { | 2 ≤ ≤ 3} =[ 2 3].
(h)Thedomainof is { | 4 ≤ ≤ 3} =[ 4 3].Estimatingthelowestpointofthegraphof ashavingcoordinates (0 0 5),therangeof isapproximately { | 0 5 ≤ ≤ 4} =[0 5 4].
5. FromFigure1inthetext,thelowestpointoccursatabout ()=(12 85).Thehighestpointoccursatabout (17 115). Thus,therangeoftheverticalgroundaccelerationis 85 ≤ ≤ 115.Writteninintervalnotation,therangeis [ 85 115]
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6. Example1: Acarisdrivenat 60 mihfor 2 hours.Thedistance traveledbythecarisafunctionofthetime .Thedomainofthe functionis { | 0 ≤ ≤ 2},where ismeasuredinhours.Therange ofthefunctionis { | 0 ≤ ≤ 120},where ismeasuredinmiles.
Example2: Atacertainuniversity,thenumberofstudents on campusatanytimeonaparticulardayisafunctionofthetime after midnight.Thedomainofthefunctionis { | 0 ≤ ≤ 24},where is measuredinhours.Therangeofthefunctionis { | 0 ≤ ≤ }, where isanintegerand isthelargestnumberofstudentson campusatonce.
Example3: Acertainemployeeispaid $8 00 perhourandworksa maximumof 30 hoursperweek.Thenumberofhoursworkedis roundeddowntothenearestquarterofanhour.Thisemployee’s grossweeklypay isafunctionofthenumberofhoursworked Thedomainofthefunctionis [0 30] andtherangeofthefunctionis
{
.Sincetheequationdeterminesexactly onevalueof foreachvalueof ,theequationdefines asafunctionof
8. Wesolve
.Sincetheequationdetermines exactlyonevalueof foreachvalueof ,theequationdefines asafunctionof
9. Wesolve 2 +( 3)
=3 ± √5 2 .Someinputvalues correspondtomorethanoneoutput .(Forinstance, =1 correspondsto =1 and to =5.)Thus,theequationdoes not define asafunctionof
(usingthequadraticformula).Someinput values correspondtomorethanoneoutput .(Forinstance, =4 correspondsto = 2 andto =25.)Thus,the equationdoes not define asafunctionof 11. Wesolve (
= 3+ 3 √2 1.Sincetheequationdeterminesexactlyonevalueof foreachvalueof ,theequationdefines asa functionof .
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12. Wesolve 2 | | =0 for : 2 | | =0 ⇔ | | =2 ⇔ = ±2.Someinputvalues correspondtomorethan oneoutput .(Forinstance, =1 correspondsto = 2 andto =2.)Thus,theequationdoes not define asafunction of
13. Theheight60in ( =60) correspondstoshoesizes7and8 ( =7 and =8).Sinceaninputvalue correspondstomore thanoutputvalue ,thetabledoes not define asafunctionof .
14. Eachyear correspondstoexactlyonetuitioncost .Thus,thetabledefines asafunctionof .
15. No,thecurveisnotthegraphofafunctionbecauseaverticallineintersectsthecurvemorethanonce.Hence,thecurvefails theVerticalLineTest.
16. Yes,thecurveisthegraphofafunctionbecauseitpassestheVerticalLineTest.Thedomainis [ 2 2] andtherange is [ 1 2].
17. Yes,thecurveisthegraphofafunctionbecauseitpassestheVerticalLineTest.Thedomainis [ 3 2] andtherange is [ 3 2) ∪ [ 1 3].
18. No,thecurveisnotthegraphofafunctionsincefor =0, ±1,and ±2,thereareinfinitelymanypointsonthecurve.
19. (a)When =1950, ≈ 13 8◦ C,sotheglobalaveragetemperaturein1950wasabout 13 8◦ C
(b)When =14 2◦ C, ≈ 1990
(c)Theglobalaveragetemperaturewassmallestin1910(theyearcorrespondingtothelowestpointonthegraph)andlargest in2000(theyearcorrespondingtothehighestpointonthegraph).
(d)When =1910, ≈ 13 5◦ C,andwhen =2000, ≈ 14 4◦ C.Thus,therangeof isabout [13 5, 14 4].
20. (a)Theringwidthvariesfromnear 0mm toabout 1 6mm,sotherangeoftheringwidthfunctionisapproximately [0 1 6]
(b)Accordingtothegraph,theearthgraduallycooledfrom 1550to1700,warmedintothelate1700s,cooledagainintothe late1800s,andhasbeensteadilywarmingsincethen.Inthemid19thcentury,therewasvariationthatcouldhavebeen associatedwithvolcaniceruptions.
21. Thewaterwillcooldownalmosttofreezingastheicemelts.Then,when theicehasmelted,thewaterwillslowlywarmuptoroomtemperature.
22. Thetemperatureofthepiewouldincreaserapidly,levelofftooven temperature,decreaserapidly,and thenlevelofftoroomtemperature.
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¤ CHAPTER1 FUNCTIONSANDMODELS
23. (a)Thepowerconsumptionat6 AM is 500MW whichisobtainedbyreadingthevalueofpower when =6 fromthe graph.At6 PM wereadthevalueof when =18 obtainingapproximately 730MW
(b)Theminimumpowerconsumptionisdeterminedby findingthetimeforthelowestpointonthegraph, =4 or 4 AM .The maximumpowerconsumptioncorrespondstothehighestpointonthegraph,whichoccursjustbefore =12 orright beforenoon.Thesetimesarereasonable,consideringthe powerconsumptionschedulesofmostindividualsand businesses.
24. RunnerAwontherace,reachingthe finishlineat 100 metersinabout 15 seconds,followedbyrunnerBwithatimeofabout 19 seconds,andthenbyrunnerCwho finishedinaround 23 seconds.Binitiallyledtherace,followedbyC,andthenA. CthenpassedBtoleadforawhile.ThenApassed firstB,andthenpassedCtotaketheleadand finish first.Finally, BpassedCto finishinsecondplace.Allthreerunnerscompletedtherace.
25. Ofcourse,thisgraphdependsstronglyonthe geographicallocation!
26. Thesummersolstice(thelongestdayoftheyear)is aroundJune21,andthewintersolstice(theshortestday) isaroundDecember22.(Exchangethedatesforthe southernhemisphere.)
27. Asthepriceincreases,theamountsolddecreases.
28. Thevalueofthecardecreasesfairlyrapidlyinitially,then somewhatlessrapidly. 29.
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30. (a) (b) (c) (d)
31. (a)
(b)9:00 AM correspondsto =9.When =9,the temperature isabout 74◦ F
32. (a)
33. ()=3 2 +2
(2)=3(2)2 2+2=12 2+2=12
( 2)=3( 2)2 ( 2)+2=12+2+2=16
()=3 2 +2. ( )=3( )2 (
2 ()=2 · ()=2(3
(b)Thebloodalcoholconcentrationrisesrapidly,thenslowly decreasestonearzero.
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Note: Wemayalsorationalizethenumerator:
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40. Thefunction ()= 2 +1 2 +4 21 isdefinedforallvaluesof exceptthoseforwhich 2 +4 21=0 ⇔ ( +7)( 3)=0 ⇔ = 7 or =3.Thus,thedomainis { ∈
∞).
41. ()= 3 √2 1 isdefinedforallrealnumbers.Infact 3 (),where () isapolynomial,isdefinedforallrealnumbers. Thus,thedomainis or (−∞ ∞)
42. ()= √3 √2+ isdefinedwhen 3
3 and 2+
≥−2.Thus,thedomainis 2 ≤ ≤ 3,or [ 2 3]
43. ()=1 4 √2 5 isdefinedwhen 2 5 0 ⇔ ( 5) 0.Notethat
=0 sincethatwouldresultin divisionbyzero.Theexpression (
5) ispositiveif
0 or 5.(SeeAppendixAformethodsforsolving inequalities.)Thus,thedomainis (−∞ 0) ∪ (5 ∞)
44. ()= +1 1+ 1 +1 isdefinedwhen
45. ()=
46. Thefunction ()= √2 4 5 isdefinedwhen 2 4 5 ≥ 0
.Thepolynomial ()= 2 4 5 maychangesignsonlyatitszeros,sowetestvaluesof ontheintervalsseparatedby = 1 and =5: ( 2)=7 0,
(0)= 5 0,and (6)=7 0.Thus,thedomainof ,equivalenttothesolutionintervals of
(
=4,so thegraphisthetophalfofacircleofradius 2 withcenterattheorigin.Thedomain is
.Fromthegraph, therangeis
48. Thefunction ()= 2 4 2 isdefinedwhen 2 =0 ⇔ =2,sothe domainis { | =2} =(−∞ 2) ∪ (2 ∞).Onitsdomain, ()= 2 4 2 = (
2)( +2) 2 = +2.Thus,thegraphof isthe line = +2 withaholeat (2 4) c ° 2021CengageLearning.AllRightsReserved.Maynotbescanned,copied,or duplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart.
16 ¤ CHAPTER1 FUNCTIONSANDMODELS
54.
55.
Sowecanwrite
59. Recallthattheslope ofalinebetweenthetwopoints (1 1 ) and (2 2 ) is = 2 1 2
1 andanequationoftheline connectingthosetwopointsis 1 = ( 1 ).Theslopeofthelinesegmentjoiningthepoints (1 3) and (5 7) is 7 ( 3) 5 1 = 5 2 ,soanequationis ( 3)= 5 2 ( 1).Thefunctionis ()= 5 2 11 2 , 1 ≤ ≤ 5
60. Theslopeofthelinesegmentjoiningthepoints
( 5)].Thefunctionis
61. Weneedtosolvethegivenequationfor
and
,soanequationis
=1 ± √ .Theexpressionwiththepositiveradicalrepresentsthetophalfoftheparabola,andtheonewiththenegative radicalrepresentsthebottomhalf.Hence,wewant ()=1 √ .Notethatthedomainis ≤ 0.
62.
.Thetophalfisgivenby thefunction
63. For 0 ≤ ≤ 3,thegraphisthelinewithslope 1 and intercept 3,thatis, = +3.For 3 ≤ 5,thegraphistheline withslope 2 passingthrough (3 0); thatis, 0=2( 3),or
()=
=2 6.Sothefunctionis
+3 if 0 ≤ ≤ 3 2 6 if 3
64. For 4 ≤ ≤−2,thegraphisthelinewithslope 3 2 passingthrough ( 2 0); thatis, 0= 3 2 [ ( 2)],or = 3 2 3.For 2 2,thegraphisthetophalfofthecirclewithcenter (0 0) andradius 2.Anequationofthecircle is 2 + 2 =4,soanequationofthetophalfis
≤ 4,thegraphisthelinewithslope 3 2 passing through (2 0); thatis, 0= 3
( 2),or =
3.Sothefunctionis
()=
65. Letthelengthandwidthoftherectanglebe and .Thentheperimeteris 2 +2 =20 andtheareais = . Solvingthe firstequationfor intermsof gives = 20 2 2 =10 .Thus, ()= (10 )=10 2 .Since lengthsarepositive,thedomainof is 0 10.Ifwefurtherrestrict tobelargerthan ,then 5 10 wouldbe thedomain.
66. Letthelengthandwidthoftherectanglebe and .Thentheareais =16,sothat =16.Theperimeteris =2 +2 ,so ()=2 +2(16)=2 +32,andthedomainof is 0,sincelengthsmustbepositive quantities.Ifwefurtherrestrict tobelargerthan ,then 4 wouldbethedomain.
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67. Letthelengthofasideoftheequilateraltrianglebe .ThenbythePythagoreanTheorem,theheight ofthetrianglesatisfies
.Usingtheformulaforthearea ofatriangle, = 1 2 (base)(height),weobtain
,withdomain 0
68. Letthelength,width,andheightoftheclosedrectangularboxbedenotedby , ,and ,respectively.Thelengthistwice thewidth,so =2 .Thevolume oftheboxisgivenby = .Since =8,wehave 8=(2 ) ⇒
2
69. Leteachsideofthebaseoftheboxhavelength ,andlettheheightoftheboxbe .Sincethevolumeis 2,weknowthat 2= 2 ,sothat =22 ,andthesurfaceareais = 2 +4.Thus, ()= 2 +4(22 )= 2 +(8),with domain 0
70. Let and denotetheradiusandtheheightoftherightcircularcylinder,respectively.Thenthevolume isgivenby = 2 ,andforthisparticularcylinderwehave 2 =25 ⇔ 2 = 25 .Solvingfor andrejectingthenegative solutiongives = 5 √ ,so = ()= 5 √ in.
71. Theheightoftheboxis andthelengthandwidthare =20 2
Thesides , ,and mustbepositive.Thus,
72. Theareaofthewindowis
Sincethelengths and mustbepositivequantities,wehave
73. Wecansummarizetheamountofthe finewitha piecewisedefinedfunction.
74. Forthe first 1200kWh, ()=10+0 06
Forusageover 1200kWh,thecostis
()=10+0 06(1200)+0 07( 1200)=82+0 07( 1200) Thus, ()= 10+006 if 0 ≤ ≤ 1200 82+0 07( 1200) if 1200
75. (a)
(b)On $14,000,taxisassessedon $4000,and 10%($4000)=$400. On $26,000,taxisassessedon $16,000,and 10%($10,000)+15%($6000)=$1000+$900=$1900
(c)Asinpart(b),thereis$1000taxassessedon$20,000ofincome,so thegraphof isalinesegmentfrom (10,000 0) to (20,000 1000). Thetaxon$30,000is$2500,sothegraphof for 20,000 is theraywithinitialpoint (20,000 1000) thatpassesthrough (30,000 2500).
76. (a)Becauseanevenfunctionissymmetricwithrespecttothe axis,andthepoint (5 3) isonthegraphofthisevenfunction, thepoint ( 5 3) mustalsobeonitsgraph.
(b)Becauseanoddfunctionissymmetricwithrespecttotheorigin,andthepoint (5 3) isonthegraphofthisoddfunction, thepoint ( 5 3) mustalsobeonitsgraph.
77. isanoddfunctionbecauseitsgraphissymmetricabouttheorigin. isanevenfunctionbecauseitsgraphissymmetricwith respecttothe axis.
78. isnotanevenfunctionsinceitisnotsymmetricwithrespecttothe axis. isnotanoddfunctionsinceitisnotsymmetric abouttheorigin.Hence, is neither evennorodd. isanevenfunctionbecauseitsgraphissymmetricwithrespecttothe axis.
79. (a)Thegraphofanevenfunctionissymmetricaboutthe axis.Wereflectthe givenportionofthegraphof aboutthe axisinordertocompleteit.
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(b)Foranoddfunction, ( )= ().Thegraphofanoddfunctionis symmetricabouttheorigin.Werotatethegivenportionofthegraphof through 180◦ abouttheorigininordertocompleteit.
80. (a)Thegraphofanevenfunctionissymmetricaboutthe axis.Wereflectthe givenportionofthegraphof aboutthe axisinordertocompleteit.
(b)Thegraphofanoddfunctionissymmetricabouttheorigin.Werotatethe givenportionofthegraphof through 180◦ abouttheorigininorderto completeit.
81. ()= 2 +1 .
Since ( )= (), isanoddfunction. 82. ()= 2 4 +1 . ( )= ( )2 ( )4 +1 = 2 4 +1 = ().
Since ( )= (), isanevenfunction.
83. ()= +1 ,so ( )= +1 = 1
Sincethisisneither () nor (),thefunction is neitherevennorodd. 84. ()= || ( )=( ) | | =( ) || = ( ||) = ()
Since ( )= (), isanoddfunction.
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85. ()=1+3 2 4
( )=1+3( )2 ( )4 =1+3 2 4 = ()
Since ( )= (), isanevenfunction.
86. ()=1+3 3 5 ,so ( )=1+3( )3 ( )5 =1+3( 3 ) ( 5 ) =1 3 3 + 5
Sincethisisneither () nor (),thefunction is neitherevennorodd.
87. (i)If and arebothevenfunctions,then ( )= () and ( )= ().Now ( + )(
isan even function.
(ii)If and arebothoddfunctions,then ( )= () and ( )= ().Now (
(
)]= (
+ )(),so + isan odd function.
(iii)If isanevenfunctionand isanoddfunction,then ( + )( )= ( )+ ( )= ()+[ ()]= () (), whichisnot ( + )() nor ( + )(),so + is neither evennorodd.(Exception:if isthezerofunction,then + willbe odd.If isthezerofunction,then + willbe even.)
88. (i)If and arebothevenfunctions,then ( )= () and ( )= ().Now ( )( )= ( ) ( )= () ()=( )(),so isan even function.
(ii)If and arebothoddfunctions,then ( )= () and ( )= ().Now ( )( )= ( ) (
isan even function.
(iii)If isanevenfunctionand isanoddfunction,then ( )( )= ( ) ( )= ()[
1.2MathematicalModels:ACatalogofEssentialFunctions
1. (a) ()= 3 +3 2 isapolynomialfunctionofdegree 3.(Thisfunctionisalsoanalgebraicfunction.)
(b) ()=cos2 sin isatrigonometricfunction.
(c) ()= √3 isapowerfunction.
(d) ()=8 isanexponentialfunction.
(e) = √ 2 +1 isanalgebraicfunction.Itisthequotientofa rootofapolynomialandapolynomialofdegree 2
(f) ()=log10 isalogarithmicfunction.
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2. (a) ()= 32 +2 isarationalfunction.(Thisfunctionisalsoanalgebraicfunction.)
(b) ( )=2 3 isanexponentialfunction.
(c) ()= √ +4 isanalgebraicfunction.Itisarootofapolynomial.
(d) = 4 +5 isapolynomialfunctionofdegree 4
(e) ()= 3 √ isarootfunction.Rewriting () as 13 ,werecognizethefunctionalsoasapowerfunction. (Thisfunctionis,further,analgebraicfunctionbecauseitisa rootofapolynomial.)
(f) = 1 2 isarationalfunction.Rewriting as 2 ,werecognizethefunctionalsoasapowerfunction. (Thisfunctionis,further,analgebraicfunction becauseitisthequotientoftwopolynomials.)
3. Wenoticefromthe figurethat and areevenfunctions(symmetricwithrespecttothe axis)andthat isanoddfunction (symmetricwithrespecttotheorigin).So(b) = 5 mustbe .Since is flatterthan neartheorigin,wemusthave
(c) = 8 matchedwith and(a) = 2 matchedwith
4. (a)Thegraphof =3 isaline(choice ).
(b) =3 isanexponentialfunction(choice ).
(c) = 3 isanoddpolynomialfunctionorpowerfunction(choice ).
(d) = 3 √ = 13 isarootfunction(choice ).
5. Thedenominatorcannotequal0,so 1 sin =0 ⇔ sin =1 ⇔ = 2 +2 .Thus,thedomainof
6. Thedenominatorcannotequal0,so 1 tan =0 ⇔ tan =1 ⇔ = 4 + .Thetangentfunctionisnotdefined if = 2 + .Thus,thedomainof
7. (a)Anequationforthefamilyoflinearfunctionswithslope 2 is = ()=2 + ,where isthe intercept.
(b) (2)=1 meansthatthepoint (2 1) isonthegraphof .Wecanusethe pointslopeformofalinetoobtainanequationforthefamilyoflinear functionsthroughthepoint (2 1) 1= ( 2),whichisequivalent to = +(1 2) inslopeinterceptform.
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(c)Tobelongtobothfamilies,anequationmusthaveslope =2,sotheequationinpart(b), = +(1 2), becomes =2 3.Itisthe only functionthatbelongstobothfamilies.
8. Allmembersofthefamilyoflinearfunctions ()=1+ ( +3) have graphsthatarelinespassingthroughthepoint ( 3 1)
9. Allmembersofthefamilyoflinearfunctions ()= havegraphs thatarelineswithslope 1.The interceptis
10. Wegraph ()=
(
) hastwo intercepts, 0 and .Thecurvehas onedecreasingportionthatbeginsorendsattheoriginandincreasesin lengthas || increases; thedecreasingportionisinquadrantIIfor 0 and inquadrantIVfor 0
11. Because isaquadraticfunction,weknowitisoftheform ()= 2 + + .The interceptis 18,so (0)=18 ⇒ =18 and ()= 2 + +18.Sincethepoints (3 0) and (4 2) lieonthegraphof ,wehave
(3)=0 ⇒ 9 +3 +18=0 ⇒ 3 + = 6 (1)
(4)=2 ⇒ 16 +4 +18=2 ⇒ 4 + = 4 (2) Thisisasystemoftwoequationsintheunknowns and ,andsubtracting (1) from (2) gives =2.From (1), 3(2)+ = 6 ⇒ = 12,soaformulafor is ()=2 2 12 +18
12. isaquadraticfunctionso ()= 2 + + .The yinterceptis 1,so (0)=1 ⇒ =1 and ()= 2 + +1 Sincethepoints ( 2 2) and (1 2 5) lieonthegraphof ,wehave
( 2)=2
(1)=
=
(2) Then (1) +2 (2) givesus 6 = 6 ⇒ = 1 andfrom (2),wehave 1+ = 3
= 25,soaformulafor is ()= 2 2 5 +1
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13. Since ( 1)= (0)= (2)=0, haszerosof 1, 0,and 2,soanequationfor is ()= [ ( 1)]( 0)( 2), or ()= ( +1)( 2).Because (1)=6,we’llsubstitute 1 for and 6 for ()
6= (1)(2)( 1) ⇒−2 =6 ⇒ = 3,soanequationfor is ()= 3( +1)( 2)
14. (a)For =0 02 +8 50,theslopeis 0 02,whichmeansthattheaveragesurfacetemperatureoftheworldisincreasingat arateof 0 02 ◦ C peryear.The interceptis 8 50,whichrepresentstheaveragesurfacetemperaturein ◦ C inthe year1900.
(b) =2100 1900=200 ⇒ =0 02(200)+8 50=12 50 ◦ C
15. (a) =200,so =0 0417 ( +1)=0 0417(200)( +1)=8 34 +8 34.Theslopeis 8 34,whichrepresentsthe changeinmgofthedosageforachildforeachchangeof1yearinage.
(b)Foranewborn, =0,so =834 mg.
16. (a)
17. (a)
(b)Theslopeof 4 meansthatforeachincreaseof 1 dollarfora rentalspace,thenumberofspacesrented decreases by 4.The interceptof 200 isthenumberofspacesthatwouldbeoccupied iftherewerenochargeforeachspace.The interceptof 50 isthe smallestrentalfeethatresultsinnospacesrented.
(b)Theslopeof 9 5 meansthat increases 9 5 degreesforeachincrease of 1◦ C.(Equivalently, increasesby 9 when increasesby 5 and decreasesby 9 when decreasesby 5.)The interceptof 32 istheFahrenheittemperaturecorrespondingtoaCelsius temperatureof 0
18. (a)JariistravelingfastersincethelinerepresentingherdistanceversustimeissteeperthanthecorrespondinglineforJade.
(b)At =0,Jadehastraveled 10miles.At =6,Jadehastraveled 16miles.Thus,Jade’sspeedis
16miles 10miles 6minutes 0minutes =1mimin.Thisis 1mile 1minute × 60minutes 1hour =60mih At =0,Jarihastraveled 0miles.At =6,Jarihastraveled 7miles.Thus,Jari’sspeedis
7miles 0miles 6minutes 0minutes = 7 6 mimin or 7miles 6minutes × 60minutes 1hour =70mih
(c)Frompart(b),wehaveaslopeof 1 (mileminute)forthelinearfunction modelingthedistancetraveledbyJadeand fromthegraphthe interceptis 10.Thus, ()=1 +10= +10.Similarly,wehaveaslopeof 7 6 milesminute for
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Jarianda interceptof 0.Thus,thedistancetraveledbyJariasafunctionoftime (inminutes)ismodeledby ()= 7 6 +0= 7 6
19. (a)Let denotethenumberofchairsproducedinonedayand theassociated cost.Usingthepoints (100 2200) and (300 4800),wegettheslope
4800 2200 300 100 = 2600 200 =13.So 2200=13( 100) ⇔ =13 +900
(b)Theslopeofthelineinpart(a)is 13 anditrepresentsthecost(indollars) ofproducingeachadditionalchair.
(c)The interceptis 900 anditrepresentsthe fixeddailycostsofoperating thefactory.
20. (a)Using inplaceof and inplaceof ,we findtheslopetobe
Soalinearequationis
(b)Letting =1500 weget = 1 4 (1500)+260=635 Thecostofdriving1500milesis$635. (c)
(d)The interceptrepresentsthe fixedcost,$260.
(e)Alinearfunctiongivesasuitablemodelinthissituationbecauseyou have fixedmonthlycostssuchasinsuranceandcarpayments,aswell ascoststhatincreaseasyoudrive,suchasgasoline,oil,andtires,and thecostoftheseforeachadditionalmiledrivenisaconstant.
Theslopeofthelinerepresentsthecostper mile, $0 25
21. (a)Wearegiven changeinpressure 10 feetchangeindepth = 4 34 10 =0 434.Using forpressureand fordepthwiththepoint ( )=(0 15),wehavetheslopeinterceptformoftheline, =0434 +15.
(b)When =100,then 100=0 434 +15 ⇔ 0 434 =85 ⇔ = 85 0 434 ≈ 195 85 feet.Thus,thepressureis 100lbin2 atadepthofapproximately 196 feet.
22. (a) ()= 2 and (0 005)=140,so 140= (0 005) 2 ⇔ =140(0 005)2 =0 0035 (b) ()=00035 2 ,soforadiameterof 0008m theresistanceis (0008)=00035(0008) 2 ≈ 547ohms.
23. If istheoriginaldistancefromthesource,thentheilluminationis ()= 2 = 2 .Movinghalfwaytothelampgives anilluminationof
),sothelightisfourtimesasbright.
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24. (a) = and =39kPa when =0 671m3 ,so 39= 0 671 ⇔ =39(0 671)=26 169
(b)When
=0916,
=26
6,sothepressureisreducedtoapproximately 286kPa
25. (a) = 3 sodoublingthewindspeed gives = (2 )3 =8( 3 ).Thus,thepoweroutputisincreasedbyafactor ofeight.
(b)Theareasweptoutbythebladesisgivenby = 2 ,where isthebladelength,sothepoweroutputis = 3 = 2 3 .Doublingthebladelengthgives
)2 3 =4(2 3 ).Thus,thepoweroutputisincreased byafactoroffour.
(c)Frompart(b)wehave = 2
,and
3 , =30m gives
For =10ms,wehave
26. (a)Wegraph ( )=(567 × 10 8 ) 4 for 100 ≤ ≤ 300:
(b)Fromthegraph,weseethatastemperatureincreases,energyincreases—slowlyat first,butthenatanincreasingrate.
27. (a)Thedataappeartobeperiodicandasineorcosine functionwouldmakethebestmodel.Amodeloftheform ()= cos()+ seemsappropriate.
(b)Thedataappeartobedecreasinginalinearfashion.Amodeloftheform ()= + seemsappropriate.
28. (a)Thedataappeartobeincreasingexponentially.Amodeloftheform ()= · or ()= · + seems appropriate.
(b)Thedataappeartobedecreasingsimilarlytothevaluesofthereciprocalfunction.Amodeloftheform ()= seemsappropriate.
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28 ¤ CHAPTER1 FUNCTIONSANDMODELS
Exercises29 – 33:Somevaluesaregiventomanydecimalplaces.Theresultsmaydependonthetechnologyused roundingislefttothereader. 29. (a)
(b)Usingthepoints (4000 14 1) and (60,000 8 2),weobtain
( 4000) or,equivalently,
Alinearmodeldoesseemappropriate.
(c)Usingacomputingdevice,weobtaintheregressionline = 0 0000997855 +13 950764
Thefollowingcommandsandscreensillustratehowto findtheregressionlineonaTI84Pluscalculator.
Enterthedataintolistone(L1)andlisttwo(L2).Press toentertheeditor.
FindtheregessionlineandstoreitinY1 .Press
Notefromthelast figurethattheregressionlinehasbeenstoredinY1 andthatPlot1hasbeenturnedon(Plot1is highlighted).YoucanturnonPlot1fromtheY=menubyplacingthecursoronPlot1andpressing orby pressing
Nowpress toproduceagraphofthedataandtheregression line.Notethatchoice9oftheZOOMmenuautomaticallyselectsawindow thatdisplaysallofthedata.
(d)When =25,000, ≈ 11456; orabout 115 per 100 population.
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(e)When =80,000, ≈ 5 968; orabouta 6% chance.
(f)When =200,000, isnegative,sothemodeldoesnotapply.
30. (a)Usingacomputingdevice,weobtaintheregressionline =0 01879 +0 30480
(b)Theregressionlineappearstobeasuitablemodelforthedata.
(c)The interceptrepresentsthepercentageoflaboratoryratsthat developlungtumorswhen not exposedtoasbestos fibers.
31. (a)
(b)Usingacomputingdevice,weobtaintheregressionline =1 88074 +82 64974
(c)When =53cm, ≈ 182 3cm.
32. (a)Seethescatterplotinpart(b).Alinearmodelseemsappropriate.
(b)Usingacomputingdevice,weobtaintheregressionline
=0 31567 +8 15578.
(c)For2005, =5 and ≈ 9 73centskWh.For2017, =17 and ≈ 13 52centskWh.
33. (a)Seethescatterplotinpart(b).Alinearmodelseemsappropriate.
(b)Usingacomputingdevice,weobtaintheregressionline
=112486 +60,11986.
(c)For2002, =17 and ≈ 79,242 thousandsofbarrelsperday. For2017, =32 and ≈ 96,115 thousandsofbarrelsperday.
34. (a) =1 000431227 1 499528750
(b)Thepowermodelinpart(a)isapproximately = 1 5 .Squaringbothsidesgivesus 2 = 3 ,sothemodelmatches Kepler’sThirdLaw, 2 = 3
35. (a)If =60,then =070 3 239,soyouwouldexpectto find2speciesofbatsinthatcave.
(b) =4 ⇒ 4=0
333 6,soweestimatethesurfaceareaofthecave tobe 334m2
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36. (a)Usingacomputingdevice,weobtainapowerfunction = ,where 3 1046 and 0 308
(b)If =291,then = 17 8,soyouwouldexpectto find18speciesofreptilesandamphibiansonDominica.
37. Wehave
1.3NewFunctionsfromOldFunctions
1. (a)Ifthegraphof isshifted 3 unitsupward,itsequationbecomes = ()+3.
(b)Ifthegraphof isshifted 3 unitsdownward,itsequationbecomes = () 3
(c)Ifthegraphof isshifted 3 unitstotheright,itsequationbecomes =
( 3)
(d)Ifthegraphof isshifted 3 unitstotheleft,itsequationbecomes = ( +3)
(e)Ifthegraphof isreflectedaboutthe axis,itsequationbecomes = ()
(f)Ifthegraphof isreflectedaboutthe axis,itsequationbecomes = ( )
(g)Ifthegraphof isstretchedverticallybyafactorof 3,itsequationbecomes =3 ()
(h)Ifthegraphof isshrunkverticallybyafactorof 3,itsequationbecomes = 1 3 ()
2. (a)Toobtainthegraphof = ()+8 fromthegraphof = (),shiftthegraph 8 unitsupward.
(b)Toobtainthegraphof = ( +8) fromthegraphof = (),shiftthegraph 8 unitstotheleft.
(c)Toobtainthegraphof =8 () fromthegraphof = (),stretchthegraphverticallybyafactorof 8
(d)Toobtainthegraphof = (8) fromthegraphof = (),shrinkthegraphhorizontallybyafactorof 8
(e)Toobtainthegraphof = () 1 fromthegraphof = (), firstreflectthegraphaboutthe axis,andthenshiftit 1 unitdownward.
(f)Toobtainthegraphof =8 ( 1 8 ) fromthegraphof = (),stretchthegraphhorizontallyandverticallybyafactor of 8.
3. (a) Graph3: Thegraphof isshifted 4 unitstotherightandhasequation = ( 4)
(b) Graph1: Thegraphof isshifted 3 unitsupwardandhasequation = ()+3
(c) Graph4: Thegraphof isshrunkverticallybyafactorof 3 andhasequation = 1 3 () (d) Graph5: Thegraphof isshifted 4 unitstotheleftandreflectedaboutthe axis.Itsequationis = ( +4) (e) Graph2: Thegraphof isshifted 6 unitstotheleftandstretchedverticallybyafactorof 2.Itsequationis
=2 ( +6)
4. (a) = () 3:Shiftthegraphof 3 unitsdown. (b) = ( +1):Shiftthegraphof 1 unittotheleft.
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(c) = 1 2 ():Shrinkthegraphof verticallybya factorof 2
5. (a)Tograph = (2) weshrinkthegraphof horizontallybyafactorof 2
SECTION1.3 NEWFUNCTIONSFROMOLDFUNCTIONS
(d) = ():Reflectthegraphof aboutthe axis.
Thepoint (4 1) onthegraphof correspondstothe point 1 2 · 4 1 =(2 1)
(c)Tograph = ( ) wereflectthegraphof about the axis.
(b)Tograph = 1 2 westretchthegraphof horizontallybyafactorof 2
Thepoint (4 1) onthegraphof correspondstothe point ( 1 4 1)=( 4 1)
Thepoint (4 1) onthegraphof correspondstothe point (2 4 1)=(8 1)
(d)Tograph = ( ) wereflectthegraphof about the axis,thenaboutthe axis.
Thepoint (4 1) onthegraphof correspondstothe point ( 1 4 1 1)=( 4 1)
6. Thegraphof = ()= √3 2 hasbeenshifted 2 unitstotherightandstretchedverticallybyafactorof 2 Thus,afunctiondescribingthegraphis
=2 ( 2)=2 3( 2) ( 2)2
7. Thegraphof = ()= √3 2 hasbeenshifted 4 unitstotheleft,reflectedaboutthe axis,andshifteddownward 1 unit.Thus,afunctiondescribingthegraphis
Thisfunctioncanbewrittenas
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32 ¤ CHAPTER1 FUNCTIONSANDMODELS
8. (a)Thegraphof =1+ √ canbeobtainedfromthegraphof = √ by shiftingitupward 1 unit.
(b)Thegraphof =sin canbeobtainedfromthegraphof =sin bycompressinghorizontallybyafactorof ,giving aperiodof 2 =2.Thegraphof =5sin isthenobtainedbystretchingverticallybyafactorof 5
9. =1+ 2 .Startwiththegraphof = 2 andshift1unitupward
10. =( +1)2 .Startwiththegraphof = 2 andshift 1 unittotheleft.
11. = | +2|.Startwiththegraphof = || andshift 2 unitstotheleft.
12. =1 3 .Startwiththegraphof = 3 ,reflectaboutthe axis,andthenshift 1 unitupward.
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13. = 1 +2.Startwiththegraphof = 1 andshift 2 unitsupward.
14. = √ 1.Startwiththegraphof = √,reflectaboutthe axis,andthenshift 1 unitdownward.
15. =sin4.Startwiththegraphof =sin andcompresshorizontallybyafactorof 4.Theperiodbecomes 24= 2.
16. =1+ 1 2 .Startwiththegraphof = 1 2 andshift 1 unitupward.
17. =2+ +1.Startwiththegraphof = √,shift 1 unittotheleft,andthenshift 2 unitsupward.
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18. = ( 1)2 +3.Startwiththegraphof = 2 ,shift 1 unittotheright,reflectaboutthe axis,andthenshift 3 units upward.
19. = 2 2 +5=( 2 2 +1)+4=( 1)2 +4.Startwiththegraphof = 2 ,shift 1 unittotheright,andthen shift 4 unitsupward.
20. =( +1)3 +2.Startwiththegraphof = 3 ,shift 1 unittotheleft,andthenshift 2 unitsupward.
21. =2 ||.Startwiththegraphof = ||,reflectaboutthe axis,andthenshift2unitsupward.
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22. =2 2cos .Startwiththegraphof =cos ,reflectaboutthe axis,stretchverticallybyafactorof 2,andthenshift 2 unitsupward.
23. =3sin 1 2 +1.Startwiththegraphof =sin ,stretchhorizontallybyafactorof 2,stretchverticallybyafactorof 3, andthenshift 1 unitupward.
24. = 1 4 tan 4 .Startwiththegraphof =tan ,shift 4 unitstotheright,andthencompressverticallybya factorof 4
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25. = |cos |.Startwiththegraphof =cos ,shrinkhorizontallybyafactorof ,andreflectallthepartsofthegraph belowthe axisaboutthe axis.
26. = √ 1 .Startwiththegraphof = √,shift 1 unitdownward,andthenreflecttheportionofthegraphbelowthe axisaboutthe axis.
27. ThisisjustlikethesolutiontoExample4excepttheamplitudeofthecurve(the30◦ NcurveinFigure9onJune21)is
14 12=2.Sothefunctionis ()=12+2sin 2 365 ( 80).March31isthe 90thdayoftheyear,sothemodelgives
(90) ≈ 12 34 h.Thedaylighttime(5:51 AM to6:18 PM )is 12 hoursand 27 minutes,or 12 45 h.Themodelvaluediffers fromtheactualvalueby 12 45 12 34 12 45 ≈ 0009,lessthan 1%
28. UsingasinefunctiontomodelthebrightnessofDeltaCepheiasafunctionoftime,wetakeitsperiodtobe 54 days,its amplitudetobe 0 35 (onthescaleofmagnitude),anditsaveragemagnitudetobe 4 0.Ifwetake =0 atatimeof averagebrightness,thenthemagnitude(brightness)asafunctionoftime indayscanbemodeledbytheformula ()=4
29. Thewaterdepth () canbemodeledbyacosinefunctionwithamplitude 12 2 2 =5m,averagemagnitude 12+2 2 =7m, andperiod 12 hours.Hightideoccurredattime6:45 AM ( =6 75h),sothecurvebeginsacycleattime =6 75h (shift 6.75unitstotheright).Thus, ()=5cos 2 12 ( 6 75) +7=5cos 6 ( 6 75) +7,where isinmetersand isthe numberofhoursaftermidnight.
30. Thetotalvolumeofair () inthelungscanbemodeledbyasinefunctionwithamplitude 2500 2000 2 =250mL,average volume 2500+2000 2 =2250mL,andperiod 4 seconds.Thus, ()=250sin 2 4 +2250=250sin 2 +2250,where isin mL and isinseconds.
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31. (a)Toobtain = (||),theportionofthegraphof = () totherightofthe axisisreflectedaboutthe axis. (b) =sin || (c) = ||
32. Themostimportantfeaturesofthegivengrapharethe interceptsandthemaximum andminimumpoints.Thegraphof =1 () hasverticalasymptotesatthe values wherethereare interceptsonthegraphof = ().Themaximumof 1 onthegraph of = () correspondstoaminimumof 11=1 on =1 ().Similarly,the minimumonthegraphof = () correspondstoamaximumonthegraphof =1 ().Asthevaluesof getlarge(positivelyornegatively)onthegraphof = (),thevaluesof getclosetozeroonthegraphof =1 ()
33. ()= √25 2 isdefinedonlywhen 25 2 ≥ 0 ⇔ 2 ≤ 25 ⇔−5 ≤ ≤ 5,sothedomainof is [ 5 5] For ()= √ +1,wemusthave +1 ≥ 0 ⇔ ≥−1,sothedomainof is [ 1 ∞)
(a) ( + )()= √25 2 + √ +1.Thedomainof + isfoundbyintersectingthedomainsof and : [ 1 5]
(b) ( )()= √25 2 √ +1.Thedomainof isfoundbyintersectingthedomainsof and : [ 1 5]
(c) ( )()= √25 2 √ +1= √ 3 2 +25 +25.Thedomainof isfoundbyintersectingthedomainsof and : [ 1 5]
(d)
inadditiontoanypreviousrestrictions. Thus,thedomainof is ( 1 5]
34. For ()= 1 1 ,wemusthave 1 =0 ⇔
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35. ()= 3 +5 and ()= 3 √.Thedomainofeachfunctionis (−∞ ∞)
+1.Wemusthave =0,sothedomainis (−∞ 0) ∪ (0 ∞).
( ◦ )()= ( ())= 1 = 1 1 = .Since requires
(d) ( ◦ )()= ( ())= (2 +1)=2(2 +1)+1=4 +3.Thedomainis (−∞ ∞)
37. ()= 1 √ and ()= +1.Thedomainof is (0 ∞).Thedomainof is (−∞ ∞).
(a) ( ◦ )()= ( ())= ( +1)= 1 √ +1 .Wemusthave +1 0,or 1,sothedomainis ( 1 ∞)
( ◦ )()= ( ())= 1
(c) ( ◦ )()= ( (
= 1 √ +1.Wemusthave 0,sothedomainis (0 ∞)
.Wemusthave 0,sothedomain is (0 ∞).
(d) ( ◦ )()= ( ())=
)=2 1.Thedomainof is (−∞ 1) ∪ ( 1 ∞).Thedomainof is (−∞ ∞). (a) ( ◦
.Wemusthave 2 =0 ⇔ =0.Thus,thedomain is (−∞ 0) ∪ (0 ∞)
(b) ( ◦ )()= (
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39. ()= 2 and ()=sin .Thedomainof is (−∞ 0) ∪ (0 ∞).Thedomainof is (−∞ ∞)
(a) ( ◦ )()= ( ())= (sin )= 2 sin =2csc .Wemusthave sin =0,sothedomainis { | = , aninteger}
(b) ( ◦ )()= ( ())= 2 =sin 2 .Wemusthave =0,sothedomainis (−∞ 0) ∪ (0 ∞) (c) ( ◦
givesadomain of
Intersectingtherestrictionson givesadomainof
.Intersectingtherestrictonson
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48.
49.
50.
53. Let
54. Let ()=tan , ()= √ +1,and ()=cos .Then (
)=
)
55. (a) ( (3))= (4)=6.(b) ( (2))= (1)=5 (c) ( ◦ )(5)= ( (5))= (3)=5.(d) ( ◦ )(5)= ( (5))= (2)=3
56. (a) ( ( (2)))= ( (3))= (4)=1.(b) ( ◦ ◦ )(1)= ( ( (1)))= ( (3))= (5)=2 (c) ( ◦ ◦ )(1)= ( ( (1)))= ( (5))= (2)=1.(d) ( ◦ ◦ )(3)= ( ( (3)))= ( (4))= (6)=2
57. (a) (2)=5,becausethepoint (2 5) isonthegraphof .Thus, ( (2))= (5)=4,becausethepoint (5 4) isonthe graphof
(b) ( (0))= (0)=3
(c) ( ◦ )(0)= ( (0))= (3)=0 (d) ( ◦ )(6)= ( (6))= (6).Thisvalueisnotdefined,becausethereisnopointonthegraphof thathas coordinate 6
(e) ( ◦ )( 2)= ( ( 2))= (1)=4 (f) ( ◦ )(4)= ( (4))= (2)= 2
58. To findaparticularvalueof ( ()),sayfor =0,wenotefromthegraphthat (0) ≈ 2 8 and (2 8) ≈−0 5.Thus,
( (0)) ≈ (2 8) ≈−0 5.Theothervalueslistedinthetablewereobtainedinasimilarfashion.
))
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59. (a)Usingtherelationship distance = rate time withtheradius asthedistance,wehave ()=60
(b) = 2 ⇒ (
2 .Thisformulagivesustheextentoftherippledarea (incm2 )atanytime
60. (a)Theradius oftheballoonisincreasingatarateof 2cms,so ()=(2cms)( s)=2 (in cm).
(b)Using
Theresult, = 32 3 3 ,givesthevolumeoftheballoon(in cm 3 )asafunctionoftime(in s).
61. (a)Fromthe figure,wehavearighttrianglewithlegs 6 and ,andhypotenuse
BythePythagoreanTheorem,
(b)Using = ,weget =(30 kmh)( hours)=30 (inkm).Thus, = ()=30.
(c) ( ◦ )()= ( ())= (30)= (30
)2 +36= √900
2 +36.Thisfunctionrepresentsthedistancebetweenthe lighthouseandtheshipasafunctionofthetimeelapsedsincenoon.
62. (a) = ⇒ ()=350
(b)ThereisaPythagoreanrelationshipinvolvingthelegswithlengths and 1 andthehypotenusewithlength :
2 +12 = 2 .Thus, ()= √ 2 +1
(c) ( ◦ )()= (())= (350)= (350)2 +1
63. (a) ()=
0 if 0 1 if ≥ 0 (b) ()= 0 if 0 120 if ≥ 0 so ()=120 () (c) Startingwiththeformulainpart(b),wereplace 120 with 240 toreflectthe differentvoltage.Also,becausewearestarting 5 unitstotherightof =0, wereplace with 5.Thus,theformulais ()=240 ( 5)
64. (a) ()= () = 0 if 0 if ≥ 0 (b) ()= 0 if 0 2 if 0 ≤ ≤ 60 so ()=2 (), ≤ 60
(c) ()= 0 if 7 4( 7) if 7 ≤ ≤ 32 so ()=4( 7) ( 7), ≤ 32
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65. If
So ◦ isalinearfunctionwithslope 1 2
66. If ()=1 04,then
Thesecompositionsrepresenttheamount oftheinvestmentafter2,3,and4years. Basedonthispattern,whenwecompose copiesof ,wegettheformula
67. (a)Byexaminingthevariabletermsin and ,wededucethatwemustsquare togettheterms
.Ifwelet
()=3 2 +3 +2=3( 2 + )+2=3( 2 + 1)+5,soweseethat ()= 2 + 1.
68. Weneedafunction sothat ( ())= ( +4)= ()=4 1=4( +4) 17.Soweseethatthefunction mustbe ()=4 17.
69. Weneedtoexamine ( ) ( )=( ◦ )( )= ( ( ))= ( ()) [because iseven] = () Because ( )= (), isanevenfunction.
70. ( )= ( ( ))= ( ()).Atthispoint,wecan’tsimplifytheexpression,sowemighttryto findacounterexampleto showthat isnotanoddfunction.Let ()= ,anoddfunction,and ()= 2 + .Then ()= 2 + whichisneither evennorodd.
Nowsuppose isanoddfunction.Then ( ())= ( ())= ().Hence, ( )= (),andso isoddif both and areodd.
Nowsuppose isanevenfunction.Then ( ())= ( ())= ().Hence, ( )= (),andso isevenif is oddand iseven.
71. (a) ()= ()+ ( ) ⇒ ( )=
).Since ( )= (), isan evenfunction. (b) ()= () (
(
) (
)]=
(
). Since ( )= (), isanoddfunction.
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(c)Foranyfunction withdomain ,definefunctions and asinparts(a)and(b).Then 1 2 iseven, 1 2 isodd,andwe showthat
asdesired.
1.4ExponentialFunctions 1.
3. (a) ()= , 0 (b) (c) (0 ∞) (d)SeeFigures4(c),4(b),and4(a),respectively.
4. (a)Thenumber isthevalueof suchthattheslopeofthetangentlineat =0 onthegraphof = isexactly 1. (b) ≈ 2 71828 (c) ()= c ° 2021CengageLearning.AllRightsReserved.Maynotbescanned,copied,or duplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart.
5. Allofthesegraphsapproach 0 as →−∞,allofthempassthroughthepoint (0 1),andallofthemareincreasingandapproach ∞ as →∞.Thelargerthe base,thefasterthefunctionincreasesfor 0,andthefasteritapproaches 0 as →−∞
Note: Thenotation“ →∞”canbethoughtofas“ becomeslarge”atthispoint. MoredetailsonthisnotationaregiveninChapter2.
6. Thegraphof isthereflectionofthegraphof aboutthe axis,andthe graphof 8 isthereflectionofthatof 8 aboutthe axis.Thegraphof 8 increasesmorequicklythanthatof for 0,andapproaches 0 faster as →−∞.
7. Thefunctionswith basegreaterthan 1 (3 and 10 )areincreasing,whilethose withbaselessthan 1 1 3 and 1 10 aredecreasing.Thegraphof 1 3 isthe reflectionofthatof 3 aboutthe axis,andthegraphof 1 10 isthereflectionof thatof 10 aboutthe axis.Thegraphof 10 increasesmorequicklythanthatof 3 for 0,andapproaches 0 fasteras →−∞.
8. Eachofthegraphsapproaches ∞ as →−∞,andeachapproaches 0 as →∞.Thesmallerthebase,thefasterthefunctiongrowsas →−∞,and thefasteritapproaches 0 as →∞
9. Westartwiththegraphof =3 (Figure15)andshift 1 unitupwardtogetthegraphof ()=3 +1
10. Westartwiththegraphof = 1 2 (Figure3)andstretchverticallybyafactorof 2 toobtainthegraphof =2 1 2 .Then weshiftthegraph 3 unitsdownwardtogetthegraphof ()=2 1 2 3
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11. Westartwiththegraphof = (Figure15)andreflectaboutthe yaxistogetthegraphof = .Thenwereflectthe graphaboutthe xaxistogetthegraphof =
12. Westartwiththegraphof =4 (Figure3)andshift 2 unitstothelefttogetthegraphof =4+2
13. Westartwiththegraphof = (Figure15)andreflectaboutthe axistogetthegraphof = .Thenwecompress thegraphverticallybyafactorof 2 toobtainthegraphof = 1 2 andthenreflectaboutthe axistogetthegraph of = 1 2 .Finally,weshiftthegraphoneunitupwardtogetthegraphof =1 1 2 .
14. Westartwiththegraphof = (Figure15)and reflecttheportionofthegraphinthe firstquadrant aboutthe axistoobtainthegraphof = ||
15. (a)To findtheequationofthegraphthatresultsfromshiftingthegraphof = twounitsdownward,wesubtract 2 fromthe originalfunctiontoget = 2
(b)To findtheequationofthegraphthatresultsfromshiftingthegraphof = twounitstotheright,wereplace with 2 intheoriginalfunctiontoget = 2 .
(c)To findtheequationofthegraphthatresultsfromreflectingthegraphof = aboutthe xaxis,wemultiplytheoriginal functionby 1 toget = .
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(d)To findtheequationofthegraphthatresultsfromreflectingthegraphof = aboutthe yaxis,wereplace with in theoriginalfunctiontoget = .
(e)To findtheequationofthegraphthatresultsfromreflectingthegraphof = aboutthe xaxisandthenaboutthe yaxis, we firstmultiplytheoriginalfunctionby 1 (toget = )andthenreplace with inthisequationto get =
16. (a)Thisreflectionconsistsof firstreflectingthegraphaboutthe axis(givingthegraphwithequation = ) andthenshiftingthisgraph 2 4=8 unitsupward.Sotheequationis = +8
(b)Thisreflectionconsistsof firstreflectingthegraphaboutthe axis(givingthegraphwithequation = ) andthenshiftingthisgraph 2 2=4 unitstotheright.Sotheequationis = ( 4)
17. (a)Thedenominatoriszerowhen
thefunction
(b)Thedenominatorisneverequaltozero,sothefunction
18. (a)Thefunction
(b)Thesineandexponentialfunctionshavedomain ,so ()=sin( 1)
19. Use = withthepoints (1
20. Use = withthepoints (
.Usingthisand thepoint
.The functionis ()=2( 2 3 ) .
21. If ()=5 ,then (
22. SupposethemonthisFebruary.Yourpaymentonthe28thdaywouldbe 228 1 =227 =134,217,728 cents,or $1,342,177.28.Clearly,thesecondmethodofpaymentresultsinalargeramountforanymonth.
23. 2 ft =24 in, (24)=242 in =576 in =48 ft. (24)=224 in =224 (12 5280) mi ≈ 265 mi
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24. Weseefromthegraphsthatfor lessthanabout 1 8, ()=5 ()= 5 ,andthennearthepoint (1 8 17 1) thecurves intersect.Then () () from ≈ 1 8 until =5.At (5 3125) thereisanotherpointofintersection,andfor 5 we seethat () ().Infact, increasesmuchmorerapidlythan beyondthatpoint.
25. Thegraphof finallysurpassesthatof at ≈ 35 8
26. Wegraph = and =1,000,000,000 anddeterminewhere
=1 × 109 .Thisseemstobetrueat ≈ 20 723,so 1 × 109 for 20 723
27. (a)
(b)Usingagraphingcalculator,weobtaintheexponential curve ()=36 89301(1 06614)
(c)UsingtheTRACEandzoomingin,we findthatthebacteriacount doublesfrom 37 to 74 inabout 1087 hours.
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28. Let =0 correspondto1900togetthemodel = ,where 80 8498 and 1 01269.Toestimatethepopulationin 1925,let =25 toobtain 111 million.Topredictthepopulationin2020,let =120 toobtain 367 million.
29. (a)Threehoursrepresents6doublingperiods (onedoublingperiodis30minutes).Thus, 500 26 =32,000.
(b)In hours,therewillbe 2 doublingperiods.Theinitialpopulationis 500, sothepopulation attime is =500 22
(c) = 40 60 = 2 3 ⇒ =500 · 22(23) ≈ 1260
(d)Wegraph 1 =500 22 and 2 =100,000.Thetwocurvesintersectat ≈ 3 82,sothepopulationreaches 100,000 inabout 3 82 hours.
30. (a)Let betheinitialpopulation.Since 18 yearsis 3 doublingperiods, · 23 =600 ⇒ = 600 8 =75.Theinitial squirrelpopluationwas 75.
(b)Aperiodof yearscorrespondsto 6 doublingperiods,sotheexpectedsquirrelpopulation yearsafterintroduction is =75 26
(c)Tenyearsfromnowwillbe 18+10=28 yearsfromintroduction.The populationisestimatedtobe =75 · 2286 ≈ 1905 squirrels.
31. Halfof 76 0 RNAcopiespermL,correspondingto =1,is 38 0 RNAcopiespermL.Usingthegraphof inFigure11,we estimatethatittakesabout 3 5 additionaldaysforthepatient’sviralloadtodecreaseto 38 RNAcopiespermL.
32. (a)Theexponentialdecaymodelhastheform ()= 1 2 1 5 ,where isthe numberofhoursaftermidnightand () istheBAC.Wearegiventhat
(0)=014,so =014,andthemodelis ()=014
(b)Fromthegraph,weestimatethattheBACis 0 08gdL when ≈ 1 2 hours.
33. Fromthegraph,itappearsthat isanoddfunction( isundefinedfor =0). Toprovethis,wemustshowthat ( )= ()
so isanoddfunction.
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34. We’llstartwith = 1 andgraph ()= 1 1+ for =0 1, 1,and 5
Fromthegraph,weseethatthereisahorizontalasymptote =0 as →−∞ andahorizontalasymptote =1 as →∞.If =1,the yinterceptis 0 1 2
As getssmaller(closeto 0),thegraphof movesleft.As getslarger,thegraph of movesright.
As changesfrom 1 to 0,thegraphof isstretchedhorizontally.As changesthroughlargenegativevalues,thegraphof iscompressedhorizontally. (Thistakescareofnegativesvaluesof .)
If ispositive,thegraphof isreflectedthroughthe yaxis.
Last,if =0,thegraphof isthehorizontalline =1(1+ )
35. Wegraphthefunction ()= 2 ( + ) for =1, 2,and 5.Because
(0)= ,the yinterceptis ,sothe yinterceptmovesupwardas increases. Noticethatthegraphalsowidens,becoming flatternearthe yaxisas increases.
1.5InverseFunctionsandLogarithms
1. (a)SeeDefinition1. (b)ItmustpasstheHorizontalLineTest.
2. (a) 1 ( )= ⇔ ()= forany in .Thedomainof 1 is andtherangeof 1 is . (b)SeethestepsinBox5.
(c)Reflectthegraphof abouttheline = .
3. isnotonetoonebecause 2 =6,but (2)=2 0= (6).
4. isonetoonebecauseitnevertakesonthesamevaluetwice.
5. Wecoulddrawahorizontallinethatintersectsthegraphinmorethanonepoint.Thus,bytheHorizontalLineTest,the functionisnotonetoone.
6. Nohorizontallineintersectsthegraphmorethanonce. Thus,bytheHorizontalLineTest,thefunctionisonetoone.
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7. Nohorizontallineintersectsthegraphmorethanonce. Thus,bytheHorizontalLineTest,thefunctionisonetoone.
8. Wecoulddrawahorizontallinethatintersectsthegraphinmorethanonepoint.Thus,bytheHorizontalLineTest,the functionisnotonetoone.
9. Thegraphof ()=2 3 isalinewithslope 2.ItpassestheHorizontalLineTest,so isonetoone.
Algebraicsolution: If 1 = 2 ,then 2
toone.
10. Thegraphof ()= 4 16 issymmetricwithrespecttothe axis.Pickany valuesequidistantfrom0to findtwoequal functionvalues.Forexample, ( 1)= 15 and (1)= 15,so isnotonetoone.
11. Nohorizontallineintersectsthegraphof ()= 3 +4 morethanonce.Thus,bytheHorizontalLineTest,thefunctionis onetoone.
Algebraicsolution: If 1 =
,then
12. Thegraphof ()= 3 √ passestheHorizontalLineTest,so isonetoone.
13. ()=1 sin (0)=1 and ( )=1,so isnotonetoone.
14. Thegraphof ()= 4 1 passestheHorizontalLineTestwhen isrestrictedtotheinterval [0,10],so isonetoone.
15. Afootballwillattaineveryheight uptoitsmaximumheighttwice:onceonthewayup,andagainonthewaydown. Thus,evenif 1 doesnotequal 2 , (1 ) mayequal (2 ),so isnot 11
16. isnot 11 becauseeventuallyweallstopgrowingandtherefore,therearetwotimesatwhichwehavethesameheight.
17. (a)Since is11, (6)=17 ⇔ 1 (17)=6. (b)Since is11, 1 (3)=2 ⇔ (2)=3
18. First,wemustdetermine suchthat ()=3.Byinspection,weseethatif =1,then (1)=3.Since is 11 ( isan increasingfunction),ithasaninverse,and 1 (3)=1.If isa 11 function,then ( 1 ())= ,so ( 1 (2))=2.
19. First,wemustdetermine suchthat ()=4.Byinspection,weseethatif =0,then ()=4.Since is 11 ( isan increasingfunction),ithasaninverse,and 1 (4)=0
20. (a) is11becauseitpassestheHorizontalLineTest.
(b)Domainof =[ 3 3]= Rangeof 1 .Rangeof =[ 1 3]= Domainof 1
(c)Since (0)=2, 1 (2)=0
(d)Since ( 1 7) ≈ 0, 1 (0) ≈−1 7
21. Wesolve = 5 9 ( 32) for : 9 5 = 32 ⇒ = 9 5 +32.Thisgivesusaformulafortheinversefunction,that is,theFahrenheittemperature asafunctionoftheCelsiustemperature ≥−45967 ⇒ 9
+32 ≥−45967 ⇒ 9 5 ≥−491 67 ⇒ ≥−273 15,thedomainoftheinversefunction.
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22.
Thisformulagivesusthespeed
23. Firstnotethat
24. Completingthesquare,wehave
oftheparticleintermsofitsmass
25. Firstwrite
Wewrite
32. = ()=1+ ⇒ = 1 ⇒− =ln( 1) ⇒ = ln( 1).Interchange and : = ln( 1)
So 1 ()= ln( 1).Fromthegraph,weseethat and 1 are reflectionsabouttheline =
33. Reflectthegraphof abouttheline = .Thepoints ( 1 2), (1 1), (2 2),and (3 3) on arereflectedto ( 2
1
34. Reflectthegraphof abouttheline =
35. (a) = ()= √1 2 (0 ≤ ≤ 1 andnotethat ≥ 0) ⇒ 2 =1 2 ⇒ 2 =1 2
1 ()= √1 2 , 0 ≤ ≤ 1.Weseethat 1 and arethesame function.
(b)Thegraphof istheportionofthecircle 2 + 2 =1 with 0 ≤ ≤ 1 and 0 ≤ ≤ 1 (quartercircleinthe firstquadrant).Thegraphof issymmetric withrespecttotheline = ,soitsreflectionabout = isitself,thatis, 1 =
36. (a) = ()= 3 √1
3 ⇒ 3 =1
3 ⇒ 3
3 ⇒ = 3 1 3 .So 1 ()= 3 √1 3 .Weseethat and 1 arethe samefunction.
(b)Thegraphof issymmetricwithrespecttotheline = ,soitsreflection about = isitself,thatis, 1 =
37. (a)Itisdefinedastheinverseoftheexponentialfunctionwithbase ,thatis, log = ⇔ = (b) (0 ∞) (c) (d)SeeFigure11.
38. (a)Thenaturallogarithmisthelogarithmwithbase ,denoted ln (b)Thecommonlogarithmisthelogarithmwithbase 10,denoted log (c)SeeFigure13.
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39. (a)
40.
41. (a)
48. (a) log3 12= ln12 ln3 ≈ 2 261860 (b) log12 6= ln6 ln12 ≈ 0 721057
49. Tographthesefunctions,weuse log1 5 = ln ln1 5 and log50 = ln ln50
Thesegraphsallapproach −∞ as → 0+ ,andtheyallpassthroughthe point (1 0).Also,theyareallincreasing,andallapproach ∞ as →∞
Thefunctionswithlargerbasesincreaseextremelyslowly,andtheoneswith smallerbasesdososomewhatmorequickly.Thefunctionswithlargebases approachthe axismorecloselyas → 0+
50. Weseethatthegraphof ln isthereflectionofthegraphof aboutthe line = ,andthatthegraphof log8 isthereflectionofthegraphof 8 aboutthesameline.Thegraphof 8 increasesmorequicklythanthatof
Alsonotethat log8 →∞ as →∞ moreslowlythan ln
51. 3 ft =36 in,soweneed suchthat log2 =36 ⇔
.Inmiles,thisis 68,719,476,736 in 1 ft 12 in 1 mi 5280 ft ≈ 1,084,587 7 mi.
,
,
,
52.
Fromthegraphs,weseethat ()= 0 1 ()=ln forapproximately 0 3 06,andthen () () for 3 06 3 43 × 1015 (approximately).Atthatpoint,thegraphof finallysurpassesthegraphof forgood.
53. (a)Shiftthegraphof =log10 fiveunitstotheleftto obtainthegraphof =log10 ( +5).Notethevertical asymptoteof = 5
=log10 =log10 ( +5)
(b)Reflectthegraphof =ln aboutthe axistoobtain thegraphof = ln =ln = ln
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54. (a)Reflectthegraphof =ln aboutthe axistoobtain thegraphof =ln( ) =ln =ln( )
55. (a)Thedomainof ()=ln +2 is 0 andtherangeis
(b) =0 ⇒ ln
(c)Weshiftthegraphof =ln twounitsupward.
(b)Reflecttheportionofthegraphof =ln totheright ofthe axisaboutthe axis.Thegraphof =ln || isthatreflectioninadditiontotheoriginalportion. =ln =ln ||
56. (a)Thedomainof ()=ln( 1) 1 is 1 andtherangeis
(b) =0 ⇒ ln( 1) 1=0 ⇒ ln( 1)=1 ⇒ 1=
(c)Weshiftthegraphof =ln oneunittotherightandoneunitdownward.
57. (a) ln(4 +2)=3
63. (a)Wemusthave
+3 0 ⇒
=ln( +3),so 1 ()=ln( +3)
3,whichistrueforanyreal ,sothedomainof 1 is
64. (a)By(9), ln300 =300 and ln( 300 )=300 (b)Acalculatorgives ln300 =300 andanerrormessagefor ln( 300 ) because 300 islargerthanmostcalculatorscan evaluate.
65. Weseethatthegraphof = ()= √3 + 2 + +1 isincreasing,so is 11 Enter = 3 + 2 + +1 anduseyourCAStosolvetheequationfor .You willlikelygettwo(irrelevant)solutions involvingimaginaryexpressions,aswell asonewhichcanbesimplifiedto
66. (a)Dependingonthesoftwareused,solving =
4 for maygivesixsolutionsoftheform =
∈ 0 4 27 . Ifwesolve = 6 + 4 for usingMaple,wegetthetworealsolutions ±
13 (
23 2
13 +4) 13 , where =108 +12 √3 (27 4),andtheinversefor = 6 + 4 ( ≥ 0)isthepositivesolution,whosedomain is 4 27 ∞
[continued]
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(a)
Mathematicaalsogivestworealsolutions,equivalenttothoseofMaple.
Thepositiveoneis
.Althoughthisexpressionalsohasdomain
,Mathematicaismysteriouslyabletoplotthesolutionforall
Formula,wecanwritethisas
bacteria(giventhenumber
68. (a)Wewrite = 0 (1 ) andsolvefor :
.UsingtheChangeofBase
.Thisfunctiontellsushowlongitwilltaketoobtain
.Thisformulagivesthetime(inseconds)neededafteradischargeto obtainagivencharge
(b)Weset =0 90 and =50 toget = 50ln
= 50ln(0 1) ≈ 115 1 seconds.Itwilltake approximately 115 seconds—justshyoftwominutes—torechargethecapacitorsto 90% ofcapacity.
69. (a) cos 1 ( 1)= because cos = 1 and isintheinterval [0 ] (therangeof cos 1 ). (b)sin 1 (05)= 6 because sin 6 =05 and 6 isintheinterval 2 2 (therangeof sin 1 ).
70. (a) tan 1 √3= 3 because tan 3 = √3 and 3 isintheinterval 2 2 (therangeof tan 1 ). (b) arctan( 1)= 4 because tan 4 = 1 and 4 isintheinterval 2 2 (therangeof arctan).
71. (a) csc 1 √2= 4 because csc
(therangeof csc 1 ). (b) arcsin1= 2 because sin 2 =1 and
(therangeof arcsin).
(a)
(b)Let =sin 1 5 13 [seethe figure]. cos2sin 1 5 13 =cos2
75. Let =sin 1 .Then 2
76. Let =sin 1 .Then sin = ,sofromthetriangle(which illustratesthecase 0),weseethat
tan(sin 1 )=tan =
77. Let =tan 1 .Then tan = ,sofromthetriangle(which illustratesthecase 0),weseethat
sin(tan 1 )=sin =
78. Let =arccos .Then cos = ,sofromthetriangle(which illustratesthecase 0),weseethat
sin(2arccos )=sin2 =2sin cos =2(
79. Thegraphof sin 1 isthereflectionofthegraphof sin abouttheline =
80. Thegraphof tan 1 isthereflectionofthegraphof tan abouttheline =
81. ()=sin 1 (3 +1)
Domain ( )= { | 1 ≤ 3 +1 ≤ 1} = { | 2 ≤ 3
Range (
82. (a) ()=sinsin 1
Sinceonefunctionundoeswhattheotheronedoes,wegetthe identityfunction, = ,ontherestricteddomain 1 ≤ ≤ 1
(b) ()=sin 1 (sin )
Thisissimilartopart(a),butwithdomain
Equationsfor onintervalsoftheform
2 + 2 + ,foranyinteger ,canbe foundusing ()=( 1) +( 1)+1
Thesinefunctionismonotoniconeachoftheseintervals,andhence,sois (butinalinearfashion).
83. (a)Ifthepoint ( ) isonthegraphof = (),thenthepoint ( ) isthatpointshifted unitstotheleft.Since is 11,thepoint () isonthegraphof = 1 () andthepointcorrespondingto ( ) onthegraphof is ( ) onthegraphof 1 .Thus,thecurve’sreflectionisshifted down thesamenumberofunitsasthecurveitselfis shiftedtotheleft.Soanexpressionfortheinversefunctionis 1 ()= 1 ()
(b)Ifwecompress(orstretch)acurvehorizontally,thecurve’sreflectionintheline = iscompressed(orstretched) vertically bythesamefactor.Usingthisgeometricprinciple,weseethattheinverseof ()= () canbeexpressedas 1 ()=(1) 1 ()
1Review
1. False.Let
2. False.Let ()= 2 .Then ( 2)=4= (2),but 2 =2
3. False.Let ()= 2 .Then
4. True.Theinversefunction 1 ofaonetoonefunction is defined by 1 ( )= ⇔ ()=
5. True.SeetheVerticalLineTest.
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6. False.Let ()= 2 and ()=2.Then (
7. False.Let ()= 3 .Then isoneto
8. True.Wecandivideby since =0 forevery
9. True.Thefunction ln isanincreasingfunctionon (0
10. False.Let
1=6 =1=(ln )6 .What is true,however, isthat ln( 6 )=6ln for 0
11. False.Let = 2 and =
=ln =1,soingeneralthestatement isfalse.What is true,however,isthat ln =ln ln
12. False.Itistruethat tan 3 4 = 1,butsincetherangeof
13. False.Forexample, tan 1 20 isdefined; sin 1 20 and cos 1 20 arenot.
14. False.Forexample,if = 3,then ( 3)2 = √9=3,not 3
1. (a)When =2, ≈ 27.Thus, (2) ≈ 27.
(b) ()=3 ⇒ ≈ 23, 56
(c)Thedomainof is 6 ≤ ≤ 6,or [ 6 6]
(d)Therangeof is 4 ≤ ≤ 4,or [ 4 4]
(e) isincreasingon [ 4 4],thatis,on 4 ≤ ≤ 4
(f) isnotonetoonebecauseitfailstheHorizontalLineTest.
(g) isoddbecauseitsgraphissymmetricabouttheorigin.
2. (a)When =2, =3.Thus, (2)=3
(b) isonetoonebecauseitpassestheHorizontalLineTest.
(c)When =2, ≈ 02.So 1 (2) ≈ 02
(d)Therangeof is [ 1 3 5],whichisthesameasthedomainof 1
(e)Wereflectthegraphof throughtheline = toobtainthegraphof 1
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4. Therewillbesomeyieldwithnofertilizer,increasingyieldswithincreasing fertilizeruse,alevelingoffofyieldsatsomepoint,anddisasterwithtoo muchfertilizeruse.
5. ()=2(3 1).Domain:
Range:allrealsexcept 0 ( =0 isthehorizontalasymptotefor .)
6.
8. = ()=3+cos2.Domain: =(−∞ ∞)
takesonallrealnumbersand,hence,therangeis
9. (a)Toobtainthegraphof = ()+5,weshiftthegraphof = ()5 unitsupward.
(b)Toobtainthegraphof = ( +5),weshiftthegraphof = ()5 unitstotheleft.
(c)Toobtainthegraphof =1+2 (),westretchthegraphof = () verticallybyafactorof 2,andthenshiftthe resultinggraph 1 unitupward.
(d)Toobtainthegraphof = ( 2) 2,weshiftthegraphof = ()2 unitstotheright(forthe“ 2”insidethe parentheses),andthenshifttheresultinggraph 2 unitsdownward.
(e)Toobtainthegraphof = (),wereflectthegraphof = () aboutthe axis.
(f)Toobtainthegraphof = 1 (),wereflectthegraphof = () abouttheline = (assuming isone–toone).
10. (a)Toobtainthegraphof = ( 8),weshiftthe graphof = () right 8 units.
(b)Toobtainthegraphof = (),wereflectthe graphof = () aboutthe axis.
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62 ¤ CHAPTER1 FUNCTIONSANDMODELS
(c)Toobtainthegraphof =2 (),wereflectthe graphof = () aboutthe axis,andthenshiftthe resultinggraph 2 unitsupward.
(d)Toobtainthegraphof = 1 2 () 1,weshrinkthe graphof = () byafactorof 2,andthenshiftthe resultinggraph 1 unitdownward.
(e)Toobtainthegraphof = 1 (),wereflectthe graphof = () abouttheline =
11. ()= 3 +2.Startwiththegraphof = 3 and shift 2 unitsupward.
(f)Toobtainthegraphof = 1 ( +3),wereflectthe graphof = () abouttheline = [seepart(e)], andthenshifttheresultinggraphleft 3 units.
12 ()=( 3)2 .Startwiththegraphof = 2 and shift 3 unitstotheright.
13. = √ +2.Startwiththegraphof = √ andshift 2 unitstotheleft.
14 =ln( +5).Startwiththegraphof =ln and shift 5 unitstotheleft.
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15. ()=1+cos2.Startwiththegraphof =cos ,compresshorizontallybyafactorof 2,andthenshift 1 unitupward.
16. ()= +2.Startwiththegraphof = ,reflectaboutthe xaxis,andthenshift 2 unitsupward.
17. ()=1+0 5 .Startwiththegraphof =0 5 = 1 2 andshift 1 unitupward.
18. ()= if 0 1 if ≥ 0
On (−∞ 0),graph = (thelinewithslope 1 and intercept 0)
withopenendpoint (0 0)
On [0 ∞),graph = 1 (thegraphof = shifted 1 unitdownward) withclosedendpoint (0 0).
19. (a) ()=2 5 3 2 +2 ⇒ ( )=2( )5
and ( ) = (), isneitherevennorodd. (b) (
()=1+sin ⇒ ( )=1+sin( )=1 sin .Now
( ) = () and ( ) = (),so isneither evennorodd.
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()=1 cos2
.Since
(
iseven.
) = () and ( ) = (), isneitherevennorodd.
20. Forthelinesegmentfrom
equivalently,
positive ).Thus,
21. ()=ln
(
0 =1,or (1 ∞) (d) ( ◦ )()= ( ())= ( 2 9)=( 2 9)2 9.Domain:
22. Let ()= + √, ()= √,and ()=1.Then ( ◦ ◦ )()= 1 + √ = ()
23.
Morethanonemodelappearstobeplausible.Yourchoiceofmodeldepends onwhetheryouthinkmedicaladvanceswillkeepincreasinglifeexpectancy,or ifthereisboundtobeanaturallevelingoffoflifeexpectancy.Alinearmodel, =0 2441 413 3960,givesusanestimateof 82 1 yearsforthe year 2030
24. (a)Let denotethenumberoftoasterovensproducedinoneweekand theassociatedcost.Usingthepoints (1000 9000) and (1500 12,000),wegetanequationofaline:
9000= 12,000 9000 1500 1000 ( 1000) ⇒
=6( 1000)+9000 ⇒ =6 +3000
(b)Theslopeof 6 meansthateachadditionaltoasterovenproducedadds $6 totheweeklyproductioncost.
(c)The interceptof 3000 representstheoverheadcost—thecostincurredwithoutproducinganything.
25. Thevalueof forwhich ()=2 +4 equals 6 willbe 1 (6).Tosolve 2 +4 =6,weeitherobservethatletting =1 givesusequality,orwegraph 1 =2 +4 and 2 =6 to findtheintersectionat =1.Since (1)=6, 1 (6)=1.
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29. (a) 2ln5 = ln52 =52 =25
(b) log
(c)Let =arcsin 4 5 ,so sin = 4 5 .Drawarighttrianglewithangle asshown inthe figure.BythePythagoreanTheorem,theadjacentsidehaslength 3, and tanarcsin 4 5 =tan = opp adj = 4 3
30. (a) ln 1 3 =ln 3 = 3 (b) sin(tan 1 1)=sin 4 = √2 2 (c) 10 3log4 =10log4 3 =4 3 = 1 43 = 1 64 31.
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37. (a)Thehalf
lifeoftheviruswiththistreatmentiseightdaysand
(b)Theviralloadishalvedevery
(c)
lives,sotheviralloadafter 24 daysis
RNAcopies
(d)Usingthefunctionfrompart(c),wehave
38. (a) Thepopulationwouldreach 900 inabout 4 4 years.
(b)
thisisthetimerequiredforthepopulationtoreachagivennumber (c)
PRINCIPLESOFPROBLEMSOLVING
1. Byusingtheareaformulaforatriangle, 1 2 (base)(height),intwoways,weseethat
2. RefertoExample1,whereweobtained = 2 100 2 .The 100 camefrom 4 timestheareaofthetriangle.Inthiscase,theareaofthetriangleis
.Thus,
3.
(Equation1)or 4
| +1| =3 (Equation2). If +1
If ≥ 3,wemusthave 1 ( 3) ≥ 5 ⇔ 2 ≥ 5,whichisfalse. Allthreecasesleadtofalsehoods,sotheinequalityhasnosolution.
5.
(
Case(i): If 0 ≤ 1,then ()= 2 4 +3
Case(ii): If 1 ≤ 3,then ()= ( 2 4 +3)= 2 +4 3
Case(iii): If 3,then ()= 2 4
Thisenablesustosketchthegraphfor ≥ 0.Thenweusethefactthat isaneven functiontoreflectthispartofthegraphaboutthe axistoobtaintheentiregraph.Or,we couldconsideralsothecases
Wewillconsidertheequation + || = + | | infourcases. (1) ≥ 0
Case 1 givesustheline = withnonnegative and .
Case 2 givesustheportionofthe axiswith negative.
Case 3 givesustheportionofthe axiswith negative.
Case 4 givesustheentirethirdquadrant. 8. | | + || | | ≤ 2 [callthisinequality()]
Case(i): ≥ ≥ 0.Then() ⇔
Case(ii): ≥ ≥ 0.Then()
Case(iii):
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Note: Insteadofconsideringcases(iv),(v),and(vi),wecouldhavenotedthat theregionisunchangedif and arereplacedby and ,sotheregionis symmetricabouttheorigin.Therefore,weneedonlydrawcases(i),(ii),and (iii),androtatethrough 180◦ abouttheorigin.
9. (a)Tosketchthegraphof ()=max { 1}, we firstgraph ()= and ()=1 onthe samecoordinateaxes.Thencreatethegraphof byplottingthelargest valueof and for everyvalueof
(b)
(c)
OntheTI84Plus,maxisfoundunder LIST,thenunder MATH. Tograph ()=max
,use Y =max( 2 max(2+ 2 ))
10. (a)If max { 2 } =1,theneither =1 and 2 ≤ 1 ≤ 1 and 2 =1.Thus,weobtainthesetofpointssuchthat =1 and ≤ 1 2 [averticallinewithhighestpoint (1 1 2 ) ≤ 1 and = 1 2 ahorizontallinewithrightmostpoint (1 1 2 )
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¤ CHAPTER1 PRINCIPLESOFPROBLEMSOLVING
(b)Thegraphof max{ 2 } =1 isshowninpart(a),andthegraphof max{ 2 } = 1 canbefoundinasimilarmanner.The inequalitiesin 1 ≤ max{ 2 } ≤ 1 giveusallthepointsonor insidetheboundaries.
(c) max{ 2 } =1 ⇔ =1 and 2 ≤ 1 [ 1
≤ 1 and 2 =1 [ = ±1].
≤ 1]
11. 1 log2 + 1 log3 + 1 log5 = 1 log log2 + 1 log log3 + 1 log log5
[ChangeofBaseformula] = log2 log + log3 log + log5 log = log2+log3+log5 log = log(2 3 5) log
[Law1ofLograithms] = log30 log = 1 log log30 = 1 log30
[ChangeofBaseformula]
12. Wenotethat 1 ≤ sin ≤ 1 forall .Thus,anysolutionof sin = 100 willhave 1 ≤ 100 ≤ 1,or 100 ≤ ≤ 100.Wenextobservethattheperiodof sin is 2 ,and sin takesoneachvalueinitsrange,exceptfor 1 and 1,twiceeachcycle.Weobservethat =0 isasolution.Finally,wenotethatbecause sin and 100 arebothodd functions,everysolutionon 0 ≤ ≤ 100 givesusacorrespondingsolutionon 100 ≤ ≤ 0
1002 ≈ 15 9,sothere 15 fullcyclesof sin on [0 100] Eachofthe 15 intervals [0 2 ], [2 4 ], , [28 30 ] must containtwosolutionsof sin = 100,asthegraphof sin willintersectthegraphof 100 twiceeachcycle.Wemust becarefulwiththenext(16th)interval [30 32 ],because 100 iscontainedintheinterval.Agraphof 1 =sin and 2 = 100 overthisintervalrevealsthattwointersectionsoccurwithintheintervalwith ≤ 100 Thus,thereare 16 2=32 solutionsof sin = 100 on [0 100].Therearealso 32 solutionsoftheequationon [ 100 0].Beingcarefultonotcountthesolution =0 twice,we findthatthereare 32+32 1=63 solutionsofthe equation sin = 100
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13. Byrearrangingterms,wewritethegivenexpressionas
Eachgroupedsumisoftheform
to
Multiplicationoftheleftandrightmembersofthisequalityby 2 givesthesumtoproductidentity
.Usingthissum
toproductidentity,wehaveeachgroupedsumequalto 0,since sin + 2 =sin =0 isalwaysafactoroftherightside.Since sin
100 =sin =0
=sin2 =0,the
sumofthegivenexpressionis 0.
Anotherapproach: Sincethesinefunctionisodd, sin( )= sin .Becausetheperiodofthesinefunctionis 2 ,wehave sin( +2 )= sin .Multiplyingeachsideby 1 andrearranging,wehave sin = sin(2 ).Thismeansthat
,andsoon,untilwehave
.Asbeforewerearrangetermstowritethegivenexpressionas
Eachsuminparenthesesis 0 sincethetwotermsareopposites,andthelasttwotermsagainreduceto sin and sin2 , respectively,eachalso 0.Thus,thevalueoftheoriginalexpressionis 0
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15. ln 2 2
Sincetheargumentmustbepositive,
.Theintersectionoftheseintervalsis
16. Assumethat log2 5 isrational.Then log2 5= fornaturalnumbers and .Changingtoexponentialformgivesus 2 =5 andthenraisingbothsidestothe
thpowergives 2 =5
.But 2 isevenand 5 isodd.Wehavearrivedata contradiction,soweconcludethatourhypothesis,that log2 5 isrational,isfalse.Thus, log2 5 isirrational.
17. Let bethedistancetraveledoneachhalfofthetrip.Let 1 and 2 bethetimestakenforthe firstandsecondhalvesofthetrip. Forthe firsthalfofthetripwehave 1 = 30 andforthesecondhalfwehave 2 = 60.Thus,theaveragespeedforthe entiretripis totaldistance totaltime =
=40.Theaveragespeedfortheentiretrip is 40 mih.
18. Let ()=sin , ()= ,and ()= .Thenthelefthandsideoftheequationis
[ ◦ ( + )]()=sin( + )=sin2 =2sin cos ; andtherighthandsideis ( ◦ )()+( ◦ )()=sin +sin =2sin .Thetwosidesarenotequal,sothegivenstatementisfalse.
19. Let bethestatementthat 7 1 isdivisibleby 6
• 1 istruebecause 71 1=6 isdivisibleby 6
• Assume istrue,thatis, 7 1 isdivisibleby 6.Inotherwords, 7 1=6 forsomepositiveinteger .Then 7+1 1=7 · 7 1=(6 +1) · 7 1=42 +6=6(7 +1),whichisdivisibleby 6,so +1 istrue.
• Therefore,bymathematicalinduction, 7 1 isdivisibleby 6 foreverypositiveinteger
20. Let bethestatementthat 1+3+5+ +(2 1)= 2 .
• 1 istruebecause [2(1) 1]=1=12
• Assume istrue,thatis, 1+3+5+ +(2 1)= 2 .Then 1+3+5+ +(2 1)+[2( +1) 1]=1+3+5+ +(2
2 whichshowsthat +1 istrue.
• Therefore,bymathematicalinduction, 1+3+5+ +(2
foreverypositiveinteger
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Thus,weconjecturethatthegeneralformulais
Toprovethis,weusethePrincipleofMathematicalInduction.Wehavealreadyveri
Assumethattheformulaistruefor
Thisshowsthattheformulafor istruefor = +1.Therefore,bymathematicalinduction,theformulaistrueforall positiveintegers
(b)Fromthegraph,wecanmakeseveralobservations:
• Thevaluesateach fixed = keepincreasingas increases.
• Theverticalasymptotegetscloserto =1 as increases.
• Thehorizontalasymptotegetscloserto =1 as increases.
• The interceptfor +1 isthevalueofthe verticalasymptotefor
• The interceptfor isthevalueofthe horizontalasymptotefor +1 .
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