Solutions Manuals for Calculus Early Transcendentals

1213 Pages·2020·24.012 MB·English

by  James Stewart, Daniel K. Clegg, Saleem Watson

1 FUNCTIONS AND MODELS
1.1 FourWaystoRepresentaFunction
1. Thefunctions()=+√2 and()=+√2 giveexactlythesameoutputvaluesforeveryinputvalue,so
− −
andareequal.
2  ( 1)
2. ()= − = − =for 1=0,so and[where()=]arenotequalbecause(1)isundefinedand
 1  1 − 6
− −
(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­valueof3,sotheonlyvalueofforwhich()=3is 4.
− −
(c)Thefunctionoutputs()arenevergreaterthan3,so() 3fortheentiredomainofthefunction.Thus,() 3for
≤ ≤
4  4(or,equivalently,ontheinterval[ 44]).
− ≤ ≤ −
(d)Thedomainconsistsofall­valuesonthegraphof:  4  4 =[ 44].Therangeofconsistsofallthe
{ |− ≤ ≤ } −
­valuesonthegraphof:  2  3 =[ 23].
{ |− ≤ ≤ } −
(e)Forany  intheinterval[02],wehave( )( ).[Thegraphrisesfrom(0 2)to(21).]Thus,()is
1 2 1 2
−
increasingon[02].
4. (a)Fromthegraph,wehave( 4)= 2and(3)=4.
− −
(b)Since( 3)= 1and( 3)=2,orbyobservingthatthegraphofisabovethegraphof at= 3,( 3)islarger
− − − − −
than( 3).
−
(c)Thegraphsof andintersectat= 2and=2,so()=()atthesetwovaluesof.
−
(d)Thegraphof liesbeloworonthegraphoffor 4  2andfor2  3.Thus,theintervalsonwhich
− ≤ ≤− ≤ ≤
() ()are[ 4 2]and[23].
≤ − −
(e)()= 1isequivalentto= 1,andthepointsonthegraphof with­valuesof 1are( 3 1)and(4 1),so
− − − − − −
thesolutionoftheequation()= 1is= 3or=4.
− −
(f)Forany  intheinterval[ 40],wehave( )( ).Thus,()isdecreasingon[ 40].
1 2 1 2
− −
(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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°
10 ¤ CHAPTER1 FUNCTIONSANDMODELS
6. Example1:Acarisdrivenat60mihfor2hours.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00perhourandworksa pay
maximumof30hoursperweek.Thenumberofhoursworkedis 240
238
roundeddowntothenearestquarterofanhour.Thisemployee’s 236
grossweeklypay isafunctionofthenumberofhoursworked.
4
Thedomainofthefunctionis[030]andtherangeofthefunctionis
2
02004002380024000 . 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
onevalueofforeachvalueof,theequationdefinesasafunctionof.
8. Wesolve32 2=5for: 32 2=5 2= 32+5 = 32 5.Sincetheequationdetermines
− − ⇔ − − ⇔ 2 − 2
exactlyonevalueofforeachvalueof,theequationdefinesasafunctionof.
9. Wesolve2+( 3)2 =5for: 2+( 3)2 =5 ( 3)2 =5 2  3= √5 2
− − ⇔ − − ⇔ − ± − ⇔
=3 √5 2.Someinputvaluescorrespondtomorethanoneoutput.(Forinstance,=1correspondsto=1and
± −
to=5.)Thus,theequationdoesnotdefineasafunctionof.
10. Wesolve2+52 =4for: 2+52 =4 52+(2) 4=0
⇔ − ⇔
2 (2)2 4(5)( 4) 2 √42+80  √2+20
= − ± − − = − ± = − ± (usingthequadraticformula).Someinput
 2(5) 10 5
valuescorrespondtomorethanoneoutput.(Forinstance,=4correspondsto= 2andto=25.)Thus,the
−
equationdoesnotdefineasafunctionof.
11. Wesolve(+3)3+1=2for: (+3)3+1=2 (+3)3 =2 1 +3= √32 1
⇔ − ⇔ − ⇔
= 3+√32 1.Sincetheequationdeterminesexactlyonevalueofforeachvalueof,theequationdefinesasa
− −
functionof.
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°
SECTION1.1 FOURWAYSTOREPRESENTAFUNCTION ¤ 11
12. Wesolve2  =0for: 2  =0  =2 = 2.Someinputvaluescorrespondtomorethan
−| | −| | ⇔ | | ⇔ ±
oneoutput.(Forinstance,=1correspondsto= 2andto=2.)Thus,theequationdoesnotdefineasafunction
−
of.
13. Theheight60in(=60)correspondstoshoesizes7and8(=7and=8).Sinceaninputvaluecorrespondstomore
thanoutputvalue,thetabledoesnotdefine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,sotheglobalaveragetemperaturein1950wasabout13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)Theringwidthvariesfromnear0mmtoabout16mm,sotherangeoftheringwidthfunctionisapproximately[016].
(b)Accordingtothegraph,theearthgraduallycooledfrom1550to1700,warmedintothelate1700s,cooledagainintothe
late1800s,andhasbeensteadilywarmingsincethen.Inthemid­19thcentury,therewasvariationthatcouldhavebeen
associatedwithvolcaniceruptions.
21. Thewaterwillcooldownalmosttofreezingastheicemelts.Then,when
theicehasmelted,thewaterwillslowlywarmuptoroomtemperature.
22. Thetemperatureofthepiewouldincreaserapidly,levelofftooven
temperature,decreaserapidly,andthenlevelofftoroomtemperature.
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°
12 ¤ CHAPTER1 FUNCTIONSANDMODELS
23. (a)Thepowerconsumptionat6AMis500 MWwhichisobtainedbyreadingthevalueofpower when=6fromthe
graph.At6PMwereadthevalueof when=18obtainingapproximately730 MW
(b)Theminimumpowerconsumptionisdeterminedbyfindingthetimeforthelowestpointonthegraph,=4or4AM.The
maximumpowerconsumptioncorrespondstothehighestpointonthegraph,whichoccursjustbefore=12orright
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.
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°
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. ()=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=32++2.
− − − −
(+1)=3(+1)2 (+1)+2=3(2+2+1)  1+2=32+6+3 +1=32+5+4.
− − − −
2()=2 ()=2(32 +2)=62 2+4.
· − −
(2)=3(2)2 (2)+2=3(42) 2+2=122 2+2.
− − −
[continued]
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14 ¤ CHAPTER1 FUNCTIONSANDMODELS
(2)=3(2)2 (2)+2=3(4) 2+2=34 2+2.
− − −
[()]2= 32 +2 2 = 32 +2 32 +2
− − −
=94 33+62 33+2 2+62 2+4=94 63+132 4+4
− − − − − −
(+)=3(+)2 (+)+2=3(2+2+2)  +2=32+6+32  +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+32+32+3,
(+) () (3+32+32+3) 3 (32+3+2)
and − = − = =32+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)isdefinedforallexceptwhen0=2 9 0=(+3)( 3) = 3or3,sothe
− − ⇔ − ⇔ −
domainis   = 33 =(  3) ( 33) (3 ).
{ ∈ | 6 − } −∞ − ∪ − ∪ ∞
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°
SECTION1.1 FOURWAYSTOREPRESENTAFUNCTION ¤ 15
2+1
40. Thefunction()= isdefinedforallvaluesofexceptthoseforwhich2+4 21=0
2+4 21 − ⇔
−
(+7)( 3)=0 = 7or=3.Thus,thedomainis   = 73 =(  7) ( 73) (3 ).
− ⇔ − { ∈ | 6 − } −∞ − ∪ − ∪ ∞
41. ()= √32 1isdefinedforallrealnumbers.Infact 3 (),where()isapolynomial,isdefinedforallrealnumbers.
−
Thus,thedomainisor(  ). 
−∞ ∞
42. ()=√3  √2+isdefinedwhen3  0  3and2+ 0  2.Thus,thedomainis
− − − ≥ ⇔ ≤ ≥ ⇔ ≥−
2  3,or[ 23].
− ≤ ≤ −
43. ()=1 √42 5 isdefinedwhen2 50 ( 5)0.Notethat2 5=0sincethatwouldresultin
− − ⇔ − − 6
divisionbyzero.Theexpression( 5)ispositiveif0or5.(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[04].
≤ ≤
46. Thefunction()=√2 4 5isdefinedwhen2 4 5 0 (+1)( 5) 0.Thepolynomial
− − − − ≥ ⇔ − ≥
()=2 4 5maychangesignsonlyatitszeros,sowetestvaluesofontheintervalsseparatedby= 1and
− − −
=5: ( 2)=70,(0)= 50,and(6)=70.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  =[ 22].Fromthegraph,
| − ≥ | ≥ { | ≥| |} −
therangeis0   2,or[02]. 
≤ ≤
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(24).
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°
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=2for 0andgraph=0(the­axis)for0
≥
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°
SECTION1.1 FOURWAYSTOREPRESENTAFUNCTION ¤ 17
+2 if +2 0
54. ()= +2 = ≥
| |  (+2) if +20
−
+2 if  2
= ≥−
  2 if  2
− − −
1 3 if 1 3 0
55. ()= 1 3 = − − ≥
| − |  (1 3) if 1 30
− − −
1 3 if  1
= − ≤ 3
3 1 if  1
− 3

56. ()= | |

Thedomainof is  =0 and  =if0,  = if0.
{ | 6 } | | | | −
Sowecanwrite

− = 1 if 0
 −
()=

 =1 if 0

 if  1
57. Tograph()= | | | |≤ ,graph=  [Figure16]
1 if  1 | |
| |
for 1  1andgraph=1for1andfor 1.
− ≤ ≤ −
1 if  1
−
 if 1 0

Wecouldrewritef as()= − − ≤ .
 if 0≤≤1
1 if 1

 1 if  1 0
58. ()=  1 = | |− | |− ≥
| |−  ( 1) if  10
  − | |− | |−
 
 
 1 if  1
= | |− | |≥
  +1 if  1
−| | | |
 1 if  1and 0  1 if  1
− | |≥ ≥ − ≥
 1 if  1and0  1 if  1
 
= − − | |≥ = − − ≤−
−+1 if ||1and≥0 −+1 if 0≤1
( )+1 if  1and0 +1 if 10
− − | |  −
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18 ¤ CHAPTER1 FUNCTIONSANDMODELS
 
59. Recallthattheslopeofalinebetweenthetwopoints(11)and(22)is= 2−1 andanequationoftheline
2 1
−
connectingthosetwopointsis  =(  ).Theslopeofthelinesegmentjoiningthepoints(1 3)and(57)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( 510)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 1and­intercept3,thatis,= +3.For3 5,thegraphistheline
≤ ≤ − − ≤
withslope2passingthrough(30);thatis, 0=2( 3),or=2 6.Sothefunctionis
− − −
+3 if 0  3
()= − ≤ ≤
2 6 if 3 5
− ≤
64. For 4  2,thegraphisthelinewithslope 3 passingthrough( 20);thatis, 0= 3[ ( 2)],or
− ≤ ≤− −2 − − −2 − −
= 3 3.For 22,thegraphisthetophalfofthecirclewithcenter(00)andradius2.Anequationofthecircle
−2 − −
is2+2 =4,soanequationofthetophalfis=√4 2.For2  4,thegraphisthelinewithslope 3 passing
− ≤ ≤ 2
through(20);thatis, 0= 3( 2),or= 3 3.Sothefunctionis
− 2 − 2 −
3 3 if 4  2
−2 − − ≤ ≤−
()=√4 2 if 22
3−3 if −2  4
2 − ≤ ≤

65. Letthelengthandwidthoftherectanglebeand.Thentheperimeteris2+2 =20andtheareais=.
20 2
Solvingthefirstequationfor intermsofgives = − =10 .Thus,()=(10 )=10 2.Since
2 − − −
lengthsarepositive,thedomainofis010.Ifwefurtherrestricttobelargerthan,then510wouldbe
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4wouldbethedomain.
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°