Instructor's Solutions Manual to Calculus: Single and Multivariable

1879 Pages·2012·23.701 MB·English

by  Deborah Hughes- Hallett and et al

1.1SOLUTIONS 1
CHAPTER ONE
Solutions forSection1.1
Exercises
1. Sincetrepresentsthenumber of yearssince1970, weseethat f(35) representsthepopulation of thecityin2005. In
2005,thecity’spopulationwas12million.
2. SinceT = f(P),weseethatf(200)isthevalueofT whenP = 200;thatis,thethicknessofpelicaneggswhenthe
concentrationofPCBsis200ppm.
3. Iftherearenoworkers,thereisnoproductivity,sothegraphgoesthroughtheorigin.Atfirst,asthenumberofworkers
increases,productivityalsoincreases.Asaresult,thecurvegoesupinitially.Atacertainpointthecurvereachesitshighest
level,afterwhichitgoesdownward;inotherwords,asthenumberofworkersincreasesbeyondthatpoint,productivity
decreases.Thismight,forexample,bedueeithertotheinefficiencyinherentinlargeorganizationsorsimplytoworkers
gettingineachother’swayastoomanyarecrammedonthesameline.Manyotherreasonsarepossible.
4. Theslopeis(1 0)/(1 0)=1.Sotheequationofthelineisy=x.
− −
5. Theslopeis(3 2)/(2 0)=1/2.Sotheequationofthelineisy=(1/2)x+2.
− −
6. Usingthepoints( 2,1)and(2,3),wehave
−
3 1 2 1
Slope = − = = .
2 ( 2) 4 2
− −
Nowweknowthaty = (1/2)x+b.Usingthepoint( 2,1),wehave1 = 2/2+b,whichyieldsb = 2.Thus,the
− −
equationofthelineisy=(1/2)x+2.
6 0
7. Slope= − =2sotheequationisy 6=2(x 2)ory=2x+2.
2 ( 1) − −
− −
5 5
8. Rewritingtheequationasy= x+4showsthattheslopeis andtheverticalinterceptis4.
−2 −2
9. Rewritingtheequationas
12 2
y= x+
− 7 7
showsthatthelinehasslope 12/7andverticalintercept2/7.
−
10. Rewritingtheequationofthelineas
2
y = − x 2
− 4 −
1
y = x+2,
2
weseethelinehasslope1/2andverticalintercept2.
11. Rewritingtheequationofthelineas
12 4
y = x
6 − 6
2
y =2x ,
− 3
weseethatthelinehasslope2andverticalintercept 2/3.
−
12. (a) is(V),becauseslopeispositive,verticalinterceptisnegative
(b) is(IV),becauseslopeisnegative,verticalinterceptispositive
(c) is(I),becauseslopeis0,verticalinterceptispositive
(d) is(VI),becauseslopeandverticalinterceptarebothnegative
(e) is(II),becauseslopeandverticalinterceptarebothpositive
(f) is(III),becauseslopeispositive,verticalinterceptis0
2 ChapterOne /SOLUTIONS
13. (a) is(V),becauseslopeisnegative,verticalinterceptis0
(b) is(VI),becauseslopeandverticalinterceptarebothpositive
(c) is(I),becauseslopeisnegative,verticalinterceptispositive
(d) is(IV),becauseslopeispositive,verticalinterceptisnegative
(e) is(III),becauseslopeandverticalinterceptarebothnegative
(f) is(II),becauseslopeispositive,verticalinterceptis0
14. Theinterceptsappeartobe(0,3)and(7.5,0),giving
3 6 2
Slope = − = = .
7.5 −15 −5
They-interceptisat(0,3),soapossibleequationforthelineis
2
y= x+3.
−5
(Answersmayvary.)
15. y c=m(x a)
− −
16. Giventhatthefunctionislinear,chooseanytwopoints,forexample(5.2,27.8)and(5.3,29.2).Then
29.2 27.8 1.4
Slope= − = =14.
5.3 5.2 0.1
−
Usingthepoint-slopeformula,withthepoint(5.2,27.8),wegettheequation
y 27.8=14(x 5.2)
− −
whichisequivalentto
y=14x 45.
−
17. y=5x 3.Sincetheslopeofthislineis5,wewantalinewithslope 1 passingthroughthepoint(2,1).Theequation
− −5
is(y 1)= 1(x 2),ory= 1x+ 7.
− −5 − −5 5
18. The line y +4x = 7 has slope 4. Therefore the parallel line has slope 4 and equation y 5 = 4(x 1) or
− − − − −
y= 4x+9.Theperpendicularlinehasslope −1 = 1 andequationy 5= 1(x 1)ory=0.25x+4.75.
− (−4) 4 − 4 −
19. Thelineparalleltoy=mx+calsohasslopem,soitsequationis
y=m(x a)+b.
−
Thelineperpendiculartoy=mx+chasslope 1/m,soitsequationwillbe
−
1
y= (x a)+b.
−m −
20. Sincethefunctiongoesfromx=0tox=4andbetweeny =0andy =2,thedomainis0 x 4andtherangeis
≤ ≤
0 y 2.
≤ ≤
21. Sincexgoesfrom1to5andygoesfrom1to6,thedomainis1 x 5andtherangeis1 y 6.
≤ ≤ ≤ ≤
22. Sincethefunctiongoesfromx= 2tox=2andfromy = 2toy =2,thedomainis 2 x 2andtherangeis
− − − ≤ ≤
2 y 2.
− ≤ ≤
23. Sincethefunctiongoesfromx=0tox=5andbetweeny =0andy =4,thedomainis0 x 5andtherangeis
≤ ≤
0 y 4.
≤ ≤
24. Thedomainisallnumbers.Therangeisallnumbers 2,sincex2 0forallx.
≥ ≥
1
25. Thedomainisallx-values,asthedenominatorisneverzero.Therangeis0<y .
≤ 2
26. Thevalueoff(t)isrealprovidedt2 16 0ort2 16.Thisoccurswheneithert 4,ort 4.Solvingf(t)=3,
− ≥ ≥ ≥ ≤−
wehave
t2 16=3
−
t2 16=9
p
−
t2 =25
1.1SOLUTIONS 3
so
t= 5.
±
4
27. WehaveV =kr3.YoumayknowthatV = πr3.
3
d
28. Ifdistanceisd,thenv= .
t
29. Forsomeconstantk,wehaveS =kh2.
30. WeknowthatEisproportionaltov3,soE =kv3,forsomeconstantk.
31. WeknowthatN isproportionalto1/l2,so
k
N = , forsomeconstantk.
l2
Problems
32. Theyear1983was25yearsbefore2008so1983correspondstot=25.Thus,anexpressionthatrepresentsthestatement
is:
f(25)=7.019
33. Theyear2008was0yearsbefore2008so2008correspondstot=0.Thus,anexpressionthatrepresentsthestatement
is:
f(0)meters.
34. Theyear1965was2008 1865 = 143yearsbefore2008so1965correspondstot = 143.Similarly,weseethatthe
−
year1911correspondstot=97.Thus,anexpressionthatrepresentsthestatementis:
f(143)=f(97)
35. Since t = 1 means one year before 2008, thent = 1 corresponds totheyear 2007. Similarly, t = 0 corresponds to
theyear2008. Thus,f(1)andf(0)aretheaverageannual sealevelvalues,inmeters,in2007and2008, respectively.
Because1millimeteristhesameas0.001meters,anexpressionthatrepresentsthestatementis:
f(0)=f(1)+0.001.
Notethatthereareotherpossibleequivalentexpressions,suchas:f(1) f(0)=0.001.
−
36. (a) Eachdate,t,hasauniquedailysnowfall,S,associatedwithit.Sosnowfallisafunctionofdate.
(b) OnDecember12,thesnowfallwasapproximately5inches.
(c) OnDecember11,thesnowfallwasabove10inches.
(d) LookingatthegraphweseethatthelargestincreaseinthesnowfallwasbetweenDecember10toDecember11.
37. (a) Whenthecaris5yearsold,itisworth$6000.
(b) Sincethevalueofthecardecreasesasthecargetsolder,thisisadecreasingfunction.ApossiblegraphisinFigure1.1:
V (thousanddollars)
(5,6)
a(years
Figure1.1
(c) TheverticalinterceptisthevalueofV whena = 0,orthevalueofthecarwhenitisnew.Thehorizontalintercept
isthevalueofawhenV =0,ortheageofthecarwhenitisworthnothing.
4 ChapterOne /SOLUTIONS
38. (a) Thestoryin(a)matchesGraph(IV),inwhichthepersonforgotherbooksandhadtoreturnhome.
(b) Thestoryin(b)matchesGraph(II),theflattirestory.Notethelongperiodoftimeduringwhichthedistancefrom
homedidnotchange(thehorizontalpart).
(c) Thestoryin(c)matchesGraph(III),inwhichthepersonstartedcalmlybutspeduplater.
Thefirstgraph(I)doesnotmatchanyofthegivenstories.Inthispicture,thepersonkeepsgoingawayfromhome,
buthisspeeddecreasesastimepasses.Soastoryforthismightbe:Istartedwalkingtoschoolatagoodpace,butsinceI
stayedupallnightstudyingcalculus,IgotmoreandmoretiredthefartherIwalked.
39. (a) f(30)=10meansthatthevalueoff att=30was10.Inotherwords,thetemperatureattimet=30minuteswas
10◦C.So,30minutesaftertheobjectwasplacedoutside,ithadcooledto10◦C.
(b) Theinterceptameasuresthevalueoff(t)whent = 0.Inotherwords,whentheobjectwasinitiallyputoutside,
ithadatemperatureofa◦C.Theinterceptbmeasuresthevalueoftwhenf(t) = 0.Inotherwords,attimebthe
object’stemperatureis0◦C.
40. (a) Theheightoftherockdecreasesastimepasses,sothegraphfallsasyoumovefromlefttoright.Onepossibilityis
showninFigure1.2.
s(meters)
t(sec)
Figure1.2
(b) Thestatementf(7)=12tellsusthat7secondsaftertherockisdropped,itis12metersabovetheground.
(c) Theverticalinterceptisthevalueofswhent=0;thatis,theheightfromwhichtherockisdropped.Thehorizontal
interceptisthevalueoftwhens=0;thatis,thetimeittakesfortherocktohittheground.
41. (a) WefindtheslopemandinterceptbinthelinearequationC =b+mw.Tofindtheslopem,weuse
∆C 12.32 8
m= = − =0.12dollarspergallon.
∆w 68 32
−
Wesubstitutetofindb:
C =b+mw
8=b+(0.12)(32)
b=4.16dollars.
ThelinearformulaisC =4.16+0.12w.
(b) Theslopeis0.12dollarspergallon.Eachadditionalgallonofwastecollectedcosts12cents.
(c) Theinterceptis$4.16.Theflatmonthlyfeetosubscribetothewastecollectionserviceis$4.16.Thisistheamount
chargedevenifthereisnowaste.
42. Wearelookingforalinearfunctiony =f(x)that,givenatimexinyears,givesavalueyindollarsforthevalueofthe
refrigerator.Weknowthatwhenx=0,thatis,whentherefrigeratorisnew,y =950,andwhenx=7,therefrigerator
isworthless,soy=0.Thus(0,950)and(7,0)areonthelinethatwearelookingfor.Theslopeisthengivenby
950
m=
7
−
Itisnegative,indicatingthatthevaluedecreasesastimepasses.Havingfoundtheslope,wecantakethepoint(7,0)and
usethepoint-slopeformula:
y y =m(x x ).
1 1
− −
So,
950
y 0= (x 7)
− − 7 −
950
y = x+950.
− 7
1.1SOLUTIONS 5
43. (a) Thefirstcompany’spriceforaday’srentalwithmmilesonitisC (m)=40+0.15m.Itscompetitor’spricefora
1
day’srentalwithmmilesonitisC (m)=50+0.10m.
2
(b) SeeFigure1.3.
C(costindollars)
150 C1(m)=40+0.15m
100
50 C2(m)=50+0.10m
0 m(miles)
200 400 600 800
Figure1.3
(c) Tofindwhichcompanyischeaper,weneedtodeterminewherethetwolinesintersect.WeletC =C ,andthus
1 2
40+0.15m =50+0.10m
0.05m =10
m=200.
Ifyouaregoingmorethan200milesaday,thecompetitorischeaper.Ifyouaregoinglessthan200milesaday,the
firstcompanyischeaper.
∆$ 55 40
44. (a) Chargepercubicfoot= = − =$0.025/cu.ft.
∆cu.ft. 1600 1000
Alternatively, ifweletc = cost,w =cub−ic feetof water, b =fixedcharge, and m = cost/cubicfeet,weobtain
c=b+mw.Substitutingtheinformationgivenintheproblem,wehave
40=b+1000m
55=b+1600m.
Subtractingthefirstequationfromthesecondyields15=600m,som=0.025.
(b) The equation is c = b+0.025w, so 40 = b+0.025(1000), which yields b = 15. Thus the equation is c =
15+0.025w.
(c) Weneedtosolvetheequation100=15+0.025w,whichyieldsw =3400.Itcosts$100touse3400cubicfeetof
water.
45. SeeFigure1.4.
drivingspeed
time
Figure1.4
46. SeeFigure1.5.
distancedriven
time
Figure1.5
6 ChapterOne /SOLUTIONS
47. SeeFigure1.6.
distancefromexit
time
Figure1.6
48. SeeFigure1.7.
distancebetweencars
distancedriven
Figure1.7
49. (a) (i) f(1985)=13
(ii) f(1990)=99
(b) Theaverageyearlyincreaseistherateofchange.
f(1990) f(1985) 99 13
Yearlyincrease = − = − =17.2billionairesperyear.
1990 1985 5
−
(c) Sinceweassumetherateofincreaseremainsconstant,weusealinearfunctionwithslope17.2billionairesperyear.
Theequationis
f(t)=b+17.2t
wheref(1985)=13,so
13=b+17.2(1985)
b= 34,129.
−
Thus,f(t)=17.2t 34,129.
−
50. (a) Thelargesttimeintervalwas2008–2009sincethepercentagegrowthrateincreasedfrom 11.7to7.3from2008to
−
2009.ThismeanstheUSconsumptionofbiofuelsgrewrelativelymorefrom2008to2009thanfrom2007to2008.
(Notethatthepercentagegrowthratewasadecreasingfunctionoftimeover2005–2007.)
(b) The largest time interval was 2005–2007 since the percentage growth rates were positive for each of these three
consecutive years.Thismeansthattheamount ofbiofuelsconsumed intheUSsteadilyincreasedduringthethree
yearspanfrom2005to2007,thendecreasedin2008.
51. (a) Thelargesttimeintervalwas2005–2007sincethepercentagegrowthratedecreasedfrom 1.9in2005to 45.4in
− −
2007.Thismeansthatfrom2005to2007theUSconsumptionofhydroelectricpowershrunkrelativelymorewith
eachsuccessiveyear.
(b) The largest time interval was 2004–2007 since the percentage growth rates were negative for each of these four
consecutiveyears.ThismeansthattheamountofhydroelectricpowerconsumedbytheUSindustrialsectorsteadily
decreasedduringthefouryearspanfrom2004to2007,thenincreasedin2008.
52. (a) Thelargesttimeinterval was2004–2006 sincethepercentage growthrateincreasedfrom 5.7 in2004 to9.7 in
−
2006. Thismeans that from 2004 to 2006 the US price per watt of a solar panel grew relatively more with each
successiveyear.
(b) The largest time interval was 2005–2006 since the percentage growth rates were positive for each of these two
consecutiveyears.ThismeansthattheUSpriceperwattofasolarpanelsteadilyincreasedduringthetwoyearspan
from2005to2006,thendecreasedin2007.
1.1SOLUTIONS 7
53. (a) Since2008correspondstot=0,theaverageannualsealevelinAberdeenin2008was7.094meters.
(b) Lookingatthetable,weseethattheaverageannualsealevelwas7.019fiftyyearsbefore2008,orintheyear1958.
Similarreasoningshowsthattheaveragesealevelwas6.957meters125yearsbefore2008,orin1883.
(c) Because125yearsbefore2008theyearwas1883,weseethatthesealevelvaluecorrespondingtotheyear1883is
6.957(thisisthesealevelvaluecorrespondingtot=125).Similarreasoningyieldsthetable:
Year 1883 1908 1933 1958 1983 2008
S 6.957 6.938 6.965 6.992 7.019 7.094
54. (a) WefindtheslopemandinterceptbinthelinearequationS =b+mt.Tofindtheslopem,weuse
∆S 66 113
m= = − = 0.94.
∆t 50 0 −
−
Whent=0,wehaveS =113,sotheinterceptbis113.Thelinearformulais
S =113 0.94t.
−
(b) WeusetheformulaS = 113 0.94t.WhenS = 20,wehave20 = 113 0.94tandsot = 98.9.Ifthislinear
− −
modelwerecorrect,theaveragemalespermcountwoulddropbelowthefertilitylevelduringtheyear2038.
55. (a) Thiscouldbealinearfunctionbecausewincreasesby5ashincreasesby1.
(b) Wefindtheslopemandtheinterceptbinthelinearequationw =b+mh.Wefirstfindtheslopemusingthefirst
twopointsinthetable.Sincewewantwtobeafunctionofh,wetake
∆w 171 166
m= = − =5.
∆h 69 68
−
Substitutingthefirstpointandtheslopem=5intothelinearequationw =b+mh,wehave166=b+(5)(68),
sob= 174.Thelinearfunctionis
−
w=5h 174.
−
Theslope,m=5,isinunitsofpoundsperinch.
(c) Wefindtheslopeandinterceptinthelinearfunctionh=b+mwusingm=∆h/∆wtoobtainthelinearfunction
h=0.2w+34.8.
Alternatively,wecouldsolvethelinearequationfoundinpart(b)forh.Theslope,m = 0.2,hasunitsinchesper
pound.
56. Wewilllet
T = amountoffuelfortake-off,
L= amountoffuelforlanding,
P = amountoffuelpermileintheair,
m= thelengthofthetripinmiles.
ThenQ,thetotalamountoffuelneeded,isgivenby
Q(m)=T +L+Pm.
57. (a) Thevariablecostsforxacresare$200x,or0.2xthousanddollars.Thetotalcost,C(againinthousandsofdollars),
ofplantingxacresis:
C =f(x)=10+0.2x.
Thisisalinearfunction.SeeFigure1.8.SinceC = f(x)increaseswithx,f isanincreasingfunctionofx.Look
atthevaluesofC showninthetable;youwillseethateachtimexincreasesby1,C increasesby0.2.BecauseC
increasesataconstantrateasxincreases,thegraphofCagainstxisaline.
8 ChapterOne /SOLUTIONS
(b) SeeFigure1.8andTable1.1.
Table1.1
Costof
planting C(thosand$)
seed
10
C=10+0.2x
x C
0 10
2 10.4 5
3 10.6
4 10.8 x(acres)
5 11 2 4 6
6 11.2 Figure1.8
(c) Theverticalinterceptof10correspondstothefixedcosts.ForC =f(x)=10+0.2x,theinterceptonthevertical
axisis10becauseC = f(0) = 10+0.2(0) = 10.Since10isthevalueofC whenx = 0,werecognizeitasthe
initialoutlayforequipment,orthefixedcost.
Theslope0.2correspondstothevariablecosts.Theslopeistellingusthatforeveryadditionalacreplanted,the
costsgoupby0.2thousanddollars.Therateatwhichthecostisincreasingis0.2thousanddollarsperacre.Thusthe
variablecostsarerepresentedbytheslopeofthelinef(x)=10+0.2x.
58. SeeFigure1.9.
Distancefrom
Kalamazoo
155
120
Time
startin arrivein arrivein
Chicago Kalamazoo Detroit
Figure1.9
59. (a) Thelinegivenby(0,2)and(1,1)hasslopem= 2−1 = 1andy-intercept2,soitsequationis
−1 −
y= x+2.
−
Thepointsofintersectionofthislinewiththeparabolay=x2aregivenby
x2 = x+2
−
x2+x 2=0
−
(x+2)(x 1)=0.
−
The solution x = 1 corresponds to the point we are already given, so the other solution, x = 2, gives the x-
−
coordinateofC.Whenwesubstitutebackintoeitherequationtogety,wegetthecoordinatesforC,( 2,4).
−
(b) Thelinegivenby(0,b)and(1,1)hasslopem= b−1 =1 b,andy-interceptat(0,b),sowecanwritetheequation
−1 −
forthelineaswedidinpart(a):
y=(1 b)x+b.
−
Wethensolveforthepointsofintersectionwithy=x2thesameway:
x2 =(1 b)x+b
−
x2 (1 b)x b=0
− − −
x2+(b 1)x b=0
− −
(x+b)(x 1)=0
−
Again,wehavethesolutionatthegivenpoint(1,1),andanewsolutionatx= b,correspondingtotheotherpoint
−
ofintersectionC.Substitutingbackintoeitherequation,wecanfindthey-coordinateforCisb2,andthusCisgiven
by( b,b2).Thisresultagreeswiththeparticularcaseofpart(a)whereb=2.
−
1.2SOLUTIONS 9
60. Looking atthegivendata,itseemsthatGalileo’shypothesiswasincorrect.Thefirsttablesuggeststhatvelocityisnot
a linear function of distance, since the increases in velocity for each foot of distance are themselves getting smaller.
Moreover, the second table suggests that velocity is instead proportional to time, since for each second of time, the
velocityincreasesby32ft/sec.
StrengthenYourUnderstanding
61. Theliney=0.5 3xhasanegativeslopeandisthereforeadecreasingfunction.
−
62. If y is directly proportional to x we have y = kx. Adding the constant 1 to give y = 2x+1 means that y is not
proportionaltox.
63. Onepossibleanswerisf(x)=2x+3.
8
64. Onepossibleanswerisq= .
p1/3
65. False. A line can be put through any two points in the plane. However, if the line is vertical, it is not the graph of a
function.
66. True.Supposewestartatx =x andincreasexby1unittox +1.Ify =b+mx,thecorrespondingvaluesofyare
1 1
b+mx andb+m(x +1).Thusyincreasesby
1 1
∆y=b+m(x +1) (b+mx )=m.
1 1
−
67. False.Forexample,lety=x+1.Thenthepoints(1,2)and(2,3)areontheline.Howevertheratios
2 3
=2 and =1.5
1 2
aredifferent.Theratioy/xisconstantforlinearfunctionsoftheformy =mx,butnotingeneral.(Otherexamplesare
possible.)
68. False.Forexample,ify=4x+1(som=4)andx=1,theny=5.Increasingxby2unitsgives3,soy=4(3)+1=
13.Thus,yhasincreasedby8units,not4+2=6.(Otherexamplesarepossible.)
69. (b)and(c).Forg(x)=√x,thedomainandrangeareallnonnegativenumbers,andforh(x)=x3,thedomainandrange
areallrealnumbers.
Solutions forSection1.2
Exercises
1. Thegraphshowsaconcaveupfunction.
2. Thegraphshowsaconcavedownfunction.
3. Thisgraphisneitherconcaveupordown.
4. Thegraphisconcaveup.
5. Initialquantity=5;growthrate=0.07=7%.
6. Initialquantity=7.7;growthrate= 0.08= 8%(decay).
− −
7. Initialquantity=3.2;growthrate=0.03=3%(continuous).
8. Initialquantity=15;growthrate= 0.06= 6%(continuousdecay).
− −
9. Sincee0.25t = e0.25 t (1.2840)t,wehaveP =15(1.2840)t.Thisisexponentialgrowthsince0.25ispositive.We
≈
canalsoseethatthisisgrowthbecause1.2840>1.
(cid:0) (cid:1)
10. Sincee−0.5t = (e−0.5)t (0.6065)t,wehaveP = 2(0.6065)t.Thisisexponentialdecaysince 0.5isnegative.We
≈ −
canalsoseethatthisisdecaybecause0.6065<1.
11. P =P (e0.2)t =P (1.2214)t.Exponentialgrowthbecause0.2>0or1.2214>1.
0 0
12. P =7(e−π)t =7(0.0432)t.Exponentialdecaybecause π<0or0.0432<1.
−
10 ChapterOne /SOLUTIONS
13. (a) LetQ=Q at.ThenQ a5 =75.94andQ a7 =170.86.So
0 0 0
Q a7 170.86
0 = =2.25=a2.
Q a5 75.94
0
Soa=1.5.
(b) Sincea=1.5,thegrowthrateisr=0.5=50%.
14. (a) LetQ=Q at.ThenQ a0.02 =25.02andQ a0.05 =25.06.So
0 0 0
Q a0.05 25.06
0 = =1.001=a0.03.
Q a0.02 25.02
0
So
100
a=(1.001) 3 =1.05.
(b) Sincea=1.05,thegrowthrateisr=0.05=5%.
15. (a) Thefunctionislinearwithinitialpopulationof1000andslopeof50,soP =1000+50t.
(b) Thisfunctionisexponentialwithinitialpopulationof1000andgrowthrateof5%,soP =1000(1.05)t.
16. (a) Thisisalinearfunctionwithslope 2gramsperdayandintercept30grams.ThefunctionisQ=30 2t,andthe
− −
graphisshowninFigure1.10.
Q(grams) Q(grams)
30 30
Q=30 2t
− Q=30(0.88)t
t(days) t(days)
15 15
Figure1.10 Figure1.11
(b) Sincethequantityisdecreasingbyaconstantpercentchange,thisisanexponentialfunctionwithbase1 0.12 =
0.88.ThefunctionisQ=30(0.88)t,andthegraphisshowninFigure1.11. −
17. ThefunctionisincreasingandconcaveupbetweenDandE,andbetweenH andI.Itisincreasingandconcavedown
between AandB,andbetweenE andF.Itisdecreasing andconcave upbetweenC andD,and betweenGand H.
Finally,itisdecreasingandconcavedownbetweenBandC,andbetweenF andG
18. (a) ItwasdecreasingfromMarch2toMarch5andincreasingfromMarch5toMarch9.
(b) FromMarch5to8,theaveragetemperatureincreased,buttherateofincreasewentdown,from12◦betweenMarch
5and6to4◦betweenMarch6and7to2◦betweenMarch7and8.
FromMarch7to9,theaveragetemperatureincreased,andtherateofincreasewentup,from2◦betweenMarch
7and8to9◦betweenMarch8and9.
Problems
19. (a) Alinearfunctionmustchangebyexactlythesameamountwheneverxchangesbysomefixedquantity.Whileh(x)
decreases by 3 whenever x increases by 1, f(x) and g(x) fail this test, since both change by different amounts
betweenx = 2andx = 1andbetweenx = 1andx = 0.Sotheonlypossiblelinearfunctionish(x),soit
− − −
willbegivenbyaformulaofthetype:h(x) = mx+b.Asnoted,m = 3.Sincethey-interceptofhis31,the
−
formulaforh(x)ish(x)=31 3x.
−
(b) Anexponential functionmust growbyexactlythesamefactor whenever xchangesbysomefixedquantity. Here,
g(x)increasesbyafactorof1.5wheneverxincreasesby1.Sincethey-interceptofg(x)is36,g(x)hastheformula
g(x)=36(1.5)x.Theothertwofunctionsarenotexponential;h(x)isnotbecauseitisalinearfunction,andf(x)
isnotbecauseitbothincreasesanddecreases.
20. TableAandTableBcouldrepresentlinearfunctionsofx.TableAcouldrepresenttheconstantlinearfunctiony = 2.2
becauseallyvaluesarethesame.TableBcouldrepresentalinearfunctionofxwithslopeequalto11/4.Thisisbecause
xvaluesthatdifferby4havecorrespondingyvaluesthatdifferby11,andxvaluesthatdifferby8havecorrespondingy
valuesthatdifferby22.InTableC,ydecreasesandthenincreasesasxincreases,sothetablecannotrepresentalinear
function.TableDdoesnotshowaconstantrateofchange,soitcannotrepresentalinearfunction.