Student Solutions Manual - Calculus Early Transcendentals

1498 Pages·0103.803 MB·English

by  Jon Rogawski

1 PRECALCULUS REVIEW
1.1 Real Numbers, Functions, and Graphs
Preliminary Questions
1. Giveanexampleofnumbersaandbsuchthata<band|a|>|b|.
solution Takea=−3andb=1.Thena<bbut|a|=3>1=|b|.
2. Whichnumberssatisfy|a|=a?Whichsatisfy|a|=−a?Whatabout|−a|=a?
solution Thenumbersa≥0satisfy|a|=aand|−a|=a.Thenumbersa≤0satisfy|a|=−a.
3. Giveanexampleofnumbersaandbsuchthat
|a+b|<|a|+|b|.
solution Takea=−3andb=1.Then
|a+b|=|−3+1|=|−2|=2, but |a|+|b|=|−3|+|1|=3+1=4.
Thus,|a+b|<|a|+|b|.
4. Aretherenumbersaandbsuchthat|a+b|>|a|+|b|?
solution No.BytheTriangleinequality,|a+b|≤|a|+|b|forallrealnumbersaandb.
5. Whatarethecoordinatesofthepointlyingattheintersectionofthelinesx =9andy =−4?
solution Thepoint(9,−4)liesattheintersectionofthelinesx =9andy =−4.
6. Inwhichquadrantdothefollowingpointslie?
(a) (1,4) (b) (−3,2) (c) (4,−3) (d) (−4,−1)
solution
(a) Becauseboththex-andy-coordinatesofthepoint(1,4)arepositive,thepoint(1,4)liesinthefirstquadrant.
(b) Becausethex-coordinateofthepoint(−3,2)isnegativebutthey-coordinateispositive,thepoint(−3,2)liesin
thesecondquadrant.
(c) Becausethex-coordinateofthepoint(4,−3)ispositivebutthey-coordinateisnegative,thepoint(4,−3)liesin
thefourthquadrant.
(d) Becauseboththex-andy-coordinatesofthepoint(−4,−1)arenegative,thepoint(−4,−1)liesinthethirdquadrant.
7. Whatistheradiusofthecirclewithequation
(x−7)2+(y−8)2=9?
solution Thecirclewithequation(x−7)2+(y−8)2=9hasradius3.
8. Theequationf(x)=5hasasolutionif(chooseone):
(a) 5belongstothedomainoff.
(b) 5belongstotherangeoff.
solution Thecorrectresponseis(b):theequationf(x)=5hasasolutionif5belongstotherangeoff.
9. Whatkindofsymmetrydoesthegraphhaveiff(−x)=−f(x)?
solution Iff(−x)=−f(x),thenthegraphoff issymmetricwithrespecttotheorigin.
10. Isthereafunctionthatisbothevenandodd?
solution Yes.Theconstantfunctionf(x)=0forallrealnumbersxisbothevenandoddbecause
f(−x)=0=f(x)
and
f(−x)=0=−0=−f(x)
forallrealnumbersx.
1
2 CHAPTER 1 PRECALCULUSREVIEW
Exercises
1. Useacalculatortofindarationalnumberrsuchthat|r−π2|<10−4.
solution r mustsatisfyπ2−10−4 <r <π2+10−4,or9.869504<r <9.869705.r =9.8696= 12337 would
1250
beonesuchnumber.
InExercWishesic3h–o8f,(eax)p–r(efs)satrheetriunetefrovraali=nte−r3masnodfabn=in2e?qualityinvolvingabsolutevalue.
3. [(−a)2,a2<] b (b) |a|<|b| (c) ab>0
1 1
solu(dt)io3na<|x3|b≤2 (e) −4a<−4b (f) <
a b
5. (0,4)
(−4,4)
solution Themidpointoftheintervalisc = (0+4)/2 = 2,andtheradiusisr = (4−0)/2 = 2;therefore,(0,4)
canbeexpressedas|x−2|<2.
7. [1,5]
[−4,0]
solution Themidpointoftheintervalisc = (1+5)/2 = 3,andtheradiusisr = (5−1)/2 = 2;therefore,the
interval[1,5]canbeexpressedas|x−3|≤2.
InExerc(i−se2s,98–)12,writetheinequalityintheforma<x <b.
9. |x|<8
solution −8<x <8
11. |2x+1|<5
|x−12|<8
solution −5<2x+1<5so−6<2x <4and−3<x <2
InExerc|3isxes−134–|1<8,2expressthesetofnumbersxsatisfyingthegivenconditionasaninterval.
13. |x|<4
solution (−4,4)
15. |x−4|<2
|x|≤9
solution Theexpression|x−4|<2isequivalentto−2<x−4<2.Therefore,2<x <6,whichrepresentsthe
interval(2,6).
17. |4x−1|≤8
|x+7|<2
solution Theexpression|4x−1|≤8isequivalentto−8≤4x−1≤8or−7≤4x ≤9.Therefore,−7 ≤x ≤ 9,
4 4
whichrepresentstheinterval[−7,9].
4 4
InExerc|3isxes+195–|2<2,1describethesetasaunionoffiniteorinfiniteintervals.
19. {x :|x−4|>2}
solution x−4>2orx−4<−2⇒x >6orx <2⇒(−∞,2)∪(6,∞)
21. {x :{x|x:2|−2x1+|>4|2>} 3} √ √
solution x2−1>2orx2−1<−2⇒x2 >3orx2 <−1(thiswillneverhappen)⇒x > 3orx <− 3⇒
√ √
(−∞,− 3)∪( 3,∞).
23. Match(a)–(f)with(i)–(vi).
{x :|x2+2x|>2} 1
(a) a>3 (b) |a−5|<
(cid:2) (cid:2) 3
(cid:2) (cid:2)
(c) (cid:2)(cid:2)a− 1(cid:2)(cid:2)<5 (d) |a|>5
3
(e) |a−4|<3 (f) 1≤a≤5
(i) aliestotherightof3.
(ii) aliesbetween1and7.
(iii) Thedistancefromato5islessthan 1.
3
(iv) Thedistancefromato3isatmost2.
(v) aislessthan5unitsfrom 1.
3
(vi) alieseithertotheleftof−5ortotherightof5.
solution
(a) Onthenumberline,numbersgreaterthan3appeartotheright;hence,a>3isequivalenttothenumberstotheright
of3:(i).
SECTION 1.1 RealNumbers,Functions,andGraphs 3
(b) |a−5|measuresthedistancefromato5;hence,|a−5|< 1 issatisfiedbythosenumberslessthan 1 ofaunitfrom
3 3
5:(iii).
(c) |a− 1|measuresthedistancefromato 1;hence,|a− 1|<5issatisfiedbythosenumberslessthan5unitsfrom
3 3 3
1:(v).
3
(d) Theinequality|a|>5isequivalenttoa>5ora<−5;thatis,eitheraliestotherightof5ortotheleftof−5:(vi).
(e) Theintervaldescribedbytheinequality|a−4|<3hasacenterat4andaradiusof3;thatis,theintervalconsists
ofthosenumbersbetween1and7:(ii).
(f) Theintervaldescribedbytheinequality1<x <5hasacenterat3andaradiusof2;thatis,theintervalconsistsof
thosenumberslessthan2unitsfrom3:(iv).
25. Describe{x :(cid:3)x2+2x <3}a(cid:4)saninterval.Hint:Ploty =x2+2x−3.
x
solutiDonescrTibheeinxeq:uxal+ity1x<2+0 2axs<an3initserevqaul.ivHailnent:tCtoonxs2id+er2txhe−si3gn<of0x.Tanhdegxr+aph1ionfdyivi=duxal2ly+. 2x−3isshown
below.Fromthisgraph,itfollowsthatx2+2x−3<0for−3<x <1.Thus,theset{x :x2+2x <3}isequivalent
totheinterval(−3,1).
y
y = x2 + 2x − 3
12
10
8
6
4
2
x
−5−4−3−2 −2 1 2 3
−4
27. ShoDwetshcaritbiefath>esbet,aonfdreaa,lbnu(cid:8)=m0b,etrhsesnatbis−f1yi>nga|x−1−,p3r|o=vid|xed−th2a|t+aa1nadsbahhaavlfe-itnhfiensaitmeeinstiegrnv.aWl. hathappensifa>0
andb<0?
solution Case1a:Ifaandbarebothpositive,thena>b⇒1> b ⇒ 1 > 1.
a b a
Case1b:Ifaandbarebothnegative,thena>b⇒1< b (sinceaisnegative)⇒ 1 > 1 (again,sincebisnegative).
a b a
Case2:Ifa>0andb<0,then 1 >0and 1 <0so 1 < 1.(SeeExercise2fforanexampleofthis).
a b b a
29. ShoWwhtihcahtxifs|aati−sfi5e|s<bo12tha|nxd−|b3−|<8|2<an12d,|txhe−n5|<1?
|(a+b)−13|<1.Hint:Usethetriangleinequality(|a+b|≤|a|+|b|).
solution
|a+b−13|=|(a−5)+(b−8)|
≤|a−5|+|b−8| (bythetriangleinequality)
1 1
< + =1.
2 2
31. Supposethat|a−6|≤2and|b|≤3.
Supposethat|x−4|≤1.
(a) Whatisthelargestpossiblevalueof|a+b|?
(a) Whatisthemaximumpossiblevalueof|x+4|?
(b) Whatisthesmallestpossiblevalueof|a+b|?
(b) Showthat|x2−16|≤9.
solution |a−6|≤2guarantees4≤a ≤8,and|b|≤3guarantees−3≤b≤3,so1≤a+b≤11.Basedonthis
information,
(a) thelargestpossiblevalueof|a+b|is11;and
(b) thesmallestpossiblevalueof|a+b|is1.
33. ExpPrreosvser1th=at0|x.2|7−a|sya|f≤ra|cxtio−n.yH|.iHnti:nt1:0A0rp1pl−ytrh1eistraiannignlteegineer.qTuahleitnyetxopyreasnsdr2x=−0y..2666... asafraction.
solution Letr1=.27.Weobservethat100r1=27.27.Therefore,100r1−r1=27.27−.27=27and
27 3
r = = .
1
99 11
Now,letr2=.2666.Then10r2=2.666and100r2=26.666.Therefore,100r2−10r2=26.666−2.666=24and
24 4
r = = .
2
90 15
35. Thetextstates:Ifthedecimalexpansionsofnumbersaandbagreetokplaces,then|a−b|≤10−k.Showthatthe
conversReeisprfeaslseen:tF1o/r7aallndk4th/e2r7eaasrerenpuematbienrgsdaeacnimdablsw.hosedecimalexpansionsdonotagreeatallbut|a−b|≤10−k.
solution Leta = 1andb = .9(seethediscussionbeforeExample1).Thedecimalexpansionsofa andbdonot
agree,but|1−.9|<10−k forallk.
Ploteachpairofpointsandcomputethedistancebetweenthem:
(a) (1,4)and(3,2) (b) (2,1)and(2,4)
(c) (0,0)and(−2,3) (d) (−3,−3)and(−2,3)
4 CHAPTER 1 PRECALCULUSREVIEW
37. Findtheequationofthecirclewithcenter(2,4):
(a) withradiusr =3.
(b) thatpassesthrough(1,−1).
solution (a)Theequationoftheindicatedcircleis(x−2)2+(y−4)2=32=9.
(b)Firstdeterminetheradiusasthedistancefromthecentertotheindicatedpointonthecircle:
(cid:5)
√
r = (2−1)2+(4−(−1))2= 26.
Thus,theequationofthecircleis(x−2)2+(y−4)2=26.
39. Determinethedomainandrangeofthefunction
Findallpointsinthexy-planewithintegercoordinateslocatedatadistance5fromtheorigin.Thenfindallpoints
withintegercoordinateslocatedatadifsta:n{cre,s5,ftr,oum}→(2,{3A).,B,C,D,E}
definedbyf(r)=A,f(s)=B,f(t)=B,f(u)=E.
solution ThedomainisthesetD={r,s,t,u};therangeisthesetR={A,B,E}.
InExercises41–48,findthedomainandrangeofthefunction.
GiveanexampleofafunctionwhosedomainDhasthreeelementsandwhoserangeRhastwoelements.Doesa
functionexistwhosedomainDhastwoelementsandwhoserangeRhasthreeelements?
41. f(x)=−x
solution D:allreals;R:allreals
43. f(x)=x3
g(t)=t4
solution D:allreals;R:allreals
45. f(x)=|x|√
g(t)= 2−t
solution D:allreals;R:{y :y ≥0}
1
47. f(x)= 1
h(s)=x2
s
solution D:{x :x (cid:8)=0};R:{y :y >0}
InExercises49–52,determinewheref isincreasing.
1
g(t)=cos
t
49. f(x)=|x+1|
solution Agraphofthefunctiony =|x+1|isshownbelow.Fromthegraph,weseethatthefunctionisincreasing
ontheinterval(−1,∞).
y
2
1
x
−3 −2 −1 1
51. f(x)=x4
f(x)=x3
solution Agraphofthefunctiony =x4 isshownbelow.Fromthegraph,weseethatthefunctionisincreasingon
theinterval(0,∞).
y
12
8
4
x
−2 −1 1 2
SECTION 1.1 RealNumbers,Functions,andGraphs 5
In Exercises 53–58, find the zeros of f and sketch its graph by plotting points. Use symmetry and increase/decrease
1
informaftio(xn)w=hereappropriate.
x4+x2+1
53. f(x)=x2−4
solution Zeros:±2
Increasing:x >0
Decreasing:x <0
Symmetry:f(−x)=f(x)(evenfunction).So,y-axissymmetry.
y
4
2
x
−2 −1 1 2
−2
−4
55. f(x)=x3−4x
f(x)=2x2−4
solution Zeros:0,±2;Symmetry:f(−x)=−f(x)(oddfunction).Sooriginsymmetry.
y
10
5
x
−2 −1−5 1 2
−10
57. f(x)=2−x3
f(x)=x3 √
solution Thisisanx-axisreflectionofx3translatedup2units.Thereisonezeroatx = 32.
y
20
10
x
−2 −1−10 1 2
−20
59. WhichofthecurvesinFigure26isthegraphofafunction?
1
f(x)=
(x−1)2+1 y y
x
x
(A) (B)
y y
x x
(C) (D)
FIGURE 26
solution (B)isthegraphofafunction.(A),(C),and(D)allfailtheverticallinetest.
61. Determinewhetherthefunctioniseven,odd,orneither.
Determinewhetherthefunctioniseven,odd,orneither.
(a) f(a()t)f=(xt)4=+x1t5+1 − t4−1t+1 (b()bg)(gt)(t=)=2tt−3−2−tt2
(c) G(c()θ)F=(t)si=nθ+1cosθ (d) H(θ)=sin(θ2)
t4+t2
6 CHAPTER 1 PRECALCULUSREVIEW
solution
(a) Because
1 1 1 1
f(−t)= − = −
(−t)4+(−t)+1 (−t)4−(−t)+1 t4−t+1 t4+t+1
(cid:6) (cid:7)
1 1
=− − =−f(t),
t4+t+1 t4−t+1
1 1
f(t)= − isanoddfunction.
t4+t+1 t4−t+1
(b) Becauseg(−t)=2−t −2−(−t)=2−t −2t =−(2t −2−t)=−g(t),g(t)=2t −2−t isanoddfunction.
(c) BecauseG(−θ)=sin(−θ)+cos(−θ)=−sinθ+cosθ equalsneitherG(θ)nor−G(θ),G(θ)=sinθ+cosθ is
neitheranevenfunctionnoranoddfunction.
(d) BecauseH(−θ)=sin((−θ)2)=sin(θ2)=H(θ),H(θ)=sin(θ2)isanevenfunction.
(cid:6) (cid:7)
1−x
63. ShoWwrtihteatff(x(x))==2lxn4−15+x3x+1is2axn2o−dd3xfu+nc4tioasn.thesumofanevenandanoddfunction.
solution Bytherulesoflogarithmicfunctions,
(cid:6) (cid:7) (cid:6) (cid:7) (cid:6) (cid:7) (cid:6) (cid:7)
1−(−x) 1+x 1−x −1 1−x
f(−x)=ln =ln =ln =−ln =−f(x).
1+(−x) 1−x 1+x 1+x
Therefore,f isanoddfunction.
InExercSitsaetse6w5h–e7t0h,elretthfefbuentchteiofnunisctiinocnresahsoinwgn,idnecFriegausrieng2,7o.rneither.
(a) Surfaceareaofasphereasafunctionofitsradius
y
(b) Temperatureatapointontheequatorasafunctionoftime
4
(c) Priceofanairlineticketasafunctionofthepriceofoil
3
(d) Pressureofthegasinapistonasafunctionofvolume
2
1
0 x
1 2 3 4
FIGURE 27
65. Findthedomainandrangeoff.
solution D:[0,4];R:[0,4]
(cid:8) (cid:9)
67. SkeStckhettchhetghreapghraspohfsyof=yf=(2fx()x,y+=2)fan12dxy,=anfd(yx)=+22f.(x).
solution Thegraphofy =f(2x)isobtainedbycompressingthegraphofy =f(x)horizontallybyafactorof2(see
thegraphbelowontheleft).Thegraphofy =f(1x)isobtainedbystretchingthegraphofy =f(x)horizontallybya
2
factorof2(seethegraphbelowinthemiddle).Thegraphofy =2f(x)isobtainedbystretchingthegraphofy =f(x)
verticallybyafactorof2(seethegraphbelowontheright).
y y y
4 4 8
3 3 6
2 2 4
1 1 2
x x x
1 2 3 4 2 4 6 8 1 2 3 4
f(2x) f(x/2) 2f(x)
69. Extendthegraphoff to[−4,4]sothatitisanevenfunction.
Sketchthegraphsofy =f(−x)andy =−f(−x).
solution Tocontinuethegraphoff(x)totheinterval[−4,4]asanevenfunction,reflectthegraphoff(x)across
they-axis(seethegraphbelow).
y
4
3
2
1
x
−4 −2 2 4
SECTION 1.1 RealNumbers,Functions,andGraphs 7
71. Supposethatf hasdomain[4,8]andrange[2,6].Findthedomainandrangeof:
Extendthegraphoff to[−4,4]sothatitisanoddfunction.
(a) y =f(x)+3 (b) y =f(x+3)
(c) y =f(3x) (d) y =3f(x)
solution
(a) f(x)+3 is obtained by shifting f(x) upward three units.Therefore, the domain remains [4,8], while the range
becomes[5,9].
(b) f(x+3)isobtainedbyshiftingf(x)leftthreeunits.Therefore,thedomainbecomes[1,5],whiletherangeremains
[2,6].
(c) f(3x)isobtainedbycompressingf(x)horizontallybyafactorofthree.Therefore,thedomainbecomes[4,8],
3 3
whiletherangeremains[2,6].
(d) 3f(x)isobtainedbystretchingf(x)verticallybyafactorofthree.Therefore,thedomainremains[4,8],whilethe
rangebecomes[6,18].
73. Supposethatthegraphoff(x) = sinx iscompressedhorizontallybyafactorof2andthenshifted5unitstothe
right. Letf(x)=x2.Sketchthegraphover[−2,2]of:
(a) W(a)hayti=sthfe(xeq+ua1ti)onforthenewgraph? (b) y =f(x)+1
(b) W(c)hayti=sthfe(5exqu)ationifyoufirstshiftby5andthencompres(sdb)yy2=? 5f(x)
(c) Verifyyouranswersbyplottingyourequations.
solution
(a) Letf(x) = sinx.Aftercompressingthegraphoff horizontallybyafactorof2,weobtainthefunctiong(x) =
f(2x)=sin2x.Shiftingthegraph5unitstotherightthenyields
h(x)=g(x−5)=sin2(x−5)=sin(2x−10).
(b) Letf(x)=sinx.Aftershiftingthegraph5unitstotheright,weobtainthefunctiong(x)=f(x−5)=sin(x−5).
Compressingthegraphhorizontallybyafactorof2thenyields
h(x)=g(2x)=sin(2x−5).
(c) The figure below at the top left shows the graphs of y = sinx (the dashed curve), the sine graph compressed
horizontallybyafactorof2(thedash,doubledotcurve)andthenshiftedright5units(thesolidcurve).Comparethislast
graphwiththegraphofy =sin(2x−10)shownatthebottomleft.
Thefigurebelowatthetoprightshowsthegraphsofy =sinx(thedashedcurve),thesinegraphshiftedtotheright
5units(thedash,doubledotcurve)andthencompressedhorizontallybyafactorof2(thesolidcurve).Comparethislast
graphwiththegraphofy =sin(2x−5)shownatthebottomright.
y y
1 1
x x
−6 −4 −2 2 4 6 −6 −4 −2 2 4 6
−1 −1
y y
1 1
x x
−6 −4 −2 2 4 6 −6 −4 −2 2 4 6
−1 −1
(cid:8) (cid:9)
75. SkeFtcighutrhee2g8rasphhowofsyth=egfra(p2hx)ofanfd(xy)==f|x|12+x 1,.wMheartechft(hxe)f=unc|xti|o+ns1(a()F–i(geu)rwe2it8h)t.heirgraphs(i)–(v).
solu(at)ioyn=Tfh(exg−ra1p)hofy = f(2x)isob(tabi)neyd=by−cfom(xp)ressingthegraphofy =(cf)(yx)=h−orfiz(oxn)ta+lly2byafactorof2
(see(tdh)egyra=phfb(exlo−w1o)n−th2eleft).Thegraph(oef)yy==ff((1xx+)is1)obtainedbystretchingthegraphofy =f(x)horizontally
2
byafactorof2(seethegraphbelowontheright).
y y
6 6
4 4
2 2
x x
−3 −2 −1 1 2 3 −3 −2 −1 1 2 3
f(2x) f(x/2)
8 CHAPTER 1 PRECALCULUSREVIEW
77. Definef(x)tobethelargerofxand2−x.Sketchthegraphoff.Whatareitsdomainandrange?Expressf(x)in
termsofFtihnedatbhseofluuntectvioanlufefwunhcotsioeng.raphisobtainedbyshiftingtheparabolay =x2 by3unitstotherightand4units
down,asinFigure29.
solution
y
2
1
x
−1 1 2 3
Thegraphofy =f(x)isshownabove.Clearly,thedomainoff isthesetofallrealnumberswhiletherangeis{y|y ≥1}.
NoticethegraphhasthestandardV-shapeassociatedwiththeabsolutevaluefunction,butthebaseoftheVhasbeen
translatedtothepoint(1,1).Thus,f(x)=|x−1|+1.
79. Showthatthesumoftwoevenfunctionsisevenandthesumoftwooddfunctionsisodd.
solutiFoonreaEcvhecnu:r(vfe+ingF)i(g−urxe)3=0,fst(a−texw)h+etgh(e−rixt)isev=seynmfm(ext)ri+cwgi(txh)r=esp(efct+togt)h(ex)y-axis,theorigin,both,orneither.
Odd:(f +g)(−x)=f(−x)+g(−x)o=dd−f(x)+−g(x)=−(f +g)(x)
81. Provethattheonlyfunctionwhosegraphissymmetricwithrespecttoboththey-axisandtheoriginisthefunction
f(x)=S0u.pposethatf andgarebothodd.Whichofthefollowingfunctionsareeven?Whichareodd?
(a) y =f(x)g(x) (b) y =f(x)3
solution Acircleofradius1withitscenterattheoriginissymmetricalbothwithrespecttothey-axisandtheorigin.
T(hce)oynl=yffu(nxc)ti−ongh(xav)ing both symmetries is f(x) = 0. F(do)r iyf=f fis(xsy)mmetric with respect to the y-axis, then
f(−x) = f(x). If f is also symmetric with respect to the origin, theng(fx()−x) = −f(x). Thus f(x) = −f(x) or
2f(x)=0.Finally,f(x)=0.
Further Insights and Challenges
83. Showthatafractionr =a/binlowesttermshasafinitedecimalexpansionifandonlyif
Provethetriangleinequality(|a+b|≤|a|+|b|)byaddingthetwoinequalities
b=2n5m forsomen,m≥0.
−|a|≤a≤|a|, −|b|≤b≤|b|
Hint:Observethatrhasafinitedecimalexpansionwhen10NrisanintegerforsomeN ≥0(andhencebdivides10N).
solution Supposer hasafinitedecimalexpansion.ThenthereexistsanintegerN ≥0suchthat10Nr isaninteger,
callitk.Thus,r =k/10N.Becausetheonlyprimefactorsof10are2and5,itfollowsthatwhenr iswritteninlowest
terms,itsdenominatormustbeoftheform2n5mforsomeintegersn,m≥0.
a a
Conversely, suppose r = a/b in lowest with b = 2n5m for some integers n,m ≥ 0. Then r = = or
b 2n5m
2n5mr = a.Ifm ≥ n,then2m5mr = a2m−n orr = a2m−n andthusr hasafinitedecimalexpansion(lessthanor
10m
equaltomterms,tobeprecise).Ontheotherhand,ifn>m,then2n5nr =a5n−morr = a5n−m andonceagainrhas
10n
afinitedecimalexpansion.
85. LetpA=fupn1ct.io.n.pfsibsesaynmimnteetgriecrwwiitthhrdeisgpitescptt1o,t.h.e.,vperst.icSahlolwinethxat=aiff(a−x)=f(a+x).
(a) Drawthegraphofafunctionthatissymmetricwithrespecttox =2.
p
(b) Showthatiff issymmetricwithrespecttox =10as,−the1n=g(0x.)p=1.f..(pxs+a)iseven.
solution
(a) TUhseertehaisretomfianndytphoesdsiebciilmitiaelse,xopnaenosfiownhoifchri=s: 2.Notethat
11
y 2 18
r = =
11 102−1
2
1
x
−1 1 2 3 4 5
y = |x − 2|
(b) Letg(x)=f(x+a).Then
g(−x)=f(−x+a)=f(a−x)
=f(a+x) symmetrywithrespecttox =a
=g(x)
Thus,g(x)iseven.
Formulateaconditionforf tobesymmetricwithrespecttothepoint(a,0)onthex-axis.
SECTION 1.2 LinearandQuadraticFunctions 9
1.2 Linear and Quadratic Functions
Preliminary Questions
1. Whatistheslopeoftheliney =−4x−9?
solution Theslopeoftheliney =−4x−9is−4,givenbythecoefficientofx.
2. Arethelinesy =2x+1andy =−2x−4perpendicular?
solution Theslopesofperpendicularlinesarenegativereciprocalsofoneanother.Becausetheslopeofy =2x+1
is2andtheslopeofy =−2x−4is−2,thesetwolinesarenotperpendicular.
3. Whenisthelineax+by =cparalleltothey-axis?Tothex-axis?
solution Thelineax+by =cwillbeparalleltothey-axiswhenb=0andparalleltothex-axiswhena=0.
4. Supposey =3x+2.Whatis(cid:4)yifxincreasesby3?
solution Becausey =3x+2isalinearfunctionwithslope3,increasingxby3willleadto(cid:4)y =3(3)=9.
5. Whatistheminimumoff(x)=(x+3)2−4?
solution Because(x+3)2≥0,itfollowsthat(x+3)2−4≥−4.Thus,theminimumvalueof(x+3)2−4is−4.
6. Whatistheresultofcompletingthesquareforf(x)=x2+1?
solution Becausethereisnoxterminx2+1,completingthesquareonthisexpressionleadsto(x−0)2+1.
Exercises
InExercises1–4,findtheslope,they-intercept,andthex-interceptofthelinewiththegivenequation.
1. y =3x+12
solution Becausetheequationofthelineisgiveninslope-interceptform,theslopeisthecoefficientofx andthe
y-interceptistheconstantterm:thatis,m=3andthey-interceptis12.Todeterminethex-intercept,substitutey =0
andthensolveforx:0=3x+12orx =−4.
3. 4x+9y =3
y =4−x
solution Todeterminetheslopeandy-intercept,wefirstsolvetheequationforytoobtaintheslope-interceptform.
This yields y = −4x + 1. From here, we see that the slope is m = −4 and the y-intercept is 1.To determine the
9 3 9 3
x-intercept,substitutey =0andsolveforx:4x =3orx = 3.
4
InExercises5–8,findtheslopeoftheline.
y−3= 1(x−6)
2
5. y =3x+2
solution m=3
7. 3x+4y =12
y =3(x−9)+2
solution Firstsolvetheequationforytoobtaintheslope-interceptform.Thisyieldsy =−3x+3.Theslopeofthe
4
lineisthereforem=−3.
4
InExerc3ixse+s94–y20=,fi−n8dtheequationofthelinewiththegivendescription.
9. Slope3,y-intercept8
solution Usingtheslope-interceptformfortheequationofaline,wehavey =3x+8.
11. Slope3,passesthrough(7,9)
Slope−2,y-intercept3
solution Usingthepoint-slopeformfortheequationofaline,wehavey−9=3(x−7)ory =3x−12.
13. Horizontal,passesthrough(0,−2)
Slope−5,passesthrough(0,0)
solution Ahorizontallinehasaslopeof0.Usingthepoint-slopeformfortheequationofaline,wehavey−(−2)=
0(x−0)ory =−2.
15. Paralleltoy =3x−4,passesthrough(1,1)
Passesthrough(−1,4)and(2,7)
solution Becausetheequationy =3x−4isinslope-interceptform,wecanreadilyidentifythatithasaslopeof3.
Parallellineshavethesameslope,sotheslopeoftherequestedlineisalso3.Usingthepoint-slopeformfortheequation
ofaline,wehavey−1=3(x−1)ory =3x−2.
Passesthrough(1,4)and(12,−3)
10 CHAPTER 1 PRECALCULUSREVIEW
17. Perpendicularto3x+5y =9,passesthrough(2,3)
solution Westartbysolvingtheequation3x+5y =9forytoobtaintheslope-interceptformfortheequationofa
line.Thisyields
3 9
y =− x+ ,
5 5
fromwhichweidentifytheslopeas−3.Perpendicularlineshaveslopesthatarenegativereciprocalsofoneanother,so
5
theslopeofthedesiredlineism⊥= 5.Usingthepoint-slopeformfortheequationofaline,wehavey−3= 5(x−2)
3 3
ory = 5x− 1.
3 3
19. Horizontal,passesthrough(8,4)
Vertical,passesthrough(−4,9)
solution Ahorizontallinehasslope0.Usingthepointslopeformfortheequationofaline,wehavey−4=0(x−8)
ory =4.
21. Findtheequationoftheperpendicularbise(cid:6)ctorofthesegm(cid:7)entjoining(1,2)and(5,4)(Figure12).Hint:Themidpoint
Slope3,x-intercept6 a+c b+d
Qofthesegmentjoining(a,b)and(c,d)is , .
2 2
y
Perpendicular
bisector
(5, 4)
Q
(1, 2)
x
FIGURE 12
solution Theslopeofthesegmentjoining(1,2)and(5,4)is
4−2 1
m= =
5−1 2
andthemidpointofthesegment(Figure12)is
(cid:6) (cid:7)
1+5 2+4
midpoint= , =(3,3)
2 2
Theperpendicularbisectorhasslope−1/m = −2andpassesthrough(3,3),soitsequationis:y−3 = −2(x−3)or
y =−2x+9.
23. Findanequationofthelinewithx-interceptx =4andy-intercepty =3.
Intercept-InterceptForm Showthatifa,b (cid:8)=0,thenthelinewithx-interceptx =aandy-intercepty =b
solution FromExercise22, x + y =1or3x+4y =12.
hasequation(Figure13) 4 3
25. Determinewhetherthereexistsaconstantcsuchthatthelinex+cy =1:
(a) hasFsilnodpey4s.uchthat(3,y)liesonthelineofslopemx=+2(ytbh)r=opu1agshse(1s,th4r)o.ugh(3,1).
a b
(c) ishorizontal. (d) isvertical.
solution
(a) Rewritingtheequationofthelineinslope-interceptformgivesy =−x + 1.Tohaveslope4requires−1 =4or
c c c
c=−1.
4
(b) Substitutingx =3andy =1intotheequationofthelinegives3+c=1orc=−2.
(c) From(a),weknowtheslopeofthelineis−1.Thereisnovalueforcthatwillmakethisslopeequalto0.
c
(d) Withc=0,theequationbecomesx =1.Thisistheequationofaverticalline.
27. SupposethatthenumberofacertaintypeofcomputerthatcanbesoldwhenitspriceisP (indollars)isgivenby
alinearAfusnsucmtioenthNat(Pth)e.nDuemtebremriNneoNfc(Pon)ciefrtNti(c1k0e0ts0)th=atc1a0n,0b0e0saonlddaNta(1p5r0ic0e)o=fP7,d5o0l0la.rWsphearttiiscktheeticshaanligneea(cid:4)rNfuninctitohne
numNbe(rPo)ffcoorm1p0u≤terPss≤old40if.tDheetperrmiceiniesNin(cPre)as(ceadllbeyd(cid:4)thPed=em1a0n0ddfoulnlacrtsio?n)ifN(10)=500andN(40)=0.Whatisthe
decrease(cid:4)N inthenumberofticketssoldifthepriceisincreasedby(cid:4)P =5dollars?
solution Wefirstdeterminetheslopeoftheline:
10000−7500 2500
m= = =−5.
1000−1500 −500
KnowingthatN(1000)=10000,itfollowsthat
N−10000=−5(P −1000), or N(P)=−5P +15000.
Becausetheslopeofthedemandfunctionis−5,a100dollarincreaseinpricewillleadtoadecreaseinthenumberof
computerssoldof5(100)=500computers.