1 PRECALCULUS REVIEW 1.1 Real Numbers, Functions, and Graphs Preliminary Questions 1. Giveanexampleofnumbersaandbsuchthata|b|. solution Takea=−3andb=1.Thena1=|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 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,writetheinequalityintheforma2} 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) Theintervaldescribedbytheinequality1esbet,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|s0} 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.

Student Solutions Manual - Calculus Early Transcendentals

Similar Student Solutions Manual - Calculus Early Transcendentals