英文導數試題及參考答案

英文導數試題及參考答案考試時間：______分鐘總分：______分姓名：______Instructions:Answerallquestionsinthespacesprovided.Showallyourworking.1.Considerthefunctionfdefinedbyf(x)=x3-6x2+9x+1.(a)Findthefirstderivative,f'(x).(b)Determinethecriticalpointsoff(x).(c)Usethefirstderivativetesttoclassifyeachcriticalpointasalocalmaximum,localminimum,orneither.(d)Findtheintervalsonwhichthefunctionf(x)isincreasinganddecreasing.2.Thepositionofaparticlemovingalongastraightlineisgivenbythefunctions(t)=t3-4t2+7,wheresismeasuredinmetersandtismeasuredinseconds.(a)Calculatethevelocityfunction,v(t)=s'(t).(b)Determinetheaccelerationfunction,a(t)=v'(t)=s''(t).(c)Findthetime(s)whentheparticleisatrest(velocityequalszero).(d)Findthetimeinterval(s)duringwhichtheparticleismovingtotheright(velocityispositive).(e)Findthetime(s)whentheparticlechangesitsdirectionofmotion.3.Findthederivativeofthefunctiony=(5x2+3x)*(2x3-x?)usingtheproductrule.4.Giventhefunctiong(x)=(x?+1)/(x-2),findg'(x)usingthequotientrule.5.Differentiatethefunctionh(x)=sin(3x2)+cos(x3)usingthechainrule.6.Findthederivativeofthefunctionk(x)=ln(x√(x+1))wherex>-1.7.Findy'ifx2y+y3=x+2byusingimplicitdifferentiation.8.Finddy/dxfortheparametricequationsx=t2-1andy=t(t-2).9.ThecostfunctionforproducingacertainitemisC(x)=0.5x2+20x+150,wherexisthenumberofitemsproduced.(a)Findtheaveragecostfunction,AC(x)=C(x)/x.(b)Determinethevalueofxthatminimizestheaveragecost.Justifyyouranswerusingthefirstderivativetestontheaveragecostfunction.10.Sketchthegraphofafunctionf(x)thatsatisfiesthefollowingconditions:*f'(x)>0forx<-1andx>2*f'(x)<0for-1<x<2*f'(x)doesnotexistatx=-1andx=2*f(x)hasalocalmaximumatx=-1*f(x)hasalocalminimumatx=2*limf(x)asx→-1?=3andlimf(x)asx→-1?=1*f(x)iscontinuousatx=2*f(2)=011.Showthatthefunctionf(x)=x3-3x+2isincreasingontheinterval(-∞,-1)and(1,∞),anddecreasingontheinterval(-1,1).12.Arectangularboxhasasquarebaseandanopentop.Thesurfaceareaoftheboxis108squareinches.Findthedimensionsoftheboxthatmaximizethevolume.試卷答案1.(a)f'(x)=3x2-12x+9(b)Criticalpointsoccurwheref'(x)=0.Solving3x2-12x+9=0givesx2-4x+3=0,whichfactorsas(x-1)(x-3)=0.Thus,criticalpointsarex=1andx=3.(c)Evaluatef'(x)aroundcriticalpoints:*Forx=0(leftof1):f'(0)=3(0)2-12(0)+9=9>0(increasing).*Forx=2(between1and3):f'(2)=3(2)2-12(2)+9=12-24+9=-3<0(decreasing).*Forx=4(rightof3):f'(4)=3(4)2-12(4)+9=48-48+9=9>0(increasing).*Therefore,x=1isalocalmaximum,x=3isalocalminimum.(d)f(x)isincreasingontheintervals(-∞,1)and(3,∞).f(x)isdecreasingontheinterval(1,3).2.(a)v(t)=s'(t)=d/dt(t3-4t2+7)=3t2-8t.(b)a(t)=v'(t)=s''(t)=d/dt(3t2-8t)=6t-8.(c)Theparticleisatrestwhenv(t)=0.Solving3t2-8t=0givest(3t-8)=0.Thus,t=0ort=8/3seconds.(d)Theparticlemovestotherightwhenv(t)>0.Wesolve3t2-8t>0.Factoringgivest(3t-8)>0.Thecriticalpointsaret=0andt=8/3.Testingintervals:*t<0:Bothfactorsarenegative,productispositive.*0<t<8/3:Firstfactorispositive,secondisnegative,productisnegative.*t>8/3:Bothfactorsarepositive,productispositive.*Therefore,theparticlemovestotherightontheintervals(-∞,0)and(8/3,∞).(e)Theparticlechangesdirectionwhenitsvelocitychangessign,whichoccursatthecriticalpointswherev(t)=0.Thus,theparticlechangesdirectionatt=0andt=8/3seconds.3.Letu=5x2+3xandv=2x3-x?.Usingtheproductrule,dy/dx=u*v'+v*u'.*u'=d/dx(5x2+3x)=10x+3.*v'=d/dx(2x3-x?)=6x2-4x3.*dy/dx=(5x2+3x)(6x2-4x3)+(2x3-x?)(10x+3)*=(30x?-20x?+18x3-12x?)+(20x?-8x?+30x3-3x?)*=(-20x?-12x?+18x3)+(20x?-8x?+30x3-3x?)*=-8x?-3x?+48x34.Letu=x?+1andv=x-2.Usingthequotientrule,dy/dx=(u*v'-v*u')/v2.*u'=d/dx(x?+1)=4x3.*v'=d/dx(x-2)=1.*dy/dx=((x?+1)(1)-(x-2)(4x3))/(x-2)2*=(x?+1-4x?+8x3)/(x-2)2*=(-3x?+8x3+1)/(x-2)25.Letu=3x2andv=sin(u).Theny=v+cos(x3).*dy/dx=d/dx(sin(3x2))+d/dx(cos(x3))*Ford/dx(sin(3x2)):Letw=3x2.Thendy/dx=cos(w)*dw/dx=cos(3x2)*(6x).*Ford/dx(cos(x3)):Letz=x3.Thendy/dx=-sin(z)*dz/dx=-sin(x3)*(3x2).*dy/dx=6xcos(3x2)-3x2sin(x3)6.y=ln(x√(x+1))=ln(x*(x+1)^(1/2))=ln(x)+ln((x+1)^(1/2))=ln(x)+(1/2)ln(x+1).dy/dx=d/dx[ln(x)+(1/2)ln(x+1)]=(1/x)+(1/2)*(1/(x+1))*d/dx(x+1)=(1/x)+(1/2)*(1/(x+1))*(1)=(1/x)+1/(2(x+1))7.Differentiatebothsidesofx2y+y3=x+2withrespecttox.Usetheproductruleforx2yandthechainrulefory3.d/dx(x2y)+d/dx(y3)=d/dx(x)+d/dx(2)(x2*dy/dx+2x*y)+3y2*dy/dx=1+0Collecttermsinvolvingdy/dx:x2*dy/dx+3y2*dy/dx=1-2x*yFactoroutdy/dx:dy/dx*(x2+3y2)=1-2xySolvefordy/dx:dy/dx=(1-2xy)/(x2+3y2)8.Givenx=t2-1andy=t(t-2)=t2-2t.Finddx/dt:dx/dt=d/dt(t2-1)=2t.Finddy/dt:dy/dt=d/dt(t2-2t)=2t-2.Usingthechainrule,dy/dx=(dy/dt)/(dx/dt).dy/dx=(2t-2)/(2t)=(2(t-1))/(2t)=(t-1)/t9.(a)AC(x)=C(x)/x=(0.5x2+20x+150)/x=0.5x+20+150/x.(b)Tominimizetheaveragecost,findthecriticalpointsofAC(x).DifferentiateAC(x)withrespecttox.d(AC)/dx=d/dx(0.5x+20+150/x)=0.5-150/x2Setthederivativeequaltozero:0.5-150/x2=0150/x2=0.5x2=150/0.5x2=300x=√300=10√3(Sincexrepresentsitemsproduced,wetakethepositiveroot).Toconfirmthisisaminimum,checkthesecondderivative:d2(AC)/dx2=d/dx(0.5-150/x2)=0-(-150*-2/x3)=-300/x3Sincex>0,-300/x3isalwaysnegative.Therefore,thefunctionAC(x)isconcavedown,andthecriticalpointx=10√3correspondstoalocalminimum,whichisalsotheglobalminimumforx>0.Theaveragecostisminimizedwhen10itemsareproduced.10.(Agraphcannotbeprovided,butthedescriptionisasfollows:)*Thegraphhasalocalmaximumatx=-1.Thisisindicatedbyapeak,andthederivativechangesfrompositivetonegativeatthispoint.*Thegraphhasalocalminimumatx=2.Thisisindicatedbyatrough,andthederivativechangesfromnegativetopositiveatthispoint.*Thederivativeisundefinedatx=-1andx=2.Thiscouldbeduetoaverticaltangent,acusp,oradiscontinuityinthederivativeatthesepoints.*Thegraphisincreasingforx<-1andx>2(abovethex-axisforthederivative).*Thegraphisdecreasingfor-1<x<2(belowthex-axisforthederivative).*Asxapproaches-1fromtheright,thefunctionvalueapproaches3.Asxapproaches-1fromtheleft,thefunctionvalueapproaches1.*Thegraphiscontinuousatx=2,meaningthereisnobreakorjumpatx=2.*Thepoint(2,0)isonthegraph.11.Letf(x)=x3-3x+2.Findthederivative:f'(x)=3x2-3=3(x2-1)=3(x-1)(x+1).Todetermineintervalsofincreaseanddecrease,findwheref'(x)=0.Solving3(x-1)(x+1)=0givescriticalpointsx=-1andx=1.Testthesignoff'(x)intheintervalsdeterminedbythecriticalpoints:(-∞,-1),(-1,1),(1,∞).*Chooseatestpointin(-∞,-1),e.g.,x=-2:f'(-2)=3((-2)-1)((-2)+1)=3(-3)(-1)=9>0.f(x)isincreasingon(-∞,-1).*Chooseatestpointin(-1,1),e.g.,x=0:f'(0)=3(0-1)(0+1)=3(-1)(1)=-3<0.f(x)isdecreasingon(-1,1).*Chooseatestpointin(1,∞),e.g.,x=2:f'(2)=3(2-1)(2+1)=3(1)(3)=9>0.f(x)isincreasingon(1,∞).Conclusion:f(x)isincreasingon(-∞,-1)and(1,∞),anddecreasingon(-1,1).12.Letthesidelengthofthesquarebasebexandtheheightbeh.ThesurfaceareaSisgivenbyS=x2+4xh=108.Solveforhintermsofx:4xh=108-x2,soh=(108-x2)/(4x).ThevolumeVoftheboxisV=basearea*height=x2h.Substitutetheexpressionforh:V(x)=x2*[(108-x2)/(4x)]=(108x-x?)/4.Tomaximizevolume,findthecriticalpointsofV(x).Sincexmustbepositive,weconsiderV(x)=(27x-x2)/2.DifferentiateV(x)withrespecttox:V'(x)=(1/2)*d/dx(27x-x2)=(1/2)*(27-2x).Setthederivativeequaltozero:(1/2)*(27-2x)=027-2x=02x=27x=27/2Toconfirmthisisamaximum,checkthesecondderivative:V''(x)=(1/2)*d/dx