英文导数试题及参考答案

英文导数试题及参考答案考试时间：______分钟总分：______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-x⁴)usingtheproductrule.4.Giventhefunctiong(x)=(x⁴+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.试卷答案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-x⁴.Usingtheproductrule,dy/dx=u*v'+v*u'.*u'=d/dx(5x²+3x)=10x+3.*v'=d/dx(2x³-x⁴)=6x²-4x³.*dy/dx=(5x²+3x)(6x²-4x³)+(2x³-x⁴)(10x+3)*=(30x⁴-20x⁵+18x³-12x⁴)+(20x⁴-8x⁶+30x³-3x⁴)*=(-20x⁵-12x⁴+18x³)+(20x⁴-8x⁶+30x³-3x⁴)*=-8x⁶-3x⁴+48x³4.Letu=x⁴+1andv=x-2.Usingthequotientrule,dy/dx=(u*v'-v*u')/v².*u'=d/dx(x⁴+1)=4x³.*v'=d/dx(x-2)=1.*dy/dx=((x⁴+1)(1)-(x-2)(4x³))/(x-2)²*=(x⁴+1-4x⁴+8x³)/(x-2)²*=(-3x⁴+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-x⁴)/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