關(guān)于導(dǎo)數(shù)的英文題目和答案

關(guān)于導(dǎo)數(shù)的英文題目和答案考試時間：______分鐘總分：______分姓名：______Instructions:Answerallquestionsinthespacesprovided.Showallyourworkingforthenumericalanswers.UseBNC(BritishNationalCorpus)styleEnglishwhereappropriate.1.Definethederivativeofafunctionf(x)atapointx=aintermsofalimit.2.Giventhefunctionf(x)=3x^3-7x^2+2x-5,findf'(x).3.Usingthequotientrule,differentiatethefunctiong(t)=(t^2+1)/(t-2)withrespecttot.4.Findthesecondderivative,y'',ofthefunctiony=x*sin(x).5.Leth(z)=z^(1/2)*e^(3z).Usetheproductruleandthechainruletofindh'(z).6.Thepositionofaparticlemovingalongastraightlineisgivenbythefunctions(t)=t^3-6t^2+9t+1,wheresisinmetersandtisinseconds.a)Findthevelocityfunctionv(t)andtheaccelerationfunctiona(t).b)Determinethetimeintervalsduringwhichtheparticleismovingtotheright(i.e.,v(t)>0).7.Explainthedifferencebetweena'criticalpoint'andan'inflectionpoint'ofadifferentiablefunction.8.Findthelocalmaximumandminimumvaluesofthefunctionf(x)=x^3-3x^2+4,ifanyexist.9.Determinetheintervalsonwhichthefunctionf(x)=x^4-8x^2+12isconcaveupandconcavedown.Identifyanypointsofinflection.10.Arectangularboxhasasquarebaseandanopentop.Thesurfaceareaoftheboxisrestrictedto108squareunits.Findthedimensionsoftheboxthatmaximizeitsvolume.11.Considerthefunctionf(x)=e^(-x^2).a)Findthefirstderivativef'(x).b)Determinetheintervalswherethefunctionisincreasing.c)Explainwhythefunctiondoesnothaveaglobalmaximumvalueontheinterval(-∞,∞).12.Amanufacturerproduceswidgets.ThecostC(x)indollarsforproducingxwidgetsisgivenbyC(x)=50x+10,000.TherevenueR(x)indollarsfromsellingxwidgetsisR(x)=75x-0.01x^2.a)FindthemarginalcostfunctionC'(x)andthemarginalrevenuefunctionR'(x).b)Determinetheproductionlevelxthatmaximizesprofit.Justifyyouranswer.13.WriteaparagraphinEnglishexplainingthesignificanceofthederivativeinunderstandingthebehaviorofafunction.Includedescriptionsofconceptssuchasincreasing/decreasingintervals,localextrema,andconcavity.14.ThetemperatureT(t)ofacoolingobjectattimet(inminutes)isgivenbyT(t)=60+100e^(-0.5t).a)Findtherateatwhichthetemperatureischangingwithrespecttotime,dT/dt.b)Whatistheinitialrateoftemperaturechange?Interpretthisvalueinpracticalterms.c)Afterhowmanyminuteswillthetemperaturebedecreasingatarateof10degreesperminute?Showyourcalculations.試卷答案1.Thederivativeofafunctionf(x)atapointx=aisdefinedasthelimitoftheaveragerateofchangeofthefunctionoveraninterval[a,a+h]astheintervalbecomesinfinitesimallysmall,representedmathematicallyas:f'(a)=lim(h→0)[f(a+h)-f(a)]/h.2.f'(x)=d/dx(3x^3-7x^2+2x-5)=3*3x^(3-1)-7*2x^(2-1)+2*1x^(1-1)-0=9x^2-14x+2.3.Usingthequotientrule((u/v)'=(u'v-uv')/v^2),whereu=t^2+1andv=t-2:u'=2tv'=1g'(t)=[(2t)(t-2)-(t^2+1)(1)]/(t-2)^2=[2t^2-4t-t^2-1]/(t-2)^2=[t^2-4t-1]/(t-2)^2.4.y=x*sin(x)Usingtheproductrule((uv)'=u'v+uv'),whereu=xandv=sin(x):u'=1v'=cos(x)y'=(1)*sin(x)+x*cos(x)=sin(x)+xcos(x)y''=d/dx(sin(x)+xcos(x))=cos(x)+[cos(x)+x(-sin(x))]=cos(x)+cos(x)-xsin(x)=2cos(x)-xsin(x).5.h(z)=z^(1/2)*e^(3z)Usingtheproductrule((uv)'=u'v+uv'),whereu=z^(1/2)andv=e^(3z):u'=(1/2)z^(-1/2)v'=3e^(3z)(usingchainruleone^(3z))h'(z)=[(1/2)z^(-1/2)]*e^(3z)+[z^(1/2)]*[3e^(3z)]=(1/2)z^(-1/2)e^(3z)+3z^(1/2)e^(3z)=e^(3z)[(1/2)z^(-1/2)+3z^(1/2)]=e^(3z)[(1+6z)/(2z^(1/2))].6.a)v(t)=s'(t)=d/dt(t^3-6t^2+9t+1)=3t^2-12t+9a)a(t)=v'(t)=d/dt(3t^2-12t+9)=6t-12b)Theparticlemovestotherightwhenv(t)>0:3t^2-12t+9>0t^2-4t+3>0(t-1)(t-3)>0Bythesignchartmethod,thesolutionist<1ort>3.Therefore,theparticleismovingtotherightontheintervals(-∞,1)and(3,∞).7.Acriticalpointofadifferentiablefunctionf(x)occursatx=cwhereeitherf'(c)=0orf'(c)isundefined.Thesepointsarepotentiallocationsforlocalmaxima,localminima,orpointsofinflection.Aninflectionpointofadifferentiablefunctionf(x)occursatx=dwheretheconcavityofthefunctionchanges(fromconcaveuptoconcavedownorviceversa).Thistypicallyhappenswherethesecondderivativef''(d)=0andchangessignaroundx=d.8.Tofindlocalextrema,findthecriticalpointsbysettingthefirstderivativetozero:f'(x)=3x^2-6xf'(x)=3x(x-2)=0Criticalpointsarex=0andx=2.Usethesecondderivativetest:f''(x)=6x-6f''(0)=6(0)-6=-6.Sincef''(0)<0,f(x)hasalocalmaximumatx=0.f''(2)=6(2)-6=6.Sincef''(2)>0,f(x)hasalocalminimumatx=2.Localmaximumvalue:f(0)=0^3-3(0)^2+4=4.Localminimumvalue:f(2)=2^3-3(2)^2+4=8-12+4=0.9.Tofindintervalsofconcavity,findthesecondderivativeanddeterminewhereitispositive(concaveup)ornegative(concavedown).f(x)=x^4-8x^2+12f'(x)=4x^3-16xf''(x)=12x^2-16Setf''(x)=0tofindpotentialinflectionpoints:12x^2-16=012x^2=16x^2=4/3x=±√(4/3)=±2/√3=±2√3/3.Testintervals:(-∞,-2√3/3),(-2√3/3,2√3/3),(2√3/3,∞).Choosetestpoints:e.g.,x=-2,x=0,x=2.f''(-2)=12(-2)^2-16=48-16=32>0.Concaveupon(-∞,-2√3/3).f''(0)=12(0)^2-16=-16<0.Concavedownon(-2√3/3,2√3/3).f''(2)=12(2)^2-16=48-16=32>0.Concaveupon(2√3/3,∞).Intervalsofconcavity:Concaveup:(-∞,-2√3/3)U(2√3/3,∞)Concavedown:(-2√3/3,2√3/3)Pointsofinflection:x=-2√3/3andx=2√3/3.10.Letxbethesidelengthofthesquarebaseandhbetheheightofthebox.ThesurfaceareaconstraintisS=x^2+4xh=108.Solveforh:h=(108-x^2)/(4x).ThevolumeVisV=basearea*height=x^2*h=x^2*[(108-x^2)/(4x)]=(108x-x^3)/4.TomaximizeV,finddV/dxandsetittozero:dV/dx=d/dx[(108x-x^3)/4]=(1/4)*[108-3x^2]=27-(3/4)x^2SetdV/dx=0:27-(3/4)x^2=0(3/4)x^2=27x^2=27*(4/3)=36x=6(sincexmustbepositivefordimensions).Findhusingthesurfaceareaconstraint:h=(108-6^2)/(4*6)=(108-36)/24=72/24=3.Thedimensionsare:Basesidelength=6units,Height=3units.11.a)f(x)=e^(-x^2)Usingthechainrule(d/dxe^u=e^u*du/dx),whereu=-x^2:f'(x)=e^(-x^2)*d/dx(-x^2)=e^(-x^2)*(-2x)=-2xe^(-x^2).b)Thefunctionf(x)isincreasingwheref'(x)>0:-2xe^(-x^2)>0Sincee^(-x^2)isalwayspositiveforallrealx,thesignoff'(x)dependssolelyonthesignof-2x.-2x>0=>x<0.Therefore,thefunctionisincreasingontheinterval(-∞,0).c)Thefunctionf(x)=e^(-x^2)doesnothaveaglobalmaximumvalueon(-∞,∞)because:-Asxapproaches±∞,e^(-x^2)approaches0.So,thefunctiongetsarbitrarilycloseto0butneverreachesitasanabsoluteminimumvalue.-Asxapproaches0,f(x)=e^(0)=1.Thisisthehighestvaluethefunctionattains.Sincethefunctionincreasestowardsx=0andthendecreasestowards0asxmovesawayfrom0,thevalueatx=0(whichis1)istheglobalmaximumvalueofthefunctionontheentirerealline.12.a)MarginalcostisthederivativeofthecostfunctionC(x):C'(x)=d/dx(50x+10,000)=50.MarginalrevenueisthederivativeoftherevenuefunctionR(x):R'(x)=d/dx(75x-0.01x^2)=75-0.02x.b)ProfitP(x)=Revenue-Cost=R(x)-C(x)=(75x-0.01x^2)-(50x+10,000)=25x-0.01x^2-10,000.Tomaximizeprofit,findthecriticalpointsofP(x)bysettingitsderivativetozero:P'(x)=d/dx(25x-0.01x^2-10,000)=25-0.02x.SetP'(x)=0:25-0.02x=00.02x=25x=25/0.02=1250.Toconfirmthisisamaximum,usethesecondderivativetest:P''(x)=d/dx(25-0.02x)=-0.02.SinceP''(x)=-0.02<0,thefunctionP(x)hasamaximumatx=1250.Theproductionlevelthatmaximizesprofitis1250widgets.13.Thederivativeofafunctionf(x),denotedasf'(x)ordy/dx,providescrucialinformationaboutthebehaviorofthefunction.Geometrically,thederivativeatapointx=arepresentstheslopeofthetangentlinetothecurvey=f(x)atthatpoint.Thisslopeindicatestherateofchangeofthefunction'svaluewithrespecttothechangeintheinputvariablex.Ifthederivativeispositiveoveraninterval,thefunctionisincreasingonthatinterval,meaningtheoutputvaluesaregettinglargerastheinputvaluesincrease.Ifthederivativeisnegativeoveraninterval,thefunctionisdecreasing,indicatingtheoutputvaluesaregettingsmallerastheinputvaluesincrease.Settingthederivativeequaltozerohelpslocatethecriticalpointsofthefunction,whicharepotentiallocationsforlocalmaximumorminimumvalues.Thesecondderivative,f''(x)ord^2y/dx^2,revealsinformationabouttheconcavityofthefunction.Whenf''(x)>0,thefunctionisconcaveuponthatinterval,meaningthegraphcurvesupwards,andanytangentlineliesbelowthecurve.Whenf''(x)<0,thefunctionisconcavedown,meaningthegraphcurvesdownwards,andanytangentlineliesabovethecurve.Pointswheretheconcavitychanges(fromconcaveuptoconcavedownorviceversa)arecalledinflectionpoints,andtheseoccurwheref''(x)=0andthesignoff''(x)changes.Insummary,thederivativeisanessentialtoolforanalyzingandunderstandingtheshape,direction,andrateofchangeofafunction's