第二學期微生一微積分Quiz(I)2009/03/18HYPERLINK\l"解答"參考解答18%StatetheExtremeValueTheorem.Letf(x)=|2x2-1|on[-2,2),findalllocalandglobalextremabythegraphoff(x).Giveanexamplethatf'(c)=0butf(x)hasneithermaximumnorminimumatc.14%StatetheMeanValueTheorem(MVT).Letf(x)=x(2-x)UsetheMVTtofindanintervalthatcontainsanumbercsothatf'(c)=0.20%Thegrowthequation(VonBertalanffyequation)L(x)=L∞-(L∞-L0)e-Kxforfishes,whereL(x)denotesthelengthatagex,L0thelengthatage0,andL∞theasymptoticmaximumattainablelength.WeassumeL∞>L0,andK(growthrate)>0.FindthefirstderivativeL'(x)andthesecondderivativeL''(x).ForL∞=8,L0=1andK=0.6,describeL(x)ofthefollowinggraph.18%Letf(x)=x3-(3/2)x2-6x+3,xRDeterminewherethefunctionisincreasingandwhereitisdecreasing.Determinewherethefunctionisconcaveupandwhereitisconcavedown.Findalllocalextremabythesecondderivativetest.Anddeterminetheglobalextrema.30%Forthefunctionf(x)=.Locatewheref(x)isincreasing,decreasingandwhereitisconcaveupanddown,findallthecriticalpoints,inflectionpoints,extrema,asymptotesandsketchthegraph.參考解答：1.18%解：(a)Iff(x)iscontinuouson[a,b],thenfhasaglobalmaximumandaglobalminimumon[a,b].(b)f(x)=|2x2-1|on[-2,2),localmaximum:atx=-2,x=0;localminimum:atglobalmaximum:atx=-2;globalminimum:at(c)Letf(x)=x3,xRf'(0)=0,howeverf(x)hasneithermaximumnorminimumatx=0.Sincef(1)=1>0,andf(-1)=-1<0.2.14%解：(a)Iff(x)iscontinuouson[a,b]anddifferentiableon(a,b),thenthereexistsatleastonec(a,b)suchthatf(b)-f(a)=f'(c)(b-a).(b)f(x)=x(2-x),f(0)=0=f(2),bytheMVTthereexitsacin(0,2)suchthatf'(c)=0.3.20%L(x)=L∞-(L∞-L0)e-Kx解：(a)L'(x)=-(L∞-L0)(-K)e-Kx=K(L∞-L0)e-Kx>0,thereforeL(x)isincreasingforxL''(x)=-K2(L∞-L0)e-Kx<0,thereforeL'(x)isdecreasingforx(b)L(x)isincreasingapproximatelytoL∞=8,howevertherateofgrowthdecreases.(魚的身長隨年齡增加而增長，但是增長的情形趨緩，最長不會超過8單位長度)4.18%Letf(x)=x3-(3/2)x2-6x+3,xRDeterminewherethefunctionisincreasingandwhereitisdecreasing.Determinewherethefunctionisconcaveupandwhereitisconcavedown.Findalllocalextremabythesecondderivativetest.