《高数双语》课件section 3-1.pptx

1、Chapter 31OverviewThe Mean Value TheoremL Hospital RulesTayors TheoremApplications of Derivatives2The Mean Value TheoremMean Value Theorem in Differential Calculus4ab1 2 xyo()yf x CSuppose and the corresponding figure is()yf x It is easy to see that there exist at least one point in the interval ,su

2、ch that(,)a b()0fx .Mean Value Theorem in Differential Calculus5See the following animation It is easy to see that when the ball runs into the right side of the line segment and begins to run back,the instantaneous velocity is zero.Mean Value Theorem in Differential Calculus60()()f xf x(Fermat lemma

3、)0:()RfU x()f xSuppose and is differentiable0 xat the point .0()f xIf is a local maximum(minimum)(),f xof the function 0()xU x that is ,we have0()()f xf x(or ).0()0fx thenwe have Proof0()()f xf x 0()xU x Suppose that ,.fSince function is differentiable0 xat the point ,00()()fxfx we have.By the defin

4、ition of derivative,0000000()()()lim,(,)()(0)xxf xf xfxxxxU xxx ,0000000()()()lim,(,)()(0)xxf xf xfxxxxU xxx .Mean Value Theorem in Differential Calculus70000000()()()lim,(,)()(0)xxf xf xfxxxxU xxx ,It is similar that this conclusion holds when 00()()0f xf xxx 0()().f xf x Proof (continued)0()()f xf

5、 x since ,we have ,00(,)xxx ,thus000(,)()xxxU x 00()()0f xf xxx When ,.Thus 0()0fx .0()0fx .And then,we have000()()()0fxfxfx.Finish.Mean Value Theorem in Differential Calculus8()0fx and solving for x.Note We cant expect to locate extreme values simply by settingOxy3yx 3()f xx Example(0)0f f has no m

6、aximum or minimum at 0.Mean Value Theorem in Differential Calculus9Note Stationary Point,Singular Point and Critical Point (1)If c is a point at which()0,f c we call c a stationary point;(2)If c is an interior point of ,Ia b where()f c fails to exist,we call c a singular point;(3)Any point of the th

7、ree types,including stationary point,singular point and end point,in the domain of a function is called a critical point of f(x).Mean Value Theorem in Differential Calculus10ab1 2 xyo()yf x C:,Rfa b (Rolles theorem)If satisfies the following conditions:,a b(1)f is continuous on the interval ;(,)a b(

8、2)f is differentiable in the interval ;()()f af b(3),(,)a b ()0f then,there exists at least one point ,such that .Mean Value Theorem in Differential Calculus11Then by the Fermats lemma,.()()f af b()0f Proof Since function f is continuous on the interval a,b,then,12,x xa b,such thatthere existMm 1)If

9、 ,then f is a constant()0f (,)a b and hence we have ,.Mm 2)If ,then at least one of M and m is not equal to f(a),since()Mf a;suppose()Mf b,so that .Then the maximum(,)a b value M must be achieved at a point .This means that(,)a b ()(,)Ua b ,and ,()()f xf ()xU such that ,.Finish.1212,()max(),()max()x

10、a bxa bf xf xMf xf xm.Mean Value Theorem in Differential Calculus12 Note If any one of the three conditions is not satisfied then the conclusion of Rolles theorem is not necessarily true.11,0,1)(),211.xxf xx ExampleyxO11Mean Value Theorem in Differential Calculus13 Prove that the equation has one an

11、d only one real root in the interval .3210 xx(1,0)denoted by .1x3()21f xxxProof Let ,1,0 then f is continuous on and(1)20f .By the zero theorem for continuous functions we know(1,0)that f has at least one zero point in the intervaland this point isWe now prove the zero point of in(-1,0)is unique.()f

12、 xSuppose it is not true,that is,2(1,0)x there is another zero point21xx(say ).12(,)(1,0),x x Then,there must be at least one point()0f such that .(1,0)x 2()320fxx This is impossible since ,and then the conclusion holds.Finish.Mean Value Theorem in Differential Calculus14:0,1Rf(1)0.f Suppose that th

13、e function is continuous on the interval0,1,(0,1)and differentiable in ,Prove that there exists at least one(0,1)c ,such that()()f cfcc .()()0.cfcf c()()(),x ccfcf cxf x Analysis0c Since ,it is enough to prove Notice thatso,(0,1)c if we can prove there has a ,()()0 x cx cFxxf xsuch that ,the stateme

14、nt can be held.Try to apply Rolles theorem to function F(x).Mean Value Theorem in Differential Calculus15Proof:(1)(1)0Ff(0)0F()()F xxf x Let .0,1Obviously,F(x)is continuous onand differentiable(0,1)in ;by assumption,and(0,1)c Hence by Rolles theorem,there exists at least one point ,such that()()0cfc

15、f c,orFinish.()()f cfcc .Mean Value Theorem in Differential Calculus16xyoABab()yf x The tangent line to the cure y=f(x)at the points C1 and C2 are parallel to line AB.2Cab1 2 xyo()yf x 1CBAMean Value Theorem in Differential Calculus171()()()f bf afba 2()()()f bf afba ab1 2 xoy)(xfy ABCDpoint A and B

16、.(,)a b (,()f()()f af b More generally,if ,by the following graph,there may have a point,such that the tangent line atparallel to the line connectedMean Value Theorem in Differential Calculus18Lagrange Formula()()()f bf afba ()()()().f bf afba or:,Rfa b (Lagranges theorem)If satisfies,a b(1)f is con

17、tinuous on the interval ;(,)a b(2)f is differentiable in the interval ;(,)a b then,there exists at least one point ,such thatMean Value Theorem in Differential Calculus19()()()f bf afba AnalysisIt is enough to prove that there exists a ,such that (,)a b ()()()0.f bf afba Notice that()()()()()()xf bf

18、 af bf afxf xbaba and is a straight line through the origin 0 and parallel to()()f bf ayxba (,()a f a(,()b f bthe line connected point and .So that the difference()()()f bf axf xba must be equal at the two end points.This analysis inspire us to construct()()()(),f bf aF xxf xxa bba Auxiliary Functio

19、nand to apply Rolles theroem.Mean Value Theorem in Differential Calculus20Proof:Auxiliary FunctionBy Rolles theorem,()()()()()()f bf aaf bbf aF aaf ababa()()()()()()f bf aaf bbf aF bbf bbabaF(x)is continuous on a,b and differentiable in(a,b).Also,Let()()()(),f bf aF xxf xxa bba .so that F(a)=F(b).()

20、()()f bf afba ()()()().f bf afba we obtain that there existssuch thator(,)a b at least one pointMean Value Theorem in Differential Calculus2101aba ()()()(),01f bf af ababa()()(),01.f xxf xfxxx (,)a b Note The Lagrange theorem has some other forms.Since ,Note When f(a)=f(b),the Lagrange theorem is Ro

21、lles theorem.So,Lagrange theorem is a generalization of Rolles theorem.aba Let ;01()abathen and .Thus the Lagrangeformula may be written as ax bxx If we choose and ,then the Lagrange formulamay also be written asMean Value Theorem in Differential Calculus220f on the interval I.is that f is a constan

22、t on I.:fIR Suppose that the function is continuous and differentiableThen the necessary and sufficient condition for on I12,x x12(,)x x 212112()()()()0,f xf xfxxx xI Proof and apply the Lagrange theorem on the interval 21,xx(or).that()f xSo that is a constant on I.The sufficiency is obvious.()0fx x

23、I ,.Suppose We now prove the necessity.1x2xI,points Take any twoThere exists at least one 21(,)xxI (or),suchMean Value Theorem in Differential Calculus23()(0)()(0).f xffx Prove the following inequalityln(1),0.1xxx xx ProofWe choose an auxiliary function()ln(1).f xxIt is easy to check that the auxili

24、ary function f(x)satisfies theconditions of the Lagrange theorem on the interval 0,xand hence there exists at least one point(0,)x such thatMean Value Theorem in Differential Calculus24ln(1)(0).1xxx xx Proof(continued)()(0)()(0).f xffx Notice that1()(0)ln(1)ln1ln(1),(),1f xfxxfxx sowhere 1ln(1)(0,).

25、1xxx ,11xxxx Sincewe haveProve the following inequalityln(1),0.1xxx xx Mean Value Theorem in Differential Calculus25Cauchy,Augustin(1789-1857)French mathematician(Cauchys theorem)(,)a b(2)both are differentiable in the interval;()()()()()()f bf afg bg ag .(,)a b Then there exists at least one point

26、following conditions:f:,ga bR,Suppose that,satisfy,the ,a b(1)both are continuous on the interval;()0g x (,)xa b (3).,such thatMean Value Theorem in Differential Calculus26()()()()()()0,f bf a gg bg af()0g x (,)xa b By the assumption,()()g bg a,we have.Therefore,such that Proof(,)a b it is enough to

27、 prove that there exists at least one ()()()()()()|0.xf bf a g xg bg af x or()()()()()()().F xf bf a g xg bg af xLet()()()()()()()0Ff bf a gg bg af()F xIt is easy to check that ,a b satisfies all the conditions of Rolles theorem on the interval,so that there,such thatand this is the equality we want.exists at least one point Finish.