工科數(shù)分一元微分學(xué)

二、典型例二、典型例題 二、典型例二、典型例題1.利 (Rolle)中值定理利用日(Lagrange)中值定理利用(Cauchy)中值定理利用零點(diǎn)定理及介值定理2證明不等式的一些方證明不等式的一些方利用微分中值定理利用函數(shù)的單調(diào)性利用曲線的凹凸性利用函數(shù)的極值利用 3.判斷方程f 的根的個(gè)數(shù)步1)求出f(x)的所有單調(diào)區(qū)間3)利用連續(xù)函數(shù)的介值定理34. 證明等式證明不等式判斷無窮小的階判斷極值求極限計(jì)算函數(shù)的近似值4二、二、典型例例1設(shè)函數(shù)f(u)在u內(nèi)可導(dǎo)且f(0)0,0xf(lnx),x,x求出f(u(u)的表達(dá)式解記ulnxxf(u) e2f(u)uu0uu00u5例2f(x)CcosxCcosx(1)Ccos122n1nnnnn證明:(1)對(duì)任意自然數(shù)nfx在區(qū)1n 中僅有一根(2)設(shè)x n 222滿f(nn)12則limx n2證明f CcosxCcosx1221)Ccosnnn1(1cosx)nnn(1)n(在2上連,,f(0)0,2nfn使fnx)2.2f(n(1cossinx0, 故f( n在 212內(nèi)單,于是滿足方程fnx2xn存在且唯一6例2例2fx1ncosxCcosx221)Ccosn1nnn證明:(1)對(duì)任意自然數(shù)nfn(x)n2中僅有一根2(2)設(shè)x n 12n(x)n2則limx n1(1cosx),1cosx1nn2nn12即xarccos(1 n1n2limxn 7例2f(x)CcosxCcosx122nnn1)Ccosn1n證明:(1)對(duì)任意自然數(shù)nfn(x)n2 中僅有一根2(2)x n 12f(x)nn2,則limx nfn(x)1(1cosx)n cosx 由f 111 1n nn1 (n故當(dāng)n充分大時(shí)f n1n f(xn fnx)0,2內(nèi)單減,故當(dāng)n充分大時(shí)xarccos1(n),于是limx2nn2n28例3例3f(x)xxx(n2,2nn證明(1)方程fnx1在[0,)內(nèi)有唯一的實(shí)根(2)求limxn證明(1)因fnx在[0,)上連續(xù)fn(0)由介值定理可知xn(0,1),fnxn1(n2,又因fn(x)12xnxn1 x即fnx)在[0,)內(nèi)嚴(yán)格單增,故xn是方程fnx在[0,)內(nèi)的唯一實(shí)根fn(1)n9例3f(x)xx2nx(n2,n證明方程fnx1在[0,)內(nèi)有唯一的實(shí)根xn(0,1),fnxn1(n2,(2)由(1)可知:0xn1,下證:xn單調(diào).若xnxn1xx2 x2xn fn(xn)fn1(xn1)由于0xxxnxn1,即數(shù)列xn}單調(diào)遞減,故limxn存在設(shè)為 1,故limxn由fxxx2 x(1xnn n n1取極a1,a1.即limx11,n12 n例4例4設(shè)函數(shù)f(x),g(x)在[a,b]上連續(xù),在(a,b)內(nèi)具有二導(dǎo)數(shù)且存在相等的最大值,f(ag(a),f(b證明(ab使得f(g(證法1h(xf(xg(x),則h(a)=0=h若函數(shù)f(x),g(x)的最大值在(a,b)內(nèi)同一點(diǎn)x0處取得則h(x0若函數(shù)f(x),g(x)的最大值在(a,b)內(nèi)不同點(diǎn)處取得則存x1x2(abx1x2,使f(x)max{f(x)}1max{g(x)}g(x2h(x1f(x1g(x10,h(x2f(x2g(x2由零點(diǎn)定理,在x1與x2之間存在x0(ab),使得h(x0)=綜合上述,存在x0(ab),使得h(x0)=例4設(shè)函數(shù)f(x),g(x)在[ab]上連續(xù),在(ab)內(nèi)具有二導(dǎo)數(shù)且存在相等的最大值,f(ag(a),f(b證明(ab使得f(g(證法1h(xf(xg(x),則h(a)=0=h綜合上述,x0(ab),使得h(x0)=由于f(x),g(x)二階可導(dǎo),由Roller定理,1(a,2(x0b),使得h(10h(20,進(jìn)而存(12(ab使得h(0,f(g(例4設(shè)函數(shù)f(x例4設(shè)函數(shù)f(x),g(x)在[ab]上連續(xù),在(ab)導(dǎo)數(shù)且存在相等的最大值,f(ag(a),f(b證明存在(ab使得f(g(證法2h(xf(xg(x),則h(a0h用反證法證明存在x0(a,b),使得h(x0)=倘若不存在x0(ab),使得h(x00.x(ah(x)0或h(x)0.不妨 , 設(shè)g(x)在x1(ab處取得最大值,則應(yīng)有f(x1g(x1),則(x)最大值必大于g(x)的最大值這與已知條件函數(shù)fg(x)在(a,b)內(nèi)有相等的最大 .故存在x0(a,例5設(shè)函數(shù)f(x)在()內(nèi)定義對(duì)任意xfx1)2fx),且當(dāng)0x1時(shí),fxx(1x2試判斷在x0處函數(shù)f(x)是否可導(dǎo)解當(dāng)1x0時(shí) 0x1f(x)f(x1)11( 222f(0)limf(x)f(0)lim(1x2)f(0)limf(x)f(0)lim1(x1)(x2) xx0f(0)f(0),故函數(shù)f(x在x0處不可導(dǎo)例6例6證明:qx0,則方程yqxy0的任一非證x1x2是原方程的一個(gè)非零解y(x)的零點(diǎn),不妨設(shè)x1x2,且在區(qū)間x1x2內(nèi)y(x)>一方面,yxqxyx)0,xx1x2即函數(shù)yx)在區(qū)間x1,x2]內(nèi)嚴(yán)格單增另一方面,(x)y(x)y(x1y(x xxxx11y(x2)limy(x)y(x2)limy(x)x這與yx在區(qū)間x1,x2]內(nèi)嚴(yán)格單xxx2x.yqxy0的任一非零解至多有一個(gè)零點(diǎn)例7設(shè)函數(shù)f(x)在(a內(nèi)有二階導(dǎo)數(shù)且f(a1)limfx)0,limfx)0.求證:在(a一點(diǎn)f(證明(1)已知f(a1)若在(a1,fx則對(duì)每個(gè)x(a1)都可以取作使f(2若在(a1,fx0,則至少存在x1(a1,fx10,fx10.f(a1)limfx0,又曲線yf(x)連續(xù) y=fOax2x1x3xx2(a1,x1使得fx2fx1x3x1),使得fx3fx1由Lagrange中值定理(x,x)f)f(x1f(x2 21x (x,xf)f(xf(x 12由于f(x)二階可導(dǎo),fx在[1,2]上連續(xù),3(1,2),使得f(3x yy=fOax2x1x3x2(a1,x1),x3(x1,),1(x2,x1),2(x1,x33(1,2),使得f(3補(bǔ)充定義:f(a)0,又f(a1)0,所以f(x)在[a,a+1]上滿足 于是存在4(a,a1),使得f(4)0.對(duì)f(x)在[4,3]使用 存在(4,3)(a),f(對(duì)于xI,有|fx|1,|fx|1,證明對(duì)于xI,有|fx|2,且有函數(shù)使得等式成立證明令I(lǐng)=[a,a+2],將f在任意點(diǎn)x[a,a2]按一階 展開,有f(t)f(x)f(x)(tx)1f()(tx)2,(a,a則ff(x)f(x)(a1f()(ax)211(a,f(a2)f(x)f(x)(a2x)1f()(a2x)2上兩式相減,2(x,a例8設(shè)f是定義于長(zhǎng)度等于2的閉區(qū)間I上的函數(shù),對(duì)于xI,有|fx)|1,|fx)|1,證明:對(duì)于xI,有|fx|2,且有函數(shù)使得等式成立1ff(x)f(x)(af)(a21,(a,1f(a2)f(x)f(x)(a2x)1f()(a2x)2上兩式相減,f(a2)f(a)2f(x)2(x,a1f)(a2x)221f()(a12f(f(a2)f(a)f()(a2x)21f()(ax)21例8例8設(shè)f是定義于長(zhǎng)度等于2的閉區(qū)間I上的函數(shù),對(duì)于xI,有|fx|1,|fx|1,證明:對(duì)于xI,有|fx|2,且有函數(shù)使得等式成立x[a,a2],1[a,x],2[x,a2f(f(a2)f(a)1f()(ax)21111(a2x)21(a2224(ax)22(ax)4(ax)(ax21(ax)24(ax)4(ax)2 故fx)例8設(shè)f是定義于長(zhǎng)度等于2的閉區(qū)間I上的函數(shù),對(duì)于xI,有|fx|1,|fx|1,證明:對(duì)于xI,有|fx)|2,且有函數(shù)使得等式成立.令I(lǐng)=[a, 取(x)使fx1,x[a,af(x)x x[a,a2則f(a2)f(x)1(xa)2|f(x)|作作業(yè)求證:方程4ax33bx22cxabc在(0,1)內(nèi)設(shè)fxgx)[a,b]上可導(dǎo),且fx)gfx)gx),證明：介于fx)的兩個(gè)零點(diǎn)x1,之間gx)的一個(gè)零點(diǎn),x1,x2均在(a,b)內(nèi)fx在區(qū)[0,1上連,(0,1)f(1)0,證明:存在(0,1),f()f() 設(shè)函f(x)在閉區(qū)間[0,1]上可微,對(duì)于閉區(qū)間[0,1]上的每一點(diǎn)x都有0f(x)1,f(x)1,證明在(0,1)內(nèi)有且僅有一個(gè)x使fx)x.5.若函f(x)二階可導(dǎo)，且在a）內(nèi)某點(diǎn)處取得最大值，又對(duì)一切x[0,a]x)m,m為常數(shù)，試證f(0)f(a)am設(shè)不恒為常數(shù)的函 (x)在[a,b]上連續(xù)，在(a,b)內(nèi)可導(dǎo)，且f(a)f(b),證明在(a,b)內(nèi)至少存在一點(diǎn)，使得f() f(x)在[a,b]上連續(xù),在(a,b)內(nèi)可導(dǎo),試證在(a,b)內(nèi)至少存在一點(diǎn)，使得f(b)f(a)f()ln(b/設(shè)fx)在[a,b]f()4f(bf(a)成立(b例10例10設(shè)g(x)在[ab]上連續(xù)在(ab)為二階可導(dǎo)gx)m0(m為常數(shù)又g(ag(b0,maxg(x)m(ba)2a8證明由g(a)=g(b)=0 中值定理,存x0(a,b),使得g(x0)0.由 ,對(duì)x[a,g(x)g(x)g()(xx)10220在x與x0之間gxgx0)1mxx2max0取xag(x)1(xa),(bx)02x200b2取xb得gxm(bx)12020maxg(x)g(x)maxm(ax),m(bx1202(ba)202028例11設(shè)函數(shù)f(x)在[ab]上連續(xù),在(ab)內(nèi)可導(dǎo)x(a,b),fx)1,求證:在曲線y=f(x),ax上存在三個(gè)點(diǎn)ABC,使得ABC(ba)3證法1設(shè)A(a,f(a)),B(b,f(b區(qū)間端點(diǎn)的函數(shù)值為0,故過A,B的弦縱坐標(biāo)與f(x)g(x)f(b)f(a)(xa)f(a)f( xb則g(x)在(a,b)內(nèi)二階可導(dǎo),g(x)f(x)g(a0g(b由上題結(jié)論,x0(a,bg(x0)1(ba)28yCBAOb例11例11設(shè)函數(shù)f(x)在[ab]上連續(xù),在(ab)內(nèi)可導(dǎo)x(a,b),fx)1,求證:在曲線y=f(x),ax上存在三個(gè)點(diǎn)ABC,使得ABC(ba)3設(shè)A(a,f(a)),B(b,fg(x)f(b)f(a)(xa)f(a)f(g(x)(ba)1b2yxCB0令Cx,fx 故ABC的面積1ABgx201ABg(x)bAOb201(ba)1(ba)21(ba)3 例11設(shè)函數(shù)f(x)在[ab]上連續(xù),在(ab)內(nèi)可導(dǎo)x(ab),fx)1,求證:在曲線y=f(x),ax上存在三個(gè)點(diǎn)ABC,使得ABC(ba)3證法2設(shè)A(a,f(a)),B(b,f(b)),Cx,fx))為曲線上的任一點(diǎn),其中xijABACb f(b)fx f(x)fk00 AOb[(ba)(f(x)f(a))(xa)(f(b)f令gx1[(bafxf(axaf(bf2則ABC的面積為|g(x)|,由于g(a0,g(bg(g(x)1[(ba)(f(x)f(a))(xa)(f(b)f2則ABC的面積為|g(x)|,由于g(a0,g(bgx)1(bafx)1(ba).x(a,b220g(x0)maxg(x)1(ba)1(ba)21(ba)3a28故取Cx0fx0則AB,C滿足要求例10設(shè)g(x)在[a,b]上連續(xù)在(a,b)為二階可導(dǎo),gx)m0(m為常數(shù)),又g(ag(b0,maxg(x)m(ba)2a 例12設(shè)函數(shù)f(x)在[0,1]上可導(dǎo)且f(0)0f(1)證明在[01]x1x2,1 f(x21證明因?yàn)楹瘮?shù)f(x)在[0,1]上連續(xù)且f(0)0,f(1)由介值定理,(0,1),f(1.由Lagrange定理f(x)f()f(0)21x1f(x2)f(1)f() (1 2(1x2(故1 22(1)2.f(x1f(x2例13例13設(shè)f(x)在閉區(qū)間[0,1]上連續(xù)在開區(qū)間(0,1)內(nèi)可導(dǎo),且f(0)0,f(1)1.試證明對(duì)于任意給定的正數(shù)a和b,在開區(qū)間(0,1)內(nèi)存在不同的和,使得 證明由介值定理,c(0,1),使得f(c)abaaa由微分中值定理f(1)f(c)f()(1c),(c,1)f(f(c)f(0)f()c,1c(ab)a1(1c)(ab)baba例14設(shè)f(x)在[0,1]上有二階導(dǎo)數(shù)f(0)f(1)f(0)f(1)證明存在(0,1f(f(證明F(x)f(x)f(x)ex,xF(0)0F(1),F(x)f(x)f(x)ex由Rolle定理存在(0,1F(即例15例15設(shè)f(x)在閉區(qū)間[ab]上連續(xù)在開區(qū)間(ab)內(nèi)可導(dǎo)0ab,證明在區(qū)間(ab)內(nèi)至少存在兩點(diǎn)1,22使f fb( 221分1f(2)tanabf(1)sinbsin 同理由Cauchy中值定理,f(b)f(a)f(1)2 11fff(cosbcos,cosbcosaf(sinbsinf(12,2f( f() 2asin2211例16已知函數(shù)f(x)在x=0的某個(gè)鄰域內(nèi)有連續(xù)導(dǎo)數(shù),limsinxf(x)試求f(0)f證明由2xsinxxo(x2f(x)f(0)f(0)xo(于是2limsinxfx)x0sinxxf(x)x2x x0x2lim(1f(0))xf(0)x2o(x2由此可知:1f(0)0,f(0)1,f(0)且limfx)1,fx0(axb),證明:fxx(ax證法由于f(x)在x0處連續(xù)及l(fā)imfx)f(0)limfxlimxfx)0.f(0)x 令g(xf(xx,則g(0)0且g(0)由于當(dāng)x(a,b時(shí),fx0,gx0,g單增,當(dāng)ax0時(shí),gxg(0)0;當(dāng)0xb時(shí),gx)g(0)0;故函數(shù)g(x在x0處取得極小值g(0)x(a,b時(shí)gx0,即fx且limfx)1,fx0(axb),證明:fx (ax證法2由于f(x)在x0處連續(xù)及l(fā)imfx)f(0)limfxlimf( 0.f(0)由 ,對(duì)任意x(a,b),2由于fx0(axb),故fxfxf(0)f(0)x1f()x2,在0與x之間例18設(shè)例18設(shè)f(x)具有連續(xù)的二階導(dǎo)數(shù),1lim1xf(x) xf(0),f(0),f(0)及l(fā)im1x0xe1 f(x)xx0 x解elim1x 11 f(x)f( limxln1xxxxelim1ln1xf(x)3limln1xf(x)x0x0xx例18設(shè)f(x)具有連續(xù)的二階導(dǎo)數(shù),lim1xf(x)試求f(0),f(0),fx0xe3及l(fā)im11xf(lim1ln1xf(x)3.limln1xf(x)x0 .x0xxlimfx)0limfx0.由f(x)的連續(xù)性xf(0)f(0)f(x)f0.由 ,f(x)f(0)f(0)x1f(0)x2o(x2 1f(0)x2o(x2),x22例18例18設(shè)f(x)具有連續(xù)的二階導(dǎo)數(shù),lim1xf(x) x試f(0),f(0),f(0)及l(fā)im1 e1f(x)lim1ln1xf(x)3.f(0)x0xx0xf(0)f(x)1f(0)x2o(x2),x3lim1ln1xf(x)lim1xf(x)x0xx0x x f(0)xo( x0 1 f12 f(0)設(shè)f(x)具有連續(xù)的二階導(dǎo)數(shù),且f(0)0.f(0)3lim1ln1xf(x)x0xlim1xf(x)xxlimx2+f(x 2=lim2+f(x)=2+f22f(0)注意：f(x)二階導(dǎo)數(shù)不連續(xù)，不能用此方法例18例18設(shè)f(x)具有連續(xù)的二階導(dǎo)數(shù),lim1xf(x)e3f(0),f(0),f(0)及l(fā)im1x0 x0f(x)1f(0)x2o(x2),xf(x1x2f(0)lim1x0f(x11xlim12xo(12xo( e22xo(x或lim11f(x)x0x=lim1xf(x)f(x)f(x)xx0xlimf(x)=limf(x)=limf(x)=f(0)2 2例19設(shè)f(x)在閉區(qū)間[0,a]上具有二階導(dǎo)數(shù),在開(0,a)內(nèi)取到最小值又|fx|Mx[0,a],證明|f(0)||f(a)|證明由題意,c(0,a),使f(c)為函數(shù)f(x)的最小值f(c0.fx)在[0c]和[ca]上使用中值定理f(0)f(c)ff(1 1f(a)f(a)f(c)f(2)(ac2|f(0)||f(1)|c|f(a)|M(a故|f(0)||f(a|例20例20設(shè)f(x)在閉區(qū)間[ab]上具有二階導(dǎo)數(shù)f(af(b0,則在(ab)內(nèi)至少存在一點(diǎn)f()4|f(b)f(a)|(bf2f(a)f1a)ba1 b22a21a22f(b)faf(b)f()ba,b a2222ab,22f(a) 18)2)b1 218f(2)f)b122f)例21設(shè)f(x)具有二階連續(xù)導(dǎo)數(shù)且f(0)0,f(0)f(0)0,在曲線yf(x)上任意取一點(diǎn)(xf(x))x作曲線的切線,此切線在x軸上的截距記作x0f(解曲線yf(x)上點(diǎn)(xf(x))Y(x)(x)(X因f(0)0,f(0)0,所以當(dāng)|x|0充分小時(shí)fxlimxf()因此切線在xxf(x)limlimf(x)limf(由x0f(f( f(x)1f(0)x2o(x2x0f(f()1f(0)2o(222x[1f(0)2o(2 x0f(x0[1f(0)x2o(x2 o(2xf(f(x2[1f(0) x0x2[1f(0)o(x2 1limf(x0 x0xf(1f(0)x2o(x2 1 122 例22設(shè)f(x)在[ab]上可微,f(a0,f(b)0,求證:在區(qū)間(ab)內(nèi)必有一點(diǎn)f(0.證明由于f(x)在[a,b]上可微,可取得最小值,又f(a)limf(x)f(a)故存在當(dāng) x1 1時(shí),fxf(a0.因此f(a)不是f在[ab]上的最小值.另一方面f(blimfxf(b)0,故存在當(dāng)x(b2,b時(shí),fxf(b0.因此f(b)不是f在[a,b]上的最小值 x2例22例22設(shè)f(x)在[ab]上可微,f(a0,f(b求證:在區(qū)間(ab)內(nèi)必有一點(diǎn)f(所以f(x)在[a,b]上連續(xù),從而[a,b]上可取得最小值,f(a)不是f(x)在[a,b]上的最小值f(b)不是f(x)在[a,b]上的最小值故函數(shù)f(x)在(a,b)內(nèi)某點(diǎn)處取得最小值,于是f()f(x)的極小值.又f(x)在可導(dǎo),f(例23設(shè)函數(shù)f(x)在[11]上三次可微,f(1)f(1)1,f(0)0.證明存在一點(diǎn)(1,1),提示:利用函數(shù)f(x)的麥克勞 ,可f(1)f(2)6.1,2例24設(shè)函數(shù)f例24設(shè)函數(shù)f(x)在[2,2]上二階可導(dǎo)且fx)1,又f2(0)f(0)24,試證:在[22]內(nèi)至少存在一點(diǎn),使得f()f()0.Fxf2xfx)2,x[2,則Fx)2fx)fxf在區(qū)間[2,0]和[0,2]上使用Lagrange中值定理,ff0f2f(b)f(2)f(0)2a f(a)F(a)2,2(b)2.F(0)4,故F(x)在[ab]b(0,f(b)FF(2)f2(2)f例24設(shè)函數(shù)f(x)在[2,2]上二階可導(dǎo)且fx)1,又f2(0)f(0)24,試證在[22]內(nèi)至少存在一點(diǎn),使得f()f()0.Fxf2xfx)2,x[2則Fx)2fx)fxfx).a(2,0),b(0,F(a2,F(b2.F(0)4,故F(x)在[ab]上的最大值必將在(a,b)內(nèi)某點(diǎn)取得,于是由于F(f2(f()2F(0)4,f()f(0,f(f(0,(a,b[2,例25例25設(shè)f(x)在[0,)上連續(xù)可導(dǎo)limfxf證明c(0,),使f(c分 (利用最值定理若函數(shù)f(x)在[0,上恒為f(0則對(duì)任意c(0,f(c0.以下設(shè)fxf(0則存在x0(0,fx0f(0),fx0f(0),limfxf(0)fx0),故存在充分大x1x0,f(x1)f(x0由于f(x)在閉區(qū)間[0,x1上連續(xù),[0,x1]存在最大值,且最大值只能在(0,x1內(nèi)點(diǎn)c取得,又f(x)在c可導(dǎo),從而f(c例26函數(shù)f(x)在[0,)可導(dǎo)0fxx證明存在0,f(11(12.分只須證存在0,1(12f(令Fx1(1x2fx),Fxxf1因?yàn)閒(0)=0,所以F(0)又因?yàn)閘imfx)0,所以limFx故函數(shù)F(x)在[0,)上的最大值或最小值必將在區(qū)(0,)內(nèi)點(diǎn)處取得.從而F(令x令xtant1tan2tdt2sec2tsec2sec2(2cos2tcos2tdt1sin2tsintcos2x11x211x2t1x例27試證:若函數(shù)f(xg(x)在閉區(qū)間[ab]上連續(xù)在開區(qū)間(ab)內(nèi)可導(dǎo)gx0,則存在(abf()f()f(a)分 g(b)g(只須證(a,b),(g(b)g())f()g()(f()f g(b)g()f()g()f()f(a)g(b)f()g()f()g()f()g()f(a))g(b)f(x)f(x)g(x)f(a)g(x)作輔助函數(shù)x)g(bfxfx)gxf(a)g由于(ag(bf(a例28例28設(shè)函數(shù)f(x)在[0,1]上有二階導(dǎo)數(shù)且f(0)f(1)試證:(0,1),使f(2f()分只須證(0,1),1(1)f()2f()(1x)f(x)f(令(x)1 x x01因(1)0,由于(0)f(0)f(0)f(0能否找到(0,1),(0.令Fx則取F(x)(1x)f( 因F(0)=0=F由Rolle定理(0,1),F(()例29設(shè)函數(shù)f(xg(x)在閉區(qū)間[ab]上存在二階導(dǎo)數(shù)且gx0,f(af(bg(ag(b0.試證(1)在開區(qū)間(ab)內(nèi),gxg( 分析對(duì)于問題(1),可以采用反證法證明對(duì)于問題(2),只須證明g(f(f()g(xgxfxfx)g(a(b0,xgxfxfx)g例29例29設(shè)函數(shù)f(xg(x)在閉區(qū)間[ab]上存在二階導(dǎo)數(shù)且gx0,f(af(bg(ag(b0.試證(1)在開區(qū)間(ab)內(nèi),gx(2)在開區(qū)間(a,b)內(nèi)至少存在一點(diǎn),使g(g()證明(1)若存在點(diǎn)c(a,b),使得g(c)0,g(ag(c)0.1(a,c),使g(1)0.2(c,b),g(20.31,2(a,b),g(30.與已知gx)0x(a,b),故gx)0.(2)作輔助函數(shù)xgx)fxfx)gx).(a)(b)0,(x)g(x)f(x)f(x)g(故(ab),(0.g(f(f()g(又當(dāng)x(a,b)時(shí),gx0,gx0.故f()f()g( 例30設(shè)函數(shù)f(x)在閉區(qū)間[0,1]上具有二階導(dǎo)數(shù)條件|fx|a,|fx|b,其中ab都是非負(fù)常數(shù)2分析本題涉及二階導(dǎo)數(shù),要利用 題中要求證明對(duì)(0,1)內(nèi)任意c都有f(c)2abc是(0,1)內(nèi)任意一點(diǎn).證明:f(c)2ab2所以要在任意點(diǎn)c處展開例30例30設(shè)函數(shù)f(x)在閉區(qū)間[0,1]上具有二階導(dǎo)數(shù),且滿足條件|fx|a,|fx|b,其中a,b都是非負(fù)常數(shù),c是(0,1)內(nèi)任意一點(diǎn).證明:f(c)2ab2證明由 ,對(duì)任意固定c(0,1),fxf