微分中值定理及其應(yīng)用

第五章微分中值定理及其應(yīng)第一分中值定方程x33xc0(c為常數(shù))在區(qū)間[0,1]內(nèi)不可能有兩個(gè)不同的實(shí)根(2)方程xnpxq0(n為正整數(shù)pq為實(shí)數(shù)),當(dāng)n為偶數(shù)時(shí)至多有兩個(gè)實(shí)根;當(dāng)n f(x)x33x值為零。那么由羅爾定理可知存在x0(x1,x2[0,1使得f'(x0但是f'(x)3x23在(0,1)內(nèi)的值域?yàn)?3,0)是不可能有零點(diǎn)的,。因此有:方x33xc0(c為常數(shù))在區(qū)間[0,1]內(nèi)不可能有兩個(gè)不同的實(shí)(2)當(dāng)n2當(dāng)n2k2時(shí),設(shè)方程xnpxq0有三個(gè)實(shí)根,即存在實(shí)數(shù)xxx f(x)xnpxq成立。那么由羅爾定理可知存在x01(x1x2x02(x2x3使得f'(x01)f'(x020,f'(x)nxn1p 再次利用羅爾定理可以知道,存在x0(x01,x02使得f''(0)0,0f''(x0)n(n1)xn20

顯然必有x00,那么就有x010,x02當(dāng)n2k12時(shí),設(shè)方程xnpxq0有三個(gè)實(shí)根,即存在實(shí)數(shù)xxxx 數(shù)f(xxnpxq0x11(x1,x2),x12(x2,x3),x13(x3,x4使得f'(x110,f'(x120,f'(x130,f'(x)nxn1p f'(x)nxn1p f'(x)nxn1p 于是就存在x21(x11x12x22(x12x13使得f''(x21)f''(x220,f''(x)n(n1)xn2 f''(x)n(n

0由于n2k12,于是此時(shí)必有x21x220;但是由于x21(x11x12x22(x12x13可知必x21x22,出現(xiàn)了因此當(dāng)n為奇數(shù)時(shí),方程xnpxq0(n為正整數(shù),pq為實(shí)數(shù))m 1證明：容易知道f(0)f(1)0,于是作為多項(xiàng)式函數(shù),必有(0,1)使得f'(0,mm1(1)nnm(1)n1由于0,10,因此整理可得m(1n,m 1sinxsinyxy,x,y(,證明：由拉格朗日中值定理可知函數(shù)f(tsint在區(qū)間[xy]上存在(x,y)f'()sinxsinyxxymaxf'(t)maxcostsinxsinyxyxtanxx( 等號成立當(dāng)且僅當(dāng)x 2證明：由拉格朗日中值定理可知函數(shù)f(ttant在區(qū)間[0,x]上存在(0,x)f'()tanxtan0tanxtanxxtanxtanx

xf'(t)

1cos21cos2,0) );即當(dāng)x0對于函數(shù)g(x)tanxx,滿足g(0)0,且有g(shù)'(x),0) );即當(dāng)x0 xtanxex1x,x證明：當(dāng)x0時(shí),由拉格朗日中值定理可知函數(shù)f(tet在區(qū)間[0,x]上,存在(0,x)f'(f(xf(0,即f'(ex1;x整理即得ex1

ex1x

f'()ee0當(dāng)x0時(shí),由拉格朗日中值定理可知函數(shù)f(tet在區(qū)間[x0]上,存在(0,x)f(0)f 1f'( 0 ,即f'(1

; 整理即得ex1綜上有ex1xxyxlnyyx,0xy;

f'()ee證明：由拉格朗日中值定理可知函數(shù)f(tlnt在區(qū)間[xy]上有(x,yf'()f(y)f(x)y1lnylnx, y

min1lnylnxmax1xty1lnylnx1 y

y xtyyxlnyyx x1

arctanxx,xf'()f(x)f(0)x即有1

arctanx,x

arctanx 故有11x2

arctanxx

0tx1t1

0tx1tx1

arctanxlimf(ah)f(ah)2f(a)f 證明：f''(alimf'(ahf limf(ah)f(a)limf(a)f(alim limf(ah)f(ah)2f(a) 因此有l(wèi)imf(ahf(ah2f(a)f

f'(xa求證：任意T0,lim[f(xt)f(x)]證明：在區(qū)間[x,xT]上有(x,xT)使f'()f(xT)f(x)T

f'()limf(xT)f(x)由于x,

limf(xT)f(x)limf'() limf(xtf(x

設(shè)函數(shù)f(x)在[ab]可導(dǎo),其中a0,證明：存在(ab2[f(b)f(a)](b2a2)f'(

F(x)x2[f(b)f(a)](b2a2)f易知F(aa2f(bb2f(aFF'()2[f(b)f(a)](b2a2)f'()整理即得2f(bf(ab2a2f'(設(shè)f(x)在(a)上可導(dǎo),且

f(x)

f(xA求證：存在(a使得f'(證明：取b0max(a0),F(t)

(b

t.f t[a,b0 b00存在t(ab0使得F'(t0,F'()(b0a)(b0t)(b0a)tf'((b0a)t) (b0 b

(b0(b0a)b0f )(b0 bt0tt0由bmax(a0)知(b0a)b00,00

t

f'((b0a)t)b0取(b0a)t,易知(a),b0設(shè)f(x)可導(dǎo),求證:f(x)的兩個(gè)零點(diǎn)之間一定有f(x)f'(x)的零點(diǎn)。證明：設(shè)x1x2是函數(shù)f(x)的兩個(gè)零點(diǎn)。構(gòu)造函數(shù)F(x)f(x)ex易知x1,x2也是函數(shù)F(x)的兩個(gè)零點(diǎn)。于是由羅爾定理可以知道,存在(x1,x2),使F'()ef()ef'()于是ef(f'(0,由于必有e0,因此f(f'(0.因此在f(x)的兩個(gè)零點(diǎn)之間一定有f(x)f'(x)的零點(diǎn)。設(shè)函數(shù)f(x)在x0點(diǎn)附近連續(xù),除x0外可導(dǎo),且limf'(xA求證：f'(x0存在,且f'(x0證明：根據(jù)拉格朗日中值定理可知存在(0,1)使f

x)f(x0x)f(x0)limf(x0x)f(x0)limf'(x 由于limf(x0xf(x0)f'(x),而limf'(xxlimf'(xA0 x0即f'(x0存在,且f'(x0若f(x)在[ab]上可導(dǎo),且f一點(diǎn)(ab使得f'(

f'(x0)limf'(x)f'(bk為介于f'(af'(b)證明：構(gòu)造函數(shù)g(xf(xkx.顯然函數(shù)g(x)在[ab]g'(a)f'(a)k,g'(b)f'(b)顯然f'(ak,f'(bk不妨設(shè)g'(af'(akA0,g'(bf'(bkB0.limg(x)g(a)A xlimg(x)g(b)B x那么由極限的保號性可知0,當(dāng)x(aa)g(xg(a)A0;當(dāng)x(bx 時(shí)有g(shù)(x)g(b)B0.于是可得g(a)g(a),g(b)g(b);那 x 間[ab]上必存在函數(shù)g(x)的一個(gè)極大值點(diǎn).于是有g(shù)'(0,即f'(k 于是存在一點(diǎn)(ab使得f'(設(shè)函數(shù)f(x)在區(qū)間(ab)內(nèi)可導(dǎo),且f'(x)單調(diào),證明：f'(x)在(ab)連續(xù)。證明：不妨設(shè)f'(x)在區(qū)間(ab)內(nèi)單調(diào)上升。那么由單調(diào)有界原理可以知道極限0xx0

f'(x)0xx0

f'(x)0xx0

f'(x)00

f'(x)f'(x0即導(dǎo)函數(shù)在x0點(diǎn)連續(xù)。由x0點(diǎn)的任意性可以知道函數(shù) '(x)在(a,b)連續(xù)設(shè)f(x)在區(qū)間(ab)上處處可導(dǎo)證明在(ab)的點(diǎn)或是f'(x)的連續(xù)點(diǎn),或是f'(x)的第二證明：由于f(x)在區(qū)間(ab)上處處可導(dǎo)那么對于x0(abf'(x0)f'(x00xx0

f(x)f(x0x拉格朗中值定limf'()(xx00xx0

xx0f'(0

x其中x0

0x0

f'()0xx0

f于是f'(x)在x0處有右極限時(shí),必有f'(x00xx0

f同理可得若f'(x)在x0處有左極限時(shí),必有f'(x00xx0點(diǎn)處,除非至少有一側(cè)f'(x)無極限,不然f'(x)

若函數(shù)f(x),g(x),h(x)在[ab]連續(xù),在(ab)可導(dǎo)證明：存在(ab)fff'(

g'(

h'(

D(x)

fff

驗(yàn)證即知有D(aD(b0,又由于函數(shù)f(xg(xh(x)在[ab]連續(xù),因此函數(shù)D(x)在[ab]上也連續(xù)。那么由羅爾定理可知道存在(ab)使得D'()0.可以求得于是存在(ab)

D'(x)ff

fff

g

h(b),hf'( g'( h'(取g(x)xh(x)1,于是存在(ab)ff

1整理即得f'(f(bf(a)b取h(x)1,于是存在ab)

f'( f'()f(bf(ag'()g(b)

fff'(

1g'( 設(shè)f(x)在()連續(xù),

f(x證明：f(x)在()證明：首先選定一個(gè)閉區(qū)間[X0,X0],可以找見f(x)在這個(gè)閉區(qū)間上的最大f(x0),(x0[X0,X0

f(x,于是對于Gmax(1,f(x00,X0,xX時(shí),有f(xG.可知對于x(,X)(X,),總有f(x)f(x0可以肯定必有[X0,X0][X,X].顯然可以找見f(x)在閉區(qū)間[X,X]的最小值點(diǎn),即有[XX使得對于x[XX]有f(xf(且有f(f(x0綜 可以 找見了一個(gè)點(diǎn)(,),對于x(,),有f(x)

設(shè)f(x)在[ab)連續(xù),

f(x)

g(x)

f

那么顯然有l(wèi)img(x)

xf(xBg(b于是g(x)在b點(diǎn)連續(xù)又因?yàn)閒(x)在[ab)于是可知道g(x)在閉區(qū)間[a,b]上連續(xù)。于是g(x)可以在[a,b]取到最大值g(由于存在點(diǎn)x1[ab)使得g(x1有[a

f(x1Bg(b于是b點(diǎn)不可能是g(x)的最大值點(diǎn)于是對于x[ab),總有f(xg(x)g(f(),因此說f(x)在[ab)上達(dá)到最大值f((2).可以:如果f(x)在[a,b)上的最大值為B,那么x1點(diǎn)即為所求;否則必存在x0[ab)使得fx00,問題化為(1).設(shè)f(x)在[a)有界,f'(x)存在,

f'(xb.求證b

f'(x)b0.那么利用極限的保號性可以知A,當(dāng)xA時(shí),f'(xb2取區(qū)間Ax],由拉格朗日中值定理可知存在滿足Ax使得f'(f(xfA)x即有f(xfAf'()(xAfAb(xA).

f(x)這與f(x) 設(shè)f(x)在[a)有界,f'(x)存在

f'(xb.求證b證明：若b0,取

b b

f'(xb,即對

0,X0,當(dāng)xXb8bbf'(x)b bb8b又因?yàn)閒'(xlimf(xxf(x),即對

0,當(dāng)0,當(dāng)0x

bf(xx)fbf(xx)f8

Xnn1,2, f(xn)f(xn)f(xn12f(xn)f(xn12

f

)f

)ff(xn)f(xn12bb bb

f'(xn1)

f

)b即b b4

f(xn)f(xn1)b2

b4bbf )(bb

)f(x)f )(bb

4 f(x)(b0

nf(x)f(x)n(bbb) b)b b若b0,那么可知lim[f(xb

)n]lim[f(x)3

當(dāng)nN時(shí),有f(x)f(x)(bb)nG.這與f(x)有 b若b0,那么可知lim[f(x)n(b )]lim[f(x)3bn];于是對G0,Nbb b當(dāng)nN時(shí),有f(xn因此必有b

f(x0)(b

)

G.這與f(x)有界 2證明：令f(xarcsinxarccosx,11f'(x) 11于是可知f(xcc又因?yàn)閒(0)0,于是可知f(xarcsinxarccosx 第二比達(dá)法 lim x0sin x0bcos2axcos 2xsin limsin x0x3sin x03x2sinxx3cos 3sinxxcos

洛比法則

1x03sinxxcos x03cosxcosxxsin

ln(1xx

1lim1 lim

cosx x0sin x0(1x)sin x0sinx(1x)cos limtanxxlimcos2

1

lim1cosxx0xsin x01cos x0

x

x0cos2 )limex1x洛比法則 ex 洛比達(dá)法則 1x0 ex x0x(ex x0ex1 x02ex asinaxcos a2cosbxbxsin lim lim lim x0lncos x0cos bsin x0b2cosaxaxsin

cos 2xsecx2x

xsin

sin

2cos2

1lim(1

x1lnx

x 洛比法則

1) x1ln x x1(x1)ln x1lnxxx

x1xlnxx x1lnx (9).lim(x)tanxlimx洛比達(dá)法則 lim2sinx

xcot x1 2sin21 limlnx洛比達(dá)法則lim(10).limx1xlimeln(x1x)lime1x ex11

xb洛比法則limbxb1洛比法 洛比達(dá)法則limb(b1)(b

[b]1b[b1]x x arctan 1lim 1

1

x1cos x(1x2)cos

11)cos(2 (2limlncx洛比達(dá)法則

clnc1x洛比法 洛比達(dá)法則

(c

x

b[c]1xbln[c1]climxblncx,(b,

lim12sinx

limlnx lim()bln

cos6洛比lim2cos

x0cott x3sin lim

clnc

limsinx t

3 1lim(1x)xe1 1

ln(1解：令y(1x)x,lnyln(1x),于是y'1 ln(1x) 因此有y'1 1 ln(11 (1x)xe洛比達(dá)法 (1

1 ln(1 1lim

lim

(1x) x(1x)ln(1 x(1x)ln(1lim (1x)xlim x2(1

x2(11 洛比達(dá)法則elim1ln(1x)1洛比達(dá)法則e 1xe x02x(1x) x02 sin limln(xsinx limsinxln

limxln limln limxsinxlim ) et e0 1 lnln ln[(ln)x xln(ln limxln(ln )xlim lim x xet

tanx

1)x2 limtanxcos2xlim( )x2lim lim

2

lim lim

lim2sin ex02x2sinxcosxex0x2sin2x 2 ex06 ex06 e3 sin2x sin2x 2sinxcosxlim( )

lim x0

sin2

x0x2sin2

lim2cos2x2lim4sin2x1

lim limsinxlnxlimxln lntlim t 對函數(shù)f(x)在[0,x]上應(yīng)用拉格朗日中值定理有f(xf(0)f'(x)x,(0,1).試證對下列函數(shù)有l(wèi)im1: (1)f(x)ln(1x);(2)f(x)證明：只要證得limf(xf(0)1

'(2

limf(x)f(0)limln(1x)ln(10)lim(2x)ln(1x) 1x f 2f(x)f

12ex

lim lim xlim

xf'( e22

limf(x2h)2f(xh)f(x)f 證明：當(dāng)h0時(shí),分子f(x2h)2f(xh)f(x)0,分母h20,因此可以對h使用limf(x2h)2f(xh)f lim2f'(x2h)2f'(x lim4f''(x2h)2f''(x lim2f''(xh)f于是有l(wèi)imf(x2h2f(xhf(x)

fx2sin x sinx2sin 2xsin1cos sin cosx2sinl x xsin1 sin x0sin limxsinxxxcosxcos極限的;而原函數(shù)是有極限的:limxsinxxxcos

1sin 2xsin x(2xsinx)esin答：這是因?yàn)楹瘮?shù) 2xsin x(2xsinx)esin

(x21)sin x1ln(1 2(x21)sin 2xsinx(x21)cos 的分子分母分別求導(dǎo)之后可得 ln(1sin

21sin2

limx1ln(1sin2第三節(jié)函數(shù)的升降、凸性和函數(shù)作 xsinxx,x(0,2證明：取函數(shù)ysinx,可求得yxcosxsinx;取函數(shù)zxcosxsinx,可以知道z(0) z'xsin 由于在區(qū)間上z0恒成立,于是函數(shù)zxcosxsin 2是有zxcosxsinxz(0)0成立那么y'

上小于零成立,因此函數(shù)y

sinx

因此有l(wèi)imy(0)y(x)y(),即1sinx22xsinxx,x(0xsinxx

3x,x0;3

證明：取函數(shù)ysinxx,可求得ycos1;在區(qū)間(0)上可以知道恒有ycos1于是函數(shù)ysinxx單調(diào)遞減,因此有ysinxxy(0)0,即得sinxxx取函數(shù)zsinxxx3,可求得zcosx1x2那么有z'(0)0,zsinxx. 區(qū)間(0)上恒有zsinxx0,因此z'滿足z'(xz'(0)0,于是函數(shù)zsinxxx36調(diào)遞減,即z(xz(0)0,于是有sinxx,x,x綜上可得xsinxx6

x3,x0.x ln(1x)x,x0;證明：取函數(shù)yln(1x)xx2,可求得y' 1x;那么有y'' 1 (1

1

1xy'(0)0;那么函數(shù)y函數(shù),于是有yln(1xx

x2y(0)0,即x

ln(1x),x(0,取函數(shù)zxln(1x),可求得z' 1

增函數(shù),即有zxln(1xz(0)0;于是ln(1xxx(0,綜上可得xx2ln(1xx,x2tanxx

,x(0, 證明：取函數(shù)ytanxx

x,可求得3y' 1x2,y'(0)3cos2y''2sinxcosx2x,y''(0)0;cos4x2(1cos4x)4sin2 cos4那么有yy''(0)0,yy'(0)0ytanxx3

y(0)0,tanxx

,x(0,2

x31,xxxxx證明：取y 31,那么有y'110在區(qū)間(1,)上恒成立,于是在此區(qū)間上xx xx有y 31y(1)0.即得 31,xxx yx3解：令y3x260,可以解得x(

(2,2)和(2,),單調(diào)遞減區(qū)間為( y 2xx22x令y' 1x 2xy2x2ln解：可以求得函數(shù)的定義域?yàn)?0,).令y4x10,解得x(0.5,x,x2y xx2 0恒成立,于是原函數(shù)的單,yxnex,n0,xnxn1exx(nnxn1exx解：令y' 0,解得x(0,n).e2x e2x y2x2sin解：可以求得函數(shù)的定義域?yàn)榱顈4xcosx0,可設(shè)這個(gè)方程的根為x0(可以大致求得x0因此原函數(shù)的遞增區(qū)間為(x0遞減區(qū)間為(,x0yxnex,n0,xnxn1exx(nnxn1exx解：令y' 0,解得x(0,n).e2x e2x yxln(1 1

,y'' (1令y' 1

0,可得x0;由于y''(0)10,因此x0是函數(shù)yxln(1x)點(diǎn),極小值為y(0)yx1x解：容易解得函數(shù)的一階和二階導(dǎo)數(shù)分別是y11y2 令y1

x14145(45(45x2)2(125x)1545x2(45x2)312 y'

3,y''(45x2令y125x0,可得x12;由于y''(120,因此x12是函數(shù)y

1143(45x2值點(diǎn),極大值為y(12) 205

(5)y2x3解：容易解得函數(shù)的一階和二階導(dǎo)數(shù)分別是y6x24x3y12x令y6x24x30,可得x0或x3;由于y''(0

0,y''(

y2x3x4的拐點(diǎn)x

(4)y

(ln;x

是函數(shù)的極大值點(diǎn),極大值為y() 2lnxln2

2x6xlnx2xln2y' ,y'' 2lnxln2 令y' 0,可得x1或xe;由于y''(1)20,y''(e

0.因此x1是函(lnx)

2)4y 的極小值點(diǎn),極小值為y(1)0;xe是函數(shù)的極大值點(diǎn),極大值為 y2x3解：容易解得函數(shù)的一階和二階導(dǎo)數(shù)分別是y6x24x3y12x令y6x24x30,可得x0或x3;由于y''(0

0,y''(

y2x3x4的拐點(diǎn)x

是函數(shù)的極大值點(diǎn),極大值為y() yarctanx1ln(1x22解：容易解得函數(shù)的一階和二階導(dǎo)數(shù)分別是y1xy

x22x11 (1x21指點(diǎn),極大值為y(1)

ln2x4sin2

x設(shè)f(x)

x證明：x0是函數(shù)的極小值點(diǎn)說明在f(x)的極小值點(diǎn)在x0處是否滿足極值的第一充分條件或第二充分條件。因?yàn)楫?dāng)x0時(shí),總有f(x)x4sin210f(0),因此x0是函數(shù)的極小值點(diǎn)。x(2)當(dāng)x0 f'(x)(x4sin21)'4x3sin21x2sin2x2(4xsin21 取x 2n

,

2n

(n1, 由于limxnlimyn0,從而對任意小的,總有xnyn(0,但是f(xn0,fyn0; 任意小的U(0,)內(nèi)f'(x)不能保持不變號,因此f(x)的極小值點(diǎn)在x0不滿足極值的第一由f(x)f

x4sin21 f'(0)lim lim limx3sin2 而

x2(4xsin21sin2)f''(0)

f'(x)f'(0)

limx(4xsin21sin2)

因此f(x)的極小值點(diǎn)在x0證明：若函數(shù)f(x)在點(diǎn)x0處有fx00,fx00,則點(diǎn)x0為f(x)證明：由于f'(xlimf(xf(x0)0,那么由極限的保號性可以知道存在001 xx01

x當(dāng)x(xx)f(xf(x0)0,由于此時(shí)xx0,因此f(xf(x0 x 0同理由于f'(xlimf(xf(x0)0,那么由極限的保號性可以知道存在002 xx02

x當(dāng)x(xx)f(xf(x0)0,由于此時(shí)xx0,因此f(xf(x0 x 0取min(1,2那么當(dāng)x(x0x0)時(shí),恒有f(x

設(shè)f(xalnxbx2x在xx

x1,x2處是極大值還是極小值。解：可求得f'(xa2bx1,令xf'(x)

f'(1)a2b10 f'(x)

f'(2)a4b1a可以解得 3

b 于是f(x2lnx1x2x,f''(x

2b2

3f''(1)10,f''(2)1 7.(1)求函數(shù)f(xaxlnx在x0上的極值;(2)求方程axlnx有兩個(gè)正實(shí)根的條件。：(1)令f'(xa10,可以解得x1.因此當(dāng)a0時(shí),函數(shù)f(x)在x0上沒有極值當(dāng)a

f(x)在x

x1;由于f''(x)1,故此 10,于是1lna是函數(shù)f(x)1

1

f''(a

f(a(2)當(dāng)a0時(shí),由于x

f(

1lna0; limf(x),limf(x)由介值定理可以知道f(x)

于是當(dāng)0a1時(shí),方程axlnx有兩個(gè)正實(shí)根。設(shè)f(x),g(x)在實(shí)軸上連續(xù)可微,

ff

g

求證：f(x)0的兩實(shí)根之間一定有g(shù)(x)證明：設(shè)ab是函數(shù)f(x)相鄰的兩個(gè)實(shí)根(那么在區(qū)間(ab)上沒有函數(shù)f(x)的根),于是必有g(shù)(a)0,g(b

f(a)f'(a)f(b)f

g'(a)g'(b)

g(a)f'(a)g(b)f'(b)

現(xiàn)在使用反證法證明f'(a),f'(b)異號,不妨設(shè)f'(a)0,f'(b)0.limf(x)f(a)limf(x) x xaxlimf(x)f(b)limf(x) x xax那么由極限的局部保號性可以知道0,當(dāng)0xa0xbf(x)0,f(x)取

ba) ,

x x分別取xax

b,

f(x1

f(x2

f(x1)0,f(x2)

數(shù)f(x)的根相。于是可知f'(af'(b)異號,那么由(*)(**)式可以知道g(ag(b)也異號,則由介值定理可以知道在區(qū)間(ab)上必有函數(shù)g(x)的一個(gè)根,即得證。y3x2解：可以求得y6x3x2y66x令y66x0,可以解得xyx21x解：可以求得y2x1y2

2令y2

0,可以解得x(

(0,).xyln(1x2解：可以求得y' 2x,y''2(1x2);令y''2(1x2)0,可以解得x1 (1x2 (1x2y1x21解：可以求得y' ,y'' ;令y'' 0,可以解得x1 (1x2 (1x2因此()是函數(shù)y1x2證明曲線yx1x2

y'

x212x(x1)(x21)2

x22x1(x2 令y0

y''

(2x36x26x2)(x2(x2

x11,x22 3,x32y13,y 3,y 84 84 得到了曲線yx1的三個(gè)拐點(diǎn)x2點(diǎn)(x1y1x2y2確定的直線84328432x

y 848432 y(1)aby''(1)6a2ba解得b

92若f(x)為下凸函數(shù)為非負(fù)實(shí)數(shù),則f(x)為下凸函數(shù)若f(x),g(x)均為下凸函數(shù)則f(x)g(x)為下凸函數(shù)若f(x)為區(qū)間I上的下凸函數(shù)g(x)為區(qū)間J上的下凸遞增函數(shù),f(IJ則為I因?yàn)閒(x)為下凸函數(shù)所以對定義域內(nèi)任意兩點(diǎn)x1,x2及任意的(0,1f(x1(1)x2)f(x1)(1)f(x2那么在上式兩邊乘上非負(fù)實(shí)數(shù)f(x1(1)x2)f(x1)(1)f(x2依定義可知道f(x)為下凸函數(shù)

f因?yàn)閒(x),g(x)均為下凸函數(shù),那么在它們的區(qū)間上的任意兩點(diǎn)x1,x2及任意的有f(x1(1)x2)f(x1)(1)f(x2g(x1(1)x2)g(x1)(1)g(x2f(x1(1)x2)g(x1(1)x2)f(x1)(1)f(x2)g(x1)(1)g(x2f(x1(1)x2)g(x1(1)x2)[f(x1)g(x1)](1)[f(x2)g(x2依定義可知道f(x)g(x)因?yàn)閒(x)為I上的下凸函數(shù)所以對I內(nèi)任意兩點(diǎn)x1,x2及任意的(0,1f(x1(1)x2)f(x1)(1)f(x2又因?yàn)間(x)為J上的遞增函數(shù),而且f(IJf(x11f(x2f(I),g(f(x1(1)x2))g(f(x1)(1)f(x2而g(x)為J上的凸函數(shù),g(f(x1)(1)f(x2))g(f(x1))(1)g(f(x2

g(f(x1(1)x2))g(f(x1))(1)g(f(x2 f(x)為下凸函數(shù)設(shè)f(x)為區(qū)間I上嚴(yán)格下凸函數(shù),證明：若x0I為f(x)的極小值點(diǎn),則x0為f(x)在I上唯一證明：假設(shè)f(x)在I上還有另一極小值點(diǎn)x1,不妨設(shè)x0x1.由定義,存在U(x0及U(x1),使得xU(x0時(shí),f(xf(x0當(dāng)xU(x1)時(shí),f(xf對任意x(xx取x1x,則01,xx1)x.據(jù)f(x)為I上的嚴(yán)格下 x 若f(x0f(x1),

f(x)f(x0)(1)ff(x)f(x1)(1)f(x1)f特別的當(dāng)xU(x1) (x0,x1)時(shí),f(x)f(x1).這與x1點(diǎn)為f(x)的極小值點(diǎn)相。若f(x0)f(x1),則f(x)f(x0)(1)f(x0)f(x0特別的當(dāng)xU(x0) (x0,x1)時(shí),f(x)f(x0).這與x0點(diǎn)為f(x)的極小值點(diǎn)相。因此x0為f(x)在I上唯一的極小值點(diǎn)。對任意實(shí)數(shù)ab對任意非負(fù)實(shí)數(shù)a,b,

e22

eb2arctanabarctanaarctan2(1)設(shè)f(xex.由于f''(xex0,故f(x)在()上位下凸函數(shù)對任意ab(取1有fab1f(af(b)), e22

eb(2)設(shè)f(x)arctanx.則f'(x) 1

,f''(x) (1x2

.那么對于任意非負(fù)實(shí)數(shù)ab有f''(x于是對于任意非負(fù)實(shí)數(shù)ab(0,)有fab1f(af(b)), 2arctanabarctanaarctan2f(x1)f(x2)f(x1x2 證明：不妨設(shè)[ab]內(nèi)任意兩點(diǎn)x1由于函數(shù)f(x)在區(qū)間[ab]上恒有f''(x0,那么在區(qū)間(ab)上也滿足f''(x0,于是在區(qū)間(ab)上函數(shù)為下凸函數(shù),因此對于任意兩點(diǎn)x1,x2(ab),總有f(x1)f(x2)f(x1x2 對于(*)式,任意取定x2,在不等式兩邊令x1趨近于a求極限,limf(x1)f(x2 x limf( 2

2由于函數(shù)的連續(xù)性可知limf(xf(alimf f(a)f(x2)2

x2fax2 f(ax2);即當(dāng)x1a時(shí),(*)式也成立。同理可得當(dāng)x2b時(shí),(*)式也成立。因此在[ab]內(nèi)任意兩點(diǎn)x1,x2,都有f(x1)f(x2)f(x1x2

yhe2h3x 4 12y' (4hx12令y0,可得x2

;由于拐點(diǎn)為,故有2

,即h 求y

x21

2y 1 ,y' ,y'' x2

x2

(x2

(x2令y''0,可得x 3;由于y(3)1, 3)1,于是拐點(diǎn) (3 3 ,)與( , 由于y'(3)33,y 3)33,那么由點(diǎn)斜式可得在點(diǎn)(3,1)與(

13,)3

yyx3

(x 3)1,y33(x 3)13 3解：可以求得函數(shù)的定義域?yàn)?由于y(x)y(x),因此為奇函數(shù)。令yx36x0,可以求得函數(shù)的零點(diǎn)x0,x 6. 2令y'3x260,可得函數(shù)兩個(gè)穩(wěn)定點(diǎn)x4,52;令y'3x260,可得函數(shù)遞增區(qū)間(,2),(2,)及遞減區(qū)間(2, 2由于y''6x,則y''(2) 0,于是穩(wěn)定點(diǎn)x42為函數(shù)極大值點(diǎn)2y''(2)6 0,于是穩(wěn)定點(diǎn)x5 令y''6x0,可得函數(shù)的上凸區(qū)間(,0),及下凸區(qū)間(0,).2綜上制表yx3x10,x2,3x4x5 (,2),(2,(2,無(,(0,無依表作圖ye(x1)2y(x1)y(x1),因此關(guān)于直線x1對稱。令ye(x1)20,無解。1令y2(x1)e(x1)20,可得穩(wěn)定點(diǎn)x1;令y3x260,可得函數(shù)遞增區(qū)間11由于y4(x1)22]e(x1)2則y''(1)20于是穩(wěn)定點(diǎn)x1為函數(shù)極大值點(diǎn)1令y''[4(x1)22]e(x1)20,可得函數(shù)的下凸區(qū)間 2), 2);令y''[4(x1)22]e(x1)20可得函數(shù)的兩個(gè)拐點(diǎn) 由于lime(x1)20,因此函數(shù)有水平漸近線y綜上制表2ye(無x1無x1無y(1 (,12),(12,x2,3 依表作圖y ;x2解：可以求得函數(shù)的定義域?yàn)閧x|x1},由于y(xy(x),因此關(guān)于直線x0令y 0,無解。令y' 0,可得穩(wěn)定點(diǎn)x0;令y' 0,可得x2 (x2 6x2

(x2由于y(x21)3,則y''(0)20,于是穩(wěn)定點(diǎn)x10為函數(shù)極大值點(diǎn)令y

6x22(x2

6x2 令y'' 0無解,沒有拐點(diǎn)。由于lim 0,因此函數(shù)有水平漸近線y(x2 xx2

x2

,

x2

,因此函數(shù)有豎直漸近線x綜上制表{x|xx0無x1無(,1),(1,(0,1,),(1y0,x(,1),(1,無yln1x;解1yln11x無無無xx依表作圖yx2arctan解：可以求得函數(shù)的定義域?yàn)?由于y(x)y(x),令yx2arctanx 可以求得函數(shù)的一個(gè)根x0;令y' 1

0,點(diǎn)x1;令y' 1

由于y''

,則y''(1)10,于是穩(wěn)定點(diǎn)x1為函數(shù)極小值點(diǎn),y''(1)10于令y令y

(1x2

,0得x0,有拐點(diǎn)x0limx2arctanxxlim(x2arctanxx;且limx2arctanx1,數(shù)有斜漸近線yx

另外y(1)10,limy,因此在區(qū)間(1)上函數(shù)有一根,同理在區(qū)間(1, yx2arctanyx2arctanx0(還有兩個(gè)xx(,1),(1,yx(,(0,x依表作圖yxex解：可以求得函數(shù)的定義域?yàn)榱顈xex0,可以求得函數(shù)的根x令yexxex0,可得穩(wěn)定點(diǎn)x由于y''2exxex,則y''(1e10,于是穩(wěn)定點(diǎn)x1為函數(shù)極大值點(diǎn),令y2exxex0,可得函數(shù)的下凸區(qū)間)及上凸區(qū)間(2令y2exxex0得x2,有拐點(diǎn)xlimxex0,因此函數(shù)有水平漸近線y綜上制表y無無xx無y(,(2,x依表作圖y

x22xx2 x22xy 無無x2點(diǎn)(x2(,2(2 5,(25,2y(0.805603,)(,6.41883),依表作圖y(x(x(xy(xx無無x無點(diǎn)無間無y1,x間(1,1),(3,(,依表作圖y(x