大一上-微积分12008ex-3答案

第三次习题讨论洛尔(Rolle)定理,研究函数零点分布与导数零点分布的关系；拉格朗日(Lagrange)定理,研究增量表示的有限形式；

ex,Sinx,Cosx,ln(1x),(1

x00（Taylor公式的练习主要在下次习题课六方面的应用:函数的性态研究与函数图形不等式证明

x[ab],f(x)0,ff(a)0,f(x)0,x[a,fk(a)0,fk1(x)0,x[a,c(a,b),f(c0,f(x)无零点x[ab];Minf(x)0(3).不定型求极限和无穷小阶的分析方程根的研究:注意f(xn1fnxfnxf(x至多有n函数的近似与值的估计LagrangeTaylor公式的灵活运用:x0和确定阶数极值问题(极大或极大Maxfs.t.x[axxx x11xln

xx

x

1x

yx

x11xln

x11xln

y0yln(y

(y1)y(ln(1y)y(1

(y1)y1(ln(1y)y(1；

1

(1ln(1y)1y

11

1

11

(121

ex2Cos2x2ex

ex2Cos2xx2ex12

ex2Cos2xx2x2=

2xex2Sin24x3

12x2ex2Cos6x =

4x36xex22Sin

1 2xex2Sin2

2xex2

2x1ex2

2xx

4x3

4x3

=

4x3

4x3

1Cos

2

6x

6x

x06x

4x36xex22Sin

6x4x

ex2Cos2xx2ex12

ex2Cos2xx2x2 x

x x 1x2

ox41 ox4=lim

xx

x x 1x2

ox4

xox4=lim

=lim 1 x x4

1cos2x1xsin ；x2(ex2解

1cos2x1xsin22

1cos2x1xsinlim

x2(ex

2sinxcosx1sin2xxcoslim 1sin2xxcoslim limcos2xcos2x2xsin2x

lim2sin2x

xf(x)

x x

x0处连续但不可导，则的取值范围是[BA0．

01．

1．

1．

f(xx0的某邻域内有定义，F(x)xf(xF(xx0要条件是[A

f(x)

f(x). （B）

f(x

f(x)在x0处连续

f(xx0f(x),g(x),xh

f(xh)

f(x)g(h)f成立，且f(0g(00,g(0f(01，求f(xf(x)

f(xh)f(x)

f(x)g(h)f(h)g(x)f(x)

f(x)(g(h)1)f(h)g(x)

f(x)(g(h)g(0))(f(h)f(0))g(x)

f(x)g(h)g(0)

g(x)f(h)f

f(x)g(0)g(x)f(0)g(x)f(x)

x

x0f(0．答案：(0){x（1）yy(xxy2y2lnx40y2d(xy2)d(ey2lnx)ey2lnxdy2lnx)xy2(2ylnxdy2

x2因而对原式两端求微分，得到(xy21)(2ylnxdyydx)0.f(x)dy 2（2）

x

x)

2xlnt

dx(costsin

dx(costsin解：et(sintcost)dt1dy y(x)dy

ycostsint

dyye(sintcosyet

d

dtdt

e

ddd

dt

d)/)t

yet)(/dt

t

t

/

设函数f(x)在[ab上可导，且导函数严格单增，若f(a)

f(b，试证对任意的xab

f(x)

f(a)

f(b证明：若xabf(x)

f(a)

f(bf(x在(abx0Lagrangef(bf(x0

f()(bx0)f(bf(x0

f()(bx0)

f(b)

f(x0)， 证毕f(xf(x0)0f()0

a0,f(00,且0

f(x)

f(x 证明f(x)0

f(x)

f(0)f(1)x

f(1)x

f21x

2x2

fn

i(0,x)(0,a]x0a使x1,f(x)0.fabR,在(ab二阶可导,f(af(b)0证明

x[a,

(a,b):f(x)

fxaxb2

g(t)f(t)Mtat确定系数M,让函数有三个零点

a,b,x

xab)这样必ab)g(0,

g()f()2M0M

2g(x)

f(x)

fxaxb2f(x)

210*．设f(x)在(abx0(a,b处可导。数列{xn},{yn}axnx0ynb,

xn

ynx0试求

fynf(xn)yn

fyn),f(xnx0Taylorf(yn)f(xn)

f(x0)f(x0)(ynx0)o(ynx0)f(x0)f(x0)(xnx0)o(xnx0)

f(x0)f(x0)(ynx0)o(ynxn)f(x0)f(x0)(xnx0)o(xnyn

f(yn)f(xn)

f(x0)(ynxn)o(ynxn)

f

00

yn

yn

f(x在[ab上连续，在(ab内二阶可导，且f(af(b0,f(a0在abf()0,由条件f(a0存在

0,使得在(aa)上

fx0x0(a,b),f(x)

f(x0

f(x0)x0显然是极值点故f(x00.由于fx0,x(a,b我们有f(x)0,x[x0再由中值定理,存在(x0,b),f(b)

f(x0)f()(bx0)

f(x0)f(x在(0,1内取到最大值，二阶可导，且f(x)1x[0,1]f(0f(1)1fx)在x0fx0|f(0)||f(1)||f(0)f(x0)||f(1)f(x0)|f(1)|x0|f(2)||1x0x0|1x0|若bae，证明abbaf(x)2nx(1x)n

Mmax{fnn

m,n0,mn

mnlnmlnnmn 证明1：为证mnlnmlnn,f(xlnmlnxmxm问题是x0,xm

f(x)0今f(m)0,f(x)110

f(x)f(m)fxm0 证明2:不等式变成1nlnmm1,令um,问题变成u 11lnuuu令u1x,问题变成

x0

1

ln1xx

x(1,)

1

ln1xxf(xxln1xg(x)ln1x在区间(1,f(0)g(0)0

1x

nnk

4nn令f(xxk1x2kknnf(x)k

kxk11x2k2kxk1x2k1nnk

kxk11x2k10

x 因f(0f(10

1为[0,13 n1k

1

n4 f(3)3

1 3

27k1 k1

2xSinxx

x

2 证明：先证x 2

2xSinx设f(x)Sinxx

x

(0,)2f(x)

xCosxSinxx2

CosxxTanx0xf(x)Sinxffx2 2 2 设x

,tSint2tMtt2 22

x0,M,x是函数(t的一个零点2这样，函数(t

[00,x 可见， 2

0,f

0

2x

,x

f(x)2x

fxx0 2

2 2Sinx2xCosxx,

xx 22

Sinx2xCosxx02 2 证明三：利用连续函数的介质性,f(x)Sinx2x

)中无2点f(x在(0x,f(x在[02 x,；从而,

,

Cos0,这是不可能的. fSin

2

10

f(x)06

x

[0,].证明四：利用函数的最小值非负性,f(x)Sinx2x(a)f(x在(0xarccos2,f(x)Sin

0,x2极大点

f(x)

有最小点，但不在

内,2f(0)

f()0,

f(x)0.由此推出x[0

f(x)02

x[0, 2xx2x若x0,xx2x1(x,x)1,且limx,x1 limx,x1 fab)R,

f(xx)f(x)

f(xxx中x(a,b)且f(x)0,有limx,x1 f(xxf(x)

k

f(x)xk

xfn(xx)x

limx,x

n1（1）

2xxx2x x,x

x0,x0

2x2xxx2x2xxx2xx xx2x22

12x2x22x

12

2xxx2x

2x1xo(x)2x

limx,x

2xxx2x

（2）f(xx)f(x)f(xx)f(x)f(x)x1fxx2ox22f(xx)

f(x)f(xx)x=f(x)f(x)fxxo(x)x1fxx2ox2=fxxo(x)x21f(x)x2ox2 x,x2

2课外练1（1）lim(12f(xsinx，其中f(x)xf(0)0,f(0)2

x0f(x)0 2f(

2limf(x)f(0)

sin

ex

sin

e221

f2f2 x

y

f(xysinx相切y故

f(x在原点处与ysinx相切，知f(0sin00f(0sinx)x0ff2 xxf2 x2lim y 2f2fx22xxfyfy2limx 2y

f(x1)满足f(x) ，x3xx3

x2解

(x1)2

f(xx

2f(3)2 证明方程2x2x2x10解：由于f(x2x2x2x1,f(x2xln24x1,f(x2xln2240f(x)0f(x)0fD[a,),

f(x存在,

f(x)证明：(法一 有界

f(x)0;x

0

f(x)x

f(x)b(法二 反证法:若

f(x)b0,设bM0xM,f(xb20f是[M,上的增函 f(Mn)f(M)f(M1k)f(Mkk k fk

bnb k

k0 f(xC1(1,1),且f(0)0,f(x)在x0连续存在，证

f(x)f(ln(1x))1f(0) 若极限x

x1

存在，求xx(1 xf(0)0f(xx0x0gx，使得f(xxg(x．证明f(xx0处可导

fxf

)(A

x

g(0)Agx

f(x)在[ab(ab内二阶可导，f(a)

f(b)0，f(a)f

0（1）（2）（3）在ab，使得f(f(f(a)0,f(b)00（1）f(