一、填空：(2-1)25

-15

=5x4,且x?(1,2)4315x=x2

+1xfi

0+

f

(0+)

=

limxsin

3x

=0f

(0-)

=

f

(0)

=a\

f

(0+)

=

f

(0-)

=

f

(0)

a

=01、f

(x)=x5在1,2]上滿足L.Th,則x

=處處連續,則a

=2、f

(x)=

a

+

x2

x

￡

0x

sin

3x

x

>

0x2

+1xxfi

0lim

sintx

=t

(利用1極限)\

f

(t)

=t2

f

'

(t)

=

2txxfi

03(1)若f

(t)

=

lim

sin

tx

t,則f

'

(t)

=

.(1xxfi

￥(2)若f

(t)

=

lim

t

1+2tx

,則f

'

(t)

=

.xxfi

￥lim(1+1

)2tx

=e2t

(利用e極限)\

f

(t)

=te2t

f

'

(t)

=e2t

+te2t

2=(2t

+1)e2t0可導極值點必為駐點：f

'(x

)=04(1)

f(x)在x=

x0處左右導存在且相等是f

'

(x

)存在的

條件.0可導 左右導數存在且相等f(x)在x=

x0處左右極限存在且相等是f

(x)在x0處連續的 條件.連續 左右極限存在且都等于函數值若f

(x)在x

=x0處可導且取得極值,則必有f

'

(x

)

=

.0

0'0

0h=

(a

-

b)

f

(x

)f

[

x

+ah+ -

f

[

x

+blimhfi

0(h)](h)]

h+若f

'

(x

)存在,a

,

b

?

R,則0Dx

f

[a+Dx+(Dx)2

]-

f

(a)

=5、設f

'(a)存在,則limDxfi

0Dxf

[a+Dx+(Dx)2

]-

f

(a)

=

f

'

(a)\

limDxfi

0x-0lim

f

(

x)-

f

(0)xfi

0二、選擇：xxfi

0=

lim

f

(

x)=lim(x2

+1)

(x2

+

n)xfi

0=

n!=

f

'

(0)1、f

(x)

=

x(x2

+1)(x2

+

2)(x2

+

n),則f

'

(0)

=

(

)A、0

B、n

C、n!

D、12、在(a,b)內,f

'

(x)

>

0,f

''

(x)

<

0,則f

(x)的圖像在(a,b)內(

)B、單減上凸

D、單增上凸A、單減上凹

C、單增上凹f

'

(x)

>

0

?,f

''(x)<0

上凸2\

lim

dy

=

1

?

0,1,￥Dxfi

0

Dx\dy與Dx同階20dy

=

f

'

(x

)Dx

=

1

Dx20

03、若f

'(x

)=1

,則Dx

fi

0時,f

(x)在x

處的B、同階無窮小

D、低階無窮小微分dy是Dx的(

)A、等價無窮小

C、高價無窮小1xcos'解：y

=

ex'cos1x,

則y

=

(

)4、設y

=ex2(-sin1)

(-

1

)x1xcos1

1

x2=

sin

eeex1x1x1x1xcos1

1

x2cos1x1xcos1x

1

x2cos1xD.

sin

eC.

sinsin

eB.

-A.

-sinx

?0

0

x

=0x

f

(x)解：

f

(0)=0\F(x)=x-0

xxfi

0

xfi

0

f

'

(0)

=limf

(x)-f

(0)

=limf

(x)

?0x\

limF(x)

=lim

f

(x)

=

f

'

(0)

?

f

(0)xfi

0

xfi

0\x

=0是F

(x)的可去間斷點x,f

(0)=0,f

'(0)?0,則

f

(x)

0

x

=0x

?05、F(x)=B、一類間斷點

D、是否連續不確定x

=0是F(x)的(

)A、連續點C、二類間斷點三、解答：只能用定義式討論導解：

f

(x)在R上可

f

(x)在x

=

0處可導

f

(x)在x

=0處ct在R上可導b

+ln(1+

x)

x

>

0x

￡

0sinax1(1)確定a,b使f

(x)=\

f

(0+)

=

f

(0-)

=

f

(0)且f

'(0)=f

'(0)+

-\

f

(0+)

=

f

(0-)

=

f

(0)

b

=0(1)+x=

lim

=

f

(0)xfi

0+

f

(x)-

f

(0)

ln(1+x)+b

'x-0limxfi

0+

f

(0+)

=

b,

f

(0-)

=

f

(0)

=

0'-=a

=

f

(0)sinax

x=

limxfi

0+

f

(x)-

f(0)

x-0limxfi

0-''(2)xb+ln(1+x)+

-=

f

(0)

a

=limxfi

0\

f

(0)聯解(1)(2)得：a

=1,b

=0.\f

'(0)=0

f

(x)在x

=0處可導\f

(x)在x

=0處ct2210

x

x

|x|x

sin

+(2)討論f

(x)=x

=

0在ln(1+

x

)

x

?

0分段點處的連續性和可導性解：

f

(0+)=f

(0-)=f

(0)=0x-0

xxfi

0–

xfi

0–

xlim

f

(x)-

f

(0)

=

lim

1

[x2

sin1

–ln(1+x2

)]12=0ln(1+x

)xxxfi

0–

xfi

0–=

lim

xsin

–

lim等價+洛必達+恒變+等價xfi

0

x2

ln(1+x)x3xfi

02(1)

lim

tan

x-x

=lim

tan

x-x3x2xfi

02=lim

sec

x-13x2xfi

02=lim

tan

xxfi

0

3x22=lim

x3=

1x2(2)lim

1+x+

1-x-2xfi

02x

1

+

-1xfi

0=lim

2

1+x

2

1-x

x

1

-

11+x

1-x4

xfi

0=

1

limx4

xfi

0

1

(

1-x-

1+x)=

1

lim

1-x2x4

xfi

0=

1

lim

1-x-

1+x2

1-x

2

1+x4

xfi

04=

1

lim(-

1

-

1

)

=-1~

x4法1：

x

fi

0時,arctan

x412x4xfi

0x2\

原式=lime

+2cosx-3

==

74次用洛必達法則2arctan

x4(3)lim

ex

+2cos

x-3xfi

0~

x42!2ex=1+

x2

+

1

(x2

)2

+(x4

)cos

x

=1-

1

x2

+

1

x4

+(x4

)2!

4!7124

4

7

12

=x4xfi

0x

+(x

)\原式==limarctan

x42(3)lim

ex

+2cos

x-3xfi

0法2：

x

fi

0時,arctanx4

1

ln

x(4)

lim(cot

x)xfi

0+xfi

0+

x2

sin

x+

lim

x-sin

xlncotx

ln

xA

=exp[lim

]xfi

0+]21

x-csc

x

cotx=exp[limxfi

0+]21

xtan

x

sin

x=exp[-

limxfi

0+2-1=

exp[-

lim

]

=

ex

xxfi

0+

1

xx3x-sinxB

=

limxfi

0+3x21-cosx=

limxfi

0+16

2

=

lim

=1

x2xfi

0+

3x2=

A+B16-1=e

+

y(0)=0且3(1)已知y

=y(x)由ex+y

-xy

=1確定,求y''(0).解：ex+y

-xy

=1ex+y

(1+y'

)-y

-xy'

=0(1)代x

=0,y

=0入(1)式得：e0+0[1+

y'

(0)]-0-0y'

(0)

=0\

y'

(0)

=-1ex+y

(1+

y')-y

-xy'

=0\

ex+y

(1+y'

)2

+ex+y

y''

-y'

-(y'

+xy''

)

=0(2)代x

=0,y

=0,y'=-1入(2)式得：e0+0(1-1)2

+e0+0

y''

(0)-(-1)-[-1+0y''

(0)]=0\

y''

(0)

=-2x-2

x-2xfi

￥

xfi

￥解：lim(

x+5

)2x+3

=lim(1+

7

)2x+3

=e14\

f

(x)

=

e14

ln(x

-

2014)\

f

(n)

(x)

=

e14

(-1)n-1(n

-1)!(x

-2014)-n(2)

f

(x)

=ln(x

-2014),2x+3

x+5x-2lim(

)xfi

￥(n)求f

(x)(3)x4

-xy

+1

y4

=1

,求y'2

2

(0,1)2(0,1)=

1

y'4(0,1)=-1

y''解：4x3

-y

-xy'+2y3

y'=0\

12x2

-y'

-y'

-xy''

+6y2

y'

y'

+2y3

y''

=0解1：x

=1

t

=1t

xx\

y

=ln

1

+3x2(-

1

)

=-

1\

dy

=

1

dx

1

x

xx2

x2dx22

1

\

d

y

=

-(-

1

)

=d

2

ytdx2x

=

1

y

=

3

+

ln

t4、已知：,求t2t

dt解2：x

=1

dx

=-

1y

=

lnt

+

3

dy

=

1dt

tdx

tt

2\

y'

=

dy

=

1

(-

1

)

=

-t'2t

2dxdx2d

y

dy=

(-1)

(-

1

)

=

t

2\

==

-1,dt'

dyt2dtdx

=-

15、求f

(x)=x3

+3x2

-24x

-32的極值解：

f

'(x)=3x2

+6x

-24\f

'(x)=0

x

=-4,x

=2且無奇點1

2f

''

(x)

=6x

+6

f

''

(x

)

<0,

f

''

(x

)

>01

2\f極大(-4)=48,f極小(2)=-60x-axfi

axfi

a=lim[2j(x)

+(x

-a)j'

(x)]

=

2j(a)xfi

a

f

(x)

=(x

-a)2j(x)\

f

'

(x)

=

2(x

-a)j(x)

+(x

-a)2j'

(x)

f

'

(a)

=0'

'\

f

''

(a)

=lim

f

(x)-

f

(a)6、設j'(x)ct,f

(x)=(x

-a)2j(x),求f

''(a)解：j'

(x)ct

\

limj'

(x)

=j'

(a)四、證明：Pf：令f

(x)=sin

x

+tan

x

-2x2

f

(0)=0且"x?

(0,p

),cosx?

(0,1)f

'

(x)

=

cos

x

+sec2

x

-

2cos2

x=cosx

+

1

-221(1)證明：x

?

(0,p

)時,sin

x

+tan

x

>2x2即"x

?

(0,p

),sin

x

+tan

x

>2x\

f

(x)

?

f

(x)

>

f

(0)

=0cos2

xf

'

(x)

=cosx

+

1

-2cos2

x>cos2

x

+

1

-2cos2

x?2

cos2

x

1

-2

=02f

'

(x)

=

g(x)

=cosx

-1+

1

x23!即"x

>

0,sin

x

>

x

-

1

x362

f

(x)

?：f

(x)

=sinx

-x

+

1

x3

>

f

(0)

=0

g(x)

?：g(x)

=cosx

-1+

1

x2

>

g(0)

=0g'

(x)

=

h(x)

=-sinx

+

xh'

(x)

=-cosx

+1>0

h(x)

?：h(x)

=-sinx

+

x

>

h(0)

=06證：令f

(x)=sin

x

-x

+1

x3

,則x

>0時3!(2)證明："x

>0,sin

x

>x

-

1

x3即

1

[xf

'(x)-f

(x)]=0Pf：令F(x)=

f

(x)

,則xF(x)?

C[a,b]

D(a,b)且F(a)=F(b)=0\$x?

(a,b)使F'(x)=0x2

xf

'

(x)

=

f

(x)2(1)設b

>a

>0,f

(x)?

C[a,b]

D(a,b),f

(a)=f

(b)=0證明：$x?

(a,b)使x

f

'(x)=f

(x)(2)

f

(x)?

C[0,b]

D(0,b),

f

(b)

=

0

$x?

(0,b)st：f

(x)

+xf

'

(x)

=

0證：令F(x)=xf

(x),則F(x)?

C[0,b]

D(0,b)且F(0)=F(b)=0F'

(x)

=

0由Rolle.Th：$x?

(0,b)使即：f

(x)+xf

'(x)=03(1)

f

(x)?

C[a,b]

D(2)

(a,b),a

<

x

<

x

<

x

<b,1

2

3f

(x1)

=

f

(x2

)

=

f

(x3)求證：$x?

(a,b)st：f

''(x)=0Pf：

f

(x)

?

C[a,

b]

D(a,

b)[x1

,

x2

] [a,

b],[x2

,

x3

] [a,

b]\

f

(x)

?

C[x1,

x2

]

D(x1,

x2

)f

(x)

?

C[x2

,

x3

]

D(x2

,

x3

)\

f

'

(x)

?

C[x

,x

]

D(x

,x

)1

2

1

2[a,

b][x1,x2

](a,b)使由Rolle.Th：$x?

(x1,x2

)f

''

(x)

=

0由R.Th：$x1

?

(x1,x2

),x2

?

(x2

,x3

)使f

'

(x

)

=

f

'

(x

)

=

01

2

f

'

(x)

?

C[a,

b]

D(a,

b)(2)f

(x)?

C[a,b]

D(2)(a,b),連接A(a,f

(a)),B(b,f

(b))的線段交y

=f

(x)于C(c,f

(c)),a

<c

<b,求證：$x?

(a,b)st：f

''(x)=0Pf：

f

(x)

?

C[a,

b]

D(a,

b)[a,

c] [a,

b],[c,

b] [a,

b]\

f

(x)

?

C[a,

c]

D(a,

c)f

(x)

?

C[c,

b]

D(c,

b)1

2

1

2\

f

'

(x)

?

C[x

,x

]

D(x

,x

)[a,

b][x1,x2

](a,b)使由Rolle.Th：$x?

(x1,x2

)f

''

(x)

=

0由L.Th：$x1

?

(a,c),x2

?

(c,b)使1

2

AB

f

'

(x)

?

C[a,

b]

D(a,

b)f

'

(x

)

=

f

'

(x

)

=

k4、證明至少存在一點x

?

(1,e)使sin1

=coslnx成立構造函數用相關定理注意寫出定理所需條件令f

(x)=sin1-cosln

x,在[1,e]上用零點Th令f

(x)=sinln

x

-ln

xsin1,在[1,e]上用洛爾Th令f

(x)=sinln

x,g(x)=ln

x，在[1,e]上用柯西Th令f

(x)

=cosln

x,

在[1,

e]上用介值Th(1)令f

(x)=sin1-cosln

x,在[1,e]上用零點ThPf：令f

(x)=sin1-cosln

x,則f

(x)?

C[1,e],且：f

(1)

=

sin1-1<

0,

f

(e)

=

sin1-cos1

>

0\由零點Th：$x

?

(1,e)使f

(x)=0即：sin1-coslnx

=0\$x

?

(1,e)使sin1

=cosln

x.#sin1

=

01

-

1x

x即：coslnx\$x

?

(1,e)使sin1

=cosln

x.#(2)令f

(x)=sinln

x

-ln

xsin1,在[1,e]上用洛爾ThPf：令f

(x)=sinln

x

-sin1

ln

x,則f

(x)?

C[1,e]

D(1,e),且：f

(1)

=

f

(e)

=

0\由洛爾Th：$x

?

(1,e)使f

'(x)=0x"

x

?

(1,

e),

g

'

(x)

=

1

?

0g'

(x)\

由Cauchy

Th：$x

?

(1,

e)使

g

(e)-g

(1)'

f

(e)-

f

(1)

=

f

(x)x

1x1

coslnx1-0sin1-sin0即：

=\$x

?

(1,e)使sin1

=cosln

x.#(3)令f

(x)=sinln

x,g(x)=ln

x，在[1,e]上用柯西ThPf：令f

(x)=sinln

x,g(x)=ln

x，則f

(x)?

C[1,e]

D(1,e),g(x)?

C[1,e]

D(1,e),且(4)令f

(x)=cosln

x,在[1,e]上用介值ThPf：令f

(x)=cosln

x,則f

(x)?

C[1,e],且：f

(1)

=1,

f

(e)

=cos1sin1?

(cos1,1)\由介值Th：$x

?

(1,e)使f

(x)=sin1即：coslnx

=sin1\$x

?

(1,e)使sin1

=cosln

x.#*(5)令f

(x)=sin

x,g(x)=ln

x,f

(x)在[0,1]上用Lagrang

Th后g(x)在[1,e]上用介值Th2*(6)sin1

=coslnx

cos(k1p

+

p

-1)

=coslnx22k2p

+

k1p

+

p

-1

=lnx取k1

=k2

=0,令g(x)=ln

x,在[1,e]上用介值Th解：據周期性有：f

(6)=f

(1),f

'(6)=f

'(1).

f

(x)連續且x

?

U

(0,d)時有：f

(1+sin

x)

-3

f

(1-sin

x)

=

8x

+(x)\

lim[

f

(1+sin

x)

-3

f

(1-sin

x)]xfi

0=

lim[8x

+(x)]xfi

0\

-2

f

(1)

=0

f

(1)

=0五、練習：1、教材P126總習題二14

f

'(1)存在且x

fi

0時有sin

x

~

xsinx

xsinxfi

0

xfi

0\

A=

lim

f

(1+sinx)-3

f

(1-sinx)

=lim8x+(x)

=8hhfi

0而A

=lim

f

(1+h)-3

f

(1-h)h

-hhfi

0

-hfi

0=lim

f

(1+h)-

f

(1)

+3

lim

f

(1-h)-

f

(1)f

(1)

=0=

f

'

(1)

+3

f

'

(1)

=4

f

'

(1)\

f

'

(1)

=

2\

f

(6)

=

f

(1)

=

0,

f

'

(6)

=

f

'

(1)

=

2.\所求切線l：y

-0

=2(x

-6)即：2

x

-y

-12

=0注1、求f(1)：利用f

(x)連續且

x

fi

0時f

(1–sin

x)fi

f

(1).注2、求f

'(1)：用到導數定義式中非零無窮小h的任意性2、教材P183總習題三19Pf

：令x0

=

(1-t)x1

+tx2

,則x0

?

(a,b)由Taylor公式：$x1,x2

?

(a,b)使f

(x

)

=

f

(x

)

+

f

'(x

)(x

-x

)

+1

f

'

(x

)(x

-x

)21

0

0

1

0

2

1

1

0f

(x

)

=

f

(x

)

+

f

'

(x

)(x

-x

)

+1