文档内容
例2.16 设 f (x)在[0,1]上连续,在(0,1)内可导,且 f (0) = 0, f (1) = 1.常数
a
a 0,b 0,证明:(1)存在(0,1)使得, f () = .
a + b
<
fill
Y floo 1 ip0
(1) =
,
,
(fIs) =
I TGE ESE (0 1)
: ,
,a b
(2)存在,(0,1), ,使得 + = a + b
f () f ()
[ F ] = =
sing
Gin
fis
,
=
-1
例2.18 设函数 f (x) = 2x−3 ，则（ A ）
（A）在其有定义的任何区间( x , x )内， f (x)必是单调减少的
1 2
（B）在点 x 及 x 处有定义，且 x x 时，必有 f (x ) f (x ) /E Ex/R]
1 2 1 2 1 2
（C）在其有定义的任何区间( x , x )内， f (x)必是单调增加的
1 2
（D）在点 x 及 x 处有定义，且 x x 时，必有 f (x ) f (x )
1 2 1 2 1 2
f(x) E X
=
( 3)
V(3
+ 2)
- 0 . ,
I
I
In 2
fix *. 20
-2 .
=
(x 35,
-
fixT ( 3) & (3 c) ↓
1 0 +
- .
,例2.19 设 f ( x ) = xex ,则 f (n) ( x ) 在点 x = - _处取极小值 -_n_.
1
- n- -
* T)
** (h1 ( X e ex ex
fin (r x ( + . = x . + n - = (x + ) .
.
= ·
6)
i X +XXt) - c , +
[f)
*
[fin T = ex + (x+H ) . ex = (x +n+ 11 . e EX = - n - 15 , = 0
0)
( - 0 , -
n+ 1)(
-
n -x +
]'
If +
-
fur
A
↓
E x2x , x 0,
例2.20 设 f ( x) = ,求 f (x)的极值.
x + 2, x 0,
elnx-e2x(nX
fix =** =
#J
① xo
,
eaxlux Y) e*xX(2(ux
fix (212x + 2x - = + 2)
= .
fix
tAJ
EX =o
=
,
El (t
+ol
(0
. , (e) e
fix
de
E
-
fix
&
-
#*
fix ↓ ↑② X= 0AE fix = X+2 fix = 10 fix
,
,
fix exx-
pu M
=
ot
x+
fix ↑ A fro
EBEI
2
it 0 =
X+2 =
,
-PERFE
fi =
3例2.21 证明：当 x 0时2sin x + ex − e−x 4x
fix
L EDA: fix = 2 SmX + e * - e- * - 4X EPIEX10 At , > O
*
fix e fia
= 210sX + + e - 4 , = 0
fi + e *, fil
-2 smX + e - = 0
=
f ex e
= - 20(X + + x - 2(3X + 2 = 2(1 - 13X)30
fixSo fir fin f
AJ : > = o
< XO ,
.
fin fir fit fix fro)
fixt
=
> = o + = o
:
,例2.22 设函数 f (x)在(0,+)内有二阶导数,且满足 f (0) = 0, f (x) 0,
0 a b,则当a x b,恒有( C ).
(A)af (x) xf (a) (B) xf (x) af (a) (C)bf (x) xf (b) (D) xf (x) bf (b)
↳
f a fiN
f(b)
El
<
x
fix-f(x)
(f) X
·
:
=
X2
fix-f(x)
/91 9(0)
35 X- = 0
- : = ,
f(x)
g'mx fix) Xf(x) f)
x
= + - =
gix 9 V
E X0Af Co
. ,
(
: 9 <910) 0 : 0
=fix-f(x)
fix-(fix-froi)
(f) X X
=
=
X2
2
X
fish
fix fix-fie)
X -X -
Ege 10 x)
=
= .
,
Xz
X
...
fin find fin fis
in co : : o
#J
Xb
: a <
flxfaka
fal #
, ,
b f(x) f(b)
X1 + x x2
例2.23 证明 x ln + cos x 1 + (−1 x 1).
1 − x 2
G EAH
:
En fix
=
X (InG
+
x)
-
Inc-X)]
+
X-1 - EPIE
/Af fix
kX 0
- ,
x)+x
-x)
fix)
fix In (HX)
(n(-X)
+ + -
smx
-
X
= 0
-
= ,
i
2
fix = ( x
+x(z)
1
& + x . - + - 10X -
c ci
1- X +
4
4
= t cosX -
-
1- X C +X) ? (
+
x)
fin
kX1 fix ↑
: - ..: > Ofix
fix flo
X0AF
- < =o fix)
: . ,
fix'
/AF fix fio i
ocx< > o
= ,
o +
-
o
im
0 I
fix
: X=
fix fio
: o
=
,b 2(b − a)
例2.24 设0 a b，求证ln .
a a + b
lnb-Ina
2 2
(
⑰ E gf(a b)
: 7
b - u
a+ b
a+
b .
a+ b i
34
2
InX-mac2(x-a
RE
⑰ E2 = o例2.25 若3a2 − 5b 0,则方程 x5 + 2ax3 + 3bx + 4c = 0( B )
(A)无实根 (B)有唯一实根 (C)有三个不同实根 (D)有五个不同实根
& fix = X + 2ax + 3bx + 4) , Xt( - a , + 6)
* Gax
fix 3) 36
= 5x + + = 5t + Ga+ + 3
(6ap
4x5x3b 36a 60b 12(2a"
5b)
- = - = - O
fix ↑ in f(-0) fitc)
: - e) = - 00 = + r
2 .
fix FE
i 0 PE --
=例2.26 讨论方程axex + b = 0(a 0)实根的情况.
12 f(x ax ex + b xz( - c , + 0)
= . ,
fix aet axex * Ex Aj fix
= + = a(n + Xe = + , =0
0)
1) (+, +
C
- x . -
fix)
+
-
↓ ↑
fix
e b)
mm(ax b
f( +
a =
=
* - u
f()
b
= - +
f(tr) M *
(axe b
=
+
=Eco 44b > Af fN ETE ⑮
Eb =0
① - ,
b
Y =
② ED AJ
6
= 0
fix
01
FE
= =
,
-
bob- 40AJ
, a
③
b
-
e
fix
02*
= X =- 1x
例2.27 函数 f (x)可导,且 f (0) = 0, f (x) 0,令 g(x) = t f (x − t)dt ,则
0
y = g(x)在(0,+)上有（ A ）
(A) 递增且为凹弧 (B) 递减且为凸弧
(C) 递减且为凹弧 (D) 递增且为凸弧
*
t( (X 1. full
(x-u) on
U = X- u) f(m) (au)
9 - =
fore ful
X/ are
future
-
=
gix 1. - finum Xfix- ) EX0AF fix fix fro
= + , , = o
*
gix) 9
: To
,
f(x)(0
q"m
9 1)
=
,例2.28 曲线 y = (x − 1)2(x − 3)2 的拐点个数为( C ).
(A)0 (B)1 (C)2 (D)3
-
35 :
-
33 7
t = x- 1 +(t
-
2)
=
t(t 4t+ 4)
=
t
-
4t
+
4t
==
X= t+ 1
y 4t 12t
ot V
= - +
y" 2
= 12t -
24+
+
8
242
4x12x870
8 =
-↑
例2.28 曲线 y = (x − 1)2(x − 3)2 的拐点个数为( C ).
(A)0 (B)1 (C)2 (D)3
35
==
⑭
127
is↑
x = t2 + 2t + 1,
例2.29 (数一、数二)求由参数方程 确定的函数 y = f (x)
y = t − ln(1 + t),
的单调区间与极值、凹凸区间与拐点.
Xa : + =( - 1 + c) X = (t +
1)
>0 XE)0 . + c)
,
# x'(t) = 2 t+2 = 2 (t+ ) 30 ..i X Bt ↑.
I
au-ayl - t
y Itt E t #,J y
= = = =0 =0
ax(at 2t + 2 2(1 + t)2
(0 b)
t ( +, 0) +
,
& SRE18 10 11 1 fix↓
X (0 1) (1 + c) . ,
, , ,
y - T (li + 3) 1 fix ↑
y ↑ P-FHE f(l)
= 0
.+)
al = til 2 (H + - t . 4 C + t)
y at
=
4(1 +)4
+
=
dX/dt
2(t+ 1)
2
2 - 2[ (1 + t)(1 - t) t
1-
= = - = = I #J Y" 0
-(t + 1)5 4(t + 1)5 1)4 , =
4(t
+
1)(l. b)
t (t + REIR 4)
X (10
, : .
(0 4) (4 + b)
↓ , E Xt(4 3)
, +
,
yY t -
FB (4 1- (n2)
. ,
Y E
