求下列函数的单调区间和极值
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求下列函数的单调区间:
解:1.函数y
=
x
3
–
3x
2
–
9x
+
14,求导可得y
’
=
3x
2
–
6x
–
9
;
1)令y
’
>
0可得3x
2
–
6x
–
9
>
0
=>
x
2
–
2x
–
3
>
0
=>
(x
+
1)(x
–
3)
>
0
=>
x
<
-1或者x
>
3,此时原函数单调递增;
2)令y
’
<
0可得3x
2
–
6x
–
9
<
0
=>
x
2
–
2x
–
3
<
0
=>
(x
+
1)(x
–
3)
<
0
=>
-1
<
x
<
3,此时原函数单调递减;
综上所述,所求函数的单调递增区间是
(-
∞,
-1)
以及
(3
,
+
∞
)
;所求函数的单调递减区间是
(-1
,
3)
;
解:2.函数y
=
x
–
ln(1
+
x),求定义域:1
+
x
>
0,即
x
>
-1
,求导可得y
’
=
1
–
1/(1
+
x)
=
[(1
+
x)
–
1]/(1
+
x)
=
x/(x
+
1)
;
1)令y
’
>
0可得x/(x
+
1)
>
0
=>
x(x
+
1)
>
0
=>
x
>
0,此时原函数单调递增;
2)令y
’
<
0可得x/(x
+
1)
<
0
=>
x(x
+
1)
<
0
=>
-1
<
x
<
0,此时原函数单调递减;
综上所述,所求函数的单调递增区间是
(0
,
+
∞
)
;所求函数的单调递减区间是
(-1
,
0)
;
解:3.函数y
=
(x
–
1)
2/3
,求导可得y
’
=
(2/3)(x
–
1)
-1/3
=
(2/3)*
3
√[1/(x
–
1)]
=
2/[3
3
√(x
–
1)]
;
1)令y
’
>
0可得2/[3
3
√(x
–
1)]
>
0
=>
3
√(x
–
1)
>
0
=>
x
–
1
>
0
=>
x
>
1,此时原函数单调递增;
2)令y
’
<
0可得2/[3
3
√(x
–
1)]
<
0
=>
3
√(x
–
1)
<
0
=>
x
–
1
<
0
=>
x
<
1,此时原函数单调递减;
综上所述,所求函数的单调递增区间是
(1
,
+
∞
)
;所求函数的单调递减区间是
(-
∞,
1)
;
求下列函数的极值点和极值:
解:f(x)
=
(1/3)x
3
–
x
2
–
3x
+
3,求导可得f
’(x)
=
x
2
–
2x
–
3,令f
’(x)
=
0可得x
2
–
2x
–
3
=
0
=>
(x
+
1)(x
–
3)
=
0
=>
x
+
1
=
0或者x
–
3
=
0,所以x
=
-1或者x
=
3,所以
-1
和
3
是极值点。
因为在x
=
-1左侧,有f
’(x)
>
0,在x
=
-1右侧,有f
’(x)
<
0,所以函数f(x)的
极大值
是f(-1)
=
(1/3)*(-1)
3
–
(-1)
2
–
3*(-1)
+
3
=
-1/3
–
1
+
3
+
3
=
14/3
;
因为在x
=
3左侧,有f
’(x)
<
0,在x
=
3右侧,有f
’(x)
>
0,所以函数f(x)的
极小值
是f(3)
=
(1/3)*3
3
–
3
2
–
3*3
+
3
=
9
–
9
–
9
+
3
=
-6
;
②y
=
3
–
2(x
+
1)
1/3
,求导可得y
’
=
(-2)*(1/3)(x
+
1)
-2/3
=
(-2/3)
3
√[1/(x
+
1)
2
]
=
-2/[3
3
√(x
+
1)
2
],恒有y
’
<
0,(函数在定义域内单调递减),因此y
’
≠
0,所以所求函数没有极值点,没有极值。
解:1.函数y
=
x
3
–
3x
2
–
9x
+
14,求导可得y
’
=
3x
2
–
6x
–
9
;
1)令y
’
>
0可得3x
2
–
6x
–
9
>
0
=>
x
2
–
2x
–
3
>
0
=>
(x
+
1)(x
–
3)
>
0
=>
x
<
-1或者x
>
3,此时原函数单调递增;
2)令y
’
<
0可得3x
2
–
6x
–
9
<
0
=>
x
2
–
2x
–
3
<
0
=>
(x
+
1)(x
–
3)
<
0
=>
-1
<
x
<
3,此时原函数单调递减;
综上所述,所求函数的单调递增区间是
(-
∞,
-1)
以及
(3
,
+
∞
)
;所求函数的单调递减区间是
(-1
,
3)
;
解:2.函数y
=
x
–
ln(1
+
x),求定义域:1
+
x
>
0,即
x
>
-1
,求导可得y
’
=
1
–
1/(1
+
x)
=
[(1
+
x)
–
1]/(1
+
x)
=
x/(x
+
1)
;
1)令y
’
>
0可得x/(x
+
1)
>
0
=>
x(x
+
1)
>
0
=>
x
>
0,此时原函数单调递增;
2)令y
’
<
0可得x/(x
+
1)
<
0
=>
x(x
+
1)
<
0
=>
-1
<
x
<
0,此时原函数单调递减;
综上所述,所求函数的单调递增区间是
(0
,
+
∞
)
;所求函数的单调递减区间是
(-1
,
0)
;
解:3.函数y
=
(x
–
1)
2/3
,求导可得y
’
=
(2/3)(x
–
1)
-1/3
=
(2/3)*
3
√[1/(x
–
1)]
=
2/[3
3
√(x
–
1)]
;
1)令y
’
>
0可得2/[3
3
√(x
–
1)]
>
0
=>
3
√(x
–
1)
>
0
=>
x
–
1
>
0
=>
x
>
1,此时原函数单调递增;
2)令y
’
<
0可得2/[3
3
√(x
–
1)]
<
0
=>
3
√(x
–
1)
<
0
=>
x
–
1
<
0
=>
x
<
1,此时原函数单调递减;
综上所述,所求函数的单调递增区间是
(1
,
+
∞
)
;所求函数的单调递减区间是
(-
∞,
1)
;
求下列函数的极值点和极值:
解:f(x)
=
(1/3)x
3
–
x
2
–
3x
+
3,求导可得f
’(x)
=
x
2
–
2x
–
3,令f
’(x)
=
0可得x
2
–
2x
–
3
=
0
=>
(x
+
1)(x
–
3)
=
0
=>
x
+
1
=
0或者x
–
3
=
0,所以x
=
-1或者x
=
3,所以
-1
和
3
是极值点。
因为在x
=
-1左侧,有f
’(x)
>
0,在x
=
-1右侧,有f
’(x)
<
0,所以函数f(x)的
极大值
是f(-1)
=
(1/3)*(-1)
3
–
(-1)
2
–
3*(-1)
+
3
=
-1/3
–
1
+
3
+
3
=
14/3
;
因为在x
=
3左侧,有f
’(x)
<
0,在x
=
3右侧,有f
’(x)
>
0,所以函数f(x)的
极小值
是f(3)
=
(1/3)*3
3
–
3
2
–
3*3
+
3
=
9
–
9
–
9
+
3
=
-6
;
②y
=
3
–
2(x
+
1)
1/3
,求导可得y
’
=
(-2)*(1/3)(x
+
1)
-2/3
=
(-2/3)
3
√[1/(x
+
1)
2
]
=
-2/[3
3
√(x
+
1)
2
],恒有y
’
<
0,(函数在定义域内单调递减),因此y
’
≠
0,所以所求函数没有极值点,没有极值。
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