已知函数f(x)=(x-1
|
∵f(x)=(x-1
(x+2) 3
,) 2
∴f′(x)=3(x-1)2•(x+2)2+2(x-1)3(x+2)
=(x-1)2(x+2)[3(x+2)+2(x-1)]
=(x-1)2(x+2)(5x+4),
由f′(x)=0,得x1=-2,x2=-
,x3=1,4 5
列表讨论,得
x | (-∞,-2) | -2 | (-2,-
| -
| (-
| 1 | (1,+∞) | ||||||
f′(x) | + | 0 | - | 0 | + | 0 | + | ||||||
f(x) | ↑ | 极大值 | ↓ | 极小值 | ↑ | ↑ |
) | 3 |
) | 2 |
故选B.