题目内容
(理)设函数f(x)=(x+1)2(x-2),则
等于( )
| lim |
| x→-1 |
| f′(x) |
| x+1 |
分析:由f′(x)=2(x+1)(x-2)+(x+1)2=3(x+1)(x-1),知
=
,由此能求出其结果.
| lim |
| x→-1 |
| f′(x) |
| x+1 |
| lim |
| x→-1 |
| 3(x+1)(x-1) |
| x+1 |
解答:解:∵f′(x)=2(x+1)(x-2)+(x+1)2
=3(x+1)(x-1),
∴
=
=
3(x-1)=-6.
故选D.
=3(x+1)(x-1),
∴
| lim |
| x→-1 |
| f′(x) |
| x+1 |
| lim |
| x→-1 |
| 3(x+1)(x-1) |
| x+1 |
=
| lim |
| x→-1 |
故选D.
点评:本题考查导数的求法和函数的极限的求法,解题时要认真审题,是基础题.
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