题目内容
已知常数a、b、c都是实数,函数
的导函数为f ′(x)
(Ⅰ)设a=f ′(2),b=f ′(1),c=f ′(0),求函数f(x)的解析式;
(Ⅱ)设 f′(x)=(x﹣γ)(x﹣β),且1<γ≤β<2,求f ′(1)
f ′(2)的取值范围.
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120818/201208181117440332758.png)
(Ⅰ)设a=f ′(2),b=f ′(1),c=f ′(0),求函数f(x)的解析式;
(Ⅱ)设 f′(x)=(x﹣γ)(x﹣β),且1<γ≤β<2,求f ′(1)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120925/20120925213818068180.png)
(Ⅰ)解:由题意可得,f ′(x)=x2+ax+b.
∴
,
解得:
.
∴
.
(II)∵f ′(x)=(x﹣γ)(x﹣β).
又 1<γ≤β<2,
∴f ′(1)=(1﹣γ)(1﹣β)>0,f ′(2)=(2﹣γ)(2﹣β)>0
∴f ′(1)
f ′(2)=(1﹣γ)(1﹣β)(2﹣γ)(2﹣β)
=[(γ﹣1)(2﹣γ)]
[(β﹣1)(2﹣β)] ![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120818/201208181117442131374.png)
∴![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120818/20120818111744256890.png)
∴
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120818/201208181117440781022.png)
解得:
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120818/20120818111744123734.png)
∴
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120818/20120818111744168903.png)
(II)∵f ′(x)=(x﹣γ)(x﹣β).
又 1<γ≤β<2,
∴f ′(1)=(1﹣γ)(1﹣β)>0,f ′(2)=(2﹣γ)(2﹣β)>0
∴f ′(1)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120925/20120925213818170180.png)
=[(γ﹣1)(2﹣γ)]
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120925/20120925213818280180.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120818/201208181117442131374.png)
∴
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20120818/20120818111744256890.png)
![](http://thumb.zyjl.cn/images/loading.gif)
练习册系列答案
相关题目