(1)f′(x)= , 令f′(x)=0,得ax2+2bx-a="0 " (*) ∵Δ=4b2+4a2>0, ∴方程(*)有两个不相等的实根,记为x1,x2(x1<x2), 则f′(x)= , 当x变化时,f′(x)与f(x)的变化情况如下表:
x
| (-∞,x1)
| x1
| (x1 ,x2)
| x2
| (x2 ,+ ∞)
| ![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018070901-30861.gif)
| -
| 0
| +
| 0
| -
| f (x)
| ![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018070902-84484.gif)
| 极小植
| ![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018070902-15257.gif)
| 极大值
| ![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018070902-61791.gif)
| 可见,f(x)的极大值点和极小值点各有一个. (2) 由(1)得![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018070902-72499.gif) 即![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018070903-66278.gif) 两式相加,得a(x1+x2)+2b=x -x . ∵x1+x2=- ,∴x -x =0, 即(x2+x1)(x2-x1)=0, 又x1<x2,∴x1+x2=0,从而b=0, ∴a(x2-1)=0,得x1=-1,x2=1, 由②得a=2. |