已知函数f(x)=ex﹣a(x2+x+1).
(1)当a=1时,证明:f(x)+x2≥0;
(2)当a
时,判断函数f(x)的单调性;
(3)若函数f(x)有三个零点,求实数a的取值范围.
(1)当a=1时,证明:f(x)+x2≥0;
(2)当a
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/52c70c0c5a061195b9941796b6a9acc4.png)
(3)若函数f(x)有三个零点,求实数a的取值范围.
更新时间:2020-03-17 11:53:11
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