题目

已知函数f(x)=xex(e为自然对数的底数).（1）求函数f(x)的单调递增区间；（2）求曲线y=f(x)在点(1,f(1))处的切线方程. 答案：解：（1）∵f(x)=xex, ∴f′(x)=ex+xex=(x+1)ex.                                                                                           令f′(x)＞0,即(x+1)ex＞0,∵ex＞0,∴x＞-1.∴函数f(x)的单调递增区间为(-1,+∞).                                                                      （2）由（1）得f′(1)=(1+1)·e=2e,f(1)=e,                                                           ∴曲线y=f(x)在点(1,f(1))处的切线方程为y-e=2e(x-1),即2ex-y-e=0.

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