题目内容

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.

解:(1)∵f(x)=xex,?

f′(x)=ex+xex=(x+1)ex.                                                                                           ?

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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