函数f(x)=(x-3)ex的单调递增区间是____.

函数f(x)=(x-3)ex的单调递增区间是____.
【正确答案】:(2,+∞)。f′(x)=(x-3)′ex+(x-3)(ex)′=(x-2)ex,令f′(x)﹥0,解得x﹥2.