Solution: ∫ [f(x) e x]' = f' (x) e x+f (x) e x = e x [(f' (x)+f (x)], x=- 1 is f (x).

∴[f(- 1)e^(- 1)]'=e^(- 1)[(f'(- 1)+f(- 1)]=0

Then f'(- 1)+f(- 1)=0.

In A and B, f'(- 1)=f(- 1)=0, which is consistent.

In c language, f'(- 1)>0, f (-1) < 0, f'(- 1)+f(- 1)=0 is possible.

In d, f'(- 1)>0, f(- 1)>0, then f'(- 1)+f(- 1)>0, which does not match.

Therefore, choose D in this question.