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How do you find the critical points for #F(x) =xe^-x#?

1 Answer

Nov 8, 2015

#f(x)# is differentiable everywhere, so the critical points will simply be the solution(s) to #f'(x)=0#

Explanation:

#f'(x)=e^-x(1-x)=0#

critical point at #1-x=0# or #x=1#

hope that helped

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