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

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BeeFree Share
Nov 8, 2015

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

#### Explanation:

$f ' \left(x\right) = {e}^{-} x \left(1 - x\right) = 0$

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

hope that helped

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