What is the local maximum value of $f(x) = x^{2}e^{-x}$?

Explanation

1. 01

   $f'(x) = 2xe^{-x} - x^{2}e^{-x} = x(2 - x)e^{-x}$, zero at $x = 0$ and $x = 2$.

2. 02

   $f'$ goes from positive to negative at 2, so the local maximum value is $f(2) = 4e^{-2} = \frac{4}{e^{2}}$.