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

Explanation

1. 01

   $f'(x) = e^{-x} - xe^{-x} = (1 - x)e^{-x}$, which is 0 at $x = 1$.

2. 02

   $f''(x) = (x - 2)e^{-x}$, and $f''(1) = -\frac{1}{e} < 0$: a maximum, with value $f(1) = \frac{1}{e}$.