Finding critical numbers of f on the interval

Using the product rule, $f'(x) = 2x e^{-x} - x^2 e^{-x} = x e^{-x}(2 - x)$. This derivative is defined for all real $x$, so the only critical numbers come from setting $f'(x) = 0$.

Since $e^{-x} \neq 0$ for all $x$, we need $x(2-x) = 0$, giving $x = 0$ or $x = 2$. Both values lie within $[-1, 5]$, so the critical numbers are $x = 0$ and $x = 2$.

The candidates for absolute extrema are the critical numbers $x = 0, 2$ and the endpoints $x = -1, 5$. Compute $f$ at each: $f(-1) = (-1)^2 e^{1} = e \approx 2.718$, $f(0) = 0$, $f(2) = 4e^{-2} \approx 0.541$, and $f(5) = 25e^{-5} \approx 0.168$.

Comparing these four values, the largest is $f(-1) = e \approx 2.718$ and the smallest is $f(0) = 0$. Therefore the absolute maximum is $e$ at $x = -1$, and the absolute minimum is $0$ at $x = 0$.