## Calculus: Early Transcendentals (2nd Edition)

The local maximum of $1/e$ at $x = 1$ is an absolute maximum. There is no absolute minimum, since the function is unbounded in the negative direction as $x→−∞$.
$f'(x) = −xe^{−x} +e^{−x} = e^{−x}(1−x)$, which is $0$ only for $x = 1$. Note that $f$ is continuous on $(−∞,∞)$ and contains only one critical point. Note that $f' > 0$ for $x < 1$ and $f' < 0$ for $x > 1$. So there is a local maximum of $f(1) = 1/e$ at $x = 1$. The local maximum of $1/e$ at $x = 1$ is an absolute maximum. There is no absolute minimum, since the function is unbounded in the negative direction as $x→−∞$.