Answer
(a). $f'(x) = e^{−x/2} + xe^{−x/2} · \frac{−1}{2}$. Since exponential function is never zero, this expression is zero only when $\frac{x}{2} = 1$, or $x = 2$. So $(2, \frac{2}{e})$ is the only critical point.
(b). We have $f(0) = 0$ and $f(2) = \frac{2}{e}
≈ .7358$, and $f(5) ≈ .410$. So the absolute maximum of $f$ on this interval is $\frac{2}{e}$ and the absolute minimum is $0$.
Work Step by Step
See the graph for explanation.
