Answer
(a) $x= \frac{5}{e}$ is the only critical point.
(b) The absolute maximum of $f$ is $0$, and the absolute minimum is $\frac{−5}{e} ≈ −1.8394$.
Work Step by Step
(a) $f'(x) = \ln(\frac{x}{5}) + x · \frac{5}{x}·\frac{1}{5} = \ln(\frac{x}{5}) + 1$. This expression is zero when $\ln(\frac{x}{5}) = −1$ or $ x =
5e^{−1} = \frac{5}{e}
≈ 1.8394$.
(b). $f(.1) ≈ −.391$, $f(\frac{5}{e})=\frac{−5}{e} ≈ −1.8394$, and $f(5) = 0$. So the absolute maximum of $f$ is $0$, and the absolute minimum is $\frac{−5}{e} ≈ −1.8394$.
