## Calculus: Early Transcendentals (2nd Edition)

(a) $x=0$ and $x = 8/5$ are two critical points. (b) $f$ has a local maximum at $x = 0$ of $f(0) = 0$ and a local minimum at $x = 8/5$ of $\frac{-48}{{5·5}^{2/3}} \approx −3.28$. (c) The absolute maximum of $f$ on $[−5, 5]$ is $\sqrt[3]{25}$ and the absolute minimum is $-9\sqrt[3]{25}$.
(a). $f'(x) = x^{2/3} + (x − 4) · \frac{2}{3}x^{−1/3} = \frac{5x−8}{3x^{1/3}}$ , which is undefined at $x = 0$ and is $0$ at $x = 8/5$. So these are the two critical points. (b). Note that $f(−1) > 0$ and $f(1) < 0$, and $f(2) > 0$ so $f$ has a local maximum at $x = 0$ of $f(0) = 0$ and a local minimum at $x = 8/5$ of $\frac{-48}{{5·5}^{2/3}} \approx −3.28$. (c). Note that $f(−5) = −9 \sqrt[3]{25} ≈ −26.32$, $f(0) = 0$, and $f(5) = \sqrt[3]{25} ≈ 2.92$, so the absolute maximum of $f$ on $[−5, 5]$ is $\sqrt[3]{25}$ and the absolute minimum is $-9\sqrt[3]{25}$.