#### Answer

(a) $x = −2$ and $x = 1$ are critical points.
(b) $f$ has a local minimum at $x = 1$ of $f(1) = −6$.
(c) The absolute maximum of $f$ on $[−2, 4]$ is $129$ and the absolute minimum is $−6$.

#### Work Step by Step

(a). $f'(x) = 6x^2 + 6x − 12 = 6(x + 2)(x − 1)$, which exists everywhere and is $0$ at $x = −2$ and $x = 1$, so those are the critical points.
(b). Note that $f'(−1.5) < 0$ and $f'(2) > 0$, so $f$ has a local minimum at $x = 1$ of $f(1) = −6$.
(c). Note that $f(−2) = 21$ and $f(4) = 129$, so the absolute maximum of $f$ on $[−2, 4]$ is $129$ and the absolute minimum is $−6$.