Calculus: Early Transcendentals (2nd Edition)

(a) $x = −1/2$ is the only critical point. (b) $f$ has a local maximum of $f(−1/2) = 9/4$ at $x = −1/2$. (c) The absolute maximum is $9/4$ and the absolute minimum is $-18$.
(a). $f'(x) = −2x − 1$, which exists everywhere and is zero only for $x = −1/2$, so that is the only critical point. (b). Note that $f'(−2) = 3 > 0$ and $f'(0) = −1 < 0$, so $f$ has a local maximum of $f(−1/2) = 9/4$ at $x = −1/2$. (c). Note that $f(−4) = −10$ and $f(4) = −18$, so the absolute maximum is $9/4$ at $x = −1/2$ and the absolute minimum is $-18$ at $x = 4$.