## Calculus: Early Transcendentals (2nd Edition)

(a) $x = 0$ is the only critical point. (b) $f$ has a local minimum of $f(0) = 3$ at $x = 0$. (c) The absolute maximum is $12$ and the absolute minimum is $3$.
(a). $f'(x) = 2x$, so $x = 0$ is the only critical point. (b). Note that $f' < 0$ for $x < 0$ and $f' > 0$ for $x > 0$, so $f$ has a local minimum of $f(0) = 3$ at $x = 0$. (c). Note that $f(−3) = 12$, $f(0) = 3$ and $f(2) = 7$, so the absolute maximum is $12$ and the absolute minimum is $3$.