Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 259: 96

$-3$ and $2$ are critical points. There is a local maximum at $t = −3$ and a local minimum at $t = 2$.

$p'(t) = 6t^2 + 6t − 36 = 6(t + 3)(t − 2)$, which is $0$ at $t = −3$ and $t = 2$. Note that $p''(t) = 12t + 6$, so $p''(−3) = −30 < 0$ and $p''(2) = 30 > 0$, so there is a local maximum at $t = −3$ and a local minimum at $t = 2$.