#### Answer

f(x) is increasing on $(-\infty, -3) \cup (2,+\infty)$ and decreasing on $(-3,2)$.
On $(-\infty, -3)$, f(x) achieves a maximum value of 101 at x=−3, and on $(2,+\infty)$ a minimum value of -24 at x=2

#### Work Step by Step

$f(x) =2x^{3}+3x^{2}-36x+20$
$f'(x)=6x^{2}+6x-36$
$f'(x)=0 \rightarrow 6x^{2}+6x-36 =0 \rightarrow x=2, x=-3$
Thus, f(x) is increasing on $(-\infty, -3) \cup (2,+\infty)$ and decreasing on $(-3,2)$
On $(-\infty, -3)$, f(x) achieves a maximum value of 101 at x=−3 and on $(2,+\infty)$ a minimum value of -24 at x=2.