#### Answer

f(x) is increasing on $(-\infty, -2) \cup (1,+\infty)$ and decreasing on $(-2,1)$.
On $(-\infty, -2)$, f(x) achieves a maximum value of 25 at x=−2 and on $(1,+\infty)$ a minimum value of -2 at x=1.

#### Work Step by Step

$f(x) =2x^{3}+3x^{2}-12x+5$
$f'(x)=6x^{2}+6x-12$
$f'(x)=0 \rightarrow 6x^{2}+6x-12 =0 \rightarrow x=1, x=-2$
Thus, f(x) is increasing on $(-\infty, -2) \cup (1,+\infty)$ and decreasing on $(-2,1)$
On $(-\infty, -2)$, f(x) achieves a maximum value of 25 at x=−2 and on $(1,+\infty)$ a minimum value of -2 at x=1