Answer
f(x) is increasing on $(-\infty, -4) \cup (2,+\infty)$ and decreasing on $(-4,2)$.
On $(-\infty, -4)$, f(x) achieves a maximum value of 82 at x=−4 and on $(2,+\infty)$ a minimum value of -26 at x=2.
Work Step by Step
$f(x) =x^{3}+3x^{2}-24x+2$
$f'(x)=3x^{2}+6x-24$
$f'(x)=0 \rightarrow 3x^{2}+6x-24=0 \rightarrow x=2, x=-4$
Thus, f(x) is increasing on $(-\infty, -4) \cup (2,+\infty)$ and decreasing on $(-4,2)$
On $(-\infty, -4)$, f(x) achieves a maximum value of 82 at x=−4 and on $(2,+\infty)$ a minimum value of -26 at x=2.
