Answer
f(x) is increasing on $(-\infty, -3) \cup (-1,+\infty)$ and decreasing on $(-3,-1)$ .
On $(-\infty, -3)$, f(x) achieves a maximum value of -8 at x=−3 and on $(-1,+\infty)$ a minimum value of -12 at x=-1.
Work Step by Step
$f(x) =x^{3}+6x^{2}+9x-8$
$f'(x)=3x^{2}+12x+9$
$f'(x)=0 \rightarrow 3x^{2}+12x+9=0 \rightarrow x=-3, x=-1$ Thus, f(x) is increasing on $(-\infty, -3) \cup (-1,+\infty)$ and decreasing on $(-3,-1)$
On $(-\infty, -3)$, f(x) achieves a maximum value of -8 at x=−3 and on $(-1,+\infty)$ a minimum value of -12 at x=-1.
