Answer
$f(x)$ is decreasing on $(-\infty,-2) \cup (-1,1) $ and increasing on $(-2,-1)\cup ( 1,\infty)$.
$f(x)$ has a local maximum at $x= -1$ and a local minimum at $x=-2,\ x=-1$.
Work Step by Step
Given $$y=3 x^{4}+8 x^{3}-6 x^{2}-24 x$$
Since
$$f'(x) =12 x^{3}+24x^{2}-12x-24 $$
Then $f(x)$ has critical points when
\begin{align*}
f'(x)&=0\\
12 x^{3}+24x^{2}-12x-24&=0\\
x^{3}+12x^{2}-x-2&=0\\
(x+2)(x-1)(x+1)&=0
\end{align*}
Then $x= -2,\ x= -1$ and $x= 1$ are critical points. To find the interval where $f$ is increasing and decreasing, choose $x=-3$, $x= -1.5,\ x=0$ and $x= 2$
\begin{align*}
f'(-3)&<0\\
f'(-1.5)&>0\\
f'(0)&<0\\
f'( 2)&>0
\end{align*}
Hence, $f(x) $ is decreasing on $(-\infty,-2) \cup (-1,1) $ and increasing on $(-2,-1)\cup ( 1,\infty) $. Hence, by using the first derivative test, $f(x)$ has a local maximum at $x= -1$ and a local minimum at $x=-2,\ x=-1$.
