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