Answer
$f(x)$ has local minima at $x= 2$ and local maxima at $x= -2$
Work Step by Step
Given $$f(x)=x^{3}-12 x-4$$
Since
$$f'(x) = 3x^2-12 $$
Then $f(x)$ has critical points when
\begin{align*}
f'(x)&=0\\
3x^2-12 &=0\\
x^2-4&=0
\end{align*}
Then $x= -2$ and $x=2$ are critical points. To find the intervals where the function is increasing and decreasing, choose $x=-3,\ x= 0$ and $x=3$
\begin{align*}
f'( -3)&>0 \\
f'(0)& <0\\
f'(3)&>0
\end{align*}
Hence, by using the first derivative test, $f(x)$ has a local minima at $x= 2$ and a local maxima at $x= -2$.
