## Calculus (3rd Edition)

$f(x)$ has local minima at $x= 2$ and local maxima at $x= -2$
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 $x= -2$.