## Calculus (3rd Edition)

$f(x)$ is increasing on $(-\infty,0) \cup (8,\infty)$ and decreasing on $(0,8)$ $f(x)$ has a local maximum at $x= 0$ and a local minimum at $x=8$
Given $$y=x^{3}-12 x^{2}$$ Since $$f'(x) =3x^{2}-24 x$$ Then $f(x)$ has critical points when \begin{align*} f'(x)&=0\\ 3x^{2}-24 x&=0\\ 3x(x-8)&=0 \end{align*} Then $x= 0$ and $x= 8$ are critical points. To find the interval where $f$ is increasing and decreasing, choose $x=-1$, $x= 1$ and $x= 9$ \begin{align*} f'(-1)&= 27>0\\ f'(1)&=-21<0\\ f'(-1)&= 27>0 \end{align*} Hence, $f(x)$ is increasing on $(-\infty,0) \cup (8,\infty)$ and decreasing on $(0,8)$. Hence, by using the first derivative test, $f(x)$ has a local maximum at $x= 0$ and a local minimum at $x=8$.