Finding critical numbers

Differentiate: $f'(x) = 4x^3 - 12x^2$. This is a polynomial, so it is defined everywhere, meaning the only critical numbers come from solving $f'(x) = 0$.

Factor: $f'(x) = 4x^2(x - 3)$. Setting this equal to zero gives $x = 0$ and $x = 3$ as the critical numbers.

The critical numbers $x=0$ and $x=3$ split the real line into three intervals: $(-\infty, 0)$, $(0,3)$, and $(3,\infty)$. Since $4x^2 \geq 0$ always, the sign of $f'(x)=4x^2(x-3)$ is determined entirely by the sign of $(x-3)$.

For $x<0$: $4x^2>0$ and $(x-3)<0$, so $f'(x)<0$. For $0<x<3$: same reasoning gives $f'(x)<0$. For $x>3$: $f'(x)>0$. Thus $f$ is decreasing on $(-\infty,0)\cup(0,3)$, which combines to decreasing on $(-\infty,3)$, and increasing on $(3,\infty)$.

At $x=0$: $f'(x)<0$ for $x$ slightly less than 0 and $f'(x)<0$ for $x$ slightly greater than 0 (since $f'$ is negative on both $(-\infty,0)$ and $(0,3)$). There is no sign change, so $x=0$ is neither a local max nor a local min.

At $x=3$: $f'(x)<0$ for $x$ slightly less than 3 and $f'(x)>0$ for $x$ slightly greater than 3, so $f'$ changes from negative to positive. By the First Derivative Test, $x=3$ is a local minimum.