Finding and classifying critical points

First compute the derivative: $f'(x) = 4x^3 - 12x^2 = 4x^2(x-3)$. Setting $f'(x) = 0$ gives $x = 0$ and $x = 3$, both of which lie in $(-1,4)$, so these are the critical numbers (note $f'$ is defined everywhere, so no other critical points arise from non-differentiability).

Testing signs of $f'(x) = 4x^2(x-3)$: for $x$ slightly less than 0 (e.g. $x=-0.5$), $f'(x) < 0$ (since $x^2>0$ and $x-3<0$); for $x$ slightly greater than 0 (e.g. $x=0.5$), $f'(x)$ is still negative since $x-3<0$. So $f'$ does not change sign at $x=0$, meaning $x=0$ is neither a max nor a min.

For $x$ slightly less than 3 (e.g. $x=2.5$), $f'(x) < 0$; for $x$ slightly greater than 3 (e.g. $x=3.5$), $f'(x) > 0$. Since $f'$ changes from negative to positive at $x=3$, this point is a relative minimum. Thus the critical numbers are $x=0$ (neither) and $x=3$ (relative minimum).