Find critical numbers

Differentiate: $f'(x) = 4x^3 - 16x = 4x(x^2 - 4)$. This is a polynomial, so it is defined for all $x$, meaning the only critical numbers come from setting $f'(x) = 0$.

Solve $4x(x^2-4) = 0$, giving $x = 0, x = 2, x = -2$. All three values lie within $(-3,3)$, so these are the critical numbers.

The sign of $f'(x) = 4x(x-2)(x+2)$ changes as follows: negative on $(-3,-2)$, positive on $(-2,0)$, negative on $(0,2)$, positive on $(2,3)$. So $f'$ goes from negative to positive at $x=-2$ (local min), positive to negative at $x=0$ (local max), and negative to positive at $x=2$ (local min).

Evaluate: $f(-2) = 16 - 32 + 3 = -13$, $f(0) = 3$, $f(2) = 16 - 32 + 3 = -13$. So there are relative minima at $x=-2$ and $x=2$ with value $-13$, and a relative maximum at $x=0$ with value $3$.

The candidates are the critical numbers $x = -2, 0, 2$ and the endpoints $x = -3, 3$. Compute $f(-3) = 81 - 72 + 3 = 12$, $f(-2) = -13$, $f(0) = 3$, $f(2) = -13$, $f(3) = 81 - 72 + 3 = 12$.

Comparing all five values $\{12, -13, 3, -13, 12\}$, the largest is $12$ (attained at $x = \pm 3$) and the smallest is $-13$ (attained at $x = \pm 2$). Thus these are the absolute maximum and minimum on $[-3,3]$.