Find critical numbers of f

Differentiate: $f'(x) = 4x^3 - 4x = 4x(x^2-1) = 4x(x-1)(x+1)$. This is a polynomial, so it is defined everywhere, meaning the only critical numbers come from setting $f'(x) = 0$.

Solving $4x(x-1)(x+1) = 0$ gives $x = 0, 1, -1$. These are the three critical numbers of $f$.

Starting from $f'(x) = 4x^3 - 4x$, differentiate term by term: the derivative of $4x^3$ is $12x^2$, and the derivative of $-4x$ is $-4$.

Therefore $f''(x) = 12x^2 - 4$.

Evaluate $f''$ at each critical number: $f''(-1) = 12(1) - 4 = 8 > 0$, $f''(0) = 12(0) - 4 = -4 < 0$, and $f''(1) = 12(1) - 4 = 8 > 0$.

Since $f''(-1) > 0$, $x=-1$ is a local minimum; since $f''(0) < 0$, $x=0$ is a local maximum; since $f''(1) > 0$, $x=1$ is a local minimum.