Intervals of increase/decrease and classification of local extrema

First differentiate: $f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x-2)(x+1)$. Setting $f'(x) = 0$ gives critical numbers $x = -1$ and $x = 2$, and $f'$ is defined everywhere else, so these are the only critical numbers.

These two points split the real line into three intervals: $(-\infty, -1)$, $(-1, 2)$, and $(2, \infty)$. Testing a point in each: at $x=-2$, $f'(-2) = 6(-4)(-1) = 24 > 0$; at $x=0$, $f'(0) = 6(-2)(1) = -12 < 0$; at $x=3$, $f'(3) = 6(1)(4) = 24 > 0$.

So $f$ is increasing on $(-\infty, -1)$ and $(2, \infty)$, and decreasing on $(-1, 2)$. Since $f'$ changes from positive to negative at $x=-1$, that point is a local maximum; since $f'$ changes from negative to positive at $x=2$, that point is a local minimum.