A function has $f'(x) = (x - 1)(x - 3)^2$. Where does it have local extrema?

Explanation

1. 01

   $(x - 3)^2 \ge 0$, so the sign of $f'$ is the sign of $x - 1$: negative before 1, positive after.

2. 02

   So $x = 1$ is a local minimum. At $x = 3$, $f'$ is positive on both sides: no extremum there.