A function has $f'(x) = \left(x - 3\right)^{2} \left(x - 1\right)$. Find the interval(s) on which $f$ is strictly increasing.

Explanation

1. 01

   $\left(x - 3\right)^{2} \ge 0$ always, so the sign of $f'$ is the sign of $x - 1$.

2. 02

   $f' > 0$ for $x > 1$ except at $x = 3$, where it is 0 for an instant; a single such point does not stop $f$ from strictly increasing.

3. 03

   So $f$ is strictly increasing on $(1, \infty)$.