Find the interval(s) on which $f(x) = 2 x^{3} + 3 x^{2} - 36 x + 10$ is strictly increasing.

Explanation

1. 01

   $f'(x) = 6 x^{2} + 6 x - 36 = 6 \left(x - 2\right) \left(x + 3\right)$, which is 0 at $x = -3$ and $x = 2$.

2. 02

   $f'(x) > 0$ on $(-\infty, -3) \cup (2, \infty)$ and $f'(x) < 0$ on $(-3, 2)$.

3. 03

   So $f$ is strictly increasing on $(-\infty, -3) \cup (2, \infty)$.