If $f(x) = x^3 - 3x$, the values of $x$ where $f'(x) = 0$ are:
- $x = 0$ only
- $x = 1, 0$
- $x = \pm \sqrt 3$
- $x = \pm 1$
Show answer and solution
Correct answer: D. $x = \pm 1$
$f'(x) = 3x^2 - 3 = 3(x^2 - 1)$.
Setting = 0: $x = \pm 1$.
These are critical points (local max at $x = -1$, local min at $x = 1$).