If $f(x) = x^3 - 3x$, the values of $x$ where $f'(x) = 0$ are:
A$x = 0$ only
B$x = 1, 0$
C$x = \pm \sqrt 3$
D$x = \pm 1$
Answer & 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 $x = 1$).
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