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The function $f(x) = x^3 - 3x + 1$ has critical points at:
A$x = 0, 1$
B$x = \pm 1$
C$x = \pm \sqrt 3$
DNone
Answer & Solution
Correct answer: B. $x = \pm 1$
$f'(x) = 3x^2 - 3 = 0$ ⇒ $x = \pm 1$. At $x = 1$ (local min, $f''(1) = 6 > 0$), at $x = -1$ (local max, $f''(-1) = -6 < 0$).
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