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The critical points of f(x) = x³ − 3x + 1 are found by solving f'(x) = 0:
A$x = ±1$ (from $3x^2 - 3 = 0$)
B$x = 0$ only (single critical)
C$x = ±2$ (incorrect calculation)
D$x = ±3$ (mistaken derivation)
Answer & Solution
Correct answer: A. $x = ±1$ (from $3x^2 - 3 = 0$)
f'(x) = 3x² − 3 = 0 → x² = 1 → x = ±1. Critical points at x = 1 (local min, f''(1) = 6 > 0) + x = -1 (local max, f''(-1) = -6 < 0).
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