\(\frac{1}{2}, 3\)
\(\frac{1}{3}, -\frac{1}{2}\)
\(1, -\frac{1}{3}\)
\(1, \frac{1}{3}\)
Correct answer is C
At turning point, \(\frac{\mathrm d y}{\mathrm d x} = 0\).
Given \(x^{3} - x^{2} - x + 6 \)
\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x - 1 = 0 \)
\(3x^{2} - 3x + x - 1 = 0 \implies (3x + 1)(x - 1) = 0\)
\(x = \frac{-1}{3}, 1\)
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