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Standard Deviation

The minimum o...

Question

The minimum of $2 x^{3} - 3 x^{2} - 12 x + 8$ occurs at

- 1

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2

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\sqrt{6}

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- \sqrt{6}

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Solution

The correct option is D $2$
$f^{'} (x) = 6 x^{2} - 6 x - 12$
$= 6 (x^{2} - x - 2)$
$f^{'} (x) = 6 (x - 2) (x + 1)$
$f^{'} (x) < 0$ If $- 1 < x < 2$
$f^{'} (x) > 0$ If $x ϵ (- \infty, - 1) \cup (2, \infty)$
So mini occur at $x = 2$
$f (x) = 16 - 12 - 24 + 8$
$= - 12$

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