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The values of x for which $x^{3} - 6 x^{2} - 36 x + 7$ increases are

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$\emptyset$

$(- 2, 6)$

$(- \infty, - 2) \cup (6, \infty)$

answer is D.

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Detailed Solution

$y = x^{3} - 6 x^{2} - 36 x + 7$

$\Rightarrow \frac{d y}{d x} = 3 x^{2} - 12 x - 36 > 0$

$3 x - 12 x - 36 > 0 \Rightarrow 3 (x^{2} - 4 x - 12) > 0$

$\Rightarrow (x + 2) (x - 6) > 0 \Rightarrow x < - 2 (o r) 6 < x$

The range of values of x is

$(- \infty, - 2) \cup (6, \infty)$

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