The values for which $f (x) = x^{3} - 6 x^{2} - 36 x + 7$ increases with $x$ are.

x < - 2 (o r) x > 6

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x > - 2 (o r) x < 6

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x = - 2 (o r) x = 6

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x = 0

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Solution

The correct option is A $x < - 2 (o r) x > 6$
$f^{'} (x) = 3 x^{2} - 12 x - 36$
$= 3 (x^{2} - 4 x - 12)$
$= 3 (x - 6) (x + 2)$
$f^{'} (x) > 0$ when $x > 6 \cup x < - 2$

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