$f (x) = x^{3} - 27 x + 5$ is strictly increasing when

x < - 3

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| x | > 3

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x \leq - 3

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| x | < 3

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Solution

The correct option is B $| x | > 3$
$f^{'} (x) = 3 x^{2} - 27$
$f^{'} (x) = 3 (x^{2} - 9)$
$0 = x^{2} - 9$
$x = \pm 3$
$f^{'} (x) > 0$ When $∣ x ∣> 3$

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