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$2 x^{3} + 18 x^{2} - 96 x + 45$ is an increasing function:

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$x \leq - 8, x \geq 2$

$x < - 2, x \geq 8$

$x \leq - 2, x \geq 8$

$0 < x \leq - 2$

answer is A.

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Detailed Solution

$f^{'} (x) = 6 x^{2} + 36 x - 96 > 0,$ for increasing

$\Rightarrow f^{'} (x) = 6 (x + 8) (x - 2) \geq 0 \Rightarrow x \geq 2, x \leq - 8 .$

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