$f (x) = x^{5} + 2 x^{3} + x$ is increasing in

(0, \infty)

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(- \infty, 0)

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(- \infty, \infty)

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\emptyset

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Solution

The correct option is B $(- \infty, \infty)$
$f^{'} (x) = 5 x^{4} + 6 x^{2}$
$f^{'} (2) > 0$ $\forall x \in R$

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There is a solution to $x^{5} - 2 x^{3} - 2 = 0$ between x=0 and x=2

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