The interval in which the function $x^{3}$ increases less rapidly than $6 x^{2} + 15 x + 15$ is

(- \infty, - 1)

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(- 2, - 1)

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(- 1, 5)

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(5, \infty)

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Solution

The correct option is C $(- 1, 5)$
Given,

$\frac{d}{d x} (x^{3}) < \frac{d}{d x} (6 x^{2} + 15 x + 15)$

$3 x^{2} < 12 x + 15$

$3 x^{2} - 12 x - 15 < 0$

$x^{2} - 4 x - 5 < 0$

$(x + 1) (x - 5) < 0$

$x \in (- 1, 5)$

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Similar questions

f (x) = x^{3}

g (x) = 6 x^{2} + 15 x + 5

f (x) = 9 x^{2} - 15 x - x^{3} + 10

(1, 5)

f (x) = 9 x^{2} - 15 x - x^{3} + 10

(1, 5)

f (x) = x^{3} - 6 x^{2} + 9 x + 15

f (x) = x^{3} - 6 x^{2} + 9 x + 10

f (x) = x^{3} - 6 x^{2} - 36 x + 2

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