The interval on which the function $f (x) = 2 x^{3} + 9 x^{2} + 12 x - 1$ is decreasing is?

[- 1, \infty)

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[- 2, - 1]

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(- \infty, - 2]

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[- 1, 1]

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Solution

The correct option is B $[- 2, - 1]$
$f (x) = 2 x^{3} + 9 x^{2} + 12 x - 1$

$f^{'} (x) = 6 x^{2} + 18 x + 12$

$= 6 (x^{2} + 3 x + 2)$

$= 6 (x + 2) (x + 1)$

$∴ f (x)$ decreases in interval $x t [- 2, - 1]$

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Similar questions

The interval on which the function f(x) = 2x³ + 9x² + 12x - 1 is decreasing is
(a) [-1, ∞) (b) [-2, -1] (c) (∞, -2] (d) [-1, 1]

f (x) = - 2 x^{3} - 9 x^{2} - 12 x + 1

2 x^{3} - 9 x^{2} + 12 x + 4

f (x) = - 2 x^{3} - 9 x^{2} - 12 x + 1

f (x)

f (x) = - 2 x^{3} - 9 x^{2} - 12 x + 1

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