Find the interval in which the following function is strictly incerasing or decreasing

$- 2 x^{3} - 9 x^{2} - 12 x + 1$

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Solution

Given, $f (x) = - 2 x^{3} - 9 x^{2} - 12 x + 1$
$\Rightarrow f^{'} (x) = - 2.3 x^{2} - 92 x - 12 = - 6 x^{2} - 18 x - 12$
On putting f'(x)=0, we get $- 6 x^{2} - 18 x - 12 = 0$
$\Rightarrow - 6 (x + 2) (x + 1) = 0 \Rightarrow x = - 2, - 1$
which divides real line into three intervals $(\infty, - 2), (- 2, - 1)$ and $(- 1, \infty)$

$\begin{matrix} I n t e r v a l s & S i g n o f f^{'} (x) & N a t u r e o f f (x) (- \infty - 2) & (-) (-) (-) = - v e & S t r i c t l y d e c r e a s i n g (- 2, - 1) & (-) (-) (+) = + v e & S t r i c t l y i n c r e a s i n g (- 1, \infty) & (-) (+) (+) = - v e & S t r i c t l y d e c r e a s i n g \end{matrix}$
Therefore, f(x) is strictly increasing in $- 2 < x < - 1$ and strictly decreasing for $x < - 2$ and $x > - 1$ .

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