Find the interval in which the following functions are strictly incerasing or decreasing

$10 - 6 x - 2 x^{2}$

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Solution

Let $f (x) = 10 - 6 x - 2 x^{2} \Rightarrow f^{'} (x) = 0 - 6 - 2 (2 x) = - 6 - 4 x$
On putting f'(x)=0, we get $- 6 - 4 x = 0 \Rightarrow x = \frac{- 3}{2}$
Which divides real line into two intervals

namely $(- \infty, - \frac{3}{2})$ and $(\frac{- 3}{2}, \infty)$
$\begin{matrix} I n t e r v a l s & S i g n o f f^{'} (x) & N a t u r e o f f (x) (- \infty, - \frac{3}{2}) & + v e & S t r i c t l y i n c r e a s i n g (\frac{- 3}{2}, \infty) & - v e & S t r i c t l y d e c e r e s i n g \end{matrix}$
Hence, f is strictly increasing for $x < - \frac{3}{2}$ and strictly decreasing for $x > - \frac{3}{2}$

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