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- (i) If $f'$ is positive on an open interval $I$. Then $f$ is strictly increasing on $I$
- (ii) If $f'$ is negative on an open interval $I$, then $f$ is strictly decreasing on $I$

$f(x)= 20-x-x^2$

Step 1:

$f'(x) =-1-2x$

$f'(x)=0=>-1-2x=0=>x =\large\frac{-1}{2}$

Step 2:

Interval $(-\infty, -\large\frac{1}{2}]$

$f'(x)=+$

$f(x)$ inc/dec in interval=$f(x)$ is increasing in $(-\infty, \large\frac{1}{2}]$

Interval $[\large\frac{1}{2}$$,\infty)$

$f'(x)=-$

$f(x)$ inc/dec in interval=$f(x)$ is decreasing in $[\large\frac{1}{2},\infty)$

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