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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 $I$

$f(x)=x+\cos x \;in\;[0,\pi]$

Step 1:

$f'(x)=1-\sin x$

Step 2:

Since $-1 \leq \sin x \leq 1$

$1-\sin x \geq 0 =>f'(x) \leq 0$

$\therefore $ the function is increasing in $[0, \pi]$

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