# Find the local maximum and local minimum value of the function. $f(x)=\sin x+\frac{1}{2}\cos 2x\;in\;[0,\pi/2]$

Toolbox:
• For local maximum or minimum $f '(x)=0$
• For maximum $f "(x)$ is negative and for minimum $f "(x)$ is positive.
Step 1
$f'(x)=cos\: x-sin2x$
$f'(x)=0$ at $x = \large\frac{\pi}{6}, \large\frac{\pi}{2}$
Step 2
$f"(x)= -sin\: x - 2cos\:2x$
is negative at $x = \large\frac{\pi}{6}$ and is positive at $\large\frac{\pi}{2}$
Step 3
At $x = \large\frac{\pi}{6}\: f(x)$ has got maxima and maximum value is $f\bigg( \large\frac{\pi}{6} \bigg)=\large\frac{3}{4}$
At $x =\large\frac{\pi}{2}\: f(x)$ has got minima and minimum value is $f \bigg( \large\frac{\pi}{2} \bigg) = \large\frac{1}{2}$