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If the solve the problem


$f(x)=\sin 2 x, 0


Given : $f(x)=\sin 2 x$

$\Rightarrow f^{\prime}(x)=2 \cos 2 x$

For a local maximum or a local minimum, we must have


$\Rightarrow 2 \cos 2 x=0$

$\Rightarrow \cos 2 x=0$

$\Rightarrow x=\frac{\pi}{4}$ or $\frac{3 \pi}{4}$

Since $f^{\prime}(x)$ changes from positive to negative when $x$ increases through $\frac{\pi}{4}, x=\frac{\pi}{4}$ is the point of maxima. The local maximum value of $f(x)$ at $x=\frac{\pi}{4}$ is given by

$\sin \left(\frac{\pi}{2}\right)=1$

Since $f^{\prime}(x)$ changes from negative to positive when $x$ increases through $\frac{3 \pi}{4}, x=\frac{3 \pi}{4}$ is the point of minima.

The local minimum value of $f(x)$ at $x=\frac{3 \pi}{4}$ is given by

$\sin \left(\frac{3 \pi}{2}\right)=-1$

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