Find the maximum and minimum values of the function $f (x) = x + s i n 2 x$

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Solution

$f (x) = x + sin 2 x f^{'} (x) = 1 + cos 2 x$

For stationary points $f^{'} (x) = 0$
$\Rightarrow 1 + 2 cos 2 x = 0 cos 2 x = - \frac{1}{2} 2 x = \frac{2 π}{3} o r 2 x = \frac{4 π}{3} \Rightarrow x = \frac{π}{3} o r x = \frac{2 π}{3}$

Now,
$f (x) = x + sin x f (\frac{π}{3}) = \frac{π}{3} + sin \frac{2 π}{3} = \frac{π}{3} + \frac{\sqrt{3}}{2} f (\frac{2 π}{3}) = \frac{2 π}{3} + sin \frac{4 π}{3} = \frac{2 π}{3} - \frac{\sqrt{3}}{2} f (2 π) = 2 π + sin 4 π = 2 π + 0 = 2 π$

Of these values the maximum value is $2 π$ and minimum value is $0$
The maximum value of $f (x)$ is $2 π$ and minimum value is $0$

Hence, this is the answer.

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