Find the maximum and minimum values of $x + \sin 2x$ on $[0,2\pi ]$.

Let $f(x) = x + \sin 2x,0 \le x \le 2\pi$
$\Rightarrow f'(x) = 1 + 2\cos 2x,0 < x < 2\pi$ .

For critical points, $f'(x) = 0$
$\Rightarrow 1 + 2\cos 2x = 0 \Rightarrow \cos 2x = - \cfrac{1}{2}$
$\Rightarrow \cos 2x = - \cos \cfrac{\pi }{3}$ $\left[ {{\rm{If}}\,\,0 < x < 2\pi } \right.$, then $\left. {0 < 2x < 4\pi } \right]$

$\Rightarrow \cos 2x = \cos \left( {\pi - \cfrac{\pi }{3}} \right),\cos \left( {\pi + \cfrac{\pi }{3}} \right),\cos \left( {3\pi - \cfrac{\pi }{3}} \right),\;\cos \left( {3\pi + \cfrac{\pi }{3}} \right)$
$\Rightarrow 2x = \cfrac{{2\pi }}{3},\cfrac{{4\pi }}{3},\cfrac{{8\pi }}{3},\cfrac{{10\pi }}{3} \Rightarrow x = \cfrac{\pi }{3},\cfrac{{2\pi }}{3},\cfrac{{4\pi }}{3},\cfrac{{5\pi }}{3}$

To find the maximum and minimum values, we evaluate$f(x)$ at $0,2\pi ,\cfrac{\pi }{3},\cfrac{{2\pi }}{3},\cfrac{{4\pi }}{3},\cfrac{{5\pi }}{3}$.

Now, $f(0) = 0 + \sin 0 = 0$
$f(2\pi ) = 2\pi + \sin 4\pi = 2\pi + 0 = 2\pi$
$f\left( {\cfrac{\pi }{3}} \right) = \cfrac{\pi }{3} + \sin \cfrac{{2\pi }}{3} = \cfrac{\pi }{3} + \sin \left( {\pi - \cfrac{\pi }{3}} \right) = \cfrac{\pi }{3} + \sin \cfrac{\pi }{3} = \cfrac{\pi }{3} + \cfrac{{\sqrt 3 }}{2}$

$f\left( {\cfrac{{2\pi }}{3}} \right) = \cfrac{{2\pi }}{3} + \sin \cfrac{{4\pi }}{3} = \cfrac{{2\pi }}{3} + \sin \left( {\pi + \cfrac{\pi }{3}} \right)$
$= \cfrac{{2\pi }}{3} - \sin \cfrac{\pi }{3} = \cfrac{{2\pi }}{3} - \cfrac{{\sqrt 3 }}{2}$

$f\left( {\cfrac{{4\pi }}{3}} \right) = \cfrac{{4\pi }}{3} + \sin \cfrac{{8\pi }}{3} = \cfrac{{4\pi }}{3} + \sin \left( {2\pi + \cfrac{{2\pi }}{3}} \right)$
$= \cfrac{{4\pi }}{3} + \sin \cfrac{{2\pi }}{3} = \cfrac{{4\pi }}{3} + \cfrac{{\sqrt 3 }}{2}$
and $f\left( {\cfrac{{5\pi }}{3}} \right) = \cfrac{{5\pi }}{3} + \sin \cfrac{{10\pi }}{3} = \cfrac{{5\pi }}{3} + \sin \left( {3\pi + \cfrac{\pi }{3}} \right)$

$= \cfrac{{5\pi }}{3} - \sin \cfrac{\pi }{3} = \cfrac{{5\pi }}{3} - \cfrac{{\sqrt 3 }}{2}$

Thus, maximum value of $f(x)$ is $2\pi$ at $x = 2\pi$ and minimum value of $f\left( x \right)$ is $0$ at $x = 0$.