The maximum value of f ( x ) = x + 2x, x ∈ [ 0, 2π ] is

Given, $f\left(x\right) = x + \sin 2x$

$⇒ f^{'} ( x ) = 1 + 2\cos 2x$

For maximum or minimum value, $f^{'}\left(x\right) = 0$

$⇒ 1 + 2\cos 2x = 0 ⇒ \cos 2x = - \dfrac{1}{2}$

$⇒ 2x = 2n\pi \pm \dfrac{2\pi}{3}$

$⇒ x = n\pi \pm \dfrac{\pi}{3}$

$⇒ x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}$

Find $f\left(0\right), f\left(\dfrac{\pi}{3}\right), f\left(\dfrac{2\pi}{3}\right), f\left(\dfrac{4\pi}{3}\right), f\left(\dfrac{5\pi}{3}\right), f ( 2\pi )$

$⇒ f\left(0\right) = 0, f\left(\dfrac{\pi}{3}\right) = \dfrac{\pi}{3} + \dfrac{3}{2}, f\left(\dfrac{2\pi}{3}\right) = \dfrac{2\pi}{3} - 0. 8$

$⇒ f\left(\dfrac{4\pi}{3}\right) = \dfrac{4\pi}{3} + 0. 8, f\left(\dfrac{5\pi}{3}\right) = \dfrac{5\pi}{3} - 0. 8$

$⇒ f\left(2\pi\right) = 2\pi + 0 = 2\pi$

$∴$ Maximum value of $f\left(x\right) = 2\pi$