# Find the maximum and minimum values of x + sin 2x on [0, 2π].

Question:

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

Solution:

Let f(x) = x + sin 2x.

$\therefore f^{\prime}(x)=1+2 \cos 2 x$

Now, $f^{\prime}(x)=0 \Rightarrow \cos 2 x=-\frac{1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3}$

$2 x=2 n \pi \pm \frac{2 \pi}{3}, n \in \mathrm{Z}$

$\Rightarrow x=n \pi \pm \frac{\pi}{3}, n \in \mathrm{Z}$

$\Rightarrow x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} \in[0,2 \pi]$

Then, we evaluate the value of $f$ at critical points $x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}$ and at the end points of the interval $[0,2 \pi]$.

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

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

$f\left(\frac{4 \pi}{3}\right)=\frac{4 \pi}{3}+\sin \frac{8 \pi}{3}=\frac{4 \pi}{3}+\frac{\sqrt{3}}{2}$

$f\left(\frac{5 \pi}{3}\right)=\frac{5 \pi}{3}+\sin \frac{10 \pi}{3}=\frac{5 \pi}{3}-\frac{\sqrt{3}}{2}$

$f(0)=0+\sin 0=0$

$f(2 \pi)=2 \pi+\sin 4 \pi=2 \pi+0=2 \pi$

Hence, we can conclude that the absolute maximum value of f(x) in the interval [0, 2π] is 2π occurring at x = 2π and the absolute minimum value of f(x) in the interval [0, 2π] is 0 occurring at x = 0.