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

Question

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

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Answer

It is given that f(x) = x + sin x, $x \in[0,2 \pi]$
f'(x) = 1 + 2cos 2x
Now, f'(x) = 0
$\Rightarrow \cos 2 \mathrm{x}=-\frac{1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3}$
$2 x=2 \pi \pm \frac{2 \pi}{3}, n \in Z$
$\Rightarrow x=n \pi \pm \frac{\pi}{3}, n \in Z$
$\Rightarrow \mathrm{x}=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} \in[0,2 \pi)$
Now, we evaluate the value of f at critical point $\mathrm{x}=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}$ and at end points of the interval $[0,2 \pi]$
$f^{\prime}\left(\frac{\pi}{3}\right)=\frac{\pi}{3}+\sin \frac{2 \pi}{3}=\frac{\pi}{3}+\frac{\sqrt{3}}{2}$
$f^{\prime}\left(\frac{2 \pi}{3}\right)=\frac{2 \pi}{3}+\sin \frac{4 \pi}{3}=\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}$,
$f^{\prime}\left(\frac{4 \pi}{3}\right)=\frac{4 \pi}{3}+\sin \frac{8 \pi}{3}=\frac{4 \pi}{3}+\frac{\sqrt{3}}{2}$
$f^{\prime}\left(\frac{5 \pi}{4}\right)=\frac{5 \pi}{3}+\sin \frac{10 \pi}{3}=\frac{5 \pi}{3}-\frac{\sqrt{3}}{2}$
f'(0) = 0 + sin 0 = 0
$f^{\prime}(2 \pi)=2 \pi+\sin 4 \pi=2 \pi+0=2 \pi$
Therefore, we have the absolute maximum value of f on $[0,2 \pi]$ is $2 \pi$ occurring at x = $2\pi$ and absolute minimum value of f(x) in the interval $[0,2 \pi]$ is 0 occurring at x = 0.

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