Question

Find the maximum and minimum values of x+sin2x on $[0,2\pi ]$

Answer 1

Let $f\left(x\right)=x+sin2x,f^{\prime }\left(x\right)=1+2cos2x$
For maxima or minima put f'(x)=0
$\Rightarrow 1+2cos2x=0\Rightarrow 2x=-\frac{1}{2}\Rightarrow cos2x=-cos\frac{\pi }{3}=cos\frac{2\pi }{3}\Rightarrow cos2x=cos(\pi -\frac{\pi }{3}),cos(\pi +\frac{\pi }{3}),cos(3\pi -\frac{\pi }{3}),cos(3\pi +\frac{\pi }{3})$
($\because$ We know that cos x is negative in second and third quadrant)
Then $2x=2n\pm \frac{2\pi }{3},nϵz\Rightarrow 2x=\frac{2\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3},\frac{10\pi }{3}\Rightarrow x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3},ϵ[0,2\pi ]$
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 $[0,2\pi ]$

$\begin{array}{cc}Atx=0& f\left(0\right)=0+sin0\\ =0Atx=\frac{\pi }{3}& f\left(\frac{\pi }{3}\right)=\frac{\pi }{3}+sin\frac{2\pi }{3}=\frac{\pi }{3}+sin\frac{\pi }{3}=\frac{\pi }{3}+\frac{\sqrt{3}}{2}\\ Atx=\frac{2\pi }{3}& f\left(\frac{2\pi }{3}\right)=\frac{2\pi }{3}+sin\frac{4\pi }{3}=\frac{2\pi }{3}-sin\frac{\pi }{3}=\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\\ Atx=\frac{4\pi }{3}& f\left(\frac{4\pi }{3}\right)=\frac{4\pi }{3}+sin\frac{8\pi }{3}=\frac{4\pi }{3}+sin\frac{2\pi }{3}=\frac{4\pi }{3}+\frac{\sqrt{3}}{2}\\ Atx=\frac{5\pi }{3}& f\left(\frac{5\pi }{3}\right)=\frac{5\pi }{3}+sin\frac{10\pi }{3}=\frac{5\pi }{3}-sin\frac{2\pi }{3}=\frac{5\pi }{3}-\frac{\sqrt{3}}{2}\\ Atx=2\pi & f\left(2\pi \right)=2\pi +sin4\pi =2\pi +0=2\pi \end{array}$
Thus, maximum value is $2\pi$ at $x=2\pi$ and minimum value is 0 at x=0