Question

Find the maximum and minimum values of \(x+1 sin2x\) on \([0,2\pi]\).

Answer 1

Let\( f(x)=x+sin2x.\)

\(f'(x)=1+2cos2x\)

Now,\(f'(x)=0\)
⇒\(cos2x=\)\(-\frac{1}{2}=\)\(-cos\frac{\pi}{3}\)

\(=cos({\pi}-\frac{\pi}{3})\)\(=cos\frac{2\pi}{3}\)

\(2x= 2\pi\pm\frac{2\pi}{3}\) \(n∈Z\)

\(⇒\)\(x=n\pi\pm\frac{\pi}{3},n∈Z\)

\(⇒x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3} ∈[0,2\pi]\)

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 \([0,2\pi]\).

\(f(\frac{\pi}{3})=\frac{\pi}{3}+sin\frac{2\pi}{3}=\frac{\pi}{3}+\frac{\sqrt{3}}{3}\)
\(f(\frac{2\pi}{3})=\frac{2\pi}{3}+sin\frac{4\pi}{3}=\frac{2\pi}{3}-\frac{\sqrt{3}}{3}\)
\(f(\frac{4\pi}{3})=\frac{4\pi}{3}+sin\frac{8\pi}{3}=\frac{4\pi}{3}+\frac{\sqrt{3}}{3}\)
\(f(\frac{5\pi}{3})=\frac{5\pi}{3}+sin\frac{10\pi}{3}=\frac{5\pi}{3}-\frac{\sqrt{3}}{3}\)
\(f(0)=0+sin0=0\)

\(f(2{\pi})=2\pi+sin 4\pi=2\pi+0=2\pi\)

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