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 $x=0$