Question

Find the maximum and minimum values of the function $f\left(x\right)=sinx+cos2x$ over the range $0<x<2\pi$.

Answer 1

$f\left(x\right)=sinx+cos2x$
$f^{\prime }\left(x\right)=cosx-2sin2x=0$
$cosx-2sin2x=0$
$cosx-2\times 2sinxcosx=0$
$cosx(1-4sinx)=0$
Either $cosx=0\Rightarrow x=(2n+1)\frac{\pi }{2},nϵZ$
or $sinx=\frac{1}{4}$
$x=n\pi +(-1{)}^{n}si{n}^{-1}(\frac{1}{4})$
$0<x<2\pi$
$x=\frac{\pi }{2},\frac{3\pi }{2},si{n}^{-1}\left(\frac{1}{4}\right),\pi -si{n}^{-1}\left(\frac{1}{4}\right)$
are the critical points and values of $f\left(x\right)$ at critical points are:
$f\left(\frac{\pi }{2}\right)=sin\frac{\pi }{2}+cos(2\times \frac{\pi }{2})=0$
$f(x=\frac{3\pi }{2})=sin\frac{3\pi }{2}+cos(2\times \frac{3\pi }{2})=-2$
$f(x=si{n}^{-1}\left(\frac{1}{4}\right))=sin\left(si{n}^{-1}\left(\frac{1}{4}\right)\right)+cos(2\times si{n}^{-1}\left(\frac{1}{4}\right))=1.125$
$f(x=\pi -si{n}^{-1}\left(\frac{1}{4}\right))=-0.625$
So, Maximum value $f\left(x\right)=1.125$
Minimum value $f\left(x\right)=-2$