Question

Find the absolute maximum and absolute minimum values of the function f given by $f\left(x\right)=si{n}^{2}x-cosx,xϵ(0,\pi )$

Answer 1

$f\left(x\right)=si{n}^{2}x-cosx{f}^{\prime }\left(x\right)=2sinx.cosx+sinx=sinx(2cosx+1)Equating{f}^{\prime }\left(x\right)tozero.f^{\prime }\left(x\right)=0sinx(2cosx+1)=0sinx=0\therefore x=0,\pi 2cosx+1=0\Rightarrow cosx=-\frac{1}{2}\therefore x=\frac{5\pi }{6}f\left(0\right)=si{n}^{2}0-cos0=-1$

$f\left(\frac{5\pi }{6}\right)=si{n}^2\left(\frac{5\pi }{6}\right)-cos\left(\frac{5\pi }{6}\right)=si{n}^2\left(\frac{\pi }{6}\right)-cos\left(\frac{\pi }{6}\right)=\frac{1}{4}-\frac{\sqrt{3}}{2}=\left(\frac{1-2\sqrt{3}}{4}\right)f\left(\pi \right)=si{n}^2\pi -cos\pi =1$

Of these values, the maximum value is 1, and the minimum value is -1.

Thus, the absolute maximum and absolute minimum values of f(x) are 1 and -1, which it attains at $x=0andx=\pi$.