Question

$\text{f}\left(x\right)=\cos x-{\cos }^{2}x$. Find range .

Answer 1

$f\left(x\right)\cos x-{\cos }^{2}x$

$\Rightarrow$ we know domain of $f\left(x\right)willbe-\infty <x<\infty$ and period of $\cos x=2\pi$

Now $f\left(x\right)=\cos x-{\cos }^{2}x$

$=f^1\left(x\right)=-\sin x+2\cos x\sin x$

For eritical point$f^1\left(x\right)=0$

$-\sin x+2\cos x\sin x=0$

$\sin x(2\cos x-1)=0$

$2\cos x-1=0\sin x=0$

$\cos x=\frac{1}{2}x=n\pi \left(nϵz\right)$

$x=\frac{\pi }{3}+3n\pi$or$\frac{-\pi }{3}+2n\pi \left(nϵz\right)$

Now $f^1\left(x\right)=\sin x+\sin 2x$

$f^{11}\left(x\right)=-\cos x+2\cos 2x$

$f^{11}\left(n\pi \right)=-\cos \left(n\pi \right)+2\cos \left(2n\pi \right)$