Question

Show that $f\left(x\right)=\cos x$ is an decreasing function on $(0,\pi )$ increasing in $(-\pi ,0)$ and neither increasing nor decreasing in $(-\pi ,\pi )$.

Answer 1

$f\left(x\right)=\cos x$

$f^{\prime }\left(x\right)=-\sin x$

(i) Now $0<x<\pi$

$-\sin x<0$

$\Rightarrow f^{\prime }\left(x\right)<0$

so $f\left(x\right)$ is decreasing on $(0,\pi )$

(ii) $\pi <x<0$

$-sinx>0$

$\Rightarrow f^{\prime }\left(x\right)>0$

so $f\left(x\right)$ is increasing on $(-\pi ,0)$

(iii) $-\pi <x<\pi$

In this interval

$-\sin x>0$ as well as $-\sin x<0$

Hence $f\left(x\right)$ is neither increasing nor decreasing in $[-\pi ,\pi ]$.