# Show that f(x)=cos x is a decreasing function

Question:

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

decreasing in $(-\pi, \pi)$.

Solution:

Given:- Function $f(x)=\cos x$

Theorem:- Let $\mathrm{f}$ be a differentiable real function defined on an open interval $(\mathrm{a}, \mathrm{b})$.

(i) If $f^{\prime}(x)>0$ for $a l l_{X} \in(a, b)$, then $f(x)$ is increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$

Algorithm:-

(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ and solve this inequation.

For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain it is decreasing.

$f(x)=\cos x$

$\Rightarrow f(x)=\frac{d}{d x}(\cos x)$

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

Taking different region from 0 to $2 \pi$

a) let $x \in(0, \pi)$

$\Rightarrow \sin (x)>0$

$\Rightarrow-\sin x<0$

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

Thus $f(x)$ is decreasing in $(0, \pi)$

b) let $x \in(-\pi, 0)$

$\Rightarrow \sin (x)<0$

$\Rightarrow-\sin x>0$

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

Thus $f(x)$ is increasing in $(-\pi, 0)$

Therefore, from above condition we find that

$\Rightarrow f(x)$ is decreasing in $(0, \pi)$ and increasing in $(-\pi, 0)$

Hence, condition for $f(x)$ neither increasing nor decreasing in $(-\pi, \pi)$