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

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

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

Solution:

Given that $f(x)=\cos x$
$\Rightarrow$
$\begin{array}{l} f(x)=\frac{d}{d x}(\cos x) \\ \Rightarrow f^{\prime}(x)=-\sin x \end{array}$
Taking different region from 0 to $2 \pi$
Suppose $\mathrm{x} \in(0, \pi)$
$\begin{array}{l} \Rightarrow \operatorname{Sin}(x)>0 \\ \Rightarrow-\sin x<0 \\ \Rightarrow f^{\prime}(x)<0 \end{array}$
Therefore $f(x)$ is decreasing in $(0, \pi)$
Suppose $x \in(-\Pi, 0)$
$\begin{array}{l} \Rightarrow \operatorname{Sin}(x)<0 \\ \Rightarrow-\sin x>0 \\ \Rightarrow f^{\prime}(x)>0 \end{array}$
Therefore $f(x)$ is increasing in $(-\pi, 0)$
Thus, from above condition we find that
$\Rightarrow f(x)$ is decreasing in $(0, \pi)$ and increasing in $(-\pi, 0)$ .
As a result, condition for $f(x)$ neither increasing nor decreasing in $(-\pi, \pi)$ ,