Question
Prove that the function $\text{f}(\text{x})=\cos\text{x}$ is:
Neither increasing nor decreasing in $(0,2\pi)$

Answer

$\text{f}(\text{x})=\cos\text{x}$
$\text{f}'(\text{x})=-\sin\text{x}$
$\Rightarrow\text{f}'(\text{x})<0,\forall\ \text{x}\in(0,\pi)\ ....(1)$
$\Rightarrow\text{f}'(\text{x})>0,\forall\ \text{x}\in(\pi,2\pi)\ ....(2)$
From eqs. (1) and (2), we get
f(x) is strictly decreasing on $(0,\pi)$ and is strictly increasing on $(\pi,2\pi).$
So, f(x) Neither increasing nor decreasing on $(0,2\pi).$

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