Question:
33 Prove that the function $f(x)=\cos x$ is :
i. strictly decreasing on $(0, \pi)$
ii. strictly increasing in $(\pi, 2 \pi)$
iii. neither increasing nor decreasing in $(0,2 \pi)$
Solution:
Given $f(x)=\cos x$
$\therefore f^{\prime}(x)=-\sin x$
(i) Since for each $x \in(0, \pi), \sin x>0$
$\Rightarrow \therefore f^{\prime}(x)<0$
So $f$ is strictly decreasing in $(0, \pi)$
(ii) Since for each $x \in(\pi, 2 \pi), \sin x<0$
$\Rightarrow \therefore f^{\prime}(x)>0$
So $f$ is strictly increasing in $(\pi, 2 \pi)$
(iii) Clearly from (1) and (2) above, $f$ is neither increasing nor decreasing in $(0,2 \pi)$