Question:
Show that the function given by f(x) = sin x is
(a) strictly increasing in $\left(0, \frac{\pi}{2}\right)$
(b) strictly decreasing in $\left(\frac{\pi}{2}, \pi\right)$
(c) neither increasing nor decreasing in $(0, \pi)$
Solution:
The given function is f(x) = sin x.
$\therefore f^{\prime}(x)=\cos x$
(a) Since for each $x \in\left(0, \frac{\pi}{2}\right), \cos x>0$, we have $f^{\prime}(x)>0$.
Hence, $f$ is strictly increasing in $\left(0, \frac{\pi}{2}\right)$.
(b) Since for each $x \in\left(\frac{\pi}{2}, \pi\right), \cos x<0$, we have $f^{\prime}(x)<0$.
Hence, $f$ is strictly decreasing in $\left(\frac{\pi}{2}, \pi\right)$.
(c) From the results obtained in (a) and (b), it is clear that f is neither increasing nor decreasing in (0, π).