Find the intervals in which the function given by $f(x) = \sin 3x, x \in \left[0, \frac{\pi}{2}\right]$ is:
$(a)$ increasing
$(b)$ decreasing.

(N/A) We have $f(x) = \sin 3x$.
Then,$f'(x) = 3 \cos 3x$.
To find the critical points,we set $f'(x) = 0$,which gives $3 \cos 3x = 0$,or $\cos 3x = 0$.
Since $x \in \left[0, \frac{\pi}{2}\right]$,we have $3x \in \left[0, \frac{3\pi}{2}\right]$.
Thus,$3x = \frac{\pi}{2}$ or $3x = \frac{3\pi}{2}$,which implies $x = \frac{\pi}{6}$ or $x = \frac{\pi}{2}$.
The point $x = \frac{\pi}{6}$ divides the interval $\left[0, \frac{\pi}{2}\right]$ into two sub-intervals: $\left[0, \frac{\pi}{6}\right)$ and $\left(\frac{\pi}{6}, \frac{\pi}{2}\right]$.
For $x \in \left[0, \frac{\pi}{6}\right)$,$0 \leq 3x < \frac{\pi}{2}$,so $\cos 3x > 0$,which means $f'(x) > 0$. Thus,$f$ is increasing on $\left[0, \frac{\pi}{6}\right]$.
For $x \in \left(\frac{\pi}{6}, \frac{\pi}{2}\right]$,$\frac{\pi}{2} < 3x \leq \frac{3\pi}{2}$,so $\cos 3x < 0$,which means $f'(x) < 0$. Thus,$f$ is decreasing on $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$.