Separate \([0, {\pi\over 2}]\) into subintervals in which f(x) = sin 3x is increasing or decreasing.

Solution :

We have, f(x) = sin 3x

\(\therefore\) f'(x) = 3cos 3x

Now, 0 < x < \(pi\over 2\) \(\implies\) 0 < 3x < \(3\pi\over 2\)

Since cosine function is positive in first quadrant and negative in the second and third quadrants. Therefore, we consider the following cases.

Case 1 : When 0 < 3x < \(\pi\over 2\) i.e. 0 < x < \(\pi\over 6\)

In this case, we have

0 < 3x < \(\pi\over 2\) \(\implies\) cos 3x > 0

\(\implies\) 3 cos 3x > 0 \(\implies\) f'(x) > 0

Thus, f'(x) > 0 for 0 < 3x < \(\pi\over 2\) i.e. 0 < x < \(\pi\over 6\)

So, f(x) is increasing on \((0, {\pi\over 6})\).

Case 2 : When \(\pi\over 2\) < 3x < \(3\pi\over 2\) i.e. \(\pi\over 6\) < x < \(\pi\over 2\)

In this case, we have

\(\pi\over 2\) < 3x < \(3\pi\over 2\) \(\implies\) cos 3x < 0

\(\implies\) 3 cos 3x < 0 \(\implies\) f'(x) < 0

Thus, f'(x) < 0 for \(\pi\over 2\) < 3x < \(3\pi\over 2\) i.e. \(\pi\over 6\) < x < \(\pi\over 2\)

So, f(x) is decreasing on \(({\pi\over 6}, {\pi\over 2})\).