# 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 $$({\pi\over 6}, {\pi\over 2})$$.

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