Question

Find the interval/intervals in which the function \( f(x) = \sin 3x - \cos 3x \), \( 0<x<\frac{\pi}{2} \) is strictly increasing.

Hint:

To find intervals of increase or decrease, differentiate the function and solve where the derivative is positive or negative.

Answer 1

To determine where \( f(x) = \sin 3x - \cos 3x \) is strictly increasing, we need to find where its derivative is positive.
Step 1: Find the derivative of \( f(x) \) We start by differentiating \( f(x) \) with respect to \( x \): \[ f'(x) = \frac{d}{dx} \left( \sin 3x - \cos 3x \right) \] Using the chain rule: \[ f'(x) = 3\cos 3x + 3\sin 3x \] Factor out the common factor of 3: \[ f'(x) = 3\left( \cos 3x + \sin 3x \right) \] Step 2: Set the derivative greater than zero to find intervals where \( f(x) \) is increasing We want \( f'(x)>0 \): \[ 3\left( \cos 3x + \sin 3x \right)>0 \quad \Rightarrow \quad \cos 3x + \sin 3x>0 \] Now, rewrite \( \cos 3x + \sin 3x \) using the trigonometric identity: \[ \cos 3x + \sin 3x = \sqrt{2} \sin \left( 3x + \frac{\pi}{4} \right) \] So the inequality becomes: \[ \sqrt{2} \sin \left( 3x + \frac{\pi}{4} \right)>0 \] Divide both sides by \( \sqrt{2} \): \[ \sin \left( 3x + \frac{\pi}{4} \right)>0 \] Step 3: Solve the inequality We know that \( \sin \theta>0 \) for \( \theta \in \left( 0, \pi \right) \), so: \[ 0<3x + \frac{\pi}{4}<\pi \] Now solve for \( x \) by first subtracting \( \frac{\pi}{4} \) from each part of the inequality: \[ -\frac{\pi}{4}<3x<\frac{3\pi}{4} \] Now divide by 3: \[ -\frac{\pi}{12}<x<\frac{\pi}{4} \] Thus, the function is strictly increasing in the interval \( \left( 0, \frac{\pi}{6} \right) \). Step 4: Conclude the final intervals Given \( 0<x<\frac{\pi}{2} \), the function \( f(x) \) is strictly increasing in two intervals: \[ x \in \left( 0, \frac{\pi}{6} \right) \cup \left( \frac{\pi}{2}, \frac{5\pi}{6} \right) \]