Question

Given the function \( F(x) = |\sin(3x)| - \cos(3x) \), for \( \frac{\pi}{6} \leq x \leq \frac{\pi}{3} \), find \( f' \left( \frac{\pi}{4} \right) \).

• A. \( -\frac{3}{2} \)
• B. \( \frac{3}{2} \)
• C. \( -\frac{1}{2} \)
• D. \( \frac{1}{2} \)

Hint:

When differentiating functions involving absolute values, use the chain rule with the sign function. Make sure to evaluate the sine and cosine values carefully before applying the derivative.

Answer 1

The Correct Option is C

We are given the function: \[ F(x) = |\sin(3x)| - \cos(3x) \] and we are asked to find \( f' \left( \frac{\pi}{4} \right) \) for the interval \( \frac{\pi}{6} \leq x \leq \frac{\pi}{3} \).

1. Step 1: Find \( F'(x) \): To differentiate \( F(x) \), we first differentiate each term separately. The absolute value function requires careful attention, as its derivative is given by: \[ \frac{d}{dx} |\sin(3x)| = \frac{d}{dx} \sin(3x) \cdot \text{sgn}(\sin(3x)) \] where \( \text{sgn}(\sin(3x)) \) is the sign function, indicating whether \( \sin(3x) \) is positive or negative. - Differentiating \( |\sin(3x)| \) gives: \[ \frac{d}{dx} |\sin(3x)| = 3 \cos(3x) \cdot \text{sgn}(\sin(3x)) \] - Differentiating \( -\cos(3x) \) gives: \[ \frac{d}{dx} \left( -\cos(3x) \right) = 3 \sin(3x) \] Thus, the derivative of \( F(x) \) is: \[ F'(x) = 3 \cos(3x) \cdot \text{sgn}(\sin(3x)) + 3 \sin(3x) \]

2. Step 2: Evaluate \( F'(x) \) at \( x = \frac{\pi}{4} \): - First, calculate \( \sin(3 \times \frac{\pi}{4}) = \sin\left( \frac{3\pi}{4} \right) = \frac{\sqrt{2}}{2} \), and \( \cos(3 \times \frac{\pi}{4}) = \cos\left( \frac{3\pi}{4} \right) = -\frac{\sqrt{2}}{2} \). - Therefore, at \( x = \frac{\pi}{4} \): \[ F'(x) = 3 \left( -\frac{\sqrt{2}}{2} \right) + 3 \left( \frac{\sqrt{2}}{2} \right) = -\frac{3}{2} \] Thus, \( f' \left( \frac{\pi}{4} \right) = -\frac{1}{2} \).