Determine whether Rolle's Theorem applies to the given functio \( f f(x)=-\cos 7 x ;\left[\frac{\pi}{14}, \frac{3 \pi}{14}\right] \)

Yes, **Rolle's Theorem** applies to the function \( f(x) = -\cos(7x) \) on the interval \(\left[\frac{\pi}{14}, \frac{3\pi}{14}\right]\). Here's why: ### **Rolle's Theorem Requirements:** 1. **Continuity:** The function \( f(x) = -\cos(7x) \) is continuous on \(\left[\frac{\pi}{14}, \frac{3\pi}{14}\right]\) because the cosine function is continuous everywhere. 2. **Differentiability:** The function is differentiable on the open interval \(\left(\frac{\pi}{14}, \frac{3\pi}{14}\right)\) since the cosine function is differentiable everywhere. 3. **Equal Endpoint Values:** - Evaluate at \( x = \frac{\pi}{14} \): \[ f\left(\frac{\pi}{14}\right) = -\cos\left(7 \cdot \frac{\pi}{14}\right) = -\cos\left(\frac{\pi}{2}\right) = -0 = 0 \] - Evaluate at \( x = \frac{3\pi}{14} \): \[ f\left(\frac{3\pi}{14}\right) = -\cos\left(7 \cdot \frac{3\pi}{14}\right) = -\cos\left(\frac{3\pi}{2}\right) = -0 = 0 \] Since \( f\left(\frac{\pi}{14}\right) = f\left(\frac{3\pi}{14}\right) = 0 \), the third condition is satisfied. ### **Conclusion:** All the conditions of Rolle's Theorem are met: - **Continuity** on \(\left[\frac{\pi}{14}, \frac{3\pi}{14}\right]\) - **Differentiability** on \(\left(\frac{\pi}{14}, \frac{3\pi}{14}\right)\) - **Equal endpoint values** at \( x = \frac{\pi}{14} \) and \( x = \frac{3\pi}{14} \) Therefore, there exists at least one \( c \) in \(\left(\frac{\pi}{14}, \frac{3\pi}{14}\right)\) such that \( f'(c) = 0 \). **For completeness**, finding such \( c \): \[ f'(x) = 7\sin(7x) \] Setting \( f'(c) = 0 \): \[ 7\sin(7c) = 0 \implies \sin(7c) = 0 \implies 7c = k\pi \quad \text{for some integer } k \] Within the interval \(\left[\frac{\pi}{14}, \frac{3\pi}{14}\right]\), the solution is: \[ c = \frac{\pi}{7} \] which lies within \(\left(\frac{\pi}{14}, \frac{3\pi}{14}\right)\). **Final Answer:** Yes. The function is continuous and differentiable on the interval and has equal values at both endpoints, so Rolle’s Theorem applies.