\( \mid \leftarrow \) Determine whether Rolle's Theorem applies to the given function on the given interval. If \[ f(x)=\cos 7 x ;\left[\frac{\pi}{14}, \frac{3 \pi}{14}\right] \]

To determine whether **Rolle's Theorem** applies to the function \( f(x) = \cos(7x) \) on the interval \(\left[ \frac{\pi}{14}, \frac{3\pi}{14} \right]\), we need to verify the three main conditions of the theorem: 1. **Continuity on the Closed Interval \([a, b]\):** - The function \( f(x) = \cos(7x) \) is a cosine function, which is continuous everywhere. - **Conclusion:** \( f(x) \) is continuous on \(\left[ \frac{\pi}{14}, \frac{3\pi}{14} \right]\). 2. **Differentiability on the Open Interval \((a, b)\):** - The function \( f(x) = \cos(7x) \) is differentiable everywhere since cosine is smooth and has a well-defined derivative. - **Conclusion:** \( f(x) \) is differentiable on \(\left( \frac{\pi}{14}, \frac{3\pi}{14} \right)\). 3. **Equal Function Values at the Endpoints:** - Evaluate \( f \) at the endpoints of the interval: \[ f\left( \frac{\pi}{14} \right) = \cos\left(7 \cdot \frac{\pi}{14}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \] \[ f\left( \frac{3\pi}{14} \right) = \cos\left(7 \cdot \frac{3\pi}{14}\right) = \cos\left(\frac{21\pi}{14}\right) = \cos\left(\frac{3\pi}{2}\right) = 0 \] - Both endpoints yield \( f(a) = f(b) = 0 \). - **Conclusion:** \( f\left( \frac{\pi}{14} \right) = f\left( \frac{3\pi}{14} \right) \). Since all three conditions of Rolle's Theorem are satisfied: - \( f(x) \) is continuous on \(\left[ \frac{\pi}{14}, \frac{3\pi}{14} \right]\), - \( f(x) \) is differentiable on \(\left( \frac{\pi}{14}, \frac{3\pi}{14} \right)\), - \( f\left( \frac{\pi}{14} \right) = f\left( \frac{3\pi}{14} \right) \), **Rolle's Theorem does apply** to the function \( f(x) = \cos(7x) \) on the interval \(\left[ \frac{\pi}{14}, \frac{3\pi}{14} \right]\). **Additional Insight:** By Rolle's Theorem, there exists at least one \( c \) in \(\left( \frac{\pi}{14}, \frac{3\pi}{14} \right)\) such that \( f'(c) = 0 \). Calculating the derivative: \[ f'(x) = -7\sin(7x) \] Setting \( f'(c) = 0 \): \[ -7\sin(7c) = 0 \quad \Rightarrow \quad \sin(7c) = 0 \] This occurs when \( 7c = k\pi \) for some integer \( k \). Within the interval, the appropriate value of \( k \) gives: \[ c = \frac{k\pi}{7} \] In the interval \(\left[ \frac{\pi}{14}, \frac{3\pi}{14} \right]\), \( k = 1 \) yields \( c = \frac{\pi}{7} \), which lies within the interval. **Final Answer:** Yes. The function f (x ) = cos 7x is continuous and differentiable on [π⁄14, 3π⁄14], and f(π⁄14) = f(3π⁄14). Thus, Rolle’s Theorem applies on this interval.