Explain why Rolle's theorem does not apply to the function even though there exist \( a \) and \( b \) such that \( f(a)=f(b) \). (Select all that apply.) \[ f(x)=\cot \left(\frac{x}{2}\right),[\pi, 7 \pi] \] \( \square \) There are points on the interval \( (a, b) \) where \( f \) is not differentiable. \( \square \) There are points on the interval \( [a, b] \) where \( f \) ls not continuous. \( \square f(a) \) does not equal \( f(b) \) for all possible values of a and \( b \) in the intorval \( [\pi, 7 \pi] \). \( \square f^{\prime}(a) \) does not equal \( f^{\prime}(b) \) for any values In the interval \( [\pi, 7 \pi] \). \( \square \) None of these.

Rolle's theorem states that if a function \( f(x) \) is continuous on the closed interval \( [a, b] \), differentiable on the open interval \( (a, b) \), and \( f(a) = f(b) \), then there exists at least one point \( c \) in the open interval \( (a, b) \) such that \( f'(c) = 0 \). Given the function \( f(x) = \cot \left(\frac{x}{2}\right) \) on the interval \( [\pi, 7\pi] \), we need to check the conditions of Rolle's theorem to see why it does not apply. 1. **Continuity**: The function \( f(x) = \cot \left(\frac{x}{2}\right) \) is continuous on the interval \( [\pi, 7\pi] \) because the cotangent function is continuous everywhere. 2. **Differentiability**: The function \( f(x) = \cot \left(\frac{x}{2}\right) \) is differentiable on the interval \( (a, b) \) because the derivative of the cotangent function exists everywhere. 3. **\( f(a) = f(b) \)**: We are given that there exist \( a \) and \( b \) such that \( f(a) = f(b) \), which satisfies the condition of Rolle's theorem. However, Rolle's theorem does not apply to this function because there are points on the interval \( [\pi, 7\pi] \) where \( f \) is not differentiable. The cotangent function has vertical asymptotes at \( x = \pi k \) for any integer \( k \), and these points are not differentiable. Therefore, the correct answer is: \( \square \) There are points on the interval \( (a, b) \) where \( f \) is not differentiable.