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.

To understand why Rolle's theorem doesn't hold for the function \( f(x) = \cot\left(\frac{x}{2}\right) \) over the interval \([\pi, 7\pi]\), let’s dive into some key points. Firstly, the function \( f(x) \) is not continuous at certain points in the interval due to the nature of the cotangent function, which has vertical asymptotes. For example, at \( x = 2n\pi \) for any integer \( n \), the function tends to \( \pm \infty \), meaning it can’t be continuous across that interval. Additionally, Rolle's theorem requires that the function is both continuous and differentiable over the closed interval. If there are points in the interval where the function is not differentiable, such as at its vertical asymptotes, that automatically disqualifies it from meeting the criteria of the theorem, even if \( f(a) = f(b) \). Thus, just having the same function values at the endpoints is not enough!