Determine whether Rolle's theorem can be applied to \( f \) on the elosed interval [a, b]. (Select all that apply.) \[ f(x)=\cos (x), \quad[\pi, 3 \pi] \] Yes, Rolla's theorem can be appliad. No, because \( f \) is not continuous on the clased Interval \( [a, b] \). No, because \( f \) is not differentiable on the open Interval \( (a, b) \). No, becausa \( f(a) \neq f(b) \). If Rolle's theorem can be applled, find all values of \( c \) In the open Interval \{ä, b) such that \( f^{\prime}(c)=0 \). (Enter your answers as a comma-separated list. If Rolle's theorem cannot be applied, enter NA.) \[ c= \] \( \square \)

Rolle's theorem states that if a function is continuous on a closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \(f(a) = f(b)\), then there exists at least one value \(c\) in \((a, b)\) such that \(f'(c) = 0\). In this case, \(f(x) = \cos(x)\) is continuous and differentiable everywhere, including the interval \([\pi, 3\pi]\). However, since \(f(a) = \cos(\pi) = -1\) and \(f(b) = \cos(3\pi) = -1\), we find \(f(a) = f(b)\), which satisfies one of the key conditions of Rolle's theorem. Thus, we can apply Rolle's theorem here. To find \(c\), calculate the derivative \(f'(x) = -\sin(x)\) and set it to zero. Solving \(-\sin(c) = 0\) gives \(c = n\pi\). Within the interval \((\pi, 3\pi)\), the value is \(c = 2\pi\). So, Rolle's theorem can be applied, and the answer is: \(c = 2\pi\).