Determine whether Rolle's theorem can be applied to fon the closed interval \( [a, b] \). (Select all that apply.) \[ f(x)=\frac{x^{2}-36}{x-5} \quad[-6,6] \] Yes, Rolle's thearem can be applied. 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, because \( f(a) \neq f(b) \). If Rolle's theorem can be applled, find all values of \( c \) in the open Interval \( (a, b) \) such that \( f^{\prime}(c)=0 \). (Enter your answers as a comma-separated list, if Rolle's theorem cannot be applied, enter NA,) \[ c= \] \( \square \)

To determine if Rolle's theorem can be applied to the function \( f(x) = \frac{x^{2}-36}{x-5} \) on the interval \( [-6, 6] \), we need to check three criteria: 1. **Continuity**: The function \( f(x) \) is a rational function, but we must check for any points where the denominator is zero. The denominator \( x-5 \) is zero at \( x = 5 \), so \( f(x) \) is not defined at \( x = 5 \). This means \( f(x) \) is not continuous on the closed interval \( [-6, 6] \). Therefore, we cannot apply Rolle's theorem here. 2. **Endpoints**: If we were to calculate the endpoints, we would find \( f(-6) \) and \( f(6) \) to check if \( f(a) \) equals \( f(b) \). However, since continuity is already violated, we don't need to proceed further. Because \( f \) is not continuous on \( [-6, 6] \) and not defined at \( x=5 \), we conclude: No, because \( f \) is not continuous on the closed interval \( [a, b] \). As for finding the values of \( c \) such that \( f'(c) = 0 \), since Rolle’s theorem cannot be applied, we enter: \[ c = NA \]