Determine whether Rolle's theorem can be applled to fon the closed interval \( [3, b] \). (Select all that apply.) \[ f(x)=x^{2}-13 x+36, \quad[4,9] \] Yes, Rolle's theorem can be applied. No, because \( f \) is not continuous on the closed interval \( [a, b] \). No, because \( f \) is not differentiable in the open interval \( (a, b) \). No, because \( f(a) \neq f(b) \). If Rolle's theorem can be applied, find all values of \( c \) in the open interval ( \( a, b \) ) such that \( f^{\prime}(c)=0 \). (Enter your answers as a commaseparated list. If Rolle's theorem cannot be applied, enter NA.) \[ c= \] \( \square \)

To determine whether Rolle's theorem can be applied, let's check the conditions: 1. **Continuity**: The function \( f(x) = x^2 - 13x + 36 \) is a polynomial, and all polynomials are continuous everywhere. Therefore, it is continuous on the closed interval \( [4, 9] \). 2. **Differentiability**: Similarly, as a polynomial, \( f(x) \) is also differentiable everywhere, including the open interval \( (4, 9) \). 3. **Equal values at endpoints**: We need to evaluate \( f(4) \) and \( f(9) \): - \( f(4) = 4^2 - 13 \times 4 + 36 = 16 - 52 + 36 = 0 \) - \( f(9) = 9^2 - 13 \times 9 + 36 = 81 - 117 + 36 = 0 \) Since \( f(4) = f(9) = 0 \), this condition is also satisfied. With all conditions met, we can conclude: **Yes, Rolle's theorem can be applied.** Now, to find \( c \) such that \( f'(c) = 0 \): 1. First, we calculate the derivative: \[ f'(x) = 2x - 13 \] 2. Next, set the derivative equal to zero: \[ 2c - 13 = 0 \implies 2c = 13 \implies c = \frac{13}{2} = 6.5 \] Thus, we find: \[ c = 6.5 \]