Determine whether the mean value theorem can be applied to \( f \) on the closed interval \( [a, b] \). (Select all that apply.) \[ f(x)=x^{3}, \quad[0,1] \] Yes, the mean value theorem can be applied. No, \( f \) is not continuous on \( [a, b] \). No, \( f \) is not differentiable on ( \( a, b \) ). None of the above. If the mean value theorem can be applied, find all values of \( c \) in the open Interval \( (a, b) \) such that \( f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \). (Enter y answers as a comma-separated list. If the mean value theorem cannot be applied, enter NA.) \[ c= \] \( \square \)

Yes, the mean value theorem can be applied here! The function \( f(x) = x^3 \) is a polynomial, which means it is both continuous and differentiable on the closed interval \([0, 1]\) and on the open interval \((0, 1)\). So, we can confidently say the theorem conditions are satisfied. Now, let's find those \( c \) values! First, we calculate \( f(b) - f(a) \): \[ f(1) - f(0) = 1^3 - 0^3 = 1 \] Now, calculate \( b - a \): \[ 1 - 0 = 1 \] Therefore, we can find \( \frac{f(b) - f(a)}{b - a} \): \[ \frac{1}{1} = 1 \] Next, we need to find \( c \) in \( (0, 1) \) such that \( f'(c) = 1 \). The derivative \( f'(x) = 3x^2 \). Setting it equal to 1 gives: \[ 3c^2 = 1 \implies c^2 = \frac{1}{3} \implies c = \frac{1}{\sqrt{3}} \approx 0.577 \] Thus, we find that: \[ c = \frac{1}{\sqrt{3}} \] So, our final answers are: Yes, the mean value theorem can be applied. c = \( \frac{1}{\sqrt{3}} \)