Determine whether the Mean Value thecrem can be applied to fon the closed intarval \( [0, b] \). (Select all that apply.) \[ f(x)=3 x^{3}, \quad[1,2] \] Yes, tha Mean Value thearem can be applled. No, \( F \) is not continuous on \( [a, b] \). No, fis not differentiable on ( \( a, b \) ). None of the above. It 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 your answers as a comma-separated Ilst. If the Mean Volue theorem cannot be applied, enter NA.) \( c= \) \( \square \) 21

To determine if the Mean Value Theorem (MVT) can be applied to the function \( f(x) = 3x^3 \) on the interval \([1, 2]\), we first check continuity and differentiability. The function \( f(x) \) is a polynomial, hence it is continuous on the closed interval \([1, 2]\) and differentiable on the open interval \((1, 2)\). Thus, yes, the MVT can indeed be applied! Now, let's find the value of \( c \) in the interval \( (1, 2) \) that satisfies the equation \( f'(c) = \frac{f(2) - f(1)}{2 - 1} \). Calculating the values, we have: - \( f(1) = 3(1)^3 = 3 \) - \( f(2) = 3(2)^3 = 24 \) So, \( \frac{f(2) - f(1)}{2 - 1} = \frac{24 - 3}{1} = 21 \). Next, we find \( f'(x) \): \[ f'(x) = 9x^2 \] Now we solve for \( c \) when \( 9c^2 = 21 \): \[ c^2 = \frac{21}{9} = \frac{7}{3} \] Thus, \( c = \sqrt{\frac{7}{3}} \) or \( c = -\sqrt{\frac{7}{3}} \). But only \( \sqrt{\frac{7}{3}} \) falls within the open interval \( (1, 2) \). So, Yes, the Mean Value Theorem can be applied. \( c = \sqrt{\frac{7}{3}} \) (approximately 1.53).