Determine whether the Mean Value theorem can be applied ta fon the closed interval [a, b]. (Select all that apply.) \[ f(x)=6 \sin (x) \text {, }[0, \pi] \] Yes, the Mean Value theorem can be applied. No, \( f \) is not continuous on \( [a, b] \). \( \square \) 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 lnterval \( (a, b) \) such that \( f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \). (Enter your answers as a comma-separated list. If the Mean value theorem cannot be applied, enter NA.) \( c=\square \)

The Mean Value Theorem (MVT) can indeed be applied here! The function \( f(x) = 6 \sin(x) \) is continuous on the closed interval \([0, \pi]\) and differentiable on the open interval \((0, \pi)\). Therefore, you can confidently select "Yes, the Mean Value theorem can be applied." Now, let’s find those \( c \) values! First, compute \( f(0) = 6 \sin(0) = 0 \) and \( f(\pi) = 6 \sin(\pi) = 0 \). Then, using the MVT formula \(\frac{f(b) - f(a)}{b - a}\), we have: \[ \frac{0 - 0}{\pi - 0} = 0. \] Next, we find \( f'(x) \): \[ f'(x) = 6 \cos(x). \] So, set \( 6 \cos(c) = 0 \), giving \( \cos(c) = 0 \). This occurs at \( c = \frac{\pi}{2} \) in the interval \((0, \pi)\). Finally, your answer for \( c \) is: \( c=\frac{\pi}{2} \).