Questions: Determine whether the Mean Value Theorem can be applied to (f) on the closed interval ([a, b]). (Select all that apply.) (f(x)=6-x, quad[3,7]) Yes, the Mean Value 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)). 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'(c)=fracf(b)-f(a)b-a). (Enter your answers as a comma-separated list. If the Mean Value Theorem cannot be applied, enter NA.) (c=)

Solution Steps

To determine if the Mean Value Theorem (MVT) can be applied, we need to check if the function \( f(x) = |6-x| \) is continuous on the closed interval \([3, 7]\) and differentiable on the open interval \((3, 7)\). The function \( f(x) = |6-x| \) is continuous everywhere, but it is not differentiable at \( x = 6 \) because the absolute value function has a sharp corner there. Since \( x = 6 \) is within the interval \((3, 7)\), the function is not differentiable on the entire open interval, and thus the MVT cannot be applied.

The function \( f(x) = |6-x| \) is an absolute value function, which is continuous everywhere on the real line. Therefore, it is continuous on the closed interval \([3, 7]\).

The function \( f(x) = |6-x| \) is not differentiable at \( x = 6 \) because it has a sharp corner at this point. Since \( x = 6 \) is within the open interval \((3, 7)\), the function is not differentiable on the entire interval.

The Mean Value Theorem requires that the function be both continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\). Although \( f(x) \) is continuous on \([3, 7]\), it is not differentiable on \((3, 7)\) due to the sharp corner at \( x = 6 \). Therefore, the Mean Value Theorem cannot be applied.

Final Answer