Final answer:
The Mean Value Theorem cannot be applied to the function f(x) = |9x - 8| on the closed interval [-1, 3] because the function has a corner at x = 8/9 and is not differentiable at that point.
Step-by-step explanation:
The Mean Value Theorem (MVT) states that if a function f is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in (a, b) such that f'(c) is equal to the average rate of change of the function over [a, b]. In the case of the function f(x) = |9x - 8|, we must check if the function is both continuous and differentiable over the interval [-1, 3].
Upon inspection, we see that f(x) is continuous everywhere because the absolute value function is continuous. However, the differentiability of f(x) may be problematic at the point where the inside of the absolute value, 9x - 8, equals zero. This point, x = 8/9, is within our interval [-1, 3]. At this point, f(x) has a corner, which means it is not differentiable. Thus, because f(x) is not differentiable over the entire open interval (-1, 3), we cannot apply the Mean Value Theorem to f(x) = |9x - 8| on the closed interval [-1, 3].