Can the Mean Value Theorem be applied to the function $ f $ on the closed interval $[a, b]$? Choose the correct option.

$ f(x) = \sqrt{1-x} $, $[-24, 1]$.

A. Yes, the Mean Value Theorem can be applied.

B. No, because $ f $ is not continuous on $[-24, 1]$.

C. No, because $ f $ is not differentiable on $[-24, 1]$.

D. No, because the endpoints are not included.

The Mean Value Theorem cannot be applied to the function f(x) = √(1-x) on the interval [-24, 1] because the function is not differentiable over the entire open interval (-24, 1).

Explanation:

To determine if the Mean Value Theorem (MVT) can be applied to the function f(x) = √(1-x) on the closed interval [-24, 1], we must verify two conditions: that f is continuous on the closed interval [a, b], and differentiable on the open interval (a, b). The function given is continuous for all x where 1-x ≥ 0, meaning x ≤ 1. Thus, it is continuous on the interval [-24, 1]. However, differentiation of f(x) yields f'(x) = -1/(2√(1-x)), which is undefined for x = 1 because it involves division by zero.

Since f'(x) is undefined at x = 1, the conditions for the Mean Value Theorem are not satisfied over the entire interval [-24, 1] because the function must be differentiable throughout the open interval (-24, 1). Therefore, the Mean Value Theorem cannot be applied to f(x) = √(1-x) on the interval [-24, 1].