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.

Question

06-12-2023
Mathematics

High School

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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.

Answer :

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jaisharma131313 jaisharma131313 · Answer 1 · 2023-09-17T10:18:25-04:00

Final answer:

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].