Question

Given a closed interval in the form [a, b], in which situation would the Mean Value Theorem not apply?

a) [0, 1]
b) [a, a]
c) [1, 0]
d) [b, a]

Answer 1

The Mean Value Theorem does not apply to the interval [a, a] because it's a degenerate interval with no length, therefore option b is correct.

Step-by-step explanation:

The Mean Value Theorem (MVT) is a fundamental result in calculus that applies to continuous functions over a closed interval. Specifically, the MVT states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in the open interval such that the instantaneous rate of change (derivative) at c is equal to the average rate of change (slope of the secant line) over the interval [a, b]. For the MVT to apply, the interval must be one where a is less than b, allowing for a nonzero length of the interval, and the function must meet the continuity and differentiability criteria.

The given options for closed intervals are:

• [0, 1]
• [a, a]
• [1, 0]
• [b, a]

Option b: [a, a] is a degenerate interval because it has no length (a = a). The MVT does not apply here because there's no open interval on which the function could be differentiable. There's no 'room' for a number c to exist where the derivative equals the average rate of change, since there's effectively no interval at all.

The correct option is b, which is the interval [a, a]. In this case, the MVT does not apply because the requirement of having a non-degenerate interval is not fulfilled.