22.7k views
5 votes
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]

User Bakaburg
by
7.9k points

1 Answer

6 votes

Final answer:

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.

User Eddie Curtis
by
8.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.