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## Differential Calculus

### Course: Differential Calculus > Unit 5

Lesson 1: Mean value theorem- Mean value theorem
- Mean value theorem example: polynomial
- Mean value theorem example: square root function
- Using the mean value theorem
- Justification with the mean value theorem: table
- Justification with the mean value theorem: equation
- Establishing differentiability for MVT
- Justification with the mean value theorem
- Mean value theorem application
- Mean value theorem review

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# Mean value theorem review

Review your knowledge of the mean value theorem and use it to solve problems.

## What is the mean value theorem?

The mean value theorem connects the average rate of change of a function to its derivative. It says that for any differentiable function f and an interval open bracket, a, comma, b, close bracket (within the domain of f), there exists a number c within left parenthesis, a, comma, b, right parenthesis such that f, prime, left parenthesis, c, right parenthesis is equal to the function's average rate of change over open bracket, a, comma, b, close bracket.

Graphically, the theorem says that for any arc between two endpoints, there's a point at which the tangent to the arc is parallel to the secant through its endpoints.

*Want to learn more about the mean value theorem? Check out this video.*

## Want to join the conversation?

- How is mean value theorem applied in real life?(10 votes)
- Mean value theorem can be used to determine the speed of something, like a policeman using a speedometer.(24 votes)

- In the question, the interval is a closed one, but in the explanation, the interval is open, thus excluding 3 as an answer. Why does the explanation have an open interval?(6 votes)
- A and B are given in a closed interval in the question because the hypotheses of the mean value theorem include that the function must be continuous on the interval [A,B] and differentiable on the interval (A,B). The function does not have to be differentiable at A and B because you are looking for a value c in between A and B that is equal to the slope of the secant line connecting A and B. Hope that helps.(11 votes)

- Sir what about roll's and lagrance's theorem

and thank u sir(1 vote) - If a=b, (the two endpoints have the same y value) can mean value theorem be applied?(0 votes)
- Yes, it would imply that the function has a point in the interval where the derivative is 0. This special case is called Rolle's Theorem.(2 votes)

- In interval the a and b are two real numbers is there ,such time how identify the roots of equation in given interval in MVT(0 votes)
- if this man stop for some minutes and continue his journey then?(0 votes)
- hello. I wanted to ask, does the application of the mean value theorem only involve velocities?(0 votes)
- The mean value theorem can involve any function as long as the interval in which you are considering satisfies the conditions.(1 vote)

- The question in practice problem 1 specifically asks for a value for c in the
**closed**interval [0,3], but marks one of the correct solutions as incorrect. I understand why it wouldn't want to accept 3 as a correct answer as it's the answer to one of the intermediate steps toward the solution it's looking for, but it*is*a correct answer to the question as it's currently posed. The question wording should be changed to ask for c in the open interval (0,3) to prevent that mistake. As Lori Feng mentions below, it's even described to be incorrect in the explanation because it assumes the question is asking for a value in the open interval rather than the closed interval listed in the question.(0 votes)- in the definition of the mean value theorem given at the top it says that the interval [a,b] is for the function, then the interval (a,b) is for the derivative, which is where c would be.

It is kinda crummy it's like that, but technically yeah, c would only be in the open interval(1 vote)

- how we can do m.v.t of f(x)= sin^-1 x, [-1,1](0 votes)
- 𝑓(𝑥) = arcsin(𝑥) ⇒ 𝑓 '(𝑥) = 1∕√(1 − 𝑥²)

𝑓(𝑥) is continuous over [−1, 1] and differentiable over (−1, 1).

Thereby the mean value theorem applies, and there exists a 𝑐 ∈ (−1, 1)

such that 𝑓 '(𝑐) equals the average rate of change of 𝑓(𝑥) over [−1, 1].

The average rate of change of 𝑓(𝑥) over [−1, 1] is

(𝑓(1) − 𝑓(−1))∕(1 − (−1)) = 𝜋∕2

1∕√(1 − 𝑐²) = 𝜋∕2 ⇒ 𝑐 = ±√(𝜋² − 4)∕𝜋(0 votes)

- It's a bit off-topic, but I failed at some of the MVT questions because of my calculator. It calculated (-6)^2 as -36. Is this normal with calculators? I'm using a T-84 Plus(0 votes)
- Sometimes, in graphical calculators, you would be presented with two (-) symbols. One represents negativity and the other represents a 'minus' In this situation it seems that you might have pressed in the wrong (-) symbol watch out for it next time!

I hope this helps :)(0 votes)