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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 $\left[a,b\right]$ (within the domain of $f$), there exists a number $c$ within $\left(a,b\right)$ such that ${f}^{\prime }\left(c\right)$ is equal to the function's average rate of change over $\left[a,b\right]$.
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.

## Check your understanding

Problem 1
$f\left(x\right)={x}^{3}-6{x}^{2}+12x$
Let $c$ be the number that satisfies the Mean Value Theorem for $f$ on the interval $\left[0,3\right]$.
What is $c$ ?

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• How is mean value theorem applied in real life?
• Mean value theorem can be used to determine the speed of something, like a policeman using a speedometer.
• 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?
• 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.
• Why c would be only in open interval? Because a and b are differentiable over open interval?
(1 vote)
• Correct. We do need to find f'(c) after all. So, c needs to be a point where the derivative exists (i.e. a differentiable point) and the interval which is differentiable is (a,b), not [a,b]
• 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.
(1 vote)
• 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
• I'm a bit confused on the graph given by the initial explanation; the MVT says that the function must be continuous on the endpoints as well, though on the graph, the right sided limit does not exist for x=6, making it discontinuous over the interval [0,6]. Is that correct or am I confusing the concept?
(1 vote)
• Yeah, you have some confusion about continuity.
A discontinuous point means that either you have no defined value at that point or it jumps to a different point. The graph you're talking doesn't show a jump at the point nor an undefined point (which is typically shown as an empty circle).
MVT doesn't care about differentiability at the endpoints as it states that it "must be differentiable over the open interval (a,b) and continuous over the closed interval [a,b]." Hope it helped.
(1 vote)
• Why do we exclude a and b when we want to find the possible values for c?
(1 vote)
• if this man stop for some minutes and continue his journey then?
(1 vote)
• Sir what about roll's and lagrance's theorem
and thank u sir
(1 vote)
• What would be an example where, if f is differentiable between a CLOSED interval of [a, b] instead of an OPEN interval (a,b), the MVT isn't valid?
(1 vote)
• (a, b) is a subset of [a, b], so if f is differentiable on [a, b], then it is automatically differentiable on (a, b) as well.

We use an open interval for that part of the statement because we generally like to use the weakest assumptions possible, so that we can apply the theorem in as many cases as possible.

If I have a function that equals sin(x) for 0<x<π and 0 otherwise, then we can apply the mean value theorem on the interval [0, π], even though our function isn't differentiable at the endpoints. If we insisted on using the closed interval in the statement of the theorem, then we couldn't apply the theorem here, and for no good reason.
(1 vote)
• If a=b, (the two endpoints have the same y value) can mean value theorem be applied?