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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition)>Unit 3

Lesson 16: Mean value theorem

# Mean value theorem (old)

An old video of Sal introducing the Mean value theorem & giving some intuition to its meaning. Created by Sal Khan.

## Want to join the conversation?

• How do you know when to use the Mean Value Theorem?
• On the AP test they pretty much ask you to use it indirectly. The word "average" should be a flag for you on whether or not you should use the MVT.
• when do you use the mean value theorem?
• MVT has LOTS of uses in math, in fact many, many important theorems in math use the MVT as part of their proof, for example: L'Hopitals Rule, the second part of the Fundamental Theorem of Calculus just to name a few of the theorems you will find here on Khan. There is much more advanced math that also relies on the MVT.
Remember, math is cumulative; every topic you learn is preparing you for something more advanced. Sometimes it may be a while before that advanced something arrives (often because you need more mathematical maturity to even begin to understand how a previous topic is applied).
You'll get there, just KEEP STUDYING!
• Doesn`t it mean that if a function is continuous, then the function is defined at any point?
• Yes. It not only means the function is defined at every point, it also means that when graphed, there are no breaks in the curve.

For instance, if there is a jump discontinuity (see http://en.wikipedia.org/wiki/Classification_of_discontinuities#Examples), the function can be defined at every point and discontinuous.
• If you take the function f(x) = x (just an example) and pick any closed interval on it, going from a - b where a and f(a) < b and f(b), there is no point "c" where f ' (c) = 0, right? So , in some functions, although they meet the requirements for the mean value theorem, the theorem "doesn´t work"?
• That the derivative of the function `ƒ` is nowhere zero, is correct. What is not correct, however, is that the mean value theorem does not hold when the requirements are met. The derivative of the function `ƒ` defined by `ƒ(x) = x` is `1` everywhere; hence nowhere zero. However, the mean value theorem does not assert that the derivative of `ƒ` is zero at some point. It asserts the following. Let `a` and `b` be two real numbers such that `a < b`. `ƒ` is clearly continuous on `[a, b]` and differentiable on `(a, b)`. By the mean value theorem, there exists some real number `c` such that `a < c < b` and `ƒ(b) - ƒ(a) = ƒ'(c)(b - a)`. But `ƒ(a) = a`, `ƒ(b) =`` b`, and `ƒ'(c) = 1` for every `c`, so this just asserts that `b - a = 1 · (b - a)`, which we know is true.
• is this for 5 th doing 7 th grade work . cause i am a fifth grader doing seventh grade work
(1 vote)
• This is college-level math, though some students in the USA (I don't know about other countries) take differential calculus in 12th grade.
• Can you post a video on Intermediate Value Theorem? Thanks!
• Sal, don't get so down on the Mean Value Theorem! It's very cool. In some states, they catch speeding cars by taking pictures of the car at two spots on the highway, calculating their average speed, and then giving them a speeding ticket if it exceeds the speed limit. Using knowledge of this theorem (MVT) you could argue that perhaps your car's position function was not differentiable at some point along that road, and therefore they cannot prove you were in fact going at a speed above the speed limit! The Mean Value Theorem could save you hundreds of dollars!
• I may be wrong, but how would you argue that your car's position function would not be differentiable at some point along the road? If you had stopped somewhere it would make it not continuous, but then you would probably not get a ticket anyway, as your average speed would be lower as it would have taken you longer to go from point A to point B. In this sense I think the MVT would be more useful for the cops to give you a ticket rather than for the driver to prove innocence. Am I right?
(1 vote)
• Why does the function need to be continuous on a closed interval but differential at an open interval?