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# Mean value theorem

The Mean Value Theorem 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 a point c in the interval (a,b) such that f'(c) is equal to the function's average rate of change over [a,b]. In other words, the graph has a tangent somewhere in (a,b) that is parallel to the secant line over [a,b]. Created by Sal Khan.

## Want to join the conversation?

• I'm having trouble thinking of when a function would be continuous over a closed period, but not differentiable over the same period. Is it just a redundant statement? If not, can anyone provide an example? •   The above answer is a good example of a function that is continuous but not differentiable at a point. If you really want to blow your mind, Google the phrase "nowhere differentiable function," and you will see examples of functions that are continuous but not differentiable at ANY point.

The Weierstrass function is perhaps the most famous example... As you might expect, these are very, very weird functions.
• Why aren't endpoints differentiable? Like let (x,f(x)) =endpoint & (x+h,f(x+h)) = some other point. Can't we use limits to figure out the derivative as h->0? Please explain! •   It would seem like we could, but to understand why it's not possible you need to return to the definition of the derivative as a limit.

We define the derivative as follows (also known as the difference quotient):

lim as h->0 of [f(x+h)-f(x)]/h

Notice that the limit is not specified as being left or right sided, so by the definition of the limit, the left sided and right sided limits as h->0 must exist and be equal for the derivative to exist.

Using this knowledge, we can see that although the limit will exist on the left side as we approach the rightmost endpoint, we cannot determine the value of the limit as we approach from the right because those values will not be included in the domain of the function f(x).

In summation: The left sided and right sided limit must exist and be equal for the derivative to exist at a given point, and by nature such two sided limits are not possible if we can only approach a point from one side. Therefore, we cannot take the derivative at the endpoints.
• At Sal says there exists a value c in the OPEN interval (a,b). In this case, why would the interval be open? • f'(c) must be on the open interval because a function defined on a closed interval is not differentiable at the endpoints because we don't have both a left hand and right hand limit to make sure that the derivative exists per the definition of a derivative. Thus, we cannot differentiate that the endpoints.

Technically speaking, we can do a one-sided limit at each of the closed interval endpoints and get what is called a one-sided derivative. But the MVT is talking about a ordinary derivative, not a one-sided derivative. Thus, x=c must be on the open interval (a,b).

There are other reasons why x=c lies on (a,b) not [a,b].
• How can the Mean Value Theorem be proved by using Rolle's Theorem? (Rolle's Theorem is a special case of the MVT; both f(a) and f(b) are equal to 0.) • Rolle's theorem states the following: suppose `ƒ` is a function continuous on the closed interval `[a, b]` and that the derivative `ƒ'` exists on `(a, b)`. Assume also that `ƒ(a) = ƒ(b)`. Then there exists a `c` in `(a, b)` for which `ƒ'(c) = 0`.

To prove the Mean Value Theorem using Rolle's theorem, we must construct a function that has equal values at both endpoints. The Mean Value Theorem states the following: suppose `ƒ` is a function continuous on a closed interval `[a, b]` and that the derivative `ƒ'` exists on `(a, b)`. Then there exists a `c` in `(a, b)` for which `ƒ(b) - ƒ(a) = ƒ'(c)(b - a)`.

Proof
Construct a new function `ß` according to the following formula:

`ß(x) = [b - a]ƒ(x) - x[ƒ(b) - ƒ(a)].`

Then `ß` is continuous on `[a, b]` and the derivative `ß'` exists on `(a, b)` (why?). We also have

`ß(a) = [b - a]ƒ(a) - a[ƒ(b) - ƒ(a)] = bƒ(a) - aƒ(b),`
`ß(b) = [b - a]ƒ(b) - b[ƒ(b) - ƒ(a)] = bƒ(a) - aƒ(b).`

Since the function `ß` satisfies the conditions of Rolle's theorem on `[a, b]`, there exists a `c` in `(a, b)` for which `ß'(c) = 0`. We have `ß'(x) = [b - a]ƒ'(x) - [ƒ(b) - ƒ(a)]`. Hence

`ß'(c) = [b - a]ƒ'(c) - [ƒ(b) - ƒ(a)] = 0`. This can be written as

`ƒ(b) - ƒ(a) = ƒ'(c)(b - a),`

and the proof is complete.
• Wouldn't the slope of the secant line of f(x)=|x| on the interval [-a,a] be 0? But there are no point where f'(x)=0. Is |x| an exception to the mean value theorem, or am I missing something? • Is this the same as Average Rate of Change? • It isn't the same, but uses the idea. MVT basically says that the average change over an interval would be equal to the instantaneous change at atleast one point on that interval.

In other words, if you covered 90 miles in 2 hours, your average speed is 45mph. There was atleast one instance where your speed at an instant was also 45mph.
• Isn't there like,a video explaining Rolle's Theorem having exercises and stuff about it?I've come across exercises that require knowledge of both MVT and Rolle's Theorem on my math book.I'm revising differntial and integral calculus for my math exam in 7 days so please if you see this answer.Yes,I've searched and searched for it and can't find it.Even the search shows nothing(which is a bit weird for such a fundamantel theorem in calculus such as Rolle's). • Our average rate of change over the interval from a to b, but what about b to a, I'm so confused.   