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Mean value theorem

AP.CALC:
FUN‑1 (EU)
,
FUN‑1.B (LO)
,
FUN‑1.B.1 (EK)
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.

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  • aqualine ultimate style avatar for user Robby Olivam
    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?
    (44 votes)
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    • hopper cool style avatar for user Mr. Jones
      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.
      (86 votes)
  • blobby green style avatar for user gyanjit.m
    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!
    (18 votes)
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    • blobby green style avatar for user rashaveraka
      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.
      (52 votes)
  • blobby green style avatar for user Ajeet Dhaliwal
    At Sal says there exists a value c in the OPEN interval (a,b). In this case, why would the interval be open?
    (10 votes)
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    • piceratops ultimate style avatar for user Just Keith
      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].
      (14 votes)
  • duskpin ultimate style avatar for user K
    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.)
    (5 votes)
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    • leaf grey style avatar for user Qeeko
      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.
      (15 votes)
  • blobby green style avatar for user zwhs.basi.dh
    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).
    (3 votes)
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  • duskpin sapling style avatar for user Alvin
    Our average rate of change over the interval from a to b, but what about b to a, I'm so confused.
    (3 votes)
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  • mr pink red style avatar for user Casper Køneke
    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?
    (4 votes)
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  • blobby green style avatar for user qudrat mommandi
    So the derivative of the function will be equal to the slope of the secant line between the two points at some point in the function?
    (2 votes)
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    • male robot hal style avatar for user Jesse
      You are correct. For a continuous function defined on a closed interval, there is a point on the interior of the interval such that the derivative at that point is the same as the slope of the line connecting the endpoints.
      (5 votes)
  • piceratops seedling style avatar for user parth
    The Mean Value Theorem called so....WHY ??
    (2 votes)
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  • aqualine seedling style avatar for user XDLAWL (Currently offline)
    I'm confused about derivatives - how come sometimes when you take a derivative you leave a dy/dx or an x' after taking chain rule, but sometimes you don't.
    For example:
    Taking the derivative of V = 4/3 * pi * r^3 leaves you with V' = 4 * pi * r^2 * r',
    But taking the derivative of something like f(x) = x^3 leaves you with f'(x) = 3x^2. Am I missing a piece of important context?
    (2 votes)
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    • leaf grey style avatar for user Alex
      Yes, and it's really a problem with the Lagrange notation. V' is not dV / dr, but dV / dt. f'(x) is still df / dx. However, for example, if you had y = x^3, then dy / dt = 3x^2 * dx / dt.
      (3 votes)

Video transcript

Let's see if we can give ourselves an intuitive understanding of the mean value theorem. And as we'll see, once you parse some of the mathematical lingo and notation, it's actually a quite intuitive theorem. And so let's just think about some function, f. So let's say I have some function f. And we know a few things about this function. We know that it is continuous over the closed interval between x equals a and x is equal to b. And so when we put these brackets here, that just means closed interval. So when I put a bracket here, that means we're including the point a. And if I put the bracket on the right hand side instead of a parentheses, that means that we are including the point b. And continuous just means we don't have any gaps or jumps in the function over this closed interval. Now, let's also assume that it's differentiable over the open interval between a and b. So now we're saying, well, it's OK if it's not differentiable right at a, or if it's not differentiable right at b. And differentiable just means that there's a defined derivative, that you can actually take the derivative at those points. So it's differentiable over the open interval between a and b. So those are the constraints we're going to put on ourselves for the mean value theorem. And so let's just try to visualize this thing. So this is my function, that's the y-axis. And then this right over here is the x-axis. And I'm going to-- let's see, x-axis, and let me draw my interval. So that's a, and then this is b right over here. And so let's say our function looks something like this. Draw an arbitrary function right over here, let's say my function looks something like that. So at this point right over here, the x value is a, and the y value is f(a). At this point right over here, the x value is b, and the y value, of course, is f(b). So all the mean value theorem tells us is if we take the average rate of change over the interval, that at some point the instantaneous rate of change, at least at some point in this open interval, the instantaneous change is going to be the same as the average change. Now what does that mean, visually? So let's calculate the average change. The average change between point a and point b, well, that's going to be the slope of the secant line. So that's-- so this is the secant line. So think about its slope. All the mean value theorem tells us is that at some point in this interval, the instant slope of the tangent line is going to be the same as the slope of the secant line. And we can see, just visually, it looks like right over here, the slope of the tangent line is it looks like the same as the slope of the secant line. It also looks like the case right over here. The slope of the tangent line is equal to the slope of the secant line. And it makes intuitive sense. At some point, your instantaneous slope is going to be the same as the average slope. Now how would we write that mathematically? Well, let's calculate the average slope over this interval. Well, the average slope over this interval, or the average change, the slope of the secant line, is going to be our change in y-- our change in y right over here-- over our change in x. Well, what is our change in y? Our change in y is f(b) minus f(a), and that's going to be over our change in x. Over b minus b minus a. I'll do that in that red color. So let's just remind ourselves what's going on here. So this right over here, this is the graph of y is equal to f(x). We're saying that the slope of the secant line, or our average rate of change over the interval from a to b, is our change in y-- that the Greek letter delta is just shorthand for change in y-- over our change in x. Which, of course, is equal to this. And the mean value theorem tells us that there exists-- so if we know these two things about the function, then there exists some x value in between a and b. So in the open interval between a and b, there exists some c. There exists some c, and we could say it's a member of the open interval between a and b. Or we could say some c such that a is less than c, which is less than b. So some c in this interval. So some c in between it where the instantaneous rate of change at that x value is the same as the average rate of change. So there exists some c in this open interval where the average rate of change is equal to the instantaneous rate of change at that point. That's all it's saying. And as we saw this diagram right over here, this could be our c. Or this could be our c as well. So nothing really-- it looks, you would say f is continuous over a, b, differentiable over-- f is continuous over the closed interval, differentiable over the open interval, and you see all this notation. You're like, what is that telling us? All it's saying is at some point in the interval, the instantaneous rate of change is going to be the same as the average rate of change over the whole interval. In the next video, we'll try to give you a kind of a real life example about when that make sense.