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### Course: Differential Calculus > Unit 1

Lesson 16: Intermediate value theorem- Intermediate value theorem
- Worked example: using the intermediate value theorem
- Using the intermediate value theorem
- Justification with the intermediate value theorem: table
- Justification with the intermediate value theorem: equation
- Justification with the intermediate value theorem
- Intermediate value theorem review

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# Intermediate value theorem

Discover the Intermediate Value Theorem, a fundamental concept in calculus that states if a function is continuous over a closed interval [a, b], it encompasses every value between f(a) and f(b) within that range. Dive into this foundational theorem and explore its connection to continuous functions and their behavior on intervals.

## Want to join the conversation?

- At3:58, Sal mentions there is an infinite number of continuous lines between point a and b. If there is a finite interval (i.e., [a, b]), wouldn't there be a limit to the number of continuous functions?(12 votes)
- These are one of the confusing things about infinities. If you've ever read The Fault in Our Stars you would know but that book isn't very mathematical. But say we think of every possible number between 1 and 2. So if we list them then we start 1...but where do we go next? Say 0.01, but obviously 0.001 should be it. But then 0.0001 is the next, and so on. There are an infinite number of numbers between 1 and 2, but lets say 1 between 1 and 3. There are more numbers between 1 and 3 than 1 and 2, even though they are both infinite. And in both these cases there is a limiting factors, for example between 1 and 2, between 1 and 3. Thus for a function constricted to the finite interval [a,b] it can do whatever it wants between a+b. An example of this is lets just say we have a function f(x) = cx with the interval x is in [a,b]. With c=any number. Now thats just a simple example as there are much more functions than just a proportional one. I think I've rambled too long now sorry(77 votes)

- Isn't the second graph in the video not a function since it does not pass the Horizontal Line Test?(6 votes)
- Vertical line checks to see if it is a function. This is because if we have multiple f(x) values for x then we don't have a function, after all, how do we determine which value we want??

The horizontal line test determines if each f(x) has a unique x value. This means that the inverse of f(x) will also be a function. Basically the horizontal line test checks if the INVERSE of f(x) is a function. The vertical line test checks if the original function is actually a function.(30 votes)

- So what if f of a equals to f of b? Doesn't it apply or something?(6 votes)
- If f(a)=f(b), then there are no values between them. So trivially, all of the values between f(a) and f(b) are achieved ("all of the none of them", as my logic teacher would say).(21 votes)

- 5:40what if you drew the graph on a ball? You could go up and around without ever having the second part of that theorem being true. Or would the math police put me in jail for that one?(6 votes)
- Assuming this is your first offence, I think you'll get off with a warning.

We really need a topologist to explain properly why you can't just "go up and around", and I'm afraid topology is above this level of mathematics. And mine.(10 votes)

- At7:12, what if L was greater than f(b)?

And if it is not possible so, why not?(6 votes)- The theorem requires 𝐿 to be between 𝑓(𝑎) and 𝑓(𝑏).

In this case 𝑓(𝑏) > 𝑓(𝑎) ⇒ 𝐿 < 𝑓(𝑏).(6 votes)

- How to prove this theorum? Sal mentioned that he is not giving a proof at7:38but it made me curious. Thanks(5 votes)
- There is a proof on the wikipedia page, https://en.wikipedia.org/wiki/Intermediate_value_theorem(7 votes)

- at3:44in the second graph, the function dives underneath f(b) does this matter to the theorem since L doesn't exist there?(4 votes)
- Not really. This theorum only applies to values between f(a) and f(b). The reason why it's ONLY those is because if a function is continuous, it MUST go over all the points in between, but it isn't guaranteed to go over points not in between them. So that doesn't affect the theorum; it's still true. Hope this helps!(5 votes)

- This to me doesn't seem like a fun theorem. It's just a rigorous definition of continuity.

Edit: typo(3 votes)- It actually isn't a rigourous definition of continuity. The more rigourous one would be the epsilon-delta definition.

Anyway, this theorem is probably not fun because well, it's kinda obvious. If I start from f(x) = 1 and end at f(x) = 5, it's obvious that I must have hit f(x) = 3 at some point, provided the function is continuous.

However, this theorem does have its uses. It can be used to prove that some polynomials have roots in an interval. For example, the polynomial f(x) = x^(4) + x - 3 is hard to factor. But, substituting x = -1 gives f(x) = -3, and substituting x = 2 gives f(x) = 15. This result directly shows that, as the graph is going from a negative f(x) value to a positive one, it must have crossed the x axis at some point. So, we can prove that the function has a root, though yes, we don't know what that root is. This theorem I used here (If a continuous function f(x) is defined as positive at some point and negative at another, it must be zero at some point between the initial two points) is actually a slightly modified version of IVT called the Bolzano's theorem.(5 votes)

- wow, i have so many questions. Is the reverse true? Like for a continuous function, for a given interval [a,b], does all the values lie between [f(a),f(b)] or ([f(b),f(a)]). Anyone??(3 votes)
- Not at all. sin(x) is continuous on [0, π], and sin(0)=sin(π)=0. If your claim was true, that would imply that sine doesn't reach any values besides 0 on that interval, which is clearly false; sine reaches every value between 0 and 1 on that interval.(4 votes)

- Why does it have to be a closed interval? Sounds like it would work for open intervals too(3 votes)
- Consider the function which is 0 for x≤0 and 1 for x>0.

On the interval (0, 1), it never reaches the value 1/2, which is between f(0) and f(1), even though the function is continuous on (0, 1).(4 votes)

## Video transcript

- [Voiceover] What we're
gonna cover in this video is the intermediate value theorem. Which, despite some of this
mathy language you'll see is one of the more intuitive theorems possibly the most
intuitive theorem you will come across in a lot of your mathematical career. So first I'll just read it out and then I'll interpret it and hopefully we'll all appreciate
that it's pretty obvious. I'm not going to prove it here. But, I think the conceptual
underpinning here is it should be straightforward. So the theorem tells us
that suppose F is a function continuous at every point of the interval the closed interval, so
we're including A and B. So it's continuous at every
point of the interval A, B. Let me just draw a couple of examples of what F could look like just based on these first lines. Suppose F is a function
continuous at every point of the interval A, B. So let me draw some axes here. So that's my Y axis. And this is my X axis. So, one situation if this is A. And this is B. F is continuous at every
point of the interval of the closed interval A and B. So that means it's got
to be for sure defined at every point. As well, as to be continuous you have to defined at every point. And the limit of the function
that is recorded at that point should be equal to the value
of the function of that point. And so the function is
definitely going to be defined at F of A. So it's definitely going to have an F of A right over here. That's right over here is F of A. Maybe F of B is higher. Although we can look at different cases. So that would be our F of B. And they tell us it is
a continuous function. It is a continuous function. So if you're trying to
imagine continuous functions one way to think about it is if we're continuous over an interval we take the value of the function at one point of the interval. And, if it's continuous
we need to be able to get to the other, the
value of the function at the other point of the interval without picking up our pencil. So, I can do all sorts of things and it still has to be a function. So, I can't do something like that. But, as long as I don't pick up my pencil this is a continuous function. So, there you go. If the somehow the graph I had to pick up my pencil. If I had to do something like this oops, I got to pick up my
pencil do something like that, well that's not continuous anymore. If I had to do something like this and oops, pick up my pencil not continuous anymore. If I had to do something like wooo. Whoa, okay, pick up my
pencil, go down here, not continuous anymore. So, this is what a continuous function that a function that is continuous over the closed interval A, B looks like. I can draw some other examples, in fact, let me do that. So let me draw one. Maybe where F of B is less than F of A. So it's my Y axis. And this is my X axis. And once again, A and B don't both have to be positive, they can both be negative. One could be, A could be negative. B could be positive. And maybe in this situation. And F of A and F of B it could also be a positive or negative. But let's take a situation where this is F of A. So that, right over there, is F of A. This right over here is F of B. F of B. And once again we're saying F is a continuous function. So I should be able to go from F of A to F of B F of B draw a function without having to pick up my pencil. So it could do something like this. Actually I want to make it go vertical. It could go like this and then go down. And then do something like that. So these are both cases and I could draw an
infinite number of cases where F is a function
continuous at every point of the interval. The closed interval, from A to B. Now, given that there's two ways to state the conclusion for the intermediate value theorem. You'll see it written in one of these ways or something close to one of these ways. And that's why I included both of these. So one way to say it is, well if this first statement is true then F will take on every value between F of A and F of B over the interval. And you see in both of these cases every interval, sorry, every every value between F of A and F of B. So every value here is being taken on at some point. You can pick some value. You can pick some value,
an arbitrary value L, right over here. Oh look. L happened right over there. If you pick L well, L happened right over there. And actually it also happened there and it also happened there. And this second bullet point describes the intermediate value
theorem more that way. For any L between the values of F and A and F of B there are exists a number C in the closed interval from A to B for which F of C equals L. So there exists at least one C. So in this case that would be our C. Over here, there's potential there's multiple candidates for C. That could be a candidate for C. That could be a C. So we could say there exists at least one number. At least one number, I'll throw that in there, at least one number C in the interval for which this is true. And, something that might amuse you for a few minutes is try to draw a function where this first statement is true. But somehow the second statement is not true. So, you say, okay, well let's say let's assume that there's an L where there isn't a C in the interval. Let me try and do that. And I'll draw it big so that we can really see how obvious that we have to take on all of the values between F and A and F of B is. So, let me draw a big axis this time. So that's my Y axis. And, that is my X axis. And I'll just do the case where just for simplicity, that is A and that is B. And let's say that this is F of A. So that is F of A. And let's say that this is F of B. Little dotted line. All right. F of B. And we assume that we we have a continuous function here. So the graph, I could draw it from F of A to F of B from this point to this point without picking up my pencil. From this coordinate A comma F of A to this coordinate B comma F of B without picking up my pencil. Well, let's assume that there is some L
that we don't take on. Let's say there's some
value L right over here. And, and we never take on this value. This continuous function
never takes on this value as we go from X equaling A to X equal B. Let's see if I can draw that. Let's see if I can get from here to here without ever essentially
crossing this dotted line. Well let's see, I could, wooo, maybe I would a little bit. But gee, how am I gonna get there? Well, without picking up my pencil. Well, well, I really need to
cross that line,all right. Well, there you go. I found, we took on the value L and it happened at C which is in that closed interval. So once again, I'm not
giving you a proof here. But hopefully you have a good intuition that the intermediate value theorem is kind of common sense. The key is you're dealing
with a continuous function. If you make its graph if you were to draw it between the coordinates A comma F of A and B comma F of B and you don't pick up your pencil, which would be true of
a continuous function. Well, it's going to take on every value between F of A and F of B.