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### Course: Calculus, all content (2017 edition)>Unit 1

Lesson 10: Intermediate value theorem

# 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?

• At , 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?
• 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
• Isn't the second graph in the video not a function since it does not pass the Horizontal Line Test?
• 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.
• So what if f of a equals to f of b? Doesn't it apply or something?
• 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).
• what 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?
• 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.
• At , what if L was greater than f(b)?
And if it is not possible so, why not?
• The theorem requires 𝐿 to be between 𝑓(𝑎) and 𝑓(𝑏).
In this case 𝑓(𝑏) > 𝑓(𝑎) ⇒ 𝐿 < 𝑓(𝑏).
• How to prove this theorum? Sal mentioned that he is not giving a proof at but it made me curious. Thanks
• at in the second graph, the function dives underneath f(b) does this matter to the theorem since L doesn't exist there?
• 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!
• This to me doesn't seem like a fun theorem. It's just a rigorous definition of continuity.

Edit: typo
• 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.