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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 1

Lesson 16: Working with the 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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# Worked example: using the intermediate value theorem

Discover how the Intermediate Value Theorem guarantees specific outcomes for continuous functions. With a given function f, where f(-2) = 3 and f(1) = 6, learn to identify the correct statement that aligns with the theorem's principles and understand its real-life applications.

## Want to join the conversation?

- I didn't quite understand why can't the function take up values that are higher or lower than the boundary values i.e. the values of f(b) and f(a)?

If the value for f(a)=5 and that for f(b)=-3 then the value of f(c) has got to be between -3 and 5, why is it so? Why Can't it be like f(c)=6 or may be f(c)= -7, where c lies between a and b ?(18 votes)- The theorem says that the function achieves every value between f(a) and f(b). It does not say that it cannot reach other values as well. This is perfectly possible, as you observe.(31 votes)

- I don't understand. If y<3 or y>6 and it's still a continous function, then how could the theorem guarantee?(11 votes)
- In this example, the theorem only guarantees that we will have all values that are greater than 3 and less than 6 (such as 4, here). We could get a y that is less than 3 or greater than 6, but that isn't guaranteed by continuity.(27 votes)

- Isn't the intermediate value theorem just basic common sense or is there more to it?(9 votes)
- It certainly feels like common sense, especially with our intuition about graphs of functions. But a truly rigorous proof of it requires a bit of work. "Common sense" is often wrong in math, which is why we must write careful proofs, even of simple-sounding statements.(20 votes)

- So if the intermediate value theorem is not followed by a function in a given interval, can we say that it's not continuous?(5 votes)
- Yes, by contraposition. All continuous functions have the intermediate value property, so anything without the intermediate value property can't be continuous.(10 votes)

- Why is the second option ruled out ? why can't f(c) be equal to 0? A function can be anything . Imagine drawing a sort of parabola which drops down from 3 to 0 and then reaches 6. why cant that be not possible?(4 votes)
- The function
*could*be zero, but it's not guaranteed to be. We're looking at what values the function*must*achieve on the given interval.(8 votes)

- About a question on the cotinutity challenge in the next lesson, I still don't get understand why is ^3√x+1 defined for all real numbers and ^4√x+1 is not? I think both are defined for all x values such that x+1≧0.(4 votes)
- The issue here is that negative numbers can have odd-numbered roots, but not even-numbered roots. The cube root of -1 is -1, because (-1)*(-1)*(-1) = -1. However, you can't take the fourth root of a negative number, just as you can't take its square root (usually).

In the problem you're referencing, I at least was given an x value of -2. If your question wasn't different, then we're looking for the cube root and fourth root of (x+1). Since 1 - 2 = -1, the cube root is defined for x = -2, but the fourth root is not. Since the value is not defined, the function is not continuous at x = -2.(5 votes)

- what is the purpose of intermediate value theorem?(3 votes)
- The intermediate value theorem can give information about the zeros (roots) of a continuous function. If, for a continuous function f, real values a and b are found such that f(a) > 0 and f(b) < 0 (or f(a) < 0 and f(b) > 0), then the function has at least one zero between a and b.

Have a blessed, wonderful day!(2 votes)

- Why can't the function go lower than 3 or higher than 6?(2 votes)
- Actually, it is very possible for the function to exceed those values in either direction, especially beyond the concerned interval. The IVT only tells us that for this case, every value between 3 and 6 is represented at some point of the function.(4 votes)

- does the graph always stays between y=3 and y = 6?(2 votes)
- No we cannot guarantee that. But it will definitely take all the values between [3,6] at least for one point between [-2,1].(4 votes)

- why at least 1 point? is it ever possible to have more than 1 point? if yes can anyone give me an example(1 vote)
- It's definitely possible to have more than 1 point. In fact, Sal draws an example graph with three c's in [-2, 1] in which f(c) = 4.

In the case of the graph of a continuous horizontal line there would be infinitely many points.(5 votes)

## Video transcript

- [Voiceover] Let F be
a continuous function on the closed interval
from negative two to one where F of negative two is equal to three and F of one is equal to six. Which of the following is guaranteed by the Intermediate Value Theorem? So before I even look at this, what do we know about the
Intermediate Value Theorem? Well it applies here, it's a continuous function
on this closed interval. We know what the value of the
function is at negative two. It's three so let me write that. F of negative two is equal to three and F of one, they tell
us right over here, is equal to six and all the Intermediate
Value Theorem tells us and if this is completely
unfamiliar to you, I encourage you to watch the video on the Intermediate Value Theorem, is that if we have a continuous function on some closed interval, then the function must take on every value between the values at the
endpoints of the interval or another way to say it is for any L between three and six, three and six, there is at least one C, there is at least one C, one C, between, or I could say
once C in the interval from negative two to
one, the closed interval, such that F of C is equal to L. This comes straight out of
the Intermediate Value Theorem and just saying it in everyday language is look this is a continuous function. Actually I'll draw it
visually in a few seconds. But it makes sense that
if it's continuous, if I were to draw the graph,
I can't pick up my pencil, well that it makes sense
that I would have to take on every value between three and six or there's at least one
point in this interval where I take on any given
value between three and six. So let's see which of these
answers are consistent with that and we only pick one. So F of C equals four. So that would be a case
where L is equal to four. So there's at least one C in this interval such that
F of C is equal to four. We could say that. But that's not exactly
what they're saying here. F of C could be four for at least one C, not in this interval,
remember the C is our X. This is our X right over here. So the C is going to be in this interval and I'll take a look at
it visually in a second so that we can validate that. We're not saying for at least one C between three and six F
of C is equal to four, we're saying for at least
one C in this interval F of C is going to be equal to four. It's important that four
is between three and six because that's the value of our function and the C needs to be
in our closed interval along the x-axis. So I'm gonna rule this out. They're trying to confuse us. Alright. F of C equals zero for at least one C between negative two and one. Well here they got the interval
along the x-axis right, that's where the C would be between, but it's not guaranteed by
the Intermediate Value Theorem that F of C is going to be equal to zero because zero is not between three and six. So I'm gonna rule that one out. I'm going to rule this one out, it's saying F of C equals zero, and let's see, we're
only left with this one so I hope it works. So F of C is equal to four, well that seems reasonable because four is between three and six, for at least one C between
negative two and one. Well yeah because that's in
this interval right over here. So I am feeling good about that and we could think about
this visually as well. The Intermediate Value Theorem when you think about it
visually makes a lot of sense. So let me draw the x-axis first actually and then let me draw my y-axis and I'm gonna draw them
at different scales 'cause my y-axis, well let's see. If this is six, this is three. That's my y-axis. This is one, this is negative one, this is negative two and so we're continuous
on the closed interval from negative two to one and F of negative two is equal to three. So let me plot that. F of negative two is equal to three. So that's right over there and F of one is equal to six. So that's right over there and so let's try to draw
a continuous function. So a continuous function
includes these points and it's continuous so an
intuitive way to think about it is I can't pick up my
pencil if I'm drawing the graph of the function,
which contains these two points. So I can't do that. That would be picking up my pencil. So it is a continuous function. So it takes on every value. As we can see, it definitely does that. It takes on every value
between three and six. It might take on other
values, but we know for sure it has to take on every
value between three and six and so if we think about
four, four is right over here. The way I drew it, it looks
like it's almost taking on that value right at the y-axis. I forgot to label my x-axis here. But you can see it took on that value in for a C in this case
between negative two and one and I could have drawn that
graph multiple different ways. I could have drawn it something
like I could have done it and actually it takes on, there's multiple times it
takes on the value four here. So this could be our C, but once again it's between the interval
negative two and one. This could be our C once again in the interval between
negative two and one or this could be our C in between the interval of negative two and one and that's just the way
I happen to draw it. I could have drawn this thing
as just a straight line. I could have drawn it like this and then it looks like
it's taking on for only one and it's doing it right around there. This isn't necessarily
true that you take on, that you become four for at least one C between three and six. Three and six aren't
even on our graph here. I would have to go all
the way to two, three. There's not guarantee that
our function takes on four for one C between three and six. We don't even know what the function does when X is between three and six.