Intermediate value theorem example | Existence theorems | AP Calculus AB | Khan Academy

Let F be a continuous function on the closed interval from -2 to 1, where F of -2 is equal to 3 and F of 1 is equal to 6. 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 -2: it's three. So let me write that F of -2 is equal to 3, and F of one, they tell us right over here, is equal to 6.

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 3 and 6, there is at least one C in the interval from -2 to 1 such that F of C is equal to L. This comes straight out of the intermediate value theorem. Just saying it in everyday language: 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 then it makes sense that I would have to take on every value between 3 and 6, or there's at least one point in this interval where I take on any given value between 3 and 6. So let's see which of these answers are consistent with that, and we only pick one.

So, F of C equals 4. That would be a case where L is equal to 4. So if there's at least one C in this interval such that F of C is equal to 4, we could say that. But they're trying to confuse us.

All right, F of C equals 0 for at least 1 C between -2 and 1. 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 0, because 0 is not between 3 and 6. So I'm going to rule that one out.

I'm going to rule this one out; it's saying F of C equals 0. And let's see, we're only left with this one, so I hope it works. So, F of C is equal to 4. Well, that seems reasonable because 4 is between 3 and 6 for at least one C between -2 and 1. Well, yeah, because that's in this interval right over here, so I am feeling good about that.

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. I'm going to draw them at different scales because my Y-axis—well, let's see, if this is 6, this is 3. That's my y-axis; this is 1, this is -1, this is -2.

And so, we're continuous on the closed interval from -2 to 1, and F of -2 is equal to 3. So let me plot that; F of -2 is equal to 3, so that's right over there, and F of one is equal to 6, so that's right over there.

So let's try to draw a continuous function. 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. I can't pick up my pencil; 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 3 and 6. It might take on other values, but we know for sure it has to take on every value between 3 and 6.

And so, when if we think about 4, 4 is right over here. The way I drew it, it actually 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 the case between -2 and 1, and I could have drawn that graph multiple different ways.

I could have drawn it like this. Actually, it takes on multiple times; it takes on the value 4 here. So this could be our C, but once again, it's between the interval -2 and 1. This could be our C, once again in the interval between -2 and 1. Or this could be our C in between the interval of -2 and 1.

That's just the way I happened 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 4 only once, and it's doing it right around there.

This isn't necessarily true that you take on—you take on that you become 4 for at least one C between 3 and 6. Three and six aren't even on our graph here. I would have to go all the way to 2, 3.

No, there's no guarantee that our function takes on 4 for one C between 3 and 6. We don't even know what the function does when X is between 3 and 6.