Conditions for IVT and EVT: table | Existence theorems | AP Calculus AB | Khan Academy

We're told this table gives a few values of function f. It tells us what f of x is equal to that x is equal to 2, 3, 4, and 5. Which condition would allow you to conclude that there exists a maximum value for f over the closed interval from 2 to 4?

So pause this video and see if you can figure it out. You might already remember if we're trying to conclude that there exists something. Anytime people are talking about there exists something over an interval, we're probably dealing with some type of an existence theorem.

And if we're talking about the existence of a maximum or a minimum value for a function over an interval, that means that we're talking about the extreme value theorem. We are likely talking about the extreme value theorem, and in order to apply the extreme value theorem (sometimes abbreviated EVT), we need to know that our function is continuous over this closed interval.

So in order to do that, we need to know that f is continuous over our closed interval here, over 2 to 4, the closed interval from 2 to 4. So let's see which of these will allow us to conclude this. Because if we're able to conclude this, then we're able to apply the extreme value theorem, which tells us that there exists a maximum value for f over that interval.

All right, choice A says f is increasing in the open interval from 2 to 3 and decreasing on the open interval from 3 to 5. Now you could complete this; this doesn't tell us that we're definitely continuous over that interval, over 2 to 4. So I would rule that out. You could definitely still be discontinuous, and statement A still being true, so I'm going to rule that one out.

F is continuous over the closed interval from 2 to 5. Well, if we're continuous over the closed interval from 2 to 5, we're definitely continuous over a subset, over that, over the closed interval from 2 to 4. And so if we're able to make that, if we know that, then we can use the extreme value theorem. We've met the condition for the extreme value theorem to say that there exists a maximum value for f over that interval.

So I like this one. Let's just look at this last one. F is differentiable over the open interval from 2 to 4 and at x is equal to 2. So this is close because differentiability does imply continuity, but it isn't telling us what's happening at x equals 4. We could still be discontinuous at x equals 4; we're just differentiable for every point up to 4 because it's talking about an open interval.

So this is close. If they said differentiable over the closed interval from 2 to 4, then this would allow us to conclude because differentiability over an interval implies continuity. But I'm going to rule this one out because it does not ensure continuity at our right boundary at x equals 4.

Let's do another one of these. H is a differentiable function, and once again they've given us our phalli, what our function h is equal to at certain sampled x points, x values. Clyde was asked whether there's a solution to h of x is equal to negative 2 on the interval from the closed interval from negative 1 to 3.

This is his solution, and then we're asked is Clyde's work correct? If not, what is his mistake? And then they give us some choices. But before even looking at the choices, let us pause this video and see if what Clyde is saying makes sense. If it doesn't make sense, try to pinpoint where he messed up. What step did he make the wrong conclusion if he made any mistake at all?

All right, so the first thing he did is he said, look, negative 2, we want to see is there an x value where h of x is equal to negative 2? So first he says, well, negative 2 is between h of negative 1 and h of 3. So let's see h of negative 1 is 2, and h of 3 is negative 5.

And he is right that negative 2 sits right in here, so if we were to… Negative 2 is right in between these two values. It's right in between h of negative 1 and h of 3. Since h is differentiable, we know it is continuous on that closed interval. That's right; we were just talking about that.

Differentiability implies continuity; continuity does not always imply differentiability. You can't make that assumption, but differentiability implies continuity. If we're differentiable over an interval, we're definitely continuous on that interval. And so he's right here that so far, step one and step two, we can conclude that we are continuous over the closed interval from negative one to three.

And then he says the extreme value theorem guarantees a solution to h of x equals negative 2 on that closed interval. So this feels a little bit fishy because the extreme value theorem says that if we can meet our conditions, and we did for even the extreme value theorem, we're continuous on that closed interval.

It says that we're going to have a well-defined maximum and minimum value on that closed interval, but we don't care about minimum and maximum values here. What we care is that we take on an intermediate value in that interval and a value in between h of negative 1 and h of 3, possibly at each of those boundaries as well.

And so he's applying the wrong theorem; it should be the intermediate value theorem. I'll just abbreviate that as IVT is what he should be saying—the intermediate value theorem—because we are continuous on the closed interval. It says that we are going to take on every value between h of negative 1 and h of 3, and negative 2 is one of those values between h of negative 1 and h of 3.

And so if he just wrote intermediate value theorem instead, he would have been correct. So let's see which choice Cora is consistent with what we just wrote. He's definitely not correct. Step one wasn't correct, wasn't incorrect; two is between h of negative one and h of three. Step two was actually correct; we know that differentiability implies continuity. Step three is incorrect: yes, he applied the wrong theorem.