Examples establishing conditions for MVT

This table gives us a few values of the function g, so we know what g of x is equal to at these values right over here: x is equal to negative 2, negative 1, 0, and 1. It says Raphael said that since g of 1 minus g of 0 over 1 minus 0 is equal to negative 5, there must be a number c in the closed interval from 0 to 1 for which g prime of c is equal to negative 5. Which condition makes Raphael's claim true? They give us four choices here, so I encourage you to pause this video and try to work through it on your own.

All right, before we even look at these choices, let's just revisit what's going on over here. Raphael is saying since this, so he's looking at g of 1 minus g of 0, so this is g of 1 minus g of 0, that's this right over here, over 1 minus 0. So what is this? This right over here is the average rate of change between x equals 0 and x is equal to 1. Another way to think about it, this is the slope of the line that connects the point (0, 8) to the point (1, 3), and that average rate of change is negative 5.

And that, he's saying there must be a number c in the interval, in the closed interval from 0 to 1, for which the derivative of that x value, when x is equal to c, is the same thing as this average rate of change. So this is what the Mean Value Theorem is all about. It's this notion that, like we plotted a little bit just as a bit of a review. So we're going from (0, 8), which is maybe right over here, to (1, 3). So if that's (1, 3), it might look something like this. So we’re seeing the average rate of change, the slope of the line that connects those two points, is like this.

What the Mean Value Theorem tells us is if we have the conditions for the Mean Value Theorem, that let's say that our function does something like this, as long as our functions meet the conditions of the Mean Value Theorem, which essentially has to be differentiable over the open interval and continuous over the closed interval, then there has to be a point c where the slope of the tangent line at point c is equal to the average rate of change.

So over here, the slope of the tangent line looks the same as our average rate of change, right? Just like that. So this is the Mean Value Theorem. But we, in order to make this claim, that there— in order to apply the Mean Value Theorem to say there must be a c in the interval where the slope of the tangent line at x equals c is the same as the average rate of change, we have to feel very confident that we are meeting the conditions for the Mean Value Theorem, namely that the function is continuous over the closed interval and differentiable over the open interval from 0 to 1.

So let's look at the choices now. g is continuous at x equals 0 and x equals 1. Well, that by itself isn't sufficient. We also have to be continuous at all the points in between, so that by itself does not make his claim true. g is continuous over the closed interval from 0 to 1. So that's nice, but that doesn't ensure that we're differentiable over the open interval from 0 to 1. We have to be continuous and differentiable— we have to be continuous over the closed interval and differentiable over the open interval.

So we can rule that one out; that's not enough. g is differentiable over the open interval from negative 1 to 1, and so that includes this right over here— includes the open interval that we care about. This includes the open interval from 0 to 1. That's what we care about. So that is a check that we’re differentiable over the open interval and continuous over the closed interval from negative 1 to 1.

Well, once again, this includes the closed interval that we care about— this includes the closed interval from 0 to 1. And so this is looking good. What we care about is that we're differentiable over this interval, and for differentiable here, we're definitely differentiable here, and then we're continuous over this interval, so I like this choice.

This right over here says g is differentiable over the open interval and at x is equal to 0. So this is getting us close, because if you are differentiable, you are continuous. So one way to interpret this is that we are differentiable over the interval that is from 0 to 1 that includes the point 0.

Now this is close. If they said that we are differentiable over the closed interval from 0 to 1, that actually would have been sufficient, because if you are differentiable, then you are also continuous. But they did not say that, so we can rule this one out as well. We would have to know this last choice still does not make us confident that we are continuous at the point 1. That's what being continuous over the closed interval would have told us.

Let's do one more example that's a little bit different than this one. So here we are told that g is a differentiable function. Once again, they've given us the function sampled at some values of x. Max was asked whether there's a solution to g prime of x is equal to 2 over on the interval, or on the closed interval from 3 to 6. And so we see 3 here, we see 3 here, so we want to care about the closed interval from 3 to 6.

So the first thing he did, he says, "Okay, let's find the average rate of change from x equals 3 to x equals 6." So this is our change in our function, so g of 6 minus g of 3 over 6 minus 3. And you'll see g of 6 is equal to 1, g of 3 is equal to negative 5. This is 3, 1 minus negative 5 over 3. That's 6 over 3; it is indeed equal to 2. So that step makes sense, and the reason I'm going through it—they say—is Max’s work correct? If not, what is his mistake?

So let's look at step two: the Mean Value Theorem. So Max is now saying, "Okay, I found the average rate of change between x equals 3 and x equals 6," and then he says, "The Mean Value Theorem guarantees a solution where g prime of x equals 2 on this interval." So we have to be careful here, because he just immediately applies the Mean Value Theorem without establishing the conditions for the Mean Value Theorem.

So I would say, even though it does say that we're differentiable over it, I would make it— I think his mistake is that he did not make this clear. He should have said his step two should have been something like, "Because g is differentiable," differentiable over—or it sounds like it's a differentiable function—so we could say for all x, it's differentiable for all x. We can say the Mean Value Theorem—the Mean Value Theorem applies over this interval. If you wanted to be really careful, you could say because g is differentiable for all x, it's definitely differentiable over this closed interval, and if it's differentiable over the closed interval, it's also continuous over the closed interval.

And so the Mean Value Theorem definitely applies, and then say the Mean Value Theorem guarantees a solution. So the reason why it's important to stress is if you're taking a test, especially things like the AP exam, they definitely—if you're going to apply one of these really any theorems—they're going to want you to put down all the conditions for the theorem to ensure that you can apply it as part of your argument.