Conditions for MVT: table | Existence theorems | AP Calculus AB | Khan Academy

So we've been given the value of h of x at a few values of x, and then we're told James said that since h of 7 minus h of 3 over 7 minus 3 is equal to 1.

So this is really the average rate of change between x is equal to 3 and x is equal to 7, between that point and that point right over there. Since that is equal to 1, there must be a number c in the closed interval from 3 to 7 for which the derivative at that value c is equal to 1.

So what James is trying to do is apply the mean value theorem, which tells us if the conditions apply for the mean value theorem. It tells us that if I'm going through two values—let's say this is a, let's say this is b—and let's say the function does something like this. If we meet the conditions for the mean value theorem, it tells us that there's some c in the closed interval from a to b where the derivative of c is equal to the average rate of change between a and b.

So the average rate of change between a and b would be the slope of the secant line right over there, and then we could just think about—well, looks like there are some points. The way I've drawn it, that point right over there seems to have the same slope, and this point right over there seems to have the same slope.

And so that's all what the mean value theorem is claiming: that there's going to be at least one c if we meet the conditions for the mean value theorem where the derivative at that point is the same as the average rate of change from the first endpoint to the second.

Now, what are the conditions for the mean value theorem to apply? And we've reviewed this in multiple videos. One way to think about it: If we're talking about the closed interval from 3 to 7, one condition is that you have to be differentiable over the open interval from 3 to 7. So that's the interval but not including the endpoints, and you have to be continuous over the entire closed interval, so including the endpoints.

And one interesting thing that we've mentioned before is that differentiability implies continuity. So, if something is differentiable over the open interval, it's also going to be continuous over this open interval. And so the second condition would just say: Well, then we also have to be continuous at the endpoints.

Now let’s look at the answers. So it says which condition makes James's claim true? So we have to feel good about these two things right over here in order to make James's claim.

Choice A: h is continuous over the closed interval from 3 to 7. So that does meet this second condition, but continuity does not imply differentiability. So that doesn't give us the confidence that we are differentiable over the open interval.

If you're differentiable, you're continuous, but if you're continuous, you're not necessarily differentiable. A classic example of that is if we have a sharp turn; something like that— we wouldn't be differentiable at that point, even though we are continuous there. So let me rule that one out.

So I'm going to rule that one out.

Choice B: h is continuous and decreasing over the closed interval from 3 to 7. No, that doesn't help us either because it still doesn’t mean you're differentiable. You could be continuous and then decreasing and still have one of these sharp turns where you're not differentiable. So we will rule this one out.

Choice C: h is differentiable over the closed interval from 3 to 7. This one feels good because if you're differentiable over the closed interval, you're definitely going to be differentiable over the open interval that does not include the endpoints. This is a subset of this right over here, and if you're differentiable over a closed interval, you're going to be continuous over it.

Differentiability implies continuity, so I like this choice right over here.

Now this last choice is the limit as x approaches five of h prime of x is equal to one. So they're saying the limit of our derivative as we approach five is equal to one.

Now that doesn't—we don't know for sure. This limit might be true, but that still does not necessarily imply that h prime of 5 is equal to 1. We still don't know that, and you know 5 is in this interval, but we still don't know just from this statement alone that there's definitely some c in the interval whose derivative is the same as the average rate of change over the interval.

So I would rule this one out as well.