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

So we're asked does the mean value theorem apply to h over the interval, and they actually give us four different intervals here. So we should separately consider them. This is the graph of y is equal to h of x.

So pause this video and see does the mean value theorem apply to h over any or all or some of these intervals. All right, now let's work through this together.

If the words mean value theorem are completely unfamiliar to you, I encourage you to look at the mean value theorem introduction or the existence theorem introductions on Khan Academy. But as a review for those of you who are familiar with the mean value theorem, the conditions are: If we're dealing with some closed interval from a to b, we have to be differentiable—I'll write just "diff" actually, just write it out: differentiable over the open interval from a to b.

And we have to be continuous over the closed interval from a to b. Or another way to think about it: If you're differentiable, you're definitely going to be continuous. So this second condition, just make sure that we're continuous at the endpoints of our closed interval. But if these conditions are met, then the mean value theorem tells us—I'll just visually draw it because this is a review here—so let's say that this is our function.

Let's say our function looks like this; this is a and this is b. And let's say that we meet these conditions over this interval. The mean value theorem tells us that there's going to be some value, let's call it c, in the interval where the derivative of c is equal to the average rate of change from a to b. So the slope of the secant line is the average rate of change.

And so you can see it visually, that looks like right about there. The slope of the tangent line would be the same, and so this would be the c that exists.

Now in this video, we're not trying to identify the c's; we're just trying to say, can we apply the mean value theorem? So let's look at this first interval from negative 5 to negative 1, the closed interval. So we're going from negative 5 to negative 1.

So are we differentiable over the interval, over the open interval? Well, we have this discontinuity over here; that's not going to make it differentiable. And of course, if we're not continuous, we're not going to be differentiable here. So this discontinuity here actually violates both of these conditions. And so for this first interval, we would say no, the mean value theorem does not apply.

Now the second one from negative one to three—so I'll do this in a slightly different color. So we're going to go from negative one right over there to three, and if you look over this interval, it looks like over that interval, over that closed interval, our function just really looks like a line.

And so it is both continuous and differentiable over that interval, and it makes sense that the mean value theorem applies. It's actually every c on this interval where the derivative is the instantaneous rate of change equal to the average rate of change because it looks linear over this interval. So the mean value theorem definitely applies over there.

Now what about from 3 to 7, the closed interval from 3 to 7? So when we look at this closed interval, we're definitely continuous over the interval, so we meet the second criteria. But are we differentiable?

Well, a good giveaway of a point that is continuous but not differentiable is right over here. Whenever you have these sharp edges, because what we don't have is a well-defined slope of a tangent line here. Maybe, you know, would it be that? Would it be that? Would it be that? And so because we aren't differentiable at that point at x is equal to 6, we aren't differentiable over this open interval from 3 to 7, and so the mean value theorem does not apply.

And you can even see that visually. The average rate of change between the endpoints of our interval—or if we want to think about the slope of the secant line— that's that right over there. And over the interval, we don't see any point where the instantaneous rate of change, where the slope of the tangent line, is equal to the slope of the secant line between the endpoints of our interval.

So once again, the mean value theorem would not apply. Now what about from negative 3 to 2? I'll do this in orange—so from negative 3 right over there to 2. Well, if we look at the open interval from negative 3 to 2, so we're not considering what happens at negative three and at two, over the open interval it does look like we are both differentiable and continuous.

So we're definitely differentiable over the open interval, but clearly, we're not continuous over the closed interval at negative 3. We actually are not continuous, and so because of that, the mean value theorem does not apply.

And you can actually even see that it would not apply because if you look at the slope of the secant line between our endpoints, that's the secant line right over there. And so you could see over that interval, there is no c between negative 3 and 2 where the slope of the tangent line or the instantaneous slope or the derivative is going to be the same as the slope of the secant line.

And that doesn't violate the mean value theorem because the mean value theorem just doesn't apply here; we haven't met the condition of being continuous over the closed interval.