 In this video, we're going to talk about the mean value theorem. And so when we say the mean value theorem, we don't mean that it's a grouch. In this situation, mean means an average, so like mean, median mode. So it's the average value theorem. That's how you should interpret this because the mean value theorem is going to make a connection between the average rate of change with the instantaneous rate of change, the instantaneous rate of change of a function being its derivative, of course. The value theorem tells us the following. If f is a function that satisfies two conditions, first, f is continuous on the closed interval a to b. And two, f is differentiable on the open interval a to b. Of course, the difference between an open interval and a closed interval is the closed interval includes the endpoints x equals a and x equals b. The open interval does not include the endpoints there. Now, of course, if a function's differentiable, that implies continuous. Continuity is a weaker condition than differentiability. But we only require differentiability between an endpoints a and b. We are okay with the weaker assumption of continuity. Now, if these assumptions look familiar at all, it's probably because we saw in the last video about Roll's theorem that this sounds a lot like the assumptions for Roll's theorem. The Roll's theorem actually has three assumptions, one, two, and three. The first two are IBC right here. The third condition for Roll's theorem, the third assumption, I should say, is that f of a equals f of b. And this is the main difference between the mean value theorem and Roll's theorem. That in the mean value theorem, we don't require that f of a equals b. But we'll see in a moment that under the right assumptions, we can view Roll's theorem as a special case theorem. The Roll's theorem is a consequence of the mean value theorem if we set x, if we have f of a equals f of b, which we don't require that. So these, again, are assumptions. If f is continuous on the interval a to b closed and chibble on the interval a to b open, then there exists some number c that sits between a and b, such that the derivative f prime of c is equal to f of b minus f of a over b minus a. I want to write this equation very slightly different. What this is telling us is that dy over dx, the derivative is equal to delta y over delta x at this value here, where of course the average rate of change is the rate of change on the interval a to b and the derivative is evaluated at x equals c. So the mean value theorem tells you in a nutshell that with some appropriate assumptions about continuity and differentiability, somewhere the function, this function's instantaneous rate of change will equal its average rate for which I want to show you this diagram that I drew all over to give you some appreciation what's going on here. So our function f is going like this. So it's a continuous, differentiable function like so. If we consider the average rate of change, that's this friend right here, delta y over delta x. The average rate of change is the slope of the secant line. So if we take the line that connects the point a comma f of a with b comma f of b, if we connect those two dots, because two dots, two points form a line, those two dots we get are so-called secant line. The slope of the secant line is the average rate of change delta y over delta x. On the other hand, if we consider a tangent line, so right here we have an example of a tangent line, tangent line where the slope of the tangent line is equal to the derivative dy over dx. We are saying that our function will have some place where the tangent line is parallel to the secant line. The secant line is fairly one secant line in consideration here. And although we're guaranteed the existence of some points, its location, we have no information about it, except it's somewhere between a and b. But somewhere, the tangent line has to be parallel to the secant line. That's what the mean value guarantees here. And that's why we call it the mean value theorem here. We're saying that the average value, the average rate of change is somewhere equal to the instantaneous rate of change. Now, if you take this equation here, f prime of c is equal to f of b minus f of a over b minus a. If you clear the denominators times both sides by b minus a, well, then this side it clears out and you get this equation right here. It's an immediate consequence and sometimes it's a useful UV error. Now, before we prove the mean value theorem, I do wanna make sort of a statement about it. Again, try to give sort of like a physical interpretation. We've seen sort of a geometric interpretation already with tangent lines and secant lines here. So think of the following. Imagine you're driving your car down a highway or the interstate and some location in the United States or wherever you're located, it doesn't really matter. And so let's say that you're driving on the highway, the speed limit, let's say is 70 miles per hour, right? So if you're driving faster than seven miles per hour, then the local highway patrol police officers could give you a ticket for speeding. That's a very common law, at least in the United States. In which case, if you're caught speeding, they'll give you a ticket. Well, if you're on a highway and there's like just flat planes, then it's pretty easy to see things far down the road. So it's not really possible for the highway patrol officers to hide because everything's just so flat, nothing to hide behind. As you're driving, you could see them, over a mile away down the road. In which case, if you're speeding like 100 miles per hour or 120 miles per hour or something, then you're gonna slow down before you hit the officer. And so when they shoot you with the speed gun, it's like, oh, I was going 70 miles per hour. And then you continue on. Well, let's say that five miles down the road, there's another police officer. You're like, what are you doing? I can totally see you. The last one didn't work. Why are you gonna think that this one's gonna work as well? Whatever. In which case, you slow down again so that your speed limits again are 70 miles per hour. And then when you pass the police officer, this one pulls you over and gives you a ticket. I have a secret for you. Law enforcement officers work together. The reason there's two is because they're timing you, right? If there's two officers that are five miles apart, think about that for a second. If the officers are five miles apart and you drive between one officer to the other one, it only takes you three minutes to get from the first one to the second one, that gives you an average rate of change, an average velocity of 100 miles per hour. I switched from minutes to hours here, but you can do the arithmetic. If you do five miles in three minutes, then your average miles per hour by the mean value theorem, that tells you that if your average speed is 100 miles per hour, at least at one moment in time, your instantaneous speed, the speed on your odometer was equal to 100 miles per hour. So even though they didn't see you speeding, the only way you can get from one point to the other so quickly is that you must have been speeding. In which case your argument is to refute the assumptions of the mean value theorem, which would have to mean that, oh, my motion function wasn't continued. I was driving down the road and some wormhole appeared and then I teleported to the other side. That's how I got here so fast, officer. So by all means, you could try to make some spatial anomaly as your argument, but this basically cause your average speed is above the speed limit. That means your instantaneous speed must have also been above the speed limit. It's a nice little trick for the police officers there. The best way to avoid it is to, well, either drive through a wormhole or just not speed, all right? Let's talk about why the mean value theorem is true. We kind of have as a physical interpretation of it now. The mean value theorem is really an immediate consequence of Roll's theorem, which we had talked about in a previous video. I actually wanna put this picture back on the screen. So let's say that the average rate of change is just M, M as if it's the slope of the secant line. And let's actually let G of X be that secant line. So let G of X equal M times X minus A plus F of A. So this is the, so G of X is actually the secant line that passes through X equals A and X equals B for our function F right here. Make that clear. And let's consider the difference between F and G. So we have a new function H of X, which equals F of X minus G of X. Now notice that F of X by assumption is a continuous function from A to B. Great. G of X will likewise be a continuous function from A to B because it's a line. Lines are continuous. Functions of continuous functions are continuous. This implies that the function H of X will be continuous from A to B. A similar argument can be made about differentiability. So F by assumption is differentiable on the interval A to B. G of X is differentiable on A to B because it's a line. And therefore the difference of differentiable functions is differentiable. So we get that H will likewise be differentiable on the interval A to B. And in fact, we can compute the derivative of H. The derivative of H, since it's the difference, if we calculate H prime of X, this is just gonna be F prime of X minus G prime of X. But as G is a line, its derivative will just be its slope, which is M. So the derivative of H is equal to F prime minus H. The average rate of change right there. And so let's evaluate this derivative at the, excuse me, let's try to evaluate the function at the values A and B, the end points here. So if we plug in A into the function H, you're gonna get F of A minus, that looks like a G, doesn't it? F of A minus G of A right here, which G of A, as the secant line that passes through A comma F of A and B comma F of B, we get that G of A is just F of A. So we get F of A minus F of A, which is of course is equal to zero. Likewise, if you take H of B, same thing's gonna happen. You're gonna get F of B minus G of B, but G of B is just B, excuse me, G of B is just F of B, and so it's gonna be zero. You can see the exact calculation there if you want to, but the point of using the secant line is that H of B is also gonna equal zero. So notice H of A is equal to zero, H of B is equal to zero, in particular, H of A equals H of B. We created this function H so that A and B, X intercepts of that function. So notice that H, well, we have these three conditions here, right? H is continuous on the closed interval, it's differentiable on the open interval, and H of A equals H of B. So it satisfies the assumptions of Roll's theorem. So Roll's theorem applies and says that H prime of C is equal to zero for some value C that sits in between A and B. So this here just means that C but sits in between A and B, like so. But remember, the derivative of H is just the difference of the derivative of F and G, where the derivative of G is just M. So if H prime of C equals F prime of C minus H, then we can just add M to both sides, and we get that F prime of C equals M, where remember, M was this arbitrary change, F of B minus F of A over B minus A. So it might seem convoluted, in some degree, it's not the mean value theorem. It generalizes Roll's theorem, but in terms of the proof, we actually can prove the mean value theorem from Roll's theorem. So the special case of Roll's theorem, which we proved in the previous video, there. Let me give you an example of what the mean value theorem is kind of measuring here. Let F of X equal X cubed minus, which you can see drawn on yellow here on the screen. Here's our function F. And let's take the value A equals zero and B equals two. So noticing where this is on the function, you have zero zero, which is on the function. You also have X equals two. The point would be here on the graph. So we get zero zero. And this point right here, you're gonna get two comma, you just put it in the function, you'll get eight minus two, which is six. So two comma six. These are the points we're gonna consider right here. So our function F is a polynomial. So polynomials are continuous. So F here is going to be continuous on the interval zero two, and that's because polynomials are continuous everywhere. And likewise, since it's polynomial, it's also differentiable. It's differentiable on the interval zero to two. So what this shows us here is that the mean value theorem applies, it applies to F on this interval zero to two. So what the mean value theorem tells us is there exists some letters C. So this backwards E is just shorthand for exist. I'll write it out here. So there is a C. Such that, well, what do we know about C? We know that C sits somewhere between zero and two. And we know that, what else do we know about C? We know that F of two minus F of zero is equal to C prime, F prime at C, excuse me, times by two minus zero. So I'm just applying the mean value theorem right here. Which F of two, we saw was six, F of zero is zero, F prime at C right here. And then we get a two. So I'm gonna put the two in front, two minus zero is two. So we get this, or in other words, we get that the derivative at C is equal to six over two, aka three. All right, so there's some point where the derivative is equal to the instantaneous rate of change. So the mean value theorem guarantees that somewhere the derivative is equal to three. Which we can actually calculate the derivative easy enough. Right, the derivative here, just by the usual power rule, we're gonna get three X squared minus one. Is this ever equal to three? Well, let's see, we can add one to both sides. And we end up with three X squared is equal to four. Divide both sides by three. We get X squared is equal to four over three. Take the square root, we're gonna get the X equals two over the square root of three. We took the square root, there's actually two of them, plus or minus two over three. Now notice we're looking for a point that's between zero and two. Negative two over the square root of three is negative, so that definitely is not larger than zero. So we are safe in assuming it was positive. And so this is the number we're looking for right here. Number, and then when we found this number, we see that the C value is in fact equal to two, two over the square root of three. Now the mean value theorem wasn't necessary to find this number, because I mean, I found it just by solving this. The value theorem does is it guaranteed the existence of such this number. Now with this example, it might seem a little bit eh, right? Why does it matter? I mean, because we can see with our picture right here that the secant line from zero, zero to two, six is this line right here. And you have this tangent line, this tangent line where the point of tangency would be about right here. This is when C equals two over the square root of three. So the mean value theorem guarantees that there's some tangent line, in fact, there's only one, two over the square root of three that's parallel to the secant line. But the mean value theorem doesn't tell you where it is, it just guarantees the existence of algebraically where that point was. So why do we need the mean value theorem if we can solve for it? Well, the kicker is that there are many situations we can't solve for it, but the guaranteed existence of said tangent line is still pretty critical. We'll see that a little bit later in this. For many students as you work through the homework, often time what the homework questions ask you to do is to say, figure out what point, what is guaranteed by the mean value theorem and then actually find that point. It's not the best example of the mean value theorem, we're gonna see later in this lecture, the next few videos in fact, why the mean value theorem is so important, at least we'll begin to see it.