 Hi everyone and welcome to this lesson on the mean value theorem. Let F be continuous on the closed interval from A to B and differentiable on the open interval from A to B where A is less than B. And if you just take a moment to take in the graph and note some of the different characteristics about it you'll notice the line that joins points A and B is a secant line. And you will notice that there are little tangent lines drawn at p sub 1, p sub 2, and p sub 3. Barring in a human error that was made in drawing this graph what do you notice about these tangent lines? Well hopefully you'll notice that they're all parallel to each other as well as to the secant line AB. So what does the fact that they are all parallel imply? It implies of course that their slopes are equal. Therefore the slope of each tangent will equal the slope of the secant line from A to B. Let's write that in terms of an equation. The slope of each tangent of course can be denoted by the derivative. And the slope of the line that connects the secant AB is simply a regular slope equation. So F of B minus F of A all over B minus A. And that brings us to the statement of the mean value theorem. So F is required to be continuous on the closed interval from A to B and differentiable on the open interval from A to B. Those are the two conditions we are going to have to make sure applies for each case. What the mean value theorem tells us then is that there will exist at least one C value in the open interval from A to B such that the derivative at C, the slope of the tangent line at C will equal the slope of the secant line connecting the endpoints. This was proven by the French mathematician Joseph Louis Lagrange. And it's used mostly to prove other theorems in calculus. Some consider it the most important theorem in calculus. And it is closely related to the fundamental theorem of calculus that you'll learn later on. Let's take a look at an example problem. Suppose we have the function F of x equals 5 minus 4 over x. And you'll see the graph of it there to the right. We are asked to find all values of C in the open interval from 1 to 4 that will satisfy the mean value theorem. Well, remember those two conditions that we need to make sure do apply. The first one is that F needs to be continuous. Well, if you consider this function, the only place it's not continuous is at x equals 0. But that's not in the interval that we're concerned with. So we can conclude that this function is otherwise continuous on the closed interval from 1 to 4. The second condition that needs to be met is that it needs to be differentiable on the open interval from 1 to 4. Now again, this function will not be differentiable at 0. But once again, 0 is not in that interval from 1 to 4. So once again, we can conclude that this function is otherwise differentiable everywhere else. Our two conditions are met, so now we can go ahead and try to find the C that is being sought. So we'll need to find the derivative. Here's the original function. The derivative simply will be 4 over x squared. Let's go ahead and find F of b minus F at a all over b minus a. So in this case, b was 4, a was 1. We will need to substitute those into the original function to determine what F of 4 and F at 1 are. And you actually do get 4 and 1 again. So our slope of the line that connects the endpoints is 1. So remember what the mean value theorem tells us. It tells us that there should be a C value at which the derivative at that C value. So all I did was substitute C in place of my x in the derivative should someplace equal the slope of the secant line connecting the endpoints. So we have 4 over C squared equals 1 to solve. And if you go ahead and solve that, you'll find that C is equal to plus or minus 2. But remember that the interval we were concerned with was the interval from 1 to 4. So only positive 2 lies in there. So positive 2 is the only C value in this scenario that will fulfill the criteria for the mean value theorem.