 Hi everyone, welcome to this lesson on Rohl's Theorem. Consider again the diagram we had in our lesson on the mean value theorem. This is a special case of the mean value theorem in which f of a is equal to f of b. If you take a look at the diagram here, it's the same one we had in our lesson on the mean value theorem just tilted to the right so that f of a is equal to f of b. 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 hopefully those conditions sound familiar to you from the mean value theorem. The difference here though compared to the mean value theorem is that secant line we have joining points a and b is now a horizontal line. You will notice once again that we have little tangent lines drawn at the points p sub 1, p sub 2, and p sub 3. And they are still parallel to each other as well as to the secant line that joins points a and b. Since the secant line that joins points a and b is a horizontal line, we know that the slope of that line as given to us by f of b minus f of a over b minus a is going to be equal to zero. But application of the mean value theorem tells us that there should be some place at which f prime of c, the slope of the tangent line at the point c is equal to the slope of the secant line that joins the endpoints. But remember in this case that secant line that joins a and b is a horizontal line. That implies that in this particular case it should be true that there is a location at which f prime of c is going to equal zero. That brings us to the statement of Rohl's theorem. Let f be a function such that f is continuous on the closed interval from a to b. f is differentiable on the open interval from a to b. And now a third condition that f of a has to equal f of b. Then there exists at least one c value in the open interval from a to b such that f prime of c is equal to zero. This was first published by the French mathematician Michel Rohl in 1691. He was originally one of the most vocal critics of the calculus claiming it yielded quote unquote erroneous results and was based on unsound reasoning. So ironic that we now have a theorem named after him. However, note that the converse of Rohl's theorem is not necessarily true. And if you compare this to the statement of the extreme value theorem remember that guarantees the existence of an absolute maximum and an absolute minimum. Rohl's theorem on the other hand guarantees the existence of at least one relative extrema on the open interval from a to b. So let's take a look at an example. We have the polynomial function f of x is equal to x to the fourth minus 2x squared and we are asked to find all values of c that satisfy Rohl's theorem on the interval from negative 2 to positive 2. As we did with the mean value theorem we need to make sure that the conditions hold true for this particular polynomial. So the first condition remember is that the function needs to be continuous on the closed interval. Well this is a polynomial function so it's going to be continuous everywhere. The second condition is that the function needs to be differentiable on the open interval. Once again we have a polynomial function so therefore it will be differentiable everywhere. Finally the last condition specific for Rohl's theorem is that it must be true that f of a is equal to f of b. In other words that the y values at the endpoints are the same. So we need to show what f of 2 and f of negative 2 are. If you substitute both of them back into the original equation you do get 8. So that condition does hold true. The three conditions for Rohl's theorem are therefore met and we can go ahead and find the values of c that satisfy Rohl's theorem on the interval from negative 2 to 2. So according to Rohl's theorem there should be a c value in that open interval from negative 2 to 2 so that f prime of c is going to equal 0. So let's go ahead and find the derivative. The derivative simply was 4x cubed minus 4x. If we substitute c in place of our x values is 4c cubed minus 4c is equal to 0. So let's go ahead and factor it. So we have a difference of two squares. The c values we get are c equals 0 from this c out here and then c equals either positive or negative 1. But remember our answers had to lie in the interval from negative 2 to 2. This is a case in which there's three different c values that are going to satisfy Rohl's theorem. 0 positive and negative 1. And if you take a look at the graph of the original equation you'll notice that it's at these x values they all have a horizontal tangent line as is the slope of the line that connects the end points at x equals 2 and x equals negative 2. Thank you.