Hello! Mr. Tarrou! Or I mean, BAM!! Anyway, we are going to be talking in this video about the Mean Value Theorem. Which is very similar to the Rolle’s Theorem, expect that the Rolle’s Theorem is specifically saying that if a function’s value is the same at two values of a and b, then that would create a secant line, which is horizontal. And it says that within that interval there is a value c, that makes the first derivative equal to zero. Basically, we are saying that there is a horizontal secant line, and somewhere within that interval, we can have a horizontal tangent line, IF we have a continuous function, and one that is differentiable. The Mean Value Theorem will say something that sounds very similar, but we don’t have to have slopes equal to zero. If f is continuous, on the closed interval [a,b], and differentiable on the open interval (a,b), then there exists a number “c” (and remember we are talking about independent variables – x values) There is this other point within that interval such that f prime of c… Remember derivative – slope. So the instantaneous slope, f prime of c, the instantaneous rate of slope, The slope of the graph at that point. Is equal to the function, f of b minus f of a, over b minus a. Which, we should be recognizing as just an average rate of change formula. So, what does that mean? That the instantaneous rate is equal to the average rate of change? Well, I’ve drawn a continous and smooth curve, going from a to b. The fact that it is smooth means it’s differentiable. I’ve got a non-example over here. It’s in orange, because my camera doesn’t pick up red that great. And i’ve got a sharp bend in here, so this would not be differentiable. So if it is smooth and continuous, then it says f of b minus f of a, over b minus a… So, we are just going to take these two x and y coordinates, find the slope of this secant line, and somewhere within that open interval, of (a,b) there is a third point of c, a third x value, such that if you plug it into the first derivative, and find the instantaneous rate of change, or the slope of the tangent line, the slope of the tangent line is going to be equal to the slope of the secant line, going through a, f of a and b, f of b. Over here, I’ve attempted to…it’s a little bit sloppy…I’ve sort of got this curve going on, but I did put a sharp bend Just to illustrate that…I drew picture. I don’t know what these actual equations are, making this function. But, if I draw a couple of points, and I mark it on the function, I get at the closed interval of [a,b] I can draw a secant line through those points. And I’ve identified a point on the function where f prime of c appears to have the same slope…the tangent line has the same slope as the secant line, but this would not fit the average of the Mean Value Theorem, because I’ve draw a function with a sharp bend. You cannot find the derivative at a sharp bend. So, within the open interval of (a,b) this function would not be differentiable, throughout that entire open interval. So, this is sort of a non-example of what this Mean Value Theorem is talking about. Let’s take a look at two examples. The first one is pretty basic. And the last example is a little bit more of an advanced question, and hopefully it will help you with a few of your homework problems. BAM! So here we have our first example. Determine whether the Mean Value Theorem can be applied to the function on the closed interval [a,b] If possible, find all values of c in the open interval (a,b) such that f prime of c, or the instantaneous rate of change, is equal to the average rate of change. So, let’s go ahead and find the derivative. If this is f of x is equal to x-squared minus 2x, minus 4. This is just a basic polynomial. We understand now that those are smooth and continuous curves. So this is going to be differentiable on the whole domain. I mean, from negative infinity to positive infinity. So, f prime of x is equal to 2x minus 2. Okay. So, that instantaneous rate of change needs to be equal to the average rate of change. Somewhere. And that value will be identified as c. So, we need to find the average rate of change. We are going to work through f of a minus f of b, over b minus a. This will be a, and this will be b – our lower and upper bounds. You want to read slope as left to right. So we want to have our larger value be the value of b. Since it will go first within the slope formula. The average rate of change formula. We have the average rate of change is equal to…5-squared, minus 2 times 5, minus 4. I’ve got a lot going on here, so I am going to use parenthesis to make sure I don’t make any sign errors. Minus, parenthesis – we are going to do the same thing with 3. So, squared, minus 2 times 3, minus 4. All over b minus a, or 5 minus 3. We have 5-squared is 25. 25 minus 10 is 15. 15 minus 4 is 11. MINUS, we’ve got 3-squared, which is 9. And we have negative 6. And 9 minus 6 is 3, and 3 minus 4 is negative one. All over 5 minus 3, which is 2. And 11 minus negative one, of course is addition. This becomes 12 divided by 2, or 6. So the average rate of change, the slope of the secant line going through the 5,11 and 3,-1… That slope of the secant line is equal to six. So, can we find the value of c, such that the slope of the the instantaneous rate of change, or the slope at that value of c is our value c, which makes the first derivative, the slope of the curve, the slope of the tangent line, equal to 6? Well, I think so! It’s a pretty basic formula here. So, f prime of x, or we are looking for c, the slope. Well, that is going to be 6 is equal to 2x minus 2. Add 2 to both sides and get 8 equals 2x. And x equals 4. So this actually becomes our “c” value. Where if I take the 4 and plug it into the derivative: 4 times 2 is 8, minus 2 is equal to 6. And that instantaneous rate of change, or the slope of the tangent line, there is a c value. That makes that derivative equal to the average rate of change. And that’s just the Mean Value Theorem. The only thing that makes these questions a bit more difficult is if you have a function that’s harder to find the derivative of, than my basic polynomial here.