 Hi everyone! We're here today to talk about applications of the mean value theorem. Hopefully you remember what the theorem actually states. If we have a function f that is continuous on the closed interval and differentiable on the open interval, we're guaranteed some x value to exist in between the endpoints of the interval so that the derivative at that x value is going to equal the slope of the secant line that connects the two endpoints. So if you think about how that might have implications for applications, we talked about, and you might remember hopefully, that a derivative gives us an instantaneous rate of change but a slope will give us an average rate of change over a time period. So what we have with the applications of the mean value theorem is a way to figure out where it is or when it is that the instantaneous rate of change is going to equal the average rate of change. So let's look at a few examples. So the first one involves the height of an object that's given by this position function and we're asked to find the average velocity during the first three seconds. So what we're trying to find is our change in position over change in time. So we're substituting into the position function. I'll leave you to make sure you're getting the right answer. So that works out in the end to a negative 14.7 and the units of measure on this would be meters per second. So remember what it means when we have a negative velocity in this case. That tells us that the object is going downward, which makes sense because we know this object was dropped from a height of 500 meters. So that very much does make sense. So what we're asked to do in the second part then is to use the mean value theorem to verify that there's going to be some point in time during those first three seconds where the instantaneous velocity is going to equal that average velocity we just found. So what we're finding is the derivative of s and I'm going to call it c. That's often the variable used to indicate that value that is guaranteed, the x value that is guaranteed by the mean value theorem. We want to see where that derivative is going to equal the average rate of change. So if you go ahead and find the derivative of the function that would give us a negative 9.8 and remember I'm calling it c instead of t. If you were to call t that's fine, no harm done. We can always just call it that's the c value in the end. So no harm. So dividing we get that c is one and a half seconds. So what that's telling us is that at the one and a half second mark the instantaneous rate of change at that particular time is equal to that average rate of change that we found over that time period from zero to three seconds. So let's look at a more business kind of application. So in this case we have a company that's selling a new product. The number of units sold is given by that s equation there and t is the time in months. And so the first thing we're asked to find is the average rate of change in the number of units sold. So that's going to be change in s over change in t. Now it's over the first year but remember we were told the time was in months. So in this case we have to do s of 12 minus s of zero. So an easy way to do that might be to use our graphing calculators and use some of the built-in features to do that and just make it a little bit easier. So let's go ahead and switch to the calculator and let me go ahead and put in that equation. And I'm going to do the up and down fraction for this. So an easy way to evaluate that remember we can use the calculators kind of built-in function notation so to speak. So let's go back to our quit screen and I'm going to pull up y-vars and my y1 because I know I want to take that function and substitute in 12. So this is essentially f of x notation y1 of 12. And I'm going to do that then at zero and subtract the two. Now because I did do that up and down fraction it probably is going to give me a fraction but that's okay. We know we have to divide that by 12. So what I could do is just divide it by 12 point. That's going to force there to be a decimal in there when it gives me the answer and now it gives me a nice decimal answer. I just thought that might be easier because remember we are talking about how many units sold so it might be easier to think of that in terms of a decimal. So what we have for here let's go back is that this is equal to 64.286 and that would be units per month. So this is telling us since it's positive the rate of change is obviously increasing at a rate of 64 just over 64 units per month so that's a good thing. It means they're selling more and more every month so that's I'm sure what they want to have happen. Now the next question again including the mean value theorem is asking us when it is during that first year that the instantaneous rate of change is going to equal the average rate of change and the number of units sold. So what we want to find is where that derivative, again I'm going to call it C, is equal to that 64.286. Now this is another great opportunity to maybe use our calculator to do some of the work for us because of course it will graph the derivative for us and we could really set that equal to this value that we found and just find the intersection. So let's go ahead and try it that way. So one thing I might want to do is take this 64 value and go ahead and store that. I'm just going to call it A because of course for as much accuracy as we can get we want to preserve that decimal. So let's go back under y equals. Alright so we want the calculator to graph the derivative of what we have under y1. So remember we can use our n derived function math 8 to do that. And I'm going to pull up reference y1 here and I want it to do the whole graph so x equals x and while I'm thinking of it I'm going to go ahead and turn off that first graph because I really don't care what the first graph looks like. I really just want to have the derivative equal to that value. So under y3 I'm going to pull up A. So it looks like that. Now I probably need to set up my window. I have a hunch if we were to do zoom 6 we're not going to get very far. So let's consider what a good window might be. I'm thinking 0 to 12 because it is during that first year that we were trying to find a value in there. So that would be 12 months. Remember that y3 we had that 64.286 so we know our y maximum has to go up at least that big. So I just set it to y max of 70. Alright so let's look at the graph. It might take a little minute to think. So here's the derivative graph and here comes the 64.286. So we want to find that intersection. So if you hit trace we'll go ahead over and trace to that. That's pretty close. So let's go ahead and run our intersection function and there's our answer 3.292. So it would be at just over three months, the three month mark, that the instant tiniest rate of change at that particular point in time is equal to the average rate of change over that one year period. And as you do more and more real life applications like this, you know using your calculator to help you get to these answers really definitely do that. That's perfectly acceptable. Really what you'd have to make sure you show is what you see here that you're setting the derivative equal to that numerical value. That's really explaining what it is you're doing. You're just then using the calculator to assist you to solve it. So let's look at another one. We're talking about the growth of a red oak tree. And the growth of the tree is approximated by this cubic polynomial. So these big decimals. And t is the age in years, two years old to 34 years old. So we first want to find the average rate of change in between years 20 and 34. So we're trying to find change in G over change in T. So I'm thinking we might want to evaluate this similar to how we did the last one and use our calculator to do that. So let's go ahead and get the equation set up under y equals. And then we can go ahead and figure this out by using the calculator's features. So let's go back under y equals. I'm going to just delete out that y1. I'm going to leave the other two for now. I might need those. So just be careful as you're typing this. So let's go ahead and find that slope value we need. So let's go back to our quit screen. Let me clear this out. So similar to what we did in the last problem, we're going to set up the numerator by using the build-in function notation. So we're doing y1 at 34 minus y1 at 20. We get that and we need to divide that by 34 minus 20. So that's 14. Nice value. We don't even have to store that. We can just use that as it is. So 1.148. So that would be feet per year. So again, make sense that it's positive because obviously the tree is going to be growing. Not shrinking as it gets older. So that very much makes sense. So our last question then is going to ask us where in there from 20 to 34 years does the instantaneous rate of change in the tree's height equal the average rate of change? So if you think of it, what it is we're trying to find, the derivative of the function at C is equal to that 1.148. So this is another one. We could use our calculator in a few different ways. We could do it like we did the last one and find the intersection. We already had some of those values on y2 was already set up for our derivative. So we could do it that way. Another option, this is a polynomial. So we could actually find the derivative by hand and then maybe solve it. I think when we do it that way, we'll have a quadratic so we could even use the quadratic formula. So maybe let's do that first and then we can confirm our answer graphically. So if we go ahead and find the derivative, we'll have a negative 0.009. And remember I'm going to use C. If you use T squared, that's okay. Plus 0.274 C plus 0.458. And that's equal to the 1.148. So we need to subtract that 1.148 over to the other side. Because remember if we're going to want to do quadratic formula, this quadratic does have to be set equal to zero. So when we subtract that away, we get a negative 0.69. So you should have the quadratic formula programmed into your calculator as a program. So let's go ahead and use that. So I'm going to go ahead and pull it up. I'm going to use the one I have called Quadrat. So remember it's asking you, it's prompting you for the coefficients. So the leading coefficient is that negative 0.009. A linear coefficient is the 0.274. And your constant is a negative. Don't forget to make a negative 0.69. And you'll notice we get two different answers. Notice only that second one falls. Remember we were trying to find somewhere in between the years, like 20 years old and 34 years old. So it's looking like that 27.674 is really the only quote unquote good answer. And the second other one is a valid answer, but it just isn't in the interval that we were given to work with. So we would say it's at 27.674 years. All right, so why don't we confirm that graphically just to double check? So let's go back to the calculator. Let's go under y equals. So what we want to do is set the derivative equal to that 1.148. So once again, I'm going to turn off my first function. Again, I don't necessarily really care what it looks like. There's y2 is already set up for my derivative, so we're good to go there. Let me change y3 to the 1.148. Might have to adjust my window. So y2 is my derivative and I'm setting that equal to the 1.148. So let's go under the window. Remember we were talking in between years 20 and 34. So why don't we just go ahead and set up our x's as that? Y min I'll keep as zero. Now remember I'm trying to see where that derivative intersects with 1.148, so I really don't have to go up too high for my maximum. So let's just maybe to go to three and just go ahead and hit graph. So here's your derivative graph coming first. And here comes the 1.148. So I want to find that point of intersection. We should better get that 27.674. There we go, got it. So a great way to confirm graphically what we were able to find algebraically. Hopefully this provides you with some really good examples of the mean value theorem. The big thing you want to take away from this is that when you are dealing with real-life applications, really what you're finding is where the instantaneous rate of change, your derivative, is equal to the average rate of change which you're getting from the slope of the secant between the endpoints.