Understanding Extrema in Calculus

which is called extrema on an interval. Okay, we'll talk about what extrema means. I think I discussed it, but now we'll get to talk about it. We'll get to use the process. All right, so this is topic 5.2. Again, this is where you'd want to go for those AP daily videos. Really good examples on there. The extreme value theorem, global versus local and critical points. It's just going to be some definitions right here in the beginning. So remember that global extrema? are your absolute extrema you're right i'm not thank you there we go thanks okay there you go so we have global extrema which is going to be the biggest our highest point on the entire graph or the lowest point on the entire graph, but it's the entire graph or over a domain. Okay. They give us a domain in the problem. Then we have to look over that domain as well. And local extrema, that could be highest and lowest points in a little neighborhood. They don't have to be the highest and lowest points of the entire graph. They just have to be the highest and lowest points in the little neighborhood, which is usually means just little area around it. Okay. And before it and little area after it and critical points. Critical points are going to be all the possible locations. So locals, another name for locals, relative. Another name for global is absolute. Critical points are going to be the possible locations for relative max and mins. The reason I say possible locations is because we have to test them. Possible locations for relative. Max and mins. And the easiest way to say max and mins together is to say the word extrema. Extrema means max and mins. So anytime you see the word extrema, that's what it means. Just think of max and mins. It's exactly what it means. And those are going to be critical points. And we'll get to discuss all this when we actually go through the process. OK, so I'll keep reading up here. The maximum and minimum values, max and min, of a function are called the extreme values or extrema. singular extremum of a function. The process used to finding them is referred to as optimization. So if you wanted to Google something like a video on something, you could Google optimization to see extra videos. There will be times when we want to find the max and min for x on the entire domain and at other times on a particular interval. In this lesson, we will learn how to use calculus to find extrema rather than a calculator. OK, so what that means, sometimes they'll give us a domain, sometimes they'll give us an interval to look at for the axis, and sometimes they won't. Sometimes we'll have to look at the entire X line and sometimes we just get to look at piece of the X axis. If a function here's a definition for global absolute extreme of a function on a normal eye. Oh, sorry, if F is a function on the interval I. Then y equals f of c is the absolute or global maximum on i, if and only if f of c is greater than or equal to f of x for all x and i. And a global min, same thing except for less than. Basically, what this is saying in terms of symbols is that if you got the highest y value over the entire interval, that's your absolute maximum. If you have the lowest y value, that's what these things are, y values. If you have the lowest y value over the entire interval, then that's called the absolute minimum or global minimum. That's all it's saying in symbols right there. Example number one. The graph of y equals h of x is shown below. Determine the extrema, max and mins, of h of x on the interval between a and f. Now, this is a closed interval. There's one way to write it. You can write it with inequalities like this. But you're going to see this written a lot of times. In fact, you've seen it already with closed brackets. And I can tell you they're closed brackets for this interval because of the equal signs. I'm going to draw those closed brackets very well. They're supposed to be closed brackets, not parentheses. And they're closed because of the equal signs. You can equal A and you can equal F. So we close them. And again, on is a key word right here. It's not at a point. It's on this interval. They even put the word interval for us. Okay. So we're going to go ahead and... Determine I already made kind of a little chart down here to get ready. I'm going to put all the relative mins, all the relative maxes, absolute mins and the absolute maxes and I'll even tell you where they're located. Okay, so the relative mins are the y values. This function is called h of x. So we'll name all the y values, like the y value for this one. Let me just give you an example. The y value for this point we'll call h of a, because that represents the y value on the graph of h when x is a. We don't know what that height is, but we'll call it h of a. So we'll call the height here for this point h of b. The height up here will be h of c. and so forth. And those y values are the relative mins, relative maxes, absolute mins, or absolute maxes. For the relative mins, we have to pick the lowest point in a little neighborhood. It has to be continuous there. It can be a sharp point like this. You see that sharp point? Even though we can't take the derivative there, you guys, that is still relative min because it is still the lowest point in a little neighborhood. It's the lowest point in a little neighborhood from the graph to the left and to the right. That's all you need for a relative. Just a little neighborhood around it. To the left and to the right of it. See, this cannot be a relative extrema. End points cannot be relative extremas because there is no neighborhood to the left of this point. And there's no neighborhood to the right of this point. So end points, these numbers right here, can never be relative extrema. Now, they can be absolute extrema. They can be absolute maximins. They can be the highest point. They don't look like they're in this case, but they can be. If they were to just draw on this point up here instead at A, then that would have been the highest point. And that would have been the absolute max. So endpoints can be absolute extrema, but they can't be relative extrema. So that's why I skip this one when I'm doing relative mins and I go right to here. This is a relative min. My relative min is going to be H of B. The relative min is H of B. Whatever the Y value is when X is B. That's going to be my relative min. And a lot of times they want to know what's the relative min and where is it located. We always give them where it's located on the x-axis. So we would say h of b or the y value, whatever that happens to be, in this case it's going to be a positive number, something like a positive 3. So the relative min is positive 3 at x equals b. We got like a negative 1. And that's 1 of the relative mins. There's another one. Okay, but notice how I wrote it. The y value at, that's my symbol for at. Everybody's symbol for at. You can write that. At x equals b. So I'm doing two things. I'm answering two questions. I'm telling them what the relative min is. It's this value right here, the y value. And I'm telling them where it's located. It's at, just look on the x-axis right here and you'll find it. Most of the time, they just ask for the relative min part. So they just want the y value. But I always tend to put both. That way I don't even have to. to think about it. Just a habit now. There's one more relative min. There's one more lowest point in the little neighborhood. Check it to the left. Check it to the right. This is the lowest point in its neighborhood, and that's going to be h of d. Whatever the y value is, when you plug a d into this function, think of the output. When you plug the input in, what's the output? At x equals d. And those are your two relative mins. Relative maxes. There's also two relative maxes in this. Two points where we have the highest in this little neighborhood. Right here at c. So we have h of c at x equals c. Whatever that y value happens to be. And right here h of e at x equals e. Now later on, we're not going to get the graph anymore, you guys. We're just going to get the function or maybe even a table, which is worse than a function, okay? But we're going to have to find these relative max and relative mins on our own. I think you might be able to guess how we can find these. These ones that are on the nice smooth parts of the curve. Think about the tangent lines at those points. What's special about the tangent lines at those points? there is zero. That's exactly right. The slope of the tangent lines is zero. So in order to find those locations, think about what we're going to do. When we get a function, we're going to find the derivative and we're going to set it equal to zero. And if we solve that equation, it'll give us these three values. And then we'll have a test to perform to see if they're relative max or relative mins. Now, the problem becomes, how do we find this one? Because the tangent line does not exist at sharp points. There is no tangent line at that point. The derivative. does not exist at b. If you were to try to plug a b in to the derivative function, you would not get an output out. There'd be some sort of problem when you try to plug a b in to the first derivative. I'm not saying to the original function because you plug it into the original function, you're going to get a number out. You'll get h of b, whatever that y value happens to be. So it's okay to plug into the original function this number. But it's not okay to plug into the first derivative. You won't get a number out when you plug that number into the first derivative. You won't get 0. You won't get negative 1. You won't get positive 1. You won't get anything. There'll be some sort of problem when you try to plug in b, like division by 0. If you try to plug this number into the first derivative, you might get a division by 0 problem. And you can't divide by 0. So this one's going to be found a different way. And now I will tell you that the most common relative maximums that we get on the AP test are these types of nice, smooth curves. And those can be found really easily by taking the first derivative and setting it equal to zero. This one's a little bit harder to think about. But I will tell you, most of them on the AP test are the nice, smooth ones. So you will find most of them this way. And we'll show you how to find this one coming up in the process. OK, now here we go. Absolute maximums. Here's the thing. Whenever you have a closed interval, you guys, like we do on this one, either from this symbol or the equal signs or these symbols with the closed brackets, we know we have a closed interval. So we're going to have endpoints. So even if this graph continued on, you guys, even if the graph continued on, we would still look at only the parts of the graph between A and F, and it would be a closed interval. It'd be the exact same graph that she gave us. Even if this graph continued on, we would still look at the exact same pieces of the graph and we wouldn't change anything. It'd still be a closed point here and a closed point there, even if the graph continues on. And I want to tell you that because we're going to see this theorem pretty soon. Whenever you have a closed interval that you're looking at, you guys, there's always going to be an absolute min and an absolute max. So what does that mean? That means whenever you have a closed interval, whenever you have a closed endpoint here and an endpoint here, that always means you're going to have a highest point on your graph and you're going to have a lowest point on your graph. You can't say that if these are open, but it's guaranteed if they're closed. If they are closed intervals, if they're equal signs on the inequalities. then there's always going to be one absolute min and one absolute max. Now, the absolute mins, I should say the absolute min, may occur at two different spots at two different x values because two of them might be the lowest point or maybe even more. But there's only going to be one y value that's the lowest and one y value that's the highest on this entire graph if you have a closed interval. That doesn't work for an open interval. And we're going to see that later on, but I just want to get you ready for it. We're going to have a theorem for that. It's coming up in a little bit. But first, let me go ahead and just get you started here. Just tell you what the absolute maximum ends are, because that's all she wants you to do right here. Again, I'm probably giving you a lot more information at this point than you need. But you know me, I like to give the story away. So my absolute min, again, is going to be the y value. And my highest point, you can see, is this right here. That's h of e. Oh, I'm looking for the absolute min. I'm sorry. I was looking at the wrong one. I was doing the absolute max. Well, I guess I could put it right here. H of E is my absolute max. And that's a train ID at X equals E. You see, an absolute max can also be a relative max, or I should say a relative max can be an absolute max. The only one of them though, that was the highest point. Now again, if we had a different, so let's say C reached the same height as E does. Then I would say A to B. Because whatever that y value is, it's still going to be the same thing for h of c. So we say whatever that y value is, let's make something up 10. So let's say these two points both reach a height of 10. So this one also went up there. Then that means that I would have still one absolute max of 10 at x equals c and e. Two spots. And that's how you would say it. h has an absolute max of 10. at x equals c and e okay and this were also to go up to this height and that height would be 10. Just a hypothetical. Okay okay in this case my lowest point I believe is right here now that I wrote all my graph let me take a look yes that's my lowest point that is going to be h of d and that's at x equals d and I know it seems kind of a redundant right now to keep putting the d and the d but remember this is going to be a number pretty soon like uh two okay two at x equals five okay so they'll be numbers pretty soon any questions about the definitions for relative max and mins or absolute max and mins or how to find them on a graph you go check the chat that's okay okay looks like we're gonna get some more definitions And I really want to get into the problems, is what I want to do. I'll show you the process, because we've already talked a little bit about this. Okay, relative or local extrema. Basically, they just got to be the highest and lowest point in their little area. Y values on the graph of a function, where the function changes from increasing to decreasing and vice versa. That's going to be important, because that's going to be what our test is about. This is describing how we're going to test for relative or local extrema. We're going to test to see that it changes from increasing to decreasing. Or in other words, the slopes change from positive to negative. If your slopes change from a positive to a negative, so all the slopes would go from positive to negative, you're going to create a relative max right there. That's going to be our test. And for vice versa, if it changes from decreasing to increasing. Then we're going to change from negative, which means decreasing is going down, to increasing. We're going to create a relative min right there. And that's where we're going to test for these relative extrema. Again, I'm probably giving away a little bit too much right now, but I kind of like to explain it before we do it. And that way you get to hear it first because some people are auditory learners. Those people tend to do very well in my class because I'm more tell the story. And then we're going to practice it. And that's where I need you guys to work on the homework to practice and then ask questions. Okay, absolute or global extrema. So these are the highest and lowest points of the entire graph. The highest or lowest y values on the graph of a function are on the specific domain of a function. So think about how we're going to do that. We're going to need to find first all the relative mins and maxes, because those could be, as we've already seen, those could be your absolute mins and maxes as well. So we have to first find all the relative extrema. And then we also have to test the endpoints for absolute extrema. Because for absolute extrema, your highest point could be at an endpoint or your lowest point could be at an endpoint. This one almost got to the lowest. So that's what's coming up to find absolute. It's a little bit more work because we have to find all the relative and local and we have to test the endpoints as well. And we have to test them by just plugging them into the original function to see which one has the highest y values out of all of them. And then we'll see which one has the lowest y value out of all of them. And that's basically how you test it. You plug it into the original function and you just compare all the y values. So you take all the x values of your relative max and mins and then the two endpoints, plug them all into your original function, and you see which one's your highest y value and you see which one's your lowest y value. And you've discovered. where your relative maximum or sorry where your absolute maximum ends are and what they are that's what's coming up this describes the test that we're going to do okay these are just simple words again it's better to see a process when you do this but some people can understand this verbally and then as we go through it it just clicks for them okay every one of those people consider yourself lucky because i was not i was not an auditory learner M-naught. All right. Definition of relative and local extrema. If there's an open interval, what I put in here, y values are the relative extreme. We already talked about that. So the relative extreme are the y values. That's basically what this is saying. If there's an open interval containing c on which f of c is a maximum, then f of c is called a relative maximum of f. If there's an open interval containing c on which f of c is a minimum, then f of c is called a relative minimum of f. So it's just saying that the y values, the y values are the relative max and mins. That's all it's saying. Okay, this one I do like, though. Some noteworthy thoughts. When we say an open interval containing C, we mean that we can find some interval, A and B, not including the endpoints such that C is in between A and B. But we don't include the endpoints, so this one is an open interval. So this one we do not need to test the endpoints, and absolute max and mins are not guaranteed when you have open intervals. All right, absolute max and mins are not guaranteed when you have open intervals. Let me give you an example of that really quick. There's an open interval. There's a graph, y equals f of x. Has an absolute min, right? It's got a lowest point on the graph right here. Whatever that y value is at that point, there's your absolute min. Where's your absolute max? It can't be this point because it doesn't exist for us. So how close do we want to get it? What if this were on the x-axis? I'll make it all positive for us. What if this were 1 right here on the x-axis? What's my max? What if that hole up here is at 2? What's my absolute max? It can't be 2 because it doesn't exist at 2. So is it 1.9? I think you can beat that. Is it 1.99? Is it 1.999? Is it 1.99999? What's the absolute max? That's why when you have an, you still may have an absolute extrema when you have an open interval. Like if I drew it. This way, there's an open interval that has an absolute min and an absolute max, but it's not guaranteed to have an absolute min and an absolute max if you have an open interval. But if your interval is closed, you're always guaranteed to have an absolute max and an absolute min. And that's going to be coming up later on. It's called the extreme value theorem. And now that I've explained it a little bit, maybe I can go a little bit faster when we actually see it coming up. That's called the extreme value theorem. And all it says is that you're going to be guaranteed extreme values, absolute max and mins, if you have a closed interval. If you have an open interval, you may have them, but you can't guarantee it. You can't be certain of it. That's called the extreme value theorem. And that's coming up here shortly. Okay, again, probably explaining more than I needed to. That is, C will be contained somewhere inside the interval and will not be either of the endpoints. That's what it's saying when C is in between some interval. It's not going to be one of the endpoints. That's all that means. A relative extrema is slightly different than an absolute extrema. Relative means it just has to be the highest in its little location. Absolute is all about being the highest of the entire graph or the entire interval. All that's necessary for a point to be relative max or min is for that point to be a maximum or minimum in some interval of x's around c. So that's what's talking about a neighborhood, a little neighborhood around it. There are going to be smaller or larger values of the function at some other location, but relative to x equals c, f of c is smaller or larger than all other function values in the neighborhood of c. So I like that part for the neighborhood. All right, now we get to do some examples. In many cases, we'll be asked to analyze both interval and closed interval functions. I'm sorry, both open interval and closed interval functions. Consider the graph of y equals g of x shown at the right. We're not given any domain on this, you guys. We're not given any intervals. So what that means is we're looking at the entire graph. See the way the arrows are pointing up? okay they don't necessarily have to have arrows you guys but if they're going to show you that it's closed they will put end points on there see these these are end points but if they leave those off you can assume that the graph is going to be going up okay they also don't have to put arrows on there to show you that it keeps going but to show you that it stops they will put end points see that but they don't necessarily have to have the arrows to show that it keeps going I know that's weird, but that's the way it is. Okay, let's get used to that. So I kind of wish you would have done this example without the arrows. I can't erase them. But without the arrows, they still, you assume them to go up. If they want to show you that they stopped there, they will put endpoints like this. Okay? Or they'll tell you the interval and they'll close it for you like this. What is this one? I'll give you this one. Zero to 12. They probably gave it to you too, but. Right there. All right? And then that way the graph does keep going. You know that if they're talking about 0 to 12, you can imagine there's an endpoint right there, and there's an endpoint right there. Whenever they give you a closed interval, you know there's going to be endpoints right there and there, and that's all you have to look at is in between. It doesn't matter that the graph keeps going. You're just going to look at an endpoint there and there. So some other situations that can occur. All right? Okay, state the relative extrema of g of x. So relative, local. So right here we have a relative extrema of negative 8. So it's going to be relative min of negative 8 at x equals negative 3. See that? That's how I write it. This one, relative max of positive two at x equals positive two. And one more relative min right here. Relative min of zero at x equals four. See how you write them? So again, the y values are the relative extrema, and the x values are the locations. Considering the domain for g, here's part b, sorry. Considering the domain for g, what are the absolute extrema of the function? Now remember, this really doesn't have a domain. You can say the domain's negative infinity to positive infinity if you want. Look at the negative x's to positive x's. You're going to look at the entire graph. You can see that this graph is going to keep going up and up forever. Even if they don't put the arrows there, it's still going to indicate it's going up and up forever because they'll either have to show you an interval like this or put the endpoints on there in the graph. So this goes up forever. What that means is that we have no highest point, you guys. There is no absolute max. No, absolute max because the graph approaches infinity. The y values approach infinity. What is this, g of x? g of x approaches infinity as x approaches positive or negative infinity. What does that say? The y values on this graph go up forever as And X goes to the left, because that's where negative infinities are, or to the right. You really only need one of them, because as long as it just goes to one side, you have no absolute max. So this graph, which we've seen before, this is a graph of next cubed. That graph definitely has no absolute max or mins. because it goes down forever in this direction, it goes up forever in that direction. This one's going to have an absolute min though. The absolute min is going to be negative eight at x equals negative three. Oops, absolute min of, put the at first, you could put the at first, I guess you just got to say it differently, of negative eight at x equals negative three. And that's it. That's it for that one. It's not too hard, but we're just learning it right now. So I got to go over every single definition. I want to try to go over every single case that they could show you. Is there any other questions that you would have from that one before we do number three? We get to see another example here that is closed. That's OK. OK, we're going good. That's what I wanted. OK. Consider the graph of Y equals F of X on the interval from zero to 12 shown below. So they did it with. Equal signs. That means closed bracket. You can also see from the graph closed circle here, closed circle there. Okay. But even if the graph didn't have closed circles, you can now put closed circles on there because of the way they wrote the interval with equal signs. And you only have to look at in between there. Any Xs that are outside that domain, you get to toss out. You don't need them. Okay. Find the relative extreme. We're going to like the toss out parts, especially when we do trig functions. Could trig functions have a lot of repeat answers? and we'll get to toss a lot of those repeat answers away and just use the ones that are in the domain. Okay. Find the relative extrema of f of x. You got it. So we're guaranteed an absolute extreme of both an absolute min and absolute max because we have endpoints. So this time we will get an absolute max. Okay. We have a relative min right here of negative two, relative min of negative two at x equals two. We have, remember, those are going to be tough to find. A little bit tougher than the nice smooth ones that we were doing before. When we actually do the functions and not the graphs anymore, these will be nice and easy to find. Those sharp points are going to be a little bit tougher. This might be like a piecewise function that's broken up into one, two, at least three pieces. Maybe even more. Maybe there's two pieces right here. Remember piecewise functions, we can have one, two, three parts to them, two parts to them. They have the, I call them the gatekeeper numbers on the sides they have to pay attention to. Okay, we'll be seeing piecewise functions coming up. So let's see, we have relative min here. That is not a relative min or max. It's not the lowest point of the entire neighborhood. It's not the highest point of the entire neighborhood, of a little neighborhood, I should say. And then this one, this one's a relative max up here. So we have relative max, I believe that's seven. at seven relative max of seven at X equals seven. We have relative men right here. That's going to be relative men of two at X equals 10. I think that's it. We got them all relative men relative max relative men now for the absolute. So this one we're going to check. all the relatives because they could be our absolutes. And we have to check the endpoints as well because the endpoints could be absolutes as well. Let's check the endpoints first. This is the highest point of the entire graph or lowest point? No, it is not. Is this the highest point or lowest point of the entire graph? No, it is not. So the endpoints are out. I always try, I always pull for the relatives anyway. Those are always the ones that I want to win, the absolutes. But sometimes the endpoints win. Okay. For this case, the absolute. is going to be right here at 2 and it is negative 2 so we have an absolute min of negative 2 at X equals 2 now is the lowest point of my entire graph and we have some absolute match right here at the relative max absolute max of 7 I believe it's at 7 at X equals 7 And that's it. So again, you are guaranteed an absolute min and absolute max if you have a closed interval. You're not guaranteed if you don't have a closed interval, even if it's an open interval. If these were open, you're not guaranteed to have it. You may still have it, but you might not. OK, there it is, the extreme value theorem. What conditions would be necessary in order to have both an absolute maximum and an absolute minimum? The extreme value theorem. If F is continuous, so here is the prerequisite. You need this. You need to say this just like we do when we do the mean value theorem. We have to say it's continuous and differentiable. Continuous on the closed interval. Differentiable on the open interval. When we do the intermediate value theorem, we have to say it's continuous through the whole. function to the whole domain to the whole control For this one, it's got to be continuous as well, just like the intermediate value theorem was. F has to be continuous on this interval, and most times because it's a polynomial, and polynomials are globally continuous everywhere, but we have to stay that before we do the extreme value theorem. Okay, if we can say that on a closed interval, then F has both a maximum value and a minimum value. Notice how they left out the word absolute. If you guys see on the AP test, just the word maximum and minimum, think of it as being the absolute maximum. If they want you to just focus on the relative, they would say relative or local. OK, but if they leave the word absolute off or global off, then you can always think you can assume it's an absolute max or an absolute min that they're talking about. So it has both an absolute max and an absolute minimum value on the interval A to B. Just like I said, and all we need to do is say that's continuous and that it's on a closed interval and that's it. And then you're guaranteed at least one. I should say you're going to have one maximum value, absolute maximum value and one absolute minimum value. They may occur at two different spots, but you're going to have one highest point and one lowest Y value. One highest Y value, one lowest Y value. Extreme values can occur at interior points. So at the relative extreme or end points of an interval. A function may have both maximum and minimum values over interval, either a maximum or a minimum value, or a minimum, or no extrema on interval. Let's see some examples below. When the hypothesis or if statement is not met, either continuity or the closed interval, there is no guarantee of the conclusion, then statement. However, a maximum minimum or both may still exist. There is no guarantee. There's just no guarantee. That's what I was talking about. When I gave you my example with little open circles, you still may have one, but it's not guaranteed to be there. It's only guaranteed if you can say it's continuous and it's on a closed interval. Then you can guarantee, yes, there's an absent max and yes, there's an absent min. I don't even need to look at it. I know there's one. Let's explore a few examples where this happens. For each graph below, explicitly state where the hypothesis of the EBT fails on the interval from zero to three. and then determine if the function has an extrema on that interval. So when I'm given an interval like this, I go automatically and put endpoints on the graph at x equals 0 and x equals 3. And that way I don't need to look at any piece of the graph that's to the left of it or any piece of the graph to the right. You probably won't need to do that on any of these because they're already marked just from 0 to 3. None of the graphs continue on in either direction. Okay, but I just want to let you know that I had graphs that continued on in either direction. When I get this interval, I put a closed circle on the graph at 0 and a closed circle on the graph at 3. It just helps me focus it. Okay. Then determine the function has an extrema on the interval. Okay. You got it. Is it guaranteed? Does the hypothesis get met for this graph? No, it doesn't. First, it's not continuous. You got the closed intervals. You got the closed circles, but you don't have the continuity. So fails. Because they didn't name the function. No, they did not. We're going to name it. y equals f of x for this one. y equals g of x for this one. y equals h of x for this one. And y equals q of x for this one. That's what I'm going to name it. Fails because f of x is not continuous. at x equals 1. And 1 is in the interval. So the extreme value theorem is not guaranteed. Now, does it have a highest point and a lowest point? Yes, it does. It has a highest point. It has an absolute max of 4. It has a closed circle there. So that's the highest point. If that was an open circle, I couldn't say it has the absolute max anymore, you guys. Because if that were the open circle, I kind of wish you would have did that one. If that were the open circle up there, then what do you say your absolute max is? You can't say four because it's open. So is it 3.9, 3.99, 3.999? How high do you go? It doesn't have an absolute max. And that's an open circle there. Because it's a closed circle, my absolute highest point is at four. See, it has an absolute max. It's just not guaranteed. Okay, lowest point, that was a closed circle as well. Yep. So we have absolute min of, I'm going to say one half, you guys. I'm going to have to guess that one. You won't have to guess on the AP test, but I'm going to say one half. That equals zero. Okay. What about B? B also fails because it's not continuous. And this one I'm just going to put fails because g of x is not continuous on the interval, because I don't know where that's at, on 0 to 3. There you go. Fails because g of x is not continuous on the interval from 0 to 3. You can call it 1 and 1 half, I guess, if you want to. But does it have a highest and lowest point? It's got a highest point. There it is. Absolute max of, shoot, I don't know, four and a half. That's a guess. At x equals zero. I know that's true. So at the end point, it's got the absolute max. What about an absolute min? Oh, here's my example right here. There is no absolute min. There's an open circle here. So it looks like it's 1, but there's an open circle there. So what is it? 1.01, 1.001, 1.001. What's your absolute min? Don't know. No absolute min there. For here, look at this. There's an open circle there. When there's an open circle, don't put a closed circle there. Only do that when you see the graph. And especially if it's continued on. And she didn't give me an example that I can show you on that. But if this graph were continued on, then you put closed circles there. But if there's already an open circle there, don't close it, you guys. Because there's an open circle. You just have to look at that. This thing is not continuous from 0 to 3. It has to be continuous at 0 and at 3. And it's both not continuous at 0 because it has a hole there. And it has a vertical absolute at 3. So it's got the two problems that we know are just content. discontinuities. A hole called a removable discontinuity and a vertical asymptote called an infinite discontinuity. So this is not continuous at the endpoints. Fails because h of x is not continuous at x equals zero or x equals three. And this has no absolute extreme at all. There is no highest point because there's an open circle there. And there's no lowest point because this thing is going to go down forever. No absolute extreme. Okay. Next one. Why does this one fail? This one fails because it's also not continuous at the end points. Now continuous at 0 and now continuous at 3. The exact same reason as this one. Fails for the same reason as C. Now continuous at the endpoints. But it does have a highest and lowest point. I don't know exactly where they are. That one looks a little bit off, but I don't think I need to name these because you've already seen how I write them. But here is your absolute max right there. And here is your absolute min right there, whatever those values are. But all this was trying to show you is that even though you can't state that there is one, there still might be one. OK, but it's not guaranteed by the extreme value theorem. It's only because it's continuous. on that closed interval, then you got an absolute max and an absolute min. Okay, guys, let me see if there's any questions about that before we move on to some of the other examples. That's good. All right, let's keep going then. The extreme value theorem gives us knowledge of when extrema occur, but how will we find them algebraically when given an equation only? Without the benefit of a graph, So our next question should be how can we identify all the bodies of X? where extreme values occur. Here we go. For each function shown below, just if I explain why the extreme value theorem can be applied or why it does not apply in the given interval. Okay, so it still hasn't had me do a problem yet. This just wants to know why could we apply it or why could we not apply the extreme value theorem? Okay, here We're going to have a division by zero problem. Whenever we try to plug in a 2 in for x, we're going to get a zero down here, which means it's not going to be continuous at 2. We have a discontinuity at x equals 2. Now, remember that we were doing these discontinuities. It could end up, division by zero could end up being a vertical asymptote if you get a non-zero number over zero, or it could end up being a whole, a removable discontinuity. But either one, it just means that's a discontinuity. There's going to be a discontinuity at x equals 2. So if I know my function is discontinuous there with either a whole or vertical asymptote, you can find out if you really want to know by plugging in the 2 in the top. But we don't need to know which type because all we need to know is that's discontinuous there. And if it's discontinuous at 2 and 2 is in between the interval that it wants me to look at, then I know the extreme value theorem will not apply. Therefore, three little dots in a triangle like that mean therefore. Therefore, the extreme value theorem does not apply. That one was easy. If you remember that division by zero is a problem and it gives you discontinuity in your graph, some sort of discontinuity, either a whole or vertical obstacle, that one wasn't bad. This one. exponential functions. Exponential functions are continuous everywhere. You can plug in any number you want into them. Their graphs look like this. And then you're just going to shift it. So all exponential graphs look like this. They go up the fastest. OK. And this one's always a one right here because anything to the zero power is always a one. So this would be like the graph of e to the x. The graph of 3 to the x, the graph of 5 to the x, the graph of 10 to the x, they all look the same, like that. They're continuous everywhere. They have no problems. They have no holes, no vertical isotopes. So this thing is going to be continuous everywhere. It doesn't matter that you're subtracting 2 from a number. You can subtract 2 from any number you want. So this one does apply. Let's say g of x is continuous on the interval from negative 5. Positive 5, the one that they wanted me to look at. Therefore, the extreme value theorem, well, I said intermediate value theorem, the extreme value theorem does apply. And for this one, last one, the thing with square roots, you have to be careful about negative numbers. So in this case, you could plug in. negative six. Yes, if you plug in a negative six into this, you get a negative four. And the square to negative four is not a real number. And in calculus, we only deal with the real numbers. So in this case, this is not going to work because this is not going to be continuous at negative six. In fact, it's going to be continuous, not going to be continuous at negative five, negative four, negative three. It'll be okay at negative two, but there's a lot of numbers where it's not continuous. So h of x. That's not even defined for those. But all you need to say is not continuous. It's not continuous on the interval negative 6 to 6. Therefore, EBT does not apply. And I think she just wanted to remind you division by zero, and then you can't take the square roots of negative numbers. because they'll be imaginary. Wasn't much to that one. I want to get up to an example where we have to do one where we don't get to look at the graph. Oh, critical number. Here's another definition for you. Okay. Remember that critical points are points on your graph that could be relative max and mins. You just have to test them. Okay. A critical number value of a function. This is actually going to show you how to find this one as well. Remember I told you. These will be easy to find, because we're going to set the derivative as equal to 0. This one will be harder. But this point right here, I should say the point B, H of B, that point right there on your graph, B, H of B right there, X value is B, Y value is H of B, that point right there is a relative min. The derivative does not exist there, but it's still a critical point. It's still a critical point. So how are we going to find that one? This is what we need to do. So to find all these critical points where these possible locations of relative maximums could be, without looking at the graph, this is what you need to do. You're going to take your derivative, and you're going to set it equal to 0. And you're going to solve that equation. And that's going to give you all of these, the nice smooth ones. But you also have to find out where your derivative does not exist. So what does that mean? That could be like division by 0. When you divide by 0, that doesn't exist. Have to check it. So this is a good example. If this were my first derivative, you guys, and not my original function, I'd want to set it equal to zero and solve it to find out all these possible locations. And then I'd also want to find out where it's undefined. And this is undefined at x equals 2. It does not exist at x equals 2 because division by zero. That would also be a critical point that I'd want to test to make sure if it's a relative max or min. So a lot of times when you're looking for where it doesn't exist, it's going to be a rational function. And when you get a derivative for a rational function, it's going to be really easy. If you want to just know steps to find out where it equals zero, that's basically the numerator. You take the top and you set it equal to zero. That's to get this part. To find out where it doesn't exist, just take the bottom and set it equal to zero. So if you're frustrated, it turns out to be a rational function like this, a numerator over denominator with X's on top and bottom. Just take the top set equals zero, take the bottom set and equal to zero. And that will give you the locations of all of your critical points. It will find the nice smooth ones produced by this. And I'll find the sharp ones produced by something like this. Now you still have to test them because they may not be relative max and mince, but at least you got. all the possible locations for them. So now all you need to do is perform the test. Okay. If X equals C is a critical value, then the point X F of C, here's my X value, here's my Y value, is called a critical point. Okay. Sometimes on the A&P test, they'll ask you to find all the critical points or all the critical values. And you just have to find out where the rate of equal zero and where it doesn't exist. Okay, theorems. Relative local extrema can only occur at a critical value on an open interval. See, I told you, that's why we need to find all the critical values first, and that will tell us where these relative maximum ends are after we test the critical values. Absolute global extrema must occur at a critical number, so it could occur at a relative extrema, or at an endpoint on an interval. So here's more to test for absolute extrema. You got to test all the critical ones and the endpoints. Using the candidates test to determine absolute or global extrema. Now we need to put these ideas together with a little direct practice. Finally, let's develop an algebraic approach to what we call the candidates test for finding extrema. That's what we're going to call candidates. There we go. Finally get to do some practice. All right. Find the absolute extrema on the function of the given interval provided the EVT is applicable, if not justify why. So we're going to find the absolute extreme, which means we need to find all the relative max and all the relative mins. And we also need to test the endpoints. We get to use a calculator. This is a polynomial. So I'm going to try to do this without a calculator. I'm not sure if it's possible, but we'll try to do without a calculator and I'll try to show you with a calculator with these because these will be trig functions. These would be definitely want to be calculator for. Okay, let's go ahead and take the first derivative. Oh, wait, let's say if this is going to be guaranteed, first of all, are we going to have an absolute max and absolute min? First of all, we got a close interval. Good. Now is it continuous on that closed interval? Yes, it is. It's continuous because it's a polynomial and polynomials are globally continuous. f of x is continuous on interval from negative one to two because f of x is a polynomial. That's all you got to say for that. Nice when they're polynomials. Okay, now that means we're going to get an absolute max and an absolute min. That could be at one of the relative max and mins, or it could be at the endpoints. All right, we're going to test them. Let's find all the relative extreme at first. So let's find the first derivative. This will give us the equation for all the slopes of the tangent lines to this graph, and that's going to be 24x squared minus 6x minus 9. And I'm going to set that equal to zero to find out where the nice smooth parts of my graph are. And I'll find these spots. But then I also want to consider if there's any x value that would make it undefined. That could also be a critical point. So two things. We're going to find out where f prime of x equals zero. So that's going to be 24x squared minus 6x minus 9 equals zero. We need to solve that equation, you guys. And we also need to consider, n is f prime of x, the first derivative. Is it ever undefined? Would there be any number that you can plug into this where you wouldn't get an output? That's no. You can plug any number you want into this and you're going to get an output because this is a polynomial, you guys. It's always defined. So what that means is there's not going to be any sharp points in this graph. that we have to check. There's just going to be the nice smooth curve. There will be no sharp points in this graph that we have to check. So this piece is out. We don't need to do this piece. Now the question becomes, do we have to write it? I probably would. I probably would at this point say that I'm looking for these two pieces, but I don't, I've never seen where you're going to be given a point for this. OK, but every time that I'm trying to find my relative maximum ends, I always do the two steps because this one is easily forgotten. People always want to take derivatives set equal to zero every single time to find all the relative extrema. And most of the time it will work to find all the relative extrema. But there's going to be some cases, you guys, especially when you get a rational function for a first derivative. There's gonna be some cases where we have to consider this as well. Not in this case because it's a polynomial. Okay, let's go ahead and solve this. I want to see if I can solve this without a calculator, and then I'll show you how to do the calculator and solve an equation. Basically, if you want to solve an equation with a calculator, you graph the left-hand side as your y1, you graph the right-hand side as your y2, and then you find out where they intersect on your calculator, and that can solve any equation. And that's going to be good for you because if you decide to take the AP test at home, because you'll get the option of taking it at home this year, you'll be able to use a calculator to solve an equation. You'll be able to use a calculator for all your problems. They can't stop you from using a calculator. So you'll be able to solve any of these. So if you're not good at factoring, if you're not good with the algebra, this may be your year for an AP test because calculators can solve these equations for you. Okay? So that's what I'm saying. Remember, if you're not good in algebra, this might be your year that you can take advantage of it because normally when you do the AP test, they take the calculator away for part of it. They can't do that when you're at home. So take advantage of that. All right, let's see. I'm going to go ahead and factor out a 3 from those numbers to make them smaller. That leaves me with 8x squared minus 2x minus 3 equals 0. You can divide both sides by 3, and the 3 will disappear. I kind of like getting rid of numbers when I'm solving equations, so I like to just factor it out. But it's okay to divide by. It wouldn't affect anything. And I'm going to try to factor. And again, if you're not good at factoring, then you would just use your calculator at this point, right? Graph this and left or y1 graph this and y2 find out where they intersect and I'll show you those when we get to the Other examples, but this one I think is factorable. So let's see what I have 8x in 3 I need to make a negative 2 so I'm gonna try 4x and 2x It's just trial and error right here and then 3 if we're gonna put here I'll make it 6 and then 1 will be a 4 and I get a 2 out of that Yes, so 3 here 1 and that I need a negative 2, so I need the 6 to be negative. and the four to be positive. There we go, and then if I solve this, this four x has to equal three in order for it to equal zero. Four x would have to equal three, because three minus three is zero, so x would equal three fourths. So your algebra one teacher always taught you steps. Little teacher, just think about what you're doing. Two x has to be a negative one in order for this to equal zero. Divide both sides by two, you get x equals negative one half. All right. So we have two critical values. We don't know the points yet because we don't know what the y values are. We want the y values on the original function, and we have to take these and plug them into x's for the original graph. There's no point in plugging them into the first grader. You're going to get zero. In fact, if you don't get zero when you plug them in the first grader, then you didn't solve it correctly. All right. So now the thing becomes, what are they? Are they relative max? Are they relative mins? If you want to test to find out what they are, you don't need to for absolute max and mins. You can go right past this, but I want to show this to you guys. Okay. Because some questions on the AP test just ask you to find relative extrema. So here's what you do to test for relative extrema. And then I'll show you the test for absolute extrema. Okay. You don't need to test for relative before you test for absolute. If we're just doing absolute like this one wanted, I don't need to do this. I don't need to tell you if these are relative maximum ends. We can just test them all into the original function to see which is the highest and lowest point. Okay. But I want to show you this test for relative extreme. Here's what you do. You make a number line. You place those two numbers on your number line. I'll do it on top. I like doing them on top actually. Okay. and then 3 fourths is after that to the right. And now just think of this as a graph so far. So we're going to basically just fill out the graph. We know that there is a nice smooth point at negative 1 half and at 3 fourths. And we know that the slope of the tangent line at negative 1 half 3 fourths is 0. Now we're going to test to see if the relative maximum ends by picking other x values to try into our first derivative. Let's say I want a number to the left of negative 1 half. How about negative 1? Pretty close, but it is to the left of negative one half on a number line. I want a number in between negative one half and three fourths to test right here. I'm going to pick zero. You don't have to pick the same numbers I do, but you got to make sure that they're in the correct intervals. You need this number to be in between these two. If you can't, always try zero somewhere. You're going to get zero somewhere. As long as it's not one of these numbers. And then another number to the right of three fourths. How about one? One's to the right of three fourths. We're going to test those three numbers into our first derivative. And all we care about is the sign of the first derivative, because the sign will tell us what the graph of the original function is doing. If we get a positive first derivative, then that means our graph is going up, because all tangent lines with positive values, with positive slopes, means that your graph is going up. If we get a negative value for this, then that means our graph is decreasing. It's going down. So we'll know what's happening right before this point. And then we're going to put this number in to the first grader to find out what's happening right after this point. Because if we have that change, we're going to create a relative extreme right here. So let's try it. Whenever I plug this in, especially into a polynomial that I've already factored, I like plugging it into the factored form. Because it's so much easier to get the signs from the factored form. You don't need the number. You just need the sign. So let's plug a negative 1 in for the x here. So you have a positive 3 that we're multiplying by. So we have a positive 3. And if I plug a negative 1 in for this x, I get 4 times negative 1, which is negative 4. Minus 3 is another negative number. If I plug a negative 1 into this x, I get 2 times negative 1, which is negative 2. Negative 2 plus 1 is a negative number. So we have a positive times a negative times a positive, which, I'm sorry, a positive times a negative times a negative is a positive. That's what I meant to say. And you know what that means since we plugged it into the first derivative. This gives us the slopes of the tangent lines for all numbers before negative one half. They're all going to be positive slopes. We know when they're zero already. These are the only places where our slopes are zero. So we know before that our slopes are positive, which means my original graph of f would be going up before it reaches negative one half. Now just because it goes up before it reaches this point, doesn't mean that it's going to go down after this point. It could go flat for a little bit and then back up. Okay, so that's why we have to test the left and the right side of these points. So now we plug a zero into the first derivative. If I plug a zero in, it's still positive three. Four times zero is zero, and zero minus three is still negative. If I plug a zero into this one, two times zero is zero, but zero plus one is a positive. So now we have a positive times a negative times a positive, and that's going to be a negative. So this means that my function, my original graph, would be decreasing between these two numbers on my x-axis. So my original graph is going to go down after this, and then it flattens out here at 3-4, and then we're going to see what it does after 3-4. It's probably going to go back up, but it doesn't necessarily have to go back up. It could keep going down. It just flattens out a little bit right here and then goes down this way. But right now, we just discovered that there's a relative max at x equals negative one half because our slopes change from positive numbers to negative numbers. And those slopes are the slopes of the tangent lines, which means this tells us that the graph goes up and when the graph goes down according to these slopes. Increasing, positive slopes, positive derivatives. Decreasing, negative slopes, negative derivatives. It's all starting to come in. Remember when we use these words to describe derivatives? We have a positive derivative. We would use the word increasing. It's increasing by this many tons per hour. We had a negative derivative. We would say decreasing. It's decreasing by this many tons per hour. It's all being connected here. Okay. And then one more to check to see if this is going to be a relative min. It's the only possibility because we're going down before it. And now we go up, we create a relative min. So let's plug in the number one. into the first derivative to see what the slope of that tangerine would be. If I plug a 1 in, that's still a positive 3. Plug a 1 in here, that's a positive because 4 minus 3 is positive. If I plug a 1 into here, that's also a positive. So I have three positives being multiplied together. That means we're going to be increasing afterwards. So it's going to be going back up, and that means we created a relative min at x equals 3 fourths. If we want to know what those min values are, Then we plug those numbers not into the first derivative because we don't want the slopes of the tangent lines. Now we want the y values on the original function. So we plug them into the original function and we figure out what those y values are. And then you can find the relative max and min. That's how you test for relative extrema. You pick numbers, not the same. You just draw a number line, pick one to the left. In between, one to the right. If there's more, do the same thing. You plug them into the first derivative. So this is called a first derivative sign chart. First derivative sign chart. That's what it's called. And it's a test for relative extrema. Okay, now that test does not have to be done for this problem. You don't need to know if these are relative max or mins in order to test for absolute. Because once you get all your critical numbers, then you're just going to go and test them all back into the original function. So here's called the candidates test. You don't say candidates test, just put candidates. You're going to have x equals negative one half. That's one of the ones that could be up for the highest or lowest point. We know from this test that it's only could be an absolute max. Because we know it's a relative max at x equals negative one half. We know that relative max can only be an absolute max. It can't, you can't have a relative max be an absolute min. It doesn't make any sense. Okay. But you don't need to know that. All you need to know that there's a critical point there. So you're going to test it. X equals three fourths is another critical point. Could be relative maximum in there, which means it could be an absolute maximum. And the other two possibilities, I call them the born-in candidates. The other two possibilities for relative maximum are the two endpoints, negative one and positive two. So if you're answering just a question about absolutes, you don't need to test for relative. You can go right to the candidate's test for all four of them. And this is where you use a calculator, especially when you have fractions like this. And you plug it into the calculator and you're going to see the Y values on all those. This is F, right? So F of negative one half, F of three fourths, F of negative one and F of two. Let's do that. Let me show you with a calculator. I got my calculator ready to go. So we'll store in the function. That's what I like doing. I'll go y1 right here. I'll get rid of these because you don't need those anymore. Back up, and we'll put in the original function, which is 8x cubed, 8x cubed. We'll put the cursor down to the right. Minus 3x squared, minus 3x squared, minus 9x. Minus 9x and then plus 2. Okay? And that's all I really need. I don't even need to see the graph because I'm just going to have this thing plug in numbers. So I'm going to quit this screen, go back to my home screen, my computation screen. I started in y1. So I got to remember that. So it's alpha t and y1 is number 1. And I'm going to have it plug in the first number, negative 1 half. So parentheses, negative 1 divided by 2. You can even put negative.5 if you want. Don't plug that number in for me. And it's 4.75. 4.75. So we know the height on the original function F. 4.75. Now we've got to compare all the other ones. So we do three-fourths next. The easiest way to do this is to do second and enter. Second, enter. Brings up your last entry. And then you just go to make changes. That way you don't have to do the alpha T again anymore. You could if you want to. Delete the negative. I'm going to replace the... One with a two, I'm sorry, the one with a three, and the two with a four. And we'll put in three fourths. Enter. Negative 3.065, negative 3.0625. Okay. Now we'll do negative one. Second, enter. And I should have probably done negative one as the second one would have been easier. Negative one. And I'll delete the four. There we go. Enter. 0. And then my last one, positive 2. Second, enter. Positive 2. Move, move. 2. Let's get rid of the negative sign. Delete. Enter. 36. All right, we got our values. Let's see the winners. So Find the absolute extreme and you got it. So the absolute extreme, so f has an absolute max of 36 at x equals 2 and an absolute min of negative 3.0625 at x equals 3 fourths on the interval. From 0 to 12, what was it? Negative 1 to 2. You know, I forgot to tell you, if any of these critical numbers were outside this interval, we could have tossed them away, but they were both inside the interval. Okay, but I did forget to mention that. If any of these critical values that we saw for, by saying they're equal to 0, were not inside this interval, we could have tossed them away, and we wouldn't have to check them. Okay. But that's how you do the test for absolute extremism. It's called the candidate's test. And you see, you don't need to know if they're relative max or mins in order to test them, because you're just going to look at the Y values and compare them. All right. How'd I do on that one? Any questions? When testing for relative achievement, how do we know the graph doesn't weirdly change in between the points where we tested? What do you mean by weirdly? You mean like a sharp point? It's not going to be a sharp point because you'd find all those by finding where the first derivative is undefined. So again, the word weirdly is a little bit ambiguous because what's weird to you might not be weird to me, but I think that's the biggest thing is you need to test to find out where it equals zero and to find out where it's undefined. A lot of times you'll get a rational function for the undefined part. A rational function for the undefined part. Okay? Okay, the last thing I want to show you though before you go is I want to show you how to do this with a calculator, because I didn't get to show you that. So let me go ahead and let's see if I can do this one really quick. So g prime of x equals 1 minus 2 cosine of x is my derivative. Set that equal to 0. I know I'm doing this quick, guys. 1 minus 2 cosine of x. I got to remember that. 1 minus 2 cosine of x. 1 minus 2 cosine of x. I'll clear that. 1 minus 2. Make sure you're in radian mode when you do this. I know I am. I'm not going to check, but make sure you are. OK, my other one is 0. So I'm going to go down, go to 0. And I have an interval from 0 to 2 pi. So let's put that in the window as well. Window from 0 to 2 pi. 0 is already my min. 2 pi is already my max. There it is right there. Zero to two pi. Let me just show it to you, though, in case you don't get the two pi. Two second, and then exponent button, two pi. And then when you move it, it puts it into decimal form. Negative five to five is pretty good. I'll just go ahead and graph with this one. And it should graph y1 and then y2. And you're going to find out where they intersect. So it's second t. Intersect is number five. And you just move the cursor to the first intersection point. You can hold it down. Move it faster. Just get it as close as you can. And then you basically hit enter three times. One, two, three. And you got your x value. You can solve any equation like that, you guys. Even the one that we just did. x equals 1.0471. This is the one good thing about distance learning is that the algebra should not bother you on here because you're going to have a calculator. And then the other one really quick. Got to do it all over again. Second t. Intersect number five and then hold it down I always like make my equations equal to zero because you get that's just the x-axis and it's easy to see the x-axis That one is five point two three five five point two three five five point two three five. I like one more nine Especially since I'm kind of in the middle of the problem. So I got those two as my critical points. Cosine of x is always defined, you guys. They get the cosine as the pretty wave. There's no discontinuities there. So it's always continuous. So I should have said that x is a polynomial. It's always continuous. Sine of x is a trig function. That's a trig function that's always continuous. So the EBT, the extreme value theorem, does apply in this interval, you guys. I forgot to mention that. But I wanted to show you the calculator part really quick before I let you go. Okay, just one more time in the video so you guys have it. Okay, let's check the roll. Sorry, we didn't get a whole lot of chance for questions. I will finish those and put them up on a video, just like I did with the rest of 5.1, okay? I'll put those up on a video. I got to the main points that I wanted to get through. Now you just need to see the examples worked out. Okay, who's my only absent today? Diego, Diego, you're my only absent today. If you're in here, let me know. Everybody else, you can go. Have a good day. I'll post the homework up. Sorry about the due date on the last one. I forgot about the three-day weekend. Have a good day, guys. I forgot to turn off the recording. Let me turn off the recording.

which is called extrema on an interval. Okay, we'll talk about what extrema means. I think I discussed it, but now we'll get to talk about it.

We'll get to use the process. All right, so this is topic 5.2. Again, this is where you'd want to go for those AP daily videos. Really good examples on there.

The extreme value theorem, global versus local and critical points. It's just going to be some definitions right here in the beginning. So remember that global extrema?

are your absolute extrema you're right i'm not thank you there we go thanks okay there you go so we have global extrema which is going to be the biggest our highest point on the entire graph or the lowest point on the entire graph, but it's the entire graph or over a domain. Okay. They give us a domain in the problem.

Then we have to look over that domain as well. And local extrema, that could be highest and lowest points in a little neighborhood. They don't have to be the highest and lowest points of the entire graph. They just have to be the highest and lowest points in the little neighborhood, which is usually means just little area around it. Okay.

And before it and little area after it and critical points. Critical points are going to be all the possible locations. So locals, another name for locals, relative.

Another name for global is absolute. Critical points are going to be the possible locations for relative max and mins. The reason I say possible locations is because we have to test them. Possible locations for relative. Max and mins.

And the easiest way to say max and mins together is to say the word extrema. Extrema means max and mins. So anytime you see the word extrema, that's what it means. Just think of max and mins.

It's exactly what it means. And those are going to be critical points. And we'll get to discuss all this when we actually go through the process.

OK, so I'll keep reading up here. The maximum and minimum values, max and min, of a function are called the extreme values or extrema. singular extremum of a function. The process used to finding them is referred to as optimization.

So if you wanted to Google something like a video on something, you could Google optimization to see extra videos. There will be times when we want to find the max and min for x on the entire domain and at other times on a particular interval. In this lesson, we will learn how to use calculus to find extrema rather than a calculator. OK, so what that means, sometimes they'll give us a domain, sometimes they'll give us an interval to look at for the axis, and sometimes they won't.

Sometimes we'll have to look at the entire X line and sometimes we just get to look at piece of the X axis. If a function here's a definition for global absolute extreme of a function on a normal eye. Oh, sorry, if F is a function on the interval I. Then y equals f of c is the absolute or global maximum on i, if and only if f of c is greater than or equal to f of x for all x and i. And a global min, same thing except for less than.

Basically, what this is saying in terms of symbols is that if you got the highest y value over the entire interval, that's your absolute maximum. If you have the lowest y value, that's what these things are, y values. If you have the lowest y value over the entire interval, then that's called the absolute minimum or global minimum.

That's all it's saying in symbols right there. Example number one. The graph of y equals h of x is shown below. Determine the extrema, max and mins, of h of x on the interval between a and f.

Now, this is a closed interval. There's one way to write it. You can write it with inequalities like this.

But you're going to see this written a lot of times. In fact, you've seen it already with closed brackets. And I can tell you they're closed brackets for this interval because of the equal signs.

I'm going to draw those closed brackets very well. They're supposed to be closed brackets, not parentheses. And they're closed because of the equal signs.

You can equal A and you can equal F. So we close them. And again, on is a key word right here.

It's not at a point. It's on this interval. They even put the word interval for us.

Okay. So we're going to go ahead and... Determine I already made kind of a little chart down here to get ready. I'm going to put all the relative mins, all the relative maxes, absolute mins and the absolute maxes and I'll even tell you where they're located.

Okay, so the relative mins are the y values. This function is called h of x. So we'll name all the y values, like the y value for this one. Let me just give you an example.

The y value for this point we'll call h of a, because that represents the y value on the graph of h when x is a. We don't know what that height is, but we'll call it h of a. So we'll call the height here for this point h of b.

The height up here will be h of c. and so forth. And those y values are the relative mins, relative maxes, absolute mins, or absolute maxes. For the relative mins, we have to pick the lowest point in a little neighborhood. It has to be continuous there.

It can be a sharp point like this. You see that sharp point? Even though we can't take the derivative there, you guys, that is still relative min because it is still the lowest point in a little neighborhood. It's the lowest point in a little neighborhood from the graph to the left and to the right. That's all you need for a relative.

Just a little neighborhood around it. To the left and to the right of it. See, this cannot be a relative extrema.

End points cannot be relative extremas because there is no neighborhood to the left of this point. And there's no neighborhood to the right of this point. So end points, these numbers right here, can never be relative extrema.

Now, they can be absolute extrema. They can be absolute maximins. They can be the highest point.

They don't look like they're in this case, but they can be. If they were to just draw on this point up here instead at A, then that would have been the highest point. And that would have been the absolute max.

So endpoints can be absolute extrema, but they can't be relative extrema. So that's why I skip this one when I'm doing relative mins and I go right to here. This is a relative min. My relative min is going to be H of B. The relative min is H of B.

Whatever the Y value is when X is B. That's going to be my relative min. And a lot of times they want to know what's the relative min and where is it located. We always give them where it's located on the x-axis. So we would say h of b or the y value, whatever that happens to be, in this case it's going to be a positive number, something like a positive 3. So the relative min is positive 3 at x equals b.

We got like a negative 1. And that's 1 of the relative mins. There's another one. Okay, but notice how I wrote it.

The y value at, that's my symbol for at. Everybody's symbol for at. You can write that. At x equals b. So I'm doing two things.

I'm answering two questions. I'm telling them what the relative min is. It's this value right here, the y value. And I'm telling them where it's located.

It's at, just look on the x-axis right here and you'll find it. Most of the time, they just ask for the relative min part. So they just want the y value. But I always tend to put both. That way I don't even have to.

to think about it. Just a habit now. There's one more relative min.

There's one more lowest point in the little neighborhood. Check it to the left. Check it to the right.

This is the lowest point in its neighborhood, and that's going to be h of d. Whatever the y value is, when you plug a d into this function, think of the output. When you plug the input in, what's the output? At x equals d.

And those are your two relative mins. Relative maxes. There's also two relative maxes in this. Two points where we have the highest in this little neighborhood.

Right here at c. So we have h of c at x equals c. Whatever that y value happens to be. And right here h of e at x equals e. Now later on, we're not going to get the graph anymore, you guys.

We're just going to get the function or maybe even a table, which is worse than a function, okay? But we're going to have to find these relative max and relative mins on our own. I think you might be able to guess how we can find these. These ones that are on the nice smooth parts of the curve. Think about the tangent lines at those points.

What's special about the tangent lines at those points? there is zero. That's exactly right.

The slope of the tangent lines is zero. So in order to find those locations, think about what we're going to do. When we get a function, we're going to find the derivative and we're going to set it equal to zero. And if we solve that equation, it'll give us these three values. And then we'll have a test to perform to see if they're relative max or relative mins.

Now, the problem becomes, how do we find this one? Because the tangent line does not exist at sharp points. There is no tangent line at that point. The derivative.

does not exist at b. If you were to try to plug a b in to the derivative function, you would not get an output out. There'd be some sort of problem when you try to plug a b in to the first derivative. I'm not saying to the original function because you plug it into the original function, you're going to get a number out. You'll get h of b, whatever that y value happens to be.

So it's okay to plug into the original function this number. But it's not okay to plug into the first derivative. You won't get a number out when you plug that number into the first derivative. You won't get 0. You won't get negative 1. You won't get positive 1. You won't get anything.

There'll be some sort of problem when you try to plug in b, like division by 0. If you try to plug this number into the first derivative, you might get a division by 0 problem. And you can't divide by 0. So this one's going to be found a different way. And now I will tell you that the most common relative maximums that we get on the AP test are these types of nice, smooth curves.

And those can be found really easily by taking the first derivative and setting it equal to zero. This one's a little bit harder to think about. But I will tell you, most of them on the AP test are the nice, smooth ones.

So you will find most of them this way. And we'll show you how to find this one coming up in the process. OK, now here we go. Absolute maximums.

Here's the thing. Whenever you have a closed interval, you guys, like we do on this one, either from this symbol or the equal signs or these symbols with the closed brackets, we know we have a closed interval. So we're going to have endpoints.

So even if this graph continued on, you guys, even if the graph continued on, we would still look at only the parts of the graph between A and F, and it would be a closed interval. It'd be the exact same graph that she gave us. Even if this graph continued on, we would still look at the exact same pieces of the graph and we wouldn't change anything. It'd still be a closed point here and a closed point there, even if the graph continues on. And I want to tell you that because we're going to see this theorem pretty soon.

Whenever you have a closed interval that you're looking at, you guys, there's always going to be an absolute min and an absolute max. So what does that mean? That means whenever you have a closed interval, whenever you have a closed endpoint here and an endpoint here, that always means you're going to have a highest point on your graph and you're going to have a lowest point on your graph. You can't say that if these are open, but it's guaranteed if they're closed.

If they are closed intervals, if they're equal signs on the inequalities. then there's always going to be one absolute min and one absolute max. Now, the absolute mins, I should say the absolute min, may occur at two different spots at two different x values because two of them might be the lowest point or maybe even more. But there's only going to be one y value that's the lowest and one y value that's the highest on this entire graph if you have a closed interval.

That doesn't work for an open interval. And we're going to see that later on, but I just want to get you ready for it. We're going to have a theorem for that.

It's coming up in a little bit. But first, let me go ahead and just get you started here. Just tell you what the absolute maximum ends are, because that's all she wants you to do right here.

Again, I'm probably giving you a lot more information at this point than you need. But you know me, I like to give the story away. So my absolute min, again, is going to be the y value.

And my highest point, you can see, is this right here. That's h of e. Oh, I'm looking for the absolute min. I'm sorry. I was looking at the wrong one.

I was doing the absolute max. Well, I guess I could put it right here. H of E is my absolute max.

And that's a train ID at X equals E. You see, an absolute max can also be a relative max, or I should say a relative max can be an absolute max. The only one of them though, that was the highest point.

Now again, if we had a different, so let's say C reached the same height as E does. Then I would say A to B. Because whatever that y value is, it's still going to be the same thing for h of c.

So we say whatever that y value is, let's make something up 10. So let's say these two points both reach a height of 10. So this one also went up there. Then that means that I would have still one absolute max of 10 at x equals c and e. Two spots. And that's how you would say it. h has an absolute max of 10. at x equals c and e okay and this were also to go up to this height and that height would be 10. Just a hypothetical.

Okay okay in this case my lowest point I believe is right here now that I wrote all my graph let me take a look yes that's my lowest point that is going to be h of d and that's at x equals d and I know it seems kind of a redundant right now to keep putting the d and the d but remember this is going to be a number pretty soon like uh two okay two at x equals five okay so they'll be numbers pretty soon any questions about the definitions for relative max and mins or absolute max and mins or how to find them on a graph you go check the chat that's okay okay looks like we're gonna get some more definitions And I really want to get into the problems, is what I want to do. I'll show you the process, because we've already talked a little bit about this. Okay, relative or local extrema. Basically, they just got to be the highest and lowest point in their little area.

Y values on the graph of a function, where the function changes from increasing to decreasing and vice versa. That's going to be important, because that's going to be what our test is about. This is describing how we're going to test for relative or local extrema. We're going to test to see that it changes from increasing to decreasing. Or in other words, the slopes change from positive to negative.

If your slopes change from a positive to a negative, so all the slopes would go from positive to negative, you're going to create a relative max right there. That's going to be our test. And for vice versa, if it changes from decreasing to increasing. Then we're going to change from negative, which means decreasing is going down, to increasing. We're going to create a relative min right there.

And that's where we're going to test for these relative extrema. Again, I'm probably giving away a little bit too much right now, but I kind of like to explain it before we do it. And that way you get to hear it first because some people are auditory learners. Those people tend to do very well in my class because I'm more tell the story. And then we're going to practice it.

And that's where I need you guys to work on the homework to practice and then ask questions. Okay, absolute or global extrema. So these are the highest and lowest points of the entire graph.

The highest or lowest y values on the graph of a function are on the specific domain of a function. So think about how we're going to do that. We're going to need to find first all the relative mins and maxes, because those could be, as we've already seen, those could be your absolute mins and maxes as well. So we have to first find all the relative extrema.

And then we also have to test the endpoints for absolute extrema. Because for absolute extrema, your highest point could be at an endpoint or your lowest point could be at an endpoint. This one almost got to the lowest.

So that's what's coming up to find absolute. It's a little bit more work because we have to find all the relative and local and we have to test the endpoints as well. And we have to test them by just plugging them into the original function to see which one has the highest y values out of all of them. And then we'll see which one has the lowest y value out of all of them.

And that's basically how you test it. You plug it into the original function and you just compare all the y values. So you take all the x values of your relative max and mins and then the two endpoints, plug them all into your original function, and you see which one's your highest y value and you see which one's your lowest y value. And you've discovered. where your relative maximum or sorry where your absolute maximum ends are and what they are that's what's coming up this describes the test that we're going to do okay these are just simple words again it's better to see a process when you do this but some people can understand this verbally and then as we go through it it just clicks for them okay every one of those people consider yourself lucky because i was not i was not an auditory learner M-naught.

All right. Definition of relative and local extrema. If there's an open interval, what I put in here, y values are the relative extreme. We already talked about that.

So the relative extreme are the y values. That's basically what this is saying. If there's an open interval containing c on which f of c is a maximum, then f of c is called a relative maximum of f. If there's an open interval containing c on which f of c is a minimum, then f of c is called a relative minimum of f.

So it's just saying that the y values, the y values are the relative max and mins. That's all it's saying. Okay, this one I do like, though.

Some noteworthy thoughts. When we say an open interval containing C, we mean that we can find some interval, A and B, not including the endpoints such that C is in between A and B. But we don't include the endpoints, so this one is an open interval. So this one we do not need to test the endpoints, and absolute max and mins are not guaranteed when you have open intervals. All right, absolute max and mins are not guaranteed when you have open intervals.

Let me give you an example of that really quick. There's an open interval. There's a graph, y equals f of x. Has an absolute min, right?

It's got a lowest point on the graph right here. Whatever that y value is at that point, there's your absolute min. Where's your absolute max?

It can't be this point because it doesn't exist for us. So how close do we want to get it? What if this were on the x-axis?

I'll make it all positive for us. What if this were 1 right here on the x-axis? What's my max?

What if that hole up here is at 2? What's my absolute max? It can't be 2 because it doesn't exist at 2. So is it 1.9?

I think you can beat that. Is it 1.99? Is it 1.999? Is it 1.99999? What's the absolute max?

That's why when you have an, you still may have an absolute extrema when you have an open interval. Like if I drew it. This way, there's an open interval that has an absolute min and an absolute max, but it's not guaranteed to have an absolute min and an absolute max if you have an open interval. But if your interval is closed, you're always guaranteed to have an absolute max and an absolute min. And that's going to be coming up later on.

It's called the extreme value theorem. And now that I've explained it a little bit, maybe I can go a little bit faster when we actually see it coming up. That's called the extreme value theorem.

And all it says is that you're going to be guaranteed extreme values, absolute max and mins, if you have a closed interval. If you have an open interval, you may have them, but you can't guarantee it. You can't be certain of it.

That's called the extreme value theorem. And that's coming up here shortly. Okay, again, probably explaining more than I needed to.

That is, C will be contained somewhere inside the interval and will not be either of the endpoints. That's what it's saying when C is in between some interval. It's not going to be one of the endpoints. That's all that means.

A relative extrema is slightly different than an absolute extrema. Relative means it just has to be the highest in its little location. Absolute is all about being the highest of the entire graph or the entire interval.

All that's necessary for a point to be relative max or min is for that point to be a maximum or minimum in some interval of x's around c. So that's what's talking about a neighborhood, a little neighborhood around it. There are going to be smaller or larger values of the function at some other location, but relative to x equals c, f of c is smaller or larger than all other function values in the neighborhood of c. So I like that part for the neighborhood.

All right, now we get to do some examples. In many cases, we'll be asked to analyze both interval and closed interval functions. I'm sorry, both open interval and closed interval functions. Consider the graph of y equals g of x shown at the right.

We're not given any domain on this, you guys. We're not given any intervals. So what that means is we're looking at the entire graph.

See the way the arrows are pointing up? okay they don't necessarily have to have arrows you guys but if they're going to show you that it's closed they will put end points on there see these these are end points but if they leave those off you can assume that the graph is going to be going up okay they also don't have to put arrows on there to show you that it keeps going but to show you that it stops they will put end points see that but they don't necessarily have to have the arrows to show that it keeps going I know that's weird, but that's the way it is. Okay, let's get used to that. So I kind of wish you would have done this example without the arrows.

I can't erase them. But without the arrows, they still, you assume them to go up. If they want to show you that they stopped there, they will put endpoints like this.

Okay? Or they'll tell you the interval and they'll close it for you like this. What is this one? I'll give you this one. Zero to 12. They probably gave it to you too, but.

Right there. All right? And then that way the graph does keep going.

You know that if they're talking about 0 to 12, you can imagine there's an endpoint right there, and there's an endpoint right there. Whenever they give you a closed interval, you know there's going to be endpoints right there and there, and that's all you have to look at is in between. It doesn't matter that the graph keeps going.

You're just going to look at an endpoint there and there. So some other situations that can occur. All right?

Okay, state the relative extrema of g of x. So relative, local. So right here we have a relative extrema of negative 8. So it's going to be relative min of negative 8 at x equals negative 3. See that? That's how I write it.

This one, relative max of positive two at x equals positive two. And one more relative min right here. Relative min of zero at x equals four.

See how you write them? So again, the y values are the relative extrema, and the x values are the locations. Considering the domain for g, here's part b, sorry.

Considering the domain for g, what are the absolute extrema of the function? Now remember, this really doesn't have a domain. You can say the domain's negative infinity to positive infinity if you want. Look at the negative x's to positive x's.

You're going to look at the entire graph. You can see that this graph is going to keep going up and up forever. Even if they don't put the arrows there, it's still going to indicate it's going up and up forever because they'll either have to show you an interval like this or put the endpoints on there in the graph.

So this goes up forever. What that means is that we have no highest point, you guys. There is no absolute max. No, absolute max because the graph approaches infinity. The y values approach infinity.

What is this, g of x? g of x approaches infinity as x approaches positive or negative infinity. What does that say? The y values on this graph go up forever as And X goes to the left, because that's where negative infinities are, or to the right. You really only need one of them, because as long as it just goes to one side, you have no absolute max.

So this graph, which we've seen before, this is a graph of next cubed. That graph definitely has no absolute max or mins. because it goes down forever in this direction, it goes up forever in that direction. This one's going to have an absolute min though.

The absolute min is going to be negative eight at x equals negative three. Oops, absolute min of, put the at first, you could put the at first, I guess you just got to say it differently, of negative eight at x equals negative three. And that's it. That's it for that one. It's not too hard, but we're just learning it right now.

So I got to go over every single definition. I want to try to go over every single case that they could show you. Is there any other questions that you would have from that one before we do number three? We get to see another example here that is closed. That's OK.

OK, we're going good. That's what I wanted. OK.

Consider the graph of Y equals F of X on the interval from zero to 12 shown below. So they did it with. Equal signs.

That means closed bracket. You can also see from the graph closed circle here, closed circle there. Okay.

But even if the graph didn't have closed circles, you can now put closed circles on there because of the way they wrote the interval with equal signs. And you only have to look at in between there. Any Xs that are outside that domain, you get to toss out. You don't need them. Okay.

Find the relative extreme. We're going to like the toss out parts, especially when we do trig functions. Could trig functions have a lot of repeat answers?

and we'll get to toss a lot of those repeat answers away and just use the ones that are in the domain. Okay. Find the relative extrema of f of x.

You got it. So we're guaranteed an absolute extreme of both an absolute min and absolute max because we have endpoints. So this time we will get an absolute max. Okay. We have a relative min right here of negative two, relative min of negative two at x equals two.

We have, remember, those are going to be tough to find. A little bit tougher than the nice smooth ones that we were doing before. When we actually do the functions and not the graphs anymore, these will be nice and easy to find. Those sharp points are going to be a little bit tougher. This might be like a piecewise function that's broken up into one, two, at least three pieces.

Maybe even more. Maybe there's two pieces right here. Remember piecewise functions, we can have one, two, three parts to them, two parts to them. They have the, I call them the gatekeeper numbers on the sides they have to pay attention to.

Okay, we'll be seeing piecewise functions coming up. So let's see, we have relative min here. That is not a relative min or max. It's not the lowest point of the entire neighborhood.

It's not the highest point of the entire neighborhood, of a little neighborhood, I should say. And then this one, this one's a relative max up here. So we have relative max, I believe that's seven.

at seven relative max of seven at X equals seven. We have relative men right here. That's going to be relative men of two at X equals 10. I think that's it. We got them all relative men relative max relative men now for the absolute.

So this one we're going to check. all the relatives because they could be our absolutes. And we have to check the endpoints as well because the endpoints could be absolutes as well. Let's check the endpoints first.

This is the highest point of the entire graph or lowest point? No, it is not. Is this the highest point or lowest point of the entire graph?

No, it is not. So the endpoints are out. I always try, I always pull for the relatives anyway.

Those are always the ones that I want to win, the absolutes. But sometimes the endpoints win. Okay. For this case, the absolute.

is going to be right here at 2 and it is negative 2 so we have an absolute min of negative 2 at X equals 2 now is the lowest point of my entire graph and we have some absolute match right here at the relative max absolute max of 7 I believe it's at 7 at X equals 7 And that's it. So again, you are guaranteed an absolute min and absolute max if you have a closed interval. You're not guaranteed if you don't have a closed interval, even if it's an open interval. If these were open, you're not guaranteed to have it. You may still have it, but you might not.

OK, there it is, the extreme value theorem. What conditions would be necessary in order to have both an absolute maximum and an absolute minimum? The extreme value theorem. If F is continuous, so here is the prerequisite.

You need this. You need to say this just like we do when we do the mean value theorem. We have to say it's continuous and differentiable. Continuous on the closed interval.

Differentiable on the open interval. When we do the intermediate value theorem, we have to say it's continuous through the whole. function to the whole domain to the whole control For this one, it's got to be continuous as well, just like the intermediate value theorem was. F has to be continuous on this interval, and most times because it's a polynomial, and polynomials are globally continuous everywhere, but we have to stay that before we do the extreme value theorem.

Okay, if we can say that on a closed interval, then F has both a maximum value and a minimum value. Notice how they left out the word absolute. If you guys see on the AP test, just the word maximum and minimum, think of it as being the absolute maximum.

If they want you to just focus on the relative, they would say relative or local. OK, but if they leave the word absolute off or global off, then you can always think you can assume it's an absolute max or an absolute min that they're talking about. So it has both an absolute max and an absolute minimum value on the interval A to B. Just like I said, and all we need to do is say that's continuous and that it's on a closed interval and that's it. And then you're guaranteed at least one.

I should say you're going to have one maximum value, absolute maximum value and one absolute minimum value. They may occur at two different spots, but you're going to have one highest point and one lowest Y value. One highest Y value, one lowest Y value. Extreme values can occur at interior points.

So at the relative extreme or end points of an interval. A function may have both maximum and minimum values over interval, either a maximum or a minimum value, or a minimum, or no extrema on interval. Let's see some examples below.

When the hypothesis or if statement is not met, either continuity or the closed interval, there is no guarantee of the conclusion, then statement. However, a maximum minimum or both may still exist. There is no guarantee. There's just no guarantee.

That's what I was talking about. When I gave you my example with little open circles, you still may have one, but it's not guaranteed to be there. It's only guaranteed if you can say it's continuous and it's on a closed interval.

Then you can guarantee, yes, there's an absent max and yes, there's an absent min. I don't even need to look at it. I know there's one. Let's explore a few examples where this happens.

For each graph below, explicitly state where the hypothesis of the EBT fails on the interval from zero to three. and then determine if the function has an extrema on that interval. So when I'm given an interval like this, I go automatically and put endpoints on the graph at x equals 0 and x equals 3. And that way I don't need to look at any piece of the graph that's to the left of it or any piece of the graph to the right. You probably won't need to do that on any of these because they're already marked just from 0 to 3. None of the graphs continue on in either direction.

Okay, but I just want to let you know that I had graphs that continued on in either direction. When I get this interval, I put a closed circle on the graph at 0 and a closed circle on the graph at 3. It just helps me focus it. Okay.

Then determine the function has an extrema on the interval. Okay. You got it.

Is it guaranteed? Does the hypothesis get met for this graph? No, it doesn't. First, it's not continuous.

You got the closed intervals. You got the closed circles, but you don't have the continuity. So fails.

Because they didn't name the function. No, they did not. We're going to name it.

y equals f of x for this one. y equals g of x for this one. y equals h of x for this one. And y equals q of x for this one. That's what I'm going to name it.

Fails because f of x is not continuous. at x equals 1. And 1 is in the interval. So the extreme value theorem is not guaranteed. Now, does it have a highest point and a lowest point? Yes, it does.

It has a highest point. It has an absolute max of 4. It has a closed circle there. So that's the highest point.

If that was an open circle, I couldn't say it has the absolute max anymore, you guys. Because if that were the open circle, I kind of wish you would have did that one. If that were the open circle up there, then what do you say your absolute max is?

You can't say four because it's open. So is it 3.9, 3.99, 3.999? How high do you go?

It doesn't have an absolute max. And that's an open circle there. Because it's a closed circle, my absolute highest point is at four. See, it has an absolute max. It's just not guaranteed.

Okay, lowest point, that was a closed circle as well. Yep. So we have absolute min of, I'm going to say one half, you guys.

I'm going to have to guess that one. You won't have to guess on the AP test, but I'm going to say one half. That equals zero. Okay. What about B?

B also fails because it's not continuous. And this one I'm just going to put fails because g of x is not continuous on the interval, because I don't know where that's at, on 0 to 3. There you go. Fails because g of x is not continuous on the interval from 0 to 3. You can call it 1 and 1 half, I guess, if you want to.

But does it have a highest and lowest point? It's got a highest point. There it is. Absolute max of, shoot, I don't know, four and a half. That's a guess.

At x equals zero. I know that's true. So at the end point, it's got the absolute max.

What about an absolute min? Oh, here's my example right here. There is no absolute min.

There's an open circle here. So it looks like it's 1, but there's an open circle there. So what is it?

1.01, 1.001, 1.001. What's your absolute min? Don't know.

No absolute min there. For here, look at this. There's an open circle there.

When there's an open circle, don't put a closed circle there. Only do that when you see the graph. And especially if it's continued on. And she didn't give me an example that I can show you on that. But if this graph were continued on, then you put closed circles there.

But if there's already an open circle there, don't close it, you guys. Because there's an open circle. You just have to look at that.

This thing is not continuous from 0 to 3. It has to be continuous at 0 and at 3. And it's both not continuous at 0 because it has a hole there. And it has a vertical absolute at 3. So it's got the two problems that we know are just content. discontinuities.

A hole called a removable discontinuity and a vertical asymptote called an infinite discontinuity. So this is not continuous at the endpoints. Fails because h of x is not continuous at x equals zero or x equals three. And this has no absolute extreme at all.

There is no highest point because there's an open circle there. And there's no lowest point because this thing is going to go down forever. No absolute extreme.

Okay. Next one. Why does this one fail?

This one fails because it's also not continuous at the end points. Now continuous at 0 and now continuous at 3. The exact same reason as this one. Fails for the same reason as C. Now continuous at the endpoints. But it does have a highest and lowest point.

I don't know exactly where they are. That one looks a little bit off, but I don't think I need to name these because you've already seen how I write them. But here is your absolute max right there. And here is your absolute min right there, whatever those values are.

But all this was trying to show you is that even though you can't state that there is one, there still might be one. OK, but it's not guaranteed by the extreme value theorem. It's only because it's continuous. on that closed interval, then you got an absolute max and an absolute min. Okay, guys, let me see if there's any questions about that before we move on to some of the other examples.

That's good. All right, let's keep going then. The extreme value theorem gives us knowledge of when extrema occur, but how will we find them algebraically when given an equation only?

Without the benefit of a graph, So our next question should be how can we identify all the bodies of X? where extreme values occur. Here we go. For each function shown below, just if I explain why the extreme value theorem can be applied or why it does not apply in the given interval. Okay, so it still hasn't had me do a problem yet.

This just wants to know why could we apply it or why could we not apply the extreme value theorem? Okay, here We're going to have a division by zero problem. Whenever we try to plug in a 2 in for x, we're going to get a zero down here, which means it's not going to be continuous at 2. We have a discontinuity at x equals 2. Now, remember that we were doing these discontinuities.

It could end up, division by zero could end up being a vertical asymptote if you get a non-zero number over zero, or it could end up being a whole, a removable discontinuity. But either one, it just means that's a discontinuity. There's going to be a discontinuity at x equals 2. So if I know my function is discontinuous there with either a whole or vertical asymptote, you can find out if you really want to know by plugging in the 2 in the top. But we don't need to know which type because all we need to know is that's discontinuous there. And if it's discontinuous at 2 and 2 is in between the interval that it wants me to look at, then I know the extreme value theorem will not apply.

Therefore, three little dots in a triangle like that mean therefore. Therefore, the extreme value theorem does not apply. That one was easy. If you remember that division by zero is a problem and it gives you discontinuity in your graph, some sort of discontinuity, either a whole or vertical obstacle, that one wasn't bad. This one.

exponential functions. Exponential functions are continuous everywhere. You can plug in any number you want into them.

Their graphs look like this. And then you're just going to shift it. So all exponential graphs look like this.

They go up the fastest. OK. And this one's always a one right here because anything to the zero power is always a one.

So this would be like the graph of e to the x. The graph of 3 to the x, the graph of 5 to the x, the graph of 10 to the x, they all look the same, like that. They're continuous everywhere.

They have no problems. They have no holes, no vertical isotopes. So this thing is going to be continuous everywhere.

It doesn't matter that you're subtracting 2 from a number. You can subtract 2 from any number you want. So this one does apply. Let's say g of x is continuous on the interval from negative 5. Positive 5, the one that they wanted me to look at.

Therefore, the extreme value theorem, well, I said intermediate value theorem, the extreme value theorem does apply. And for this one, last one, the thing with square roots, you have to be careful about negative numbers. So in this case, you could plug in.

negative six. Yes, if you plug in a negative six into this, you get a negative four. And the square to negative four is not a real number. And in calculus, we only deal with the real numbers.

So in this case, this is not going to work because this is not going to be continuous at negative six. In fact, it's going to be continuous, not going to be continuous at negative five, negative four, negative three. It'll be okay at negative two, but there's a lot of numbers where it's not continuous.

So h of x. That's not even defined for those. But all you need to say is not continuous.

It's not continuous on the interval negative 6 to 6. Therefore, EBT does not apply. And I think she just wanted to remind you division by zero, and then you can't take the square roots of negative numbers. because they'll be imaginary. Wasn't much to that one. I want to get up to an example where we have to do one where we don't get to look at the graph.

Oh, critical number. Here's another definition for you. Okay.

Remember that critical points are points on your graph that could be relative max and mins. You just have to test them. Okay.

A critical number value of a function. This is actually going to show you how to find this one as well. Remember I told you. These will be easy to find, because we're going to set the derivative as equal to 0. This one will be harder.

But this point right here, I should say the point B, H of B, that point right there on your graph, B, H of B right there, X value is B, Y value is H of B, that point right there is a relative min. The derivative does not exist there, but it's still a critical point. It's still a critical point. So how are we going to find that one? This is what we need to do.

So to find all these critical points where these possible locations of relative maximums could be, without looking at the graph, this is what you need to do. You're going to take your derivative, and you're going to set it equal to 0. And you're going to solve that equation. And that's going to give you all of these, the nice smooth ones. But you also have to find out where your derivative does not exist.

So what does that mean? That could be like division by 0. When you divide by 0, that doesn't exist. Have to check it.

So this is a good example. If this were my first derivative, you guys, and not my original function, I'd want to set it equal to zero and solve it to find out all these possible locations. And then I'd also want to find out where it's undefined.

And this is undefined at x equals 2. It does not exist at x equals 2 because division by zero. That would also be a critical point that I'd want to test to make sure if it's a relative max or min. So a lot of times when you're looking for where it doesn't exist, it's going to be a rational function. And when you get a derivative for a rational function, it's going to be really easy.

If you want to just know steps to find out where it equals zero, that's basically the numerator. You take the top and you set it equal to zero. That's to get this part.

To find out where it doesn't exist, just take the bottom and set it equal to zero. So if you're frustrated, it turns out to be a rational function like this, a numerator over denominator with X's on top and bottom. Just take the top set equals zero, take the bottom set and equal to zero.

And that will give you the locations of all of your critical points. It will find the nice smooth ones produced by this. And I'll find the sharp ones produced by something like this. Now you still have to test them because they may not be relative max and mince, but at least you got.

all the possible locations for them. So now all you need to do is perform the test. Okay.

If X equals C is a critical value, then the point X F of C, here's my X value, here's my Y value, is called a critical point. Okay. Sometimes on the A&P test, they'll ask you to find all the critical points or all the critical values.

And you just have to find out where the rate of equal zero and where it doesn't exist. Okay, theorems. Relative local extrema can only occur at a critical value on an open interval. See, I told you, that's why we need to find all the critical values first, and that will tell us where these relative maximum ends are after we test the critical values. Absolute global extrema must occur at a critical number, so it could occur at a relative extrema, or at an endpoint on an interval.

So here's more to test for absolute extrema. You got to test all the critical ones and the endpoints. Using the candidates test to determine absolute or global extrema. Now we need to put these ideas together with a little direct practice. Finally, let's develop an algebraic approach to what we call the candidates test for finding extrema.

That's what we're going to call candidates. There we go. Finally get to do some practice. All right.

Find the absolute extrema on the function of the given interval provided the EVT is applicable, if not justify why. So we're going to find the absolute extreme, which means we need to find all the relative max and all the relative mins. And we also need to test the endpoints.

We get to use a calculator. This is a polynomial. So I'm going to try to do this without a calculator.

I'm not sure if it's possible, but we'll try to do without a calculator and I'll try to show you with a calculator with these because these will be trig functions. These would be definitely want to be calculator for. Okay, let's go ahead and take the first derivative. Oh, wait, let's say if this is going to be guaranteed, first of all, are we going to have an absolute max and absolute min? First of all, we got a close interval.

Good. Now is it continuous on that closed interval? Yes, it is.

It's continuous because it's a polynomial and polynomials are globally continuous. f of x is continuous on interval from negative one to two because f of x is a polynomial. That's all you got to say for that. Nice when they're polynomials. Okay, now that means we're going to get an absolute max and an absolute min.

That could be at one of the relative max and mins, or it could be at the endpoints. All right, we're going to test them. Let's find all the relative extreme at first.

So let's find the first derivative. This will give us the equation for all the slopes of the tangent lines to this graph, and that's going to be 24x squared minus 6x minus 9. And I'm going to set that equal to zero to find out where the nice smooth parts of my graph are. And I'll find these spots.

But then I also want to consider if there's any x value that would make it undefined. That could also be a critical point. So two things. We're going to find out where f prime of x equals zero.

So that's going to be 24x squared minus 6x minus 9 equals zero. We need to solve that equation, you guys. And we also need to consider, n is f prime of x, the first derivative.

Is it ever undefined? Would there be any number that you can plug into this where you wouldn't get an output? That's no. You can plug any number you want into this and you're going to get an output because this is a polynomial, you guys.

It's always defined. So what that means is there's not going to be any sharp points in this graph. that we have to check.

There's just going to be the nice smooth curve. There will be no sharp points in this graph that we have to check. So this piece is out. We don't need to do this piece. Now the question becomes, do we have to write it?

I probably would. I probably would at this point say that I'm looking for these two pieces, but I don't, I've never seen where you're going to be given a point for this. OK, but every time that I'm trying to find my relative maximum ends, I always do the two steps because this one is easily forgotten.

People always want to take derivatives set equal to zero every single time to find all the relative extrema. And most of the time it will work to find all the relative extrema. But there's going to be some cases, you guys, especially when you get a rational function for a first derivative. There's gonna be some cases where we have to consider this as well.

Not in this case because it's a polynomial. Okay, let's go ahead and solve this. I want to see if I can solve this without a calculator, and then I'll show you how to do the calculator and solve an equation. Basically, if you want to solve an equation with a calculator, you graph the left-hand side as your y1, you graph the right-hand side as your y2, and then you find out where they intersect on your calculator, and that can solve any equation.

And that's going to be good for you because if you decide to take the AP test at home, because you'll get the option of taking it at home this year, you'll be able to use a calculator to solve an equation. You'll be able to use a calculator for all your problems. They can't stop you from using a calculator. So you'll be able to solve any of these.

So if you're not good at factoring, if you're not good with the algebra, this may be your year for an AP test because calculators can solve these equations for you. Okay? So that's what I'm saying.

Remember, if you're not good in algebra, this might be your year that you can take advantage of it because normally when you do the AP test, they take the calculator away for part of it. They can't do that when you're at home. So take advantage of that. All right, let's see.

I'm going to go ahead and factor out a 3 from those numbers to make them smaller. That leaves me with 8x squared minus 2x minus 3 equals 0. You can divide both sides by 3, and the 3 will disappear. I kind of like getting rid of numbers when I'm solving equations, so I like to just factor it out. But it's okay to divide by. It wouldn't affect anything.

And I'm going to try to factor. And again, if you're not good at factoring, then you would just use your calculator at this point, right? Graph this and left or y1 graph this and y2 find out where they intersect and I'll show you those when we get to the Other examples, but this one I think is factorable.

So let's see what I have 8x in 3 I need to make a negative 2 so I'm gonna try 4x and 2x It's just trial and error right here and then 3 if we're gonna put here I'll make it 6 and then 1 will be a 4 and I get a 2 out of that Yes, so 3 here 1 and that I need a negative 2, so I need the 6 to be negative. and the four to be positive. There we go, and then if I solve this, this four x has to equal three in order for it to equal zero. Four x would have to equal three, because three minus three is zero, so x would equal three fourths. So your algebra one teacher always taught you steps.

Little teacher, just think about what you're doing. Two x has to be a negative one in order for this to equal zero. Divide both sides by two, you get x equals negative one half. All right.

So we have two critical values. We don't know the points yet because we don't know what the y values are. We want the y values on the original function, and we have to take these and plug them into x's for the original graph. There's no point in plugging them into the first grader. You're going to get zero.

In fact, if you don't get zero when you plug them in the first grader, then you didn't solve it correctly. All right. So now the thing becomes, what are they? Are they relative max? Are they relative mins?

If you want to test to find out what they are, you don't need to for absolute max and mins. You can go right past this, but I want to show this to you guys. Okay. Because some questions on the AP test just ask you to find relative extrema.

So here's what you do to test for relative extrema. And then I'll show you the test for absolute extrema. Okay. You don't need to test for relative before you test for absolute. If we're just doing absolute like this one wanted, I don't need to do this.

I don't need to tell you if these are relative maximum ends. We can just test them all into the original function to see which is the highest and lowest point. Okay.

But I want to show you this test for relative extreme. Here's what you do. You make a number line.

You place those two numbers on your number line. I'll do it on top. I like doing them on top actually. Okay.

and then 3 fourths is after that to the right. And now just think of this as a graph so far. So we're going to basically just fill out the graph. We know that there is a nice smooth point at negative 1 half and at 3 fourths. And we know that the slope of the tangent line at negative 1 half 3 fourths is 0. Now we're going to test to see if the relative maximum ends by picking other x values to try into our first derivative.

Let's say I want a number to the left of negative 1 half. How about negative 1? Pretty close, but it is to the left of negative one half on a number line.

I want a number in between negative one half and three fourths to test right here. I'm going to pick zero. You don't have to pick the same numbers I do, but you got to make sure that they're in the correct intervals. You need this number to be in between these two. If you can't, always try zero somewhere.

You're going to get zero somewhere. As long as it's not one of these numbers. And then another number to the right of three fourths.

How about one? One's to the right of three fourths. We're going to test those three numbers into our first derivative.

And all we care about is the sign of the first derivative, because the sign will tell us what the graph of the original function is doing. If we get a positive first derivative, then that means our graph is going up, because all tangent lines with positive values, with positive slopes, means that your graph is going up. If we get a negative value for this, then that means our graph is decreasing.

It's going down. So we'll know what's happening right before this point. And then we're going to put this number in to the first grader to find out what's happening right after this point. Because if we have that change, we're going to create a relative extreme right here.

So let's try it. Whenever I plug this in, especially into a polynomial that I've already factored, I like plugging it into the factored form. Because it's so much easier to get the signs from the factored form. You don't need the number. You just need the sign.

So let's plug a negative 1 in for the x here. So you have a positive 3 that we're multiplying by. So we have a positive 3. And if I plug a negative 1 in for this x, I get 4 times negative 1, which is negative 4. Minus 3 is another negative number. If I plug a negative 1 into this x, I get 2 times negative 1, which is negative 2. Negative 2 plus 1 is a negative number.

So we have a positive times a negative times a positive, which, I'm sorry, a positive times a negative times a negative is a positive. That's what I meant to say. And you know what that means since we plugged it into the first derivative. This gives us the slopes of the tangent lines for all numbers before negative one half. They're all going to be positive slopes.

We know when they're zero already. These are the only places where our slopes are zero. So we know before that our slopes are positive, which means my original graph of f would be going up before it reaches negative one half. Now just because it goes up before it reaches this point, doesn't mean that it's going to go down after this point. It could go flat for a little bit and then back up.

Okay, so that's why we have to test the left and the right side of these points. So now we plug a zero into the first derivative. If I plug a zero in, it's still positive three. Four times zero is zero, and zero minus three is still negative. If I plug a zero into this one, two times zero is zero, but zero plus one is a positive.

So now we have a positive times a negative times a positive, and that's going to be a negative. So this means that my function, my original graph, would be decreasing between these two numbers on my x-axis. So my original graph is going to go down after this, and then it flattens out here at 3-4, and then we're going to see what it does after 3-4.

It's probably going to go back up, but it doesn't necessarily have to go back up. It could keep going down. It just flattens out a little bit right here and then goes down this way.

But right now, we just discovered that there's a relative max at x equals negative one half because our slopes change from positive numbers to negative numbers. And those slopes are the slopes of the tangent lines, which means this tells us that the graph goes up and when the graph goes down according to these slopes. Increasing, positive slopes, positive derivatives. Decreasing, negative slopes, negative derivatives.

It's all starting to come in. Remember when we use these words to describe derivatives? We have a positive derivative.

We would use the word increasing. It's increasing by this many tons per hour. We had a negative derivative.

We would say decreasing. It's decreasing by this many tons per hour. It's all being connected here. Okay.

And then one more to check to see if this is going to be a relative min. It's the only possibility because we're going down before it. And now we go up, we create a relative min.

So let's plug in the number one. into the first derivative to see what the slope of that tangerine would be. If I plug a 1 in, that's still a positive 3. Plug a 1 in here, that's a positive because 4 minus 3 is positive.

If I plug a 1 into here, that's also a positive. So I have three positives being multiplied together. That means we're going to be increasing afterwards. So it's going to be going back up, and that means we created a relative min at x equals 3 fourths.

If we want to know what those min values are, Then we plug those numbers not into the first derivative because we don't want the slopes of the tangent lines. Now we want the y values on the original function. So we plug them into the original function and we figure out what those y values are.

And then you can find the relative max and min. That's how you test for relative extrema. You pick numbers, not the same. You just draw a number line, pick one to the left. In between, one to the right.

If there's more, do the same thing. You plug them into the first derivative. So this is called a first derivative sign chart.

First derivative sign chart. That's what it's called. And it's a test for relative extrema. Okay, now that test does not have to be done for this problem.

You don't need to know if these are relative max or mins in order to test for absolute. Because once you get all your critical numbers, then you're just going to go and test them all back into the original function. So here's called the candidates test.

You don't say candidates test, just put candidates. You're going to have x equals negative one half. That's one of the ones that could be up for the highest or lowest point.

We know from this test that it's only could be an absolute max. Because we know it's a relative max at x equals negative one half. We know that relative max can only be an absolute max.

It can't, you can't have a relative max be an absolute min. It doesn't make any sense. Okay.

But you don't need to know that. All you need to know that there's a critical point there. So you're going to test it.

X equals three fourths is another critical point. Could be relative maximum in there, which means it could be an absolute maximum. And the other two possibilities, I call them the born-in candidates.

The other two possibilities for relative maximum are the two endpoints, negative one and positive two. So if you're answering just a question about absolutes, you don't need to test for relative. You can go right to the candidate's test for all four of them. And this is where you use a calculator, especially when you have fractions like this. And you plug it into the calculator and you're going to see the Y values on all those.

This is F, right? So F of negative one half, F of three fourths, F of negative one and F of two. Let's do that.

Let me show you with a calculator. I got my calculator ready to go. So we'll store in the function.

That's what I like doing. I'll go y1 right here. I'll get rid of these because you don't need those anymore. Back up, and we'll put in the original function, which is 8x cubed, 8x cubed.

We'll put the cursor down to the right. Minus 3x squared, minus 3x squared, minus 9x. Minus 9x and then plus 2. Okay? And that's all I really need. I don't even need to see the graph because I'm just going to have this thing plug in numbers.

So I'm going to quit this screen, go back to my home screen, my computation screen. I started in y1. So I got to remember that. So it's alpha t and y1 is number 1. And I'm going to have it plug in the first number, negative 1 half.

So parentheses, negative 1 divided by 2. You can even put negative.5 if you want. Don't plug that number in for me. And it's 4.75. 4.75. So we know the height on the original function F.

4.75. Now we've got to compare all the other ones. So we do three-fourths next.

The easiest way to do this is to do second and enter. Second, enter. Brings up your last entry.

And then you just go to make changes. That way you don't have to do the alpha T again anymore. You could if you want to. Delete the negative. I'm going to replace the...

One with a two, I'm sorry, the one with a three, and the two with a four. And we'll put in three fourths. Enter. Negative 3.065, negative 3.0625.

Okay. Now we'll do negative one. Second, enter. And I should have probably done negative one as the second one would have been easier. Negative one.

And I'll delete the four. There we go. Enter. 0. And then my last one, positive 2. Second, enter.

Positive 2. Move, move. 2. Let's get rid of the negative sign. Delete.

Enter. 36. All right, we got our values. Let's see the winners. So Find the absolute extreme and you got it.

So the absolute extreme, so f has an absolute max of 36 at x equals 2 and an absolute min of negative 3.0625 at x equals 3 fourths on the interval. From 0 to 12, what was it? Negative 1 to 2. You know, I forgot to tell you, if any of these critical numbers were outside this interval, we could have tossed them away, but they were both inside the interval.

Okay, but I did forget to mention that. If any of these critical values that we saw for, by saying they're equal to 0, were not inside this interval, we could have tossed them away, and we wouldn't have to check them. Okay. But that's how you do the test for absolute extremism.

It's called the candidate's test. And you see, you don't need to know if they're relative max or mins in order to test them, because you're just going to look at the Y values and compare them. All right. How'd I do on that one?

Any questions? When testing for relative achievement, how do we know the graph doesn't weirdly change in between the points where we tested? What do you mean by weirdly? You mean like a sharp point? It's not going to be a sharp point because you'd find all those by finding where the first derivative is undefined.

So again, the word weirdly is a little bit ambiguous because what's weird to you might not be weird to me, but I think that's the biggest thing is you need to test to find out where it equals zero and to find out where it's undefined. A lot of times you'll get a rational function for the undefined part. A rational function for the undefined part. Okay? Okay, the last thing I want to show you though before you go is I want to show you how to do this with a calculator, because I didn't get to show you that.

So let me go ahead and let's see if I can do this one really quick. So g prime of x equals 1 minus 2 cosine of x is my derivative. Set that equal to 0. I know I'm doing this quick, guys.

1 minus 2 cosine of x. I got to remember that. 1 minus 2 cosine of x. 1 minus 2 cosine of x.

I'll clear that. 1 minus 2. Make sure you're in radian mode when you do this. I know I am. I'm not going to check, but make sure you are.

OK, my other one is 0. So I'm going to go down, go to 0. And I have an interval from 0 to 2 pi. So let's put that in the window as well. Window from 0 to 2 pi.

0 is already my min. 2 pi is already my max. There it is right there. Zero to two pi. Let me just show it to you, though, in case you don't get the two pi.

Two second, and then exponent button, two pi. And then when you move it, it puts it into decimal form. Negative five to five is pretty good.

I'll just go ahead and graph with this one. And it should graph y1 and then y2. And you're going to find out where they intersect.

So it's second t. Intersect is number five. And you just move the cursor to the first intersection point. You can hold it down. Move it faster.

Just get it as close as you can. And then you basically hit enter three times. One, two, three. And you got your x value. You can solve any equation like that, you guys.

Even the one that we just did. x equals 1.0471. This is the one good thing about distance learning is that the algebra should not bother you on here because you're going to have a calculator. And then the other one really quick. Got to do it all over again.

Second t. Intersect number five and then hold it down I always like make my equations equal to zero because you get that's just the x-axis and it's easy to see the x-axis That one is five point two three five five point two three five five point two three five. I like one more nine Especially since I'm kind of in the middle of the problem. So I got those two as my critical points. Cosine of x is always defined, you guys.

They get the cosine as the pretty wave. There's no discontinuities there. So it's always continuous.

So I should have said that x is a polynomial. It's always continuous. Sine of x is a trig function. That's a trig function that's always continuous.

So the EBT, the extreme value theorem, does apply in this interval, you guys. I forgot to mention that. But I wanted to show you the calculator part really quick before I let you go.

Okay, just one more time in the video so you guys have it. Okay, let's check the roll. Sorry, we didn't get a whole lot of chance for questions. I will finish those and put them up on a video, just like I did with the rest of 5.1, okay? I'll put those up on a video.

I got to the main points that I wanted to get through. Now you just need to see the examples worked out. Okay, who's my only absent today?

Diego, Diego, you're my only absent today. If you're in here, let me know. Everybody else, you can go.

Have a good day. I'll post the homework up. Sorry about the due date on the last one. I forgot about the three-day weekend.

Have a good day, guys. I forgot to turn off the recording. Let me turn off the recording.

Transcript for:Understanding Extrema in Calculus

Transcript for:
Understanding Extrema in Calculus