Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Extreme values of functions. Let F be a function with domain D Then F has an absolute maximum value on D at a Point C if for all F of XS it's less than or equal to F of C for all X in D The same kind of idea occurs. For an absolute minimum, we would have to have some location where all F of XS were greater than that single location. Absolute maxism and min's are called absolute extrema. We have an extreme value theorem. If F is continuous on a closed interval AB, then F attains both an absolute maximum and an absolute minimum value in that closed interval. With F of X1 equaling little mie. The little Ms. going to stand for our minimum and F of X2 equaling capital M. The capital M is going to be our maximum, and M is less than or equal to F of X, less than or equal to M for every other X in the interval. So basically, if we have a maximum and a minimum, every other value has to fall between it. The first derivative test doesn't have to be a closed interval anymore, and it's for local extreme values. Those at local extreme values are also known as relative extrema. So if we look at just any old curve, then we can see that by the 1st derivative test, maximums and minimums occur when the first derivative equals 0. So if F has a local Max or min at an interior Point C of its domain, and if F prime is defined at C, then F prime C = 0. So whenever the first derivative equals zero, we have a Max or a min. Building on that, what if it doesn't equal 0 but it still has a Max or a min? What happens here is that then it can be undefined for the derivative. So if we have a critical point, an interior point of the domain of function F is where F prime is 0 or undefined. So back here, if F has a local Max or min at an interior Point C of its domain, and if the derivative is defined at C, then the first derivative has to equal 0 for a critical point. A critical point could also occur if the first derivative is 0 or the 1st derivative is undefined. So how we find absolute extrema of a continuous function? The word continuous is important here. On a finite closed interval, we're going to evaluate F at all the critical points and at the end points. So to find the critical points, you take the first derivative and find where that first derivative is 0 or undefined. You're going to take the largest, which will be the maximum, and the smallest, which will be the minimum F values. The only places that extrema can occur are at end points where F prime is 0 or where F prime is undefined. Let's look at some examples. The first one are going to involve graphs. So we want to look at the graphs and figure out whether the function has any absolute extreme on AB and if so, where. So when we look at this one, the smallest point is going to be C2. So our min is going to occur at F of C2. Whatever value that is our highest point for number 1 is going to be at this B. So our Max is going to be at actually it told us HS for this function. So H of C2 and H of B. Looking at this next one, we have a highest location right here. So our capital M is going to be F of C Our lowest or our smallest M is going to be F of B. We have local extrema here too. Local Max on number one would occur at HC1A. Local min would occur at H of A. On this new one, we have a local min at F of A, and what a local means is just all of the points immediately around it, or the closest points it's the highest or lowest value for. If we look at #3 over here, we're going to have an absolute Max at F of C, We're not going to have any M's. And it's because it wasn't a closed continuous interval. These points were open. So there is no lowest point, there's only the highest #4 we're actually not going to have any Max and we're not going to have any min because there was no highest point or lowest point. The highest points and lowest points are open dots where they're undefined. 7 through 10. We're looking at graphs. Again, we just want to find the absolute extreme values and where they occur. So the highest point is an open dot. The lowest point is an open dot. There is no absolutes. However, there is going to be some locals. A local minimum is going to occur at that -1 because that's the smallest point for all the points immediately around it. We're going to have a local maximum at 1:00 because all of the points immediately around it are less than that Y value at 1:00. When we look at #8, we have an absolute at X = 0, sorry, an absolute Max. We have two absolute mins here at X equal -2 and X equal 2 #9. We're going to have an absolute Max at X = 0. The maximum value is really 5. So we could actually say for all of these, we have an absolute maximum at X = 0 of 2, we have an absolute minimum at X equal -2 of 0 and X equal to of 0. This last one we're going to have an absolute Max at X equal 1 of the height 2. We're going to have an absolute min at X = 0 of -1. We're going to have a local Max at -3 X equal -3 of 0 and a local min at X equal 2 of 0. Looking at some more examples, this next type they're going to tell us what our derivative is, and they want us to match it with the graphs. So at A we want a derivative of 0. That means that we have to have a horizontal line or a Max or a min at A. At B, we want a derivative of 0, so we have to have a horizontal line at B. So A&B are going to both be Max's and min's. And at Point C we want our slope to be positive. So our original function needs to be increasing. So when we look at this, I needed horizontal line at A and AB and at CI needed a positive slope. So #11 is going to be matched with graph C If we look at 12, I need a horizontal line at both A&B again, and now I need my C to have a negative slope. So here's a horizontal at both A&B and my C needed to be a negative slope. So we're going to have that be graph B 13. The slope at A does not exist. It's going to be a cusp or corner. So it's got to be a sharp instead of smooth continuous. At B, we're going to have a horizontal line or a Max or a min. And at C we need a negative slope. So here, because of that cusp, it's not going to exist at B on part A, it's not going to exist either because it's a sharp point. So if we look down here at D, we have a cusp, so not defined, but at BA smooth continuous curve. So we do have a horizontal and at CA negative slope. So this one's going to be D14 does not exist for a does not exist for B&C has a negative slope. So a cusp, another cusp, and a negative slope. That one's going to be our letter A. Let's look at some that aren't graphs. Let's look at some now that have equations. So if we're given F of X equaling negative X -, 4, if we want to figure out and we're given an interval -4 less than or equal to X, less than or equal to four or less than or equal to 1. So our first thing is going to be to find the endpoints F and -4 and F of one. So F and -4 the opposite of -4 -, 4 is going to give us 0. The opposite of 1 -, 4 is going to give us -5. Now we also need to find the first derivative. So the first derivative is just -1 and we need to know what constant in the interval is going to give us out -1. So we're going to say F prime of C is -1. So this is going to occur for all C's in the interval. The first derivative is -1. So if we really thought about this, this is just a graph of a line. So we have the point -4 zero and we have the point one -5. So do we agree that every point along the way has a slope of -1? So F prime of C is never ever ever equal to 0. So there can't be any extreme points in between. So this is going to be an absolute Max and an absolute min. Looking at some others, if we have F of X equaling 4 -, X ^2 on the interval -3 less than or equal to X less than or equal to one, first thing we're going to do is we're going to find F and -3 and F of 1. So F and negative three 4 - -3 ^2 is -5 four -1 ^2 is 3. Then we have to find the first derivative. The first derivative is going to equal -2 X. We're going to set that first derivative equal to 0 and solve for X. And then we're going to say what is F of that location? So F of 0 is going to give us 4. When I look at those Y values, that tells me this is an absolute min. This is an absolute Max, and something's happening here. So if we look at this graph, this graph's going to look like this. So this one's going to be a local min where this was negative three -5 this was 1-3 and that was 04. If we look at another example, F of T equaling the absolute value of t - 5 from 4 less than or equal to T less than or equal to 7:00. So we start by finding F of four and F of seven, F of four. The absolute value of 4 -, 5 is going to give us one. The absolute value of 7 - 5 is going to give us 2. Now we need to find the first derivative, and we're going to start by thinking about our F of T. We have to figure out how to get rid of that absolute value. And there's a couple different ways to do it. But one of the standard methods is to take the square root of the quantity squared. And the reason is that if I have the square root of something, it always has to come out positive. And if I took that original something and squared it, it's always going to be positive. So we could think of this as t -, 5 ^2 to the one half power. And then if we do the chain rule, so we're going to bring down the 1/2 to the one less power. And then we're going to times that by the derivative of the inside piece. So then we're going to bring down the power again and take it to the one less exponent times the derivative of the inside piece. So we're going to get F prime of T equaling this 1/2 and that too we're going to cancel. So we're going to get the derivative of the inside was just one. Simplifying this up, the 1/2 and the two cancel. We have t -, 5 on top. That negative exponent says we're going to bring it to the denominator. And then because we used the fact that the square root of a square is the absolute value, we could rewrite this as as the absolute value. Now some of you may have said, well, we could see from the original that we were going on four to seven, hence we should just use positive. And the problem is that when we stick in four, we actually get a negative value. So it's important that we understand what I just showed and that we can understand and do it ourselves, yourselves, everybody can do it. So here we're now going to set that derivative equal to 0. If we cross multiply, the denominator goes away and we're going to get T equaling 5. So I need to figure out what F of five is. F of five. The absolute value of 5 -, 5 is 0. So our three Y values, we're looking at 1-2 and zero. So this is going to be my absolute min because it's the smallest of the three. This is going to be my absolute Max because it was the biggest of them. And then we're going to have a local Max here. If we thought about what this graph looks like, this graph at 4:00 would have a value of one at 0, no at 5 would have a value of 0, and that seven would have a value of 2, something like that. So we have critical points zero critical point at 50, a critical point at 41 and a critical point at 7 two. Looking at another example, Y equal X ^3 -, 2 X +4. So this time they're not going to give us any intervals. We're just going to do this on any interval. So if we have Y prime, we're going to get three X ^2 -, 2. We're going to set the first derivative equal to 0 and solve. So we're going to get 2/3 equaling X ^2. Our X is going to be positive negative sqrt 2 thirds. Our X is positive negative root 6 / 3. So we're going to have F of root 6 / 3. And when we plug this in, we're going to get root 6 / 3 ^3 -, 2 root 6 / 3 + 4. Root 6 ^3 is going to be 6 root sixes 3 ^3 is going to be 27 two root 6 / 3 + 4. So simplifying this up, we're going to get 2 root 6 / 9 minus. If I got a common denominator, I'd have 6 root 6 / 9 + 4 or -4 root 6 / 9 + 4. If I then turn around and do the F of negative root 6 / 3, when I put in the negative, a negative cubed is going to come out as a negative. When I put in a negative, a negative times a negative where that second term is going to turn out to be a positive. So when we combine these, we get 4 root 6 / 9 + 4. So we can see that this is going to be a local min, and this one's a local Max. And the reason it's a local min is this is a negative value. Wait, let's look at the graph here, OK, it's an X ^3. So we're going to have it going something like this. So the positive value is a local min, and the negative value was a local Max. OK, so the -4 root 6 / 9 + 4 made it a local min in part. We're doing our leading coefficient test. If it's an X ^3, we know it's got to go positive off this direction because the leading coefficient was a positive and it was a cube. So at our root 6 / 3, we've got to have a lowest point. At our negative root 6 / 3, we've got to have the highest point. It's a cube. So as X goes to negative Infinity, Y goes to negative Infinity that look at those two locations. Our first derivative equaled 0, so we had horizontal lines. Let's do one more Y equal sqrt 3 plus two X -, X ^2. So Y prime is going to equal the derivative of the whole thing times the derivative of the inside. So the derivative of that inside is going to equal 2 - 2 X. We could have the twos cancel and we're going to have it equaling 0. So when I cross multiply here, the denominator is always going to go away. So we know that X equaling 1 is going to be a critical point now because we also know that the inside of a square root has got to always be greater than or equal to 0. This actually may not be negative Infinity to Infinity for our domain. So if we factored this, we'd get 3 -, X and 1 + X, and we need it when it's greater than or equal to 0. If we thought about this on a graph at -1 and at three, this would be a parabola going down. So I want the values that are greater than or equal to 0. So I need the YS that are positive. So I want the YS in here that are positive. Those Y values we can see are positive. So we're actually going to test Y of -1 Y of one, and Y of three because we have end points where it's defined. So our domain here is just -1 to 3:00. So at -1 sticking that into the original square root 3 + 2 * -1 - -1 ^2, we're going to get out zero at one. We get sqrt 3 + 2 * 1 - 1 ^2 and that's going to give us sqrt 4 or two, and at three we're going to get 3 + 2 * 3 - 3 ^2 or zero again. So we're going to have absolute min, absolute Max, and absolute min. You might want to see an example of a piecewise. If we have Y equaling 3 -, X three plus two X -, X ^2, where X is less than 0 and X is greater than or equal to 0. When we take our derivative, we're going to take the derivative of each one separately. So we're going to get -1 for X less than 0 and 2 - 2 X for X greater than or equal to zero. Well, this first one, the derivative is never going to equal 0. There's no constant where zero would equal -1. The second one, zero 2 - 2 X is going to occur when X equal 1 and X equal 1 is within the interval. So we know that we're going to have X equaling 1 as a critical point. So we'd have 3 + 2 * 1 - 1 ^2 or four. Now when we look at X = 0, what happens when X = 0 for the derivative from one side? So the limit from one side, the derivative of 0 from the left would be -1. But the derivative from zero from the right is going to be two. So the derivative at 0 does not exist. So if the derivative is not defined or if it doesn't exist, that's a potential critical point also. So if I think about Y of 0, going to stick 0IN and I'm going to get out 3 + 2 * 0 - 0 ^2 or three. So if we actually look at the graph, the equation from X less than 0 is just a line. And if we put in zero, we'd get out three. If we put in -1, we'd get out four. So it's going to go off like this and it's going to be an open dot there. If we look at the second one, we have a parabola. When X is 0, we have 3. So we're going to fill that dot and then we're going to go something like this. So we can see that this .14 is going to be a local Max and at 03 it's going to be a local min. And it's obvious that the derivative there doesn't occur because it's not a smooth and continuous curve. Thank you. And have a.