WEBVTT 00:00:02.000 --> 00:00:07.901 The last thing we need to learn how to do before we can find maximum profit, minimum cost, and 00:00:07.903 --> 00:00:13.401 solve other optimization problems, which is what those problems are called, is to find out how to find 00:00:13.403 --> 00:00:23.901 absolute extrema on a closed interval. Let's take a look at what that means. The value f of c is an 00:00:23.903 --> 00:00:34.402 absolute maximum of f if for all x in the domain of f, this value f of c is greater than or equal to f of x. 00:00:34.404 --> 00:00:41.204 And that's exactly what you would expect an absolute max to be. It would be the value of y that is greater than 00:00:41.206 --> 00:00:50.603 or equal to all other y-values of the function. The value of f of c is an absolute min of f if for all x in the 00:00:50.605 --> 00:00:57.000 domain, f of c is less than or equal to f of x. 00:00:57.002 --> 00:01:07.300 On a closed interval, we get the extreme value theorem. It states that if a function is continuous on a 00:01:07.302 --> 00:01:17.906 closed interval from a to b, then the function must have, or take on, both an absolute max and an 00:01:17.908 --> 00:01:26.904 absolute min on that closed interval. This kind of situation comes up repeatedly in optimization problems. 00:01:26.906 --> 00:01:41.900 Let's look at the possibilities. Here we have a closed interval from a to b. We have an absolute min, 00:01:41.902 --> 00:01:49.602 but it's going to be found at a relative min, and we have an absolute max that's going to be found at a 00:01:49.604 --> 00:01:59.205 relative max. So obviously we have to look at the relative extrema in the interval, in the closed interval. 00:01:59.207 --> 00:02:08.302 But there are other possibilities. In this closed interval and this graph, we have here, which is going to be 00:02:08.304 --> 00:02:19.802 at a relative min, we have an absolute min, but this absolute max appears at an endpoint of the interval. 00:02:19.804 --> 00:02:26.302 That tells us that when we look for these absolute extrema, we're going to have to look not only at the 00:02:26.304 --> 00:02:35.500 relative extrema, but also at the endpoints of the interval. Now this graph is similar to this one because 00:02:35.502 --> 00:02:44.800 we have an absolute min at an endpoint, and we have an absolute max at a relative max. So again we're going 00:02:44.802 --> 00:02:55.606 to have to check endpoints and relative extrema. In this situation we have the absolute max at an endpoint 00:02:55.608 --> 00:03:03.803 and the absolute min at an endpoint. So when we have a closed interval we're going to develop a test in order 00:03:03.805 --> 00:03:14.301 to find the absolute extrema, and that test is very simple. You find the critical values. You find the endpoints, 00:03:14.303 --> 00:03:21.702 or you are given a closed interval, or you have to define it in an optimization problem, and then you simply test, 00:03:21.704 --> 00:03:29.903 and whatever y-value, that's your f of c, whatever the y-value is that is the largest is going to give you 00:03:29.905 --> 00:03:38.303 the absolute max over that closed interval. Whichever one is the smallest is going to be the absolute min 00:03:38.305 --> 00:03:46.802 over the closed interval. We're going to check, once again, critical values and endpoints. Alright, let's 00:03:46.804 --> 00:04:01.800 take a look at an example. So we're given the function f of x equals x cubed plus 3 x squared minus 9x minus 7, 00:04:01.802 --> 00:04:13.302 where this function is defined on the closed interval from negative 2 to 2. We're going to find the critical 00:04:13.304 --> 00:04:20.500 values, and they have to be in this interval, of course. We will check those, and we will check the endpoints. 00:04:20.502 --> 00:04:34.604 First derivative would be 3 x squared plus 6x minus 9. We look for values here where the first derivative is 0 00:04:34.604 --> 00:04:46.103 or not defined. We can divide through by 3 here. We could have factored it just like it was, but it's a little 00:04:46.105 --> 00:04:58.400 simpler, and now we factor, carefully. Watch your signs. Let's check this before we go any further. This would 00:04:58.402 --> 00:05:08.005 be positive 2 minus 1. I mean positive 3 minus 1 x, so that's 2x. The signs are correct. We get x equals 00:05:08.007 --> 00:05:17.503 negative 3, and the other one, x is 1. These are critical values, and they are possible absolute extrema. 00:05:17.505 --> 00:05:24.499 But they're not the only possible ones. The other thing you have to watch for is these are possible absolute 00:05:24.501 --> 00:05:34.205 extrema for this function, but we've got to look at the closed interval. What do you notice about this? Well, 00:05:34.205 --> 00:05:41.600 this critical value is not in this interval, so we're not going to even look at this one. So now we have to 00:05:41.602 --> 00:05:53.803 check this critical value and these two endpoints. This is going to be just a little work here with making some 00:05:53.805 --> 00:06:02.803 sort of table. We usually put the first endpoint here, and then this helps you to check because if you hadn't already 00:06:02.805 --> 00:06:09.401 noticed that negative 3 was not in the interval you would catch it at this point. Then we're going to put 1, 00:06:09.403 --> 00:06:16.602 which is the critical value in the interval, and then the other endpoint. We go through, and we evaluate 00:06:16.604 --> 00:06:22.302 the original function, not the derivative but the original function, and we just want to know which one 00:06:22.304 --> 00:06:28.105 of these is the largest number. It's the absolute max over the closed interval. The one that's the smallest 00:06:28.107 --> 00:06:35.803 number is the absolute min. Alright, let's do this. Negative 2 into here would be, and I'm sure I don't have to 00:06:35.805 --> 00:06:45.103 tell you here be careful. Take your time. Negative 2 into here is negative 8. Negative 2 here is squared. 00:06:45.105 --> 00:06:59.907 That's negative 2 squared, so that's 4 times 3 is 12, minus 9 times negative 2 is positive 18, minus 7. 00:06:59.909 --> 00:07:16.303 So what did we get? We got 30 minus 15 which is 15. That goes with the negative 2. Alright, 00:07:16.305 --> 00:07:26.102 1 would be 1 plus 3. Now there are a lot of numbers up here so be sure you go back to the original function. 00:07:26.104 --> 00:07:35.704 Sometimes on this problem I do like this to remind myself that that's where I'm going to get these values. 00:07:35.706 --> 00:07:50.602 Ok, so where was I? 1, 3, minus 9, minus 7. So I have 4 minus 16, and so that's negative 12. So that means 00:07:50.604 --> 00:08:11.602 f of 1 is negative 12. 2 would be 8 plus 12 minus 18 minus 7. So what do I have here? That would be 20 minus 18 00:08:11.604 --> 00:08:25.101 is 2, minus 7 is negative 5. Now we compare the values, 15, negative 12, and negative 5. Well it looks like 00:08:25.103 --> 00:08:33.704 15 is the largest of the three, so over the closed interval the max would be, the absolute max would be found at 00:08:33.706 --> 00:08:45.100 negative 2, and it would have the value 15. So this is the absolute maximum value of the function over the 00:08:45.102 --> 00:08:56.303 closed interval, 15. Now negative 12 is smaller than negative 5, so this value, absolute value, let's 00:08:56.305 --> 00:09:09.002 go back and let's call it the absolute max, and we're going to call this one the absolute min. 00:09:09.004 --> 00:09:18.003 Alright and let's check what we have done. We have found that 15 is the absolute max on the closed interval. 00:09:18.005 --> 00:09:29.604 Negative 12 is the absolute min. One of these occurs at an endpoint. The other occurs at a relative extremum. 00:09:29.606 --> 00:09:47.804 Ok, let's take a look at another example. f of x equals x cubed minus 12x on an interval, closed, from negative 4 00:09:47.806 --> 00:09:56.499 to 3. Ok the process is the same, so we can pick up the pace a little bit. We take the first derivative, and 00:09:56.501 --> 00:10:09.206 we get 3 x squared minus 12. We're looking for critical values, so we divide by 3. We set this equal to 0, and 00:10:09.208 --> 00:10:17.300 we get x is 2, x is negative 2. So this time we got two critical values. Are they both in the interval? 00:10:17.302 --> 00:10:34.403 Check them. 2 ? Yes. Negative 2? Yes. Ok, so we now go to our little t-chart right here. Negative 4 goes 00:10:34.405 --> 00:10:41.700 first, and it's really useful if you follow a pattern every time you do this. Then you check negative 2. Then 00:10:41.702 --> 00:10:55.701 you check 2, and then you check 3. Ok, negative 4. Negative 4 cubed would be negative. Then 4 times 4 is 00:10:55.703 --> 00:11:05.704 16 times 4 is 64, and you know I'm going to do this again. I think this was quite helpful. So that's this one, 00:11:05.706 --> 00:11:16.903 and then I have minus 12 times negative 4, so that's positive 48, and so what I got here is negative 16. 00:11:16.905 --> 00:11:31.903 Negative 2 into here would be negative 8 plus 24, so this time I got positive 16. 2 would be 8 minus 24, and 00:11:31.905 --> 00:11:40.301 I got negative 16. Now that's interesting. I got the same value I got here. Then when I put 3 into here 00:11:40.303 --> 00:11:55.300 I get 3 cubed is 27 minus 12 times 3 is 36, and so what I get here is negative 9. Alright, let's evaluate what 00:11:55.302 --> 00:12:05.500 we've got in here. Well we have gotten negative 16 to be the absolute min. It does happen that it occurs 00:12:05.502 --> 00:12:13.203 twice, and this is fine. What this means is on the graph you have an endpoint down at negative 16, and then you 00:12:13.205 --> 00:12:25.001 have a relative extrema that's also at negative 16, but this is still the absolute [min]. It's fine that 00:12:25.003 --> 00:12:31.600 it occurs twice. It occurs at an endpoint. Alright which one of these is the greatest? That would be 16, and 00:12:31.602 --> 00:12:43.501 so what we have here is an absolute [max]. And so we now know how to find absolute extrema on a closed interval. 00:12:43.503 --> 00:12:54.102 Now let's see what we have here. I think we wrote these backwards. It looks like this one should be the min. 00:12:54.104 --> 00:13:04.905 So let's fix that. I have a couple of warnings for you. That's one of them. Always go back and check. 00:13:04.907 --> 00:13:12.702 16 is the max. Negative 16 is the min. That's a common error. Let me tell you a couple of other common errors 00:13:12.704 --> 00:13:25.204 you really have to watch on this. In other processes you evaluate a critical value in the second derivative. 00:13:25.206 --> 00:13:32.802 Now stop and think about where that is. If you're doing the second derivative test for relative extrema and 00:13:32.804 --> 00:13:38.507 you're looking at concavity, you would take this critical value, and you would plug it into the second derivative 00:13:38.509 --> 00:13:45.702 to evaluate the concavity to determine whether you've got a relative min or a relative max. That test is 00:13:45.704 --> 00:13:52.207 very easy to confuse with this test, so you have to stop and think about what you're doing here. 00:13:52.209 --> 00:13:59.805 You are trying to find absolute extrema on a closed interval. You have to check the endpoints, and you have to 00:13:59.807 --> 00:14:06.904 check the critical values in the interval, but you check them in the original function because all you want 00:14:06.906 --> 00:14:13.302 to know is what is the largest value of the function and the smallest value of the function. 00:14:13.304 --> 00:14:20.802 That now completes the last tool we need to move on to one of the most important applications of the derivative 00:14:20.804 --> 00:14:26.000 which is learning how to solve optimization problems.