WEBVTT 00:00:02.000 --> 00:00:08.500 Maximum profit and minimum cost. That should get your attention. In topic seven we begin 00:00:08.502 --> 00:00:15.001 our quest to find these quantitities. The first step is to learn how to take the equation of a function, 00:00:15.003 --> 00:00:23.501 analyze it and sketch the graph. We begin to make observations by a graph that we already have. 00:00:23.503 --> 00:00:31.800 Let's take a look at the graph of this function f. Let's label some of these quantitities. 00:00:31.802 --> 00:00:48.101 I'm going to label this value of x, x-one, and then the corresponding value of y, f of x-one. 00:00:48.103 --> 00:01:02.700 Then I'm going to label this value of x, x-two, and the corresponding value of y, f of x-sub-two. 00:01:02.702 --> 00:01:17.903 So that means this point right here, it's the one I'm trying to name, is (x-sub-one, f of x-sub-one), and this 00:01:17.905 --> 00:01:34.200 point right here is (x-sub-two, f of x-sub-two). Let's make these observations about the graph. First we want 00:01:34.202 --> 00:01:44.701 to state the intervals on x where this function is increasing. We notice that as we move from left to right, 00:01:44.703 --> 00:01:51.001 which is the way we always read increasing and decreasing, as you move from left to right, 00:01:51.003 --> 00:01:57.100 even though this arrow points down, don't let that lead you astray. As you go from left to right, 00:01:57.102 --> 00:02:05.800 the values of the function are increasing until you get here, and then the y-values decrease 00:02:05.802 --> 00:02:16.800 as x goes from x-one to x-two. Then from x-two on out to infinity the y-values are increasing. So, 00:02:16.802 --> 00:02:24.802 let's write this in interval notation. The function is increasing, and these are always values of x, 00:02:24.804 --> 00:02:44.501 increasing from negative infinity to x-one and then again from x-two out to positive infinity. 00:02:44.503 --> 00:02:51.800 Again, x-values. This is where the function is increasing, from here to here. 00:02:51.804 --> 00:02:57.801 And then increasing again from here to here. Alright, then the function is decreasing. You can see 00:02:57.803 --> 00:03:09.900 these y-values are going down as x goes from x-one to x-two. You can also notice that since there is 00:03:09.902 --> 00:03:17.300 no place where the function is constant, this interval, this interval, and this interval covers the domain 00:03:17.302 --> 00:03:28.600 with the exception of these actual points, which we call relative or also we often use the word local, 00:03:28.602 --> 00:03:36.200 in place of relative extrema. So we want to identify the relative extrema. Now the relative extrema 00:03:36.202 --> 00:03:43.801 are going to be either here, where we have a local max, or here where we have a local min. 00:03:43.803 --> 00:03:51.100 The actual extrema is the y-value so we're not going to name the point here. We're simply going to name 00:03:51.102 --> 00:04:02.901 the highest point in this local area, which is f of x-sub-one. That's going to be a local max, for maximum. 00:04:02.903 --> 00:04:13.801 And then, this is a local minimum. So we just put min, and this function has a min of f of x-two. 00:04:13.803 --> 00:04:20.901 Now it's about time to figure out what this has to do with calculus. So let's take a look. 00:04:20.903 --> 00:04:26.401 Over the interval from negative infinity to [x-one], we've identified that the function is increasing. 00:04:26.403 --> 00:04:36.701 So let's take a look at the tangent lines. So from negative infinity to x-one let's just pick a point here 00:04:36.703 --> 00:04:43.400 and let's look at the tangent line, because remember the derivative is the slope of the tangent line. 00:04:43.402 --> 00:04:52.600 Notice here that the [slope of the] tangent line is positive. Ok, let's just pick another point in the interval. 00:04:52.602 --> 00:04:56.900 We can notice here that the [slope of the] tangent line is positive again. We can keep doing this over the 00:04:56.902 --> 00:05:03.600 entire interval from negative infinity to x-one, and we can see here that all of the slopes of the tangent 00:05:03.602 --> 00:05:14.201 lines are positive. Right here the slope of the tangent line is 0 at this point, and then on the other side, 00:05:14.203 --> 00:05:20.902 in the interval from x-one to x-two, the slopes of the tangent lines are all negative. So we make a 00:05:20.904 --> 00:05:29.001 generalization here that where the slope of the tangent line changes from positive to negative 00:05:29.003 --> 00:05:36.699 we have a local max. That's exactly the case we have here. Alright, the same thing happens in this case. 00:05:36.701 --> 00:05:43.600 We have, except of course it's the reverse. The tangent lines are negative on this side of x-two, and then 00:05:43.602 --> 00:05:51.400 the tangent line at any point to f on this side would all have a positive [slope]. So when the tangent 00:05:51.402 --> 00:06:00.101 lines change from having negative to positive [slopes], we have a local min. Alright, once again, 00:06:00.103 --> 00:06:05.900 the slope of the tangent line is the derivative so let's take it a step further. When the first derivative is 00:06:05.902 --> 00:06:14.201 positive and then changes to negative, at that point where the change takes place we have a relative max. 00:06:14.203 --> 00:06:20.599 And then here we have a relative min, where the first derivative changes from negative to positive. 00:06:20.601 --> 00:06:29.802 Alright, let's go a step further. Notice here that the slope of the tangent line is 0. So this max right here 00:06:29.804 --> 00:06:45.601 occurs where f prime at x-one equals 0. That's the slope of the tangent line. This extrema results from 00:06:45.603 --> 00:06:53.502 here, a place where the first derivative is not defined. There is no distinct tangent line here and so the 00:06:53.504 --> 00:07:05.201 first derivative at x-two does not exist. Ok, so we generalize from this. What if we don't have the sketch? 00:07:05.203 --> 00:07:09.700 We want the sketch of the graph of the function and we want to analyze the function, and all we have 00:07:09.702 --> 00:07:17.300 is the equation. What we're going to need to do is figure out a way to find out how to break up or 00:07:17.302 --> 00:07:26.301 partition the domain in order to find these extrema. Alright let's look at some generalizations that we 00:07:26.303 --> 00:07:32.702 have at this point. What we're going to do is we're going to call these numbers that we're looking for, 00:07:32.704 --> 00:07:41.500 where we break the axis, partition numbers. In particular we're going to call these numbers, some of them, 00:07:41.502 --> 00:07:50.600 critical values. Now we have to look at the difference. Ok, this is partition number. This is critical value. 00:07:50.602 --> 00:07:57.101 We notice that we will partition the axis or just think of it as breaking it up, anywhere where the 00:07:57.103 --> 00:08:01.900 function is discontinuous. Now there is no example of this on this graph. That would be something 00:08:01.902 --> 00:08:09.701 like a vertical asymptote or a hole. That does not qualify as a critical value. It cannot be an extrema 00:08:09.703 --> 00:08:14.200 because the function is not defined where there is either an asymptote or a hole in the function. So you 00:08:14.202 --> 00:08:19.602 use it as a partition point, but it is not going to be an extrema. So you can see now a definition developing 00:08:19.604 --> 00:08:26.701 where the critical value is going to be if there is an extrema. This is going to be where it is. 00:08:26.701 --> 00:08:34.901 So back to the partitions. We partition the axis at the values of x where the function is discontinuous, 00:08:34.903 --> 00:08:41.800 where the first derivative is 0, or where the first derivative is not defined. Ok, one more time. 00:08:41.802 --> 00:08:49.101 Let's go back. Partition numbers: discontinuous, first derivative zero, first derivative not defined. 00:08:49.103 --> 00:08:55.982 Critical values: only first derivative zero and first derivative not defined. Alright, some more generalizations. 00:08:55.984 --> 00:09:06.181 Increasing and decreasing. So if you have a function f of x on an interval from left to right, reading 00:09:06.183 --> 00:09:13.381 from left to right, we generalize that if the first derivative is positive, that's the slope of the tangent line, 00:09:13.383 --> 00:09:22.680 then the function itself is increasing. So first derivative positive, function is increasing. And if the 00:09:22.682 --> 00:09:31.581 first derivative is negative, the function is decreasing. First derivative, slope of the tangent line, is 00:09:31.583 --> 00:09:44.181 negative, function is decreasing. Now putting this all together. This is how we identify relative extrema. 00:09:44.183 --> 00:09:50.482 We let c be a critical value. Now that's not a partition number, it's a critical value. So this comes from a 00:09:50.484 --> 00:09:57.081 value of x where the function is either equal to zero or... not the function, the derivative is either 00:09:57.083 --> 00:10:07.400 equal to zero or not defined. So which is which? Well if the function is increasing, let's look at this. 00:10:07.403 --> 00:10:18.400 If the function is increasing on the left side of c and decreasing on the right side of c, then we have 00:10:18.402 --> 00:10:25.703 a relative max. And then the second bullet is, if the function is decreasing on the left side and increasing 00:10:25.705 --> 00:10:32.300 on the right side we have a relative, or local, min. Alright let's take a look at an example of a function 00:10:32.302 --> 00:10:40.302 where all we have is the equation and see if we can figure out some of this. So let's take a function that 00:10:40.304 --> 00:10:47.100 we're somewhat familiar with but that we couldn't really sketch the graph of just using some basic 00:10:47.102 --> 00:10:57.901 algebra techniques like finding the intercepts. So let's take an example, and let's look at the function 00:10:57.903 --> 00:11:12.300 f of x equals negative x to the fourth plus 4 x-cubed minus 5. And all we want to do here with this, 00:11:12.302 --> 00:11:29.602 We're not going to try to sketch the graph at this time. We're just going to try to identify relative extrema. 00:11:29.604 --> 00:11:38.901 Ok, so what do we do? Well the solution would be to use calculus. Identify partition numbers, 00:11:38.903 --> 00:11:46.900 critical values, test them using increasing and decreasing and see if we have any extrema. Ok, so 00:11:46.902 --> 00:12:02.700 we take the derivative so we get negative 4 x-cubed plus 12 x-squared and the derivative of 5 is 0. 00:12:02.702 --> 00:12:11.200 Now we want to identify the values of x where this derivative is either discontinuous, or values of 00:12:11.202 --> 00:12:16.000 even the function. But let's back up, let's look at the function. Since it's a polynomial function, it's 00:12:16.002 --> 00:12:21.001 continuous everywhere. So we actually don't have to worry about that part on a polynomial function. 00:12:21.003 --> 00:12:25.900 There are no discontinuous points on this function. So we're really only looking for places where the 00:12:25.902 --> 00:12:34.200 first derivative is either equal to zero or not defined. Well, this function is going to be defined for all 00:12:34.202 --> 00:12:41.600 real numbers. So all we really want to know is where is the first derivative equal to zero? 00:12:41.602 --> 00:12:48.701 Ok, so we're looking for those values. So we're going to solve this equation. Alright, let's see. 00:12:48.703 --> 00:12:56.100 Let's factor out negative 4 x-squared. Now be careful here, because if you make a sign error here 00:12:56.102 --> 00:13:04.498 it makes everything else a mess. So when I factor out negative 4 x-squared, I still have an x. This 00:13:04.500 --> 00:13:13.301 sign changes to negative 3, and then I've taken out the x-squared. So now I use the zero factor property. 00:13:13.303 --> 00:13:24.600 I set each part equal to zero. So I get negative 4 x-squared equals 0, and x minus 3 equals 0. 00:13:24.602 --> 00:13:28.901 After you practice this a few times you get pretty quick at this and you can go from here to pulling out 00:13:28.903 --> 00:13:36.300 the critical values right away. Let's show all of it right now. So x equals 3. We divide both sides by 00:13:36.302 --> 00:13:43.401 negative 4. We take the square root of both sides and we get x is 0. So this is a big step. 00:13:43.403 --> 00:13:51.200 We have now determined that x equals 0 and x equals 3 are not only partition numbers, but they are 00:13:51.202 --> 00:13:59.401 also critical values of the function. That means that if there are any extrema they will occur 00:13:59.403 --> 00:14:08.700 at these points. Ok, now we need to analyze the sign of the function in each partition. Ok. 00:14:08.702 --> 00:14:18.100 Let's leave this down here, and let's do a sign chart here to see what our function looks like. 00:14:18.102 --> 00:14:27.600 We'll just keep our critical values down here for the moment. We also need, we probably need to have... 00:14:27.602 --> 00:14:38.902 Let me put this back. We probably need at least that much, the derivative of the function. 00:14:38.904 --> 00:14:46.500 Then let's work up here. Alright, I am going to partition the x-axis. The first interval will be 00:14:46.502 --> 00:14:54.800 negative infinity to 0. Let's just write this out here to help you to see this as an interval. The next one 00:14:54.802 --> 00:15:05.201 will be 0 to 3, and then the third one will be 3 to infinity. Now we're going to take the function, 00:15:05.203 --> 00:15:12.301 I mean the derivative, this derivative, and we're going to analyze the sign of f prime. It may not 00:15:12.303 --> 00:15:18.100 seem important now, but it does turn out to be important later, that you label this f-prime because, 00:15:18.102 --> 00:15:24.500 just trust me. You need to do this. Alright what we're going to do is we're going to pick a test 00:15:24.502 --> 00:15:29.099 value in each of these intervals, the easiest number we can work with. So here I'm going to 00:15:29.101 --> 00:15:35.299 pick negative 1. Here I'm going to pick 1, and here I'm going to pick 4. Obviously we want to use 00:15:35.301 --> 00:15:40.501 whole numbers. They're easier to work with, and we want the smallest numbers we can possibly find. 00:15:40.503 --> 00:15:47.400 Now we want to know the sign of the first derivative. We don't care how positive it is, we just want to 00:15:47.402 --> 00:15:53.901 know if it's positive or negative. So we don't actually have to have a numerical value. We just need a sign. 00:15:53.903 --> 00:16:00.300 So take negative 1 and plug it into the derivative. Ok, when you do you would get, and let's just write 00:16:00.302 --> 00:16:12.399 write it down here, negative 1. So we're going to do f prime of negative 1. So that would be, and this is 00:16:12.401 --> 00:16:24.400 another place where you have to be really careful. A sign error here is a mess. So negative 1 into here 00:16:24.400 --> 00:16:32.500 and again all we really want is the sign. So negative 1 squared is 1, times negative 4. So this factor is 00:16:32.502 --> 00:16:43.501 negative. This factor is also negative. Negative times negative would be a positive number, so over 00:16:43.503 --> 00:16:52.400 this interval the first derivative is positive. Now what does that mean? Well that means that the function 00:16:52.402 --> 00:17:00.600 itself over this interval is increasing, because the slopes of the tangent lines are positive. This way and 00:17:00.602 --> 00:17:07.200 that's kind of why I draw the arrow here. Alright, let's pick a number in this interval. F prime of 1 00:17:07.202 --> 00:17:14.904 and I substitute into here. Now this would be negative 4 times 1. So that's negative and then 1 minus 00:17:14.906 --> 00:17:26.098 negative 3. So what we have here is negative times negative. So again we get positive and so the 00:17:26.100 --> 00:17:32.300 function is increasing. Now that's the sign of the derivative. That's the sign of the function. 00:17:32.302 --> 00:17:39.500 Well, that's the direction of the function. The function is increasing. Alright, and then we pick 4 here. 00:17:39.502 --> 00:17:47.900 F prime evaluated at 4. 4 into here would be positive times negative 4, and we can start to do this 00:17:47.902 --> 00:17:56.200 without maybe quite so much arithmetic here. That's going to be negative. And then 4 into this factor 00:17:56.202 --> 00:18:04.901 would be 4 minus 3 which is 1. That's obviously positive, and so the product here is negative. 00:18:04.903 --> 00:18:10.700 So that means the first derivative is negative. The slope of the tangent line is negative, and the function 00:18:10.702 --> 00:18:19.101 itself is decreasing. So it's a lot of work here, and you have to be careful not to lose sight of what it is 00:18:19.103 --> 00:18:26.500 you're trying to do. We are looking for relative extrema. Did we find any? Well extrema occur where the 00:18:26.502 --> 00:18:33.100 function changes from increasing to decreasing or decreasing to increasing. So at f of 0 there is not going 00:18:33.102 --> 00:18:40.401 to be an extrema. But right here you can see the function is increasing and then it's decreasing. So at 3 00:18:40.403 --> 00:18:54.101 you're going to have a relative or local max, and the actual max is f of 3. Let's go ahead and see. We could 00:18:54.103 --> 00:19:02.300 calculate f of 3. Notice that this is not f prime of 3. This is f of 3. So we have to go back to the original 00:19:02.302 --> 00:19:12.901 function. Ok the original function was negative x to the fourth plus 4, and I'm subbing in the 3 for x, 00:19:12.903 --> 00:19:21.100 cubed minus 5. One of the classic mistakes on this problem is to get all the way to this point and 00:19:21.102 --> 00:19:26.101 actually sub the 3 back into the derivative, because of course that's what you do to get these signs. 00:19:26.103 --> 00:19:37.200 Ok so what do we have here? This is going to be negative 81. This is going to be 27 times 4. So that's 108 00:19:37.202 --> 00:19:51.801 and that's positive, and that's negative 5. So what we have here is 108 minus 86 which is positive 22. 00:19:51.803 --> 00:20:04.301 So all of that work identified any relative extrema. Here's our final answer. Maximum. We have a relative 00:20:04.301 --> 00:20:19.799 maximum, relative max, and the value is 22. There's a little shadow right there. And so, that is how you use 00:20:19.801 --> 00:20:28.300 calculus to identify a relative extrema. Now what we'd like to be able to do at this point is sketch the graph. 00:20:28.302 --> 00:20:34.000 The problem is we're missing some information. We aren't sure exactly what happens here. 00:20:34.002 --> 00:20:38.701 We know something happens. There's something going on here, but it is not an extrema. 00:20:38.703 --> 00:20:45.201 So this is the motivation for the next topic. The next topic is going to tell us how to use 00:20:45.203 --> 00:20:52.620 the second derivative to glean information in order to sketch the graph of this function.