In this video, we're going to be looking at how curves behave and how to find specific points on those curves. Let's begin by looking at a sketch of a curve. Now we've got a curve. There's some noticeable things about it. First of all here. The curve turns if we draw tangent to the curve. It's level parallel to the X axis. Here I've exaggerated it a little bit. The curve is flat and the tangent is again parallel to the X axis. Here the curve turns and again a tangent is parallel to the X axis. Now, one of the things that we do know is that the gradient of a curve is given by DY by The X. What's clear is that each of these points here. Here and here. Because the tangent is flat parallel to the X axis, the gradient is zero and so D why by DX is 0. Now for points like these where DY by the X is zero, we have a special name, we call them. Stationary points. If you like the curve, sort of isn't moving, it isn't changing. At that point. The why by DX, which is the rate of change, is 0. Now two of these I discribed the curve as turning. I said there the curb was turning and here again the curve is turning so that's going from coming up to going down and this one is from going down to coming up. This one isn't this one is very different so these two. Are special types of stationary points. These two. Are cold turning points. So at turning points divided by DX is 0. Thing to realize though, is that if the wide by the X is zero, that doesn't mean we've got a turning point. We might have one of these, that's what makes life just a little bit tricky, but nevertheless thing to remember is that turning points divided by DX is 0. Now let's have a look at the nature of this particular turning point. In the immediate area of it. The curve has no bigger value. We go up, we reach a value and then we go down again. So this point. Is called a local maximum. Because in the immediate vicinity. There is no bigger value. Now usually we forget about the local and we just call it a maximum. Similarly, this point is called a local. Minimum. Because again, in this particular region of it, the curve goes no lower, so this is the least value in this area are local minimum. Again, we forget the word local quite often. Unjust use minimum. In essence, this video is going to be about finding these points. Maxima and minima finding these two points Maxima and minima. And distinguishing between them because being able to find them isn't a great deal of use unless we can tell them apart. So we need to be able to distinguish between a maximum and the minimum. So let's begin by exploring what happens around a minimum point. Draw a minimum. Let's take the Tangent. At the minimum point, we know that the gradient of this tangent is 0 because it's parallel to the X axis. So DY by DX is 0. Let's take a point a little bit before. The minimum point and take a tangent at that point. And here we see that the Tangent. Is negative, it's got a negative slope and so D. Why by DX is negative. I'm using this as a shorthand for negative, not actually bothered about the value of the why by DX just it's sign. Let's take a point here a little bit after. What's happened there? Well, we can see we've got a line that's got a positive slope to it, which means that the why by the X has got to be positive. So what's happening as we pass through this point as X increases? What's happening to do, why by DX? Well, as X increases, the why by DX goes from negative. 0. To positive. So divide by the ex is increasing as X increases, so let's write that down deep. Why by DX is increasing. As X. Increases. What does that mean? DY by DX is increasing well that means that the rate of change. Of the why by DX? Is positive. IE D2Y by DX squared is greater than 0. So if D2Y by the X squared is greater than 0. And divide by the X is equal to 0. We know that we have got a minimum point. Is that true for all minimum points? Little trick of logic here, but just write down what we've got if. DY by DX equals 0. And. D2Y by DX squared is greater than or positive then. We have a. Minimum. Does it work the other way around? If we have a minimum? Do these two both apply? Well, certainly that one does that if we have a minimum DY by the X will be 0, but not necessarily this one. Let's have a look at the curve Y equals X to the 4th. If I sketch this curve. It looks. Like that? Here we've clearly got our turning point. Our minimum point. Let's have a look what's happening. DY by the X equals well. Why was X to the 4th so DY by DX is 4X cubed, and we know that for our stationary points we put it equal to 0, which tells us X equals 0 and X equals 0 tells us that Y equals 0, so we know that this is a stationary. Point What kind of stationary point is it? Let's do our second derivative. D2 why by DX squared is 12 X squared, but it equals 0. When X equals 0. So this doesn't work in reverse. If the why by DX is zero and Y2Y by the X squared is greater than not, then we know we have a minimum. But it doesn't work going the other way. What that means is that we have to be especially careful whenever D2, why? By DX squared comes out to be 0 when we're doing the test, so to speak. We need to look at it in a slightly different way. In fact, what we need to do is go back to basics and I'll show you an example later where we do go back to basics and how that works out. So having had a look at a minimum point and looked at it in some detail, let's now move on and look at a maximum point and see what happens there. So here's our maximum point. And at the point itself, we know that the tangent is parallel to the X axis and so its gradient. And hence the why by DX is 0. Let's take a point a little bit before that there. We can see that we've got a line which has a positive slope to it, and so DY by the X is positive there. Now let's take somewhere just past the point. And if we have a look at this week and see we've got a tangent. Where the slope is negative and so divide by the X is negative. So what's happening as we pass through this maximum point as X is increasing? What's happening today, why by DX? Well, divide by the axe goes from being positive to being zero to being negative. So as X increases, divided by DX decreases. As X increases. Why by DX decreases? Does that mean it must mean that the rate of change of the why by DX? Is. Negative. As X increases as we go through this point at this point. An the derivative divide by the X is negative. In other words, D2Y by the X squared is negative. So let's just write down again what we've got. If. Divide by DX equals 0. And. D2 why by DX squared. Is less than 0. Then we have a maximum point. Again, just the same argument applies. Does it work back the same way? Answer no, it doesn't. And the problem is what do we do about D2? Why? By DX squared being zero? Well, as I said, will show a way of dealing with that in a moment. It's time now to actually look at an example or two and see if we can make use of this test. In order to sort out the turning points on a curve. So the first example we're going to have a look at is Y equals X cubed, minus three X +2. The first thing we need to do is differentiated. Why by DX equals if we differentiate this we have three X squared. We multiply by the index and subtract 1 prom it minus the derivative of the three axis, just three, and this is equal to 0 for stationary. Points. We can factorize this. There's a common factor of three leaves us with X squared minus one, and this is now recognizable as the difference of two squares, so it factorizes us three times, X minus 1X plus one. And that's equal to 0. Now what we can see is that. Either this bracket is 0 or this bracket is 0, or they're both 0, so we have X equals 1 or X equals minus one. Now this is just given us the X coordinates of the points that were interested in what we need now or the Y coordinates, and so let's work those out, X equals 1, Y equals and will substitute it in on the very first line. So that's one cubed is just 1 - 3 times by one is minus 3 + 2. Altogether, that gives us 0. And so our point, our stationary point is 1 zero. If we take X equals minus one and Y is equal to. And we put minus one in there well minus one cubed is just minus 1 - 3 times by minus one is plus three and plus two at the end, and altogether 3 + 2 is 5 takeaway, one is 4 and so our other stationary point is minus 1 four. Now we need to decide. All these points, Maxima, minima or that funny one where it was flat. So in order to sort that out, what we need to have a look at is the 2nd derivative. So I'm going to write the equation of the curve down again, and then I'm going to write down this first derivative, then the points, and then the second derivative. So we've Y equals X cubed, minus three X +2. We've got the first derivative D. Why by DX3 X squared minus three and the points that were interested in with the points one not and minus 1 four. So now. Let's take the second derivative D2Y by DX squared equals. So we differentiate this. The derivative of the first term we multiply by the index. So two, freezer 6 and take one of the index that's X and the derivative of three is just zero. So let's now look at the points and. It's the values of X we're interested in. So X equals 1. D2Y by DX squared equals 6. So this is positive. What do we know? We know that if divided by DX is zero and if D2, why by DX squared is positive, then the point we're looking at is a minimum turning point, so therefore 10 is a minimum mean for short. X equals minus one. This is our second point. D2Y by the X squared is equal to minus 6. And that is negative. The value of these doesn't matter. It's their sign that matters and therefore, what do we know? We know that if the Wi-Fi DX is zero and Y2Y by DX squared is negative, then we have got a maximum, so minus 1 four is a Max. One of the things that this does help us do is when we do know the turning points. We can use this to help us gain a picture of the graph to help us sketch a picture. So let's just plot these two points. And see if by plotting them we can build up a picture of the graph. Now this won't be an exact picture. The scales will be different. They'll be a little bit cramped in some cases, expanded in others because what we're trying to show the important points on the graph, so we've one and minus one. Where are two important values of X, and we know that one 0. Was one of our turning points and if we just go back, we know that one zero was a minimum turning point, so we know that at this point the curve looks like that. The other point was minus 1 four there. And we know that at that point the curve was like that. It was a maximum. The other thing we know is that there are no turning points anywhere else. These are the only ones, and so effectively we can join up the curve and so that we get a rough picture that it looks something like that. We can find this point of course by substituting X equals 0. Remember, our curve was Y equals X cubed minus 3X, plus 2X is 0. Then this must be 2 here. This we can find bike waiting why to zero and solving but we're not interested in that at the moment. It's the idea of using this maximum and minimum turning points in order to help us gain a picture of the shape of the curve. We're going to have a look at an example now where the method we use is one that enables us to go back to 1st principles, and this is also the method that we would use if D2. Why? By DX squared turned out to be 0. So it's a very useful method to have at your fingertips. So let's begin. We're going to be looking at finding the turning points of this particular curve. X minus one all squared over X. Well it's a U over a V. It's a quotient, so we need to be able to differentiate as though it were a quotient. So let's remember how we do that. We take the bottom the X. And we multiply X by the derivative of U what's on top? So that's two times X minus one. Times the derivative of what's inside a one and then minus. And we take UX minus one 4 squared and we multiply it by the derivative of E. What's on the bottom? And that's the derivative of X. That's just one, and then it's all over X squared. And for our stationary points we put this equal to 0. Now. Let's have a look at this. We need to tidy it up. It looks a bit of a mess, so let's have a look for some common factors. Well, here we've got an X minus one and here as well. So let's take that X minus one out as a factor. Let's have a look what we're left with. Well, here we've got two and X giving us 2X and one times by one. That's still 2X. Minus and here we've got one times by now we just took out X minus one as a factor, so it's minus X minus one and notice I've kept it in the bracket because it's the whole of X minus one that I'm taking away. Close that bracket all over X squared so this is still DY by DX. Now I can simplify what's in the bracket and have DY by the X is equal to X minus one times by now have two X takeaway X that's just an X, but then I've also to take away minus one and so that gives me plus one. All over X squared and this is to be equal to 0 for my stationary points. Now when we have an expression like that that's equal to 0, it's only the numerator that we're interested in. It's only the numerator that will make this expression 0. So got is X minus 1X plus one equals 0. And that tells us that either this bracket is equal to 0 or that bracket is equal to 0. In other words, X minus one equals 0 or X Plus one equals 0. So X equals 1. Four X equals minus one. Now it's the points where after so let's now calculate the values of Y, remembering that Y is equal to X minus one 4 squared all over X. So let's take X equals 1 and see what that gives us for why? We put one in there. We have 1 - 1 all squared over one. Well, that's just zero because 1 - 1 is 0 square at that still 0 / 1. The answer is still 0. So one of our points is one Nord. That's one of our stationary points. Now let's do X equals minus one. Why is equal to minus 1 - 1 all squared all over minus 1 - 1 - 1 is minus 2 and we square it to that's four but we divide by minus one, so that is minus four. So other point is minus 1 - 4. So we've got two points which are stationary points and we need to find out are they turning points and if they are turning points, are they Maxima or minima? So let's write down again. Why do you? Why by DX and our points. So we had why that was X minus one or squared over X. We had DY by the X&YYY DX was over X squared and let's just go back so we can see what. The numerator was it was X minus one times by X plus one. X minus one times by X Plus One and our two points. Are the points one note and minus 1 - 4? Now. In order to sort out if these are turning points and if they are, which is a maximum and which is a minimum, then we need to differentiate this again and that is very very fearsome. Very fearsome indeed. Is it really worth differentiating differentiating it again? We might make a mistake. The algebra might go wrong. So rather than differentiating it again, let's do what we did before. Let's look at the gradient a little bit before X equals 1. At X equals 1 and a little bit after X equals 1 and form a picture from doing that of whether this is a maximum or minimum point, and this method that we're going to have a look at also works. Wendy to why? By DX squared is 0, so hence it's a good method and it's well worth adding it to your Armory of techniques for looking at stationary points. So let's set this up and the way we're going to set it up is too. Have a little table. So here's the Y by The X. And here is the point where interested in at X equals 1. I want to look a little bit before X equals 1, so I'm going to say this is X equals 1 minus Epsilon and all. I mean by Ipsilon is a little positive bit of X. North Point One North Point North one if you like, and I'm also going to look at the gradient. A little bit after X equals 1, so let me call that one plus Epsilon. Now I'm not interested in the value of the why by DX. I'm only interested in its sign whether it's positive or negative. I know that when X equals 1 divided by DX is 0. What happens if I am a little bit before X equals 1? X is a little bit less than one, so I have something a little bit less than one, and I'm taking one away. So the answer to that is negative. This bracket here is negative. However, I've got something that's a little bit less than one plus one that's positive. So I have a negative times by a positive and I've got X squared on the bottom here and X squared is positive, so I'm negative times a positive divided by a positive is a negative. Let's have a look when X is a little bit more than one little bit bigger than one 1.911. Take away one. Well, a little bit more than one. We set. Take away one that's positive. One plus a little bit more than one that's. Positive again, and this is definitely positive in the denominator, so we've a positive times by a positive divided by a positive gives us a positive. So if we sketch. The slopes of the tangents. We have a negative slope aflat slope under positive slope and So what we can see is that this defines the shape of a minimum and so we can conclude that one zero is a minimum. Point notice what we've done is go back to basics. This is how we arrived at our test for D2. Why? By DX squared. Now let's do it again for this point, minus 1 - 4. So we've got DYIDX is equal to X minus one times by X Plus one all over X squared, and the point we're having a look at is minus 1 - 4. So let's set up our table again. We're looking at the sign of DY by The X. My looking at it at the point X equals minus one. As we pass through that point as X increases. So we want to know what's happening alittle bit before that minus one minus Epsilon. And we want to know what's happening alittle bit after that minus one plus Epsilon. Well, we know that when X is minus one DY by DX is 0. Because this factor here is 0. What happens if X is a little bit less than minus one? Something like minus 1.1? Well you got minus 1.1 takeaway. One that's definitely negative. Minus 1.1 at on. One that's minus .1. That's definitely negative as well. However, on the bottom here, we've got a number and we're squaring it, so it must be positive. So we've a negative times a negative, which is a positive divided by a positive. So the answer must be positive. Now let's have a look a little bit after X equals minus 1 - 1, plus a little bit. So the little bit might be not .1 and minus one plus a little bit would then be minus N .9. So we've got minus 9.9 - 1 - 1.9 it's negative. I've got minus not .9 plus one. Well, that's not .1, it's positive, so have a negative times a positive and this term X squared is always positive 'cause it's a square. So we've got a negative times by a positive and divided by a positive. So I'll answer. Must be negative. So let's sketch the shape. We've got a positive slope, aflat slope and negative slope, and this defines the shape of a maximum, and so we can say minus 1 - 4 is a maximum. Finally, let's just see how this helps us to sketch the curve. We recall the equation of the curve Y equals X minus one all squared over X, and we had a maximum point at minus 1 four. And we had a minimum point at one 0. So some axes, let's mark the .1 and the point minus one goes on the X axis. Now 10 is here and we know it's a minimum, so we know it looks something like that there minus 1 - 4 down here somewhere. And we know that that's a maximum, so it kind of looks like that. How can we join these up? Well, clearly there's something odd going on around 0 here because the curve seems to go that way and that way it seems to go in opposite directions. What's happening here? What's going on? Well, let's just have a little think about this as X approaches 0 from above. And this is always going to be positive, but this is going to be. Positive as well, and were divided by something very very small. Our answer is going to be very, very big, so we're going to have something up there when X is negative but near to 0. This is still going to be positive because we're squaring it, but we're dividing it by something which is getting very very small, but negative. So the answer is going to be very big, but negative, and so it will be somewhere down here. So what we can see. Is that the curve is going to look something? Like That and they will in fact be this almost what looks like a sort of hole in the middle of it. Bang on this value X equals 0. So what have we done? We have found out how to find stationary points. Stationary points occur on a curve. Wendy why by DX is 0. If the why by DX is 0? Then some of those points are what we call turning points, Maxima or minima. We can find those. In one of two ways, if the why by DX is zero and D2, why by DX squared is greater than not, then we know for sure that we have got a minimum point. If you divide by the X is zero and D2 why by DX squared is negative, then we know for sure we've got a maximum point. If however, D2 why by DX squared is 0, then we have no information whatsoever and we need to look very closely using the methods that I've just shown here. In order to determine what kind of stationary point we've got, whether we've got a maximum, whether we've got a minimum or whether we've got one of those odd ones that looked a little bit flat. A kink in the curve.