- [Instructor] Well, we're going back into our motivation. Way back on day one we had this picture, and we said, look, we can find tangent lines. And how do we find tangent lines? Well, we find tangent lines, so the tangent line is this red line, which is tangent. Now the word tangent means to touch, so it's just touching the curve at a single point. And we can say, well, the problem is you only have one point to work with, this point on the curve. But secant lines, and with seeking you cross, you've gotten two points to work with. So secant line, you have an A and a B, and you draw a line. And the philosophy is, oh, if we let the two points get closer and closer, so you have this notion that B is getting closer to A, then the secant lines, which might not be very good approximations if our two points are far away, become good approximations as you get close and become great approximations as you get even closer. And so we have the following: If we now say using our wonderful terminology, this toolbox that we have, this idea of B is getting close to A, it's just another way of saying we're taking a limiting process. There's a limit involved. So limits are I'm trying to get close to something. So B is trying to get close to A. And now we are trying to find our tangent line. Well, how do you find the line? You need two pieces of information. You need the point, and that's a freebie. So the point of our tangent line, which is this point up here, is always gonna be A comma F of A. Now we need something that's going to give us our slope. So this F prime of A, this is sometimes called our instantaneous rate of change, this is the slope of our tangent line. So the derivative tells us how our tangent line is changing. Now remember that we said slopes correspond to rates of change. And so when we had a secant line crossing through two points, that was an average rate of change. So it's the change in the output, F of B minus F of A over the change in the input, B minus A. And now as we take this limit, we get the slope of the tangent line. That's the instantaneous rate of change, how the function is behaving at that point. All right. Well, what do we have? A lot of the things are what we just said. So again, this F prime of A, it's our instantaneous rate of change, it has a couple different interpretations. So you might see things like find the instantaneous rate of change or you might see find the derivative. You know, that's the same thing. It might say what's happening at that point. That's again, what you're trying to capture is we're trying to express the same thing in in many different ways. So understand what is F prime of A? Well, we call it the derivative of F of X at X equals A. It's how fast the function is changing at X equals A. It's our slope of our tangent line. So we have all these wonderful things. Now there's a different variation that we oftentimes use for our definition of derivative. And this says, okay, let's think of B not as a number, but let's think of B as we took A, and then we perturbed it a little bit. So in other words, B is A plus a small change. So what we're thinking of here is we're thinking of this small gap as being H. So to get from A to B, you went H. So now I can say, okay, what changes to the formula? Well, F of B minus F of A, that's F of A plus H minus F of A, and then the B minus A, that's our change. That's H. And so we can write it as F of A plus H minus F of A all over H. Now I will say that they are both equivalent in that you'll get the same answer. In terms of practical application when you actually have to apply the definition, and we will get to the point where we don't apply the definition anymore. It'll be about a week from now when we don't have to go back to the limit definition. In practical application this second one is easier to apply. And the reason I say it's easier to apply is think of the philosophy we had when we were trying to deal with limits. More or less, when we dealt with a challenging limit which required more work. So something of the form zero over zero. Then what we wanted to do is we wanted to somehow rewrite the expression so that we got a cancellation of zeros. And so you'll see we do get zero over zero because if you put an H equals zero, you get F of A minus F of A on top. And if you put an H equal zero, you get zero on the bottom. So this is a zero over zero. So we are in that more work setting. So we wanna get cancellation. The reason why it's easier when we just have an H downstairs is now the goal is how do we get an H out of this F of A plus H minus F of A? It's a little bit easier to figure that out. Whereas here we have to say, okay, to get to cancellation, I need to figure out where's my B minus A. That's a little bit harder to extract. So if you're going to use the limit definition, you can use either, but I strongly encourage you to use this one. All right. So let's do an example. For the quadratic function F of X equals something find F prime of A for any A. So we haven't actually written down the quadratic, that's the second order polynomial. So does anyone have a favorite quadratic they'd like us to do? - [Student] Five X squared. - [Instructor] Five X squared. We need a quadratic. So we need a square. We can have more stuff. - [Student] Plus seven X. - [Instructor] Plus seven X, and? - [Student] Plus two. - [Instructor] Plus two. Five X squared plus seven X plus two. So I wanna find F prime of A for any A. So let's see what happens. So we have the formula that we just had. So F prime of A, that's the limit. H goes to zero of F of A plus H minus F of A, all of that divided by H. Now we run into one of the biggest challenges when it comes to taking derivatives, and we should take a second here and pause, just make sure we understand what's going on. So here's our original function. F of X equals five X square plus seven X plus two. Let's just do a couple things. What is F of one? Well, five times one squared plus seven times one plus two. That's what F of one is. You plug it in. And then you say, okay, after you plug it in, you can do algebra, you can do arithmetic, you can simplify. So you get down to 14. What if I plugged in, say something exotic like negative two? Alright, well five times negative two squared plus seven times negative two plus two. Okay, so far, all right, you're probably thinking easy, easy. All right, let's try something a little bit more interesting. Suppose I plugged in F of a cat. (students laugh) What would you get? - [Student] Cat squared. - [Instructor] Five times your cat squared, plus? - [Student] Seven times cat. - [Instructor] Seven times your cat, plus two. Now you might be saying, "But that's nonsense!" Yes it is. But that's the rule. We follow the rule. So whatever the rule is, we make sure we follow it. Alright, so as another example, if I just plugged in say A, what would I get out? - [Student] Five A squared. - [Instructor] Five A squared plus? - Seven A. - Seven A. - Plus two. - Plus two. So here's the point, what do you get when you plug an F of A plus H? What do you do? - [Student] Five times (indistinct). - [Instructor] Five times A plus H squared plus seven times A plus H plus two. A lot of people will say, oh F of A plus H, they're like, well that's the easy, right? Isn't it the case if I want F of A plus H, that's just F of A plus H? No, it's not. Be careful. So make sure you follow the rule. So once you have a function, whenever you're putting in your A plus H, just everywhere where it was there was an X before becomes an A plus H. So this is our limit as H goes to zero five times A plus H squared plus seven times A plus H plus two. And I'm gonna use parentheses 'cause parentheses help save points because it helps us avoid mistakes. Now what I mean by that is whenever you have a minus, you run into a danger of forgetting that a minus distributes through. So when in doubt put parentheses and just be careful. So I replace what F of A plus H is, I replace what F of A is, what happens next? (indistinct) Okay, so start doing algebra. What are some algebra things we can do to this? - [Student] Distribute. - [Instructor] We can distribute, we can expand, we can start gathering like terms. So there should be no real surprises. This is a type of of a problem where you just sort of carefully follow through. So A plus H squared would be A squared. So that'll give five A squared, two times A times H. Then there's another five, so that'd be 10 times A times H, and then H squared. So five H squared. That's the first term. Next term seven A plus seven H. Make sure you get that seven distributing through, then you have a plus two. Then over here the minus distributes through, minus five A squared minus seven A minus two, all over H. Now if we're good people, what should happen? - [Student] Things should cancel. - [Instructor] Things should cancel. And if they're not canceling, you should be very suspicious. You should go back and check your work. In our case, an awful lot cancels. Now sometimes you have to do a little bit more algebraic manipulation, but you're gonna get to a point where it's like, okay, I get a lot of cancellation. Let's look at what remains. There's a 10 A times H, a five H squared, and a seven times H. Do they all have anything in common? They all have H. So we can pull that part out. That's the limit as H goes to zero of H times 10 times A plus five times H plus seven, all over H. And now we can cancel off our Hs. See, this is a beautiful thing. We're trying to get to the point where we can actually do a limit. Initially when we do a limit, if we don't do anything, we're gonna get zero over zero. It's guaranteed. So you gotta find a way to cancel. And in particular we know that the H downstairs is causing a problem. So we want to get a place where there's an H upstairs and now we'd have that, the H is canceled. So our limit becomes just 10 A plus five H plus seven. So how do we finish by taking this limit? What do we do? (students speaking indistinctly) Yeah, we can just plug in. Now what do we plug in for? - [Student] H. - [Instructor] Yeah, we plug in H, 'cause you know this is one of those times where's like, wait, which one is, 'cause there's two things, there's like A and there's H. So you have to think about which one is the one that you use. And the answer is what's the limit telling you to do? Follow the limit. The limit says H goes to zero. So you put an H equals zero. So we end up with 10 times A plus seven. And that's the answer. So if you gimme any A at all, lo and behold, I can tell you what the slope of the tangent line is at that particular point, 10 A plus seven. Now for those of you who have seen this before, you're like, that's a really long way to get to the answer. And yes, we're gonna get to the point where you can do this in your sleep. Let's make an observation here. Did this plus two end up actually mattering? - [Students] No. - [Instructor] No. Now, can we think about a good reason why plus two wouldn't matter? (students speaking indistinctly) So if you think about the plus two, if we were to graph our function here, the plus two shifts it up. But if I have a slope of my tangent line and I shift it up, what happens to the slope of the tangent line? - [Student] It doesn't change. - [Instructor] It doesn't change, right? It remains the same. So somehow you know, there are things where it's like oh, this won't affect the derivative, and one of them is adding a constant. Alright, well let's do another example. So suppose I have F of X equals absolute value of X. Here's a question, what is F prime of zero? Well let's see if we can figure this out. So according to the rule F prime of zero, that's equal to the limit as H goes to zero of F of zero plus H minus F of zero, all over H. I'm just putting in the definition so far. Now, zero plus H is another way to say this, just simply H. And the F of H would be absolute value of H. Absolute value of zero is zero. So I get the limit of H goes to zero of absolute value of H over H. Now that limit's not so obvious right away what it is, so let's try to think about it in two sides. So suppose I break it up into the left and the right. Let's start from the right. In other words, I'm a little bit above zero. What happens to absolute value of H if I'm a little bit above zero? (students speaking indistinctly) It's the same thing, it's just H. If I'm above zero, the absolute value doesn't have to do anything. So it'd be H over H, if I'm above zero, also known as one. What's the limit of one? - [Student] One? - [Instructor] It's one. If I'm a little bit below zero, what does the absolute value do? - [Student] Opposite. - [Instructor] So someone says it's gonna be opposite, right? 'Cause you know, if I'm a little bit below zero, my input is a negative thing, but I have to come out with a positive thing. So I better put a negative. So it's negative H over H. Well negative H over H that becomes, that's below, that's negative one. What's the limit of negative one? Negative one. Now we seem to have a challenge here. From the right it looks like our answer should be positive one. From the left, our answer looks like it should be negative one. So what should we do? Should we say split the difference and say life is good? (students speaking indistinctly) No, we don't do that. What do we do? - [Student] Doesn't exist. - [Instructor] Doesn't exist. Yeah. So what is the derivative of zero? The answer is it does not exist because the limit doesn't exist. See, the derivative is built off a limit and so the limit tells us it doesn't exist. Now let's think about this in terms of our sort of intuition about what's going on with limits. Now when I say our intuition, remember what did we do? We said sketch your function, so you know how to sketch absolute value X, and now zoom in. Okay. So if you zoom in, what are you gonna see? You're gonna see something that looks like that. So what do you do? Zoom in some more. What's gonna happen? You are gonna see something that looks like that. And what do you do? Keep zooming. But the thing is, no matter how much you zoom, what's gonna happen at zero? (students speaking indistinctly) You're always gonna see that V shape coming in. Now when we talk about derivatives, we're saying hey, that derivative is talking to us about that line, that line that we would see when we zoom in. But what happens is at this point at zero there's no line that you see, you never can zoom in far enough so it becomes flat, it'll always have that sharp turn and therefore there is no derivative because there's no tangent line. It doesn't become flat. So that's okay. It happens, it happens. Now. Some side notes here. The tangent line, that part that you see when you zoom in, it's really important to to get the following idea it. It's what the function looks like. And so a nice way to think about what the tangent line is, is it's the best line, and the best in the following sense: It's the line which is best mimicking the function. So that is, again, the intuition, you keep zooming in, and after you keep zooming in close enough, what you see is flat. And so that line is that flat thing, it's the best approximation. So we have a couple things to know. Of course we know the point on the line, we know the slope of the line, but we'll get to the point, this is foreshadowing of what's to come, is well, okay, so let me, let me sidetrack here. So common ways to write the tangent line. So if you know your point, Y minus your Y coordinate is M, your slope, times X minus your X coordinate. And you can rewrite this by adding F of A to both sides. You get Y equals F of A plus F prime of A times X minus A. And you can also rewrite this in sort of slope intercept form. Y equals F prime of A times X plus F of A minus A F, prime of A. But the moral here, that that punchline, I just wanna point out, is when we talk about saying that our tangent line is a good approximation, you see this Y right now is a tangent line. So we're gonna come back and say, hey, F of X is about the tangent line when I'm near A, because the tangent line mimics the function and therefore that tangent line says that F of X is about the function at A plus F prime of A times X minus A. So this is gonna be a really important idea as we go into the next part of the course. This is part two of the course. So let's see what we can do. Suppose we're told the following: Given that Y equals three X minus seven is tangent to Y equals F of X at X equals four, what can we say about the function at four? In particular, what is the value of the function at four, and what is the value of the derivative at four? So which one of these is easy to find? (student speaking indistinctly) Okay. The derivative might actually be the faster one to find because what does the derivative correspond to in terms of the line? The derivative corresponds to the slope. Can we read off the slope? Yeah, we can read it off. It's right there. It's three. So F prime of four, that's your slope, and therefore that's equal to three. How do you find the function at four? Is the answer negative seven? (student speaking indistinctly) The answer is not negative seven, because negative seven, that's where you hit the Y axis. But you may not be looking at the tangent line when you want to look at where you hit the Y axis. Alright, so how do you find F of four? - [Student] Plug four in. - [Instructor] The answer is you plug four in. Because what's happening? You see the tangent line mimics the function. So that means that the tangent line has to match the function. So if you think about this, this three X minus seven, this is a function and it's saying that at X equals four, this new function three X minus seven has to match the function F of X. So they're gonna have the same value, so you plug four in. Three times four minus seven. And what does that give us? - [Student] Five. - [Instructor] It gives us five. And that's our answer. So to find our derivative, read off the slope. To find the value of the function, evaluate. And so they're both fairly easy. So there's a very convenient way for us, if we have our tangent line, we can reconstruct information about our function. And then if we have information about the function, we can get our tangent line. So some of the problems that you'll see on maybe a quiz or possibly an exam, we'll be like, oh here's the tangent line, find this other tangent line. What's actually happening is it's saying here's the tangent line, translate that information into what's the value of the function, what's the value of the derivative. Get that to find new information about a different value of a function in a different derivative and then translate that back into a tangent line. So be comfortable going back and forth from tangent line to data, data to tangent line. Alright. Find the tangent line to Y equals sine of four X at X equals zero. Well let's see. What do we need? Well we need to know the value of the function at zero. So let's do that. So this sine of four X, this is our F of X. So what can we say about F of zero? F of zero, that sine at four times zero, which is zero and sine of zero is zero. And now we say great news, we're passing through the 0.00. That's the point on our tangent line. What's the next thing we need? - [Student] Slope. - [Instructor] We need slope. Okay, so slope means I need F prime of zero. So according to the rule, that's the limit, as H goes to zero of F of zero plus H minus F of zero, all over H, using our definition. So F of H, well that would be sine of four H. F of zero, we already saw that with zero, all divided by H. Now this is good because we say, oh wait a second, this kind of reminds us of something. Do you remember what this reminds? What does this remind us of? - [Student] Sine of blah over blah. - [Instructor] Sine of blah over blah. Wow, someone even remembers the technical terms. Right. Good. So if you have sine of blah over blah. Now the trick is they have to match. See we don't match yet. You have a four H, but that's just an H. We need to get a four there. Okay, I'll put a four there. Am I allowed to just do that? - [Student] No. - [Instructor] No. I gotta compensate. What else do I need to do? - [Student] Multiply by four. - [Instructor] Yeah, multiply by four. I can put a four, I'll put it over here just 'cause there's more space. Just you know, I modified it. So first off, the modification doesn't change the value, I'm just modifying the way I'm expressing it, but I'm putting it so that the inside of the sine matches what's below the sine. Now we're of the right form. So this sine of four H over four H, what happens to that? That part goes to one. And the four, what happens to that? It's gonna go to four, 'cause in fact it was always four. Okay, so now we have the derivative. Are we done? - [Student] No. - [Instructor] No, we're not done because we always remember the question asks us for something, we make sure we get what it's asked. It asks for the tangent line. So now what do we have? We have the slope, and we have the point. So Y minus the Y coordinate is equal to the slope coming from the derivative X minus the X coordinate. And that simplifies to Y equals four times X. And there you go. All right. Let's try another one. Find the tangent line to Y equals one over the squared of X at X equals four. Alright, well this should be fun. So first off, let's talk about the point. How did we get the point? (student speaking indistinctly) Yeah, we already have the X coordinate. So for the Y coordinate, we plug it in one over the squared to four is also known as? (student speaking indistinctly) One over two or a half. All right, what's next? - [Student] Find the derivative. - [Instructor] Right. We need to find the derivative to help us find the slope. So derivative at four. So we're gonna have F prime at four. That's gonna be the limit, as H goes to zero of F of four plus H minus F of four, all over H. So this is the limit as H goes to zero, F of four plus H. Well, what will F of four plus H look like? (student speaking indistinctly) One over the square root of? - [Student] Four plus H. - [Instructor] Four plus H, inside the square root. So think of the four plus H as just one thing, you know. So everywhere where there was an X before, boom, it's a four plus H. Minus F of four. Well we actually computed F of four, that's a half and that's all over H. Okay, that's a big fraction there. Alright, we're still getting zero over zero, which means that we have to do more work. But what's the more work we should be doing here? What's our next step? (students speaking indistinctly) So there's a couple things that you can do. And when I say there's a couple things, there's sort of two different orders. One thing you can do is you can actually do a conjugate here. You can say, hey, I'll multiply by one over squared of four plus H plus one over two. And that will help get you closer. You get rid of that square root. Before I would do that though, I would say I see here fractions within fractions, and that's tough. Even just a single fraction is tough. But when I see fractions within fractions, it unsettles me. I'm not comfortable with that. So I wanna get rid of the idea of fractions within fractions. So what can I do to help me get rid of that situation? Yeah, common denominator upstairs, add those two things together. So we're gonna get the limit as H goes to zero. Here the common denominator would be if you multiply the two of them together. So you'd end up with needing to multiply the first term by two and the second term by squared of four plus H. Then your common denominator would be the square root of four plus H times two. And that's all over H. Now here's the nice thing is you can think of this all over H as being the same as one over H multiplied there. So now you don't have a fraction within a fraction. And now we say okay, what's our problem? Well, downstairs we still have that H, we want to cancel it. Upstairs, you have that two minus squared of four plus H. Now what do you do? Well. Any idea? Conjugate. See the conjugate is a useful tool because it's gonna help us get a square. If we square a square root, that's gonna help get us closer to the truth. So applying that, we're gonna get the limit as H goes to zero. Upstairs we get two minus squared of four plus H and two plus squared of four plus H, which is gonna become four, two squared minus parenthesis four plus H. Downstairs, well, we have a whole bunch of stuff. We have a squared of four plus H, A two, an H, and a two plus squared of four plus H. All right, upstairs we have something nice. What happens? The fours cancel. So you're left with negative H. Now this is great that we have a negative H on top because what does negative H have in common? Well in fact it's just one term, it has negative H. But what else do we see? - [Student] H on the bottom. - [Instructor] There's an H on the bottom H downstairs. So I can actually cancel this H with that H downstairs. So if we were to do that, it's kind of two steps we did, but we can handle that, we're left with negative one on the top over squared of four plus H times two times two plus squared of four plus H. Now the important thing is that when we canceled those Hs, we canceled something which was causing zeros on both top and bottom. So the next thing we should do after we've canceled something which was causing zeros, is to do what? - [Student] Plug it in. - [Instructor] Plug it in, and maybe we're done. And in fact not just maybe we're done, we know we're done. How do we know we're done before we even evaluate? - [Student] Zero on the top. - [Instructor] Because we have negative one on the top. Negative one means it's game over man. Game over. Whatever happens, it happens right now. Okay, so now you just have to evaluate the bottom. Squared of four, two, two, two, two, still two, square of four, still two. All right, so you get two times two times two plus two, which gives us a grand total of 16. So minus one over 16. So that's our slope. So we're not done because again this was a find the tangent line question. But we're close because we have our slope and we have our point, and we put the two of them together. Let's see if we can fit it in here. So Y minus the Y coordinate, which is one half, is equal to our slope minus one over 16 times X minus the X coordinate, which is four. And unless we're told to simplify, that's our answer. We don't have to go any further. All right. So given the parabola Y equals, so does anyone have a favorite parabola? (student speaking indistinctly) What? - [Student] X (indistinct). - [Instructor] X squared, that's your favorite one? All right. I guess you know, we all start out with simple parabolas in life. Okay And the line Y equals? (student speaking indistinctly) What? - [Student] X. (students laugh) - [Instructor] X. (students laugh) Find all tangent lines to the parabola which are parallel to the given line. Also find all tangent lines to the parabola which are perpendicular to the given line. Now in the interest of time and the fact that we sort of have seen this type of argument before. So if this X squared thing that says our F of X, if we went through, if I ask you what is F prime of A, what would it turn out to be? - [Student] Two A. - [Instructor] It'd be two A. Yeah, derivative of X squared is two X. This will make more sense as we go on, but it's the same process. We've done it quadratic before, there's no real surprises. We wanna get to the fun stuff. Okay, so our derivative at any given particular point, we just double the value where we're at. Now we're told to do two things. We're told to, first off, find tangent lines which are parallel, and then we're gonna do perpendicular. So when are two lines parallel? (students speaking indistinctly) They have to have the same slope, right? Okay, that's fine. So what's the slope of this line? - [Students] One. - [Instructor] One. So for parallel, what we need is slope has to equal one. So we take our slope of our tangent line, that's two A, we set that equal to one. What does that tell us about what A should be? (student speaking indistinctly) A should be one half. So now we know where our tangent should occur. Our tangent line should be at equals a half. Well that's not really a point. That's really our X coordinate. What's our Y coordinate? One fourth, because our function is we square. So if our X coordinate is a half, you get a Y coordinate, we square that, we get a fourth. And therefore our line, which is parallel, Y minus our Y coordinate is our slope X minus our X coordinate. And if you wanted to, you could clean that up. You could say Y equals X minus one fourth. 'Cause you can add a fourth to both sides. Okay, so that's the parallel. There's a second part, perpendicular. When are two lines perpendicular? (students speaking indistinctly) Yeah, so there's that there that sort of phrasing. Now... The moral here is that perpendicular means negative reciprocal. So for example, if I have slope M, then I would flip it upside down and put a negative in front. So for our case, our original slope is one. So if I want perpendicular, what should my slope be to achieve that? - [Students] Negative one. - [Instructor] Negative one. So we do the same process. So I need to have that two times A, that's my slope at my tangent line has equal negative one. What should that tell me about A? (students speaking indistinctly) Negative half. So what's the point? Negative one half coma one fourth, 'cause we squared. So Y minus our Y value is equal to the slope, negative one, X minus r X value. And again, you can clean this up, you can say Y should equal, this would be negative X. And let's see, this would be plus a half, but then you have a minus a half, so it's really minus a half plus a fourth. So it's negative X minus a fourth. They're usually not quite so nice and symmetrical, but somebody wanted some nice functions. That's all right. We'll save the interesting functions for the exam. (students laugh) Alright. So last problem. And you might notice, hey, there's a star on here, that means it's a good one, this is a good one. Find all lines which are tangent to both F of X equals two X squared plus four X plus two and G of X, which is negative X squared plus two X minus one. Now we don't ask you to graph it, but we could graph this one. It's actually not too bad. Because if you look at what this function is, if you pull out a two X squared plus two X plus one, do you recognize X squared plus two X plus one? What is it? - [Student] Squared. - [Instructor] It's X plus one squared. Over here you could pull out a negative, X squared minus two X plus one. And then what would you have? X minus one squared. So in terms of what you have, you can see, all right, I have one parabola that's been shifted over so that the vertex here is at negative one and then it's pulled up a little bit. So here's my two times X plus one squared. Then I have another parabola which has been shifted over here to plus one but it opens down. Alright. And so our goal is to find all lines which are tangent to both curves. I claim that there's one easy answer and then there's another answer which is gonna take us about 10 minutes, which is the easy answer? (students speaking indistinctly) - [Student] X axis. - [Instructor] The X axis. The line Y equals zero is tangent to both of them because you can see that for Y equals zero, it touches both curves at a single point, it's not crossing them. Alright. So at the end of the process we should be able to recover Y equals zero. And if we didn't get Y equals zero we'd be in trouble. Now what's the subtlety about this? Well, the thing is that you can even see it here when you have Y equals zero, the tangent places, you know the locations where you have a tangent line are not the same. (instructor sighs) All right. So this is gonna make it interesting. So now we have to start thinking. How do we figure this out? Well, let's do the following. Let's say we're gonna have our tangent line that we're after, maybe we'll we'll say it's intersecting here at A and maybe it intersects at this one down here, it's B. This is obviously not the right ones because those lines are not the same. But alright so I have a tangent line to the first function, it's gonna hit at A. The tangent line for the second function's gonna hit it B. Now let's ask a really very simple sounding question. So be ready to give a simple answer. When are two lines the same? - [Students] Slopes. - [Instructor] Slopes are the same. - [Students] Intercepts. - [Instructor] Intercepts are the same. That's an easy answer to an easy question, but we can use that. Okay, so here's the idea. Let's actually figure out, alright, so I have this function F of X and what would be F prime at A? So you go through, take the derivative, what would you get? - [Students] Four (indistinct). - [Instructor] Four, A and four. If you did it for G, G prime at B, what would it be? - [Student] Negative two. - [Instructor] Negative two B, or is it not two B? Well, it's negative, so it's probably the not two B. And then plus two. So that would be if you ran through the limit definition. After you do about two or three of these with quadratics, you're like okay, you always see the same thing over and over. In the interest of time we're gonna skip that. So here's something that we have to have. We need to have that four A plus four had better be equal to negative two B plus two. Or we can rearrange things, four A plus two had better be equal to negative two or A, sorry, two A plus B had better equal to negative one. So there's one fact that we need. That's not enough. Alright, so we said the slopes have to match. The other thing we said is that the intercepts have to match. And we wrote this down and sort of like didn't pay much attention to it, but we actually have a formula for the intercepts. So the slope is the derivative. The intercept, I'm gonna copy it down here. That would be F of A minus A times F prime at A. Okay, that's our intercept. So for us, well, F of A two A squared plus four A plus two, subtract A times four A plus four. So cleaning this up, that's two A squared plus four A plus two minus four A squared plus, sorry, whoops, minus four A 'cause the minus goes through, the four A and the minus four A cancel. So that makes two A squared plus two. That's the Y intercept for the tangent line here at A. Well that's gotta equal the Y intercept for the other tangent line at B. So we're gonna run through it. Do the same thing, G of B minus B squared plus two B minus one minus B times minus two B plus two, and that's minus B squared plus two B minus one plus two B squared minus two B. The two Bs cancel. So this becomes B squared minus one. So the conclusion is that we also need to have minus two A squared plus two had better be equal to B squared minus one. So there's our second equation. We now have two equations and two unknowns. So when you have two equations and two unknowns, what should we do? We should solve. How do we solve? Well, in this case we can solve by substitution. So let's solve for B. I can say that B is minus one minus two A, and now I can put that information in here. So minus two A squared plus two is going to be equal to minus one minus two A squared minus one. All right. This is good. We're getting closer. Square that out. One cross terms will be plus four A and then plus four A squared and then minus one. And now where we say okay, cool, move everything to one side. We're gonna get six A squared 'cause there's four A squared plus two A squared, six a squared, then we get plus four A and then we get minus two equals zero. Which we could also say three a squared plus two A minus one equals zero. Now if the problem were nice, what should be true? It should factor. Do you think the problem is going to be nice? - [Student] No. - [Instructor] What? You're supposed to say, "Of course it is, "because our instructor loves us!" In fact it has to be nice. In fact, we know a factor without even working. Can you see what one of the factors has to be? (students speaking indistinctly) Do you see this right here? That's a tangent line that works. Where's a tangent at? (student speaking indistinctly) At negative one. So a better factor is a plus one, 'cause A equals negative one has to be a solution, and then there should be something else. So what's the something else? (students speaking indistinctly) Three A minus one. Now you can go back and check. Three a squared minus a plus three A, that's two A and minus one. So after all that we get to the following thrilling point. We have that the tangent line will happen at A equals negative one or at a equals? (students speaking indistinctly) One third. Now we actually already know the the tangent line for equals negative one. So that one was the one we already said before we began. So this is going to produce the line Y equals zero. So now we say aha, there's one more line. One more line, which is tangent to both. So at a equals one third, what's gonna be the slope of that line? Well, plug in a third. Four thirds plus four, also known as 16 thirds. That's our slope. And what else do we know? Well, what's the point? Well, we can take it and say the X one is one third, the Y coordinate would be two ninths, 'cause two times a third squared, plus four thirds plus two, which of course that's one third, we can think of this as two ninths plus 12 ninths plus 18 ninths, which is 32 ninths. And therefore bring this all together. Ah, okay, I need a bigger piece of paper. Alright, Y minus the Y coordinate is equal to the slope 16 thirds X minus the X coordinate one third. And there's that line. You wouldn't have guessed that, but we didn't have to guess it 'cause we've got math and that's far better than guessing. Alright, done. Done for today.