WEBVTT mathematics/pre-calculus/selhorst-jones 00:00:00.000 --> 00:00:02.200 Hi--welcome back to Educator.com. 00:00:02.200 --> 00:00:05.200 Today, we are going to talk about inverse functions. 00:00:05.200 --> 00:00:09.100 A function does a transformation on an input; we have talked about functions for a while now. 00:00:09.100 --> 00:00:12.400 But what if there was some way to reverse that transformation? 00:00:12.400 --> 00:00:21.200 This is the idea behind an inverse function: it is a way to reverse a transformation, reverse the process that another function is doing. 00:00:21.200 --> 00:00:24.300 It is a way to get back to our original input. 00:00:24.300 --> 00:00:29.400 By way of analogy, let's imagine a factory where, if you give them a pile of parts, they will make you a car. 00:00:29.400 --> 00:00:34.500 Now, if you take that car down the road to this other factory, you can give them that car, 00:00:34.500 --> 00:00:38.200 and they will give you back the original pile of parts you started with. 00:00:38.200 --> 00:00:42.500 There is one factory where they make cars out of parts, but then there is a second factory 00:00:42.500 --> 00:00:46.400 where they take cars and break them down into the original parts that were used to make them. 00:00:46.400 --> 00:00:50.800 There is one process, but there is also an inverse process that gets you back to where you started. 00:00:50.800 --> 00:00:54.800 If you follow one process with the other immediately, it ends up as if you haven't done anything. 00:00:54.800 --> 00:00:58.800 If you bring the pile of parts to the first factory, and then take that car to the second factory, 00:00:58.800 --> 00:01:03.200 and they give you back the pile of parts, it is like you just started with a pile of parts and didn't do anything to it. 00:01:03.200 --> 00:01:11.000 This is the idea behind an inverse function: it reverses a process--it reverses a transformation and gets you back to where you started. 00:01:11.000 --> 00:01:13.500 We have used the analogy of a function as a machine before; 00:01:13.500 --> 00:01:19.100 and it is a good image for being able to get across what is going on with inverse functions, as well. 00:01:19.100 --> 00:01:23.800 So, a function machine takes inputs, and it transforms them to outputs by some rule. 00:01:23.800 --> 00:01:32.000 So, what we are used to is: we plug in x into the function f, and it gives out f acting on x, f(x). 00:01:32.000 --> 00:01:37.200 Now, we could plug it into another one; we could plug it into an inverse machine, an inverse to f; 00:01:37.200 --> 00:01:42.800 and that would be called "f inverse," the f with the little -1 in the corner. 00:01:42.800 --> 00:01:48.200 f^-1 denotes the inverse of f; we call it "f inverse." 00:01:48.200 --> 00:01:54.000 We plug f(x) into f^-1, into that machine; we get right back to our original x. 00:01:54.000 --> 00:01:59.100 It is as if we hadn't done anything; the first machine does something, but then the second machine 00:01:59.100 --> 00:02:03.300 reverses that process and gets us back to where we started. 00:02:03.300 --> 00:02:08.700 So, is there a requirement for reversing--can we make an inverse out of all functions out there? 00:02:08.700 --> 00:02:12.700 No; let's see why by analogy first. 00:02:12.700 --> 00:02:18.900 Imagine a factory where, if you give them a pile of wood or a pile of metal, they give you a basketball in return. 00:02:18.900 --> 00:02:24.700 The basketball is the exact same, whether you started with wood or metal; it is always the exact same basketball. 00:02:24.700 --> 00:02:28.100 It doesn't matter what you gave them; it is just a basketball. 00:02:28.100 --> 00:02:31.800 Now, let's say you walk down the road to another factory; you give them that basketball; 00:02:31.800 --> 00:02:35.800 you tell them to reverse the process; and then you walk away--you give them no other information. 00:02:35.800 --> 00:02:44.400 Can that factory take a basketball and transform it back into a pile of wood or a pile of metal, if all they have is a basketball and no other information? 00:02:44.400 --> 00:02:47.700 No, they have no idea what you started with. 00:02:47.700 --> 00:02:52.000 Maybe they have wood; maybe they have metal; but the point is that they have no way 00:02:52.000 --> 00:02:55.100 to be able to know which one they are supposed to give you at this point. 00:02:55.100 --> 00:03:00.300 They don't have the information; the only person who has the information is you, when you brought the original wood pile, 00:03:00.300 --> 00:03:07.800 or brought the original metal pile--because all they have is the basketball, and the basketball could indicate wood, or it could indicate metal. 00:03:07.800 --> 00:03:11.200 They have no way to know what you started with; there is no way to figure it out. 00:03:11.200 --> 00:03:17.300 The information about what you originally had has been destroyed (although you would know it, because you brought it to the factory). 00:03:17.300 --> 00:03:22.300 But assuming you forgot, then the information has been destroyed--no one has the information anymore. 00:03:22.300 --> 00:03:26.500 Another way to think about it would be if you took a piece of paper, and you burned that piece of paper. 00:03:26.500 --> 00:03:27.800 You would be left with a pile of ashes. 00:03:27.800 --> 00:03:32.200 Now, someone could come along and think, "Oh, a pile of ashes--it used to be a piece of paper." 00:03:32.200 --> 00:03:37.300 But if you take two pieces of paper, and you write two completely different things on the two pieces of paper, 00:03:37.300 --> 00:03:40.700 and then you burn the two of them, a person could come along and think, "Two piles of ashes..." 00:03:40.700 --> 00:03:43.700 And they would know it was paper, but they wouldn't know what was written on them. 00:03:43.700 --> 00:03:47.300 They wouldn't be able to get that information back; the information has been destroyed. 00:03:47.300 --> 00:03:51.200 They know it was paper, but they don't know what was written on the paper. 00:03:51.200 --> 00:03:54.900 The basketball one...you have given them wood; you have given them metal; you get the same thing. 00:03:54.900 --> 00:03:59.200 The information about what you started with has been lost; the information has been destroyed, 00:03:59.200 --> 00:04:04.800 unless you come along and also say, "Oh, by the way, that basketball came from ____." 00:04:04.800 --> 00:04:12.200 The issue, in this scenario, is that we have two inputs providing the same output--whether it is metal or wood, you get a basketball. 00:04:12.200 --> 00:04:16.500 So, if we try to have a reverse on that, we have no way to know which one to go back to. 00:04:16.500 --> 00:04:19.300 We don't know if we are going to go back to metal; we don't know if we are going to go back to wood, 00:04:19.300 --> 00:04:21.800 because we don't know what the basketball is representing. 00:04:21.800 --> 00:04:31.200 So, to be a reversible process--for it to be possible to reverse something--the process has to have a different output for every input. 00:04:31.200 --> 00:04:37.100 If you give them metal, they have to give you one color of basketball; and if you give them wood, they give you a totally different color of basketball. 00:04:37.100 --> 00:04:41.400 Then, the second factory would say, "Oh, that is a wood basketball" or "Oh, that is a metal basketball." 00:04:41.400 --> 00:04:44.000 And they would be able to know what to do at that point. 00:04:44.000 --> 00:04:51.500 So, for a function f to have an inverse, it has to be that, for any a and b in the domain, any a or b that we could use in f normally, 00:04:51.500 --> 00:04:56.700 where a is not the same thing as b (where a and b are distinct from each other--they are different), 00:04:56.700 --> 00:05:02.100 then f(a) is different than f(b); f(a) does not equal f(b). 00:05:02.100 --> 00:05:08.000 So, if a and b are different, then the function's outputs on a and b are different, as well. 00:05:08.000 --> 00:05:16.600 So, different inputs going into a function have to produce different outputs; we call this property one-to-one. 00:05:16.600 --> 00:05:22.700 If this function has a property where whatever you put in, as long as it is different from something else going in, 00:05:22.700 --> 00:05:28.500 it means the two things will be different, that is called one-to-one: different inputs produce different outputs. 00:05:28.500 --> 00:05:33.800 You give them metal; you get one color of basketball; you give them wood--you get a different color of basketball. 00:05:33.800 --> 00:05:36.300 Here are some examples, so we can see this in a diagram. 00:05:36.300 --> 00:05:44.200 Here is an example that is one-to-one: a goes to 2; b goes to 1; c goes to 3. 00:05:44.200 --> 00:05:50.000 They each go to different things: different inputs each have different places that they end up going. 00:05:50.000 --> 00:05:54.400 Something that is not one-to-one: a goes to 1 and b goes to 1. 00:05:54.400 --> 00:06:04.900 It doesn't matter that c goes to 3, because a and b have both gone to the same thing, so they have different inputs that are producing the exact same output. 00:06:04.900 --> 00:06:08.900 a and b are different things, but they both produce a 1; so it is not one-to-one. 00:06:08.900 --> 00:06:13.600 We have that copy; we are putting in wood, and putting in metal, and we get basketball in both cases. 00:06:13.600 --> 00:06:18.400 And then, finally, just to remind us: this one over here (hopefully you remember this) is not a function. 00:06:18.400 --> 00:06:24.400 And it is not a function, because b is able to produce two outputs at once; and that is something that is not allowed for a function. 00:06:24.400 --> 00:06:31.300 If a function takes in one input, it is only allowed to produce one output; it can't produce multiple outputs from a single input. 00:06:31.300 --> 00:06:35.200 So, why do we call it one-to-one--why are we using this word, "one-to-one"? 00:06:35.200 --> 00:06:42.500 Well, we can think of it as being because a has one partner, and b has one partner, and c has one partner. 00:06:42.500 --> 00:06:48.100 Everybody gets a partner, and nobody has shared partners; everybody gets their partner, and that partner is theirs. 00:06:48.100 --> 00:06:50.400 They don't have to share it with anybody else. 00:06:50.400 --> 00:06:57.200 It is one-to-one: this thing is matched to this thing, and there is nobody else who is going to match up to that one: one-to-one. 00:06:57.200 --> 00:07:00.200 All right, how can we test for this? 00:07:00.200 --> 00:07:05.000 One way to test for this, to test if a function is one-to-one: we know, if we are going to be one-to-one, 00:07:05.000 --> 00:07:09.400 that every input must have a unique output; that was what it meant to be one-to-one. 00:07:09.400 --> 00:07:12.000 If we have different inputs, we have different outputs. 00:07:12.000 --> 00:07:17.700 So, if we draw a horizontal line on the graph, it can intersect the graph only once, or not at all. 00:07:17.700 --> 00:07:23.200 Remember, if we have some picture on a graph--like if we have this point--then what that means 00:07:23.200 --> 00:07:29.000 is that this is the input, and this over here is the output. 00:07:29.000 --> 00:07:37.700 We make it a point: (input,output); that is why it is (x,f(x)). 00:07:37.700 --> 00:07:45.100 If it is f(x) = x², then we plug in 3, and we get (3,9), (3,3²). 00:07:45.100 --> 00:07:49.600 The input is our horizontal location; the output is our vertical location. 00:07:49.600 --> 00:07:57.400 The horizontal line test is a way to test if the function has the same output for multiple inputs. 00:07:57.400 --> 00:08:04.800 We draw a horizontal line across, because wherever an output hits the graph, we know that there must be an input directly below it. 00:08:04.800 --> 00:08:10.600 If a function is not one-to-one, you will be able to draw a horizontal line that will intersect it twice, or maybe even more. 00:08:10.600 --> 00:08:15.700 Let's look at some examples: first, here is one that is one-to-one, because whenever we draw 00:08:15.700 --> 00:08:19.100 any sort of horizontal line, it is only going to cut it once. 00:08:19.100 --> 00:08:22.700 The only place that might seem a little confusing is if we draw it near here. 00:08:22.700 --> 00:08:25.900 It might make you think, "Well, doesn't it look like those are stacked?" 00:08:25.900 --> 00:08:31.400 Well, remember, we can't draw perfectly what is being represented by the mathematics. 00:08:31.400 --> 00:08:36.700 We have to give our lines some thickness; in reality, the line is infinitely thin. 00:08:36.700 --> 00:08:42.200 So, while it looks like they are kind of getting stacked, it is actually still moving through that zone; it is not constant. 00:08:42.200 --> 00:08:45.600 It is increasing just a little bit, but it isn't constant. 00:08:45.600 --> 00:08:52.200 Let's look at one that is not one-to-one: over here, this horizontal line (or many horizontal lines that we could make)-- 00:08:52.200 --> 00:08:57.500 it cuts it in two places; so we know that, here and here, there are two inputs. 00:08:57.500 --> 00:09:01.200 We can produce the same output from two different locations. 00:09:01.200 --> 00:09:05.400 We have two inputs making one output; so that means we are not one-to-one, 00:09:05.400 --> 00:09:09.800 because this one is partner to that height, but this one is also partner to this height. 00:09:09.800 --> 00:09:14.500 So, we are not one-to-one, because we have to share an output. 00:09:14.500 --> 00:09:21.200 Now that we have all of these ideas in mind, we are finally ready to define an inverse function; we can really talk about them and sink our teeth into them. 00:09:21.200 --> 00:09:26.200 Given some function f that is one-to-one (it has to be one-to-one for this to happen), 00:09:26.200 --> 00:09:33.900 there exists and inverse function, which we denote as f^-1 such that, for all x in the domain of f, 00:09:33.900 --> 00:09:38.200 any x that could normally go into f, any value that could normally be input, 00:09:38.200 --> 00:09:43.000 f^-1 acting on f(x) becomes just x. 00:09:43.000 --> 00:09:51.500 So, we have f acting on x like normal; and then, f^-1 acts on that whole thing. 00:09:51.500 --> 00:09:56.300 And it breaks the action that was done by f and returns us back to our original input. 00:09:56.300 --> 00:10:03.900 In other words, when f^-1 operates on the output of f, it gives the original input that went into f. 00:10:03.900 --> 00:10:12.400 Caution: I want to warn you about something: f^-1 means the inverse of f, not 1/f. 00:10:12.400 --> 00:10:16.900 This can be confusing, because, if you have taken algebra and remember your exponents 00:10:16.900 --> 00:10:22.600 (you might have forgotten them, but we will talk about them later in this course), -1 can mean a reciprocal for numbers. 00:10:22.600 --> 00:10:29.900 So, 7^-1 becomes 1/7; x^-1 becomes 1/x. 00:10:29.900 --> 00:10:36.100 But f is not "to the -1"; it is just a symbol that says inverse--"This is a function inverse." 00:10:36.100 --> 00:10:43.400 So, in general, for the most part, f^-1, "f inverse," is not the same thing as 1/f. 00:10:43.400 --> 00:10:49.000 The inverse of f is normally not the reciprocal of f, 1/f. 00:10:49.000 --> 00:10:55.900 This exponent thing, where 7^-1 is 1/7, is not the case for functions. 00:10:55.900 --> 00:11:01.300 On a function, the -1 does not represent an exponent; it is not an exponent. 00:11:01.300 --> 00:11:13.100 But it instead tells us function inverse; it is a way of saying, "This is an inverse function"; that is what it is telling us, not "flip it to the reciprocal." 00:11:13.100 --> 00:11:15.600 How can we get domain and range for f^-1? 00:11:15.600 --> 00:11:27.300 We can figure these out by looking at f; remember, the set of all outputs from f is its range. 00:11:27.300 --> 00:11:35.400 The things that x can get mapped to by f, what f is able to map x to, is the range of f. 00:11:35.400 --> 00:11:41.500 The domain of f is everything that x can be, everything that we can plug into f. 00:11:41.500 --> 00:11:44.600 And then, the range of f is everything that can come out of f. 00:11:44.600 --> 00:11:49.500 Now, f^-1 has to be able to take any output of f. 00:11:49.500 --> 00:11:53.100 It is not very good at reversing if there are some numbers that it is not allowed to reverse. 00:11:53.100 --> 00:11:56.300 So, it has to be able to reverse anything from f. 00:11:56.300 --> 00:12:02.800 If it is able to reverse anything from f, then that means the range of f has to be everything that we can put into f^-1. 00:12:02.800 --> 00:12:07.700 So, the domain of f^-1, the things we can put into f^-1, is the range of f. 00:12:07.700 --> 00:12:12.000 The domain of f^-1 is the range of f. 00:12:12.000 --> 00:12:18.400 Likewise, because f^-1 then breaks that f(x) and turns it back into original inputs, 00:12:18.400 --> 00:12:24.700 we must be able to turn it back into all of the original inputs, because all of the original outputs have to be over here. 00:12:24.700 --> 00:12:28.600 So, anything that we can make it to, we have to be able to make it back from. 00:12:28.600 --> 00:12:35.300 So, since we are able to get back everywhere, that means that we can output all the possible inputs for f. 00:12:35.300 --> 00:12:39.900 Since we can output all of the possible inputs, because we can reverse any of the processes, 00:12:39.900 --> 00:12:46.800 then it must mean that we are able to get the range of f^-1 from the domain of f. 00:12:46.800 --> 00:12:52.300 So, the domain of f tells us the range of f^-1, what we are allowed to output with f^-1. 00:12:52.300 --> 00:12:56.200 And the domain of f^-1 tells us the range of f. 00:12:56.200 --> 00:13:04.400 So, the domain of f^-1 is the things that f is able to output; and the range of f^-1, 00:13:04.400 --> 00:13:10.400 the things that f^-1 is able to output, is what we can put into f, the domain of f. 00:13:10.400 --> 00:13:12.700 This idea is going to let us prove something later on. 00:13:12.700 --> 00:13:17.500 Now we can get to that proof--the inverse of an inverse: what is the inverse of an inverse? 00:13:17.500 --> 00:13:25.900 In symbols, what is (f^-1)^-1--what do we do if we are going to take the inverse of something that is already doing inverses? 00:13:25.900 --> 00:13:31.800 Now, it might seem a little surprising, but it turns out that the inverse of f^-1 is just f. 00:13:31.800 --> 00:13:39.900 The inverse of f^-1 is f; it seems a little surprising, maybe, but it makes a sort of intuitive sense. 00:13:39.900 --> 00:13:44.600 If you do the opposite of an opposite, you get to the original thing. 00:13:44.600 --> 00:13:48.000 If you do an action, but then you are going to do the opposite of that action, 00:13:48.000 --> 00:13:53.200 but then you do the opposite of the opposite of the action, then you must be back at your original action. 00:13:53.200 --> 00:13:59.800 So, we might be able to believe this on an intuitive level; it makes sense, intuitively, that two opposites gets us back to where we started. 00:13:59.800 --> 00:14:04.700 But let's see a proof of this fact, formally; let's see it in formal mathematics. 00:14:04.700 --> 00:14:09.600 So, how do we get this started? Well, by definition, f^-1 is the function where, 00:14:09.600 --> 00:14:16.000 for any x, f^-1 acting on f(x) is going to just give us our original x. 00:14:16.000 --> 00:14:21.900 If f acts on an input, and then f^-1 comes and acts on that, we get back to our original input. 00:14:21.900 --> 00:14:28.900 Now, consider (f^-1)^-1: by this definition of the way inverses work, it must be that f inverse, inverse, 00:14:28.900 --> 00:14:34.600 when it acts on the thing that it is an inverse of...f inverse, inverse, is an inverse of f inverse... 00:14:34.600 --> 00:14:41.300 I know it is complicated to say...but this one right here is going to be the opposite action of f^-1. 00:14:41.300 --> 00:14:47.300 So, if we take any y (don't get too worried about x and y; remember, they are just placeholders for inputs), 00:14:47.300 --> 00:14:52.000 similarly, for any y, (f^-1)^-1, acting on f^-1(y), 00:14:52.000 --> 00:14:54.900 is going to just get us right back to our original y. 00:14:54.900 --> 00:15:01.500 It is the same structure as what is going on here with f^-1(f(x)) = x: we are just reversing a process. 00:15:01.500 --> 00:15:07.600 So, it doesn't matter that one of the processes is already a reversed process, because we are reversing this other reversed process. 00:15:07.600 --> 00:15:10.500 So, we get back to our original thing. 00:15:10.500 --> 00:15:17.200 Now, we know that y is in the domain of f^-1, because we are allowed to put it into f^-1. 00:15:17.200 --> 00:15:22.500 It is allowed to go into f^-1; now, we know, from our thing that we were just talking about, 00:15:22.500 --> 00:15:30.200 that the domain of f^-1 is the range of f; so there has to be... 00:15:30.200 --> 00:15:39.900 If f^-1 is the range of f (the domain of f^-1 is the range of f), 00:15:39.900 --> 00:15:44.500 if you are in the range, then that means that there is something out there that can produce this. 00:15:44.500 --> 00:15:49.600 That means that, if you are in the range of f, there must be some x in the domain of f; 00:15:49.600 --> 00:15:54.600 there has to be some way to get to that place in the range, so that f(x) is equal to y. 00:15:54.600 --> 00:15:59.900 There is some x out there in the domain of our original f, that f(x) is equal to y. 00:15:59.900 --> 00:16:06.800 So now, we have what we need: we can use this f(x) = y, and we can just plug it in right here and here. 00:16:06.800 --> 00:16:09.900 We will plug it in for the two y's up here, and we will see what happens. 00:16:09.900 --> 00:16:13.700 Thus, f inverse, inverse, acting on f^-1(f(x))... 00:16:13.700 --> 00:16:19.400 because remember, we know that there has to be some way to get an x such that f(x) = y, 00:16:19.400 --> 00:16:25.000 because of this business about domain and range; so we plug that in here, and we plug that in here. 00:16:25.000 --> 00:16:33.300 And we have that f inverse, inverse, on f inverse of f of x, must be equal to this over here on the right, as well. 00:16:33.300 --> 00:16:34.800 We are just doing substitution. 00:16:34.800 --> 00:16:41.700 But we know, by the definition of f^-1(f(x)), that this just turns out to be x. 00:16:41.700 --> 00:16:46.700 This whole thing right here just comes down to x--it simplifies right out to x. 00:16:46.700 --> 00:16:51.700 So, it must be the case that f inverse, inverse, of x is the same thing as f(x). 00:16:51.700 --> 00:17:02.900 If f(x) is the exact same thing as f inverse, inverse of x, it must be the case that (f^-1)^-1 is just the same thing as f. 00:17:02.900 --> 00:17:08.300 And our proof is finished; great. 00:17:08.300 --> 00:17:12.100 How can we interpret this graphically?--there is a great way to interpret inverses through graphs. 00:17:12.100 --> 00:17:14.900 First, let's consider f(x) = x³ + 1. 00:17:14.900 --> 00:17:20.100 Now, we know that this one has to be one-to-one, because it passes the horizontal line test. 00:17:20.100 --> 00:17:25.500 We come along and try to cut this with any horizontal line; they are only going to be able to cut in one place. 00:17:25.500 --> 00:17:30.800 Even here, where we have sort of seemed to flatten out, it is still moving, because we know it is x³ + 1. 00:17:30.800 --> 00:17:34.900 And it never actually stops going up; it just slows down how fast it is going up. 00:17:34.900 --> 00:17:39.300 And our lines have to have thickness, so while it kind of looks like they are stacked, they are not really. 00:17:39.300 --> 00:17:43.800 So, we see that it passed the horizontal line test; so it must be one-to-one. 00:17:43.800 --> 00:17:49.300 If it is one-to-one, we know it has to have an inverse; that is how we talked about this, right from the beginning. 00:17:49.300 --> 00:17:53.100 Now, notice that the graph, any graph, is made up of the points (x,f(x)). 00:17:53.100 --> 00:18:01.900 We talked about this before: 0 gets mapped to 1 when we plug it in as f(0) = 1; so that gives us the point (0,1). 00:18:01.900 --> 00:18:04.600 That is how we make up our original graph for f. 00:18:04.600 --> 00:18:08.200 Now, the graph of f^-1 will swap these coordinates. 00:18:08.200 --> 00:18:16.500 It takes in outputs and gives out inputs, in a way; so its input will swap these two things. 00:18:16.500 --> 00:18:24.800 It takes in f(x), and it gives out x; so the points of f^-1 will be the reverse of what we had for the points of the other one. 00:18:24.800 --> 00:18:29.000 So, (f(x),x) is what we get for f^-1. 00:18:29.000 --> 00:18:37.100 Now, visually, what that means is that f^-1 is going to be the mirror of y = x; and that is our line right here, y = x. 00:18:37.100 --> 00:18:41.800 Why is that the case? Well, look: we swap x and y coordinates if we go across this, 00:18:41.800 --> 00:18:57.900 because (-3,0) swaps to (0,-3); if we are going over y = x, if we are mirroring across this, we will swap the locations. 00:18:57.900 --> 00:19:05.100 And so, if we mirror over y = x, we are going to swap x and y, y and x; we will swap the order of our points, 00:19:05.100 --> 00:19:13.600 because y = x is sort of a way of saying, "Let's pretend for today that I am you and you are me." 00:19:13.600 --> 00:19:15.800 y is going to pretend that it is x, and x is going to pretend that it is y. 00:19:15.800 --> 00:19:18.800 They are sort of swapping places when we do a mirror over it. 00:19:18.800 --> 00:19:24.200 So, that means our picture, mirroring f over y = x...we get the graph of f^-1. 00:19:24.200 --> 00:19:30.900 So, we look at this; we mirror over it; and we get places like this. 00:19:30.900 --> 00:19:39.100 All right, we see how we are just sort of bouncing across it. 00:19:39.100 --> 00:19:43.400 And this is going to happen with any of our inverses graphically. 00:19:43.400 --> 00:19:50.000 So, any time f^-1 is being looked at, we know it is going to be a bounce, a reflection through, a mirror over; 00:19:50.000 --> 00:19:53.400 it is going to be symmetric to f with respect to the line y = x. 00:19:53.400 --> 00:20:01.500 Since f^-1 is swapping outputs and inputs, it is going to be sort of reversing the placements of these. 00:20:01.500 --> 00:20:06.300 So, the graph of f^-1 will always be symmetric to f, with respect to the line y = x. 00:20:06.300 --> 00:20:11.500 It will bounce across, because when you bounce across y = x, you swap your coordinates. 00:20:11.500 --> 00:20:15.400 Now, there are many ways to say this; I am saying "bounce across"; that is not really formal. 00:20:15.400 --> 00:20:20.100 But we can formally say that it is symmetric to f, with respect to the line y = x. 00:20:20.100 --> 00:20:24.300 You could also say that it reflects through y = x, or it reflects over y = x. 00:20:24.300 --> 00:20:27.300 You could also say that it mirrors over, or it mirrors through, y = x. 00:20:27.300 --> 00:20:32.000 There are many ways to say it; but in all of these things, the same idea is that we are going to bounce across, 00:20:32.000 --> 00:20:34.700 and that that point will now show up at that same distance here. 00:20:34.700 --> 00:20:40.400 So, let's see what it looks like: we bring them in, and indeed, they pop into those places. 00:20:40.400 --> 00:20:46.800 They pop into being a nice, symmetric-to-the-line, y = x; and that makes sense. 00:20:46.800 --> 00:20:51.700 We replaced the inputs with outputs and outputs with inputs; they have swapped locations. 00:20:51.700 --> 00:21:03.900 We look at this one here, and the point (3,0) on the inverse is connected to (0,3) on the original function--the same sort of thing on both of them. 00:21:03.900 --> 00:21:10.600 All right, so we have talked a lot about what is going on; we have a really great understanding of the mechanics behind an inverse. 00:21:10.600 --> 00:21:13.000 But how do we actually find an inverse? 00:21:13.000 --> 00:21:16.500 Now that we understand them, we are ready to actually go and find them. 00:21:16.500 --> 00:21:22.000 How do we turn an algebraic function like f(x) = x³ + 1 into a formula? 00:21:22.000 --> 00:21:30.400 Before we do this formula for f^-1, consider that f^-1 is taking the output of f(x); and it is transforming that into x. 00:21:30.400 --> 00:21:38.000 To find a formula for f^-1, we want a formula that gives x, if we know f(x). 00:21:38.000 --> 00:21:43.500 Normally, f(x) = x³ + 1, for example--normally we have x. 00:21:43.500 --> 00:21:52.400 We know x, and from that, we get our f(x); you plug in an x into a function, and it gives out f(x). 00:21:52.400 --> 00:21:58.400 So, f^-1 is the reverse of that; we know f(x), and we want to get x out of it. 00:21:58.400 --> 00:22:03.100 So, to be able to do this, we are solving f(x) = x³ + 1 in reverse. 00:22:03.100 --> 00:22:18.000 f(x) = x³ + 1; well, we move that over: f(x) - 1 = x³; so now we have ³√(f(x) - 1) = x. 00:22:18.000 --> 00:22:22.600 If we know what f(x) is, we can figure out that that is what the original x that did it is. 00:22:22.600 --> 00:22:25.600 We are solving it in reverse; we have reversed the function. 00:22:25.600 --> 00:22:31.100 As opposed to solving f(x) in terms of x, we are solving x in terms of f(x). 00:22:31.100 --> 00:22:37.500 Now, this is a little bit of a confusing idea; so instead, I am going to show you a method to do this. 00:22:37.500 --> 00:22:44.800 The idea of reversing is really what is behind inverses; it can be a little hard to understand what to do on a step-by-step basis. 00:22:44.800 --> 00:22:52.200 We are normally used to solving for f(x) in terms of x, having f(x) just on its own on one side, and having a bunch of stuff involving x on the other side. 00:22:52.200 --> 00:22:56.200 So, at this point, it might be a little bit confusing for you to try to do it the other way. 00:22:56.200 --> 00:23:01.700 And it would work; but let's learn a method that makes some of that confusion go away, and do things we are more used to doing. 00:23:01.700 --> 00:23:04.500 Here is one step-by-step guide for finding inverse functions. 00:23:04.500 --> 00:23:08.300 The very first thing we have to do: we have to check that the function is one-to-one. 00:23:08.300 --> 00:23:11.300 It has to be one-to-one for us to be able to find an inverse at all. 00:23:11.300 --> 00:23:16.800 Now, f(x) = x³ + 1...we just saw its graph; remember, it looked something like this. 00:23:16.800 --> 00:23:21.500 So, we already know that it passes the horizontal line test; it does a great job; it is a great function. 00:23:21.500 --> 00:23:25.800 It is one-to-one; great--we have already passed that part for this. 00:23:25.800 --> 00:23:30.000 Next, we swap f(x) for y; this is going to be a little bit easier for us in solving. 00:23:30.000 --> 00:23:35.500 We are used to solving for y's; we are not really used to solving for f(x)'s; so this will make it a little bit less confusing. 00:23:35.500 --> 00:23:43.500 We switch out f(x) for y; great; in the next step, we interchange the x and the y. 00:23:43.500 --> 00:23:47.200 In this one, we have x in its normal place and y in its normal place. 00:23:47.200 --> 00:23:56.600 What we do on step 3 is swap their places: y takes the place of all of the x's, and x takes the place of y. 00:23:56.600 --> 00:24:03.600 We swap x and y, interchange x and y; every time you had an x in step 2, you are now going to have a y; 00:24:03.600 --> 00:24:08.400 every time you had a y (which is probably just the one time, since it was from a function), you are going to now have an x. 00:24:08.400 --> 00:24:12.800 That is how we are doing this step that is the reversing step. 00:24:12.800 --> 00:24:20.500 Solve for y: at this point, we have x = y³ + 1; so if x = y³ + 1, we solve for it. 00:24:20.500 --> 00:24:28.700 We just move that 1 over: x - 1 = y³; and we have ³√(x - 1) = y. 00:24:28.700 --> 00:24:33.100 And you will notice that this actually looks pretty much the exact same as what we just did on the previous slide-- 00:24:33.100 --> 00:24:35.900 but perhaps a little less confusing, because it is what we are used to seeing. 00:24:35.900 --> 00:24:46.100 So, we have y = ³√(x - 1); and finally, just like we replaced f(x) with y, we now do a reverse replacement. 00:24:46.100 --> 00:24:51.800 But we are now going to f^-1; so y now becomes f^-1(x). 00:24:51.800 --> 00:24:59.900 f^-1(x) is equal to ³√(x - 1); f^-1(input) = ³√(input - 1). 00:24:59.900 --> 00:25:06.800 Great; now, while this method will produce the inverse if followed correctly, it is not perfect. 00:25:06.800 --> 00:25:15.000 Now, remember steps #2 and #3; in that, we had to swap f(x) for y, and then we were told to interchange x and y. 00:25:15.000 --> 00:25:20.600 Remember, they swapped places; now notice, these equations are completely different. 00:25:20.600 --> 00:25:30.000 They are totally, totally different from one another; yet they are still using the same x and y. 00:25:30.000 --> 00:25:34.300 Technically, it is not possible to have both of these equations be true with the same x and y. 00:25:34.300 --> 00:25:40.700 x and y can't possibly fulfill both of these equations at the same time, because they are completely different equations. 00:25:40.700 --> 00:25:48.000 So, what is going on here? When we swap in step #3, we are really creating a new, different y. 00:25:48.000 --> 00:25:53.100 When we have "swap f(x) for y," it is really red y or something here. 00:25:53.100 --> 00:26:00.300 But then, when we do the interchange, it becomes a totally different color of y; it becomes like blue y here. 00:26:00.300 --> 00:26:07.200 So, we are creating a new, different y; the y when we first swap is different than the meaning of the second y. 00:26:07.200 --> 00:26:12.500 The swapping y is a different y from our first time that we replaced f(x). 00:26:12.500 --> 00:26:18.500 The first one is standing in for f(x); that was our red y. 00:26:18.500 --> 00:26:24.100 And then, the second one stands in for f^-1(x); that is really taking the place there. 00:26:24.100 --> 00:26:28.100 This implicit difference between y's can be confusing; so be careful. 00:26:28.100 --> 00:26:33.400 I would recommend making a note on your paper; make a note when you are working that says where you swap. 00:26:33.400 --> 00:26:39.500 Use a note to see that swap of x and y, so that you can see the switch over to this inverse world, 00:26:39.500 --> 00:26:43.000 where you are now in an inverse world, and you can solve for an answer. 00:26:43.000 --> 00:26:49.100 This is a bit confusing; so why are we learning this method, if it has this hidden, confusing 00:26:49.100 --> 00:26:52.300 implicit difference, when we really think about what is going on? 00:26:52.300 --> 00:26:56.100 In short, the reason we are doing this is because everybody else does. 00:26:56.100 --> 00:27:03.200 That is not because it is perfect; it is because everyone else out there pretty much learns this method for solving inverses. 00:27:03.200 --> 00:27:06.700 Most textbooks, and almost all of the teachers out there, teach this method. 00:27:06.700 --> 00:27:11.700 So, it is important to learn, not because it is absolutely, perfectly correct, but because it is standard-- 00:27:11.700 --> 00:27:15.100 so that you can talk to other people, and talk about inverses, and they will understand 00:27:15.100 --> 00:27:18.900 what you are talking about, because they are doing the same method that you are doing. 00:27:18.900 --> 00:27:21.200 If you do something different, they might get confused. 00:27:21.200 --> 00:27:24.700 If they are really clever, or they really understand what is going on, they will think, "Oh, yes, that makes perfect sense." 00:27:24.700 --> 00:27:28.700 But we want to go with the standard method, so that other people will understand what we are doing. 00:27:28.700 --> 00:27:32.700 And if we are taking a course at the same time as we are watching this course, 00:27:32.700 --> 00:27:37.000 the teacher will think that is correct, as opposed to being confused by what you are doing and marking your grade down. 00:27:37.000 --> 00:27:43.600 But the important idea here, the really important idea inside of this thing, that is confusing, is the reversal. 00:27:43.600 --> 00:27:51.100 That is what the moment is all about--that moment between #2 and #3--the #3 step where we reverse, and we create this new y. 00:27:51.100 --> 00:27:58.200 We reverse the places; instead of solving for an output, we are solving for input. 00:27:58.200 --> 00:28:05.000 We are reversing the places, so we can do this directly; I did that with f(x), where I did f(x) = y; 00:28:05.000 --> 00:28:07.400 and I solved directly for if we know what f(x) is. 00:28:07.400 --> 00:28:14.200 I'm sorry, f(x) equals stuff involving x; I solved it directly for f(x)... 00:28:14.200 --> 00:28:25.700 We had f(x) = x³ + 1, and we figured out that it also is the same thing that the cube root of f(x) - 1 is equal to x. 00:28:25.700 --> 00:28:28.800 We figured that out; so there is this direct way of being able to do this. 00:28:28.800 --> 00:28:36.500 We can do this directly; but lots of students find this difficult or confusing, so we have this method of swapping x and y. 00:28:36.500 --> 00:28:42.700 And also, it has just become the standard way to do things; so it is good to practice this way, even though it is not absolutely perfect. 00:28:42.700 --> 00:28:45.200 It is not a perfect method, but it does the job. 00:28:45.200 --> 00:28:48.800 As long as you are careful and you pay attention to what you are doing--you closely follow its steps-- 00:28:48.800 --> 00:28:53.300 you will be able to get the answer, and you will be able to find the inverse function. 00:28:53.300 --> 00:28:57.300 Taking inverses can be difficult; it seemed a little bit confusing from what I have been saying so far. 00:28:57.300 --> 00:29:03.300 And it is an easy one to make a mistake on; this means it is really important to check your work. 00:29:03.300 --> 00:29:06.400 You really want to make sure that you check your work on this. 00:29:06.400 --> 00:29:12.200 How do you do this? Well, remember: by definition, f^-1(f(x)) is equal to x. 00:29:12.200 --> 00:29:16.300 That means, if we know what f^-1 is (we have figured out its formula), and we know what f(x) is 00:29:16.300 --> 00:29:20.200 (we were probably told f(x), we can just compose them. 00:29:20.200 --> 00:29:22.900 We know how to compose them from our lesson Composite Functions. 00:29:22.900 --> 00:29:28.400 If you didn't check out Composite Functions, you will have to watch that before you are able to compose them and do this check. 00:29:28.400 --> 00:29:35.100 But if it is really the inverse, you will get x; if you compose f^-1 with f(x), it has to come out to be x, 00:29:35.100 --> 00:29:39.600 because that is the definition of how we are creating this stuff, right from the beginning. 00:29:39.600 --> 00:29:45.200 Furthermore, we also know that f^-1, inverse, was just f; 00:29:45.200 --> 00:29:50.300 so it also must be the case that f acting on f^-1(x) will give us x, as well. 00:29:50.300 --> 00:29:56.500 You can compose them in either order when you are doing a check; and you will end up being able to get it correct. 00:29:56.500 --> 00:30:02.700 Let's see a quick example: for example, if f(x) = x³ + 1, and f^-1(x) = ³√(x - 1) 00:30:02.700 --> 00:30:04.800 (the ones we have been working with), how do we check this? 00:30:04.800 --> 00:30:14.100 Well, let's start with f^-1(f(x)); we compose this: we plug in f(x) = x³ + 1. 00:30:14.100 --> 00:30:20.700 So, f^-1 acting on x³ + 1...now, remember, we are going to plug that into f^-1(x). 00:30:20.700 --> 00:30:25.300 But it is f^-1(input); whatever is in the box just goes to the box over here. 00:30:25.300 --> 00:30:32.500 So, it is going to be that f^-1 will become cube root...where does the box go? 00:30:32.500 --> 00:30:41.600 x³ + 1...that is our box...minus 1; so the cube root of x³ + 1 - 1... 00:30:41.600 --> 00:30:47.300 + 1 - 1 cancels; the cube root of x³ equals x; great--that checks out. 00:30:47.300 --> 00:30:51.900 What about if we did it the other way--if we did it as f(f^-1(x))? 00:30:51.900 --> 00:30:54.500 Hopefully, this will work out, as well (and it will). 00:30:54.500 --> 00:31:04.200 So, what is f^-1(x)? f^-1(x) is the cube root of x - 1, so f(³√(x - 1))... 00:31:04.200 --> 00:31:09.300 what is going to happen over here?--we know that you plug in the box; you plug in the box. 00:31:09.300 --> 00:31:15.000 So, f(³√(x - 1))...we are going to take that, and we are going to plug it in right here. 00:31:15.000 --> 00:31:25.600 It is going to be ³√(x - 1), the quantity cubed, because it has to go in as the box; plus 1--finish out that function. 00:31:25.600 --> 00:31:35.400 The cube root, cubed...those are going to cancel each other; we will get x - 1 + 1, which is just equal to x; and it checks out. 00:31:35.400 --> 00:31:41.700 So, we can check it as f^-1(f(x)) or f(f^-1(x)); sometimes it might be easier for us to do it one way or the other. 00:31:41.700 --> 00:31:46.800 We could also do both ways, if you want to check and be absolutely, doubly sure that we really got our work correct. 00:31:46.800 --> 00:31:49.600 All right, let's move on to some examples. 00:31:49.600 --> 00:31:54.400 Using these graphs for assistance, which of the following functions are one-to-one? 00:31:54.400 --> 00:32:01.200 The first one is f(x) = 1/x; we do the horizontal line test--it is going to pass any high horizontal lines. 00:32:01.200 --> 00:32:08.200 What about as we get lower? Well, we know that 1/x continues to move--it never freezes and becomes constant. 00:32:08.200 --> 00:32:11.800 Does it ever cross this x-axis, though? No, it doesn't. 00:32:11.800 --> 00:32:16.200 We haven't formally talked about asymptotes yet; we will talk about asymptotes in a later lesson. 00:32:16.200 --> 00:32:24.800 But 1/x...as we go positive (f of a positive), 1 over a positive is going to also have to be positive. 00:32:24.800 --> 00:32:30.900 So, it never crosses the x-axis; the same thing goes with the negatives--f of a negative is going to be a negative. 00:32:30.900 --> 00:32:38.100 So, when it goes to the left, it never manages to cross this x-axis; as it goes to the right, it never manages to cross this x-axis. 00:32:38.100 --> 00:32:40.700 And it keeps changing; so the two things never cross over each other. 00:32:40.700 --> 00:32:48.500 So, yes, this is one-to-one. 00:32:48.500 --> 00:32:53.500 What about the blue one, g(x) = x³ - 2x² - x + 1? 00:32:53.500 --> 00:33:01.700 It is easy to say it fails: we cross lots of places in the middle here, and it is able to have multiple points at the same time. 00:33:01.700 --> 00:33:12.000 So, any one of these hits here and here and here; there are three points that all give the same output of 0; so it fails the horizontal line test. 00:33:12.000 --> 00:33:18.400 It is not one-to-one. 00:33:18.400 --> 00:33:27.100 Finally, (2x - 1) and (x² + 1); 2x - 1 is just a line that is going to keep going on this way forever and ever and ever. 00:33:27.100 --> 00:33:33.100 2x - 1, when x is less than or equal to 1...this is from piecewise functions; if you haven't checked out piecewise functions, this might be a little confusing. 00:33:33.100 --> 00:33:35.300 But hopefully, you have watched that lesson already. 00:33:35.300 --> 00:33:40.300 2x - 1 is x ≤ -1; it is just going to keep going on down and down and down, to the left and left and left. 00:33:40.300 --> 00:33:47.000 And x² + 1 is the right side of the parabola; if we plug in higher and higher numbers, it just keeps curving up and up and up to the right. 00:33:47.000 --> 00:33:52.900 So, that means that we are never going to cross; the parabola is never going to double back and manage to touch itself again. 00:33:52.900 --> 00:33:56.800 The parabola might eventually do this, but that part isn't on it. 00:33:56.800 --> 00:34:02.400 And the line is never going to be able to go down to have itself crossed horizontally. 00:34:02.400 --> 00:34:14.900 So, if we do any horizontal line crossing on this, it is never going to hit twice; so it is one-to-one. 00:34:14.900 --> 00:34:20.600 One thing I would like to make a special comment on: notice that right here there is an empty space. 00:34:20.600 --> 00:34:24.300 There is this gap where it jumps; is that a problem for a horizontal line test? 00:34:24.300 --> 00:34:29.700 No, it is not a problem at all, because the horizontal line test is allowed to hit no points, as well. 00:34:29.700 --> 00:34:36.000 It is allowed to hit one point or zero points; in this case, if it goes through that gap, it hits no points; but that is OK. 00:34:36.000 --> 00:34:39.700 We are only worried about having multiple inputs for the same output. 00:34:39.700 --> 00:34:42.600 It is OK if there are no inputs to make an output; the important thing 00:34:42.600 --> 00:34:46.800 is that there are no double sets of inputs that all make the same output. 00:34:46.800 --> 00:34:52.500 Like, in the blue one, where we had multiple different places where we could plug in some number-- 00:34:52.500 --> 00:34:56.600 plug in different numbers, but they would all produce zeroes. 00:34:56.600 --> 00:35:02.300 All right, let's actually find an f^-1: f(x) = -3x/(x + 3). 00:35:02.300 --> 00:35:08.300 They told us, right from the beginning, that it is one-to-one; so we can jump right to figuring it out: what is f^-1(x)? 00:35:08.300 --> 00:35:10.800 And then, after it, we need to check our answer. 00:35:10.800 --> 00:35:15.100 OK, so what is f^-1(x)? Remember all of our steps, one by one. 00:35:15.100 --> 00:35:22.600 f(x) = -3x/(x + 3): they told us, right from the beginning, that it is one-to-one, so we are already checked out. 00:35:22.600 --> 00:35:24.500 We have already checked out the first one. 00:35:24.500 --> 00:35:30.800 The next step: we swap y for f(x): y = -3x/(x + 3). 00:35:30.800 --> 00:35:35.800 Now, that is not the important part of when we reverse, though; we reverse into inverse world. 00:35:35.800 --> 00:35:44.200 So, here is when we go into inverse world; we reverse the place of x and y. 00:35:44.200 --> 00:35:50.900 So now, it is x where y was, and it is -3y/(y + 3). 00:35:50.900 --> 00:35:54.300 Now, we just go about this, and we solve it for y like we normally would. 00:35:54.300 --> 00:36:09.900 Multiply both sides by y; we get x times (y + 3) equals -3y; let's distribute this out: xy + 3x...let's also move the 3y over, so + 3y = 0. 00:36:09.900 --> 00:36:15.900 OK, at this point, we will pull out the y's from these two things; we will move them together, so we can see it a little bit easier at first. 00:36:15.900 --> 00:36:25.700 xy + 3y + 3x = 0; let's subtract that 3x to move it over; -3x, -3x here. 00:36:25.700 --> 00:36:35.500 So, then we will pull out the y's to the right; so we have x + 3, times quantity y, equals -3x. 00:36:35.500 --> 00:36:42.200 Finally, we divide by that x + 3, and we get y = -3x/(x + 3). 00:36:42.200 --> 00:36:55.700 And now, finally, we can plug in f^-1 for this y; so we plug it in, and we get f^-1(x) = -3x/(x + 3). 00:36:55.700 --> 00:36:59.100 Great; now, let's check and make sure that we got this right. 00:36:59.100 --> 00:37:11.800 We check this in red; here is our check--let's check it by plugging f into f^-1. 00:37:11.800 --> 00:37:18.900 So, we want this to come out to be x; it should be x, if we got everything right. 00:37:18.900 --> 00:37:24.700 So, f^-1(f(x)); what is f(x)? f(x) is this; and here is something funny to notice. 00:37:24.700 --> 00:37:39.700 Notice -3x/(x + 3); amazingly, it just so happens that for -3x/(x + 3), f(x) and f^-1(x) are the exact same thing--kind of impressive. 00:37:39.700 --> 00:37:53.700 We plug this in; we have f^-1(f(x)); f(x) is -3x/(x + 3); now, over here, we plug it in; what is in the box? 00:37:53.700 --> 00:38:01.300 The box shows up here; the box shows up here; it shows up twice, so it is f^-1 on -3x/(x + 3). 00:38:01.300 --> 00:38:21.500 It is going to be -3...what is in the box?...-3x/(x + 3), over (-3x/(x + 3)) + 3. 00:38:21.500 --> 00:38:27.300 Great; so the first thing that is going to be confusing is that we have this x + 3, and we have this x + 3 here. 00:38:27.300 --> 00:38:32.100 So, let's take that out by multiplying the whole thing by (x + 3)/(x + 3). 00:38:32.100 --> 00:38:37.000 We can get away with that, because it is just the same thing as 1: (x + 3)/(x + 3) is just 1. 00:38:37.000 --> 00:38:44.300 So, (x + 3)/(x + 3)...multiply that here; the (x + 3) will cancel out here and cancel out here. 00:38:44.300 --> 00:38:50.200 But remember, it also has to distribute to the other part, because they are not connected through multiplication on that; they are connected through addition. 00:38:50.200 --> 00:39:01.600 So, we have -3, -3x, over -3x plus 3 times x plus 3. 00:39:01.600 --> 00:39:10.500 These two negatives cancel out; so we have 3 times 3x on the top, -3x plus 3(x + 3)...so we have 9x on the top, 00:39:10.500 --> 00:39:21.700 divided by -3x plus 3x plus 9; -3x plus 3x...they cancel each other out; we have 9x/9. 00:39:21.700 --> 00:39:27.000 9 over 9...those cancel out, and we have just x. 00:39:27.000 --> 00:39:30.200 So, that checks out--great, we have the answer. 00:39:30.200 --> 00:39:35.200 All right, the next one: we have, this time, a piecewise function. 00:39:35.200 --> 00:39:38.300 This is a little confusing: we didn't talk about this formula, but we will see how to do it. 00:39:38.300 --> 00:39:46.900 f(x) = -x + 1 when x < 0, and -√x when x ≥ 0. 00:39:46.900 --> 00:39:53.100 This is confusing; we don't know what to do about the different pieces of the piecewise function. 00:39:53.100 --> 00:39:57.400 We don't know what to do about these two different categories: we have x < 0 and x ≥ 0. 00:39:57.400 --> 00:40:02.600 We didn't learn that when we learned how to do inverses; but we could still figure out these two. 00:40:02.600 --> 00:40:07.000 We could figure out what is the inverse of -x + 1 and what is the inverse of -√x. 00:40:07.000 --> 00:40:12.900 We were told, explicitly, that this is one-to-one; so we can go ahead and do this, and then we will think about it. 00:40:12.900 --> 00:40:22.200 First, we will do inverses on these two rules; and then we will figure out how they fit together--what are the categories for these two rules? 00:40:22.200 --> 00:40:39.400 So, first, -x + 1; we will have y = -x + 1; we swap them, so we now get into our inverse world. 00:40:39.400 --> 00:40:46.000 Swap their locations; we interchange them, and we are now at negative...sorry, not -x; the negative does not swap. 00:40:46.000 --> 00:40:58.100 We are at x = -y + 1; we move the y over and move the x over; we get -x + 1; so y = -x + 1, 00:40:58.100 --> 00:41:04.200 which is going to give us f^-1 for at least the first rule here. 00:41:04.200 --> 00:41:06.800 Now, what about the other one?--let's do that, as well. 00:41:06.800 --> 00:41:22.600 So, y = -√x; we go into inverse mode; we reverse their locations; and we are now at x = -√y. 00:41:22.600 --> 00:41:27.600 So, how do we solve for y? Well, we move this negative over: -x = √y. 00:41:27.600 --> 00:41:37.700 Square both sides; we get (-x)² = y; and then (-x)²...the negatives will cancel out, so we get just x² = y. 00:41:37.700 --> 00:41:41.700 And so, this is the inverse rule for this part. 00:41:41.700 --> 00:41:52.700 Now, here is the part where we start thinking: we know that f^-1 is going to break into a piecewise function using these two different things. 00:41:52.700 --> 00:41:59.700 y = -x + 1...so it will be -x + 1 for the first rule, and then x² for the second rule. 00:41:59.700 --> 00:42:04.300 But the question is that we don't know what the categories are. 00:42:04.300 --> 00:42:06.000 How do we figure out what the categories are? 00:42:06.000 --> 00:42:23.000 Well, remember: if f goes from its domain to its range, let's call that a to b, then f^-1 does the reverse of that. 00:42:23.000 --> 00:42:28.500 f^-1 goes from b to a; it does the reverse. 00:42:28.500 --> 00:42:37.300 What that means is that the domain...the thing that determined which rule we used...we need to do the range to determine which way to get back. 00:42:37.300 --> 00:42:51.600 The range on these two rules...now we are back to using f, so range on f...for -x + 1: 00:42:51.600 --> 00:42:56.700 well, -x + 1 was x < 0; that was the category, so it has to be within those. 00:42:56.700 --> 00:43:03.800 So, what can it go to? Well, if we plug in a really big negative number, like, say, -100, we will get -(-100) + 1; so we get 101. 00:43:03.800 --> 00:43:07.600 So, as long as we keep plugging in more and more negative numbers, we get bigger and bigger numbers. 00:43:07.600 --> 00:43:12.300 We are able to get all the way out to positive infinity, as we are really far in negative numbers. 00:43:12.300 --> 00:43:20.700 What is the lowest that we can get to? Well, we could get really close to 1, as we plug in -0.00000001. 00:43:20.700 --> 00:43:24.700 We are really close to being to 1, so we can get right up to 1; but we can't actually touch it. 00:43:24.700 --> 00:43:26.900 We have to exclude it, so we use parentheses. 00:43:26.900 --> 00:43:34.600 So, the range for the first rule is this: -x + 1 becomes this. 00:43:34.600 --> 00:43:38.800 So, I will put a red dot on that, because that matches to this rule here. 00:43:38.800 --> 00:43:41.800 Now, what about the range for the other rule? 00:43:41.800 --> 00:43:50.100 The range on this rule is -√x; it has x ≥ 0 as its domain; what are the numbers we can get out of this? 00:43:50.100 --> 00:43:52.300 What is the largest number we can get out of it? 00:43:52.300 --> 00:43:59.100 The largest number we can get out of it is actually 0; why?--because, when we plug in any reasonably large positive number, 00:43:59.100 --> 00:44:09.300 like, say, 100, then -√100 is -10; so as we get bigger and bigger positive numbers that we plug in, 00:44:09.300 --> 00:44:10.900 we actually get more and more negative. 00:44:10.900 --> 00:44:15.500 So, we can actually go to any negative number we want; we can go all the way down to negative infinity. 00:44:15.500 --> 00:44:20.300 Can we actually reach 0? Yes, we actually can reach 0, because it is greater than or equal to. 00:44:20.300 --> 00:44:27.700 So, if we plug in x = 0, we get -√0, which is just 0; and we put a bracket to indicate that we are actually allowed to do it. 00:44:27.700 --> 00:44:37.400 This one is the range for -√x, that rule; it is going to get a green dot on it, because it matches to the green rule. 00:44:37.400 --> 00:44:44.600 That means that -x + 1 is allowed to take in...what values? It is allowed to take in the range values. 00:44:44.600 --> 00:44:48.700 It is allowed to do a reverse on anything that shows up in the range (1,∞). 00:44:48.700 --> 00:44:55.600 Also notice: these two ranges, (1,∞) and (-∞,0], don't have any intersection. 00:44:55.600 --> 00:45:00.600 They don't overlap at all, so we don't have any worries about pulling from one versus pulling from the other. 00:45:00.600 --> 00:45:03.400 They will never get in each other's way. 00:45:03.400 --> 00:45:11.400 So, for this one, -x + 1, if it is going to be allowed to go from 1 to infinity, then that means we can plug in anything into f^-1, 00:45:11.400 --> 00:45:19.500 where x is greater than 1, which is to say input; it is not the same x that was up here. 00:45:19.500 --> 00:45:22.500 It is now just saying "placeholder--anything that we are plugging in." 00:45:22.500 --> 00:45:27.600 What about x²? Well, that was the green dot--that was allowed to go from negative infinity up to 0. 00:45:27.600 --> 00:45:35.500 So, it is allowed to have x ≤ 0; it is allowed to go all the way up to negative infinity, but it can only just get to touching 0. 00:45:35.500 --> 00:45:41.100 It is allowed to actually have 0, though; x > 1 is not actually allowed to touch 1, but it is able to get as close as it wants. 00:45:41.100 --> 00:45:46.700 And there is our piecewise inverse function. 00:45:46.700 --> 00:45:50.600 It is a little bit difficult, but if you think about it, you do each of the inverses, and then you think about 00:45:50.600 --> 00:46:00.000 "How do I get the domain for the inverse? I get it from the domain of f becoming the range of our inverse, 00:46:00.000 --> 00:46:06.500 and the range of our f becoming the domain of our inverse." 00:46:06.500 --> 00:46:12.700 So, what the original function was able to output to is what the inverse is allowed to take in. 00:46:12.700 --> 00:46:19.100 And that is how we figured out these rules, these categories--what the categories were for these two different transformations. 00:46:19.100 --> 00:46:30.800 All right, the final example: f and g are one-to-one functions; now, prove that f composed with g, inverse, is equal to g inverse composed with f inverse. 00:46:30.800 --> 00:46:34.900 This might be a little daunting at first; these are weird symbols; we are not used to using these sorts of things. 00:46:34.900 --> 00:46:45.900 So, if that is the case, let's remind ourselves: from composition, f composed with g, acting on x, is equal to f(g(x)). 00:46:45.900 --> 00:46:51.400 Now, I said before: it makes things always, always, always easier to see it in that format. 00:46:51.400 --> 00:47:03.600 What we want to show is that g^-1 composed with f^-1 (which would be g^-1(f^-1(x)))... 00:47:03.600 --> 00:47:09.400 we want to show that this one here is an inverse to that one over there. 00:47:09.400 --> 00:47:12.900 That is what we are trying to prove, that f composed with g inverse... 00:47:12.900 --> 00:47:17.200 We know, by the definition of how this symbol works, by how inverses work... 00:47:17.200 --> 00:47:25.300 f composed with g^-1, acting on f composed with g, on x, is going to just leave us as if we had done nothing, 00:47:25.300 --> 00:47:28.200 because we are putting an inverse on something. 00:47:28.200 --> 00:47:34.000 So, we want to show that this means the exact same thing as this right here. 00:47:34.000 --> 00:47:50.700 So, let's just try it out: we will set it up like this: f composed with g^-1, acting on f composed with g, acting on x. 00:47:50.700 --> 00:48:01.400 OK, so what does that become? Well, we know that f composed with g, acting on x, is the same thing as f(g(x)). 00:48:01.400 --> 00:48:06.800 All right, what is f composed with g^-1? Well, we know (from what we did over here) 00:48:06.800 --> 00:48:10.700 that we can bring that into g^-1 acting on f^-1, acting on whatever is going into it. 00:48:10.700 --> 00:48:28.600 What is going into it here is this whole thing; so, it is going to be g^-1, acting on f^-1, acting on f, acting on g, acting on x. 00:48:28.600 --> 00:48:31.700 And then, we close up all of those parentheses. 00:48:31.700 --> 00:48:42.300 That is a little bit confusing; but we are seeing inverses right next to functions: f^-1 acting on f, acting on whatever is in there. 00:48:42.300 --> 00:48:46.300 It just cancels out and gets us right back to what we originally had in there. 00:48:46.300 --> 00:48:53.400 So, f^-1 acting on f...that cancels out, and we get g^-1, acting on whatever was in there, which was g(x). 00:48:53.400 --> 00:49:00.200 So, g^-1 acting on g(x)...the exact same thing: we get down to x; so we have proved it. 00:49:00.200 --> 00:49:07.700 g^-1 composed with f^-1 is how we create f composed with g, inverse. 00:49:07.700 --> 00:49:08.800 Great; we have proved it. 00:49:08.800 --> 00:49:12.200 All right, I hope you have a much better idea of how inverses work at this point. 00:49:12.200 --> 00:49:16.200 They can be a little bit confusing, but you have that method to be able to guide you through it. 00:49:16.200 --> 00:49:18.300 Just follow it really carefully, step-by-step. 00:49:18.300 --> 00:49:23.200 The danger is if you break from those steps and do something else; that is where you can make mistakes. 00:49:23.200 --> 00:49:26.200 If you really understand what is going on, you don't even have to use that method. 00:49:26.200 --> 00:49:32.100 But it really is the standard method, so it is a good idea to stick with it, just because it is what a lot of other people are used to using. 00:49:32.100 --> 00:49:34.300 And you can find it in a lot of textbooks. 00:49:34.300 --> 00:49:37.000 All right, we will see you at Educator.com later--goodbye!