WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about inverse functions.
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A function does a transformation on an input; we have talked about functions for a while now.
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But what if there was some way to reverse that transformation?
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This is the idea behind an **inverse function**: it is a way to reverse a transformation, reverse the process that another function is doing.
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It is a way to get back to our original input.
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By way of analogy, let's imagine a factory where, if you give them a pile of parts, they will make you a car.
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Now, if you take that car down the road to this other factory, you can give them that car,
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and they will give you back the original pile of parts you started with.
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There is one factory where they make cars out of parts, but then there is a second factory
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where they take cars and break them down into the original parts that were used to make them.
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There is one process, but there is also an inverse process that gets you back to where you started.
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If you follow one process with the other immediately, it ends up as if you haven't done anything.
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If you bring the pile of parts to the first factory, and then take that car to the second factory,
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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.
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This is the idea behind an inverse function: it reverses a process--it reverses a transformation and gets you back to where you started.
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We have used the analogy of a function as a machine before;
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and it is a good image for being able to get across what is going on with inverse functions, as well.
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So, a function machine takes inputs, and it transforms them to outputs by some rule.
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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).
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Now, we could plug it into another one; we could plug it into an inverse machine, an inverse to f;
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and that would be called "f inverse," the f with the little -1 in the corner.
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f^-1 denotes the inverse of f; we call it "f inverse."
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We plug f(x) into f^-1, into that machine; we get right back to our original x.
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It is as if we hadn't done anything; the first machine does something, but then the second machine
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reverses that process and gets us back to where we started.
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So, is there a requirement for reversing--can we make an inverse out of all functions out there?
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No; let's see why by analogy first.
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Imagine a factory where, if you give them a pile of wood or a pile of metal, they give you a basketball in return.
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The basketball is the exact same, whether you started with wood or metal; it is always the exact same basketball.
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It doesn't matter what you gave them; it is just a basketball.
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Now, let's say you walk down the road to another factory; you give them that basketball;
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you tell them to reverse the process; and then you walk away--you give them no other information.
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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?
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No, they have no idea what you started with.
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Maybe they have wood; maybe they have metal; but the point is that they have no way
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to be able to know which one they are supposed to give you at this point.
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They don't have the information; the only person who has the information is you, when you brought the original wood pile,
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or brought the original metal pile--because all they have is the basketball, and the basketball could indicate wood, or it could indicate metal.
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They have no way to know what you started with; there is no way to figure it out.
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The information about what you originally had has been destroyed (although you would know it, because you brought it to the factory).
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But assuming you forgot, then the information has been destroyed--no one has the information anymore.
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Another way to think about it would be if you took a piece of paper, and you burned that piece of paper.
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You would be left with a pile of ashes.
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Now, someone could come along and think, "Oh, a pile of ashes--it used to be a piece of paper."
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But if you take two pieces of paper, and you write two completely different things on the two pieces of paper,
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and then you burn the two of them, a person could come along and think, "Two piles of ashes..."
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And they would know it was paper, but they wouldn't know what was written on them.
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They wouldn't be able to get that information back; the information has been destroyed.
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They know it was paper, but they don't know what was written on the paper.
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The basketball one...you have given them wood; you have given them metal; you get the same thing.
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The information about what you started with has been lost; the information has been destroyed,
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unless you come along and also say, "Oh, by the way, that basketball came from ____."
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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.
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So, if we try to have a reverse on that, we have no way to know which one to go back to.
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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,
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because we don't know what the basketball is representing.
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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.
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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.
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Then, the second factory would say, "Oh, that is a wood basketball" or "Oh, that is a metal basketball."
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And they would be able to know what to do at that point.
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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,
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where a is not the same thing as b (where a and b are distinct from each other--they are different),
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then f(a) is different than f(b); f(a) does not equal f(b).
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So, if a and b are different, then the function's outputs on a and b are different, as well.
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So, different inputs going into a function have to produce different outputs; we call this property **one-to-one**.
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If this function has a property where whatever you put in, as long as it is different from something else going in,
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it means the two things will be different, that is called one-to-one: different inputs produce different outputs.
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You give them metal; you get one color of basketball; you give them wood--you get a different color of basketball.
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Here are some examples, so we can see this in a diagram.
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Here is an example that is one-to-one: a goes to 2; b goes to 1; c goes to 3.
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They each go to different things: different inputs each have different places that they end up going.
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Something that is not one-to-one: a goes to 1 and b goes to 1.
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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.
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a and b are different things, but they both produce a 1; so it is not one-to-one.
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We have that copy; we are putting in wood, and putting in metal, and we get basketball in both cases.
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And then, finally, just to remind us: this one over here (hopefully you remember this) is not a function.
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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.
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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.
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So, why do we call it one-to-one--why are we using this word, "one-to-one"?
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Well, we can think of it as being because a has one partner, and b has one partner, and c has one partner.
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Everybody gets a partner, and nobody has shared partners; everybody gets their partner, and that partner is theirs.
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They don't have to share it with anybody else.
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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.
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All right, how can we test for this?
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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,
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that every input must have a unique output; that was what it meant to be one-to-one.
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If we have different inputs, we have different outputs.
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So, if we draw a horizontal line on the graph, it can intersect the graph only once, or not at all.
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Remember, if we have some picture on a graph--like if we have this point--then what that means
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is that this is the input, and this over here is the output.
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We make it a point: (input,output); that is why it is (x,f(x)).
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If it is f(x) = x², then we plug in 3, and we get (3,9), (3,3²).
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The input is our horizontal location; the output is our vertical location.
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The horizontal line test is a way to test if the function has the same output for multiple inputs.
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We draw a horizontal line across, because wherever an output hits the graph, we know that there must be an input directly below it.
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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.
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Let's look at some examples: first, here is one that is one-to-one, because whenever we draw
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any sort of horizontal line, it is only going to cut it once.
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The only place that might seem a little confusing is if we draw it near here.
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It might make you think, "Well, doesn't it look like those are stacked?"
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Well, remember, we can't draw perfectly what is being represented by the mathematics.
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We have to give our lines some thickness; in reality, the line is infinitely thin.
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So, while it looks like they are kind of getting stacked, it is actually still moving through that zone; it is not constant.
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It is increasing just a little bit, but it isn't constant.
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Let's look at one that is not one-to-one: over here, this horizontal line (or many horizontal lines that we could make)--
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it cuts it in two places; so we know that, here and here, there are two inputs.
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We can produce the same output from two different locations.
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We have two inputs making one output; so that means we are not one-to-one,
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because this one is partner to that height, but this one is also partner to this height.
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So, we are not one-to-one, because we have to share an output.
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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.
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Given some function f that is one-to-one (it has to be one-to-one for this to happen),
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there exists and inverse function, which we denote as f^-1 such that, for all x in the domain of f,
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any x that could normally go into f, any value that could normally be input,
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f^-1 acting on f(x) becomes just x.
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So, we have f acting on x like normal; and then, f^-1 acts on that whole thing.
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And it breaks the action that was done by f and returns us back to our original input.
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In other words, when f^-1 operates on the output of f, it gives the original input that went into f.
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Caution: I want to warn you about something: f^-1 means the *inverse* of f, not 1/f.
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This can be confusing, because, if you have taken algebra and remember your exponents
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(you might have forgotten them, but we will talk about them later in this course), -1 can mean a reciprocal for numbers.
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So, 7^-1 becomes 1/7; x^-1 becomes 1/x.
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But f is not "to the -1"; it is just a symbol that says inverse--"This is a function inverse."
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So, in general, for the most part, f^-1, "f inverse," is not the same thing as 1/f.
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The inverse of f is normally not the reciprocal of f, 1/f.
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This exponent thing, where 7^-1 is 1/7, is not the case for functions.
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On a function, the -1 does not represent an exponent; it is not an exponent.
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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."
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How can we get domain and range for f^-1?
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We can figure these out by looking at f; remember, the set of all outputs from f is its range.
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The things that x can get mapped to by f, what f is able to map x to, is the range of f.
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The domain of f is everything that x can be, everything that we can plug into f.
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And then, the range of f is everything that can come out of f.
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Now, f^-1 has to be able to take any output of f.
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It is not very good at reversing if there are some numbers that it is not allowed to reverse.
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So, it has to be able to reverse anything from f.
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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.
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So, the domain of f^-1, the things we can put into f^-1, is the range of f.
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The domain of f^-1 is the range of f.
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Likewise, because f^-1 then breaks that f(x) and turns it back into original inputs,
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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.
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So, anything that we can make it to, we have to be able to make it back from.
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So, since we are able to get back everywhere, that means that we can output all the possible inputs for f.
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Since we can output all of the possible inputs, because we can reverse any of the processes,
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then it must mean that we are able to get the range of f^-1 from the domain of f.
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So, the domain of f tells us the range of f^-1, what we are allowed to output with f^-1.
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And the domain of f^-1 tells us the range of f.
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So, the domain of f^-1 is the things that f is able to output; and the range of f^-1,
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the things that f^-1 is able to output, is what we can put into f, the domain of f.
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This idea is going to let us prove something later on.
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Now we can get to that proof--the inverse of an inverse: what is the inverse of an inverse?
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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?
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Now, it might seem a little surprising, but it turns out that the inverse of f^-1 is just f.
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The inverse of f^-1 is f; it seems a little surprising, maybe, but it makes a sort of intuitive sense.
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If you do the opposite of an opposite, you get to the original thing.
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If you do an action, but then you are going to do the opposite of that action,
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but then you do the opposite of the opposite of the action, then you must be back at your original action.
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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.
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But let's see a proof of this fact, formally; let's see it in formal mathematics.
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So, how do we get this started? Well, by definition, f^-1 is the function where,
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for any x, f^-1 acting on f(x) is going to just give us our original x.
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If f acts on an input, and then f^-1 comes and acts on that, we get back to our original input.
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Now, consider (f^-1)^-1: by this definition of the way inverses work, it must be that f inverse, inverse,
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when it acts on the thing that it is an inverse of...f inverse, inverse, is an inverse of f inverse...
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I know it is complicated to say...but this one right here is going to be the opposite action of f^-1.
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So, if we take any y (don't get too worried about x and y; remember, they are just placeholders for inputs),
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similarly, for any y, (f^-1)^-1, acting on f^-1(y),
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is going to just get us right back to our original y.
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It is the same structure as what is going on here with f^-1(f(x)) = x: we are just reversing a process.
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So, it doesn't matter that one of the processes is already a reversed process, because we are reversing this other reversed process.
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So, we get back to our original thing.
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Now, we know that y is in the domain of f^-1, because we are allowed to put it into f^-1.
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It is allowed to go into f^-1; now, we know, from our thing that we were just talking about,
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that the domain of f^-1 is the range of f; so there has to be...
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If f^-1 is the range of f (the domain of f^-1 is the range of f),
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if you are in the range, then that means that there is something out there that can produce this.
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That means that, if you are in the range of f, there must be some x in the domain of f;
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there has to be some way to get to that place in the range, so that f(x) is equal to y.
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There is some x out there in the domain of our original f, that f(x) is equal to y.
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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.
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We will plug it in for the two y's up here, and we will see what happens.
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Thus, f inverse, inverse, acting on f^-1(f(x))...
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because remember, we know that there has to be some way to get an x such that f(x) = y,
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because of this business about domain and range; so we plug that in here, and we plug that in here.
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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.
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We are just doing substitution.
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But we know, by the definition of f^-1(f(x)), that this just turns out to be x.
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This whole thing right here just comes down to x--it simplifies right out to x.
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So, it must be the case that f inverse, inverse, of x is the same thing as f(x).
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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.
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And our proof is finished; great.
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How can we interpret this graphically?--there is a great way to interpret inverses through graphs.
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First, let's consider f(x) = x³ + 1.
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Now, we know that this one has to be one-to-one, because it passes the horizontal line test.
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We come along and try to cut this with any horizontal line; they are only going to be able to cut in one place.
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Even here, where we have sort of seemed to flatten out, it is still moving, because we know it is x³ + 1.
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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!