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# In mathematics, a function (or map) f from a set X to a set Y is a rule which assigns to each element x of X a unique element y of Y, the value of f at x, such that the following conditions are met: 1) For every x in X there is exactly one y in Y, the value of f at x; 2) If x and y are in X, then f(x) = y; 3) If x and y are in X, then f(x) = f(y) implies x = y; 4) For every x in X, there exists a y in Y such that f(x) = y.

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

Okay, so we want to determine whether this function is once one. And if it is, find the inverse. If it's not one toe one, then we'll find a domain restriction for which the function is 1 to 1, and then we'll find the embers of that function. So with this first function, Okay, so the first thing to notice is, Well, the domain is all real numbers. Okay, so is it 1 to 1 on its natural domain? Well, to do that, we want to set f of X equal to f of Z, not assuming that X and Z are the same and actually show that X has two equals E so f of X is three x plus five and then f of Z is three z plus five. And if you look, we can subtract five from both sides, so three X has to equal three Z. And if we divide both sides by three, we have X is equal to Z, which is what we want to show. Therefore, F is 1 to 1. Okay, so it has an inverse. And how do we find the inverse? Well, this is something that you may have seen before. What we're going to dio to find the inverse is we're going Thio say Okay, the function. Why is my deep in it variable? It's my output and all we want to do is switched the input and the output. So we have X is now going to be my input of my inverse and my output is going to be white. Okay. And then just saw for why, in terms of X for a new function. So I subtract five. So we'll have three y minus. Okay, three y equals X minus five. And if I divided by three, I have, why is equal to X minus five all divided by three. Okay, so this is the inverse function, so f in verse X is just X minus five, divided by three. And notice that if I compose these two functions, if I take f off F members, I get X. And if I take f members of f, I get exactly X. So right off the bat here we see the dysfunction is not 1 to 1. And that's because f of to is equal to f of four, which is equal to negative one. Okay, so because two is not equal to four, but two and four give me the same output. This function is not 1 to 1. So what we need to do is restrict to a domain for which f is one. The one. And how do we do that? We'll notice that X squared if we require X to be greater than or equal to zero is one the one. So what we're gonna dio is we're going to require here X minus three squared X minus three be greater than or equal to zero or, in other words, x greater than or equal to three. So we're going to. So in that case, X minus three square is 11 subjecting to isn't going to change anything. So the domain restriction that we're going to use is three to infinity. We could also use a negative infinity to three, and actually, um, that would still be correct. That would be like the negative when we take the square root would be like taking a negative square root. This is gonna be like taking the positive square root. But this is the domain restriction that we're going to use. And so in this domain restriction, we can invert the function. And so remember that we do that by saying Okay, well, my output why is given by X minus three squared minus two. And I'm just going to switch the input and the output and solve for the new output. So I'm going to add to so I'll have why minus three squared is equal to X plus two. Then I'm going to take a square root. So that's going to be why minus three is equal to square root of X plus two. And now here's what I'm saying. I'm taking the positive square root by choosing this domain restriction. So here, when I take the square root, I'm really gonna be taking the positive square root and then I can add three. So why is going to be X plus two plus three? So, in other words, my inverse function here is going to be square root of X plus two plus three. You can verify that if I compose f with F Emerse, I'm going to get X. And if I compose f members with F, I'm also going to get X Okay, so the first thing you notice is that this function is 1 to 1. And why is that the case? Well, if I set ffx equal to F of Z, if F is 121 it better be the case that X is equal to Z. So this is one over X plus two. This is one over Z plus two. And if I cross multiply here, I'll get Z Plus two equals X plus two, subtracting two from both sides. I see that X is equal to Z, so F is among the one. So now it's just a matter of finding the inverse. And we do that by switching the input and output. So for an output, why? For our function with input X one over X plus two. Let's just see what happens if I If I want my input to be my output, then I'll have y plus two. And so notice that negative two is not in the domain of dysfunction. Okay, so let's see will have we cross multiply again will have x times y plus two is equal to one. Then we will see we can divide by X here and now. So I said the negative to was not in the domain of this function, also noticed that zero is not in the range of this function. And so the only way I'm going to be able to get zero is if the numerator is zero. But the numerator is never equal to zero. So that's why it's actually okay that I'm dividing by X here. Because remember, X is going to be the input, my inverse function, meaning it's an output from my function. So zero is not in the range of my function, so it's not in the domain of my inverse function, and then I can just subtract two. So then my inverse function is just one over X minus two. That's the inverse function, and again you can verify. F of F members will give me X in that members of death will also give me X

Georgia Southern University

#### Topics

Limits

Derivatives

Differentiation

Applications of the Derivative

Integrals

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University