 This video will talk about how to use a function to find an inverse function with restrictions. So this one says it's not 1 to 1. If we look at this graph we can graph this one by hand because we know our transformations. It's just x squared which has a vertex of 0, 0, but it's going to go up 3. So now it has a graph that looks like this and this is 3. But we need to restrict the domain because if I do my horizontal line test it doesn't work. Not 1 to 1, but if I could use only half of this graph then I would have a 1 to 1 function. So I want to look at this part of the graph only. So to restrict that I have to say that my domain now is going to be x is an element of starting at 0 and going to infinity instead of all reals like the green graph was. So it's going to start at 0 and include 0 and go to infinity. That means that my range values then, I'm looking at this graph over here. Remember this is 3 so it starts at 3. Why is an element that starts at 3 and increases from there to infinity? Now it says find the inverse function. And I don't know which way you liked better. I can do them both again, I guess. The first step would be you start with x and the first thing we did to it was that we squared it. Cure it to. And the second thing we did was we added 3. So inverse, the first thing I'm going to do is I'm going to subtract 3 and I'm going to square root instead of squaring it would be the opposite function. So my f inverse of x is going to start with x, subtract 3 and then all of that is going to get my square root. Again the other way you would have said the x was equal to y squared plus 3 and solve for y. So x minus 3 is equal to y squared. And now we're ready to take the square root of both sides to get rid of the square because the square and the square root cancel. So we had the square root of x minus 3 equal to y. So f inverse of x is equal to the square root of x minus 3. Remember y and f of x, even if it's f inverse of x, they mean the same thing. So state the domain and range of this function. Okay, if I think about the domain here, x has to be greater than 3 or else I'm going to have negatives underneath there. So I'm going to say x is an element going from 3, including 3 because it could be 0 underneath the radical, to infinity. And if you notice the range up here with my function was 3 to infinity. Now my inverse function domain is 3 to infinity. And then my range, well if I start there, if I let this be 3, then I have the square root of 0 which makes that 0. So I start at 0 and go to infinity. And again, if you look at my domain, it is the range of my inverse function. And the range of my function is the domain of my inverse function, just like we thought. All right, now it says that we want to prove these inverse functions of each other. If they're really inverse functions of each other, the result will be x because everything else will cancel because they're opposite operations and opposite numbers. So we should just be left with our x. So let's try this composition. Remember f is the outside function. So 3 times my g function minus 4. And my g function is x plus 4 over 3. Well when I multiply here, I have, this is 3 over 1 times x plus 4 over 3. So those 3 are going to cancel out. And I'm left with x plus 4 and then my minus 4, which of course is equal to x because these two cancel. Now if we do, it has to work both ways. So we do the g function. And the g function says x, whatever we had for f of x is going to go in there, plus 4 divided by 3. And we put our f function in here, which is 3x minus 4. Well we've got to simplify. And I can simplify my numerator. So that gives me just 3x because the x and then minus 4, the 4 and the negative 4 cancel each other out. So 3x over 3, the 3s cancel each other out. And we're left with x. So we just proved that these are really inverse functions of each other. All right, so let's try again with a different, more interesting looking graph. State the domain of this function. Well, that says that 2x minus 5, if you don't know, you could do it this way. We know that radicals have to be greater than or equal to 0 underneath. So we take what's underneath and set it greater than or equal to 0. 2x is greater than or equal to 5. And so x is greater than or equal to 5 over 2, or 2.5. So find the inverse function. All they wanted to know here was the domain. If we wanted to know the range, if we put 5 halves in here, it's going to start at 0. So we would say that y is greater than or equal to 0. That would be our range. Find the inverse function. All right, let's do this algebraically. So x is equal to the square root of 2y minus 5. I have to clear the square root. So that means I'm going to square both sides. So x squared will be equal to the square in the square root, cancel each other out. So 2y minus 5, add 5. And then we're finally going to divide by 2. And we have x squared plus 5 all over 2 is equal to y. So f inverse of x is equal to that x squared plus 5 over 2. So state the new domain and range. Well, the new domain, if we let x work in here, it should be. We would be able to put any x in here, add 5 to it and divide by 2. So you would think that the domain would be all reels. And the range would be, no matter what I put in here for x squared, add 5 and divide by 2. It's always going to be a real number. But it can't be that. Because my function was restricted on both the x and the y, then I really come back over here and I say, okay, my domain is x is greater than 0. I remember this will be my domain. And my range will be y is greater than or equal to 5 halves if I do it with inequalities. And then it asks us to prove that they're inverse functions. And I didn't leave myself much room here, did I? But let's try. The f function is, so we're doing f and f inverse. We want to do f of f inverse. So that'll be the square root of 2 times my inverse function minus 5. And the inverse function happens to be x squared plus 5 over 2. Well, this is 2 over 1, so those two's cancel. So we have the square root of x squared plus 5 and then minus 5. But those two are going to cancel each other out because they're opposites. So we have the square root of x squared, which equals x. And if we do it the other way, f inverse of f of x, we have our f function that needs to go in here and be squared plus 5 over 2. And inside there, we have our square root of 2x minus 5. The square is going to cancel the square root, so we're going to have 2x minus 5 plus 5 over 2. These two cancel and these two cancel, so it's equal to x. So in summary, the domain, the domain of f of x is equal to the range of f inverse of x. And the range of f of x is equal to the domain of our inverse function because the function has x, y, and the inverse function has y, x. How do you find the inverse function? Well, table method says list the order of operations and then do opposite order with opposite operations. And algebraically, we're going to say that you switch x and y, solve for y, and then replace y with the inverse notation. And stating the domain and range of the inverse function, we really talked about that already. Figure out what the domain and range were for your function. And even if it doesn't seem like it, that's what it should be for the inverse function. You automatically know that the most restricted domain and range will be your domain and range for the inverse function.