Finding inverse functions: quadratic (example 2)

We have the function f of x is equal to x minus 1 squared minus 2. And they've constrained the domain to x being less than or equal to 1. So we have the left half of a parabola right here. They've constrained so that it's not a full U parabola. And I'll let you think about why that would make finding the inverse difficult. But let's try to find the inverse here. And a good place to start -- let's just set y being equal to f of x. You could say y is equal to f of x or we could just write that y is equal to x minus 1 squared minus 2. We know it's for x is less than or equal to 1. But right now we have y solved for in terms of x. Or we've solved for y, to find the inverse we're going to want to solve for x in terms of y. And we're going to constrain y similarly. We could look at the graph and we could say, well, in this graph right here, this is defined for y being greater than or equal to negative 2. So we can maybe put in parentheses y being greater than or equal to negative 2. Because this is going to be -- right now this is our range. But when we swap the x's and y's, this is going to be our domain. So let's just keep that in parentheses right there. So let's solve for x. So that's all you have to do, to find the inverse. Solve for x and make sure you keep track of the domains and the ranges. So, let's see. We could add 2 to both sides of this equation. We get y plus 2 is equal to x minus 1 squared. Minus 2, plus 2, so those cancel out. And then, I'm just going to switch to the y constraint. Because now it's not clear what we're -- whether x is the domain or the range. But we know by the end of this problem, y is going to be the domain. So let's just swap this here. So, 4 for y is greater than or equal to negative 2. And we could also say, in parentheses, x is less than 1. This is -- we haven't solved it explicitly for either one, so we'll keep both of them around right now. Now, to solve for x, you might be tempted to just take the square root of both sides here. And you wouldn't be completely wrong. But we have to be very, very, very careful here. And this might not be something that you've ever seen before. So this is an interesting point here. We want the right side to just be x minus 1. That's our goal here, in taking the square root of both sides. We want to just have an x minus 1 over there. Now, is x minus 1 a positive or a negative number? Well, we've constrained our x's to being less than 1. So we're dealing only in a situation where x is less than or equal to 1. So if x is less than or equal to 1, this is negative. This is negative. So we want to take the negative square root. Let me just be very clear here. If I take negative 3 -- I take negative 3 and I were to square it, that is equal to 9. Now, if we take the square root of 9 -- if we take the square root -- let's say we take the square root of both sides of this equation. And our goal is to get back to negative 3. If we take the positive square root. If we just take the principle root of both sides of that, we would get 3 is equal to 3. But that's not our goal. We want to get back to negative 3. So we want to take the negative square root of our square. So because this expression is negative and we want to get back to this expression, we want to get back to this x minus 1, we need to take the negative square root of both sides. You can always -- every perfect square has a positive or a negative root. The principle root is a positive root. But here we want to take the negative root because this expression right here is going to be negative. And that's what we want to solve for. So let's take the negative root of both sides. So you get the negative square root of y plus 2 is equal to -- and I'll just write this extra step here, just so you realize what we're doing. Is equal to the negative square root, the negative square root of x minus 1 squared. For y is greater than or equal to negative 2. And x is less than or equal to 1. That's why the whole reason we're going to take the negative square there. And then this expression right here -- so let me just write the left again. Negative square root of y plus 2 is equal to the negative square root of x minus 1 squared is just going to be x minus 1. It's just going to be x minus one. x minus 1 squared is some positive quantity. The negative root is the negative number that you have to square to get it. To get x minus 1 squared. So that just becomes x minus 1. Hopefully that doesn't confuse you too much. We just want to get rid of this squared sign. We want to make sure we get the negative version. We don't want the positive version, which would have been 1 minus x. Don't want to confuse you. So here, we now just have to solve for x. Add 1. And let me write the 4. y is greater than or equal to negative 2. Add 1 to both sides. You get negative square root of y plus 2 plus 1 is equal to x for y is greater than or equal to negative 2. Or, if we want to rewrite it, we could say that x is equal to the negative square root of y plus 2 plus 1 for y is greater than or equal to negative 2. Or if we want to write it in terms, as an inverse function of y, we could say -- so we could say that f inverse of y is equal to this, or f inverse of y is equal to the negative square root of y plus 2 plus 1, for y is greater than or equal to negative 2. And now, if we wanted this in terms of x. If we just want to rename y as x we just replace the y's with x's. So we could write f inverse of x -- I'm just renaming the y here. Is equal to the negative square root of x plus 2 plus 1 for, I'm just renaming the y, for x is greater than or equal to negative 2. And if we were to graph this, let's see. If we started at x is equal to negative 2, this is 0. So the point negative 2, 1 is going to be on our graph. So negative 2, 1 is going to be on our graph. Let's see, if we go to negative 1, negative 1 this will become a negative 1. Negative 1 is 0 on our graph. Negative 1, 0 is on our graph. And then, let's see. If we were to do, if we were to put x is equal to 2 here. So x is equal to 2 is 4. 4 square root, principle root is 2. It becomes a negative 2 So it becomes 2, negative 1. So that's on our graph right there. So the graph is going to look something like this, of f inverse. It's going to look something like that right there. As you can see, it is a reflection of our original f of x along the line y is equal to x. Along the line y is equal to x. Because we've essentially just swapped the x and the y. This is about as hard of an inverse problem that I expect you to see. Especially in a precalculus class because it really is tricky to realize that you have to take the negative square root here. Because the way our domain was constrained, this value right here is going to be negative. So to solve for it, you want to have the negative square root.