Inverting Functions with Restricted Domains

On the previous page, we saw that this function, absent the domain restriction, is not invertible. The only difference with this function is that the domain has been restricted to include only the negative half of the x-axis; that is, the function has been restricted to just the left-hand half of the parabola. This restriction makes the graph look like this:

This function passes the Horizontal Line Test, so it will have an inverse that is also a function.

(fyi: Just about any time they give you a problem where they've taken the trouble to restrict the domain, they've also almost always given you something whose inverse will also be a function. However, you should take care with the algebra and draw a nice picture, because it may require some extra effort to show that the inverse is a function. And, if nothing else, the restricted domain on the function will require a restricted range on the inverse function.)

In this case, since the domain is x ≤ 0 and the range (from the graph) is y ≥ 1, then the inverse will have a domain of x ≥ 1 and a range of y ≤ 0.

I start by solving for "x =":

I'd figured out the domain and range of the inverse function before I went and found that function. By figuring out the domain and range of the inverse in advance, I know that I have to choose the negative sign for the square root on my inverse function to be the correct inverse of the original function:

Now I'll switch the x and y; the new "y =" is the inverse:

(The "x ≥ 1" restriction is confirmed by the fact that the expression x − 1 is inside a square root. The only way for the square root to be defined is for x − 1 ≥ 0.)

So my answer is:

For this function, the domain restriction comes from the fact that x is inside a square root. Usually I wouldn't bother writing down "x ≥ 2", because I know that x-values less than 2 would give me negative numbers inside the square root. But the restriction is useful in this case because, together with the graph, it will help me determine the domain and range on the inverse:

The domain (given in the exercise statement) is x ≥ 2; the range (found from the graph) is y ≤ 0. Then the domain of the inverse will be x ≤ 0; the range will be y ≥ 2.

Now I need to find my inverse:

This inverse function graphs as a parabola, where I'd started with half of a (sideways) parabola. Since I already figured out the domain and range — in particular, the domain of the inverse is x ≤ 0 — I know that I need to take the negative half of the parabola. So my answer is:

With the domain restriction, the graph looks like this:

From what I know about graphing quadratics, the vertex is at (x, y) = (1.5, −0.25), so this graph is the left-hand half of the parabola. And this half of the parabola passes the Horizontal Line Test, so the (restricted) function is invertible. But how to solve for the inverse? I start as usual:

Now I solve for "x=" by using the Quadratic Formula:

Since x ≤ 1.5, then I want the *negative* square root:

Then, after renaming, the inverse is given by: