The restricted function
passes the Horizontal Line Test, so the inverse will now be a function.
Since the domain of the original function is –2
<x< 0 and
the range is –2
<y< 0,
then the domain of the inverse will be –2
<x< 0 and
the range will be –2
<y< 0. Yes,
the domains and the ranges are identical. Here's the algebra:

The
original function:

I
rename "f(x)"
as "y":

Then
I solve for "x
=":

Since
I already figured out the domain and range, I know that I
have to pick the NEGATIVE square root here:

Now
I switch x
and y:

And
rename "y"
as "f-inverse".

So the
inverse is the exact same function I started with!

If
you think about it, this makes perfect sense. The original function
was one quarter of the circle centered at the origin and having
radius r
= 2. Considering
where the reflecting line y
= x goes,
and the fact that the points on either side of the reflecting
line are neatly symmetric, then the inverse couldn't be anything
other than what we got.

If
you use the upper-right quarter of the circle you'll get the same result;
the inverse will be identical to the function. However, if you take either
of the other two quarters, you'll get the remaining quarter as the inverse: