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:
"f(x)" as "y":
I solve for "x
already figured out the domain and range, I know that I have to pick the NEGATIVE square
Now I switch
x and y:
"y" as "f-inverse".
inverse is the exact same function I started with!
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.
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: