Finding the Inverse of a Function: Yet More Examples

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Finding the Inverse of a Function (page 6 of 7)

Sections: Definition / Inverting a graph, Is the inverse a function?, Finding inverses, Proving inverses

Find the inverse of f(x) = –sqrt[ 4 – x² ], –2 < x < 0

Without the domain restriction, the graph looks like this:

This clearly fails the Horizontal Line Test, so the inverse, without the domain restriction, would not be a function.

However, with the domain restriction, I get this:

y = -sqrt(4 - x^2), -2 <= x <= 0

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.

y = -sqrt(4 - x^2), -2 <= x <= 0

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:

Function... ...and: ...and inverse function.

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Cite this article as:

Stapel, Elizabeth. "Finding the Inverse of a Function." Purplemath. Available from
http://www.purplemath.com/modules/invrsfcn6.htm. Accessed

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