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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.


    y = –sqrt(4 – x^2)

       Copyright Elizabeth Stapel 2000-2011 All Rights Reserved

    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:

      f(x) = –sqrt(4 – x^2)

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

      y = –sqrt(4 – x^2)

      Then I solve for "x =":

      y^2 = 4 – x^2

      y^2 – 4 = –x^2

      x^2 = 4 – y^2

      x =  sqrt(4 – y^2)

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

      x = –sqrt(4 – y^2)

      Now I switch x and y:

      y = –sqrt(4 – x^2)

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

      inverse function

    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 Accessed


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