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Finding the Inverse of a Function (page 5 of 7)

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


  • Find the inverse f(x) = (x 2) / (x + 2)where x does not equal 2.
    Is the inverse a function?
       

      

    First, I recognize that f(x) is a rational function. Here's its graph:

      

    y = (x - 2) / (x + 2)

    The restriction on the domain comes from the fact that I can't divide by zero, so x can't be equal to 2. I usually wouldn't bother writing down the restriction, but it's helpful here because I need to know the domain and range of the inverse. Note from the picture (and recalling the concept of horizontal asymptotes) that y will never equal 1. Then the domain is "x is not equal to 2" and the range is " y is not equal to 1". For the inverse, they'll be swapped: the domain will be "x is not equal to 1" and the range will be "y is not equal to 2". Here's the algebra:

      The original function:

      f(x) = (x - 2)/(x + 2)

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

      y = (x - 2)/(x + 2)

      Then I solve for "x =":

       

      y(x + 2) = x - 2

      xy + 2y = x - 2

      I get the x-stuff on one side:

      xy - x = -2y - 2

      Here's the trick: I factor out the x!

      x(y - 1) = -2y - 2

       

      x = (-2y - 2)/(y - 1)

      Then I switch x and y:

      y = (-2x - 2)/(x - 1)

      And rename "y" as "f-inverse"; the domain restriction comes from the fact that this is a rational function.

      inverse function

      

    Since the inverse is just a rational function, then the inverse is indeed a function.

      

    Here's the graph:

      

      

    inverse function

    Then the inverse is  y = (2x 2) / (x 1), and the inverse is also a function, with domain of all x not equal to 1 and range of all y not equal to 2.

  • Find the inverse of  f(x) = x2 3x + 2,  x < 1.5
       

      

    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.

       

    graph of f(x) = x^2 ? 3x + 2, x <= 1.5

    This half of the parabola passes the Horizontal Line Test, so the (restricted) function is invertible. But how to solve for the inverse?  Copyright Elizabeth Stapel 2000-2011 All Rights Reserved

      The original function:

      f(x) = x2 3x + 2

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

      y = x2 3x + 2

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

        

      0 = x2 3x + 2 y
      0 = x2 3x + (2 y)

      x = [3  sqrt(1 + 4y)]/2

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

      x = [3 – sqrt(1 + 4y)]/2

      Now I switch x and y:

      y = [3 - sqrt(1 + 4x)]/2

      And rename "y" as "f-inverse"; the domain restriction comes from the fact that this is a rational function.

      f^(–1)(x) = [3 - sqrt(1 + 4x)]/2

    Then the inverse is given by:

        f^(-1)(x) = [3 - sqrt(1 + 4x)]/2, x greater than or equal to -1/4

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Cite this article as:

Stapel, Elizabeth. "Finding the Inverse of a Function." Purplemath. Available from
    http://www.purplemath.com/modules/invrsfcn5.htm. Accessed
 

 



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