Return to the Purplemath home page

 


powered by FreeFind

 

Print-friendly page

 

 

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?
       

      

    You should recognize that f(x) is a rational function.   Here's the graph:

      

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

    The restriction on the domain comes from the fact that you can't divide by zero, so x can't be equal to –2. You usually wouldn't bother writing down the restriction, but it's helpful here, where you 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, then, 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)
      Rename "f(x)" as "y": y = (x - 2)/(x + 2)
      Solve for "x =":

       

      		 y(x + 2) = x - 2
      xy + 2y = x - 2
      Get the x-stuff on one side: xy - x = -2y - 2
      Here's the trick:   Factor out the x! x(y - 1) = -2y - 2
        x = (-2y - 2)/(y - 1)
      Switch x and y: y = (-2x - 2)/(x - 1)
      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  x is not equal to 1 and range y is 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:

       

    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 2006-2008 All Rights Reserved

      The original function: f(x) = x2 – 3x + 2
      Rename "f(x)" as "y": y = x2 – 3x + 2
      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 we want the negative square root: x = [3 – sqrt(1 + 4y)]/2
      Switch x and y: y = [3 - sqrt(1 + 4x)]/2
      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 f^(-1)(x) = [3 - sqrt(1 + 4x)]/2, x greater than or equal to -1/4.

<< Previous  Top  |  1 | 2 | 3 | 4 | 5 | 6 | 7  |  Return to Index  Next >>

Cite this article as:

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

 

Lessons index

Lessons CD

Purplemath:
  Linking to this site
  Printing pages
  Donating
  School licensing

Reviews of
Internet Sites:
   Free Help
   Practice
   Et Cetera

The "Homework
   Guidelines"

Study Skills Survey

Tutoring ($$)

This lesson may be printed out for your personal use.

Content copyright protected by Copyscape website plagiarism search
  

  Copyright © 2006-2008  Elizabeth Stapel   |   About   |   Terms of Use

 

 Feedback   |   Error?