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Simplifying Inverse Functions

  1. Jul 6, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex] f(x) = 3x^2-6 [/itex]

    We are asked to solve for the inverse function of the above function.

    2. Relevant equations

    3. The attempt at a solution
    [itex] y=3x^2-6 [/itex]
    [itex] x=3y^2-6 [/itex]
    [itex] \frac{3y^2=x+6}{3} [/itex]
    [itex] y^2 = \frac{x+6}{3} [/itex]
    [itex] \sqrt{y^2}= \frac{\sqrt{x+6}}{3} [/itex]
    [itex] y= \frac{\sqrt{x+6}}{\sqrt{3}} [/itex]
    [itex] y= \frac{\sqrt{x+6}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} [/itex]
    [itex] y= \frac{\sqrt{3x+18}}{3} [/itex]
    [itex] f^{-1}(x) = \frac{\sqrt{3x+18}}{3} [/itex]

    however, my teacher marked me wrong. I don't know why. I might ask her tomorrow but it's our mastery test and she might not explain it to me. Anyone care to tell me what wrong I did? Is it the rationalization? Is it supposed to have a radical denominator? Thank!
  2. jcsd
  3. Jul 6, 2011 #2


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    What was the exact wording of the question? Since [itex]3x^2- 6[/itex] is not one-to-one, strictly speaking it does not have an inverse.

    If we restrict the original function to [itex]f(x)= 3x^2- 6[/itex] for [itex]x\ge 0[/itex] and not defined for x< 0, then
    [tex]f^{-1}(x)= \sqrt{\frac{x+ 6}{3}}= \frac{\sqrt{x+ 6}}{\sqrt{3}}= \frac{\sqrt{3x+ 18}}{3}[/tex]

    If we restrict the original function to [itex]f(x)= 3x^2- 6[/itex] for [itex]x\le 0[/tex]
    and not defined for x> 0, then
    [tex]f^{-1}(x)= -\sqrt{\frac{x+ 6}{3}}= -\frac{\sqrt{x+ 6}}{\sqrt{3}}= -\frac{\sqrt{3x+ 18}}{3}[/tex]

    Your teacher might accept
    [tex]f^{-1}(x)= \pm\frac{\sqrt{x+ 6}}{\sqrt{3}}= \pm\frac{\sqrt{3x+ 18}}{3}[/tex]
    Last edited by a moderator: Jul 6, 2011
  4. Jul 6, 2011 #3
    oooh. Her instructions was that we need to change a list of functions into inverse functions, nothing more. I think she was exercising us in changing functions to inverse functions but did not bother to tell us that functions need to be one-to-one to have an inverse. She didn't also provide any other information like restrictions [itex] x \leq 0[/itex] or [itex]x \geq 0[/itex]
    so I guess she'll have to accept my answer. Thanks for helping! :)
  5. Jul 7, 2011 #4
    just wanna say that she accepted it. :) Thanks!
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