1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook