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!

Inverse function

  1. Feb 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Prove/Disprove following function being one-to-one.If yes,find its inverse.

    g(x)=x-[itex]\frac{1}{x}[/itex],x>0

    2. Relevant equations



    3. The attempt at a solution
    My tutor said that it is one-to-one,but I found that the are two solutions for g-1(x).
    Are there any mistakes?
    g(x)=x-[itex]\frac{1}{x}[/itex],x>0
    x=g-1(x)-[itex]\frac{1}{g^{-1}(x)}[/itex]
    [g-1(x)]2-xg-1(x)-1=0
    g-1(x)=[itex](x\pm\sqrt{x^2-4(1)(-1)})/2[/itex]
    THANKS
     
    Last edited by a moderator: Feb 15, 2014
  2. jcsd
  3. Feb 15, 2014 #2

    lurflurf

    User Avatar
    Homework Helper

    x>0
    so
    $$\frac{x\pm\sqrt{x^2+4}}{2}$$
    reduces to
    $$\frac{x+\sqrt{x^2+4}}{2}$$
    keep in mind f is increasing that is
    $$\mathrm{f}(x+h)-\mathrm{f}(x)=h \left( 1+\frac{1}{x(x+h)}\right)>0$$
    so
    f(x+h)=f(x)
    implies x=x+h
     
  4. Feb 15, 2014 #3

    Mark44

    Staff: Mentor

    I fixed the broken LaTeX near the bottom of post #1. It didn't render correctly because tags were mixed in some itex script.
     
  5. Feb 15, 2014 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    How are the domain of a function and the range of the inverse of that function related ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Inverse function
  1. Inverse of a function (Replies: 2)

  2. Inverse of a function (Replies: 6)

  3. Inverse of a Function (Replies: 2)

  4. Inverse Functions (Replies: 2)

Loading...