Results 1 to 3 of 3

Thread: Inverse functions

  1. #1
    Newbie
    Joined
    Jul 2008
    Posts
    21

    Inverse functions

    Show that $\displaystyle f\left(g^{-1}(x)\right)=g\left(f^{-1}(x)\right) \Leftrightarrow f^{-1}\left(g(x)\right)=g^{-1}\left(f(x)\right)$

    If we substitute $\displaystyle x=f(u)$, we get

    $\displaystyle f\left(g^{-1}\left(f(u)\right)\right)=g^{-1}\left(f\left(f^{-1}(u)\right)\right)$
    $\displaystyle \Updownarrow$
    $\displaystyle f\left(g^{-1}\left(f(u)\right)\right)=g(u)$

    This means that $\displaystyle g^{-1}\left(f(u)\right)=f^{-1}\left(g(u)\right)$, because $\displaystyle f\left(f^{-1}\left(g(u)\right)\right)=g(u)$.

    Now, if the above equality holds for $\displaystyle u$, it must also hold for $\displaystyle x$, and we have that

    $\displaystyle g^{-1}\left(f(x)\right)=f^{-1}\left(g(x)\right)$

    and that

    $\displaystyle f\left(g^{-1}(x)\right)=g\left(f^{-1}(x)\right) \Leftrightarrow f^{-1}\left(g(x)\right)=g^{-1}\left(f(x)\right)$

    Which was to be shown.


    Does this hold, or have I overlooked something?

    Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by espen180 View Post
    Does this hold, or have I overlooked something?
    That is good. The only issue I have with this is that we do not know much about $\displaystyle f,g$ themselves.
    If these were bijections on $\displaystyle \mathbb{R}$ then what you did is okay.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2008
    Posts
    21
    Yes, I assumed that was the case with f and g. But what if they were bijections on $\displaystyle \mathbb{C}$? Would it not hold then?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. inverse functions
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Nov 22nd 2009, 04:58 PM
  2. Replies: 2
    Last Post: Oct 19th 2009, 02:47 AM
  3. inverse one-one functions
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Sep 12th 2009, 04:06 PM
  4. inverse functions
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 27th 2009, 07:54 AM
  5. Functions and Inverse
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 16th 2009, 01:16 AM

Search Tags


/mathhelpforum @mathhelpforum