Results 1 to 8 of 8

Thread: f(g(x))=g(f(x))

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    55

    f(g(x))=g(f(x))

    Is it possible and how to transform f(g(x))=g(f(x)) into f(x) = g^(-1)(x) . I saw that on f(g(x))=g(f(x)&#41 ; - Wolfram|Alpha
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Oct 2010
    Posts
    13
    Quote Originally Posted by Garas View Post
    Is it possible and how to transform f(g(x))=g(f(x)) into f(x) = g^(-1)(x) . I saw that on f(g(x))=g(f(x)&#41 ; - Wolfram|Alpha
    Actually what you posted is the definition of an inverse function. To find an inverse you swap the positions of x and y and solve for y.

    Ex) if f(x) = 6x+17
    then to find $\displaystyle f^{-1}(x)$ write y=6x+17

    swap x and y's placement x = 6y+17

    solve for y, $\displaystyle y = \frac{x-17}{6}$

    and so $\displaystyle f^{-1}(x) = \frac{x-17}{6}$

    You can check that the second function, that may be referred to as g(x), is an inverse to f(x) by seeing that g(f(x))=f(g(x)).

    Using the above example to see this (and calling [tex]f^{-1}(x) by the name g(x)):

    $\displaystyle f(g(x)) = 6(\frac{x-17}{6})+17$

    $\displaystyle f(g(x)) = x-17+17 = x$

    $\displaystyle g(f(x)) = \frac{(6x+17)-17}{6}$

    $\displaystyle g(f(x)) = \frac{6x}{6} = x$

    and so g(x) is indeed the inverse function of f(x)

    Now we can see that $\displaystyle f^{-1}(x) = g(x)$ just to show the reverse that would follow your question better:

    $\displaystyle g(x) = \frac{x-17}{6}$

    write $\displaystyle y = \frac{x-17}{6}$

    swap x and y placement $\displaystyle x = \frac{y-17}{6}$

    6x = y-17
    6x+17 = y

    $\displaystyle g^{-1}(x) = 6x+17 = f(x)$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,769
    Thanks
    3027
    From f(g(x))= g(f(x)), it does NOT follow that $\displaystyle f^{-1}(x)= g(x)$. For example, if f(x)= x+ 3 and g(x)= x+ 4, then f(g(x))= (x+4)+ 3= x+ 7= (x+3)+ 4= g(f(x)) but those surely are not inverse to one another.

    What is true is that if f(g(x))= g(f(x))= x then f and g are inverse functions to one another.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2010
    Posts
    55
    How to transform f(g(x))=g(f(x)) into f(x) = g^(-1)(x) without using examples?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,769
    Thanks
    3027
    I guess I'll have to say it a second time: you can't, it isn't true!

    As I said before, f(g(x))= g(f(x)) isn't enough. You need instead, f(g(x))= g(f(x))= x. From that, by taking $\displaystyle g^{-1}$ of both sides of g(f(x))= x, we get $\displaystyle g^{-1}(g(f(x))= f(x)= g^{-1}(x)$.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,577
    Thanks
    790
    Hmm, I think HallsofIvy showed that this is impossible.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jan 2010
    Posts
    55
    OK, but how to solve this then:
    Element x(belongs to A) is a fixed point of function f: A → A if f(x)=x. S is a set of all fixed points of function f. If g: A→A and f(g(x))=g(f(x)) prove that g(S) ⊆ S.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,577
    Thanks
    790
    This is proved in the standard direct way. You need to show that for all x in S, g(x) is in S. Fix some x in S. What does it mean that g(x) is in S? By definition of S, this means that f(g(x)) = g(x). Now finish the proof.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum