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Math Help - f(g(x))=g(f(x))

  1. #1
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    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
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  2. #2
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    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 f^{-1}(x) write y=6x+17

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

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

    and so 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)):

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

    f(g(x)) = x-17+17 = x

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

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

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

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

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

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

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

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

    g^{-1}(x) = 6x+17 = f(x)
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  3. #3
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    From f(g(x))= g(f(x)), it does NOT follow that 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.
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  4. #4
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    How to transform f(g(x))=g(f(x)) into f(x) = g^(-1)(x) without using examples?
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  5. #5
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    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 g^{-1} of both sides of g(f(x))= x, we get g^{-1}(g(f(x))= f(x)= g^{-1}(x).
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  6. #6
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    Hmm, I think HallsofIvy showed that this is impossible.
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  7. #7
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    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.
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  8. #8
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    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.
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