1. ## 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&#40;g&#40;x&#41;&#41;&#61;g&#40;f&#40;x&#41;&#41 ; - Wolfram|Alpha

2. Originally Posted by Garas
Is it possible and how to transform f(g(x))=g(f(x)) into f(x) = g^(-1)(x) . I saw that on f&#40;g&#40;x&#41;&#41;&#61;g&#40;f&#40;x&#41;&#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

$\displaystyle g^{-1}(x) = 6x+17 = f(x)$

3. 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.

4. How to transform f(g(x))=g(f(x)) into f(x) = g^(-1)(x) without using examples?

5. 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 $\displaystyle g^{-1}(g(f(x))= f(x)= g^{-1}(x)$.

6. Hmm, I think HallsofIvy showed that this is impossible.

7. 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.

8. 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.