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)) ; - Wolfram|Alpha

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- Oct 26th 2010, 02:00 PM #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)) ; - Wolfram|Alpha

- Oct 26th 2010, 03:07 PM #2

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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

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

- Oct 26th 2010, 04:38 PM #3

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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.

- Oct 27th 2010, 07:56 AM #4

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- Oct 27th 2010, 08:32 AM #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 $\displaystyle g^{-1}$ of both sides of g(f(x))= x, we get $\displaystyle g^{-1}(g(f(x))= f(x)= g^{-1}(x)$.

- Oct 27th 2010, 08:32 AM #6

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- Oct 27th 2010, 08:57 AM #7

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- Oct 27th 2010, 10:27 AM #8

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