# proving one G(X)=F^-1(X), ending up with lXl, for both compositions.

• Aug 1st 2011, 01:44 AM
StudentMCCS
proving one G(X)=F^-1(X), ending up with lXl, for both compositions.
I just got through a lession @ purple math about functions.

The last example, in the proving that two functions are inverses of each other gives a specific example where F(g(X)) would equal lxl, except that a domain restriction forces the answer to be greater or equal to zero, so the absolute value can be removed.

And G(f(x))=lxl

so, F(G(X))=x, and G(F(X))=lxl and are therefore not inverses of each other.

My question is, what if the domain restriction forcing F(G(X)) to be greater than or equal to 0, had not existed? You would end up solving for lxl for both compositions. They would still not be inverses of each other right?
• Aug 1st 2011, 02:08 AM
chisigma
Re: proving one G(X)=F^-1(X), ending up with lXl, for both compositions.
The condition that a function is the inverse of itself is equivalent to write...

$y^{'}= \frac{1}{y^{'}} \implies y^{'\ 2}=1$ (1)

The (1) is a [very symple...] differential equation the solutions of which, with the condition $y(0)=0$ are $y=x$ and $y=-x$. Both these solutions are defined in the whole domain of x and no more functions exist that are inverse of themselves...

Kind regards

$\chi$ $\sigma$
• Aug 1st 2011, 02:21 AM
HallsofIvy
Re: proving one G(X)=F^-1(X), ending up with lXl, for both compositions.
Yes, that is correct. In order that F and G be inverses, we must have F(G(x))= x and G(F(x))= x, not |x|.

For example, take $F(x)= x^2$ and $G(x)= \sqrt{x}$ if $x\ge 0$, $-\sqrt{-x}$ if x< 0. F and G are both defined for all x. Now, for all real numbers, x, $F(x)\ge 0$ so G(F(x))= \sqrt{x^2}= |x|. If $x\ge 0$, $F(G(x))= (\sqrt{x})^2= x= |x|$ and if x< 0, $F(G(x))= (-\sqrt{-x})^2= -x= |x|$. But G and F are not inverse functions since $F(G(-3))= (-\sqrt{3})^2= 3$, not -3.

(What chisigma says is true but I don't see how it is related to this question.)