f(x)= a/x + b , x is real, x does not equal 0, b or -a/b

ff(x) = f^-1(x)

prove that a + b^2 = 0

Would really appreciate a hint!

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- December 3rd 2012, 08:09 AMkinhew93Proof involving functions
f(x)= a/x + b , x is real, x does not equal 0, b or -a/b

ff(x) = f^-1(x)

prove that a + b^2 = 0

Would really appreciate a hint! - December 3rd 2012, 02:24 PMkinhew93Re: Proof involving functions
Does anyone know how to do this?

- December 3rd 2012, 03:06 PMPlatoRe: Proof involving functions
- December 3rd 2012, 03:30 PMtopsquarkRe: Proof involving functions
- December 4th 2012, 05:41 AMkinhew93Re: Proof involving functions
Sorry I didn't realise that commenting on this would send it back up the list! - It was just that I suspected that people were also unable to see a proof and was hoping that someone would confirm this for me.

The question is designed to require more 'creative thinking' however, so I assume that there is more to it than just setting ff(x) = f^-1(x)?

Can anyone see another relationship in here somewhere? fff(x) = x and ffff(x) = f(x) Will they help?

There are other questions of this vein in the book so I think that there is a proof. - December 4th 2012, 06:26 AMPlatoRe: Proof involving functions
- December 4th 2012, 06:38 AMkinhew93Re: Proof involving functions
Ok sorry I think that the british notation I use is slightly different:

I would write f(x) . f(x) . f(x) = (f(x))^3

and

fff(x) = fofof(x) = f(f(f(x))) = f(^3)(x)

so by ff(x) I actually mean f(f(x)) = f o f(x) - December 4th 2012, 07:01 AMHallsofIvyRe: Proof involving functions
Okay, so have you determined what [tex]f^{-1}(x)[/itex] is? . Set that equal to , "cross multiply" so that you get two polynomials equal to each other, for all x. What does the fact that they are "equal for all x" tell you about the coefficients?

- December 4th 2012, 07:14 AMkinhew93Re: Proof involving functions
Thanks a lot HallsofIvy!

Just one question though; Why is it okay to crossmultiply if the expressions are 'exactly equal' when a and b are constants?