# Proof involving functions

• Dec 3rd 2012, 09:09 AM
kinhew93
Proof 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!
• Dec 3rd 2012, 03:24 PM
kinhew93
Re: Proof involving functions
Does anyone know how to do this?
• Dec 3rd 2012, 04:06 PM
Plato
Re: Proof involving functions
Quote:

Originally Posted by kinhew93
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

Quote:

Originally Posted by kinhew93
Does anyone know how to do this?

It is against forum rules to bump. Don't do it.

I did not choose to help because I think the question is flawed.

If it is $f(x)=\frac{a}{x}+b$ then $f^{-1}(x)=\frac{a}{x-b}$.

BUT WHAT IS $f(f(x))~?$
• Dec 3rd 2012, 04:30 PM
topsquark
Re: Proof involving functions
Quote:

Originally Posted by kinhew93
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!

I agree with Plato. I did the problem and didn't come up with a proof that $a + b^2 = 0$ The defining equation that you have,
$f(f(x)) = f^{-1}(x)$

gives a relationship between x, a, and b, not just a and b. Something isn't right.

-Dan
• Dec 4th 2012, 06:41 AM
kinhew93
Re: 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.
• Dec 4th 2012, 07:26 AM
Plato
Re: Proof involving functions
Quote:

Originally Posted by kinhew93
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.

Here is the point: we do not know the notation.

$fff(x)$ is meaningless.

Is it $f\circ f\circ f(x)=f(f(f(x)))~?$ or $fff(x)=f^3(x)=f(x)\cdot f(x)\cdot f(x)~?$.
• Dec 4th 2012, 07:38 AM
kinhew93
Re: 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)
• Dec 4th 2012, 08:01 AM
HallsofIvy
Re: Proof involving functions
Okay, so have you determined what [tex]f^{-1}(x)[/itex] is? $f(f(x))= \frac{a}{\frac{a}{x}+ b}+ b= \frac{ax}{a+bx}$. Set that equal to $f^{-1}(x)$, "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?
• Dec 4th 2012, 08:14 AM
kinhew93
Re: 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?