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!
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.
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)
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?