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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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!
Does anyone know how to do this?
Quote:
Originally Posted by kinhew93
Quote:
Originally Posted by kinhew93
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 isthen
.
BUT WHAT IS
I agree with Plato. I did the problem and didn't come up with a proof thatQuote:
Originally Posted by kinhew93
The defining equation that you have,
gives a relationship between x, a, and b, not just a and b. Something isn't right.
-Dan
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.
Quote:
Originally Posted by kinhew93
Here is the point: we do not know the notation.
is meaningless.
Is itor
.
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?
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?