# Proof involving functions

• Dec 7th 2012, 11:37 AM
kinhew93
Proof involving functions
f(x) = (ax+b)/(cx+d) , x is real, x=/= -d/c , b=/=0 , c=/=0

Prove that if

a + d =/= 0

and

(a-d)^2 +4bc = 0

then y=f(x) and y=f^-1(x) intersect at exactly one point

So far I've tried setting f(x) = f^(x) and writing as a quadratic in x. I then tried to find the discriminant and show that it was (a-d)^2 +4bc but I just got into a big mess.

Is this method correct or am I missing something? Is there another approach?
• Dec 7th 2012, 12:09 PM
HallsofIvy
Re: Proof involving functions
Have you found f^(-1)(x)? What is it?

(I hope you understand that f^{-1} is NOT the reciprocal "1/f(x)" but the inverse function.)
• Dec 7th 2012, 12:36 PM
kinhew93
Re: Proof involving functions
f^-1(x) = (b-dx)/(cx-a)
• Dec 7th 2012, 04:10 PM
topsquark
Re: Proof involving functions
Quote:

Originally Posted by kinhew93
f^-1(x) = (b-dx)/(cx-a)

Good. Now how to find where they intersect:
y = f(x) = f^{-1}(x)

Given that they can only intersect at one point (that's why I set them equal to each other) what do a, b, c, and d have to be?

-Dan
• Dec 10th 2012, 07:48 AM
kinhew93
Re: Proof involving functions
So

(ax+b)/(cx+d) = (b-dx)/(cx-a)

(ax+b)(cx-a) = (b-dx)(cx+d)

then all I can think to do is write as a polynamial equal to zero, and find the discriminant. This didn't get me (a-d)^2 +4bc.

Quote:

Originally Posted by topsquark
Good. Now how to find where they intersect:
y = f(x) = f^{-1}(x)

Given that they can only intersect at one point (that's why I set them equal to each other) what do a, b, c, and d have to be?

-Dan

I don't know :(