# Problem involving functions and inverses

• Dec 23rd 2012, 02:48 PM
kinhew93
Problem involving functions and inverses
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.

This is what I did:

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

(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.

Is this method correct or am I missing something? Is there another approach?
• Dec 23rd 2012, 02:51 PM
chiro
Re: Problem involving functions and inverses
Hey kinhew93.

Can you solve for the inverse function? [Hint: exchange the x's and y's and solve for y = f^(-1)[x]).
• Dec 23rd 2012, 02:58 PM
kinhew93
Re: Problem involving functions and inverses
Quote:

Originally Posted by chiro
Hey kinhew93.

Can you solve for the inverse function? [Hint: exchange the x's and y's and solve for y = f^(-1)[x]).

Yeah if I understand you correctly I've already done that, but from there I am totally stuck...
• Dec 23rd 2012, 04:03 PM
skeeter
Re: Problem involving functions and inverses
$(ax+b)(cx-a) = (b-dx)(cx+d)$

$acx^2 + (bc-a^2)x - ab = -cdx^2 + (bc-d^2)x + bd$

$(ac+cd)x^2 + (bc - a^2 - bc + d^2)x - (ab + bd) = 0$

$(ac+cd)x^2 + (d^2 - a^2)x - (ab + bd) = 0$

$c(a+d)x^2 + (d-a)(d+a)x - b(a+d) = 0$

since $a+d \ne 0$

$cx^2 + (d-a)x - b = 0$

finish it ...