be satisfactory f(f(x)) = x for all x.

Question

For which values a, b, c, d the f(x) =(ax+b)/(cx+d) satisfy f(f(x))=x for all x.

Answer 1

This is a tough one and will look kind of nasty. Honestly it is too much to type up like this, need a whiteboard. 

 

f(x) = (ax + b) / (cx+d)

 

f(f(x)) = (a*f(x) + b) / (c*f(x)+d)

 

so f(f(x)) = (a*[(ax + b)/(cx+d)] + b) / (c*[(ax + b)/(cx+d)]+d)

 

= (a(ax + b)/(cx+d) + b) / (c(ax + b)/(cx+d) + d)              

 

Now we need to make things have a common denominator so we can add them

 

= (a(ax + b)/(cx+d) + b(cx+d)/(cx+d)) / (c(ax+b)/(cx+d) + d(cx+d)/(cx+d))

 

= ( a(ax+b) + b(cx+d) )/(cx+d) )   /   ( c(ax+b) + d(cx+d) )/(cx+d) )

 

= ( a(ax+b) + b(cx+d) )/(cx+d) )   *  ( (cx+d) / c(ax+b) + d(cx+d) )

 

=  ( a(ax+b) + b(cx+d) ) / ( c(ax+b) + d(cx+d) )

 

= xa^2 + ab + bcx + bd /  cax + bc + dcx + d^2

 

now we just rearrange

 

= (x(a^2 + bc) + b(a+d)) / (x(ca + dc) + bc + d^2)

 

So f(f(x)) =  (x(a^2 + bc) + b(a+d)) / (x(ca + dc) + bc + d^2) and we need f(f(x) = x for all x. So set that = x.

 

(x(a^2 + bc) + b(a+d)) / (x(ca + dc) + bc + d^2) = x

 

Multiply the denominator over to the right side...

 

 

x(a^2 + bc) + b(a+d) = x * (x(ca + dc) + bc + d^2)

 

Now just distribute everything and simplify, bringing everything to one side. This is too much to type out... but go through this and you will end up with something of this form:

 

____x^2 + ______x + _______ = 0    , where each _____ is some expression involving a, b, c, and d.

 

Now, the only way a polynomial like that is equal to 0 for all x, is if it is just always 0. So each of those _____ must be 0. That is your answer.