Solving A Fun Functional Equation

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[frendlyleaf]

In fact for any
f((ax+b)/(bx+a)) = g(x) with {a≠b}
f(x) = g((ax-b)/(a-bx))
But what's cool is that the intersection of the x, y asymptotes of f(x) are at the point
the point ((a/b), g(a/b)) if g(x) is an even function
and at ((a/b), - g(a/b)) if g(x) is an odd function

[musicforall]

If you want f(1) u just need to replace x=1 in the first equation
U get f(a+b/a+b) = 1^2 =1

You dont have to go through all this calculation

[ramanraghuwanshi]

An even easier way is
substitute ax+b/bx+a = t
Find x in terms of t
Put this expression of x in above equation (LHS and RHS)
You'll get f(t)=(at-b/a-bt)^2
Just replace t=x.

[Blaqjaqshellaq]

Notice that the curve f(x) approaches the asymptotes x=a/b and y=a^2/b^2 (assuming b isn't zero).

[mega_mango]

(ax + b)/(bx+a) = n
nbx + na = ax + b
x(nb - a) = b - na
x = (b-na)/(nb-a)
f(n) = (b-na)²/(nb-a)²

[doctorb]

The first thing I did was plug in 1 to get f(1) = 1. Did not see a way to easily get other values.

[j.pachelbel]

I thought we were supposed to find a and b. But yeah it's not possible