# What is f(x), f(y) in x*f(y)+y*f(x)=(x+y)*f(x)*f(y)?

Posted on

You should consider the number `x_0 != 0`  as a root of f(x) such that `f(x_0)=0` .

Hence, plugging `x = x_0`  in equation `x*f(y)+y*f(x)=(x+y)*f(x)*f(y)`  yields:

`x_0*f(y)+y*f(x_0)=(x_0+y)*f(x_0)*f(y) =gt x_0*f(y) = 0`

Since `x_0 != 0`  then`f(y) = 0`  that is a contradiction, hence there is no value for x such that `f(x)=0 lt=gt f(x) != 0.` Considering x=y yields:

`x*f(x)+x*f(x)=(x+x)*f(x)*f(x) =gt 2xf(x) = 2xf^2(x)`

Subtracting `2xf(x)`  both sides yields:

`2xf^2(x) - 2xf(x) = 0`

Factoring out `2xf(x)`  yields:

`2xf(x)(f(x) - 1) = 0`

Since`f(x) != 0 =gt f(x) - 1 = 0 =gt f(x) = 1`

Hence, the function f(x) may have two forms: if `x!=0 =gt f(x)=1;`  if `x = 0 =gt f(x) = a`  (a denotes a constant in R).

Posted on

Since f(x)=y and f(y)=x , then  x^2 + y^2 = y x^2 + x y^2

So  f(x) = y = x(-x^2-sqrt(x^2+4x-4))/(2(x-1)) , -4.9<x<1  ...

AND  f(x) = y = (x sqrt(x^2+4x-4)-x^2)/(2(x-1)) , -4.9<x<1

Finding f(y) is the same grind.