# Question concerning functions

• steve B. 98
In summary, the conversation discusses finding all functions f : R → R that satisfy the equation f(f(x+y)-f(x-y))=xy for all real x,y. The participants suggest considering the domain and range of the function, the use of derivatives, and potential functions such as f(x) = x^√2 / A.

#### steve B. 98

Member warned about posting with no template
I'm trying to solve this problem from a high school math competition:
Find all functions f : R → R such that, f(f(x+y)-f(x-y))=xy, for all real x,y.
Any ideas of how to approach it.
I have found that f(0)=0, if x=y f(f(2x))=x^2

Some thoughts:
I would note that the domain and range of the function must be all real numbers. This should eliminate many of the trig functions and exponentials.
Based on what you have, it seems like f(x) might incorporate some improper exponent...
Say...##f(x) = \frac{x^\sqrt {2} }{A}## where A scales 2 to 1 over two iterations.
I am not 100% sure how you would expand this out for the sums, and I don't think that function is defined for all x,y in the real numbers.

Another option,
Think of the derivatives:
##\frac{\partial}{\partial x} f ( f( x+ y) - f(x-y) ) = \frac{\partial}{\partial x} xy ##
##f ' ( f( x+ y) - f(x-y) ) * (f'(x+y)-f'(x-y)) = y ##
##\frac{\partial}{\partial y} f ( f( x+ y) - f(x-y) ) = \frac{\partial}{\partial y} xy ##
##f ' ( f( x+ y) - f(x-y) ) * (f'(x+y)+f'(x-y)) = x ##
And second derivatives are all zero.