341We can prove that no other solutions exist. Let f(x) = 2x+g(x) be a solution to the given functional equation.
Then LHS = f(x+f(y)) = f(x+2y+g(y)) = 2(x+2y)+2g(y) + g(g(y))
RHS = f(x+y) + f(y) = 2(x+y)+g(x+y)+2y+g(y) = 2(x+2y)+g(x+y)+g(y)
This implies that g(x+y) = g(y) + g(g(y) is independent of x
In other words g(x) is a constant = c. Putting it back in the equation above, we have c = 0
Hence f(x) = 2x is the unique solution satisfying the given conditions
f(x) = 0 doesnt qualify as it is given f(x):R+→R+
