The function f : R -> R satisfies
f(x)f(y) = f(x + y) + f(xy)
for all real numbers x and y. Find all possible functions f.
+9519 When x = y = 0, \((f(0))^2 = 2f(0)\).
This implies f(0) = 0 or f(0) = 2.
Case 1: f(0) = 2.
When x = 0, 2 f(y) = f(y) + 2
f(y) = 2 for all real numbers y.
The function f(x) = 2 is a solution.
Case 2: f(0) = 0.
When x = 0, 0 = f(y) for all real numbers y.
The function f(x) = 0 is a solution.
Therefore, the possible functions are the constant functions f(x) = 0 and f(x) = 2.
