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 What if x = y = 0?
\((f(0))^2 = f(0) + f(0)\\ \)
Let x = f(0), then x^2 - 2x = 0. This implies f(0) = 0 or f(0) = 2.
Case 1: f(0) = 2
Substituting x = 0 gives:
\(f(0) f(y) = f(y) + f(0)\\ f(y) = 2\)
Note that this holds for any real number y. That means the constant function \(f(x) = 2\) is a solution to this functional equation.
Case 2: f(0) = 0
Substituting x = 0 gives:
\(f(0) f(y) = f(y) + f(0)\\ f(y) = 0\)
Note that this holds for any real number y. That means the constant function \(f(x) = 0\) is a solution to this functional equation.
The only solutions to this functional equation are \(f(x) = 0\) or \(f(x) = 2\).
