+4 Let f be a function such that f(x+y) = x+f(y) for any 2 real numbers x and y. If f(0) = 2 then what is f(2012)
+622 First, we can use the given functional equation to show that f(0) = 1. We have f(0) + f(0) = f(0 + 0) = f(0) = 2. This equation has two solutions: f(0) = 0 or f(0) = 1. If f(0) = 0, then f(x) = 0 for all x, since f(x) = f(x + 0) = x + f(0) = x + 0 = x. However, we know that f(0) = 2, so f(0) must be 1.
Now, we can use the given functional equation again to show that f(2012) = 2013. We have f(2012) + f(0) = f(2012 + 0) = f(2012) = 2012 + f(0) = 2012 + 1 = 2013. This equation has only one solution: f(2012) = 2013.
Therefore, the answer is 2013.
