150 Followers Question

It appears a linear function $f(x) = Ax + B$ will do the trick here. If this functional equation holds true for all $x, y \in \mathbb{R}$ , then at $y = 0$ we obtain $f(0) = 0$ as another boundary condition. Coupling this condition along with $f(5) = 3$ yields $\Rightarrow f(x) = \frac{3}{5}x$ and $f(2015) = \boxed{1209}.$