Function Busters

We have $f(x)f(y) = f(2xy+3) + 3f(x+y) - 3f(x) + 6x$

Switching $x$ and $y$ , we also have $f(x)f(y) = f(2xy+3) + 3f(x+y) - 3f(y) + 6y$

It follows that $6x-3f(x)$ is a constant; say (for convenience) $6x-3f(x)=3c$ . So $f(x)=2x-c$ .

Substituting into the original functional equation, we find $c=-3$ , or $f(x)=2x+3$ , so in particular $f(2009)=\boxed{4021}$ .

Putting $y=0$ we see that $\begin{aligned} f(x)f(0) & = \; f(3) + 3f(x) - 3f(x) + 6x \\ f(0)f(x) & = \; f(3) + 6x \end{aligned}$ for all $x$ . Thus $f(3) = f(0)^2$ . If $f(0) = 0$ we would deduce that $6x = 0$ for all $x$ . Thus $f(0) \neq 0$ , and so $f(x)$ is a nonconstant linear function.

Putting $x=0$ we obtain $\begin{aligned} f(0)f(y) & = \; f(3) + 3f(y) - 3f(0) \\ [f(0)-3]f(y) & = \; f(0)[f(0)-3] \end{aligned}$ Since $f$ is not constant, we deduce that $f(0) = 3$ , so that $f(x) = 3 + 2x$ for all $x$ , and hence $f(2009) = \boxed{4021}$ .

Given $f(x)f(y)= f(2xy+3)+3f(x+y) - 3f(x) + 6x$ , then

$\begin{cases} \begin{aligned} f(0)^2 & = f(3) + 3f(0) - 3f(0) + 0 = f(3) & ...(1) \\ f(x)f(0) & = f(3) + 3f(x) - 3f(x) + 6x = f(3) + 6x & ...(2) \\ f(0)f(x) & = f(3) + 3f(x) - 3f(0) & ...(3) \end{aligned} \end{cases}$

$\begin{aligned} (2)=(3): \quad f(x) & = 2x + f(0) \\ \implies f(3) & = 6 + f(0) \\ f^2(0) & = 6 + f(0) \\ f^2(0) - f(0) - 6 & = 0 \\ (f(0) - 3)(f(0) + 2) & = 0 \end{aligned}$

$\implies \begin{cases} f(0) = 3 & \implies f(x) = 2x + 3 & \implies f(x)f(y) \blue = f(2xy+3)+3f(x+y) - 3f(x) + 6x & \small \blue{\text{acceptable}} \\ f(0) = -2 & \implies f(x) = 2x -2 & \implies f(x)f(y) \red \ne f(2xy+3)+3f(x+y) - 3f(x) + 6x & \small \red{\text{unacceptable}} \end{cases}$

Therefore $f(x) = 2x + 3 \implies f(2009) = \boxed{4021}$ .