How would you solve for a function when given only the relationships between its transformations?

Question

By that I mean, if you're given $f(x+3)=2f(x)$ is true for all x, is there some way to find out f(x), and if so, how would one go about doing that? And if it's not possible, why?

Answer 1

Even if you require continuity you can get an infinite number of solutions.

For each real number $c$,

$$ {f(x)=c(\sqrt[3]{2})^x} $$

is a solution since

\begin{eqnarray} {f(x+3)}&=&{c(\sqrt[3]{2})^{x+3}}\\ &=&{c(\sqrt[3]{2})^x(\sqrt[3]{2})^3}\\ &=&{2c(\sqrt[3]{2})^x}\\ &=&{2f(x)} \end{eqnarray}

Answer 2

$f(x)=0$ satisfies your definition.

So does the function $$f(x) = \begin{cases} 1, & 0\le x<3 \\ 2, & 3\le x<6 \\ 4, & 6\le x<9 \\ 8, & 9\le x<12 \\ 16, & 12\le x<15 \\ ... \end{cases}$$

In fact, you can see how defining $f(x)$ to be whatever you want on the interval $[0,3)$ leads to a unique $f$ that satisfies $\forall x\in\mathbb{R}:f(x+3)=2f(x)$. But your equation does not uniquely define a function. In general, recursive definitions require some initial conditions for uniqueness.