Find all continuous functions $f:\mathbb{R}\to\mathbb{R}$ satisfying $\frac{f(x+3)}{3+f(x)}=\frac{4+x^2}{x^2}$

Question

Find all continuous $f:\mathbb{R}\to\mathbb{R}$ satisfying $$\frac{f(x+3)}{3+f(x)}=\frac{4+x^2}{x^2}.$$

I believe the original question was $$\frac{f(x)}{3+f(x)}=\frac{4+x^2}{x^2},$$ which has a simple solution when you simplify.

Answer 1

There are many solutions. Let $h\colon[0,3]\to\mathbb R\setminus\{0\}$ be any continuous function such that $$\tag1h(3)=\frac13+\frac49h(0).$$ Now define $g\colon[0,3]\to\mathbb R$, $$ \tag2g(x)=x^2h(x)-3.$$ We can define $f\colon\mathbb R\to\mathbb R$ by letting $f(x)=g(x)$ for $x\in[0,3]$ and otherwise definig recursively $$\tag3f(x)=\frac{4+(x-3)^2}{(x-3)^2}(f(x-3)+3)\qquad\text{for $x>3$} $$ and $$\tag4f(x)=\frac{x^2}{4+x^2}f(x+3)-3\qquad\text{for $x<0$}.$$ Continuity for $f$ follows, it need only be checked at $x=0$ from the left and at $x=3$ from the right. And indeed, from $(4)$ we have $$\lim_{x\to 0^-}f(x)=-3=f(0)$$ and $$\begin{align}\lim_{x\to 3^+}f(x)&\stackrel{(2)}=\lim_{\delta\to 0^+}\left(\frac{4+\delta^2}{\delta^2}(g(\delta)+3)\right)\\&\stackrel{(2)}=\lim_{\delta\to 0^+}(4+\delta^2)h(\delta)\\&=4h(0)\stackrel{(1)}=9h(3)-3\stackrel{(2)}=g(3)=f(3).\end{align}$$