Find the continuous function that satisfies $f(x+1) + 3x^2 + 5x = f(2x+1), \forall x \in \mathbb{R}$

Question

Find the continuous function that satisfies $$f(x+1) + 3x^2 + 5x = f(2x+1), \forall x \in \mathbb{R}$$ The only hint I have is that this can be solved by making both sides have a pattern, and turn it into a "iconic" function, and then calculate its derivative, and prove that it is monotonic. I kinda understand the hint, but the "iconic" function I found, $g(x)=f(x+1)-x^2-5x$, I can't find anything to do with it. Or maybe I should find the limit first?

Answer 1

Let $f(x)=g(x)+x^{2}+3x$, putting this in the equation we get that $g(x+1)=g(2x+1)$. By putting repeatedly $x \to x/2$ gives $g(x/2+1)=g(x+1)$ ,$g(x/4+1)=g(x/2+1)$...By continuity we conclude $g(x+1)=g(1)=c$. So $f(x)=c+3x+x^{2}$.

Answer 2

You can find one solution using the Ansatz $f(t)=at^2+bt+c.$

Then you can try to show that if $f_1$ and $f_2$ are two solutions then $f_1-f_2$ is constant.

Conclude that all solutions are on the form $f(t)=at^2+bt+c$ where $a,b$ have fixed values while $c$ is arbitrary.