Functional equations elementary

Question

Let $f:\mathbb R \to \mathbb R$ be a continuous function such that $f(0)=1$ and $f(x)-f(\frac{x}{2})=\frac{3x^2}{4}+x$. Find $f(3)$. How to proceed please help?

Answer 1

Let $f(x) = ax^2+bx+c$. Then we have $f(\frac{x}{2}) = \frac{ax^2}{4}+\frac{bx}{2}+c$. So $$f(x)-f\bigg(\frac{x}{2}\bigg) = \frac{3ax^2}{4}+\frac{bx}{2} = \frac{3x^2}{4}+x \implies a = 1,\ b = 2$$ Using the initial condition, we have $$1 = f(0) = c \implies c = 1$$ Therefore, we have $f(x) = x^2+2x+1$ and $f(3) = 16$.

Answer 2

$$f(3)=f\left(\frac{3}{2}\right)+\frac{3}{4}\cdot3^2+3=f\left(\frac{3}{2^2}\right)+\frac{3}{4}\cdot3^2+3+\frac{3}{4}\left(\frac{3}{2}\right)^2+\frac{3}{2}=...$$ $$=\lim_{n\rightarrow+\infty}f\left(\frac{3}{2^n}\right)+\frac{3}{4}\left(3^2+\frac{3^2}{2^2}+\frac{3^2}{2^4}...\right)+3\left(1+\frac{1}{2}+\frac{1}{2^2}+...\right)=$$ $$=1+\frac{27}{4}\cdot\frac{1}{1-\frac{1}{4}}+\frac{3}{1-\frac{1}{2}}=16$$