The problem that I have to solve is:
If the following function is valid for every value of $x$
$f(3x + 1) = 9x^2 + 3x$
find the function $f(x)$ and prove that for every $x\in\mathbb R$ the following is valid: $f(2x) - 4f(x) = 2x$
The problem that I have to solve is:
If the following function is valid for every value of $x$
$f(3x + 1) = 9x^2 + 3x$
find the function $f(x)$ and prove that for every $x\in\mathbb R$ the following is valid: $f(2x) - 4f(x) = 2x$
Hint: $3x+1=y \Rightarrow x= \frac{y-1}{3}$. So replace $x$ by $\frac{y-1}{3}$ and magic happens.
$\rm \dfrac{f(3x\!+\!1)}{3x\!+\!1} = 3x\:\Rightarrow \dfrac{f(z)}z = z\!-\!1\:\Rightarrow \dfrac{f(2x)\!-\!4f(x)}{2x} = \color{#0A0}{\dfrac{f(2x)}{2x}}- 2 \color{#C00}{\dfrac{f(x)}x} = \color{#0A0}{2x\!-\!1} - 2(\color{#C00}{x\!-\!1}) = 1$
Remark $\ $ The point of presenting it like this is to emphasize how exploiting the innate linear structure serves to simplify the calculations (from nonlinear to linear). In less trivial problems this can yield much greater simplifications (e.g. in operator calculus with $q$-difference operators).
Here $f(3x+1)=3x(3x+1)=((3x+1)-1)(3x+1)$ $\implies f(x)=(x-1)x=x^2-x$ $\implies f(2x)-4f(x)=4x^2-2x-4x^2+4x=2x$
In general, just let $3x+1=t$ and express $x$ in terms of $t$ and substitute in $f(t)$