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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: $f(2x) - 4f(x) = 2x$

3 Answers 3

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Hint: $3x+1=y \Rightarrow x= \frac{y-1}{3}$. So replace $x$ by $\frac{y-1}{3}$ and magic happens.

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$\rm \dfrac{f(3x\!+\!1)}{3x\!+\!1} = 3x\:\Rightarrow \dfrac{f(z)}z = z\!-\!1\:\Rightarrow \dfrac{f(2x)\!-\!4f(x)}{2x} = \color{#0A0}{\dfrac{f(2x)}{2x}}- 2 \color{#C00}{\dfrac{f(x)}x} = \color{#0A0}{2x\!-\!1} - 2(\color{#C00}{x\!-\!1}) = 1$

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).

  • 1
    @TheChaz Yes, that's probably the most straightforward way to proceed. But I thought it might prove of interest to emphasize that there is additional structure that simplifies matters. Such structure comes to the fore when one studies $q$-difference operators and related operator calculus.2012-08-09
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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 $f(t)$

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    +1 to both of you for the above comments (OP showing work and avatar following up)2012-08-09