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

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up vote 7 down vote accepted

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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Thanks, that's also what my theory says thanks for the solving the starting process :) – Chris Aug 9 '12 at 18:16
I did the math $$ x = ( w - 1 ) / 3 $$ and i have concluded to this $$ f(w) = w^2 - 2w + 1 + ( 3w - 3 ) / 3 $$ however there is an f(x) missing what should i do ?? – Chris Aug 9 '12 at 18:40
$f(w)=w^2-2w+1+\frac{3(w-1)}{3}=w^2-2w+1+w-1=w^2-w$.Here, $w$ is just a dummy variable , so you can substitute $x$ for $w$ which gives $f(x)=x^2-x$. Keep in mind this new $x$ is different from previous $x$ – Aang Aug 9 '12 at 18:44
+1 to both of you for the above comments (OP showing work and avatar following up) – The Chaz 2.0 Aug 9 '12 at 19:17

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

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).

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Why would this be downvoted? – Pedro Tamaroff Aug 9 '12 at 19:12
@Downvoter If something is not clear then please feel welcome to ask for an explanation. – Bill Dubuque Aug 9 '12 at 19:12
I like this, but it took me quite a while to grasp! (Of course, I'm rather low on the "totem pole"...) – The Chaz 2.0 Aug 9 '12 at 19:16
@TheChaz I've added some color to help follow the logic. – Bill Dubuque Aug 9 '12 at 19:18
@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. – Bill Dubuque Aug 9 '12 at 19:27

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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Thanks for your answer. – Chris Aug 9 '12 at 18:16

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