$3f(x) + f(2-x) = (x)^2 , f(x)=$?

Question

$3f(x) + f(2-x) = (x)^2 , f(x)=$?

I saw someone solved it like this: She first takes that $x = u$, then equation looks like this $3f(u) + f(2-u) = (u)^2$ Then she takes that $x = 2-u$, then equation looks like this $3f(2-u) + f(u) = (2-u)^2$ Then she solves these two equations like system, and finally she gets that $f(x) = [x^2 + 2x -2] / 4$

But I'm confused with this method. First she takes that $x = u$, then she takes that $x = (2-u)$ and then she merges these two in system, how's that possible that $x$ can be two things at same time?

Answer 1

It is probably assumed that the relation holds for all $x\in\mathbb{R}$.

For example the relation holds for $x=8$ which implies $$ 3 f(8) + f(-6) = 64 $$ Moreover, the relation also holds for $x=-6$ which implies $$ 3 f(-6) + f(8) = 36 $$ It's not too difficult now to solve for $f(8)$ and $f(-6)$ in this case by looking at the induced linear system.

In general you could just fill in $x=u$ as an example and you could fill in $x=2-u$ in the equation just as I did above in my short example.

I hope this helps!

Answer 2

After replecing $x$ in $3f\left( x \right) +f\left( 2-x \right) ={ x }^{ 2 }$ with the $2-x$ you will get a new equation $\\ 3f\left( 2-x \right) +f\left( x \right) ={ \left( 2-x \right) }^{ 2 }\\ $ then you can solve two system equations $$\\ \\ \begin{cases} 3f\left( x \right) +f\left( 2-x \right) ={ x }^{ 2 } \\ 3f\left( 2-x \right) +f\left( x \right) ={ \left( 2-x \right) }^{ 2 } \end{cases}\Rightarrow \begin{cases} -9f\left( x \right) -3f\left( 2-x \right) =-3{ x }^{ 2 } \\ 3f\left( 2-x \right) +f\left( x \right) ={ \left( 2-x \right) }^{ 2 } \end{cases}\Rightarrow f\left( x \right) =\frac { { x }^{ 2 }+2x-2 }{ 4 } $$