Let $f:\mathbb R \to \mathbb R$ be a function, such as : $3f(x+1)-2f(2-x) = x^2 + 14x - 5$. Find $f(x)$. Let $f:\mathbb R \to \mathbb R$ be a function, such as :
$$3f(x+1)-2f(2-x) = x^2 + 14x - 5$$
Find f(x)
Attempt :
Let $x+1 := y \Leftrightarrow x = y-1$. Then, we get : 
$$3f(y) - 2f(3-y) = (y-1)^2 + 14(y-1) - 5$$
so I got $f(y)$ in the expression but still have to simplify it further.
I also observed that the two arguments of the functions $f$ become equal at :
$$y = 3-y \Leftrightarrow y = \frac{1}{2}$$
It should be an easy one since I found it across an Elementary Leaflet but my head has been stuck for now.
Any help would be appreciated !
 A: I'd just assume an ansatz: $f(x)=ax^2+bx+c$.
This gives
$$ax^2 + (14a + 5b)x+(-5a-b+c) = x^2 +14x - 5.$$
Clearly, $a=1$ from the quadratic term, and then $b=0$ from the linear term. Then, $c=0$ from the constant term.
So it looks like $f(x)=x^2.$
A: Let $z=1-x$ & we have
\begin{eqnarray*}
3f(2-z)-2f(1+z)=z^2-16z+10.
\end{eqnarray*}
Now let $z=x$ in the equation above & multiply this by $3$ and add $2$ times the first equation and we have 
\begin{eqnarray*}
5f(2-x)=5x^2-20x+20.
\end{eqnarray*}
So $\color{blue}{f(x)=x^2}$.
A: Note that $x+1=2-x$ if $x = 1/2$.  Thus for $x = 1/2$ the equation says $f(3/2) = 9/4$.  You can assign arbitrary values to $f(x)$ for $x > 3/2$, while for $x < 3/2$, 
$$f \left( x \right) =\frac13\,{x}^{2}+4\,x-6+\frac23\,f \left( 3-x \right) $$
A: Hint The substitution $x = 1-y$ gives
$$3 f(2-y) - 2 f(1+y) = (1-y)^2+ 14(1-y) - 5$$
A: We can make the two $f(a)$s equal by setting $x=1/2$
$$3f(x+1) -2f(2-x) = x^2 + 14x - 5$$
$$3f(3/2) - 2f(3/2) = (1/2)^2 + 14 \cdot 1/2-5$$
$$f(3/2) = 9/4$$
From which we can guess that $f(x) = x^2$
