Question 1165361
Let's make {{{system(a=f(x), b=f(2-x))}}}
 
When the variable is x, the equation is
{{{x*f(2-x)-2*f(x)=x^2+1}}} , which we write in abbreviated form as
{{{x*b-2a=x^2+1}}} (1)
 
When the variable is 2-x, the equation is
{{{(2-x)*f(2-(2-x))-2*f(2-x)=(2-x)^2+1}}} , which we want to simplify as
{{{(2-x)*f(x)-2*f(2-x)=4-4x+x^2+1}}} , and
{{{(2-x)*f(x)-2*f(2-x)=x^2-4x+5}}} , and abbreviate as
{{{(2-x)*a-2b=x^2-4x+5}}} (2)
 
In (1) and (2), we have a system of equations on {{{a}}} and {{{b}}} :
{{{system(x*a-2b=x^2-4x+5,-2a+x*b=x^2+1)}}} . Let's solve for {{{a}}} .
 
Multiplying (2) times {{{x/2}}} we get {{{((2x-x^2)/2)a-(x/2)2b=(x/2)(x^2-4x+5)}}} ,
which we simplify to
{{{((-x^2+2x)/2)a-x*b=(x^3-4x^2+5x)/2}}} ,
and add it to (1) to get
{{{((-x^2+2x)/2)a-2a=(x^3-4x^2+5x)/2+x^2+1}}} ,
{{{((-x^2+2x-4)/2)a=(x^3-4x^2+5x+2x^2+2)/2}}}
{{{-x^2+2x-4=-(x^2-2x+4)}}} is not zero for any real number,
so we can divide by that expression to solve for {{{a}}}
{{{(-x^2+2x-4)a=(x^3-2x^2+5x+2)}}}
{{{a=-(x^3-2x^2+5x+2)/(x^2-2x+4)}}}
 
The function is
{{{highlight(f(x)=-(x^3-2x^2+5x+2)/(x^2-2x+4))}}} .