Find $f(x)$ given that $xf(x)+f(-x)=x$

Question

$$xf(x)+f(-x)=x$$ $$f(x)= ?$$

I saw this question in a group and couldn't find a way out of it. I've tried various things one of which is down below but these don't seem to work out.

$$xf(x) + f(-x) = x$$ $$xf(-x)-f(x) = x$$ $$xf(x)+f(-x) = xf(-x) -f(x)$$ $$f(x)+f(-x)=x\left(f(-x)-f(x)\right)$$

I actually did proceed from here but it really was not useful at all. I even tried taking derivatives etc. but none of them worked.

I searched this functional equation question on Web but didn't find anything related to this. Any help would be greatly appreciated.

Answer 1

For every $x$ you have two equations: $$\begin{cases} x f(x) + f(-x) = x \\ f(x) - x f(-x) = -x \end{cases}$$ where the second equation is obtained by plugging $-x$ in the place of $x$.

This gives you a linear system with two equations and two unknowns $f(x), f(-x)$. The solution is $$f(x)= \frac{x^2-x}{x^2+1}$$ Now, you can easily check that such a solution is well defined for all $x \in \Bbb R$ and satisfies the functional equation.

Answer 2

$$ xf(x)+f(-x)=x~~~~(1)$$ Change $x\to -x$ to get $$-xf(-x)+f(x)=-x~~~~(2)$$ Let $f(x)=A, f(-x)=B$, then (1) and (2) become $$xA+B=x ~ \& ~-xB+A=-x$$ Solve these two as linear simultaneous equations in $A$ and $B$, to get $$B=\frac{x^2+x}{1+x^2}=f(-x)$$ and $$A=\frac{x^2-x}{1+x^2}=f(x)$$