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I am trying to find the functional equation of $f(x)$, where $f$ satisfies $$ f(x) +xf(-x)=x+1.\tag{1} $$ There is no further information.

I attempted inverting the equation $f^{-1}(f(x)) = x = f^{-1}(x(1-f(-x))+1)$ but it doesn't seem to lead anywhere. I also tried plugging random $x$ values in order to understand the problem better, but still I could find nothing interesting.

Any ideas?

Nathanael Skrepek
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HelloWorld
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3 Answers3

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By replacing $x$ with $-x$ in (1) we get $$f(-x)-xf(x)=-x+1\Leftrightarrow f(-x)=x(f(x)-1)+1.$$ Therefore, by plugging it into (1), we find $$f(x) +x^2(f(x)-1)+x=x+1$$ that is $$(x^2+1)(f(x)-1)=0$$ and we may conclude that $f(x)=1$ for all $x\in \mathbb{R}$.

Robert Z
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  • Presumably this is for real $x$. It seems to be possible for $f(i)$ to take any value in which case $f(-i) = if(i)-i+1$ – Henry Jul 08 '20 at 11:09
  • @Henry Yes, If $x\not=\pm i$ then $f(x)=1$. Given the value of $f(i)$ then $f(-i)=i(f(i)-1)+1$. Thanks for pointing out. – Robert Z Jul 08 '20 at 12:10
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Plug in $-x$ to get $$\tag2f(-x)-xf(x)=-x+1.$$ Now you have two linear equations in two unkowns $f(x)$ and $f(-x)$.

Hagen von Eitzen
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Change $x$ to $-x$ to get $f(-x)-xf(x)=1-x$. Multiply by $x$ to get $xf(-x)-x^{2}f(x)=x(1-x)$. Substitute for $xf(-x)$ from the original equation and solve for $f(x)$.

Kavi Rama Murthy
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