What is $f(x)$ if $f(x) +xf(-x)=x+1$?

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 $x\in \mathbb{R}$.


Plug in $-x$ to get $$\tag2f(-x)-xf(x)=-x+1.$$ Now you have two linear equations in two unkowns $f(x)$ and $f(-x)$.


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 $f(x)$.

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Calculus

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