$\frac{1}{x}f(x) = xf(\frac{1}{x})$ and $f(x+y)+1 = f(x)+f(y)$

Question

Determine all $ f \colon \mathbb{R}\to \mathbb{R} $ such that $\frac{1}{x}f(x) = xf(\frac{1}{x})$ and $f(x+y)+1 = f(x)+f(y)$

Answer 1

Let $g(x)=f(x)-1$. Then by the second condition, $$g(x+y)=g(x)+g(y). $$ It is well know that this implies $g(ax)=ag(x)$ whenever $a\in\Bbb Q$, so here $$ f(ax)=af(x)-a+1.$$ In case $a:=x^2$ is a positive rational, we conclude from the first condition $$ xf(\tfrac 1x)=\frac 1xf(x)=\frac1xf(x^2\cdot\tfrac 1x)=\frac1x\left(x^2f(\tfrac1x)-x^2+1\right)=xf(\tfrac1x)-x+\frac 1x$$ and therefore $x=\frac 1x$, which is absurd.

Answer 2

Hint: Let $ g(x) = f(x) - 1$, then $ g(x+y) = g(x) + g(y)$.

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