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#1 h Apr 21, 2025, 2:22 PM

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f:R\{0} ->R\{0}
f(x/y+y/x)=f(x)/f(y)+f(y)/f(x)
f(xy)=f(x).f(y)

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#2 h Apr 21, 2025, 3:07 PM

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Musashi123 wrote:

f:R\{0} ->R\{0}
f(x/y+y/x)=f(x)/f(y)+f(y)/f(x)
f(xy)=f(x).f(y)

$f(x)=f(x\times 1)=f(x)f(1)$ and so $f(1)=1$
$1=f(1)=f((-1)(-1))=f(-1)^2$ and so $f(-1)=\pm 1$
If $x>0$ , $f(x)=f(\sqrt x^2)=f(\sqrt x)^2>0$

Let $g(x)=\ln f(e^x)$ so that $f(x)=e^{g(\ln x)}$ $\forall x>0$
$f(xy)=f(x)f(y)$ implies $g(\ln x+\ln y)=g(\ln x)+g(\ln y)$ $\forall x,y>0$ and so $g(x+y)=g(x)+g(y)$ $\forall x,y\in\mathbb R$

$f(\frac xy+\frac yx)=\frac{f(x)}{f(y)}+\frac{f(y)}{f(x)}$ and so $f(x)\ge 2$ $\forall x\ge 2$ and so $g(x)\ge\ln 2$ $\forall x\ge\ln 2$
So $g(x)$ is additive and lowerbounded over $[\ln 2,+\infty)$ and so is linear
So $g(x)=ax$ for some $a>0$ and $f(x)=x^a$ $\forall x>0$

And $\left(\frac xy+\frac yx\right)^a=\frac{x^a}{y^a}+\frac{y^a}{x^a}$ $\forall x,y>0$ implies $a=1$ and $f(x)=x$ $\forall x>0$

If $f(-1)=-1$ , this implies $\boxed{\text{S1 : }f(x)=x\quad\forall x\ne 0}$ , which indeed fits

If $f(-1)=1$ , this implies $\boxed{\text{S2 : }f(x)=|x|\quad\forall x\ne 0}$ , which indeed fits

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