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Question Number 202536 by cortano12 last updated on 29/Dec/23

Answered by taguim001 last updated on 29/Dec/23

$$$$$$f\left(xy\right)f\left(x\right)-f\left(xy\right)f\left(y\right)=xf\left(x\right)f\left(y\right)-yf\left(x\right)f\left(y\right)$$$$\{{}_{f\left(xy\right)f\left(y\right)=yf\left(x\right)f\left(y\right)}^{f\left(xy\right)f\left(x\right)=xf\left(x\right)f\left(y\right)}$$$$\{{}_{f\left(xy\right)=yf\left(x\right)}^{f\left(xy\right)=xf\left(y\right)}$$$$\iff xf\left(y\right)=yf\left(x\right)$$$$\iff \frac{f\left(x\right)}{x}=\frac{f\left(y\right)}{y}$$$$\iff \frac{f\left(x\right)}{x}=k\left(kϵ\mathbb{R}\right)$$$$\iff f\left(x\right)=kx\left(kϵ\mathbb{R}\right)$$$$f:\mathbb{R}\to \mathbb{R}$$$$x\mid \to kx\left(kϵ\mathbb{R}\right)$$$$$$

Answered by witcher3 last updated on 30/Dec/23

$$\mathrm{f}\left(\mathrm{y}\right)(\mathrm{f}\left(1\right)-\mathrm{f}\left(\mathrm{y}\right))=(1-\mathrm{y})\mathrm{f}\left(1\right)\mathrm{f}\left(\mathrm{y}\right)$$$$\Rightarrow \mathrm{f}\left(\mathrm{y}\right)(\mathrm{f}\left(1\right)-\mathrm{f}\left(\mathrm{y}\right)-\mathrm{f}\left(1\right)(1-\mathrm{y}))=0$$$$\mathrm{f}\left(\mathrm{y}\right)=0$$$$\mathrm{or}\mathrm{f}\left(\mathrm{y}\right)=\mathrm{f}\left(1\right)\mathrm{y}$$$$\mathrm{f}\left(\mathrm{y}\right)=\mathrm{ay}$$$$\mathrm{f}\left(\mathrm{y}\right)=0\in \{\mathbb{R}\stackrel{\mathrm{f}}{\to }\mathbb{R}\mid \mathrm{\forall }\mathrm{x}\in \mathbb{R}\mathrm{\exists }!\mathrm{a}\in \mathbb{R}\mid \mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}\}$$$$\mathrm{f}\left(\mathrm{y}\right)=\mathrm{ay};\mathrm{a}\in \mathbb{R}$$

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