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Question Number 176391 by cortano1 last updated on 18/Sep/22

Commented by mr W last updated on 18/Sep/22

$$f\left(x\right)=x^3$$

Answered by TheHoneyCat last updated on 18/Sep/22

$${\mathrm{Here}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{proof}:$$$$$$$$yf\left(2x\right)-xf\left(2y\right)=8xy(x^2-y^2)$$$$$$$$\mathrm{On}{\left({\mathbb{R}}^{\ast }\right)}^2$$$$\frac{f\left(2x\right)}{x}-\frac{f\left(2y\right)}{y}=8(x^2-y^2)=2({\left(2x\right)}^2-{\left(2y\right)}^2)$$$$\mathrm{thus}\mathrm{denoting}:\{\begin{array}{l}a=2x\\ b=2y\\ g\left(x\right)=\frac{f\left(x\right)}{x}\end{array}$$$$g\left(a\right)-g\left(b\right)=a^2-b^2$$$$\mathrm{fixing}b\mathrm{to}\mathrm{any}\mathrm{value},\mathrm{we}\mathrm{get}\mathrm{that}$$$$\mathrm{\exists }c\in \mathbb{R}/g:\{\begin{array}{lll}{\mathbb{R}}^{\ast }& \to & \mathbb{R}\\ x& & x^2+c\end{array}$$$$\mathrm{now}\mathrm{comming}\mathrm{back}\mathrm{to}g\left(a\right)-g\left(b\right)=a^2-b^2$$$$\mathrm{fixing}b=-a$$$$g\left(a\right)-g(-a)=0$$$$\mathrm{thus}\mathrm{\forall }x\in {\mathbb{R}}^{\ast }g\left(x\right)=g(-x)$$$$\mathrm{hence}c=0$$$$\mathrm{so}:$$$$g:\{\begin{array}{lll}{\mathbb{R}}^{\ast }& \to & \mathbb{R}\\ x& & x^2\end{array}$$$$$$$$\mathrm{thus}{f}_{\mid {\mathbb{R}}^{\ast }}:\{\begin{array}{lll}{\mathbb{R}}^{\ast }& \to & \mathbb{R}\\ x& & x^3\end{array}$$$$$$$$\mathrm{the}\mathrm{only}\mathrm{value}\mathrm{left}\mathrm{to}\mathrm{find}\mathrm{is}f\left(0\right)$$$$\mathrm{choosing}x=0\mathrm{and}y=1(\mathrm{or}\mathrm{anything}\mathrm{such}\mathrm{that}y\ne 0)$$$$\mathrm{we}\mathrm{get}\mathrm{that}:$$$$1\times f\left(0\right)-0\times f\left(2\right)=8\times 0\times 1\times (0^2-1^2)$$$$ief\left(0\right)=0$$$$$$$$\mathrm{Thefore},\mathrm{indeed},$$$$f:\{\begin{array}{lll}\mathbb{R}& \to & \mathbb{R}\\ x& & x^3\end{array}$$$$◻$$

Answered by mr W last updated on 19/Sep/22

$$lety=\frac{x}{2}$$$$\frac{xf\left(2x\right)}{2}-xf\left(x\right)=4x^2\times \frac{3x^2}{4}$$$$f\left(2x\right)-2f\left(x\right)=6x^3$$$$letf\left(x\right)={ax}^3$$$$8{ax}^3-2{ax}^3=6x^3$$$$\Rightarrow a=1$$$$\Rightarrow f\left(x\right)=x^3$$

Commented by Tawa11 last updated on 20/Sep/22

$$\mathrm{Great}\mathrm{sir}$$

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