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Question Number 122074 by bemath last updated on 13/Nov/20

$$Givenf:R\to Rsuchthat$$$$x^{2}f\left(x\right)+f(1-x)=2x-x^4$$$$findf\left(x\right).$$

Answered by bobhans last updated on 14/Nov/20

$$\left(1\right)x^{2}f\left(x\right)+f(1-x)=2x-x^4$$$$replacingxby1-xgive$$$$\to {(1-x)}^{2}f(1-x)+f\left(x\right)=2(1-x)-{(1-x)}^4$$$$multiplyeq\left(1\right)with{(1-x)}^{2}give$$$$\left(2\right)\to {(1-x)}^{2}f(1-x)+{(1-x)}^2{x}^{2}f\left(x\right)={(1-x)}^2(2x-x^4)$$$$substract\left(1\right)by\left(2\right)give$$$$\{1-x^2{(1-x)}^2\}f\left(x\right)=(2-2x)-(1-x^4)-{(1-x)}^2(2x-x^4)$$$$\because f\left(x\right)\frac{(2-2x)-(1-x^4)-{(1-x)}^2(2x-2x^4)}{1-x^2{(1-x)}^2}$$

Answered by mathmax by abdo last updated on 14/Nov/20

$$\Rightarrow {\mathrm{x}}^2\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}(1-\mathrm{x})=-{\mathrm{x}}^4+2\mathrm{x}$$$$\mathrm{let}\mathrm{change}\mathrm{x}\mathrm{by}1-\mathrm{x}\Rightarrow {(1-\mathrm{x})}^2\mathrm{f}(1-\mathrm{x})+\mathrm{f}\left(\mathrm{x}\right)=2(1-\mathrm{x})-{(1-\mathrm{x})}^4$$$$\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{system}\{\begin{array}{l}{\mathrm{x}}^2\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}(1-\mathrm{x})=2\mathrm{x}-{\mathrm{x}}^4\\ \mathrm{f}\left(\mathrm{x}\right)+{(1-\mathrm{x})}^2\mathrm{f}(1-\mathrm{x})=2(1-\mathrm{x})-{(1-\mathrm{x})}^4\end{array}$$$${\mathrm{\Delta }}_{\mathrm{s}}={\mathrm{x}}^2{(1-\mathrm{x})}^2-1\Rightarrow {(\mathrm{x}-{\mathrm{x}}^2)}^2-1={\mathrm{x}}^2-{2\mathrm{x}}^3+{\mathrm{x}}^4-1$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\left|\begin{array}{c}2\mathrm{x}-{\mathrm{x}}^{4}1\\ 2(1-\mathrm{x})-{(1-\mathrm{x})}^4{(1-\mathrm{x})}^2\end{array}\right|}{{\mathrm{x}}^4-{2\mathrm{x}}^3+{\mathrm{x}}^2-1}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{(2\mathrm{x}-{\mathrm{x}}^4){(1-\mathrm{x})}^2-2(1-\mathrm{x})+{(1-\mathrm{x})}^4}{{\mathrm{x}}^4-{2\mathrm{x}}^3+{\mathrm{x}}^2-1}$$

Answered by ajfour last updated on 14/Nov/20

$$x^{2}f\left(x\right)+f(1-x)=2x-x^4$$$${(1-x)}^{2}f(1-x)+f\left(x\right)=2(1-x)-{(1-x)}^4$$$$f\left(x\right)=\frac{{(1-x)}^2(2x-x^4)-2(1-x)+{(1-x)}^4}{x^2{(1-x)}^2-1}$$$$=\frac{(1-x)\{(1-x)(2x-x^4)-2+{(1-x)}^3\}}{\{x(1-x)-1\}\{x(1-x)+1\}}$$$$=\frac{(1-x)(x^5-x^4-x^3+x^2-x-1)}{\{x(1-x)-1\}\{x(1-x)+1\}}$$$$=\frac{(1-x)(x^3+1)(x^2-x-1)}{(x^2-x+1)(x^2-x-1)}$$$$f\left(x\right)=\frac{(1-x)(1+x)(x^2-x+1)}{x^2-x+1}$$$$\Rightarrow f\left(x\right)=1-x^2★$$

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