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Question Number 192084 by sciencestudentW last updated on 07/May/23

$$f\left(x\right)+x\cdot f(-x)=x^2+1$$$$f\left(\sqrt{2}\right)=?$$

Answered by deleteduser1 last updated on 07/May/23

$$x^2+1=-(1-x)(1+x)+2=-(1+x-x(1+x))+2$$$$=1-x+x(1+x)=f\left(x\right)+x\cdot f(-x)$$$$\Rightarrow f\left(x\right)=1-x\Rightarrow f\left(\sqrt{2}\right)=1-\sqrt{2}$$

Answered by York12 last updated on 07/May/23

$$f\left(x\right)+xf(-x)=x^2+1\to xf\left(x\right)+x^{2}f(-x)=x^3+x\to \left(I\right)$$$$f(-x)-xf\left(x\right)=x^2+1\to \left(II\right)$$$$summingIandII$$$$weget:$$$$f(-x)=\frac{(x^4-1)}{(x-1)(x^2+1)}=x+1\to f\left(x\right)=1-x$$$$\therefore f\left(\sqrt{2}\right)=(1-\sqrt{2})\to \left({That}^{\prime }sit\right)$$$$\left\{\mathcal{BY}YORK\right\}$$$$$$

Answered by mehdee42 last updated on 07/May/23

$$x=\sqrt{2}\to f\left(\sqrt{2}\right)+\sqrt{2}f(-\sqrt{2})=3\left(i\right)$$$$x=-\sqrt{2}\to f(-\sqrt{2})-\sqrt{2}f(\sqrt{2)}=3\stackrel{\times -\sqrt{2}}{\Rightarrow }-\sqrt{2}f(-\sqrt{2})+2f\left(\sqrt{2}\right)=-3\sqrt{2}\left(ii\right)$$$$\left(i\right)+\left(ii\right)\to 3f\left(\sqrt{2}\right)=3-3\sqrt{2}\Rightarrow f\left(\sqrt{2}\right)=1-\sqrt{2}✓$$

Answered by gatocomcirrose last updated on 08/May/23

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}+\mathrm{b}$$$$\mathrm{ax}+\mathrm{b}-{\mathrm{ax}}^2+\mathrm{bx}={\mathrm{x}}^2+1$$$$\Rightarrow \mathrm{a}=-1,\mathrm{b}=1$$$$\mathrm{f}\left(\mathrm{x}\right)=-\mathrm{x}+1\Rightarrow \mathrm{f}\left(\sqrt{2}\right)=1-\sqrt{2}$$

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