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Question Number 140470 by EDWIN88 last updated on 08/May/21

$${\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2.\mathrm{f}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)=64\mathrm{x},\mathrm{\forall }\mathrm{x}\in \mathrm{D}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=?$$

Answered by benjo_mathlover last updated on 08/May/21

$$\left(1\right){\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2.\mathrm{f}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)=64\mathrm{x}$$$$\left(2\right)\mathrm{replace}\mathrm{x}\to \frac{1-\mathrm{x}}{1+\mathrm{x}}$$$$\Rightarrow {\left(\mathrm{f}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)\right)}^2.\mathrm{f}\left(\mathrm{x}\right)=\frac{64(1-\mathrm{x})}{1+\mathrm{x}}$$$$\mathrm{eq}\left(1\right)\Rightarrow \mathrm{squaring}$$$$\because {\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^4.{\left(\mathrm{f}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)\right)}^2={\left(64\mathrm{x}\right)}^2$$$$\Rightarrow \frac{{\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^4.\cancel{{\left(\mathrm{f}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)\right)}^2}}{\mathrm{f}\left(\mathrm{x}\right).\cancel{{\left(\mathrm{f}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)\right)}^2}}=\frac{{64}^{\cancel{2}}{\mathrm{x}}^2(1+\mathrm{x})}{\cancel{64}(1-\mathrm{x})}$$$$\Rightarrow {\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^3=\frac{{64\mathrm{x}}^2(1+\mathrm{x})}{1-\mathrm{x}}$$$$\therefore \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=4\sqrt[3]{\frac{{\mathrm{x}}^3+{\mathrm{x}}^2}{1-\mathrm{x}}}$$

Answered by mathmax by abdo last updated on 09/May/21

$$\frac{1-\mathrm{x}}{1+\mathrm{x}}=\mathrm{t}\Rightarrow 1-\mathrm{x}=\mathrm{t}+\mathrm{tx}\Rightarrow (1+\mathrm{t})\mathrm{x}=1-\mathrm{t}\Rightarrow \mathrm{x}=\frac{1-\mathrm{t}}{1+\mathrm{t}}\Rightarrow$$$${\mathrm{f}}^2\left(\frac{1-\mathrm{t}}{1+\mathrm{t}}\right).\mathrm{f}\left(\mathrm{t}\right)=64.\frac{1-\mathrm{t}}{1+\mathrm{t}}\mathrm{and}{\mathrm{f}}^2\left(\mathrm{t}\right).\mathrm{f}\left(\frac{1-\mathrm{t}}{1+\mathrm{t}}\right)=64\mathrm{t}\Rightarrow$$$$\mathrm{f}\left(\frac{1-\mathrm{t}}{1+\mathrm{t}}\right)=\frac{64\mathrm{t}}{{\mathrm{f}}^2\left(\mathrm{t}\right)}\Rightarrow {\left(\frac{64\mathrm{t}}{{\mathrm{f}}^2\left(\mathrm{t}\right)}\right)}^2.\mathrm{f}\left(\mathrm{t}\right)=64\frac{1-\mathrm{t}}{1+\mathrm{t}}\Rightarrow$$$$\frac{{64\mathrm{t}}^2}{{\mathrm{f}}^3\left(\mathrm{t}\right)}=\frac{1-\mathrm{t}}{1+\mathrm{t}}\Rightarrow \frac{{\mathrm{f}}^3\left(\mathrm{t}\right)}{{64\mathrm{t}}^2}=\frac{1+\mathrm{t}}{1-\mathrm{t}}\Rightarrow {\mathrm{f}}^3\left(\mathrm{t}\right)=64\frac{{\mathrm{t}}^2+{\mathrm{t}}^3}{1-\mathrm{t}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{t}\right){=}^3\sqrt{64.\frac{{\mathrm{t}}^2+{\mathrm{t}}^3}{1-\mathrm{t}}}$$

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