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Question Number 117934 by Ar Brandon last updated on 14/Oct/20

$$\mathrm{Let}f:[1,\mathrm{\infty })\to [2,\mathrm{\infty })\mathrm{be}\mathrm{the}\mathrm{function}\mathrm{defined}\mathrm{by}$$$$f\left(x\right)=x+\frac{1}{x}$$$$\mathrm{If}\mathrm{g}:[2,\mathrm{\infty })\to [1,\mathrm{\infty }),\mathrm{is}\mathrm{a}\mathrm{function}\mathrm{such}\mathrm{that}(\mathrm{g}\circ f)\left(x\right)=x$$$$\mathrm{for}\mathrm{all}x⩾1.\mathrm{Show}\mathrm{that}\mathrm{g}\left(t\right)=\frac{t+\sqrt{t^2-4}}{2}$$

Answered by Lordose last updated on 14/Oct/20

$$\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{x}$$$$\mathrm{g}(\mathrm{x}+\frac{1}{\mathrm{x}})=\mathrm{x}$$$$\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{y}$$$${\mathrm{x}}^2-\mathrm{yx}+1=0$$$${\mathrm{x}}^2-\mathrm{yx}+{(-\frac{\mathrm{y}}{2})}^2=-1+\frac{{\mathrm{y}}^2}{4}$$$$(\mathrm{x}-\frac{\mathrm{y}}{2})=\pm \sqrt{\frac{{\mathrm{y}}^2-4}{4}}$$$$\mathrm{x}=\frac{1}{2}(\mathrm{y}\pm \sqrt{{\mathrm{y}}^2-4})$$$$\mathrm{g}\left(\mathrm{y}\right)=\frac{1}{2}(\mathrm{y}\pm \sqrt{{\mathrm{y}}^2-4})$$$$\mathrm{g}\left(\mathrm{t}\right)=\frac{1}{2}(\mathrm{t}\pm \sqrt{{\mathrm{t}}^2-4})$$

Commented by Ar Brandon last updated on 14/Oct/20

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