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Question Number 139029 by bramlexs22 last updated on 21/Apr/21

$$\mathrm{Given}\mathrm{f}(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})=\frac{\mathrm{x}}{\mathrm{x}+1}$$$$\mathrm{find}\mathrm{f}\left(\mathrm{x}\right)=?$$

Answered by EDWIN88 last updated on 21/Apr/21

$$\mathrm{u}=\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}$$$${\mathrm{u}}^2-2\mathrm{ux}+{\mathrm{x}}^2=1+{\mathrm{x}}^2$$$$\mathrm{x}=\frac{{\mathrm{u}}^2-1}{2\mathrm{u}}\Rightarrow \mathrm{f}\left(\mathrm{u}\right)=\frac{\frac{{\mathrm{u}}^2-1}{2\mathrm{u}}}{\frac{{\mathrm{u}}^2-1}{2\mathrm{u}}+1}=\frac{{\mathrm{u}}^2-1}{{\mathrm{u}}^2+2\mathrm{u}-1}$$$$\mathrm{then}\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^2+2\mathrm{x}-1}$$

Answered by mathmax by abdo last updated on 22/Apr/21

$$\mathrm{let}\mathrm{x}=\mathrm{sht}\Rightarrow \mathrm{t}=\mathrm{argshx}=\log (\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})$$$$\mathrm{f}(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})=\mathrm{f}(\mathrm{sht}+\mathrm{cht})=\frac{\mathrm{sht}}{1+\mathrm{sht}}\Rightarrow \mathrm{f}\left({\mathrm{e}}^{\mathrm{t}}\right)=\frac{\frac{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}{2}}{1+\frac{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}{2}}$$$$=\frac{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}{2+{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}\mathrm{let}{\mathrm{e}}^{\mathrm{t}}=\mathrm{x}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}-{\mathrm{x}}^{-1}}{2+\mathrm{x}-{\mathrm{x}}^{-}}=\frac{{\mathrm{x}}^2-1}{2\mathrm{x}+{\mathrm{x}}^2-1}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^2+2\mathrm{x}-1}$$

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