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Question Number 126712 by help last updated on 23/Dec/20

Commented by liberty last updated on 24/Dec/20

$$(\bullet )x+\frac{1}{x}=w\Rightarrow x^2+\frac{1}{x^2}=w^2-2$$$${(x-\frac{1}{x})}^2+2=w^2-2$$$$x-\frac{1}{x}=\pm \sqrt{w^2-4}$$$$f(x+\frac{1}{x})=x-\frac{1}{x}$$$$f\left(w\right)=\pm \sqrt{w^2-4}$$$$f\left(x\right)=\pm \sqrt{x^2-4}$$$$$$$$$$

Answered by Dwaipayan Shikari last updated on 23/Dec/20

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

Answered by mathmax by abdo last updated on 24/Dec/20

$$\mathrm{let}\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{t}\Rightarrow {\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}+2={\mathrm{t}}^2\Rightarrow {\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}={\mathrm{t}}^2-2$$$${\mathrm{f}}^2(\mathrm{x}+\frac{1}{\mathrm{x}})={(\mathrm{x}-\frac{1}{\mathrm{x}})}^2\Rightarrow {\mathrm{f}}^2\left(\mathrm{t}\right)={\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}-2={\mathrm{t}}^2-4\Rightarrow \mathrm{f}\left(\mathrm{t}\right)=\stackrel{-}{+}\sqrt{{\mathrm{t}}^2-4}$$

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