Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

Commented by liberty last updated on 24/Dec/20

(•) x+(1/x) = w ⇒x^2 +(1/x^2 ) = w^2 −2 (x−(1/x))^2 +2 = w^2 −2 x−(1/x) = ± (√(w^2 −4)) f(x+(1/x))=x−(1/x) f(w)=±(√(w^2 −4)) f(x)= ± (√(x^2 −4))

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

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

f^2 (x+(1/x))=(x−(1/x))^2 f^2 (x+(1/x))=(x+(1/x))^2 −4 f^2 (x)=x^2 −4⇒f(x)=(√(x^2 −4))

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

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

let x+(1/x)=t ⇒x^2 +(1/x^2 ) +2=t^2 ⇒x^2 +(1/x^2 )=t^2 −2 f^2 (x+(1/x))=(x−(1/x))^2 ⇒f^2 (t)=x^2 +(1/x^2 )−2 =t^2 −4 ⇒f(t)=+^− (√(t^2 −4))

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com