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Question Number 154133 by qaz last updated on 14/Sep/21
Given f(x)=((x+(√(1+x^2 ))))^(1/3) +((x−(√(1+x^2 ))))^(1/3) Find f^(−1) (x)=?
$$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt[{\mathrm{3}}]{\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}+\sqrt[{\mathrm{3}}]{\mathrm{x}−\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{Find}\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=? \\ $$
Answered by EDWIN88 last updated on 14/Sep/21
⇔y−((x+(√(1+x^2 ))))^(1/3) −((x−(√(1+x^2 ))))^(1/3) =0 ⇒y^3 −(x+(√(1+x^2 )))−(x−(√(1+x^2 )))=3y (((x+(√(1+x^2 )))(x−(√(1+x^2 )))))^(1/3) ⇒y^3 −2x= 3y ((x^2 −(1+x^2 )))^(1/3) ⇒y^3 −2x = −3y ⇒2x=y^3 +3y ⇒x =(1/2)(y^3 +3y) ⇒f^(−1) (x)=(1/2)(x^3 +3x)
$$\Leftrightarrow{y}−\sqrt[{\mathrm{3}}]{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}−\sqrt[{\mathrm{3}}]{{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:=\mathrm{0} \\ $$$$\Rightarrow{y}^{\mathrm{3}} −\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)−\left({x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)=\mathrm{3}{y}\:\sqrt[{\mathrm{3}}]{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\left({x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)} \\ $$$$\Rightarrow{y}^{\mathrm{3}} −\mathrm{2}{x}=\:\mathrm{3}{y}\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} −\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow{y}^{\mathrm{3}} −\mathrm{2}{x}\:=\:−\mathrm{3}{y} \\ $$$$\Rightarrow\mathrm{2}{x}={y}^{\mathrm{3}} +\mathrm{3}{y} \\ $$$$\Rightarrow{x}\:=\frac{\mathrm{1}}{\mathrm{2}}\left({y}^{\mathrm{3}} +\mathrm{3}{y}\right) \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{3}} +\mathrm{3}{x}\right) \\ $$
Answered by mr W last updated on 14/Sep/21
f(x)=y=((x+(√(1+x^2 ))))^(1/3) +((x−(√(1+x^2 ))))^(1/3) y=((−(−x)+(√(1^3 +(−x)^2 ))))^(1/3) −(((−x)+(√(1^3 +(−x)^2 ))))^(1/3) that means y is the root of cubic eqn. with respect to t: t^3 +3t−2x=0 i.e. y^3 +3y−2x=0 ⇒x=((y(y^2 +3))/2) ⇒f^(−1) (x)=((x(x^2 +3))/2)
$$\mathrm{f}\left(\mathrm{x}\right)={y}=\sqrt[{\mathrm{3}}]{\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}+\sqrt[{\mathrm{3}}]{\mathrm{x}−\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }} \\ $$$${y}=\sqrt[{\mathrm{3}}]{−\left(−\mathrm{x}\right)+\sqrt{\mathrm{1}^{\mathrm{3}} +\left(−{x}\right)^{\mathrm{2}} }}−\sqrt[{\mathrm{3}}]{\left(−\mathrm{x}\right)+\sqrt{\mathrm{1}^{\mathrm{3}} +\left(−{x}\right)^{\mathrm{2}} }} \\ $$$${that}\:{means}\:{y}\:{is}\:{the}\:{root}\:{of}\:{cubic}\:{eqn}. \\ $$$${with}\:{respect}\:{to}\:{t}: \\ $$$${t}^{\mathrm{3}} +\mathrm{3}{t}−\mathrm{2}{x}=\mathrm{0} \\ $$$${i}.{e}. \\ $$$${y}^{\mathrm{3}} +\mathrm{3}{y}−\mathrm{2}{x}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{{y}\left({y}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{2}} \\ $$

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