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Question Number 85238 by naka3546 last updated on 20/Mar/20

f(x)  =  x^3  + 3x  f^(−1) (x)  =  ?

$${f}\left({x}\right)\:\:=\:\:{x}^{\mathrm{3}} \:+\:\mathrm{3}{x} \\ $$$${f}\:^{−\mathrm{1}} \left({x}\right)\:\:=\:\:? \\ $$

Answered by mr W last updated on 20/Mar/20

y=x^3 +3x  x^3 +3x−y=0  ⇒x=(((√(1+(y^2 /4)))+(y/2)))^(1/3) −(((√(1+(y^2 /4)))−(y/2)))^(1/3)   ⇒f^(−1) (x)=(((√(1+(x^2 /4)))+(x/2)))^(1/3) −(((√(1+(x^2 /4)))−(x/2)))^(1/3)

$${y}={x}^{\mathrm{3}} +\mathrm{3}{x} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}−{y}=\mathrm{0} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{1}+\frac{{y}^{\mathrm{2}} }{\mathrm{4}}}+\frac{{y}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{1}+\frac{{y}^{\mathrm{2}} }{\mathrm{4}}}−\frac{{y}}{\mathrm{2}}} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}}+\frac{{x}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}}−\frac{{x}}{\mathrm{2}}} \\ $$

Commented by jagoll last updated on 20/Mar/20

cardano?

$$\mathrm{cardano}? \\ $$

Commented by mr W last updated on 20/Mar/20

yes

$${yes} \\ $$

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