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Let-f-x-x-3-2x-6-f-1-x-




Question Number 124580 by Boucatchou last updated on 04/Dec/20
          Let  f(x)=x^3 +2x−6,    f^(−1) (x)=?
$$ \\ $$$$\:\:\:\:\:\:\:\:{Let}\:\:{f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{6},\:\:\:\:{f}^{−\mathrm{1}} \left({x}\right)=? \\ $$
Answered by MJS_new last updated on 04/Dec/20
x=y^3 +2y−6  y^3 +2y−(x+6)=0  D=(p^3 /(27))+(q^2 /4)=(8/(27))+(((x+6)^2 )/4)>0 ∀x∈R ⇒ we need  Cardano′s solution  f^(−1) (x)=(((x/2)+3+(√((x^2 /2)+6x+((494)/(27))))))^(1/3) −((−(x/2)−3+(√((x^2 /2)+6x+((494)/(27))))))^(1/3)
$${x}={y}^{\mathrm{3}} +\mathrm{2}{y}−\mathrm{6} \\ $$$${y}^{\mathrm{3}} +\mathrm{2}{y}−\left({x}+\mathrm{6}\right)=\mathrm{0} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{8}}{\mathrm{27}}+\frac{\left({x}+\mathrm{6}\right)^{\mathrm{2}} }{\mathrm{4}}>\mathrm{0}\:\forall{x}\in\mathbb{R}\:\Rightarrow\:\mathrm{we}\:\mathrm{need} \\ $$$$\mathrm{Cardano}'\mathrm{s}\:\mathrm{solution} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\sqrt[{\mathrm{3}}]{\frac{{x}}{\mathrm{2}}+\mathrm{3}+\sqrt{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{6}{x}+\frac{\mathrm{494}}{\mathrm{27}}}}−\sqrt[{\mathrm{3}}]{−\frac{{x}}{\mathrm{2}}−\mathrm{3}+\sqrt{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{6}{x}+\frac{\mathrm{494}}{\mathrm{27}}}} \\ $$

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