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Question Number 123987 by mnjuly1970 last updated on 29/Nov/20
              ...nice   mathematics...     if    f(x)=x^3 +x  then           find   f^( −1) (x)=?
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:\:{mathematics}… \\ $$$$\:\:\:{if}\:\:\:\:{f}\left({x}\right)={x}^{\mathrm{3}} +{x}\:\:{then} \\ $$$$\:\:\:\:\:\:\:\:\:{find}\:\:\:{f}^{\:−\mathrm{1}} \left({x}\right)=? \\ $$$$\:\:\:\: \\ $$
Answered by MJS_new last updated on 29/Nov/20
y^3 +y−x=0  D=(1/(27))+(x^2 /4)>0∀x∈R  ⇒  f^(−1) (x)=(((x/2)+((√(3(27x^2 +4)))/(18))))^(1/3) −((−(x/2)+((√(3(27x^2 +4)))/(18))))^(1/3)
$${y}^{\mathrm{3}} +{y}−{x}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{1}}{\mathrm{27}}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}>\mathrm{0}\forall{x}\in\mathbb{R} \\ $$$$\Rightarrow \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\sqrt[{\mathrm{3}}]{\frac{{x}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}\left(\mathrm{27}{x}^{\mathrm{2}} +\mathrm{4}\right)}}{\mathrm{18}}}−\sqrt[{\mathrm{3}}]{−\frac{{x}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}\left(\mathrm{27}{x}^{\mathrm{2}} +\mathrm{4}\right)}}{\mathrm{18}}} \\ $$
Commented by john_santu last updated on 30/Nov/20
Cardano
$${Cardano} \\ $$
Commented by MJS_new last updated on 30/Nov/20
yes
$$\mathrm{yes} \\ $$

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