Question Number 154725 by SLVR last updated on 21/Sep/21
$${If}\:{f}\left({x}\right)={x}^{\mathrm{3}} +{x}+\mathrm{1}\:{then}\:{invrse}\:{function} \\ $$$${of}\:{f}\left({x}\right)\:{is}??? \\ $$
Answered by mr W last updated on 21/Sep/21
$${y}={x}^{\mathrm{3}} +{x}+\mathrm{1} \\ $$$${x}^{\mathrm{3}} +{x}+\mathrm{1}−{y}=\mathrm{0} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} +\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)^{\mathrm{2}} }−\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)}−\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} +\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)^{\mathrm{2}} }+\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{\left({y}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}}+\frac{{y}−\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{\left({y}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}}−\frac{{y}−\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}}+\frac{{x}−\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}}−\frac{{x}−\mathrm{1}}{\mathrm{2}}} \\ $$
Commented by SLVR last updated on 21/Sep/21
$${Great}...{dear}\:{professor}...{mr}.{W} \\ $$$${I}\:{need}\:{to}\:{know}\:{how}\:{x}=\:{in}\:\mathrm{3}{rdstep}? \\ $$$${kindly}...{explain} \\ $$
Commented by mr W last updated on 21/Sep/21
$${Cardano}'{s}\:{formula}\:{for}\:{cubic}\:{equation}. \\ $$$${see}\:{Q}\mathrm{89687} \\ $$
Commented by SLVR last updated on 21/Sep/21
$${Thank}\:{you}\:\:\:...{sir}...{so}..\:{kind}\:{of}\:{you} \\ $$