Question Number 196464 by mokys last updated on 25/Aug/23
$${if}\:{f}\left({x}\right)=\:{x}^{\mathrm{3}} +{x}^{\mathrm{2}} \:\:,\:{g}\left({x}\right)=\:{f}^{−\mathrm{1}} \left({x}\right)\:,\:{find}\:{g}^{'} \left(\mathrm{3}\right)\:?\:\: \\ $$
Answered by Frix last updated on 25/Aug/23
![y=x^3 +x^2 dy=x(3x+2)dx g′(x)=(dx/dy)=(1/(x(3x+2))) x^3 +x^2 =3 ⇒ x=−(1/3)+((((79)/(54))+((√(77))/6)))^(1/3) +((((79)/(54))−((√(77))/6)))^(1/3) x≈1.17455941 g′(x)≈.154133358 [This is the real solution of r^3 +(r/(231))−(1/(231))=0]](https://www.tinkutara.com/question/Q196468.png)
$${y}={x}^{\mathrm{3}} +{x}^{\mathrm{2}} \\ $$$${dy}={x}\left(\mathrm{3}{x}+\mathrm{2}\right){dx} \\ $$$${g}'\left({x}\right)=\frac{{dx}}{{dy}}=\frac{\mathrm{1}}{{x}\left(\mathrm{3}{x}+\mathrm{2}\right)} \\ $$$${x}^{\mathrm{3}} +{x}^{\mathrm{2}} =\mathrm{3}\:\Rightarrow\:{x}=−\frac{\mathrm{1}}{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{79}}{\mathrm{54}}+\frac{\sqrt{\mathrm{77}}}{\mathrm{6}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{79}}{\mathrm{54}}−\frac{\sqrt{\mathrm{77}}}{\mathrm{6}}} \\ $$$${x}\approx\mathrm{1}.\mathrm{17455941} \\ $$$${g}'\left({x}\right)\approx.\mathrm{154133358} \\ $$$$\left[\mathrm{This}\:\mathrm{is}\:\mathrm{the}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{of}\:{r}^{\mathrm{3}} +\frac{{r}}{\mathrm{231}}−\frac{\mathrm{1}}{\mathrm{231}}=\mathrm{0}\right] \\ $$