Question Number 208909 by lepuissantcedricjunior last updated on 26/Jun/24
![soit la fonction f(x)=x^3 +x definie sur R on note g(x)=f^(−1) (x) alors que la primitive G(x)=∫_0 ^x g(t)dt](https://www.tinkutara.com/question/Q208909.png)
$$\:\:\:\:\:\boldsymbol{{soit}}\:\boldsymbol{{la}}\:\boldsymbol{{fonction}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{x}}\:\:\boldsymbol{{definie}} \\ $$$$\boldsymbol{{sur}}\:\mathbb{R}\:\boldsymbol{{on}}\:\boldsymbol{{note}}\:\boldsymbol{{g}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{f}}^{−\mathrm{1}} \left(\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{alors}}\:\boldsymbol{{que}}\:\:\boldsymbol{{la}}\:\boldsymbol{{primitive}}\:\boldsymbol{{G}}\left(\boldsymbol{{x}}\right)=\int_{\mathrm{0}} ^{\boldsymbol{{x}}} \boldsymbol{{g}}\left(\boldsymbol{{t}}\right)\boldsymbol{{dt}} \\ $$
Commented by Ghisom last updated on 27/Jun/24
![G(x)=α where α is the real solution of α^3 −(1/2)α^2 +((18x^2 +1)/(16))α−((x^2 (27x^2 +2))/(64))=0](https://www.tinkutara.com/question/Q208910.png)
$${G}\left({x}\right)=\alpha\:\mathrm{where}\:\alpha\:\mathrm{is}\:\mathrm{the}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{of} \\ $$$$\alpha^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}\alpha^{\mathrm{2}} +\frac{\mathrm{18}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{16}}\alpha−\frac{{x}^{\mathrm{2}} \left(\mathrm{27}{x}^{\mathrm{2}} +\mathrm{2}\right)}{\mathrm{64}}=\mathrm{0} \\ $$