Question Number 178349 by doline last updated on 15/Oct/22
$${etudier}\:{la}\:{fonction}\:{suivante}\:\mathrm{4}{x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{1} \\ $$$${calculer}\:{les}\:{limites}\:{puis}\:{dresser}\:{son}\:{tableau}\:{de}\:{variation} \\ $$
Answered by Acem last updated on 16/Oct/22
![lim_(x→∓∞) f(x)= ∓∞ lim_(x→∓∞) [f(x)−y_Δ ]= ∓∞ il n′y a pas d′asymptote oblic pour C_f f^( ′) (x)= 12x^2 −18x+ 6, f^( ′) (x)= 0 ⇒ p_1 ((1/2), 2.75) , p_2 (1, 2) determinant ((( x),(−∞),( (1/2)),,1,( +∞)),((f(x)),( + ),( 0),−,0,( +)),((f(x)),( _(−∞ ) ↗),,( ↘),,( ↗ ^(+∞) ))) Points d′aide: p_3 (0, 1) , p_4 (−0.1365, 0)_(de la calculatrice graphique)](https://www.tinkutara.com/question/Q178361.png)
$$ \\ $$$$\underset{{x}\rightarrow\mp\infty} {\mathrm{lim}}{f}\left({x}\right)=\:\mp\infty \\ $$$$\:\underset{{x}\rightarrow\mp\infty} {\mathrm{lim}}\left[{f}\left({x}\right)−{y}_{\Delta} \right]=\:\mp\infty\:{il}\:{n}'{y}\:{a}\:{pas}\:{d}'{asymptote}\:{oblic}\:{pour}\:{C}_{{f}} \\ $$$$\:{f}^{\:'} \left({x}\right)=\:\mathrm{12}{x}^{\mathrm{2}} −\mathrm{18}{x}+\:\mathrm{6},\:{f}^{\:'} \left({x}\right)=\:\mathrm{0} \\ $$$$\:\Rightarrow\:{p}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{2}.\mathrm{75}\right)\:,\:{p}_{\mathrm{2}} \left(\mathrm{1},\:\mathrm{2}\right) \\ $$$$\underline{ }\begin{array}{|c|c|c|}{\:\:\:{x}}&\hline{−\infty}&\hline{\:\:\frac{\mathrm{1}}{\mathrm{2}}}&\hline{}&\hline{\mathrm{1}}&\hline{\:\:\:\:\:\:\:\:+\infty}\\{{f}\left({x}\right)}&\hline{\:\:\:\:\:\:\:\:\:\:\:\:+\:}&\hline{\:\:\:\mathrm{0}}&\hline{−}&\hline{\mathrm{0}}&\hline{\:\:\:+}\\{{f}\left({x}\right)}&\hline{\:_{−\infty\:\:} \:\nearrow}&\hline{}&\hline{\:\searrow}&\hline{}&\hline{\:\nearrow\:\:\:^{+\infty} }\\\hline\end{array} \\ $$$$ \\ $$$$\:{Points}\:{d}'{aide}: \\ $$$$\:{p}_{\mathrm{3}} \left(\mathrm{0},\:\mathrm{1}\right)\:\:,\:\:{p}_{\mathrm{4}} \left(−\mathrm{0}.\mathrm{1365},\:\mathrm{0}\right)_{{de}\:{la}\:{calculatrice}\:{graphique}} \\ $$$$ \\ $$