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Question Number 77340 by BK last updated on 05/Jan/20

Answered by mind is power last updated on 05/Jan/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{t}\right)={\mathrm{t}}^3+{\mathrm{t}}^2+16\mathrm{t}+60$$$$\mathrm{z}=\mathrm{f}\left(\mathrm{x}\right)$$$$\mathrm{y}=\mathrm{fof}\left(\mathrm{x}\right)$$$$\mathrm{x}=\mathrm{fofo}\left(\mathrm{x}\right)$$$$\Rightarrow \mathrm{x}=\mathrm{fofo}\left(\mathrm{x}\right),\mathrm{x}\mathrm{is}\mathrm{a}\mathrm{fix}\mathrm{point}\mathrm{of}\mathrm{f}$$$${\mathrm{f}}^{\prime }\left(\mathrm{t}\right)={3\mathrm{t}}^2+2\mathrm{t}+16,\mathrm{\forall }\mathrm{t}\in \mathbb{R}\mid {\mathrm{f}}^{\prime }\left(\mathrm{t}\right)>0\Rightarrow \mathrm{f}\mathrm{is}\mathrm{increse}\mathrm{fonction}$$$$\Rightarrow \mathrm{fofof}\mathrm{increas}\mathrm{function}$$$$\mathrm{fofof}\left(\mathrm{x}\right)=\mathrm{x}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}$$$$\Rightarrow {\mathrm{x}}^3+{\mathrm{x}}^2+16\mathrm{x}+60=\mathrm{x}\Rightarrow {\mathrm{x}}^3+{\mathrm{x}}^2+15\mathrm{x}+60=0\mathrm{use}\mathrm{cardan}$$$$\mathrm{x},\mathrm{z}=\mathrm{f}\left(\mathrm{x}\right),\mathrm{y}=\mathrm{f}\left(\mathrm{z}\right)$$$$$$$$$$

Commented by BK last updated on 05/Jan/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}\mathrm{great}!$$

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