Question Number 210701 by noraell last updated on 17/Aug/24
Answered by Berbere last updated on 18/Aug/24
$${f}'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}\geqslant\mathrm{1}\:{Increase}\:{steictly}\: \\ $$$${f}\:{is}\:{bijection}\:\mathbb{R}\rightarrow\mathbb{R}\Rightarrow{f}^{−} \:{existe} \\ $$$$\mathrm{2}{f}\left({x}\right)+\mathrm{3}{f}^{−} \left({x}\right)=\mathrm{10} \\ $$$${since}\:{f}\:{is}\:{increasing}\:\Rightarrow{f}^{−} \:{increase}\: \\ $$$$\Rightarrow\mathrm{2}{f}\left({x}\right)+\mathrm{3}{f}^{−} \left({x}\right)={g}\left({x}\right)\:{increase} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{2}\Rightarrow{f}^{−} \left(\mathrm{2}\right)=\mathrm{2} \\ $$$$\mathrm{2}.{f}\left(\mathrm{2}\right)+\mathrm{3}{f}^{−} \left(\mathrm{2}\right)=\mathrm{10}={g}\left(\mathrm{2}\right) \\ $$$${since}\:{g}\:{increase}\:{x}=\mathrm{2}\:{is}\:{unique}\:{solution} \\ $$$${of}\:\mathrm{2}{f}\left({x}\right)+\mathrm{3}{f}^{−} \left({x}\right)=\mathrm{10} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$