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Question-210701




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

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