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Question Number 207935 by meo last updated on 31/May/24

Answered by Frix last updated on 01/Jun/24

$$\mathrm{\forall }x\in \mathbb{R}:F\left(x\right)>0$$$$F^{\prime }\left(x\right)=0$$$$2x-1-\frac{4x}{{(x^2+1)}^2}=0$$$$2x^5-x^4+4x^3-2x^2-2x-1=0$$$$(x-1)(2x^4+x^3+5x^2+3x+1)=0$$$$x\in \mathbb{R}\Rightarrow x=1$$$$F^{″}\left(x\right)=2+\frac{4(3x^2-1)}{{(x^2+1)}^3};F^{″}\left(1\right)=3>0\Rightarrow$$$$F\left(1\right)=2025\mathrm{is}\mathrm{the}\mathrm{absolute}\mathrm{minimum}$$

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