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Question-207935




Question Number 207935 by meo last updated on 31/May/24
Answered by Frix last updated on 01/Jun/24
∀x∈R: F(x)>0  F′(x)=0  2x−1−((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∈R ⇒ x=1  F′′(x)=2+((4(3x^2 −1))/((x^2 +1)^3 )); F′′(1)=3>0 ⇒  F(1)=2025 is the absolute minimum
$$\forall{x}\in\mathbb{R}:\:{F}\left({x}\right)>\mathrm{0} \\ $$$${F}'\left({x}\right)=\mathrm{0} \\ $$$$\mathrm{2}{x}−\mathrm{1}−\frac{\mathrm{4}{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{5}} −{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}\in\mathbb{R}\:\Rightarrow\:{x}=\mathrm{1} \\ $$$${F}''\left({x}\right)=\mathrm{2}+\frac{\mathrm{4}\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} };\:{F}''\left(\mathrm{1}\right)=\mathrm{3}>\mathrm{0}\:\Rightarrow \\ $$$${F}\left(\mathrm{1}\right)=\mathrm{2025}\:\mathrm{is}\:\mathrm{the}\:\mathrm{absolute}\:\mathrm{minimum} \\ $$

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