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Question Number 112624 by Aina Samuel Temidayo last updated on 09/Sep/20

$$\mathrm{Minimum}\mathrm{value}\mathrm{of}{\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2+1}-3\mathrm{is}$$

Answered by MJS_new last updated on 09/Sep/20

$$x^2+1+\frac{1}{x^2+1}-4$$$$x^2+1=t⩾1$$$$t+\frac{1}{t}-4$$$$\frac{d}{dt}[t+\frac{1}{t}-4]=1-\frac{1}{t^2}=0\Rightarrow t=1\Rightarrow x^2+1=1\Rightarrow x=0$$$$\Rightarrow \mathrm{answer}\mathrm{is}-2$$

Answered by ajfour last updated on 09/Sep/20

$$(x^2+1)+\frac{1}{(x^2+1)}⩾2$$$$\Rightarrow min(x^2+\frac{1}{x^2+1}-3)+4=2$$$$\Rightarrow min(x^2+\frac{1}{x^2+1}-3)=-2$$

Commented by Aina Samuel Temidayo last updated on 09/Sep/20

$$\mathrm{Thanks}.$$

Answered by 1549442205PVT last updated on 09/Sep/20

$$\mathrm{P}={\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2+1}-3={\mathrm{x}}^2+1+\frac{1}{{\mathrm{x}}^2+1}-4$$$$={(\sqrt{{\mathrm{x}}^2+1}-\frac{1}{\sqrt{{\mathrm{x}}^2+1}})}^2-2⩾-2$$$$\mathrm{The}\mathrm{equality}\mathrm{ocurrs}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}$$$$\sqrt{{\mathrm{x}}^2+1}=\frac{1}{\sqrt{{\mathrm{x}}^2+1}}\iff {\mathrm{x}}^2+1=1\iff \mathrm{x}=0$$$$\mathrm{Therefore},{\mathrm{P}}_{\min }=-2\mathrm{when}\mathrm{x}=0$$

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