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Question Number 57012 by rahul 19 last updated on 28/Mar/19

$$Findthelocalminimumvalueoff\left(x\right)$$$$wheref\left(x\right)=(x-\frac{1}{x})+\left(\frac{2}{x-\frac{1}{x}}\right)?$$

Answered by mr W last updated on 29/Mar/19

$$lett=x-\frac{1}{x}$$$$ift>0:$$$$f\left(x\right)=t+\frac{2}{t}⩾2\sqrt{t\times \frac{2}{t}}=2\sqrt{2}=localmin$$$$ift<0:$$$$f\left(x\right)=t+\frac{2}{t}=-(-t+\frac{2}{-t})⩽-2\sqrt{(-t)\times \frac{2}{(-t)}}=-2\sqrt{2}=localmax$$

Commented by rahul 19 last updated on 30/Mar/19

thank you sir!

Commented by rahul 19 last updated on 30/Mar/19

$$Sir,howtodothesameproblemif$$$$f\left(x\right)=(x+\frac{1}{x})+\left(\frac{2}{x+\frac{1}{x}}\right).$$

Commented by mr W last updated on 31/Mar/19

$$lett=x+\frac{1}{x}$$$$forx>0:t⩾2$$$$forx<0:t⩽-2$$$$$$$$f\left(t\right)=t+\frac{2}{t}$$$$f^{\prime }\left(t\right)=1-\frac{2}{t^2}>0\Rightarrow f\left(t\right)isstrictlyincreasing$$$$$$$$ift⩾2,i.e.x>0:$$$$sincsf\left(t\right)isstrictlyincreasing$$$$\Rightarrow min.=f(t=2)=3$$$$$$$$ift⩽-2,i.e.x<0:$$$$sincsf\left(t\right)isstrictlyincreasing$$$$\Rightarrow max.=f(t=-2)=-3$$

Commented by rahul 19 last updated on 31/Mar/19

$$Sir,inthiscasewecannotapply$$$$A.M⩾G.Mbecauset⩾2forx>0$$$$unliketheaboveQ.wherelocalminimum$$$$occuredatt=\sqrt{2}?$$

Commented by mr W last updated on 31/Mar/19

$${that}^{\prime }sthereason,$$$$youareabsolutelycorrect!$$

Answered by MJS last updated on 28/Mar/19

$$f\left(x\right)=\frac{x^4+1}{x(x^2-1)}$$$$f^{\prime }\left(x\right)=\frac{x^6-3x^4-3x^2+1}{x^2{(x^2-1)}^2}$$$$x^6-3x^4-3x^2+1=0$$$$x=\sqrt{y}$$$$y^3-3y^2-3y+1=0$$$$y=z+1$$$$z^3-6z-4=0$$$$\mathrm{trying}\mathrm{factors}\mathrm{of}-4\mathrm{we}\mathrm{find}$$$$z_1=-2\Rightarrow y_1=-1\Rightarrow x_1=\pm \mathrm{i}$$$$(z+2)(z^2-2z-2)=0$$$$z_2=1-\sqrt{3}\Rightarrow y_2=2-\sqrt{3}\Rightarrow x_2=\pm (\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2})$$$$z_3=1+\sqrt{3}\Rightarrow y_3=2+\sqrt{3}\Rightarrow x_3=\pm (\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2})$$$$f^{″}\left(x\right)=\frac{2(x^6+9x^4-3x^2+1)}{x^3{(x^2-1)}^3}$$$$f^{″}(-\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2})<0$$$$f^{″}(-\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2})>0\Rightarrow \min ;f\left(x\right)=2\sqrt{2}$$$$f^{″}(\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2})<0$$$$f^{″}(\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2})>0\Rightarrow \min \Rightarrow f\left(x\right)=2\sqrt{2}$$

Commented by rahul 19 last updated on 30/Mar/19

thank you sir!

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