Let f(x)=x^2+frac{1}{x^2} and g(x)=x-frac{1}{ | vedclass

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(B) Given $f(x)=x^2+\frac{1}{x^2}$ and $g(x)=x-\frac{1}{x}$.We can write $f(x)$ in terms of $g(x)$ as $f(x)=\left(x-\frac{1}{x}\right)^2+2$.Let $t=x-\

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$2 \sqrt{2}$

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(B) Given $f(x)=x^2+\frac{1}{x^2}$ and $g(x)=x-\frac{1}{x}$.We can write $f(x)$ in terms of $g(x)$ as $f(x)=\left(x-\frac{1}{x}\right)^2+2$.Let $t=x-\frac{1}{x}$. Then $f(x)=t^2+2$ and $g(x)=t$.We define $h(t)=\frac{f(x)}{g(x)}=\frac{t^2+2}{t}=t+\frac{2}{t}$.To find the local minimum,we find the derivative $h'(t)=1-\frac{2}{t^2}$.Setting $h'(t)=0$,we get $1-\frac{2}{t^2}=0$,which implies $t^2=2$,so $t=\pm \sqrt{2}$.Using the second derivative test,$h''(t)=\frac{4}{t^3}$.For $t=\sqrt{2}$,$h''(\sqrt{2})=\frac{4}{(\sqrt{2})^3} > 0$,so $t=\sqrt{2}$ is a point of local minimum.The local minimum value is $h(\sqrt{2})=\sqrt{2}+\frac{2}{\sqrt{2}}=\sqrt{2}+\sqrt{2}=2\sqrt{2}$.

Let $f(x)=x^2+\frac{1}{x^2}$ and $g(x)=x-\frac{1}{x}$ for $x \in R-\{-1,0,1\}$,then the local minimum of $\frac{f(x)}{g(x)}$ is

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