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

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(C) Given $h(x) = \frac{x^2 + \frac{1}{x^2}}{x - \frac{1}{x}}$.We can rewrite the numerator as $(x - \frac{1}{x})^2 + 2$.So,$h(x) = \frac{(x - \frac{1

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$2\sqrt{2}$

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(C) Given $h(x) = \frac{x^2 + \frac{1}{x^2}}{x - \frac{1}{x}}$.We can rewrite the numerator as $(x - \frac{1}{x})^2 + 2$.So,$h(x) = \frac{(x - \frac{1}{x})^2 + 2}{x - \frac{1}{x}} = (x - \frac{1}{x}) + \frac{2}{x - \frac{1}{x}}$.Let $t = x - \frac{1}{x}$. Since $x \in R - \{-1, 1, 0\}$,$t$ can take any real value except $0$.Then $h(t) = t + \frac{2}{t}$.For $t > 0$,by the $AM$-$GM$ inequality,$t + \frac{2}{t} \ge 2\sqrt{t \cdot \frac{2}{t}} = 2\sqrt{2}$.The equality holds when $t = \frac{2}{t}$,i.e.,$t^2 = 2$,so $t = \sqrt{2}$ (since $t > 0$).For $t  0$. Then $h(t) = -u - \frac{2}{u} = -(u + \frac{2}{u}) \le -2\sqrt{2}$.Thus,the local minimum value of $h(x)$ is $2\sqrt{2}$.

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

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