Let f(x)=x2+1x2 and g(x)=x−1x ,x∈R−{−1,0,1} . If h(x)=f(x)g(x) , then the local minimum value of h(x) is

h(x)=x2+1x2x−1x=(x−1x)2+2x−1x A.M ≥ G.M (x−1x)+(2x−1x)2≥2 h(x)≥22

Let f(x)=x2+1x2 and g(x)=x−1x ,x∈R−{−1,0,1} . If h(x)=f(x)g(x) , then the local minimum value of h(x) is

h(x)=x2+1x2x−1x=(x−1x)2+2x−1x A.M ≥ G.M (x−1x)+(2x−1x)2≥2 h(x)≥22

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

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$2 \sqrt{2}$

$- 3$

$- 2 \sqrt{2}$

answer is C.

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$h (x) = \frac{x^{2} + \frac{1}{x^{2}}}{x - \frac{1}{x}} = \frac{{(x - \frac{1}{x})}^{2} + 2}{x - \frac{1}{x}} A . M \geq G . M \frac{(x - \frac{1}{x}) + (\frac{2}{x - \frac{1}{x}})}{2} \geq \sqrt{2} h (x) \geq 2 \sqrt{2}$

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Let f(x)=x2+1x2 and g(x)=x−1x ,x∈R−{−1,0,1} . If h(x)=f(x)g(x) , then the local minimum value of h(x) is

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