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 :
$-3$
$-2 \sqrt{2}$
$2 \sqrt{2}$
3
Correct Option: , 3
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