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 :

Question

( A ) $2 \sqrt{2}$
( B ) $-3$
( C ) $-2 \sqrt{2}$
( D ) $3$