Solve this following
Question:

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 :

1. $-3$

2. $-2 \sqrt{2}$

3. $2 \sqrt{2}$

4. 3

Correct Option: , 3

Solution: