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:

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