If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$$
$$h(x)=\dfrac{x^2+x^2-2+2}{x-\dfrac{1}{x}}$$
$$h(x)=x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}$$
Case $1$: $x-\dfrac{1}{x}>0$
$$\dfrac{x^2-1}{x}>0$$
$$x\in(-1,0) \cup (1,\infty)$$
$$AM\ge GM$$
$$\dfrac{x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}}{2}>\sqrt{2}$$
$$h(x)\ge 2\sqrt{2}$$
$$x-\dfrac{1}{x}=\dfrac{2}{x-\dfrac{1}{x}}$$
$$x^2+\dfrac{1}{x^2}-2=2$$
$$x^2+\dfrac{1}{x^2}=4$$
$$x^4-4x^2+1=0$$
$$x^2=2\pm\sqrt{3}$$
Only $x=\sqrt{2+\sqrt{3}},-\sqrt{2-\sqrt{3}}$ are the valid solutions.
Case $2$: $x-\dfrac{1}{x}<0$
$$x\in(-\infty,-1) \cup (0,1)$$
$$h(x)=-\left(\dfrac{1}{x}-x+\dfrac{2}{\dfrac{1}{x}-x}\right)$$
By $AM\ge GM$, $h(x)\ge-2\sqrt{2}$
We will get this minimum value at $-\sqrt{2+\sqrt{3}},\sqrt{2-\sqrt{3}}$
So answer should have been $-2\sqrt{2}$ but actual answer is $2\sqrt{2}$. What am I missing here.

Assuming that h(x)= f(x)/g(x), I would not actually calculate h itself.  Instead, use the "quotient rule": $\frac{h'(x)= f'(x)g(x)- f(x)g'(x)}{g^2(x)}$.  That will be 0 if and only if the numerator is 0: $f'(x)g(x)- f(x)g'(x)= 0$.  
$f(x)= x^2+ x^{-2}$ so $f'(x)= 2x+ 2x^{-3}$, $g(x)= x- x^{-1}$ so $g'(x)= 1- x^{-2}$.
So $f'(x)g(x)- f(x)g'(x)= (2x+ 2x^{-3})(x- x^{-1})- (x^2+ x^{-2})(1- x^{-2})= 0$
Seeing that "$x^{-3}$ and $x^{-1}$ I would now multiply both sides by $x^4$ (of course x cannot be 0):
$(2x^4+ 2)(x^2- 2)- (x^4+ 1)(x^2- 1)= 2x^6- 4x^4+ 2x^2- 4- x^6+ x^4- x^2- 1= -3x^4+ x^3- 5= 0$.
