<s>[INST] Finding the Domain? The set ${{x:\left|x+\frac{1}{x}\right|>6}}$ equals the set
My Approach
We know,IF $|f(x)| \geqslant a, \quad ;a>0$
then $f(x) \geqslant a \quad \cup \quad f(x) \leqslant-a$
$\frac{x^{2}+1}{x}> 6$ & $\frac{x^{2}+1}{x}<-6$
$\frac{x^{2}-6 x+1}{x}>0..........(1)$ 
$\frac{x^{2}+6 x+1}{x}<0$............(2)
$Using$ $Wavy$ $curve$ $method,$
From (1) I can find,
$x \in(0,3-2 \sqrt{2}) \cup(3+2 \sqrt{2}, \alpha)$
From (2) ,I can find,
$x \in(-\alpha,-3-2 \sqrt{2}) \cup(-3+2 \sqrt{2},0)$
Next I am unable to find any common between them.
 [/INST] We have  $x\neq0$ and  $$x^2+1>6|x|$$ or
$$|x|^2-6|x|+1>0,$$ which is
$$|x|>3+2\sqrt2$$ or
$$|x|<3-2\sqrt2,$$ which gives the answer:
$$(-\infty,-3-2\sqrt2)\cup(-3+2\sqrt2,0)\cup(0,3-2\sqrt2)\cup(3+2\sqrt2,+\infty)$$
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