<s>[INST] Finding domain of $\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$ How can I find the domain of:
$$\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$
I think the hard part will be to find: 
$$\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} \ge 0$$
So far I have: not sure how to preceed: 
$$\sqrt{ \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)} }$$
For $\sqrt{g(x)}$ to be valid, $g(x) \ge 0$
For $f(x)$ to be valid, $(x^2-2)(x^2-4)(x^2-6) \ne 0$
Thus, $x \ne \sqrt 2, 2, \sqrt 3$
$$g(x) = { \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)}}  \ge 0$$
 [/INST] +1 for your handwriting :-)


*

*$\frac{f(x)}{g(x)} \ge 0$ is not so different from $f(x)\times g(x) \ge 0$ (except for zeros of $g$.)

*With $(x^2-1) = (x-1)(x+1)$ etc, your problem reduces to the form of $(x-a)(x-b)(x-c)(x-d)...(x-z) \ge 0$


Edit: oops I only read the hand-written part!  Anyways thanks to the monotonicity of $x^5-1$ etc, you can still use similar argument.
Edit 2: Plot it in google.
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