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, $x \ne \sqrt 2, 2, \sqrt 3$
$$g(x) = { \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)}}  \ge 0$$
 A: The first thing you should note is that the expression 
$$\Phi=\frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)}  $$
is undefined at any zero of the denominator. 
So, the points $x=\pm \sqrt 2$, $\pm 2$, and $\pm\sqrt 6$ are 
not in the domain of $\sqrt\Phi$.
Now, to find the domain of $\sqrt\Phi$, we need to find when $\Phi$ is nonnegative.
Towards solving $\Phi\ge0$,
the fundamental observation to make is that
the only points "across which" $\Phi$ can change signs are at the zeros of its numerator or at the zeroes of its denominator.
The zeroes of the numerator are: $$\pm 1, \pm \sqrt 3, \pm \sqrt 5$$ 
and  the zeroes of the denominator are
$$
\pm \sqrt 2, \pm 2 \pm\sqrt 6.
$$
As already mentioned, the only points across which $\Phi$ can change sign are at one of the zeroes above.
Let's also note the expression $\Phi$ is "even": if $\Phi(x)\ge 0$, then $\Phi(-x)\ge 0$. So, let's find the $x$ values on the nonnegative $x$-axis that satisfy $\Phi(x)\ge0$.  Then by "reflection" we'll obtain the points on the negative $x$-axis where $\Phi(x)\ge0$.
So, draw a number line with  the  zeroes listed above:

The zeroes subdivide the nonnegative $x$-axis into intervals, the endpoints of which, except the endpoint 0, are the zeroes of either the numerator or the denominator of $\Phi$.  In each subinterval, if you pick a point $x_0$ (not an endpoint, save for 0), evaluate $\Phi(x_0)$, and note its sign, then across
that entire subinterval, the expression $\Phi$ will have that sign.
For example in $[0,1)$, picking $x=1/2$ it is easy to see that the sign of $\Phi(1/2)$ is
$$\frac{((1/2)^2-1)((1/2)^2-3)((1/2)^2-5)}{((1/2)^2-2)((1/2)^2-4)((1/2)^2-6)} 
={ (-)(-)(-)\over(-)(-)(-)}  
\ge 0$$
So on all of $[0,1)$, the expression $\Phi$ is positive. (Note, you only need to find the sign, there is no need to do the actual arithmetic.)
The complete "sign chart" is shown below:

And we see that $\Phi>0$ on
$$
[0, 1)\cup(\sqrt2,\sqrt3)\cup(2,\sqrt5)\cup(\sqrt6,\infty).
$$ 
By symmetry, $\Phi>0$ on 
$$
 ( -\infty,-\sqrt6)\cup(-\sqrt5,2)\cup (-\sqrt3,-\sqrt2)\cup (-1,0].
$$
So, the domain of $\sqrt\Phi$ is the union of the two sets above, together with
the zeroes of the numerator of $\Phi$ that aren't zeros of the denominator of $\Phi$: $x=\pm1$, $x=\pm \sqrt3$, and $x=\pm \sqrt5$.

Below is a plot of $y=\Phi$ and $y=\sqrt\Phi$. Note that the zeroes of the denominator of $\Phi$ are vertical asymptotes of the graph of $y=\Phi$ and the zeroes of the numerator of $\Phi$ are the zeroes of $\Phi$.

A: +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.
A: $(x^2-1)$ is negative iff $x\in(-1,1)$.  Similarly, $(x^2-2)$ is negative for $x\in(-\sqrt{2},\sqrt{2})$, $(x^2-3)$ for $x\in(-\sqrt{3},\sqrt{3}), \dots,(x^2-6)$ for $x\in(-\sqrt{6},\sqrt{6})$.
Consider the open intervals $(-\infty,-\sqrt{6}), (-\sqrt{6},-\sqrt{5}),(-\sqrt{5},-2),\dots$ separately. Don't forget to look at how $f(x)$ behaves on the boundaries. E.g $x=\pm\sqrt{n}$ for $n=1,2,3,4,5,6$.
