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, $x \ne \sqrt 2, 2, \sqrt 3$

$$g(x) = { \frac{(x^2-1)(x^2-3)(x^2-5)}{(x^2-2)(x^2-4)(x^2-6)}} \ge 0$$

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A friendly note: You have posting pictures in which you presumably had written the Math. I'd suggest you learn $\TeX$. Use the community to learn the commands. You can see the basic codes by clicking on them as seeing Math as $\TeX$ commands here. It will pay you in the long run and you'll become effective without having to write those crappy pics. Will you? (I like the way write your questions showing us what you have done! +1 for that!) – user21436 Feb 23 '12 at 12:09
@KannappanSampath, I know TeX but it takes longer to type them :) perhaps I need more practice. I also have put the more important parts of the question in TeX, the not as important stuff, I thought providing an image will suffice – Jiew Meng Feb 23 '12 at 12:12
I see you already know $\TeX$ from the first part of your question. Why don't you do the same thing for the Math in the pic as well? – user21436 Feb 23 '12 at 12:13
Is the numerator $(x^2-1)(x^2-3)(x^2-5)$ or $(x^2-1)(x^3-1)(x^5-1)$? – anthus Feb 23 '12 at 12:13
Its the 1st, $(x^2-1)(x^2-3)(x^2-5)$ – Jiew Meng Feb 23 '12 at 12:14

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 $\Phi$.

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1. $\frac{f(x)}{g(x)} \ge 0$ is not so different from $f(x)\times g(x) \ge 0$ (except for zeros of $g$.)
2. 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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+1 for the link. I didn't know Google could plot. – user23211 Feb 23 '12 at 13:37

$(x^2-1)$ 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 $n=1,2,3,4,5,6$.

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Ok, now I get the domain as $-\sqrt{5} \lt x \lt -2$, $-\sqrt{3} \lt x \lt -\sqrt{2}$, $-1 \lt x \lt 1$ ... but how do I find out without graph about what happens $x \lt -\sqrt{6}$ and $x \gt \sqrt{6}$ – Jiew Meng Feb 23 '12 at 14:23
When x>$\sqrt{6}$ all the factors in the numerator and denominator are positive, so the whole expression is positive. The situation is the same for $x<−\sqrt{6}$ – anthus Feb 23 '12 at 14:49