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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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    Never mind, learn to write $\TeX$ quicker. Why don't you edit this post and make the picture into $\TeX$ take your own time!2012-02-23

3 Answers 3

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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:

enter image description here

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:

enter image description here

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 $\Phi$. enter image description here

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+1 for your handwriting :-)

  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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$(x^2-1)$ 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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    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}2012-02-23