Find the domain of a radical function

Question

I'm having trouble with this function $f(x)=\sqrt{x\sqrt{x^2-4}}$ for which I have to find the domain.

To find it I first set $x\sqrt{x^2-4}\geq0$ because the argument of the radical must be non negative. Now we have to study the sign of this product, so we study the signs of the single factors.

$x>0$

$\sqrt{x^2-4}>0\rightarrow x^2-4>0\rightarrow x<-2\vee x>2$

And now we have to make the table of signs.

$-2\quad0\quad2$

$-|-|+|+\quad(x>0)$

$+|-|-|+\quad(\sqrt{x^2-4}>0)$

$-|+|-|+\quad(result)$

Since the original inequality is $\geq$ the result is

$-2\le x\le0\vee x\geq2$

but I notice that for $x\in(-2,0]$ the term $\sqrt{x^2-4}$ is not defined (in $\mathbb{R}$), and so the solution should be

$[-2,0]\cup[2,+\infty)\setminus(-2,0]=\{-2\}\cup[2,+\infty)$

so I made a mistake in the previous calculus since from the table of signs I firstly got $-2\le x\le0\vee x\geq2$ which is wrong, but I cannot understand why.

Answer 1

@HagenvonEitzen's answer is very good, you should mark that as the correct answer. However, from your comment, I thought I would point out where you must have made your mistake. You correctly find the domain of $\sqrt{x^2-4}$. Looking at the negative part of that domain, you have $x\le-2$. Then you look at the expression $x\sqrt{x^2-4}$. When $x\le-2$, this expression is negative, which would make $f(x)$ undefined. This is correct. But then it seems that you decided that for $-2<x<0$, $f(x)$ would be defined, apparently because a negative times a negative is a positive. However, we have already determined that $\sqrt{x^2-4}$ is undefined $(-2, 0)$, so $f(x)$ would have to be undefined as well. My guess is you mixed the meanings of negative and undefined when you were reading the table of signs.

Answer 2

$\sqrt{x^2-4}$ is defined (and automatically non-negative) if $x^2\ge 4$, i.e., if $x\le -2$ or $x\ge 2$. We can ignore $(-2,2)$ immediately because $\sqrt{x^2-4}$ is not defined there (even if some smart persons might argue that $x\sqrt{x^2-4}$ ought to be considered defined and zero when $x=0$).

For those points where $\sqrt{x^2-4}=0$, i.e., for $x\in\{-2,2\}$, the value of $x$ does not matter, we will certainly have $x\sqrt{x^2-4}=0\ge 0$ and hence $\sqrt{x\sqrt{x^2-4}}$ defined.

For the other points, i.e., when $\in(-\infty,-2)\cup(2,\infty)$, we need $x\ge 0$. This leaves us only with $(2,\infty)$.

In summary, the domain in question is $\{-2,2\}\cup (2,\infty)=\{-2\}\cup[2,\infty)$.