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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.

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@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.

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  • $\begingroup$ Yes Hagen explanation is very good but it did not answer my question, as you did. Thank you so I understand now where I made the mistake $\endgroup$ – sound wave Dec 6 '17 at 23:12
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$\sqrt{x^2-4}$ 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 $\{-2,2\}\cup (2,\infty)=\{-2\}\cup[2,\infty)$.

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  • $\begingroup$ Thank you for the answer, I understand but I don't get where my calculus are wrong. I usually solve equations using intersection of solutions (if I have a system of equations) or with the table of signs, as in the case of the product of 2 or more factors. But in this case with the "mechanical" method I get a wrong solution. I know that is better to not use the same method over and over, but it is just a personal curiosity. $\endgroup$ – sound wave Dec 5 '17 at 22:53
  • $\begingroup$ I have a little question, you said $\sqrt{x^2-4}$ is defined (and automatically non-negative) if $x^2-4\geq0$, but about the non-negative affirmation, I have a doubt because for example $\sqrt{4}=+2$ or $-2$ right? $\endgroup$ – sound wave Dec 8 '17 at 11:55
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    $\begingroup$ @soundwave No, $\sqrt4=2$ never $-2$ because $\sqrt 4$ is defined as the non-negative number $x$ for which $x^2=4$ this is because we want the square root to be a function which means it must give exactly one value (on it's domain). $\endgroup$ – kingW3 Dec 8 '17 at 12:57
  • $\begingroup$ @kingW3 ok I understand, the square root has to be a function so every value in the domain has to be linked with only one value in the codomain, right ? so at the start of the exercise I could define a square root for which the codomain contains only negative values (i.e. $\sqrt{4}=-2$) and solve the exercise in this different way ? $\endgroup$ – sound wave Dec 8 '17 at 13:10
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    $\begingroup$ @soundwave You could define it that way, but that's not the definition of the square root. You would be the only one using that definition, unless explicitly stated otherwise the definition of $\sqrt \cdot$ is understood to return only non-negative values. $\endgroup$ – kingW3 Dec 8 '17 at 13:21

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