For example $n(x)=\sqrt{x-2}\sqrt{4-x}$

My attempt,



and $4-x\ge0$

$4\ge x$


So the domain is $2\le x\le4$. Am I correct?

How about $f(x)=\frac{1}{\sqrt{x-1}\sqrt{x+3}}$ or even $f(x)=\sqrt{\frac{x+1}{x-1}}$ this kind of function?

  • $\begingroup$ You have to change $\le$ and $\ge$ to < and >. $\endgroup$
    – zoli
    May 31, 2015 at 14:33
  • $\begingroup$ @zoli 2 and 4 are allowed values for $n(x)$ $\endgroup$
    – wythagoras
    May 31, 2015 at 14:34
  • 1
    $\begingroup$ @zoli: Why? There's nothing wrong with $\sqrt 0$ -- it's $0$. $\endgroup$ May 31, 2015 at 14:34
  • 1
    $\begingroup$ You are correct! For the other two just remember that you can't have zero on the denominator and take a similar approach. $\endgroup$ May 31, 2015 at 14:35
  • $\begingroup$ @Mathxx: I answered the "How about" part. Sorry, if it was misleading. My comment should have been this: Yes, your are correct but ... $\endgroup$
    – zoli
    May 31, 2015 at 14:38

1 Answer 1


Your solution is correct.

For the second, it goes almost the same way, but we can't have $x=1$ or $x=-3$ to avoid division by zero.

For the third, notice we can't have have $x=1$. Further, we want $x+1$ and $x-1$ to be both positive or both negative. This gives $x \leq -1$ or $x \geq 1$, but we won't want $x=1$. So the third has domain:

$$x \leq -1 \vee x > 1$$

in the notation with $x$ or in the interval notation:

$$(\leftarrow,-1] \cup (1,\rightarrow) $$


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