What is the problem in finding domain of $ \frac{x-3}{(x+3)(\sqrt{x^2-4})} $?

Question

I'm having

$$ f(x)=\frac{x-3}{(x+3)(\sqrt{x^2-4})} $$

Now, if I want to find its domain, I should write:

$$ x+3 ≠ 0 \ \ \ \ \ \ \ \text{AND} \ \ \ \ \ \ \ \ \ (x+2)(x-2) >0 $$

$ \text{as the denominator must be non-vanishing} \\ $

• So I get

$ (-∞,-3) \cup (-3,-2) \cup (2,∞) $

Which is of course correct

But.

If I use

$ (x+2)>0 $ AND $ (x-2)>0 $

separately, I don't get $ (-∞,-3) \cup (-3,-2) $ .

Why? Can't I do it separately?

Answer 1

Note, in the denominator :

$x \not = -3$ and the expression under the root must be positive, i.e.

$(x-2)(x+2)>0.$

1) The product is positive if both factors are positive:

$x-2 >0$ and $x+2 >0,$

$x>2$ and $x>-2$, hence $x>2.$

2) The product is positive if both factors are negative:

$x-2<0$ and $x+2<0$,

$x<2$ and $x<-2$, hence $x < -2.$

Answer 2

Recall that

$$x^2-4>0 \iff (x-2)(x+2)>0 \iff \begin{cases}x-2>0 \land x+2>0\\\\x-2<0 \land x+2<0 \end{cases}$$