Question 1208404
<font color=black size=3>
Problem 1.


x^2-9 = 0 had been correctly solved to get x = -3 and x = 3.
And you are correct that x cannot equal either of these values; otherwise, we get a division by zero error.


Another way to solve is:
x^2-9 = 0
x^2-3^2 = 0
(x-3)(x+3) = 0 .... difference of squares rule
x-3 = 0 or x+3 = 0
x = 3 or x = -3


The domain in set-builder notation is 
*[Tex \Large \left\{\text{x} | \text{x} \in \mathbb{R}, \ \text{x} \not= -3, \text{x} \not= 3 \right\}]
Or you can slightly condense things to get
*[Tex \Large \left\{\text{x} | \text{x} \in \mathbb{R}, \ \text{x} \not= \pm3 \right\}]
The *[Tex \large \text{x} \in \mathbb{R}] portion translates to "x is in the set of real numbers" i.e. "x is a real number".



The interval notation would look like this *[Tex \Large \left(-\infty,-3\right) \cup \left(-3,3\right) \cup  \left(3,\infty\right) ]


This is basically us starting with the entire number line *[Tex \Large \left(-\infty,\infty\right) ] and then poking two holes at x = -3 and x = 3 on the number line. 
Each "U" refers to a set union joining or gluing together the disjoint intervals.
Curved parenthesis exclude each endpoint. 


--------------------------------------------------------------------------


Problem 2.


You have the correct scratch work. I'll assume your second step should have read x^3+x=0.



The domain in set-builder notation is 
*[Tex \Large \left\{\text{x} | \text{x} \in \mathbb{R}, \text{x} \not= 0 \right\}]
In words we can say "x is any nonzero real number".



The domain as interval notation would be *[Tex \Large \left(-\infty,0\right) \cup  \left(0,\infty\right) ]
It's the same idea as earlier, except we only have one hole this time. 
</font>