Answer:
[tex]\displaystyle \mathbb{R}-\left \{ -6,\ 6 \right \}[/tex]
Step-by-step explanation:
Domain of Functions
Given a function h(x), we call the domain of h, all the values x can take so h exists. It's commonly easier to find the values of x for which h does NOT exist and then exclude them from the general domain of the real numbers.
The question provides us with the following functions
[tex]\displaystyle f(x)=x^2-3x-18[/tex]
[tex]\displaystyle g(x)=x^2-36[/tex]
[tex]\displaystyle h(x)=\frac{x^2-3x-18}{x^2-36}[/tex]
We called h(x) to the newly created function f/g(x)
To investigate the possible points of nonexistence for h, we factor h(x)
[tex]\displaystyle h(x)=\frac{(x-6)(x+3)}{(x-6)(x+6)}[/tex]
We can clearly see the denominator has two zeros at
[tex]\displaystyle x=6,\ x=-6[/tex]
A rational function cannot exist where the denominator is zero, regardless of the possible simplification with the numerator, as in this case. For example, for x=6 the function gets the value 0/0 and it's not defined there. We cannot simplify the common factor because they are part of the original function.
So, the domain of h is
[tex]\displaystyle (-\infty,\ -6 )\bigcup (-6,\ 6)\bigcup (6,+\infty )[/tex]
Or equivalently
[tex]\displaystyle \mathbb{R}-\left \{ -6,\ 6 \right \}[/tex]
