Question

(b) $g(x) = \frac{x - 4}{x^2 - 9}$
the domain is all real numbers except for those that make the denominator zero.
the variable $x$ has the restriction that $x^2 - 9 \
eq 0$. therefore, $x \
eq \pm \square$.
domain: $\square$

Explanation:

Step1: Solve denominator for zero

$x^2 - 9 = 0$
$x^2 = 9$
$x = \pm 3$

Step2: Identify excluded values

$x
eq \pm 3$

Step3: Write domain in interval form

Combine intervals of all real numbers except $x=3$ and $x=-3$

Answer:

First blank: $3$
Domain: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$