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

$$x^2 - 9 = 0$$
$$x^2 = 9$$
$$x = \pm\sqrt{9} = \pm3$$

Step2: State excluded x-values

$x
eq \pm3$

Step3: Write domain in interval form

Combine all real numbers except $3$ and $-3$:
$$(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$$

Answer:

The missing value is $3$.
Domain: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$