Refer to functions $$ s $$ and $$ t $$. Find the indicated function and write the domain in interval notation. $$ s(x)=\frac{x-1}{x^{2}-9} $$ Part: $$ 0 / 2 $$ Part 1 of 2

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Find the function $$ s(x)=\frac{x-1}{x^{2}-9} $$.
Simplify the expression by following steps:
- step0: Solution:
  $$\frac{x-1}{x^{2}-9}$$
The function $$ s(x) $$ is given by $$ s(x) = \frac{x-1}{x^{2}-9} $$.

To find the domain of the function, we need to determine the values of $$ x $$ for which the function is defined. The function is defined as long as the denominator $$ x^{2}-9 $$ is not equal to zero.

To find the values of $$ x $$ that make the denominator zero, we set $$ x^{2}-9 = 0 $$ and solve for $$ x $$:

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

Therefore, the function is not defined at $$ x = 3 $$ and $$ x = -3 $$.

The domain of the function $$ s(x) $$ is all real numbers except $$ x = 3 $$ and $$ x = -3 $$, which can be written in interval notation as:

$$ (-\infty, -3) \cup (-3, 3) \cup (3, \infty) $$
The domain of $$ s(x) = \frac{x-1}{x^{2}-9} $$ is all real numbers except $$ x = 3 $$ and $$ x = -3 $$, which is written as $$ (-\infty, -3) \cup (-3, 3) \cup (3, \infty) $$.

Refer to functions \( s \) and \( t \). Find the indicated function and write the domain in interval notation. \[ s(x)=\frac{x-1}{x^{2}-9} \] Part: \( 0 / 2 \) Part 1 of 2

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