To determine the domain of the function [tex]\( \frac{f(x)}{g(x)} \)[/tex] where [tex]\( f(x) = \frac{1}{x+1} \)[/tex] and [tex]\( g(x) = x-2 \)[/tex], we need to consider where both functions are well-defined and where the overall expression is defined.
1. Identify the domain of [tex]\( f(x) \)[/tex]:
[tex]\( f(x) = \frac{1}{x+1} \)[/tex]. This function is undefined where the denominator is zero:
[tex]\[
x + 1 = 0 \implies x = -1
\][/tex]
Therefore, [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = -1 \)[/tex]. So the domain of [tex]\( f(x) \)[/tex] is all real numbers except [tex]\( x = -1 \)[/tex]:
[tex]\[
\text{Domain of } f(x) = \{ x \in \mathbb{R} : x \neq -1 \}
\][/tex]
2. Identify the domain of [tex]\( g(x) \)[/tex]:
[tex]\( g(x) = x - 2 \)[/tex]. This is a linear function and it is defined for all real numbers:
[tex]\[
\text{Domain of } g(x) = \mathbb{R}
\][/tex]
3. Identify where [tex]\( g(x) = 0 \)[/tex]:
[tex]\[
x - 2 = 0 \implies x = 2
\][/tex]
For the function [tex]\( \frac{f(x)}{g(x)} = \frac{\frac{1}{x+1}}{x-2} = \frac{1}{(x+1)(x-2)} \)[/tex], it will be undefined not only where [tex]\( f(x) \)[/tex] or [tex]\( g(x) \)[/tex] are undefined, but also where [tex]\( g(x) = 0 \)[/tex].
4. Combine the constraints:
We need [tex]\( x \neq -1 \)[/tex] and [tex]\( x \neq 2 \)[/tex]. So, combining these, the domain of [tex]\( \frac{f(x)}{g(x)} \)[/tex] is:
[tex]\[
\{ x \in \mathbb{R} : x \neq -1 \text{ and } x \neq 2 \}
\][/tex]
In interval notation, this is expressed as:
[tex]\[
(-\infty, -1) \cup (-1, 2) \cup (2, \infty)
\][/tex]
Therefore, the correct answer is:
[tex]\[
\boxed{B. (-\infty, -1), (-1, 2), \text{ and } (2, \infty)}
\][/tex]
