To determine the domain of the function [tex]\( h(x) = \frac{4}{x-5} \)[/tex], we need to identify all values of [tex]\( x \)[/tex] for which the function is defined.
1. Identify the issue points: The function [tex]\( h(x) \)[/tex] is a rational function and is undefined where the denominator is zero. Therefore, we need to find the values of [tex]\( x \)[/tex] that make the denominator zero.
The denominator of [tex]\( h(x) \)[/tex] is [tex]\( x - 5 \)[/tex]. So we set:
[tex]\[
x - 5 = 0
\][/tex]
Solving for [tex]\( x \)[/tex], we get:
[tex]\[
x = 5
\][/tex]
This means that [tex]\( x = 5 \)[/tex] will make the denominator zero, and thus the function [tex]\( h(x) \)[/tex] is undefined at [tex]\( x = 5 \)[/tex].
2. State the domain: Since the function is undefined at [tex]\( x = 5 \)[/tex], [tex]\( x = 5 \)[/tex] must be excluded from the domain. The domain of the function includes all real numbers except [tex]\( x = 5 \)[/tex].
3. Express the domain: We can express this using interval notation. The domain includes all real numbers less than 5 and all real numbers greater than 5.
In interval notation, this is represented as:
[tex]\[
(-\infty, 5) \cup (5, \infty)
\][/tex]
Therefore, the domain of [tex]\( h(x) = \frac{4}{x-5} \)[/tex] is:
[tex]\[
(-\infty, 5) \cup (5, \infty)
\][/tex]
This means the function is defined for all real numbers except [tex]\( x = 5 \)[/tex].
