Certainly! We are given the functions [tex]\( f(x) = \frac{x-6}{x-3} \)[/tex] and [tex]\( g(x) = 4x - 9 \)[/tex]. We need to find the composition [tex]\( (f \circ g)(x) \)[/tex], which is the same as [tex]\( f(g(x)) \)[/tex], and determine its domain.
To begin, we substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(g(x)) = f(4x - 9) \][/tex]
Now, substitute [tex]\( 4x - 9 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(4x - 9) = \frac{(4x - 9) - 6}{(4x - 9) - 3} \][/tex]
Simplify the numerator and the denominator:
[tex]\[ f(4x - 9) = \frac{4x - 15}{4x - 12} \][/tex]
So, the composition is:
[tex]\[ (f \circ g)(x) = \frac{4x - 15}{4x - 12} \][/tex]
Next, we need to determine the domain of the composition [tex]\( f(g(x)) \)[/tex].
The function [tex]\( g(x) = 4x - 9 \)[/tex] is defined for all real numbers since it is a linear function. However, we need to ensure the composition [tex]\( \frac{4x - 15}{4x - 12} \)[/tex] is also defined, which requires the denominator to be nonzero. Thus, we must exclude any [tex]\( x \)[/tex] that makes [tex]\( 4x - 12 = 0 \)[/tex].
Solve for [tex]\( x \)[/tex]:
[tex]\[ 4x - 12 = 0 \][/tex]
[tex]\[ 4x = 12 \][/tex]
[tex]\[ x = 3 \][/tex]
So, the function is undefined at [tex]\( x = 3 \)[/tex].
Therefore, the domain of [tex]\( (f \circ g)(x) \)[/tex] is all real numbers except [tex]\( x = 3 \)[/tex]. In interval notation, this is:
[tex]\[ (-\infty, 3) \cup (3, \infty) \][/tex]
Summarizing:
[tex]\[ (f \circ g)(x) = \frac{4x - 15}{4x - 12} \][/tex]
Domain: [tex]\( (-\infty, 3) \cup (3, \infty) \)[/tex]
Thus, the full solution is:
[tex]\[ (f \circ g)(x) = \frac{4x - 15}{4x - 12} \][/tex]
[tex]\[ \text{Domain: } (-\infty, 3) \cup (3, \infty) \][/tex]
