To solve the problem, we will go through each part step-by-step.
### Part a: Finding [tex]\((f \circ g)(x)\)[/tex]:
1. Understand the Composite Function Notation:
[tex]\((f \circ g)(x)\)[/tex] means [tex]\(f(g(x))\)[/tex].
2. Substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
Given [tex]\(f(x) = \frac{9}{x+6}\)[/tex] and [tex]\(g(x) = \frac{7}{x}\)[/tex], we need to find [tex]\(f(g(x))\)[/tex].
3. Calculate [tex]\(g(x)\)[/tex]:
[tex]\(g(x) = \frac{7}{x}\)[/tex].
4. Substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[
f(g(x)) = f\left(\frac{7}{x}\right)
\][/tex]
Since [tex]\(f(x) = \frac{9}{x+6}\)[/tex], we substitute [tex]\(\frac{7}{x}\)[/tex] for [tex]\(x\)[/tex]:
[tex]\[
f\left(\frac{7}{x}\right) = \frac{9}{\left(\frac{7}{x}\right) + 6}
\][/tex]
5. Simplify the Expression:
[tex]\[
f\left(\frac{7}{x}\right) = \frac{9}{\frac{7}{x} + 6}
\][/tex]
To combine the terms in the denominator, get a common denominator:
[tex]\[
\frac{7}{x} + 6 = \frac{7 + 6x}{x}
\][/tex]
So the expression is:
[tex]\[
f\left(\frac{7}{x}\right) = \frac{9}{\frac{7 + 6x}{x}} = \frac{9x}{7 + 6x}
\][/tex]
Thus,
[tex]\[
(f \circ g)(x) = \frac{9x}{7 + 6x}
\][/tex]
### Part b: Finding the Domain of [tex]\(f \circ g\)[/tex]:
1. Identify the Restrictions:
- The function [tex]\(g(x) = \frac{7}{x}\)[/tex] is not defined when [tex]\(x = 0\)[/tex] because division by zero is not allowed.
- Additionally, [tex]\(f(x) = \frac{9}{x+6}\)[/tex] is not defined when its denominator is zero: [tex]\(x + 6 \neq 0 \implies x \neq -6\)[/tex].
- We substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]: [tex]\(f(g(x)) = f\left(\frac{7}{x}\right)\)[/tex] requires that [tex]\(\frac{7}{x} + 6 \neq 0\)[/tex]
[tex]\[
\frac{7}{x} + 6 = 0 \implies \frac{7}{x} = -6 \implies 7 = -6x \implies x = -\frac{7}{6}
\][/tex]
So, [tex]\(x \neq -\frac{7}{6}\)[/tex].
2. Combine the Restrictions:
Taking all restrictions into account:
- [tex]\(x \neq 0\)[/tex]
- [tex]\(x \neq -\frac{7}{6}\)[/tex]
The domain of [tex]\(f \circ g\)[/tex] is:
[tex]\[
(-\infty, -\frac{7}{6}) \cup (-\frac{7}{6}, 0) \cup (0, \infty)
\][/tex]
Therefore, the answers are:
a. [tex]\((f \circ g)(x) = \frac{9x}{7 + 6x}\)[/tex]
b. The domain of [tex]\(f \circ g\)[/tex] is
D. [tex]\(\left(-\infty, -\frac{7}{6}\right) \cup \left(-\frac{7}{6}, 0\right) \cup (0, \infty)\)[/tex]
