To find the domain of the composite function [tex]\((f \circ g)(x)\)[/tex], we need to determine where both functions are defined and where the composite function itself is defined. Let's break this problem into steps.
1. Identify the domain of [tex]\(g(x)\)[/tex]:
The function [tex]\(g(x) = \frac{1}{-6x + 5}\)[/tex] is defined for all [tex]\(x\)[/tex] except where the denominator is zero. Set the denominator equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[
-6x + 5 = 0 \implies x = \frac{5}{6}
\][/tex]
Therefore, [tex]\(g(x)\)[/tex] is defined for all [tex]\(x \neq \frac{5}{6}\)[/tex].
2. Identify the domain of [tex]\(f(g(x))\)[/tex]:
The function [tex]\(f(x) = \frac{1}{-10x - 7}\)[/tex] requires that [tex]\(x \neq -\frac{7}{10}\)[/tex] to be defined. Therefore, [tex]\(f\)[/tex] is defined for all [tex]\(x \neq -\frac{7}{10}\)[/tex]. In our case, we need this to [tex]\(f(g(x))\)[/tex].
To find where [tex]\(f(g(x))\)[/tex] is defined, we substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[
f(g(x)) = f\left( \frac{1}{-6x + 5} \right) = \frac{1}{-10 \left( \frac{1}{-6x + 5} \right) - 7}
\][/tex]
3. Determine when [tex]\(f(g(x))\)[/tex] is undefined:
The function [tex]\(f(g(x))\)[/tex] becomes undefined when its denominator is zero:
[tex]\[
-10 \left( \frac{1}{-6x + 5} \right) - 7 = 0
\][/tex]
Simplify inside the denominator:
[tex]\[
- \frac{10}{-6x + 5} - 7 = 0
\][/tex]
Rearrange to solve for [tex]\(x\)[/tex]:
[tex]\[
- \frac{10}{-6x + 5} = 7 \implies \frac{10}{6x - 5} = 7 \implies 10 = 7(6x - 5) \implies 10 = 42x - 35 \implies 42x = 45 \implies x = \frac{45}{42} = \frac{15}{14}
\][/tex]
Therefore, the domain of the composite function [tex]\((f \circ g)(x)\)[/tex] is all real numbers except where the original [tex]\(g(x)\)[/tex] or the intermediate steps of [tex]\(f(g(x))\)[/tex] become undefined.
Given the calculations:
- [tex]\(g(x)\)[/tex] is undefined at [tex]\(x = \frac{5}{6}}.
- \(f(g(x))\)[/tex] is undefined at [tex]\(x = \frac{15}{14}}.
Thus, the domain of \((f \circ g)(x)\)[/tex] is all real numbers except [tex]\(x = \frac{5}{6}\)[/tex] and [tex]\(x = \frac{15}{14}\)[/tex].
The correct answer is:
All real numbers except [tex]\(\frac{5}{6}\)[/tex] and [tex]\(\frac{15}{14} \)[/tex].
Therefore, the correct option is:
#### All real numbers except [tex]\(\frac{5}{6}\)[/tex] and [tex]\(\frac{15}{14}\)[/tex].
