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Given the functions below, find the domain of [tex]\( (f \circ g)(x) \)[/tex].

[tex]\( f(x) = \frac{1}{-10x-7} \)[/tex]
[tex]\( g(x) = \frac{1}{-6x+5} \)[/tex]

Select the correct answer below:
A. All real numbers except [tex]\( \frac{5}{6} \)[/tex] and [tex]\( \frac{15}{14} \)[/tex]
B. All real numbers except [tex]\( \frac{5}{6} \)[/tex] and [tex]\( \frac{47}{42} \)[/tex]
C. All real numbers except [tex]\( \frac{5}{6} \)[/tex] and [tex]\( \frac{41}{42} \)[/tex]
D. All real numbers except [tex]\( \frac{5}{6} \)[/tex] and [tex]\( \frac{13}{14} \)[/tex]
E. All real numbers except [tex]\( \frac{5}{6} \)[/tex] and [tex]\( \frac{53}{42} \)[/tex]
F. All real numbers except [tex]\( \frac{5}{6} \)[/tex] and [tex]\( -\frac{7}{10} \)[/tex]

Sagot :

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].