Answer :
To answer the given question, we need to perform the following steps:
### Part (a): Find [tex]\((f \circ g)(x)\)[/tex]
To find the composition of the functions [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex], we substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
1. Define the functions:
[tex]\[ f(x) = \frac{2}{x + 7} \][/tex]
[tex]\[ g(x) = \frac{1}{x} \][/tex]
2. Substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ (f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) \][/tex]
3. Replace [tex]\(x\)[/tex] in [tex]\(f(x)\)[/tex] with [tex]\(\frac{1}{x}\)[/tex]:
[tex]\[ f\left(\frac{1}{x}\right) = \frac{2}{\frac{1}{x} + 7} \][/tex]
4. Simplify the expression:
We need a common denominator to combine the terms in the denominator:
[tex]\[ \frac{1}{x} + 7 = \frac{1 + 7x}{x} \][/tex]
5. Simplify the fraction:
[tex]\[ f\left(\frac{1}{x}\right) = \frac{2}{\frac{1 + 7x}{x}} = 2 \cdot \frac{x}{1 + 7x} = \frac{2x}{1 + 7x} \][/tex]
Therefore, the function [tex]\((f \circ g)(x) = \frac{2}{7 + \frac{1}{x}} = \frac{2}{\frac{7x + 1}{x}} = \frac{2x}{7x + 1} \)[/tex].
However given the result above true is:
[tex]\[ (f \circ g)(x) = \frac{2}{7 + \frac{1}{x}} \][/tex]
### Part (b): Determine the domain of [tex]\(f \circ g\)[/tex]
To determine the domain of the composition [tex]\(f \circ g\)[/tex], we need to consider the domains of both [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex] and identify any restrictions:
1. Domain of [tex]\(g(x) = \frac{1}{x}\)[/tex]:
- [tex]\(g(x)\)[/tex] is defined for all [tex]\(x \neq 0\)[/tex]. Therefore, the domain of [tex]\(g(x)\)[/tex] is all real numbers except [tex]\(0\)[/tex].
2. Considering the composition [tex]\(f(g(x))\)[/tex]:
- Next, we need to identify where [tex]\(f\left(\frac{1}{x}\right)\)[/tex] is defined.
- We must have [tex]\(x \neq 0\)[/tex] initially for [tex]\(g(x)\)[/tex].
- Then, for [tex]\(f(x)\)[/tex], we need to ensure that the expression inside is not causing any restrictions. For [tex]\(f(x) = \frac{2}{x + 7}\)[/tex], [tex]\(x + 7 \neq 0\)[/tex], which means [tex]\(x \neq -7\)[/tex].
- Thus, for [tex]\(f\left(g(x)\right) = f\left(\frac{1}{x}\right)\)[/tex], the composite function [tex]\(f(g(x)) = \frac{2}{7 + \frac{1}{x}}\)[/tex] needs to check if any more restrictions in domain arise. Most importantly here, One would identify that input to [tex]\(g\)[/tex] fail to satisfy condition for [tex]\(x = g^{-1}(-7)\)[/tex]. This means, one need to exclude like above , and impacts original domain subset.
Combining these, we need to exclude both [tex]\(x = 0\)[/tex] and solutions that caused potential issue with inclusion like [tex]\(x = 0(-7x{\textstyle domain } )\)[/tex]. Thus the composite function ends up with no practical value unless clarified, hence Empty set derived, correctly [tex]\( i.e. \boxed{\emptyset }\)[/tex].
### Part (a): Find [tex]\((f \circ g)(x)\)[/tex]
To find the composition of the functions [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex], we substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
1. Define the functions:
[tex]\[ f(x) = \frac{2}{x + 7} \][/tex]
[tex]\[ g(x) = \frac{1}{x} \][/tex]
2. Substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ (f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) \][/tex]
3. Replace [tex]\(x\)[/tex] in [tex]\(f(x)\)[/tex] with [tex]\(\frac{1}{x}\)[/tex]:
[tex]\[ f\left(\frac{1}{x}\right) = \frac{2}{\frac{1}{x} + 7} \][/tex]
4. Simplify the expression:
We need a common denominator to combine the terms in the denominator:
[tex]\[ \frac{1}{x} + 7 = \frac{1 + 7x}{x} \][/tex]
5. Simplify the fraction:
[tex]\[ f\left(\frac{1}{x}\right) = \frac{2}{\frac{1 + 7x}{x}} = 2 \cdot \frac{x}{1 + 7x} = \frac{2x}{1 + 7x} \][/tex]
Therefore, the function [tex]\((f \circ g)(x) = \frac{2}{7 + \frac{1}{x}} = \frac{2}{\frac{7x + 1}{x}} = \frac{2x}{7x + 1} \)[/tex].
However given the result above true is:
[tex]\[ (f \circ g)(x) = \frac{2}{7 + \frac{1}{x}} \][/tex]
### Part (b): Determine the domain of [tex]\(f \circ g\)[/tex]
To determine the domain of the composition [tex]\(f \circ g\)[/tex], we need to consider the domains of both [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex] and identify any restrictions:
1. Domain of [tex]\(g(x) = \frac{1}{x}\)[/tex]:
- [tex]\(g(x)\)[/tex] is defined for all [tex]\(x \neq 0\)[/tex]. Therefore, the domain of [tex]\(g(x)\)[/tex] is all real numbers except [tex]\(0\)[/tex].
2. Considering the composition [tex]\(f(g(x))\)[/tex]:
- Next, we need to identify where [tex]\(f\left(\frac{1}{x}\right)\)[/tex] is defined.
- We must have [tex]\(x \neq 0\)[/tex] initially for [tex]\(g(x)\)[/tex].
- Then, for [tex]\(f(x)\)[/tex], we need to ensure that the expression inside is not causing any restrictions. For [tex]\(f(x) = \frac{2}{x + 7}\)[/tex], [tex]\(x + 7 \neq 0\)[/tex], which means [tex]\(x \neq -7\)[/tex].
- Thus, for [tex]\(f\left(g(x)\right) = f\left(\frac{1}{x}\right)\)[/tex], the composite function [tex]\(f(g(x)) = \frac{2}{7 + \frac{1}{x}}\)[/tex] needs to check if any more restrictions in domain arise. Most importantly here, One would identify that input to [tex]\(g\)[/tex] fail to satisfy condition for [tex]\(x = g^{-1}(-7)\)[/tex]. This means, one need to exclude like above , and impacts original domain subset.
Combining these, we need to exclude both [tex]\(x = 0\)[/tex] and solutions that caused potential issue with inclusion like [tex]\(x = 0(-7x{\textstyle domain } )\)[/tex]. Thus the composite function ends up with no practical value unless clarified, hence Empty set derived, correctly [tex]\( i.e. \boxed{\emptyset }\)[/tex].
