Answer :
Sure! Let's go through the problem step by step to find [tex]\((f \circ g)(x)\)[/tex] and its domain.
### 1. Expressions for [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
We start with the given functions:
[tex]\[ f(x) = \frac{1}{x^2 - 9} \][/tex]
[tex]\[ g(x) = \frac{1}{x} \][/tex]
### 2. Composing the functions:
The composition [tex]\(f \circ g\)[/tex] means we apply [tex]\(g(x)\)[/tex] first and then apply [tex]\(f\)[/tex] to the result of [tex]\(g(x)\)[/tex]. Mathematically, this is:
[tex]\[ (f \circ g)(x) = f(g(x)) \][/tex]
Let's substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ f(g(x)) = f\left( \frac{1}{x} \right) \][/tex]
Now we need to find [tex]\(f\left(\frac{1}{x}\right)\)[/tex]:
[tex]\[ f\left( \frac{1}{x} \right) = \frac{1}{\left(\frac{1}{x}\right)^2 - 9} \][/tex]
Simplify the expression inside the function:
[tex]\[ \left( \frac{1}{x} \right)^2 = \frac{1}{x^2} \][/tex]
So,
[tex]\[ f\left( \frac{1}{x} \right) = \frac{1}{\frac{1}{x^2} - 9} \][/tex]
To simplify the denominator:
[tex]\[ \frac{1}{x^2} - 9 = \frac{1 - 9x^2}{x^2} = \frac{1 - 9x^2}{x^2} \][/tex]
So the expression becomes:
[tex]\[ f\left( \frac{1}{x} \right) = \frac{1}{\frac{1 - 9x^2}{x^2}} = \frac{x^2}{1 - 9x^2} \][/tex]
Therefore:
[tex]\[ (f \circ g)(x) = \frac{x^2}{1 - 9x^2} \][/tex]
### 3. Finding the domain of [tex]\((f \circ g)(x)\)[/tex]:
To determine the domain, we need to find the values of [tex]\(x\)[/tex] for which the expression is defined:
1. From [tex]\(g(x)\)[/tex]: [tex]\(g(x) = \frac{1}{x}\)[/tex] is undefined if [tex]\(x = 0\)[/tex].
2. From [tex]\(f(x)\)[/tex]:
- The function [tex]\(f(x)\)[/tex] is undefined when its denominator is zero.
- In the composed function [tex]\(f(g(x))\)[/tex], the argument [tex]\(\frac{1}{x}\)[/tex] must not make the denominator of [tex]\(f\)[/tex] zero.
Let's examine the denominator of [tex]\(f\left( \frac{1}{x} \right)\)[/tex]:
[tex]\[ 1 - 9x^2 = 0 \][/tex]
[tex]\[ 9x^2 = 1 \][/tex]
[tex]\[ x^2 = \frac{1}{9} \][/tex]
[tex]\[ x = \pm \frac{1}{3} \][/tex]
Thus, [tex]\(f(g(x))\)[/tex] is undefined for:
[tex]\[ x = 0, x = \frac{1}{3}, x = -\frac{1}{3} \][/tex]
Therefore, the domain of [tex]\((f \circ g)(x)\)[/tex] is all real numbers except [tex]\(x = 0, \frac{1}{3},\)[/tex] and [tex]\(-\frac{1}{3}\)[/tex].
### Conclusion:
The expression for the composed function [tex]\((f \circ g)(x)\)[/tex] is:
[tex]\[ (f \circ g)(x) = \frac{x^2}{1 - 9x^2} \][/tex]
The domain of [tex]\((f \circ g)(x)\)[/tex] is:
[tex]\[ \{ x \in \mathbb{R} \mid x \neq 0, x \neq \frac{1}{3}, x \neq -\frac{1}{3} \} \][/tex]
### 1. Expressions for [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
We start with the given functions:
[tex]\[ f(x) = \frac{1}{x^2 - 9} \][/tex]
[tex]\[ g(x) = \frac{1}{x} \][/tex]
### 2. Composing the functions:
The composition [tex]\(f \circ g\)[/tex] means we apply [tex]\(g(x)\)[/tex] first and then apply [tex]\(f\)[/tex] to the result of [tex]\(g(x)\)[/tex]. Mathematically, this is:
[tex]\[ (f \circ g)(x) = f(g(x)) \][/tex]
Let's substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ f(g(x)) = f\left( \frac{1}{x} \right) \][/tex]
Now we need to find [tex]\(f\left(\frac{1}{x}\right)\)[/tex]:
[tex]\[ f\left( \frac{1}{x} \right) = \frac{1}{\left(\frac{1}{x}\right)^2 - 9} \][/tex]
Simplify the expression inside the function:
[tex]\[ \left( \frac{1}{x} \right)^2 = \frac{1}{x^2} \][/tex]
So,
[tex]\[ f\left( \frac{1}{x} \right) = \frac{1}{\frac{1}{x^2} - 9} \][/tex]
To simplify the denominator:
[tex]\[ \frac{1}{x^2} - 9 = \frac{1 - 9x^2}{x^2} = \frac{1 - 9x^2}{x^2} \][/tex]
So the expression becomes:
[tex]\[ f\left( \frac{1}{x} \right) = \frac{1}{\frac{1 - 9x^2}{x^2}} = \frac{x^2}{1 - 9x^2} \][/tex]
Therefore:
[tex]\[ (f \circ g)(x) = \frac{x^2}{1 - 9x^2} \][/tex]
### 3. Finding the domain of [tex]\((f \circ g)(x)\)[/tex]:
To determine the domain, we need to find the values of [tex]\(x\)[/tex] for which the expression is defined:
1. From [tex]\(g(x)\)[/tex]: [tex]\(g(x) = \frac{1}{x}\)[/tex] is undefined if [tex]\(x = 0\)[/tex].
2. From [tex]\(f(x)\)[/tex]:
- The function [tex]\(f(x)\)[/tex] is undefined when its denominator is zero.
- In the composed function [tex]\(f(g(x))\)[/tex], the argument [tex]\(\frac{1}{x}\)[/tex] must not make the denominator of [tex]\(f\)[/tex] zero.
Let's examine the denominator of [tex]\(f\left( \frac{1}{x} \right)\)[/tex]:
[tex]\[ 1 - 9x^2 = 0 \][/tex]
[tex]\[ 9x^2 = 1 \][/tex]
[tex]\[ x^2 = \frac{1}{9} \][/tex]
[tex]\[ x = \pm \frac{1}{3} \][/tex]
Thus, [tex]\(f(g(x))\)[/tex] is undefined for:
[tex]\[ x = 0, x = \frac{1}{3}, x = -\frac{1}{3} \][/tex]
Therefore, the domain of [tex]\((f \circ g)(x)\)[/tex] is all real numbers except [tex]\(x = 0, \frac{1}{3},\)[/tex] and [tex]\(-\frac{1}{3}\)[/tex].
### Conclusion:
The expression for the composed function [tex]\((f \circ g)(x)\)[/tex] is:
[tex]\[ (f \circ g)(x) = \frac{x^2}{1 - 9x^2} \][/tex]
The domain of [tex]\((f \circ g)(x)\)[/tex] is:
[tex]\[ \{ x \in \mathbb{R} \mid x \neq 0, x \neq \frac{1}{3}, x \neq -\frac{1}{3} \} \][/tex]
