Get expert insights and community support for your questions on IDNLearn.com. Discover prompt and accurate responses from our experts, ensuring you get the information you need quickly.
Sagot :
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]
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. Thank you for visiting IDNLearn.com. We’re here to provide clear and concise answers, so visit us again soon.
