Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Get detailed and precise answers to your questions from a dedicated community of experts on our Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Sure! Let's break down and solve each part of this question step-by-step.
### Part (a) Finding [tex]\((f \circ g)(x)\)[/tex]:
The notation [tex]\((f \circ g)(x)\)[/tex] represents the composition of the functions [tex]\(f\)[/tex] and [tex]\(g\)[/tex], defined as [tex]\((f \circ g)(x) = f(g(x))\)[/tex]. This means we first apply the function [tex]\(g\)[/tex] to [tex]\(x\)[/tex] and then apply the function [tex]\(f\)[/tex] to the result.
Given:
[tex]\[ f(x) = \frac{5}{x+2} \][/tex]
[tex]\[ g(x) = \frac{1}{x} \][/tex]
First, we find [tex]\(g(x)\)[/tex]:
[tex]\[ g(x) = \frac{1}{x} \][/tex]
Next, we substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{1}{x}\right) = \frac{5}{\left(\frac{1}{x}\right) + 2} \][/tex]
To simplify, we need a common denominator:
[tex]\[ f\left(\frac{1}{x}\right) = \frac{5}{\frac{1 + 2x}{x}} \][/tex]
Recall that dividing by a fraction is the same as multiplying by its reciprocal:
[tex]\[ f\left(\frac{1}{x}\right) = 5 \cdot \frac{x}{1 + 2x} = \frac{5x}{1 + 2x} \][/tex]
Therefore,
[tex]\[ (f \circ g)(x) = \frac{5x}{1 + 2x} \][/tex]
### Part (b) Finding the domain of [tex]\(f \circ g\)[/tex]:
To determine the domain of [tex]\(f \circ g\)[/tex], we need to consider the domains of both [tex]\(f\)[/tex] and [tex]\(g\)[/tex] individually and how they interact in the composition.
1. Domain of [tex]\(g(x) = \frac{1}{x}\)[/tex]:
- [tex]\(g(x)\)[/tex] is defined for all real numbers except [tex]\(x = 0\)[/tex] (since division by zero is undefined).
- Thus, the domain of [tex]\(g\)[/tex] is [tex]\(\mathbb{R} \setminus \{0\}\)[/tex] (all real numbers except [tex]\(0\)[/tex]).
2. Domain of [tex]\(f(x) = \frac{5}{x+2}\)[/tex]:
- [tex]\(f(x)\)[/tex] is defined for all real numbers except [tex]\(x = -2\)[/tex] (since division by zero is undefined).
- In the composition [tex]\(f(g(x))\)[/tex], we need to ensure that the output of [tex]\(g(x)\)[/tex] is within the domain of [tex]\(f\)[/tex].
- So, we require [tex]\(g(x) \neq -2\)[/tex].
3. Find values of [tex]\(x\)[/tex] for which [tex]\(g(x) = -2\)[/tex]:
- Set [tex]\(g(x) = -2\)[/tex]:
[tex]\[ \frac{1}{x} = -2 \][/tex]
- Solve for [tex]\(x\)[/tex]:
[tex]\[ x = -\frac{1}{2} \][/tex]
Therefore, [tex]\(x = -\frac{1}{2}\)[/tex] makes the output of [tex]\(g(x) = -2\)[/tex], which is not in the domain of [tex]\(f\)[/tex].
4. Combining the constraints:
- [tex]\(x\)[/tex] must be in the domain of [tex]\(g\)[/tex], which is [tex]\(x \neq 0\)[/tex].
- [tex]\(x\)[/tex] must also not make [tex]\(g(x) = -2\)[/tex], which means [tex]\(x \neq -\frac{1}{2}\)[/tex].
Therefore, the domain of [tex]\(f \circ g\)[/tex] is:
[tex]\[ \boxed{\text{All real numbers except } 0 \text{ and } -\frac{1}{2}} \][/tex]
### Part (a) Finding [tex]\((f \circ g)(x)\)[/tex]:
The notation [tex]\((f \circ g)(x)\)[/tex] represents the composition of the functions [tex]\(f\)[/tex] and [tex]\(g\)[/tex], defined as [tex]\((f \circ g)(x) = f(g(x))\)[/tex]. This means we first apply the function [tex]\(g\)[/tex] to [tex]\(x\)[/tex] and then apply the function [tex]\(f\)[/tex] to the result.
Given:
[tex]\[ f(x) = \frac{5}{x+2} \][/tex]
[tex]\[ g(x) = \frac{1}{x} \][/tex]
First, we find [tex]\(g(x)\)[/tex]:
[tex]\[ g(x) = \frac{1}{x} \][/tex]
Next, we substitute [tex]\(g(x)\)[/tex] into [tex]\(f(x)\)[/tex]:
[tex]\[ f(g(x)) = f\left(\frac{1}{x}\right) = \frac{5}{\left(\frac{1}{x}\right) + 2} \][/tex]
To simplify, we need a common denominator:
[tex]\[ f\left(\frac{1}{x}\right) = \frac{5}{\frac{1 + 2x}{x}} \][/tex]
Recall that dividing by a fraction is the same as multiplying by its reciprocal:
[tex]\[ f\left(\frac{1}{x}\right) = 5 \cdot \frac{x}{1 + 2x} = \frac{5x}{1 + 2x} \][/tex]
Therefore,
[tex]\[ (f \circ g)(x) = \frac{5x}{1 + 2x} \][/tex]
### Part (b) Finding the domain of [tex]\(f \circ g\)[/tex]:
To determine the domain of [tex]\(f \circ g\)[/tex], we need to consider the domains of both [tex]\(f\)[/tex] and [tex]\(g\)[/tex] individually and how they interact in the composition.
1. Domain of [tex]\(g(x) = \frac{1}{x}\)[/tex]:
- [tex]\(g(x)\)[/tex] is defined for all real numbers except [tex]\(x = 0\)[/tex] (since division by zero is undefined).
- Thus, the domain of [tex]\(g\)[/tex] is [tex]\(\mathbb{R} \setminus \{0\}\)[/tex] (all real numbers except [tex]\(0\)[/tex]).
2. Domain of [tex]\(f(x) = \frac{5}{x+2}\)[/tex]:
- [tex]\(f(x)\)[/tex] is defined for all real numbers except [tex]\(x = -2\)[/tex] (since division by zero is undefined).
- In the composition [tex]\(f(g(x))\)[/tex], we need to ensure that the output of [tex]\(g(x)\)[/tex] is within the domain of [tex]\(f\)[/tex].
- So, we require [tex]\(g(x) \neq -2\)[/tex].
3. Find values of [tex]\(x\)[/tex] for which [tex]\(g(x) = -2\)[/tex]:
- Set [tex]\(g(x) = -2\)[/tex]:
[tex]\[ \frac{1}{x} = -2 \][/tex]
- Solve for [tex]\(x\)[/tex]:
[tex]\[ x = -\frac{1}{2} \][/tex]
Therefore, [tex]\(x = -\frac{1}{2}\)[/tex] makes the output of [tex]\(g(x) = -2\)[/tex], which is not in the domain of [tex]\(f\)[/tex].
4. Combining the constraints:
- [tex]\(x\)[/tex] must be in the domain of [tex]\(g\)[/tex], which is [tex]\(x \neq 0\)[/tex].
- [tex]\(x\)[/tex] must also not make [tex]\(g(x) = -2\)[/tex], which means [tex]\(x \neq -\frac{1}{2}\)[/tex].
Therefore, the domain of [tex]\(f \circ g\)[/tex] is:
[tex]\[ \boxed{\text{All real numbers except } 0 \text{ and } -\frac{1}{2}} \][/tex]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.