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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]
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