Answer :
Sure, let's go through each of the composite functions one by one and evaluate their values with an example at [tex]\(x = 0\)[/tex]. We'll also describe the domain of each resulting function.
Given:
[tex]\[ f(x) = 3 - 2x \][/tex]
[tex]\[ g(x) = 6 + x^2 \][/tex]
### a. [tex]\( f \circ g \)[/tex]
The composite function [tex]\( f \circ g \)[/tex] is defined as:
[tex]\[ (f \circ g)(x) = f(g(x)) \][/tex]
First, compute [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = 6 + x^2 \][/tex]
Then, substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(g(x)) = f(6 + x^2) \][/tex]
Using the definition of [tex]\( f(x) \)[/tex]:
[tex]\[ f(6 + x^2) = 3 - 2(6 + x^2) \][/tex]
[tex]\[ = 3 - 12 - 2x^2 \][/tex]
[tex]\[ = -9 - 2x^2 \][/tex]
So, the composite function [tex]\( f \circ g \)[/tex] is:
[tex]\[ (f \circ g)(x) = -9 - 2x^2 \][/tex]
Domain of [tex]\( f \circ g \)[/tex]:
The domain of [tex]\( f(x) \)[/tex] is all real numbers. The domain of [tex]\( g(x) \)[/tex] is also all real numbers. Therefore, the domain of [tex]\( f \circ g \)[/tex] is all real numbers.
Example Calculation at [tex]\( x = 0 \)[/tex]:
[tex]\[ (f \circ g)(0) = -9 \][/tex]
### b. [tex]\( g \circ f \)[/tex]
The composite function [tex]\( g \circ f \)[/tex] is defined as:
[tex]\[ (g \circ f)(x) = g(f(x)) \][/tex]
First, compute [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 3 - 2x \][/tex]
Then, substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(f(x)) = g(3 - 2x) \][/tex]
Using the definition of [tex]\( g(x) \)[/tex]:
[tex]\[ g(3 - 2x) = 6 + (3 - 2x)^2 \][/tex]
[tex]\[ = 6 + (9 - 12x + 4x^2) \][/tex]
[tex]\[ = 6 + 9 - 12x + 4x^2 \][/tex]
[tex]\[ = 15 - 12x + 4x^2 \][/tex]
So, the composite function [tex]\( g \circ f \)[/tex] is:
[tex]\[ (g \circ f)(x) = 15 - 12x + 4x^2 \][/tex]
Domain of [tex]\( g \circ f \)[/tex]:
The domain of [tex]\( f(x) \)[/tex] is all real numbers. The domain of [tex]\( g(x) \)[/tex] is also all real numbers. Therefore, the domain of [tex]\( g \circ f \)[/tex] is all real numbers.
Example Calculation at [tex]\( x = 0 \)[/tex]:
[tex]\[ (g \circ f)(0) = 15 \][/tex]
### c. [tex]\( f \circ f \)[/tex]
The composite function [tex]\( f \circ f \)[/tex] is defined as:
[tex]\[ (f \circ f)(x) = f(f(x)) \][/tex]
First, compute [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 3 - 2x \][/tex]
Then, substitute [tex]\( f(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(f(x)) = f(3 - 2x) \][/tex]
Using the definition of [tex]\( f(x) \)[/tex]:
[tex]\[ f(3 - 2x) = 3 - 2(3 - 2x) \][/tex]
[tex]\[ = 3 - 6 + 4x \][/tex]
[tex]\[ = -3 + 4x \][/tex]
So, the composite function [tex]\( f \circ f \)[/tex] is:
[tex]\[ (f \circ f)(x) = -3 + 4x \][/tex]
Domain of [tex]\( f \circ f \)[/tex]:
The domain of [tex]\( f(x) \)[/tex] is all real numbers. Therefore, the domain of [tex]\( f \circ f \)[/tex] is all real numbers.
Example Calculation at [tex]\( x = 0 \)[/tex]:
[tex]\[ (f \circ f)(0) = -3 \][/tex]
### d. [tex]\( g \circ g \)[/tex]
The composite function [tex]\( g \circ g \)[/tex] is defined as:
[tex]\[ (g \circ g)(x) = g(g(x)) \][/tex]
First, compute [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = 6 + x^2 \][/tex]
Then, substitute [tex]\( g(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(g(x)) = g(6 + x^2) \][/tex]
Using the definition of [tex]\( g(x) \)[/tex]:
[tex]\[ g(6 + x^2) = 6 + (6 + x^2)^2 \][/tex]
[tex]\[ = 6 + (36 + 12x^2 + x^4) \][/tex]
[tex]\[ = 6 + 36 + 12x^2 + x^4 \][/tex]
[tex]\[ = 42 + 12x^2 + x^4 \][/tex]
So, the composite function [tex]\( g \circ g \)[/tex] is:
[tex]\[ (g \circ g)(x) = 42 + 12x^2 + x^4 \][/tex]
Domain of [tex]\( g \circ g \)[/tex]:
The domain of [tex]\( g(x) \)[/tex] is all real numbers. Therefore, the domain of [tex]\( g \circ g \)[/tex] is all real numbers.
Example Calculation at [tex]\( x = 0 \)[/tex]:
[tex]\[ (g \circ g)(0) = 42 \][/tex]
Given:
[tex]\[ f(x) = 3 - 2x \][/tex]
[tex]\[ g(x) = 6 + x^2 \][/tex]
### a. [tex]\( f \circ g \)[/tex]
The composite function [tex]\( f \circ g \)[/tex] is defined as:
[tex]\[ (f \circ g)(x) = f(g(x)) \][/tex]
First, compute [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = 6 + x^2 \][/tex]
Then, substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(g(x)) = f(6 + x^2) \][/tex]
Using the definition of [tex]\( f(x) \)[/tex]:
[tex]\[ f(6 + x^2) = 3 - 2(6 + x^2) \][/tex]
[tex]\[ = 3 - 12 - 2x^2 \][/tex]
[tex]\[ = -9 - 2x^2 \][/tex]
So, the composite function [tex]\( f \circ g \)[/tex] is:
[tex]\[ (f \circ g)(x) = -9 - 2x^2 \][/tex]
Domain of [tex]\( f \circ g \)[/tex]:
The domain of [tex]\( f(x) \)[/tex] is all real numbers. The domain of [tex]\( g(x) \)[/tex] is also all real numbers. Therefore, the domain of [tex]\( f \circ g \)[/tex] is all real numbers.
Example Calculation at [tex]\( x = 0 \)[/tex]:
[tex]\[ (f \circ g)(0) = -9 \][/tex]
### b. [tex]\( g \circ f \)[/tex]
The composite function [tex]\( g \circ f \)[/tex] is defined as:
[tex]\[ (g \circ f)(x) = g(f(x)) \][/tex]
First, compute [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 3 - 2x \][/tex]
Then, substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(f(x)) = g(3 - 2x) \][/tex]
Using the definition of [tex]\( g(x) \)[/tex]:
[tex]\[ g(3 - 2x) = 6 + (3 - 2x)^2 \][/tex]
[tex]\[ = 6 + (9 - 12x + 4x^2) \][/tex]
[tex]\[ = 6 + 9 - 12x + 4x^2 \][/tex]
[tex]\[ = 15 - 12x + 4x^2 \][/tex]
So, the composite function [tex]\( g \circ f \)[/tex] is:
[tex]\[ (g \circ f)(x) = 15 - 12x + 4x^2 \][/tex]
Domain of [tex]\( g \circ f \)[/tex]:
The domain of [tex]\( f(x) \)[/tex] is all real numbers. The domain of [tex]\( g(x) \)[/tex] is also all real numbers. Therefore, the domain of [tex]\( g \circ f \)[/tex] is all real numbers.
Example Calculation at [tex]\( x = 0 \)[/tex]:
[tex]\[ (g \circ f)(0) = 15 \][/tex]
### c. [tex]\( f \circ f \)[/tex]
The composite function [tex]\( f \circ f \)[/tex] is defined as:
[tex]\[ (f \circ f)(x) = f(f(x)) \][/tex]
First, compute [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 3 - 2x \][/tex]
Then, substitute [tex]\( f(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(f(x)) = f(3 - 2x) \][/tex]
Using the definition of [tex]\( f(x) \)[/tex]:
[tex]\[ f(3 - 2x) = 3 - 2(3 - 2x) \][/tex]
[tex]\[ = 3 - 6 + 4x \][/tex]
[tex]\[ = -3 + 4x \][/tex]
So, the composite function [tex]\( f \circ f \)[/tex] is:
[tex]\[ (f \circ f)(x) = -3 + 4x \][/tex]
Domain of [tex]\( f \circ f \)[/tex]:
The domain of [tex]\( f(x) \)[/tex] is all real numbers. Therefore, the domain of [tex]\( f \circ f \)[/tex] is all real numbers.
Example Calculation at [tex]\( x = 0 \)[/tex]:
[tex]\[ (f \circ f)(0) = -3 \][/tex]
### d. [tex]\( g \circ g \)[/tex]
The composite function [tex]\( g \circ g \)[/tex] is defined as:
[tex]\[ (g \circ g)(x) = g(g(x)) \][/tex]
First, compute [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = 6 + x^2 \][/tex]
Then, substitute [tex]\( g(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(g(x)) = g(6 + x^2) \][/tex]
Using the definition of [tex]\( g(x) \)[/tex]:
[tex]\[ g(6 + x^2) = 6 + (6 + x^2)^2 \][/tex]
[tex]\[ = 6 + (36 + 12x^2 + x^4) \][/tex]
[tex]\[ = 6 + 36 + 12x^2 + x^4 \][/tex]
[tex]\[ = 42 + 12x^2 + x^4 \][/tex]
So, the composite function [tex]\( g \circ g \)[/tex] is:
[tex]\[ (g \circ g)(x) = 42 + 12x^2 + x^4 \][/tex]
Domain of [tex]\( g \circ g \)[/tex]:
The domain of [tex]\( g(x) \)[/tex] is all real numbers. Therefore, the domain of [tex]\( g \circ g \)[/tex] is all real numbers.
Example Calculation at [tex]\( x = 0 \)[/tex]:
[tex]\[ (g \circ g)(0) = 42 \][/tex]
