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Sagot :
To determine whether the given functions are even, odd, or neither, we need to use the definitions and properties of even and odd functions.
### Definitions:
- A function [tex]\( f(x) \)[/tex] is even if [tex]\( f(-x) = f(x) \)[/tex] for all [tex]\( x \)[/tex] in the domain of [tex]\( f \)[/tex].
- A function [tex]\( g(x) \)[/tex] is odd if [tex]\( g(-x) = -g(x) \)[/tex] for all [tex]\( x \)[/tex] in the domain of [tex]\( g \)[/tex].
Given:
1. [tex]\( f(x) \)[/tex] is even.
2. [tex]\( g(x) \)[/tex] is odd.
### Part 1: Determine [tex]\((f \cdot g)(x)\)[/tex]
Let's denote the product of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] as [tex]\( h(x) = (f \cdot g)(x) \)[/tex]. We want to determine if [tex]\( h(x) \)[/tex] is even, odd, or neither.
To do so, we analyze [tex]\( h(-x) \)[/tex]:
[tex]\[ h(-x) = (f \cdot g)(-x) = f(-x) \cdot g(-x) \][/tex]
Using the properties of even and odd functions:
- Since [tex]\( f(x) \)[/tex] is even, we have [tex]\( f(-x) = f(x) \)[/tex].
- Since [tex]\( g(x) \)[/tex] is odd, we have [tex]\( g(-x) = -g(x) \)[/tex].
Substituting these into the expression for [tex]\( h(-x) \)[/tex]:
[tex]\[ h(-x) = f(-x) \cdot g(-x) \][/tex]
[tex]\[ h(-x) = f(x) \cdot (-g(x)) \][/tex]
[tex]\[ h(-x) = - (f(x) \cdot g(x)) \][/tex]
[tex]\[ h(-x) = -h(x) \][/tex]
Since [tex]\( h(-x) = -h(x) \)[/tex], the function [tex]\( h(x) = (f \cdot g)(x) \)[/tex] is odd.
### Part 2: Determine [tex]\((g \cdot g)(x)\)[/tex]
Let's denote the product of [tex]\( g(x) \)[/tex] with itself as [tex]\( k(x) = (g \cdot g)(x) \)[/tex]. We want to determine if [tex]\( k(x) \)[/tex] is even, odd, or neither.
To do so, we analyze [tex]\( k(-x) \)[/tex]:
[tex]\[ k(-x) = (g \cdot g)(-x) = g(-x) \cdot g(-x) \][/tex]
Using the property of odd functions:
- Since [tex]\( g(x) \)[/tex] is odd, we have [tex]\( g(-x) = -g(x) \)[/tex].
Substituting this into the expression for [tex]\( k(-x) \)[/tex]:
[tex]\[ k(-x) = (-g(x)) \cdot (-g(x)) \][/tex]
[tex]\[ k(-x) = g(x) \cdot g(x) \][/tex]
[tex]\[ k(-x) = k(x) \][/tex]
Since [tex]\( k(-x) = k(x) \)[/tex], the function [tex]\( k(x) = (g \cdot g)(x) \)[/tex] is even.
### Conclusion:
- [tex]\((f \cdot g)(x)\)[/tex] is odd.
- [tex]\((g \cdot g)(x)\)[/tex] is even.
Thus, we have:
[tex]\[ \begin{array}{l} (f \cdot g)(x)=\text{odd} \quad \\ (g \cdot g)(x)=\text{even} \end{array} \][/tex]
### Definitions:
- A function [tex]\( f(x) \)[/tex] is even if [tex]\( f(-x) = f(x) \)[/tex] for all [tex]\( x \)[/tex] in the domain of [tex]\( f \)[/tex].
- A function [tex]\( g(x) \)[/tex] is odd if [tex]\( g(-x) = -g(x) \)[/tex] for all [tex]\( x \)[/tex] in the domain of [tex]\( g \)[/tex].
Given:
1. [tex]\( f(x) \)[/tex] is even.
2. [tex]\( g(x) \)[/tex] is odd.
### Part 1: Determine [tex]\((f \cdot g)(x)\)[/tex]
Let's denote the product of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] as [tex]\( h(x) = (f \cdot g)(x) \)[/tex]. We want to determine if [tex]\( h(x) \)[/tex] is even, odd, or neither.
To do so, we analyze [tex]\( h(-x) \)[/tex]:
[tex]\[ h(-x) = (f \cdot g)(-x) = f(-x) \cdot g(-x) \][/tex]
Using the properties of even and odd functions:
- Since [tex]\( f(x) \)[/tex] is even, we have [tex]\( f(-x) = f(x) \)[/tex].
- Since [tex]\( g(x) \)[/tex] is odd, we have [tex]\( g(-x) = -g(x) \)[/tex].
Substituting these into the expression for [tex]\( h(-x) \)[/tex]:
[tex]\[ h(-x) = f(-x) \cdot g(-x) \][/tex]
[tex]\[ h(-x) = f(x) \cdot (-g(x)) \][/tex]
[tex]\[ h(-x) = - (f(x) \cdot g(x)) \][/tex]
[tex]\[ h(-x) = -h(x) \][/tex]
Since [tex]\( h(-x) = -h(x) \)[/tex], the function [tex]\( h(x) = (f \cdot g)(x) \)[/tex] is odd.
### Part 2: Determine [tex]\((g \cdot g)(x)\)[/tex]
Let's denote the product of [tex]\( g(x) \)[/tex] with itself as [tex]\( k(x) = (g \cdot g)(x) \)[/tex]. We want to determine if [tex]\( k(x) \)[/tex] is even, odd, or neither.
To do so, we analyze [tex]\( k(-x) \)[/tex]:
[tex]\[ k(-x) = (g \cdot g)(-x) = g(-x) \cdot g(-x) \][/tex]
Using the property of odd functions:
- Since [tex]\( g(x) \)[/tex] is odd, we have [tex]\( g(-x) = -g(x) \)[/tex].
Substituting this into the expression for [tex]\( k(-x) \)[/tex]:
[tex]\[ k(-x) = (-g(x)) \cdot (-g(x)) \][/tex]
[tex]\[ k(-x) = g(x) \cdot g(x) \][/tex]
[tex]\[ k(-x) = k(x) \][/tex]
Since [tex]\( k(-x) = k(x) \)[/tex], the function [tex]\( k(x) = (g \cdot g)(x) \)[/tex] is even.
### Conclusion:
- [tex]\((f \cdot g)(x)\)[/tex] is odd.
- [tex]\((g \cdot g)(x)\)[/tex] is even.
Thus, we have:
[tex]\[ \begin{array}{l} (f \cdot g)(x)=\text{odd} \quad \\ (g \cdot g)(x)=\text{even} \end{array} \][/tex]
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