Sure! Let's analyze the properties of the functions given:
First, recall the definitions:
- A function [tex]\( f(x) \)[/tex] is even if [tex]\( f(-x) = f(x) \)[/tex] for all [tex]\( x \)[/tex].
- A function [tex]\( g(x) \)[/tex] is odd if [tex]\( g(-x) = -g(x) \)[/tex] for all [tex]\( x \)[/tex].
### 1. Determining [tex]\((f \cdot g)(x)\)[/tex]:
Consider the function [tex]\((f \cdot g)(x)\)[/tex]. Here, we multiply the even function [tex]\( f(x) \)[/tex] by the odd function [tex]\( g(x) \)[/tex].
To determine if [tex]\((f \cdot g)(x)\)[/tex] is even, odd, or neither:
1. Evaluate [tex]\((f \cdot g)(-x)\)[/tex]:
[tex]\[
(f \cdot g)(-x) = f(-x) \cdot g(-x)
\][/tex]
2. Since [tex]\( f(x) \)[/tex] is even, [tex]\( f(-x) = f(x) \)[/tex]. Since [tex]\( g(x) \)[/tex] is odd, [tex]\( g(-x) = -g(x) \)[/tex]. Substitute these identities in:
[tex]\[
(f \cdot g)(-x) = f(x) \cdot (-g(x)) = - f(x) \cdot g(x)
\][/tex]
This shows that:
[tex]\[
(f \cdot g)(-x) = - (f \cdot g)(x)
\][/tex]
Therefore, the function [tex]\((f \cdot g)(x)\)[/tex] satisfies the condition for being odd. So:
[tex]\[
(f \cdot g)(x) = \text{odd}
\][/tex]
### 2. Determining [tex]\((g \cdot g)(x)\)[/tex]:
Next, consider the function [tex]\((g \cdot g)(x)\)[/tex]. Here, we multiply the odd function [tex]\( g(x) \)[/tex] by itself.
To determine if [tex]\((g \cdot g)(x)\)[/tex] is even, odd, or neither:
1. Evaluate [tex]\((g \cdot g)(-x)\)[/tex]:
[tex]\[
(g \cdot g)(-x) = g(-x) \cdot g(-x)
\][/tex]
2. Since [tex]\( g(x) \)[/tex] is odd, [tex]\( g(-x) = -g(x) \)[/tex]. Substitute this identity:
[tex]\[
(g \cdot g)(-x) = (-g(x)) \cdot (-g(x)) = g(x) \cdot g(x)
\][/tex]
This shows that:
[tex]\[
(g \cdot g)(-x) = (g \cdot g)(x)
\][/tex]
Therefore, the function [tex]\((g \cdot g)(x)\)[/tex] satisfies the condition for being even. So:
[tex]\[
(g \cdot g)(x) = \text{even}
\][/tex]
### Summary:
[tex]\[
(f \cdot g)(x) = \text{odd}
\][/tex]
[tex]\[
(g \cdot g)(x) = \text{even}
\][/tex]
