To determine which of the given functions are even, we need to check whether each function satisfies the property of even functions: [tex]\( f(-x) = f(x) \)[/tex]. Let's analyze each function one by one.
### Analysis of [tex]\( f(x) \)[/tex]
The function [tex]\( f(x) \)[/tex] is defined as:
[tex]\[ f(x) = \begin{cases}
x^3 & \text{if } x \geq 0 \\
x & \text{if } x < 0
\end{cases} \][/tex]
To check if [tex]\( f(x) \)[/tex] is even, we need to compare [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex].
1. If [tex]\( x \geq 0 \)[/tex]:
[tex]\[
f(-x) = \begin{cases}
(-x)^3 = -x^3 & \text{if } -x \geq 0 \text{ (i.e., } x \leq 0) \\
-x & \text{if } -x < 0 \text{ (i.e., } x > 0)
\end{cases}
\][/tex]
Since [tex]\( x \geq 0 \)[/tex], we are in the [tex]\( x > 0 \)[/tex] case:
[tex]\[
f(x) = x^3 \quad \text{and} \quad f(-x) = -x
\][/tex]
2. If [tex]\( x < 0 \)[/tex]:
[tex]\[
f(-x) = (-x)^3 = -x^3 \quad \text{(since } -x > 0 \text{)}
\][/tex]
Since [tex]\( x < 0 \)[/tex]:
[tex]\[
f(x) = x
\][/tex]
From both cases, it's clear that [tex]\( f(-x) \neq f(x) \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] is not an even function.
### Analysis of [tex]\( f(|x|) \)[/tex]
The function [tex]\( f(|x|) \)[/tex] can be calculated by taking the absolute value of [tex]\( x \)[/tex] before applying [tex]\( f \)[/tex]:
[tex]\[ f(|x|) = f(\text{abs}(x)) = \begin{cases}
|x|^3 & \text{if } |x| \geq 0 \\
|x| & \text{if } |x| < 0
\end{cases} \][/tex]
Since [tex]\(|x| \geq 0\)[/tex] for all [tex]\( x \)[/tex], we always have [tex]\( f(|x|) = |x|^3 \)[/tex].
Since [tex]\( |x| = |-x| \)[/tex]:
[tex]\[ f(|x|) = f(|-x|) = (|x|)^3 = (|-x|)^3 \][/tex]
This simplifies to:
[tex]\[ f(|x|) = f(|-x|) = |x|^3 \][/tex]
Thus, [tex]\( f(|x|) \)[/tex] is an even function.
### Analysis of [tex]\( |f(x)| \)[/tex]
The function [tex]\( |f(x)| \)[/tex] takes the absolute value of [tex]\( f(x) \)[/tex]:
[tex]\[ |f(x)| = \begin{cases}
|x^3| & \text{if } x \geq 0 \\
|x| & \text{if } x < 0
\end{cases} \][/tex]
Since [tex]\( |x^3| = (|x|)^3 \)[/tex]:
[tex]\[ |f(x)| = \begin{cases}
x^3 & \text{if } x \geq 0 \\
-x & \text{if } x < 0
\end{cases} \][/tex]
Let's check [tex]\( |f(-x)| \)[/tex]:
1. If [tex]\( x \geq 0 \)[/tex]:
[tex]\[
|f(-x)| = \begin{cases}
|-x^3| = x^3 & \text{if } -x \geq 0 \text{ (i.e., } x \leq 0) \\
|-x| = x & \text{if } -x < 0 \text{ (i.e., } x > 0)
\end{cases}
\][/tex]
Since [tex]\( x \geq 0 \)[/tex] falls in the [tex]\( x > 0 \)[/tex] case:
[tex]\[
|f(x)| = |x^3| = x^3 \quad \text{and} \quad |f(-x)| = |-x| = x
\][/tex]
2. If [tex]\( x < 0 \)[/tex]:
[tex]\[
|f(-x)| = \begin{cases}
|-x^3| = x^3 & \text{if } -x \geq 0 \text{ (i.e., x < 0) \\
|-x| = x & \text{if } -x < 0 \text{ (i.e., } x > 0)}
\end{cases}
\][/tex]
Since [tex]\( x < 0 \)[/tex], we are in the [tex]\(-x \geq 0\)[/tex] case:
[tex]\[
|f(x)| = |x| = -x \quad \text{and} \quad |f(-x)| = |x^3| = -x
\][/tex]
So, making above clear: for analysis we have
[tex]\[ |f(x)| = |f(-x)| \][/tex]
Therefore, [tex]\(|f(x)|\)[/tex] is an even function.
### Conclusion
- [tex]\( f(x) \)[/tex]: Not even
- [tex]\( f(|x|) \)[/tex]: Even
- [tex]\( |f(x)| \)[/tex]: Even
So, the correct answer is [tex]\( \text{C. II only} \)[/tex].
Therefore, the correct choice is [tex]\( \text{B. II only} \)[/tex].
