Determining Even Functions Involving A Piecewise Function

In mathematics, understanding the properties of functions is crucial, especially when dealing with piecewise functions. Piecewise functions are defined by different expressions over different intervals of their domain. This article delves into a specific piecewise function and explores which related functions are even. We will analyze the given function, ${ f(x) }$, and then examine three functions derived from it to determine their evenness. An even function is defined as a function that satisfies the condition ${ f(x) = f(-x) }$ for all ${ x }$ in its domain. This means the graph of an even function is symmetric with respect to the y-axis. Let’s dive into the problem and systematically determine which of the given functions are even.

Problem Statement

Given the piecewise function: ${ f(x) = \begin{cases} x^3 & \text{if } x \geq 0 \\ x & \text{if } x < 0 \end{cases} }$

Determine which of the following functions is even:

I. ${ f(x) }$ II. ${ f(|x|) }$ III. ${ |f(x)| }$

We will analyze each option individually to check if it satisfies the condition for an even function.

Analysis of I: ${ f(x) }$

To determine if ${ f(x) }$ is even, we need to check if ${ f(x) = f(-x) }$ for all ${ x }$. We will consider two cases: ${ x \geq 0 }$ and ${ x < 0 }$.

Case 1: ${ x \geq 0 }$

In this case, ${ f(x) = x^3 }$. Now we need to find ${ f(-x) }$. Since ${ x \geq 0 }$, then ${ -x \leq 0 }$, so we use the second part of the piecewise function: ${ f(-x) = -x }$

For ${ f(x) }$ to be even, we must have ${ x^3 = -x }$ for all ${ x \geq 0 }$. This is only true for ${ x = 0 }$ and ${ x = 1 }$, but not for all ${ x \geq 0 }$. For example, if ${ x = 2 }$, then ${ f(2) = 2^3 = 8 }$ and ${ f(-2) = -2 }$, so ${ f(2) \neq f(-2) }$.

Case 2: ${ x < 0 }$

In this case, ${ f(x) = x }$. Now we need to find ${ f(-x) }$. Since ${ x < 0 }$, then ${ -x > 0 }$, so we use the first part of the piecewise function: ${ f(-x) = (-x)^3 = -x^3 }$

For ${ f(x) }$ to be even, we must have ${ x = -x^3 }$ for all ${ x < 0 }$. This is only true for ${ x = -1 }$ and ${ x = 0 }$, but not for all ${ x < 0 }$. For example, if ${ x = -2 }$, then ${ f(-2) = -2 }$ and ${ f(2) = 2^3 = 8 }$, so ${ f(-2) \neq f(2) }$.

Since ${ f(x) \neq f(-x) }$ for all ${ x }$, function I, ${ f(x) }$, is not even.

Analysis of II: ${ f(|x|) }$

To determine if ${ f(|x|) }$ is even, we need to check if ${ f(|x|) = f(|-x|) }$ for all ${ x }$. Let’s first find ${ f(|x|) }$ using the given piecewise function. The absolute value ${ |x| }$ is always non-negative, so we use the first part of the definition of ${ f(x) }$ for ${ |x| }$: ${ f(|x|) = (|x|)^3 = |x|^3 }$

Now we need to find ${ f(|-x|) }$. Since ${ |-x| = |x| }$, we have: ${ f(|-x|) = f(|x|) = |x|^3 }$

Since ${ f(|x|) = f(|-x|) }$ for all ${ x }$, function II, ${ f(|x|) }$, is even. This is because substituting the absolute value of ${ x }$ into the function results in the same value regardless of the sign of ${ x }$, thus satisfying the condition for an even function.

Analysis of III: ${ |f(x)| }$

To determine if ${ |f(x)| }$ is even, we need to check if ${ |f(x)| = |f(-x)| }$ for all ${ x }$. We will use the piecewise definition of ${ f(x) }$ to find ${ |f(x)| }$ and ${ |f(-x)| }$.

Case 1: ${ x \geq 0 }$

In this case, ${ f(x) = x^3 }$, so ${ |f(x)| = |x^3| = x^3 }$ since ${ x^3 }$ is non-negative for ${ x \geq 0 }$. For ${ f(-x) }$, since ${ -x \leq 0 }$, we have ${ f(-x) = -x }$, so ${ |f(-x)| = |-x| = |x| }$.

Case 2: ${ x < 0 }$

In this case, ${ f(x) = x }$, so ${ |f(x)| = |x| }$. For ${ f(-x) }$, since ${ -x > 0 }$, we have ${ f(-x) = (-x)^3 = -x^3 }$, so ${ |f(-x)| = |-x^3| = |x^3| }$.

Now, let's express ${ |f(x)| }$ as a piecewise function: ${ |f(x)| = \begin{cases} |x^3| = x^3 & \text{if } x \geq 0 \\ |x| = -x & \text{if } x < 0 \end{cases} }$

And ${ |f(-x)| }$ as a piecewise function: ${ |f(-x)| = \begin{cases} |-x^3| = x^3 & \text{if } -x \geq 0 \text{ (or } x \leq 0 ext{)} \\ |-x| = x & \text{if } -x < 0 \text{ (or } x > 0 ext{)} \end{cases} }$

${ |f(-x)| = \begin{cases} x^3 & \text{if } x \leq 0 \\ x & \text{if } x > 0 \end{cases} }$

We need to check if ${ |f(x)| = |f(-x)| }$ for all ${ x }$. Let's compare the piecewise functions:

${ |f(x)| = \begin{cases} x^3 & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} }$

${ |f(-x)| = \begin{cases} x & \text{if } x > 0 \\ x^3 & \text{if } x \leq 0 \end{cases} }$

For ${ x \geq 0 }$, we compare ${ x^3 }$ (from ${ |f(x)| }$) and ${ x }$ (from ${ |f(-x)| }$). They are not equal for all ${ x \geq 0 }$. For example, if ${ x = 2 }$, then ${ |f(2)| = 2^3 = 8 }$ and ${ |f(-2)| = |-2| = 2 }$, so ${ |f(2)| \neq |f(-2)| }$.

However, if we look at ${ |f(-x)| = \begin{cases} x^3 & \text{if } x \leq 0 \\ x & \text{if } x > 0 \end{cases} }$

There's a mix up here, we should have: ${ |f(-x)| = \begin{cases} |(-x)^3| = |-x^3| = x^3 & \text{if } -x \geq 0 \Leftrightarrow x \leq 0 \\ |-x| = x & \text{if } -x < 0 \Leftrightarrow x > 0 \end{cases} }$

So ${ |f(-x)| = \begin{cases} x^3 & \text{if } x \leq 0 \\ x & \text{if } x > 0 \end{cases} }$ Now compare with ${ |f(x)| = \begin{cases} x^3 & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} }$

If we consider ${ x = -1 }$, ${ |f(-1)| = |-1| = 1 }$ and ${ |f(1)| = |1^3| = 1 }$. If we consider ${ x = 2 }$, ${ |f(2)| = |2^3| = 8 }$ and ${ |f(-2)| = |-2| = 2 }$. So in summary: ${ |f(x)| = \begin{cases} x^3 & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} }$

${ |f(-x)| = \begin{cases} (-x)^3 & \text{if } -x \geq 0 \Leftrightarrow x \leq 0 \\ -(-x) & \text{if } -x < 0 \Leftrightarrow x > 0 \end{cases} }$ ${ |f(-x)| = \begin{cases} -x^3 & \text{if } x \leq 0 \\ x & \text{if } x > 0 \end{cases} }$

This function is not even since ${ |f(x)| \neq |f(-x)| }$.

Since ${ |f(x)| \neq |f(-x)| }$ for all ${ x }$, function III, ${ |f(x)| }$, is not even. This is evident when considering different values of ${ x }$; the absolute value of the function at ${ x }$ and ${ -x }$ do not always match.

Conclusion

After analyzing each function, we conclude that only function II, ${ f(|x|) }$, is even. The original function ${ f(x) }$ and the absolute value of the function ${ |f(x)| }$ do not satisfy the condition ${ f(x) = f(-x) }$ for all ${ x }$ in their domain.

Therefore, the correct answer is:

II. ${ f(|x|) }$

This detailed analysis provides a clear understanding of how to determine even functions, especially when dealing with piecewise functions and absolute values. This is a fundamental concept in mathematics, and mastering it enhances problem-solving skills in various mathematical contexts.

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