Piecewise Function is even or odd?

Advanced Set Theory

In Advanced Set Theory, we study many important ideas related to the size and structure of sets. We begin by discussing equivalent sets and understanding the difference between countable and uncountable sets, along with their examples. We explore cardinal numbers as a way to compare set sizes and learn about Cantor’s theorem which shows that some infinities are bigger than others. We also look at partially ordered sets, chains, and lattices, which help us understand the order and arrangement of elements. The Cantor-Bernstein theorem helps in proving when two sets have the same size. We learn how to add, multiply, and take powers of cardinal numbers. Important logical tools like the Axiom of Choice and Zorn’s Lemma are also studied, along with their equivalence. Finally, we discuss well-ordered sets, ordinal numbers, and how we can perform addition and multiplication with them. All these concepts help us deeply understand infinite sets and the foundations of modern mathematics. Notes ●...

Analysis of Function

Analysis of Function:

\( f(x) = \begin{cases} x + 5, & -2 < x < 0 \\[6pt] -x + 5, & 0 \le x < 2 \end{cases} \)

1. Symmetric Domain Check:

The domain of this function is \( (-2, 0) \cup [0, 2) = (-2, 2) \).
This domain is symmetric about the origin. So, the first condition is met.

2. Functional Property Check:

Case 1: \( x \in (0, 2) \)

\( f(x) = -x + 5 \)
Now consider \( -x \). Since \( x \in (0, 2) \), then \( -x \in (-2, 0) \).
For \( -x \in (-2, 0) \), \( f(-x) = (-x) + 5 = -x + 5 \).
Is \( f(-x) = f(x) \)? Is \( -x + 5 = -x + 5 \)? Yes. This suggests it might be even.

Case 2: \( x \in (-2, 0) \)

\( f(x) = x + 5 \)
Now consider \( -x \). Since \( x \in (-2, 0) \), then \( -x \in (0, 2) \).
For \( -x \in (0, 2) \), \( f(-x) = -(-x) + 5 = x + 5 \).
Is \( f(-x) = f(x) \)? Is \( x + 5 = x + 5 \)? Yes. This also suggests it might be even.

Check at \( x = 0 \):

\( f(0) = -(0) + 5 = 5 \).
For an even function, \( f(0) \) can be any value. For an odd function, \( f(0) \) must be 0. Since \( f(0) = 5 \neq 0 \), it cannot be an odd function.

Conclusion: This function is even.

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Piecewise Function is even or odd?

Analysis of Function:

1. Symmetric Domain Check:

2. Functional Property Check:

Case 1: \( x \in (0, 2) \)

Case 2: \( x \in (-2, 0) \)

Check at \( x = 0 \):