Can a function be both even and odd? Give reasons for your answer.

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To determine if a function can be both even and odd, we need to understand the definitions of even and odd functions. An even function satisfies the condition \( f(x) = f(-x) \) for all \( x \) in its domain, while an odd function satisfies \( f(x) = -f(-x) \) for all \( x \) in its domain.

Consider a function \( f(x) \) that is both even and odd. This means it must satisfy both \( f(x) = f(-x) \) and \( f(x) = -f(-x) \) simultaneously.

If we set \( f(x) = f(-x) \) equal to \( f(x) = -f(-x) \), we get \( f(x) = -f(x) \).

The equation \( f(x) = -f(x) \) implies that \( 2f(x) = 0 \), which simplifies to \( f(x) = 0 \).

Therefore, the only function that is both even and odd is the zero function, \( f(x) = 0 \), which is trivially both even and odd because it satisfies both conditions for all \( x \).

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Even Functions

A function is considered even if it satisfies the condition f(x) = f(-x) for all x in its domain. This means that the graph of the function is symmetric with respect to the y-axis. Common examples include f(x) = x² and f(x) = cos(x). Even functions exhibit a specific type of symmetry that influences their behavior and properties.

Odd Functions

A function is classified as odd if it meets the condition f(-x) = -f(x) for all x in its domain. This indicates that the graph of the function is symmetric with respect to the origin. Examples of odd functions include f(x) = x³ and f(x) = sin(x). The oddness of a function implies a certain type of rotational symmetry around the origin.

Mutual Exclusivity of Even and Odd Functions

A function cannot be both even and odd simultaneously. If a function is even, it must satisfy f(x) = f(-x), while being odd requires f(-x) = -f(x). The only function that can be classified as both is the zero function, f(x) = 0, since it satisfies both conditions. This mutual exclusivity is fundamental in understanding the nature of function symmetry.