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Is the given function even, odd, or neither? \( f(x)=\frac{x}{x^{2}+4} \)

Ask by Boyd Best. in the United States
Jan 23,2025

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Answer

The function is odd.

Solution

To determine whether the function \( f(x) = \frac{x}{x^{2}+4} \) is even, odd, or neither, we can use the following definitions: - **Even Function**: A function \( f \) is even if \( f(-x) = f(x) \) for all \( x \) in the domain. - **Odd Function**: A function \( f \) is odd if \( f(-x) = -f(x) \) for all \( x \) in the domain. Let's evaluate \( f(-x) \): \[ f(-x) = \frac{-x}{(-x)^{2} + 4} = \frac{-x}{x^{2} + 4} = -\frac{x}{x^{2} + 4} = -f(x) \] Since \( f(-x) = -f(x) \), the function \( f(x) \) satisfies the condition for being an **odd function**. **Answer:** The function is odd.

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To determine if the function \( f(x) = \frac{x}{x^2 + 4} \) is even, odd, or neither, we can evaluate \( f(-x) \). Replacing \( x \) with \( -x \) gives us: \[ f(-x) = \frac{-x}{(-x)^2 + 4} = \frac{-x}{x^2 + 4}. \] Now, notice that: \[ f(-x) = -\frac{x}{x^2 + 4} = -f(x). \] Since \( f(-x) = -f(x) \), this means the function is odd. As a fun nugget, the concept of even and odd functions was introduced in the 17th century and is key in various fields such as calculus and physics, especially when discussing symmetry. Odd functions, like this one, have rotational symmetry about the origin, which can be pretty neat when grappling with graphs! For practical application, odd functions often help in simplifying integrals over symmetric intervals like \([-a, a]\), where the integral of an odd function will always result in zero. So, knowing whether a function is odd can save time and effort in calculations!

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