To determine whether the function $f(x) = \frac{x}{x^2 + 2x}$ is even, odd, or neither, let's analyze it based on the definitions of even and odd functions.

Step 1: Definitions

1. Even Function: $f(x)$ is even if $f(-x) = f(x)$ for all $x$ in the domain of $f$.
2. Odd Function: $f(x)$ is odd if $f(-x) = -f(x)$ for all $x$ in the domain of $f$.

Step 2: Calculate $f(-x)$

Given: $f(x) = \frac{x}{x^2 + 2x}$ we need to substitute $-x$ in place of $x$: $f(-x) = \frac{-x}{(-x)^2 + 2(-x)}$ This simplifies to: $f(-x) = \frac{-x}{x^2 - 2x}$

Step 3: Compare $f(-x)$ and $f(x)$

Now we have:

• $f(x) = \frac{x}{x^2 + 2x}$
• $f(-x) = \frac{-x}{x^2 - 2x}$

Since $f(-x) \neq f(x)$ and $f(-x) \neq -f(x)$, the function $f(x)$ is neither even nor odd.

Conclusion

The function $f(x) = \frac{x}{x^2 + 2x}$ is neither even nor odd.

Would you like further explanation on any part of this solution?

Here are some related questions for further exploration:

1. How can you determine the domain of the function $f(x) = \frac{x}{x^2 + 2x}$?
2. What happens to the behavior of $f(x)$ near points where the denominator is zero?
3. How would you graph this function to visualize its symmetry properties?
4. Can the concept of even and odd functions apply to functions with asymptotes?
5. How would adding a constant term affect the parity (even/odd nature) of $f(x)$?

Tip: Testing for even or odd symmetry is often easier if you simplify the expression first, where possible.