Question

Determine whether the function is even, odd, or neither. $$ f(x)=\frac{x}{x^{2}+1} $$

Short answer

The function is odd since \(f(-x) = -f(x)\) with \(f(x) = \frac{x}{x^2+1}\) and \(f(-x) = \frac{-x}{x^2+1}\).

Step-by-step solution

Check Even Property

Substitute \(-x\) for \(x\) in the function, then compare it to the original function. If they are equal, the function is even. Function with \(-x\) substituted for \(x\): \(f(-x) = \frac{-x}{(-x)^2 + 1}\) Since \((-x)^2 = x^2\), the expression becomes: \(f(-x) = \frac{-x}{x^2 + 1}\) Comparing this to the given function \(f(x) = \frac{x}{x^2+1}\), observe that they are not equal. Therefore, the function is not even.

Check Odd Property

Substitute \(-x\) for \(x\) in the function, then compare it to the negation of the original function. If they are equal, the function is odd. Negation of the original function: \(-f(x) = -\frac{x}{x^2+1}\) Comparing this to the expression we found in Step 1, \(f(-x) = \frac{-x}{x^2 + 1}\), we can see that they are equal. Therefore, the function is odd.