Chapter 13.3, Problem 15WE

To determine

To find:Whether the function is even, odd or neither.

Answer to Problem 15WE

The function is even.

Explanation of Solution

Given information: The function is $f\left(x\right)=\frac{x^2}{x^2+4}$

Calculation:

Consider the function.

  $f\left(x\right)=\frac{x^2}{x^2+4}$

Substitute $x=-x$ in above function.

  $\begin{array}{l}f\left(-x\right)=\frac{{\left(-x\right)}^2}{{\left(-x\right)}^2+4}\\ f\left(-x\right)=\frac{x^2}{x^2+4}\\ f\left(-x\right)=f\left(x\right)\end{array}$

Thus, the function is even.