Chapter 0.2, Problem 22E

To determine

To state:

Whether the given function is even, odd, or neither.

Answer to Problem 22E

The given function is neither even nor odd.

Explanation of Solution

Given:

The given function $y=x+x^2$ .

Concepts Used:

If $f\left(-x\right)=f\left(x\right)$ then the function $f$ is said to be an even function.

If $f\left(-x\right)=-f\left(x\right)$ then the function $f$ is said to be an odd function.

Calculations:

Let $y=f\left(x\right)$ , thus $f\left(x\right)=x+x^2$ .

Calculate $f\left(-x\right)$ and check if it equals $f\left(x\right)$ or $-f\left(x\right)$ .

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

Conclusion:

Since $f\left(-x\right)$ equals neither $f\left(x\right)$ nor $-f\left(x\right)$ for $f\left(x\right)=x+x^2$ , the given function is neither even nor odd.