Chapter 2, Problem 34RE

To determine

To find: whether $H$ is even, odd, or neither.

Answer to Problem 34RE

  $H\left(x\right)$ is neither even nor odd function.

Explanation of Solution

Given information:

Given function

  $H\left(x\right)=1+x+x^2$

Concept used:

A function $f\left(x\right)$ is even if $f\left(-x\right)=f\left(x\right)$ .

A function $f\left(x\right)$ is odd if $f\left(-x\right)=-f\left(x\right)$ .

Otherwise neither even nor odd.

Calculation:

Replacing $x$ by $-x$ in the given function $H\left(x\right)=1+x+x^2$ , we get

  $\begin{array}{c}H\left(-x\right)=1+\left(-x\right)+{\left(-x\right)}^2\\ =1-x+x^2\end{array}$

Since $H\left(-x\right)\ne -H\left(x\right)$ or $H\left(-x\right)\ne H\left(x\right)$ , therefore $H\left(x\right)$ is neither even nor odd function.