Q. 20 - Precalculus Enhanced with Graphing Utilities

Question

Determine(algebraically)whethere the given function is even, odd,or neither.

$G\left(x\right)=1-x+x^2$

Step-by-Step Solution

Answer

The given function $G\left(x\right)=1-x+x^2$is neither even nor odd.

Step 1.Given information

The given function $G\left(x\right)=1-x+x^2$

Step 2. Determine the given function is even,odd or neither

The function is even if $G(-x)=G\left(x\right)$ and odd  if $G(-x)=-G\left(x\right)$ .If none of these conditions are satisfied.

Find $G(-x)$ by replacing -x for x in the given function.
$$\begin{gathered} G(-x)=1-(-x)+{(-x)}^3 \\ =1+x-x^3 \end{gathered}$$

We can see that $G(-x)\ne G\left(x\right) and G(-x)\ne -G\left(x\right)$.Therefore the given function is neither even nor odd