Q.39 - Precalculus Enhanced with Graphing Utilities

Question

Determine algebraically whether each function is even, odd, or neither.

$f\left(x\right)=x+\left|x\right|$

Step-by-Step Solution

Answer

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

Step 1: Given information.

The function, $f\left(x\right)=x+\left|x\right|$.

Step 2: Write definition of even function and odd function.

• A function $f\left(x\right)$ is said to be an even function if $f\left(-x\right)=f\left(x\right)$.
• A function $f\left(x\right)$ is said to be an odd function if $f\left(-x\right)=-f\left(x\right)$.

Step 3: Calculate f - x .

Replace $x$ by $-x$ into the function $f\left(x\right)=x+\left|x\right|$.

$$\begin{gathered} f\left(-x\right)=-x+\left|-x\right| \\ = -x+\left|x\right| \end{gathered}$$

Clearly, $f\left(-x\right)\ne f\left(x\right) and f\left(-x\right)\ne -f\left(x\right)$

So, by definition: $f\left(x\right)$ is neither even nor odd function.