Q.40 - Precalculus Enhanced with Graphing Utilities

Question

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

$f\left(x\right)=\sqrt[3]{2x^2+1}$

Step-by-Step Solution

Answer

$f\left(x\right)$ is an even function.

Step 1: Given information.

The function, $f\left(x\right)=\sqrt[3]{2x^2+1}$.

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 $f\left(x\right)=\sqrt[3]{2x^2+1}$.

width="147" style="max-width: none; vertical-align: -42px;" $$\begin{gathered} f\left(-x\right)=\sqrt[3]{2{\left(-x\right)}^2+1} \\ = \sqrt[3]{2x^2+1} \\ =f\left(x\right) \end{gathered}$$

Clearly, $f\left(-x\right)=f\left(x\right)$

So by definition, $f\left(x\right)$ is an even function.