Determine if Odd, Even, or Neither cube root of 6x+1

$\sqrt[3]{6 x + 1}$

Write $\sqrt[3]{6 x + 1}$ as a function.

$f (x) = \sqrt[3]{6 x + 1}$

Find $f (- x)$ .

Find $f (- x)$ by substituting $- x$ for all occurrence of $x$ in $f (x)$ .

$f (- x) = \sqrt[3]{6 (- x) + 1}$

Multiply $- 1$ by $6$ .

$f (- x) = \sqrt[3]{- 6 x + 1}$

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

Check if $f (- x) = f (x)$ .

$\sqrt[3]{- 6 x + 1}$ $\neq$ $\sqrt[3]{6 x + 1}$

Since $\sqrt[3]{- 6 x + 1}$ $\neq$ $\sqrt[3]{6 x + 1}$ , the function is not even.

The function is not even

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

Multiply $- 1$ by $\sqrt[3]{6 x + 1}$ .

$- f (x) = - \sqrt[3]{6 x + 1}$

Since $\sqrt[3]{- 6 x + 1}$ $\neq$ $- \sqrt[3]{6 x + 1}$ , the function is not odd.

The function is not odd

The function is neither odd nor even