Determine if Odd, Even, or Neither f(x) = cube root of 7x

$f (x) = \sqrt[3]{7 x}$

Step 1

Find $f (- x)$ .

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

$f (- x) = \sqrt[3]{7 (- x)}$

Multiply $- 1$ by $7$ .

$f (- x) = \sqrt[3]{- 7 x}$

Rewrite $- 7 x$ as ${({(- 1)}^{3})}^{3} \cdot (7 x)$ .

Rewrite $- 7$ as $- 1 (7)$ .

$f (- x) = \sqrt[3]{- 1 \cdot 7 x}$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$f (- x) = \sqrt[3]{{(- 1)}^{3} \cdot (7 x)}$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$f (- x) = \sqrt[3]{{({(- 1)}^{3})}^{3} \cdot (7 x)}$

Add parentheses.

$f (- x) = \sqrt[3]{{({(- 1)}^{3})}^{3} \cdot (7 x)}$

Pull terms out from under the radical.

$f (- x) = {(- 1)}^{3} \sqrt[3]{7 x}$

Simplify the expression.

Raise $- 1$ to the power of $3$ .

$f (- x) = - 1 \sqrt[3]{7 x}$

Rewrite $- 1 \sqrt[3]{7 x}$ as $- \sqrt[3]{7 x}$ .

$f (- x) = - \sqrt[3]{7 x}$

Step 2

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

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

Since $- \sqrt[3]{7 x}$ $\neq$ $\sqrt[3]{7 x}$ , the function is not even.

The function is not even

Step 3

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

Multiply $- 1$ by $\sqrt[3]{7 x}$ .

$- f (x) = - \sqrt[3]{7 x}$

Since $- \sqrt[3]{7 x} = - \sqrt[3]{7 x}$ , the function is odd.

The function is odd

Step 4