Determine if Odd, Even, or Neither 2x^5-x^3

$2 x^{5} - x^{3}$

Step 1

Write $2 x^{5} - x^{3}$ as a function.

$f (x) = 2 x^{5} - x^{3}$

Step 2

Find $f (- x)$ .

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

$f (- x) = 2 {(- x)}^{5} - {(- x)}^{3}$

Simplify each term.

Apply the product rule to $- x$ .

$f (- x) = 2 ({(- 1)}^{5} x^{5}) - {(- x)}^{3}$

Raise $- 1$ to the power of $5$ .

$f (- x) = 2 (- x^{5}) - {(- x)}^{3}$

Multiply $- 1$ by $2$ .

$f (- x) = - 2 x^{5} - {(- x)}^{3}$

Apply the product rule to $- x$ .

$f (- x) = - 2 x^{5} - ({(- 1)}^{3} x^{3})$

Multiply $- 1$ by ${(- 1)}^{3}$ by adding the exponents.

Move ${(- 1)}^{3}$ .

$f (- x) = - 2 x^{5} + {(- 1)}^{3} \cdot (- 1 x^{3})$

Multiply ${(- 1)}^{3}$ by $- 1$ .

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

$f (- x) = - 2 x^{5} + {(- 1)}^{3} \cdot ((- 1) x^{3})$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (- x) = - 2 x^{5} + {(- 1)}^{3 + 1} x^{3}$

Add $3$ and $1$ .

$f (- x) = - 2 x^{5} + {(- 1)}^{4} x^{3}$

Raise $- 1$ to the power of $4$ .

$f (- x) = - 2 x^{5} + 1 x^{3}$

Multiply $x^{3}$ by $1$ .

$f (- x) = - 2 x^{5} + x^{3}$

Step 3

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

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

Since $- 2 x^{5} + x^{3}$ $\neq$ $2 x^{5} - x^{3}$ , the function is not even.

The function is not even

Step 4

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

Find $- f (x)$ .

Multiply $2 x^{5} - x^{3}$ by $- 1$ .

$- f (x) = - (2 x^{5} - x^{3})$

Apply the distributive property.

$- f (x) = - (2 x^{5}) + x^{3}$

Multiply $2$ by $- 1$ .

$- f (x) = - 2 x^{5} + x^{3}$

Since $- 2 x^{5} + x^{3} = - 2 x^{5} + x^{3}$ , the function is odd.

The function is odd

Step 5

Determine if Odd, Even, or Neither 2x^5-x^3

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