Determine if Odd, Even, or Neither h(x)=-2x^5+3x^3

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

Step 1

Find $f (- x)$ .

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

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

Simplify each term.

Apply the product rule to $- x$ .

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

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

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

Multiply $- 1$ by $- 2$ .

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

Apply the product rule to $- x$ .

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

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

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

Multiply $- 1$ by $3$ .

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

Step 2

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

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

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

The function is not even

Step 3

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

Find $- f (x)$ .

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

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

Apply the distributive property.

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

Multiply.

Multiply $- 2$ by $- 1$ .

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

Multiply $3$ by $- 1$ .

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

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

The function is odd

Step 4