Determine if Odd, Even, or Neither f(x)=-2x^5+7x^2

$f (x) = - 2 x^{5} + 7 x^{2}$

Step 1

Find $f (- x)$ .

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

$f (- x) = - 2 {(- x)}^{5} + 7 {(- x)}^{2}$

Simplify each term.

Apply the product rule to $- x$ .

$f (- x) = - 2 ({(- 1)}^{5} x^{5}) + 7 {(- x)}^{2}$

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

$f (- x) = - 2 (- x^{5}) + 7 {(- x)}^{2}$

Multiply $- 1$ by $- 2$ .

$f (- x) = 2 x^{5} + 7 {(- x)}^{2}$

Apply the product rule to $- x$ .

$f (- x) = 2 x^{5} + 7 ({(- 1)}^{2} x^{2})$

Raise $- 1$ to the power of $2$ .

$f (- x) = 2 x^{5} + 7 (1 x^{2})$

Multiply $x^{2}$ by $1$ .

$f (- x) = 2 x^{5} + 7 x^{2}$

Step 2

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

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

Since $2 x^{5} + 7 x^{2}$ $\neq$ $- 2 x^{5} + 7 x^{2}$ , 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 $- 2 x^{5} + 7 x^{2}$ by $- 1$ .

$- f (x) = - (- 2 x^{5} + 7 x^{2})$

Apply the distributive property.

$- f (x) = - (- 2 x^{5}) - (7 x^{2})$

Multiply.

Multiply $- 2$ by $- 1$ .

$- f (x) = 2 x^{5} - (7 x^{2})$

Multiply $7$ by $- 1$ .

$- f (x) = 2 x^{5} - 7 x^{2}$

Since $2 x^{5} + 7 x^{2}$ $\neq$ $2 x^{5} - 7 x^{2}$ , the function is not odd.

The function is not odd

Step 4

The function is neither odd nor even

Step 5