Determine if Odd, Even, or Neither -x^6+5x^2

$- x^{6} + 5 x^{2}$

Step 1

Write $- x^{6} + 5 x^{2}$ as a function.

$f (x) = - x^{6} + 5 x^{2}$

Step 2

Find $f (- x)$ .

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

$f (- x) = - {(- x)}^{6} + 5 {(- x)}^{2}$

Simplify each term.

Apply the product rule to $- x$ .

$f (- x) = - ({(- 1)}^{6} x^{6}) + 5 {(- x)}^{2}$

Multiply $- 1$ by ${(- 1)}^{6}$ by adding the exponents.

Move ${(- 1)}^{6}$ .

$f (- x) = {(- 1)}^{6} \cdot (- 1 x^{6}) + 5 {(- x)}^{2}$

Multiply ${(- 1)}^{6}$ by $- 1$ .

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

$f (- x) = {(- 1)}^{6} \cdot ((- 1) x^{6}) + 5 {(- x)}^{2}$

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

$f (- x) = {(- 1)}^{6 + 1} x^{6} + 5 {(- x)}^{2}$

Add $6$ and $1$ .

$f (- x) = {(- 1)}^{7} x^{6} + 5 {(- x)}^{2}$

Raise $- 1$ to the power of $7$ .

$f (- x) = - x^{6} + 5 {(- x)}^{2}$

Apply the product rule to $- x$ .

$f (- x) = - x^{6} + 5 ({(- 1)}^{2} x^{2})$

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

$f (- x) = - x^{6} + 5 (1 x^{2})$

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

$f (- x) = - x^{6} + 5 x^{2}$

Step 3

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

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

Since $- x^{6} + 5 x^{2} = - x^{6} + 5 x^{2}$ , the function is even.

The function is even

Step 4