Determine if Odd, Even, or Neither -5x^10+7x^6+3x^2-1

$- 5 x^{10} + 7 x^{6} + 3 x^{2} - 1$

Step 1

Write $- 5 x^{10} + 7 x^{6} + 3 x^{2} - 1$ as a function.

$f (x) = - 5 x^{10} + 7 x^{6} + 3 x^{2} - 1$

Step 2

Find $f (- x)$ .

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

$f (- x) = - 5 {(- x)}^{10} + 7 {(- x)}^{6} + 3 {(- x)}^{2} - 1$

Simplify each term.

Apply the product rule to $- x$ .

$f (- x) = - 5 ({(- 1)}^{10} x^{10}) + 7 {(- x)}^{6} + 3 {(- x)}^{2} - 1$

Raise $- 1$ to the power of $10$ .

$f (- x) = - 5 (1 x^{10}) + 7 {(- x)}^{6} + 3 {(- x)}^{2} - 1$

Multiply $x^{10}$ by $1$ .

$f (- x) = - 5 x^{10} + 7 {(- x)}^{6} + 3 {(- x)}^{2} - 1$

Apply the product rule to $- x$ .

$f (- x) = - 5 x^{10} + 7 ({(- 1)}^{6} x^{6}) + 3 {(- x)}^{2} - 1$

Raise $- 1$ to the power of $6$ .

$f (- x) = - 5 x^{10} + 7 (1 x^{6}) + 3 {(- x)}^{2} - 1$

Multiply $x^{6}$ by $1$ .

$f (- x) = - 5 x^{10} + 7 x^{6} + 3 {(- x)}^{2} - 1$

Apply the product rule to $- x$ .

$f (- x) = - 5 x^{10} + 7 x^{6} + 3 ({(- 1)}^{2} x^{2}) - 1$

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

$f (- x) = - 5 x^{10} + 7 x^{6} + 3 (1 x^{2}) - 1$

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

$f (- x) = - 5 x^{10} + 7 x^{6} + 3 x^{2} - 1$

Step 3

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

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

Since $- 5 x^{10} + 7 x^{6} + 3 x^{2} - 1 = - 5 x^{10} + 7 x^{6} + 3 x^{2} - 1$ , the function is even.

The function is even

Step 4