Algebra Examples

$- 6 x^{5} + 3 x^{2}$

Step 1

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

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

Step 2

Find $f (- x)$ .

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Step 2.1

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

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

Step 2.2

Simplify each term.

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Step 2.2.1

Apply the product rule to $- x$ .

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $- 1$ by $- 6$ .

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

Step 2.2.4

Apply the product rule to $- x$ .

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 3

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

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Step 3.1

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

Step 3.2

Since $6 x^{5} + 3 x^{2}$ $\neq$ $- 6 x^{5} + 3 x^{2}$ , the function is not even.

The function is not even

Step 4

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

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Step 4.1

Find $- f (x)$ .

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Step 4.1.1

Multiply $- 6 x^{5} + 3 x^{2}$ by $- 1$ .

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

Step 4.1.2

Apply the distributive property.

$- f (x) = - (- 6 x^{5}) - (3 x^{2})$

Step 4.1.3

Multiply.

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Step 4.1.3.1

Multiply $- 6$ by $- 1$ .

$- f (x) = 6 x^{5} - (3 x^{2})$

Step 4.1.3.2

Multiply $3$ by $- 1$ .

$- f (x) = 6 x^{5} - 3 x^{2}$

Step 4.2

Since $6 x^{5} + 3 x^{2}$ $\neq$ $6 x^{5} - 3 x^{2}$ , the function is not odd.

The function is not odd

Step 5

The function is neither odd nor even

Step 6