Precalculus Examples

$g (x) = - 6 x^{5} + 7 x^{3}$

Step 1

Find $f (- x)$ .

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Step 1.1

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

$g (- x) = - 6 {(- x)}^{5} + 7 {(- x)}^{3}$

Step 1.2

Simplify each term.

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Step 1.2.1

Apply the product rule to $- x$ .

$g (- x) = - 6 ({(- 1)}^{5} x^{5}) + 7 {(- x)}^{3}$

Step 1.2.2

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

$g (- x) = - 6 (- x^{5}) + 7 {(- x)}^{3}$

Step 1.2.3

Multiply $- 1$ by $- 6$ .

$g (- x) = 6 x^{5} + 7 {(- x)}^{3}$

Step 1.2.4

Apply the product rule to $- x$ .

$g (- x) = 6 x^{5} + 7 ({(- 1)}^{3} x^{3})$

Step 1.2.5

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

$g (- x) = 6 x^{5} + 7 (- x^{3})$

Step 1.2.6

Multiply $- 1$ by $7$ .

$g (- x) = 6 x^{5} - 7 x^{3}$

Step 2

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

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Step 2.1

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

Step 2.2

Since $6 x^{5} - 7 x^{3}$ $\neq$ $- 6 x^{5} + 7 x^{3}$ , the function is not even.

The function is not even

Step 3

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

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Step 3.1

Find $- f (x)$ .

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Step 3.1.1

Multiply $- 6 x^{5} + 7 x^{3}$ by $- 1$ .

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

Step 3.1.2

Apply the distributive property.

$- f (x) = - (- 6 x^{5}) - (7 x^{3})$

Step 3.1.3

Multiply.

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Step 3.1.3.1

Multiply $- 6$ by $- 1$ .

$- f (x) = 6 x^{5} - (7 x^{3})$

Step 3.1.3.2

Multiply $7$ by $- 1$ .

$- f (x) = 6 x^{5} - 7 x^{3}$

Step 3.2

Since $6 x^{5} - 7 x^{3} = 6 x^{5} - 7 x^{3}$ , the function is odd.

The function is odd

Step 4