Precalculus Examples

$f (x) = \frac{x^{2} - 7}{5 x^{3} + x}$

Step 1

Factor $x$ out of $5 x^{3} + x$ .

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Step 1.1

Factor $x$ out of $5 x^{3}$ .

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

Step 1.2

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

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

Step 1.3

Factor $x$ out of $x^{1}$ .

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

Step 1.4

Factor $x$ out of $x (5 x^{2}) + x \cdot 1$ .

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

Step 2

Find $f (- x)$ .

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Step 2.1

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

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

Step 2.2

Simplify the numerator.

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Step 2.2.1

Apply the product rule to $- x$ .

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

Simplify the denominator.

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Step 2.3.1

Apply the product rule to $- x$ .

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.4

Move the negative in front of the fraction.

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

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 $- \frac{x^{2} - 7}{x (5 x^{2} + 1)}$ $\neq$ $\frac{x^{2} - 7}{x (5 x^{2} + 1)}$ , 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

Multiply $- 1$ by $\frac{x^{2} - 7}{x (5 x^{2} + 1)}$ .

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

Step 4.2

Since $- \frac{x^{2} - 7}{x (5 x^{2} + 1)} = - \frac{x^{2} - 7}{x (5 x^{2} + 1)}$ , the function is odd.

The function is odd

Step 5