Calculus Examples

$f (x) = \frac{x^{2} + 9}{x}$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x^{2} + 9$ and $g (x) = x$ .

$f' (x) = \frac{x \frac{d}{d x} (x^{2} + 9) - (x^{2} + 9) \frac{d}{d x} x}{x^{2}}$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $x^{2} + 9$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [9]$ .

$f' (x) = \frac{x (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (9)) - (x^{2} + 9) \frac{d}{d x} x}{x^{2}}$

Step 1.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f' (x) = \frac{x (2 x + \frac{d}{d x} (9)) - (x^{2} + 9) \frac{d}{d x} x}{x^{2}}$

Step 1.1.2.3

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$f' (x) = \frac{x (2 x + 0) - (x^{2} + 9) \frac{d}{d x} x}{x^{2}}$

Step 1.1.2.4

Add $2 x$ and $0$ .

$f' (x) = \frac{x (2 x) - (x^{2} + 9) \frac{d}{d x} x}{x^{2}}$

Step 1.1.3

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

$f' (x) = \frac{2 (x \cdot x) - (x^{2} + 9) \frac{d}{d x} x}{x^{2}}$

Step 1.1.4

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

$f' (x) = \frac{2 (x \cdot x) - (x^{2} + 9) \frac{d}{d x} x}{x^{2}}$

Step 1.1.5

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

$f' (x) = \frac{2 x^{1 + 1} - (x^{2} + 9) \frac{d}{d x} x}{x^{2}}$

Step 1.1.6

Add $1$ and $1$ .

$f' (x) = \frac{2 x^{2} - (x^{2} + 9) \frac{d}{d x} x}{x^{2}}$

Step 1.1.7

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f' (x) = \frac{2 x^{2} - (x^{2} + 9) \cdot 1}{x^{2}}$

Step 1.1.8

Multiply $- 1$ by $1$ .

$f' (x) = \frac{2 x^{2} - (x^{2} + 9)}{x^{2}}$

Step 1.1.9

Simplify.

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Step 1.1.9.1

Apply the distributive property.

$f' (x) = \frac{2 x^{2} - x^{2} - 1 \cdot 9}{x^{2}}$

Step 1.1.9.2

Simplify the numerator.

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Step 1.1.9.2.1

Multiply $- 1$ by $9$ .

$f' (x) = \frac{2 x^{2} - x^{2} - 9}{x^{2}}$

Step 1.1.9.2.2

Subtract $x^{2}$ from $2 x^{2}$ .

$f' (x) = \frac{x^{2} - 9}{x^{2}}$

Step 1.1.9.3

Simplify the numerator.

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Step 1.1.9.3.1

Rewrite $9$ as $3^{2}$ .

$f' (x) = \frac{x^{2} - 3^{2}}{x^{2}}$

Step 1.1.9.3.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

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

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{(x + 3) (x - 3)}{x^{2}}$ .

$\frac{(x + 3) (x - 3)}{x^{2}}$

Step 2

Set the first derivative equal to $0$ then solve the equation $\frac{(x + 3) (x - 3)}{x^{2}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$\frac{(x + 3) (x - 3)}{x^{2}} = 0$

Step 2.2

Set the numerator equal to zero.

$(x + 3) (x - 3) = 0$

Step 2.3

Solve the equation for $x$ .

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Step 2.3.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 3 = 0$

$x - 3 = 0$

Step 2.3.2

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 2.3.2.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 2.3.2.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.3.3

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 2.3.3.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 2.3.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.3.4

The final solution is all the values that make $(x + 3) (x - 3) = 0$ true.

$x = - 3, 3$

Step 3

The values which make the derivative equal to $0$ are $- 3, 3$ .

$- 3, 3$

Step 4

Find where the derivative is undefined.

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Step 4.1

Set the denominator in $\frac{(x + 3) (x - 3)}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Step 4.2

Solve for $x$ .

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Step 4.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Step 4.2.2

Simplify $\pm \sqrt{0}$ .

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Step 4.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 4.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Step 4.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 5

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the derivative $0$ or undefined.

$(- \infty, - 3) \cup (- 3, 0) \cup (0, 3) \cup (3, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - 3)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 4$ in the expression.

$f' (- 4) = \frac{((- 4) + 3) ((- 4) - 3)}{{(- 4)}^{2}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify the numerator.

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Step 6.2.1.1

Add $- 4$ and $3$ .

$f' (- 4) = \frac{- 1 (- 4 - 3)}{{(- 4)}^{2}}$

Step 6.2.1.2

Subtract $3$ from $- 4$ .

$f' (- 4) = \frac{- 1 \cdot - 7}{{(- 4)}^{2}}$

Step 6.2.2

Simplify the expression.

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Step 6.2.2.1

Raise $- 4$ to the power of $2$ .

$f' (- 4) = \frac{- 1 \cdot - 7}{16}$

Step 6.2.2.2

Multiply $- 1$ by $- 7$ .

$f' (- 4) = \frac{7}{16}$

Step 6.2.3

The final answer is $\frac{7}{16}$ .

$\frac{7}{16}$

Step 6.3

At $x = - 4$ the derivative is $\frac{7}{16}$ . Since this is positive, the function is increasing on $(- \infty, - 3)$ .

Increasing on $(- \infty, - 3)$ since $f' (x) > 0$

Step 7

Substitute a value from the interval $(- 3, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $- \frac{3}{2}$ in the expression.

$f' (- \frac{3}{2}) = \frac{((- \frac{3}{2}) + 3) ((- \frac{3}{2}) - 3)}{{(- \frac{3}{2})}^{2}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify the numerator.

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Step 7.2.1.1

To write $3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (- \frac{3}{2}) = \frac{(- \frac{3}{2} + 3 \cdot \frac{2}{2}) (- \frac{3}{2} - 3)}{{(- \frac{3}{2})}^{2}}$

Step 7.2.1.2

Combine $3$ and $\frac{2}{2}$ .

$f' (- \frac{3}{2}) = \frac{(- \frac{3}{2} + \frac{3 \cdot 2}{2}) (- \frac{3}{2} - 3)}{{(- \frac{3}{2})}^{2}}$

Step 7.2.1.3

Combine the numerators over the common denominator.

$f' (- \frac{3}{2}) = \frac{\frac{- 3 + 3 \cdot 2}{2} \cdot (- \frac{3}{2} - 3)}{{(- \frac{3}{2})}^{2}}$

Step 7.2.1.4

Simplify the numerator.

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Step 7.2.1.4.1

Multiply $3$ by $2$ .

$f' (- \frac{3}{2}) = \frac{\frac{- 3 + 6}{2} \cdot (- \frac{3}{2} - 3)}{{(- \frac{3}{2})}^{2}}$

Step 7.2.1.4.2

Add $- 3$ and $6$ .

$f' (- \frac{3}{2}) = \frac{\frac{3}{2} \cdot (- \frac{3}{2} - 3)}{{(- \frac{3}{2})}^{2}}$

Step 7.2.1.5

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (- \frac{3}{2}) = \frac{\frac{3}{2} \cdot (- \frac{3}{2} - 3 \cdot \frac{2}{2})}{{(- \frac{3}{2})}^{2}}$

Step 7.2.1.6

Combine $- 3$ and $\frac{2}{2}$ .

$f' (- \frac{3}{2}) = \frac{\frac{3}{2} \cdot (- \frac{3}{2} + \frac{- 3 \cdot 2}{2})}{{(- \frac{3}{2})}^{2}}$

Step 7.2.1.7

Combine the numerators over the common denominator.

$f' (- \frac{3}{2}) = \frac{\frac{3}{2} \cdot \frac{- 3 - 3 \cdot 2}{2}}{{(- \frac{3}{2})}^{2}}$

Step 7.2.1.8

Simplify the numerator.

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Step 7.2.1.8.1

Multiply $- 3$ by $2$ .

$f' (- \frac{3}{2}) = \frac{\frac{3}{2} \cdot \frac{- 3 - 6}{2}}{{(- \frac{3}{2})}^{2}}$

Step 7.2.1.8.2

Subtract $6$ from $- 3$ .

$f' (- \frac{3}{2}) = \frac{\frac{3}{2} \cdot \frac{- 9}{2}}{{(- \frac{3}{2})}^{2}}$

Step 7.2.1.9

Move the negative in front of the fraction.

$f' (- \frac{3}{2}) = \frac{\frac{3}{2} \cdot (- 1 (\frac{9}{2}))}{{(- \frac{3}{2})}^{2}}$

Step 7.2.1.10

Combine exponents.

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Step 7.2.1.10.1

Factor out negative.

$f' (- \frac{3}{2}) = \frac{- (\frac{3}{2} \cdot \frac{9}{2})}{{(- \frac{3}{2})}^{2}}$

Step 7.2.1.10.2

Multiply $\frac{3}{2}$ by $\frac{9}{2}$ .

$f' (- \frac{3}{2}) = \frac{- \frac{3 \cdot 9}{2 \cdot 2}}{{(- \frac{3}{2})}^{2}}$

Step 7.2.1.10.3

Multiply $3$ by $9$ .

$f' (- \frac{3}{2}) = \frac{- \frac{27}{2 \cdot 2}}{{(- \frac{3}{2})}^{2}}$

Step 7.2.1.10.4

Multiply $2$ by $2$ .

$f' (- \frac{3}{2}) = \frac{- \frac{27}{4}}{{(- \frac{3}{2})}^{2}}$

Step 7.2.2

Simplify the denominator.

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Step 7.2.2.1

Apply the product rule to $- \frac{3}{2}$ .

$f' (- \frac{3}{2}) = \frac{- \frac{27}{4}}{{(- 1)}^{2} {(\frac{3}{2})}^{2}}$

Step 7.2.2.2

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

$f' (- \frac{3}{2}) = \frac{- \frac{27}{4}}{1 {(\frac{3}{2})}^{2}}$

Step 7.2.2.3

Apply the product rule to $\frac{3}{2}$ .

$f' (- \frac{3}{2}) = \frac{- \frac{27}{4}}{1 (\frac{3^{2}}{2^{2}})}$

Step 7.2.2.4

Raise $3$ to the power of $2$ .

$f' (- \frac{3}{2}) = \frac{- \frac{27}{4}}{1 (\frac{9}{2^{2}})}$

Step 7.2.2.5

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

$f' (- \frac{3}{2}) = \frac{- \frac{27}{4}}{1 (\frac{9}{4})}$

Step 7.2.2.6

Multiply $\frac{9}{4}$ by $1$ .

$f' (- \frac{3}{2}) = \frac{- \frac{27}{4}}{\frac{9}{4}}$

Step 7.2.3

Multiply the numerator by the reciprocal of the denominator.

$f' (- \frac{3}{2}) = - \frac{27}{4} \cdot \frac{4}{9}$

Step 7.2.4

Cancel the common factor of $9$ .

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Step 7.2.4.1

Move the leading negative in $- \frac{27}{4}$ into the numerator.

$f' (- \frac{3}{2}) = \frac{- 27}{4} \cdot \frac{4}{9}$

Step 7.2.4.2

Factor $9$ out of $- 27$ .

$f' (- \frac{3}{2}) = \frac{9 (- 3)}{4} \cdot \frac{4}{9}$

Step 7.2.4.3

Cancel the common factor.

$f' (- \frac{3}{2}) = \frac{9 \cdot - 3}{4} \cdot \frac{4}{9}$

Step 7.2.4.4

Rewrite the expression.

$f' (- \frac{3}{2}) = \frac{- 3}{4} \cdot 4$

Step 7.2.5

Cancel the common factor of $4$ .

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Step 7.2.5.1

Cancel the common factor.

$f' (- \frac{3}{2}) = \frac{- 3}{4} \cdot 4$

Step 7.2.5.2

Rewrite the expression.

$f' (- \frac{3}{2}) = - 3$

Step 7.2.6

The final answer is $- 3$ .

$- 3$

Step 7.3

At $x = - \frac{3}{2}$ the derivative is $- 3$ . Since this is negative, the function is decreasing on $(- 3, 0)$ .

Decreasing on $(- 3, 0)$ since $f' (x) < 0$

Step 8

Substitute a value from the interval $(0, 3)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $\frac{3}{2}$ in the expression.

$f' (\frac{3}{2}) = \frac{((\frac{3}{2}) + 3) ((\frac{3}{2}) - 3)}{{(\frac{3}{2})}^{2}}$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify the numerator.

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Step 8.2.1.1

To write $3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (\frac{3}{2}) = \frac{(\frac{3}{2} + 3 \cdot \frac{2}{2}) (\frac{3}{2} - 3)}{{(\frac{3}{2})}^{2}}$

Step 8.2.1.2

Combine $3$ and $\frac{2}{2}$ .

$f' (\frac{3}{2}) = \frac{(\frac{3}{2} + \frac{3 \cdot 2}{2}) (\frac{3}{2} - 3)}{{(\frac{3}{2})}^{2}}$

Step 8.2.1.3

Combine the numerators over the common denominator.

$f' (\frac{3}{2}) = \frac{\frac{3 + 3 \cdot 2}{2} \cdot (\frac{3}{2} - 3)}{{(\frac{3}{2})}^{2}}$

Step 8.2.1.4

Simplify the numerator.

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Step 8.2.1.4.1

Multiply $3$ by $2$ .

$f' (\frac{3}{2}) = \frac{\frac{3 + 6}{2} \cdot (\frac{3}{2} - 3)}{{(\frac{3}{2})}^{2}}$

Step 8.2.1.4.2

Add $3$ and $6$ .

$f' (\frac{3}{2}) = \frac{\frac{9}{2} \cdot (\frac{3}{2} - 3)}{{(\frac{3}{2})}^{2}}$

Step 8.2.1.5

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (\frac{3}{2}) = \frac{\frac{9}{2} \cdot (\frac{3}{2} - 3 \cdot \frac{2}{2})}{{(\frac{3}{2})}^{2}}$

Step 8.2.1.6

Combine $- 3$ and $\frac{2}{2}$ .

$f' (\frac{3}{2}) = \frac{\frac{9}{2} \cdot (\frac{3}{2} + \frac{- 3 \cdot 2}{2})}{{(\frac{3}{2})}^{2}}$

Step 8.2.1.7

Combine the numerators over the common denominator.

$f' (\frac{3}{2}) = \frac{\frac{9}{2} \cdot \frac{3 - 3 \cdot 2}{2}}{{(\frac{3}{2})}^{2}}$

Step 8.2.1.8

Simplify the numerator.

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Step 8.2.1.8.1

Multiply $- 3$ by $2$ .

$f' (\frac{3}{2}) = \frac{\frac{9}{2} \cdot \frac{3 - 6}{2}}{{(\frac{3}{2})}^{2}}$

Step 8.2.1.8.2

Subtract $6$ from $3$ .

$f' (\frac{3}{2}) = \frac{\frac{9}{2} \cdot \frac{- 3}{2}}{{(\frac{3}{2})}^{2}}$

Step 8.2.1.9

Move the negative in front of the fraction.

$f' (\frac{3}{2}) = \frac{\frac{9}{2} \cdot (- 1 (\frac{3}{2}))}{{(\frac{3}{2})}^{2}}$

Step 8.2.1.10

Combine exponents.

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Step 8.2.1.10.1

Factor out negative.

$f' (\frac{3}{2}) = \frac{- (\frac{9}{2} \cdot \frac{3}{2})}{{(\frac{3}{2})}^{2}}$

Step 8.2.1.10.2

Multiply $\frac{9}{2}$ by $\frac{3}{2}$ .

$f' (\frac{3}{2}) = \frac{- \frac{9 \cdot 3}{2 \cdot 2}}{{(\frac{3}{2})}^{2}}$

Step 8.2.1.10.3

Multiply $9$ by $3$ .

$f' (\frac{3}{2}) = \frac{- \frac{27}{2 \cdot 2}}{{(\frac{3}{2})}^{2}}$

Step 8.2.1.10.4

Multiply $2$ by $2$ .

$f' (\frac{3}{2}) = \frac{- \frac{27}{4}}{{(\frac{3}{2})}^{2}}$

Step 8.2.2

Simplify the denominator.

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Step 8.2.2.1

Apply the product rule to $\frac{3}{2}$ .

$f' (\frac{3}{2}) = \frac{- \frac{27}{4}}{\frac{3^{2}}{2^{2}}}$

Step 8.2.2.2

Raise $3$ to the power of $2$ .

$f' (\frac{3}{2}) = \frac{- \frac{27}{4}}{\frac{9}{2^{2}}}$

Step 8.2.2.3

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

$f' (\frac{3}{2}) = \frac{- \frac{27}{4}}{\frac{9}{4}}$

Step 8.2.3

Multiply the numerator by the reciprocal of the denominator.

$f' (\frac{3}{2}) = - \frac{27}{4} \cdot \frac{4}{9}$

Step 8.2.4

Cancel the common factor of $9$ .

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Step 8.2.4.1

Move the leading negative in $- \frac{27}{4}$ into the numerator.

$f' (\frac{3}{2}) = \frac{- 27}{4} \cdot \frac{4}{9}$

Step 8.2.4.2

Factor $9$ out of $- 27$ .

$f' (\frac{3}{2}) = \frac{9 (- 3)}{4} \cdot \frac{4}{9}$

Step 8.2.4.3

Cancel the common factor.

$f' (\frac{3}{2}) = \frac{9 \cdot - 3}{4} \cdot \frac{4}{9}$

Step 8.2.4.4

Rewrite the expression.

$f' (\frac{3}{2}) = \frac{- 3}{4} \cdot 4$

Step 8.2.5

Cancel the common factor of $4$ .

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Step 8.2.5.1

Cancel the common factor.

$f' (\frac{3}{2}) = \frac{- 3}{4} \cdot 4$

Step 8.2.5.2

Rewrite the expression.

$f' (\frac{3}{2}) = - 3$

Step 8.2.6

The final answer is $- 3$ .

$- 3$

Step 8.3

At $x = \frac{3}{2}$ the derivative is $- 3$ . Since this is negative, the function is decreasing on $(0, 3)$ .

Decreasing on $(0, 3)$ since $f' (x) < 0$

Step 9

Substitute a value from the interval $(3, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 9.1

Replace the variable $x$ with $4$ in the expression.

$f' (4) = \frac{((4) + 3) ((4) - 3)}{{(4)}^{2}}$

Step 9.2

Simplify the result.

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Step 9.2.1

Simplify the numerator.

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Step 9.2.1.1

Add $4$ and $3$ .

$f' (4) = \frac{7 (4 - 3)}{4^{2}}$

Step 9.2.1.2

Subtract $3$ from $4$ .

$f' (4) = \frac{7 \cdot 1}{4^{2}}$

Step 9.2.2

Simplify the expression.

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Step 9.2.2.1

Raise $4$ to the power of $2$ .

$f' (4) = \frac{7 \cdot 1}{16}$

Step 9.2.2.2

Multiply $7$ by $1$ .

$f' (4) = \frac{7}{16}$

Step 9.2.3

The final answer is $\frac{7}{16}$ .

$\frac{7}{16}$

Step 9.3

At $x = 4$ the derivative is $\frac{7}{16}$ . Since this is positive, the function is increasing on $(3, \infty)$ .

Increasing on $(3, \infty)$ since $f' (x) > 0$

Step 10

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- \infty, - 3), (3, \infty)$

Decreasing on: $(- 3, 0), (0, 3)$

Step 11