Calculus Examples

$f (x) = \frac{- x^{2} - 25}{x}$

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 $f (x) = - x^{2} - 25$ and $g (x) = x$ .

$\frac{x \frac{d}{d x} [- x^{2} - 25] - (- x^{2} - 25) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $- x^{2} - 25$ with respect to $x$ is $\frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 25]$ .

$\frac{x (\frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 25]) - (- x^{2} - 25) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.2.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$\frac{x (- \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 25]) - (- x^{2} - 25) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.2.3

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

$\frac{x (- (2 x) + \frac{d}{d x} [- 25]) - (- x^{2} - 25) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.2.4

Multiply $2$ by $- 1$ .

$\frac{x (- 2 x + \frac{d}{d x} [- 25]) - (- x^{2} - 25) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.2.5

Since $- 25$ is constant with respect to $x$ , the derivative of $- 25$ with respect to $x$ is $0$ .

$\frac{x (- 2 x + 0) - (- x^{2} - 25) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.2.6

Add $- 2 x$ and $0$ .

$\frac{x (- 2 x) - (- x^{2} - 25) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.3

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

$\frac{- 2 (x^{1} x) - (- x^{2} - 25) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.4

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

$\frac{- 2 (x^{1} x^{1}) - (- x^{2} - 25) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.5

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

$\frac{- 2 x^{1 + 1} - (- x^{2} - 25) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.6

Add $1$ and $1$ .

$\frac{- 2 x^{2} - (- x^{2} - 25) \frac{d}{d x} [x]}{x^{2}}$

Step 1.1.7

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

$\frac{- 2 x^{2} - (- x^{2} - 25) \cdot 1}{x^{2}}$

Step 1.1.8

Multiply $- 1$ by $1$ .

$\frac{- 2 x^{2} - (- x^{2} - 25)}{x^{2}}$

Step 1.1.9

Simplify.

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Step 1.1.9.1

Apply the distributive property.

$\frac{- 2 x^{2} - - x^{2} - - 25}{x^{2}}$

Step 1.1.9.2

Simplify the numerator.

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Step 1.1.9.2.1

Simplify each term.

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Step 1.1.9.2.1.1

Multiply $- - x^{2}$ .

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Step 1.1.9.2.1.1.1

Multiply $- 1$ by $- 1$ .

$\frac{- 2 x^{2} + 1 x^{2} - - 25}{x^{2}}$

Step 1.1.9.2.1.1.2

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

$\frac{- 2 x^{2} + x^{2} - - 25}{x^{2}}$

Step 1.1.9.2.1.2

Multiply $- 1$ by $- 25$ .

$\frac{- 2 x^{2} + x^{2} + 25}{x^{2}}$

Step 1.1.9.2.2

Add $- 2 x^{2}$ and $x^{2}$ .

$\frac{- x^{2} + 25}{x^{2}}$

Step 1.1.9.3

Simplify the numerator.

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Step 1.1.9.3.1

Rewrite $25$ as $5^{2}$ .

$\frac{- x^{2} + 5^{2}}{x^{2}}$

Step 1.1.9.3.2

Reorder $- x^{2}$ and $5^{2}$ .

$\frac{5^{2} - x^{2}}{x^{2}}$

Step 1.1.9.3.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5$ and $b = x$ .

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

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{(5 + x) (5 - x)}{x^{2}}$ .

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

Step 2

Set the first derivative equal to $0$ then solve the equation $\frac{(5 + x) (5 - x)}{x^{2}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$\frac{(5 + x) (5 - x)}{x^{2}} = 0$

Step 2.2

Set the numerator equal to zero.

$(5 + x) (5 - x) = 0$

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 $0$ .

$5 + x = 0$

$5 - x = 0$

Step 2.3.2

Set $5 + x$ equal to $0$ and solve for $x$ .

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Step 2.3.2.1

Set $5 + x$ equal to $0$ .

$5 + x = 0$

Step 2.3.2.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.3.3

Set $5 - x$ equal to $0$ and solve for $x$ .

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Step 2.3.3.1

Set $5 - x$ equal to $0$ .

$5 - x = 0$

Step 2.3.3.2

Solve $5 - x = 0$ for $x$ .

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Step 2.3.3.2.1

Subtract $5$ from both sides of the equation.

$- x = - 5$

Step 2.3.3.2.2

Divide each term in $- x = - 5$ by $- 1$ and simplify.

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Step 2.3.3.2.2.1

Divide each term in $- x = - 5$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 5}{- 1}$

Step 2.3.3.2.2.2

Simplify the left side.

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Step 2.3.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 5}{- 1}$

Step 2.3.3.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 5}{- 1}$

Step 2.3.3.2.2.3

Simplify the right side.

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Step 2.3.3.2.2.3.1

Divide $- 5$ by $- 1$ .

$x = 5$

Step 2.3.4

The final solution is all the values that make $(5 + x) (5 - x) = 0$ true.

$x = - 5, 5$

Step 3

The values which make the derivative equal to $0$ are $- 5, 5$ .

$- 5, 5$

Step 4

Find where the derivative is undefined.

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Step 4.1

Set the denominator in $\frac{(5 + x) (5 - x)}{x^{2}}$ 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.

$(- \infty, - 5) \cup (- 5, 0) \cup (0, 5) \cup (5, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - 5)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 6$ in the expression.

$f' (- 6) = \frac{(5 - 6) (5 - (- 6))}{{(- 6)}^{2}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Remove parentheses.

$f' (- 6) = \frac{(5 - 6) (5 - (- 6))}{{(- 6)}^{2}}$

Step 6.2.2

Simplify the numerator.

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Step 6.2.2.1

Subtract $6$ from $5$ .

$f' (- 6) = \frac{- 1 (5 - (- 6))}{{(- 6)}^{2}}$

Step 6.2.2.2

Multiply $- 1$ by $- 6$ .

$f' (- 6) = \frac{- 1 (5 + 6)}{{(- 6)}^{2}}$

Step 6.2.2.3

Add $5$ and $6$ .

$f' (- 6) = \frac{- 1 \cdot 11}{{(- 6)}^{2}}$

Step 6.2.3

Simplify the expression.

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Step 6.2.3.1

Raise $- 6$ to the power of $2$ .

$f' (- 6) = \frac{- 1 \cdot 11}{36}$

Step 6.2.3.2

Multiply $- 1$ by $11$ .

$f' (- 6) = \frac{- 11}{36}$

Step 6.2.3.3

Move the negative in front of the fraction.

$f' (- 6) = - \frac{11}{36}$

Step 6.2.4

The final answer is $- \frac{11}{36}$ .

$- \frac{11}{36}$

Step 6.3

At $x = - 6$ the derivative is $- \frac{11}{36}$ . Since this is negative, the function is decreasing on $(- \infty, - 5)$ .

Decreasing on $(- \infty, - 5)$ since $f' (x) < 0$

Step 7

Substitute a value from the interval $(- 5, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $- \frac{5}{2}$ in the expression.

$f' (- \frac{5}{2}) = \frac{(5 - \frac{5}{2}) (5 - (- \frac{5}{2}))}{{(- \frac{5}{2})}^{2}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Remove parentheses.

$f' (- \frac{5}{2}) = \frac{(5 - \frac{5}{2}) (5 - (- \frac{5}{2}))}{{(- \frac{5}{2})}^{2}}$

Step 7.2.2

Simplify the numerator.

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Step 7.2.2.1

To write $5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (- \frac{5}{2}) = \frac{(5 \cdot \frac{2}{2} - \frac{5}{2}) (5 + \frac{5}{2})}{{(- \frac{5}{2})}^{2}}$

Step 7.2.2.2

Combine $5$ and $\frac{2}{2}$ .

$f' (- \frac{5}{2}) = \frac{(\frac{5 \cdot 2}{2} - \frac{5}{2}) (5 + \frac{5}{2})}{{(- \frac{5}{2})}^{2}}$

Step 7.2.2.3

Combine the numerators over the common denominator.

$f' (- \frac{5}{2}) = \frac{\frac{5 \cdot 2 - 5}{2} \cdot (5 + \frac{5}{2})}{{(- \frac{5}{2})}^{2}}$

Step 7.2.2.4

Simplify the numerator.

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Step 7.2.2.4.1

Multiply $5$ by $2$ .

$f' (- \frac{5}{2}) = \frac{\frac{10 - 5}{2} \cdot (5 + \frac{5}{2})}{{(- \frac{5}{2})}^{2}}$

Step 7.2.2.4.2

Subtract $5$ from $10$ .

$f' (- \frac{5}{2}) = \frac{\frac{5}{2} \cdot (5 + \frac{5}{2})}{{(- \frac{5}{2})}^{2}}$

Step 7.2.2.5

Multiply $- - \frac{5}{2}$ .

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Step 7.2.2.5.1

Multiply $- 1$ by $- 1$ .

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

Step 7.2.2.5.2

Multiply $\frac{5}{2}$ by $1$ .

$f' (- \frac{5}{2}) = \frac{\frac{5}{2} \cdot (5 + \frac{5}{2})}{{(- \frac{5}{2})}^{2}}$

Step 7.2.2.6

To write $5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (- \frac{5}{2}) = \frac{\frac{5}{2} \cdot (5 \cdot \frac{2}{2} + \frac{5}{2})}{{(- \frac{5}{2})}^{2}}$

Step 7.2.2.7

Combine $5$ and $\frac{2}{2}$ .

$f' (- \frac{5}{2}) = \frac{\frac{5}{2} \cdot (\frac{5 \cdot 2}{2} + \frac{5}{2})}{{(- \frac{5}{2})}^{2}}$

Step 7.2.2.8

Combine the numerators over the common denominator.

$f' (- \frac{5}{2}) = \frac{\frac{5}{2} \cdot \frac{5 \cdot 2 + 5}{2}}{{(- \frac{5}{2})}^{2}}$

Step 7.2.2.9

Simplify the numerator.

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Step 7.2.2.9.1

Multiply $5$ by $2$ .

$f' (- \frac{5}{2}) = \frac{\frac{5}{2} \cdot \frac{10 + 5}{2}}{{(- \frac{5}{2})}^{2}}$

Step 7.2.2.9.2

Add $10$ and $5$ .

$f' (- \frac{5}{2}) = \frac{\frac{5}{2} \cdot \frac{15}{2}}{{(- \frac{5}{2})}^{2}}$

Step 7.2.3

Simplify the denominator.

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Step 7.2.3.1

Apply the product rule to $- \frac{5}{2}$ .

$f' (- \frac{5}{2}) = \frac{\frac{5}{2} \cdot \frac{15}{2}}{{(- 1)}^{2} {(\frac{5}{2})}^{2}}$

Step 7.2.3.2

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

$f' (- \frac{5}{2}) = \frac{\frac{5}{2} \cdot \frac{15}{2}}{1 {(\frac{5}{2})}^{2}}$

Step 7.2.3.3

Apply the product rule to $\frac{5}{2}$ .

$f' (- \frac{5}{2}) = \frac{\frac{5}{2} \cdot \frac{15}{2}}{1 (\frac{5^{2}}{2^{2}})}$

Step 7.2.3.4

Raise $5$ to the power of $2$ .

$f' (- \frac{5}{2}) = \frac{\frac{5}{2} \cdot \frac{15}{2}}{1 (\frac{25}{2^{2}})}$

Step 7.2.3.5

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

$f' (- \frac{5}{2}) = \frac{\frac{5}{2} \cdot \frac{15}{2}}{1 (\frac{25}{4})}$

Step 7.2.3.6

Multiply $\frac{25}{4}$ by $1$ .

$f' (- \frac{5}{2}) = \frac{\frac{5}{2} \cdot \frac{15}{2}}{\frac{25}{4}}$

Step 7.2.4

Combine fractions.

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Step 7.2.4.1

Multiply $\frac{5}{2}$ by $\frac{15}{2}$ .

$f' (- \frac{5}{2}) = \frac{\frac{5 \cdot 15}{2 \cdot 2}}{\frac{25}{4}}$

Step 7.2.4.2

Multiply.

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Step 7.2.4.2.1

Multiply $5$ by $15$ .

$f' (- \frac{5}{2}) = \frac{\frac{75}{2 \cdot 2}}{\frac{25}{4}}$

Step 7.2.4.2.2

Multiply $2$ by $2$ .

$f' (- \frac{5}{2}) = \frac{\frac{75}{4}}{\frac{25}{4}}$

Step 7.2.5

Multiply the numerator by the reciprocal of the denominator.

$f' (- \frac{5}{2}) = \frac{75}{4} \cdot \frac{4}{25}$

Step 7.2.6

Cancel the common factor of $25$ .

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Step 7.2.6.1

Factor $25$ out of $75$ .

$f' (- \frac{5}{2}) = \frac{25 (3)}{4} \cdot \frac{4}{25}$

Step 7.2.6.2

Cancel the common factor.

$f' (- \frac{5}{2}) = \frac{25 \cdot 3}{4} \cdot \frac{4}{25}$

Step 7.2.6.3

Rewrite the expression.

$f' (- \frac{5}{2}) = \frac{3}{4} \cdot 4$

Step 7.2.7

Cancel the common factor of $4$ .

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Step 7.2.7.1

Cancel the common factor.

$f' (- \frac{5}{2}) = \frac{3}{4} \cdot 4$

Step 7.2.7.2

Rewrite the expression.

$f' (- \frac{5}{2}) = 3$

Step 7.2.8

The final answer is $3$ .

$3$

Step 7.3

At $x = - \frac{5}{2}$ the derivative is $3$ . Since this is positive, the function is increasing on $(- 5, 0)$ .

Increasing on $(- 5, 0)$ since $f' (x) > 0$

Step 8

Substitute a value from the interval $(0, 5)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $\frac{5}{2}$ in the expression.

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

Step 8.2

Simplify the result.

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Step 8.2.1

Remove parentheses.

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

Step 8.2.2

Simplify the numerator.

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Step 8.2.2.1

To write $5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (\frac{5}{2}) = \frac{(5 \cdot \frac{2}{2} + \frac{5}{2}) (5 - \frac{5}{2})}{{(\frac{5}{2})}^{2}}$

Step 8.2.2.2

Combine $5$ and $\frac{2}{2}$ .

$f' (\frac{5}{2}) = \frac{(\frac{5 \cdot 2}{2} + \frac{5}{2}) (5 - \frac{5}{2})}{{(\frac{5}{2})}^{2}}$

Step 8.2.2.3

Combine the numerators over the common denominator.

$f' (\frac{5}{2}) = \frac{\frac{5 \cdot 2 + 5}{2} \cdot (5 - \frac{5}{2})}{{(\frac{5}{2})}^{2}}$

Step 8.2.2.4

Simplify the numerator.

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Step 8.2.2.4.1

Multiply $5$ by $2$ .

$f' (\frac{5}{2}) = \frac{\frac{10 + 5}{2} \cdot (5 - \frac{5}{2})}{{(\frac{5}{2})}^{2}}$

Step 8.2.2.4.2

Add $10$ and $5$ .

$f' (\frac{5}{2}) = \frac{\frac{15}{2} \cdot (5 - \frac{5}{2})}{{(\frac{5}{2})}^{2}}$

Step 8.2.2.5

To write $5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (\frac{5}{2}) = \frac{\frac{15}{2} \cdot (5 \cdot \frac{2}{2} - \frac{5}{2})}{{(\frac{5}{2})}^{2}}$

Step 8.2.2.6

Combine $5$ and $\frac{2}{2}$ .

$f' (\frac{5}{2}) = \frac{\frac{15}{2} \cdot (\frac{5 \cdot 2}{2} - \frac{5}{2})}{{(\frac{5}{2})}^{2}}$

Step 8.2.2.7

Combine the numerators over the common denominator.

$f' (\frac{5}{2}) = \frac{\frac{15}{2} \cdot \frac{5 \cdot 2 - 5}{2}}{{(\frac{5}{2})}^{2}}$

Step 8.2.2.8

Simplify the numerator.

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Step 8.2.2.8.1

Multiply $5$ by $2$ .

$f' (\frac{5}{2}) = \frac{\frac{15}{2} \cdot \frac{10 - 5}{2}}{{(\frac{5}{2})}^{2}}$

Step 8.2.2.8.2

Subtract $5$ from $10$ .

$f' (\frac{5}{2}) = \frac{\frac{15}{2} \cdot \frac{5}{2}}{{(\frac{5}{2})}^{2}}$

Step 8.2.3

Simplify the denominator.

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Step 8.2.3.1

Apply the product rule to $\frac{5}{2}$ .

$f' (\frac{5}{2}) = \frac{\frac{15}{2} \cdot \frac{5}{2}}{\frac{5^{2}}{2^{2}}}$

Step 8.2.3.2

Raise $5$ to the power of $2$ .

$f' (\frac{5}{2}) = \frac{\frac{15}{2} \cdot \frac{5}{2}}{\frac{25}{2^{2}}}$

Step 8.2.3.3

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

$f' (\frac{5}{2}) = \frac{\frac{15}{2} \cdot \frac{5}{2}}{\frac{25}{4}}$

Step 8.2.4

Combine fractions.

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Step 8.2.4.1

Multiply $\frac{15}{2}$ by $\frac{5}{2}$ .

$f' (\frac{5}{2}) = \frac{\frac{15 \cdot 5}{2 \cdot 2}}{\frac{25}{4}}$

Step 8.2.4.2

Multiply.

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Step 8.2.4.2.1

Multiply $15$ by $5$ .

$f' (\frac{5}{2}) = \frac{\frac{75}{2 \cdot 2}}{\frac{25}{4}}$

Step 8.2.4.2.2

Multiply $2$ by $2$ .

$f' (\frac{5}{2}) = \frac{\frac{75}{4}}{\frac{25}{4}}$

Step 8.2.5

Multiply the numerator by the reciprocal of the denominator.

$f' (\frac{5}{2}) = \frac{75}{4} \cdot \frac{4}{25}$

Step 8.2.6

Cancel the common factor of $25$ .

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Step 8.2.6.1

Factor $25$ out of $75$ .

$f' (\frac{5}{2}) = \frac{25 (3)}{4} \cdot \frac{4}{25}$

Step 8.2.6.2

Cancel the common factor.

$f' (\frac{5}{2}) = \frac{25 \cdot 3}{4} \cdot \frac{4}{25}$

Step 8.2.6.3

Rewrite the expression.

$f' (\frac{5}{2}) = \frac{3}{4} \cdot 4$

Step 8.2.7

Cancel the common factor of $4$ .

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Step 8.2.7.1

Cancel the common factor.

$f' (\frac{5}{2}) = \frac{3}{4} \cdot 4$

Step 8.2.7.2

Rewrite the expression.

$f' (\frac{5}{2}) = 3$

Step 8.2.8

The final answer is $3$ .

$3$

Step 8.3

At $x = \frac{5}{2}$ the derivative is $3$ . Since this is positive, the function is increasing on $(0, 5)$ .

Increasing on $(0, 5)$ since $f' (x) > 0$

Step 9

Substitute a value from the interval $(5, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 9.1

Replace the variable $x$ with $6$ in the expression.

$f' (6) = \frac{(5 + 6) (5 - (6))}{{(6)}^{2}}$

Step 9.2

Simplify the result.

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Step 9.2.1

Remove parentheses.

$f' (6) = \frac{(5 + 6) (5 - (6))}{{(6)}^{2}}$

Step 9.2.2

Simplify the numerator.

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Step 9.2.2.1

Add $5$ and $6$ .

$f' (6) = \frac{11 (5 - (6))}{6^{2}}$

Step 9.2.2.2

Multiply $- 1$ by $6$ .

$f' (6) = \frac{11 (5 - 6)}{6^{2}}$

Step 9.2.2.3

Subtract $6$ from $5$ .

$f' (6) = \frac{11 \cdot - 1}{6^{2}}$

Step 9.2.3

Simplify the expression.

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Step 9.2.3.1

Raise $6$ to the power of $2$ .

$f' (6) = \frac{11 \cdot - 1}{36}$

Step 9.2.3.2

Multiply $11$ by $- 1$ .

$f' (6) = \frac{- 11}{36}$

Step 9.2.3.3

Move the negative in front of the fraction.

$f' (6) = - \frac{11}{36}$

Step 9.2.4

The final answer is $- \frac{11}{36}$ .

$- \frac{11}{36}$

Step 9.3

At $x = 6$ the derivative is $- \frac{11}{36}$ . Since this is negative, the function is decreasing on $(5, \infty)$ .

Decreasing on $(5, \infty)$ since $f' (x) < 0$

Step 10

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

Increasing on: $(- 5, 0), (0, 5)$

Decreasing on: $(- \infty, - 5), (5, \infty)$

Step 11