Calculus Examples

$f (x) = \frac{x^{2} - 63}{x - 8}$

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 $f (x) = x^{2} - 63$ and $g (x) = x - 8$ .

$f' (x) = \frac{(x - 8) \frac{d}{d x} (x^{2} - 63) - (x^{2} - 63) \frac{d}{d x} (x - 8)}{{(x - 8)}^{2}}$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $x^{2} - 63$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 63]$ .

$f' (x) = \frac{(x - 8) (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 63)) - (x^{2} - 63) \frac{d}{d x} (x - 8)}{{(x - 8)}^{2}}$

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 $n = 2$ .

$f' (x) = \frac{(x - 8) (2 x + \frac{d}{d x} (- 63)) - (x^{2} - 63) \frac{d}{d x} (x - 8)}{{(x - 8)}^{2}}$

Step 1.1.2.3

Since $- 63$ is constant with respect to $x$ , the derivative of $- 63$ with respect to $x$ is $0$ .

$f' (x) = \frac{(x - 8) (2 x + 0) - (x^{2} - 63) \frac{d}{d x} (x - 8)}{{(x - 8)}^{2}}$

Step 1.1.2.4

Simplify the expression.

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Step 1.1.2.4.1

Add $2 x$ and $0$ .

$f' (x) = \frac{(x - 8) (2 x) - (x^{2} - 63) \frac{d}{d x} (x - 8)}{{(x - 8)}^{2}}$

Step 1.1.2.4.2

Move $2$ to the left of $x - 8$ .

$f' (x) = \frac{2 \cdot ((x - 8) x) - (x^{2} - 63) \frac{d}{d x} (x - 8)}{{(x - 8)}^{2}}$

Step 1.1.2.5

By the Sum Rule, the derivative of $x - 8$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 8]$ .

$f' (x) = \frac{2 (x - 8) x - (x^{2} - 63) (\frac{d}{d x} (x) + \frac{d}{d x} (- 8))}{{(x - 8)}^{2}}$

Step 1.1.2.6

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

$f' (x) = \frac{2 (x - 8) x - (x^{2} - 63) (1 + \frac{d}{d x} (- 8))}{{(x - 8)}^{2}}$

Step 1.1.2.7

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$f' (x) = \frac{2 (x - 8) x - (x^{2} - 63) (1 + 0)}{{(x - 8)}^{2}}$

Step 1.1.2.8

Simplify the expression.

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Step 1.1.2.8.1

Add $1$ and $0$ .

$f' (x) = \frac{2 (x - 8) x - (x^{2} - 63) \cdot 1}{{(x - 8)}^{2}}$

Step 1.1.2.8.2

Multiply $- 1$ by $1$ .

$f' (x) = \frac{2 (x - 8) x - (x^{2} - 63)}{{(x - 8)}^{2}}$

Step 1.1.3

Simplify.

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Step 1.1.3.1

Apply the distributive property.

$f' (x) = \frac{(2 x + 2 \cdot - 8) x - (x^{2} - 63)}{{(x - 8)}^{2}}$

Step 1.1.3.2

Apply the distributive property.

$f' (x) = \frac{2 x \cdot x + 2 \cdot (- 8 x) - (x^{2} - 63)}{{(x - 8)}^{2}}$

Step 1.1.3.3

Apply the distributive property.

$f' (x) = \frac{2 x \cdot x + 2 \cdot (- 8 x) - x^{2} + 63}{{(x - 8)}^{2}}$

Step 1.1.3.4

Simplify the numerator.

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Step 1.1.3.4.1

Simplify each term.

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Step 1.1.3.4.1.1

Multiply $x$ by $x$ by adding the exponents.

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Step 1.1.3.4.1.1.1

Move $x$ .

$f' (x) = \frac{2 (x \cdot x) + 2 \cdot (- 8 x) - x^{2} + 63}{{(x - 8)}^{2}}$

Step 1.1.3.4.1.1.2

Multiply $x$ by $x$ .

$f' (x) = \frac{2 x^{2} + 2 \cdot (- 8 x) - x^{2} + 63}{{(x - 8)}^{2}}$

Step 1.1.3.4.1.2

Multiply $2$ by $- 8$ .

$f' (x) = \frac{2 x^{2} - 16 x - x^{2} + 63}{{(x - 8)}^{2}}$

Step 1.1.3.4.1.3

Multiply $- 1$ by $- 63$ .

$f' (x) = \frac{2 x^{2} - 16 x - x^{2} + 63}{{(x - 8)}^{2}}$

Step 1.1.3.4.2

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

$f' (x) = \frac{x^{2} - 16 x + 63}{{(x - 8)}^{2}}$

Step 1.1.3.5

Factor $x^{2} - 16 x + 63$ using the AC method.

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Step 1.1.3.5.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $63$ and whose sum is $- 16$ .

$- 9, - 7$

Step 1.1.3.5.2

Write the factored form using these integers.

$f' (x) = \frac{(x - 9) (x - 7)}{{(x - 8)}^{2}}$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{(x - 9) (x - 7)}{{(x - 8)}^{2}}$ .

$\frac{(x - 9) (x - 7)}{{(x - 8)}^{2}}$

Step 2

Set the first derivative equal to $0$ then solve the equation $\frac{(x - 9) (x - 7)}{{(x - 8)}^{2}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$\frac{(x - 9) (x - 7)}{{(x - 8)}^{2}} = 0$

Step 2.2

Set the numerator equal to zero.

$(x - 9) (x - 7) = 0$

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 $0$ .

$x - 9 = 0$

$x - 7 = 0$

Step 2.3.2

Set $x - 9$ equal to $0$ and solve for $x$ .

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Step 2.3.2.1

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Step 2.3.2.2

Add $9$ to both sides of the equation.

$x = 9$

Step 2.3.3

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 2.3.3.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 2.3.3.2

Add $7$ to both sides of the equation.

$x = 7$

Step 2.3.4

The final solution is all the values that make $(x - 9) (x - 7) = 0$ true.

$x = 9, 7$

Step 3

The values which make the derivative equal to $0$ are $9, 7$ .

$9, 7$

Step 4

Find where the derivative is undefined.

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Step 4.1

Set the denominator in $\frac{(x - 9) (x - 7)}{{(x - 8)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(x - 8)}^{2} = 0$

Step 4.2

Solve for $x$ .

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Step 4.2.1

Set the $x - 8$ equal to $0$ .

$x - 8 = 0$

Step 4.2.2

Add $8$ to both sides of the equation.

$x = 8$

Step 5

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

$(- \infty, 7) \cup (7, 8) \cup (8, 9) \cup (9, \infty)$

Step 6

Substitute a value from the interval $(- \infty, 7)$ 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.

$f' (6) = \frac{((6) - 9) ((6) - 7)}{{((6) - 8)}^{2}}$

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

Subtract $9$ from $6$ .

$f' (6) = \frac{- 3 (6 - 7)}{{(6 - 8)}^{2}}$

Step 6.2.1.2

Subtract $7$ from $6$ .

$f' (6) = \frac{- 3 \cdot - 1}{{(6 - 8)}^{2}}$

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

Subtract $8$ from $6$ .

$f' (6) = \frac{- 3 \cdot - 1}{{(- 2)}^{2}}$

Step 6.2.2.2

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

$f' (6) = \frac{- 3 \cdot - 1}{4}$

Step 6.2.3

Multiply $- 3$ by $- 1$ .

$f' (6) = \frac{3}{4}$

Step 6.2.4

The final answer is $\frac{3}{4}$ .

$\frac{3}{4}$

Step 6.3

At $x = 6$ the derivative is $\frac{3}{4}$ . Since this is positive, the function is increasing on $(- \infty, 7)$ .

Increasing on $(- \infty, 7)$ since $f' (x) > 0$

Step 7

Substitute a value from the interval $(7, 8)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $\frac{15}{2}$ in the expression.

$f' (\frac{15}{2}) = \frac{((\frac{15}{2}) - 9) ((\frac{15}{2}) - 7)}{{((\frac{15}{2}) - 8)}^{2}}$

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 $- 9$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (\frac{15}{2}) = \frac{(\frac{15}{2} - 9 \cdot \frac{2}{2}) (\frac{15}{2} - 7)}{{(\frac{15}{2} - 8)}^{2}}$

Step 7.2.1.2

Combine $- 9$ and $\frac{2}{2}$ .

$f' (\frac{15}{2}) = \frac{(\frac{15}{2} + \frac{- 9 \cdot 2}{2}) (\frac{15}{2} - 7)}{{(\frac{15}{2} - 8)}^{2}}$

Step 7.2.1.3

Combine the numerators over the common denominator.

$f' (\frac{15}{2}) = \frac{\frac{15 - 9 \cdot 2}{2} \cdot (\frac{15}{2} - 7)}{{(\frac{15}{2} - 8)}^{2}}$

Step 7.2.1.4

Simplify the numerator.

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Step 7.2.1.4.1

Multiply $- 9$ by $2$ .

$f' (\frac{15}{2}) = \frac{\frac{15 - 18}{2} \cdot (\frac{15}{2} - 7)}{{(\frac{15}{2} - 8)}^{2}}$

Step 7.2.1.4.2

Subtract $18$ from $15$ .

$f' (\frac{15}{2}) = \frac{\frac{- 3}{2} \cdot (\frac{15}{2} - 7)}{{(\frac{15}{2} - 8)}^{2}}$

Step 7.2.1.5

Move the negative in front of the fraction.

$f' (\frac{15}{2}) = \frac{- \frac{3}{2} \cdot (\frac{15}{2} - 7)}{{(\frac{15}{2} - 8)}^{2}}$

Step 7.2.1.6

To write $- 7$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (\frac{15}{2}) = \frac{- \frac{3}{2} \cdot (\frac{15}{2} - 7 \cdot \frac{2}{2})}{{(\frac{15}{2} - 8)}^{2}}$

Step 7.2.1.7

Combine $- 7$ and $\frac{2}{2}$ .

$f' (\frac{15}{2}) = \frac{- \frac{3}{2} \cdot (\frac{15}{2} + \frac{- 7 \cdot 2}{2})}{{(\frac{15}{2} - 8)}^{2}}$

Step 7.2.1.8

Combine the numerators over the common denominator.

$f' (\frac{15}{2}) = \frac{- \frac{3}{2} \cdot \frac{15 - 7 \cdot 2}{2}}{{(\frac{15}{2} - 8)}^{2}}$

Step 7.2.1.9

Simplify the numerator.

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Step 7.2.1.9.1

Multiply $- 7$ by $2$ .

$f' (\frac{15}{2}) = \frac{- \frac{3}{2} \cdot \frac{15 - 14}{2}}{{(\frac{15}{2} - 8)}^{2}}$

Step 7.2.1.9.2

Subtract $14$ from $15$ .

$f' (\frac{15}{2}) = \frac{- \frac{3}{2} \cdot \frac{1}{2}}{{(\frac{15}{2} - 8)}^{2}}$

Step 7.2.1.10

Combine exponents.

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Step 7.2.1.10.1

Multiply $\frac{1}{2}$ by $\frac{3}{2}$ .

$f' (\frac{15}{2}) = \frac{- \frac{3}{2 \cdot 2}}{{(\frac{15}{2} - 8)}^{2}}$

Step 7.2.1.10.2

Multiply $2$ by $2$ .

$f' (\frac{15}{2}) = \frac{- \frac{3}{4}}{{(\frac{15}{2} - 8)}^{2}}$

Step 7.2.2

Simplify the denominator.

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Step 7.2.2.1

To write $- 8$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (\frac{15}{2}) = \frac{- \frac{3}{4}}{{(\frac{15}{2} - 8 \cdot \frac{2}{2})}^{2}}$

Step 7.2.2.2

Combine $- 8$ and $\frac{2}{2}$ .

$f' (\frac{15}{2}) = \frac{- \frac{3}{4}}{{(\frac{15}{2} + \frac{- 8 \cdot 2}{2})}^{2}}$

Step 7.2.2.3

Combine the numerators over the common denominator.

$f' (\frac{15}{2}) = \frac{- \frac{3}{4}}{{(\frac{15 - 8 \cdot 2}{2})}^{2}}$

Step 7.2.2.4

Simplify the numerator.

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Step 7.2.2.4.1

Multiply $- 8$ by $2$ .

$f' (\frac{15}{2}) = \frac{- \frac{3}{4}}{{(\frac{15 - 16}{2})}^{2}}$

Step 7.2.2.4.2

Subtract $16$ from $15$ .

$f' (\frac{15}{2}) = \frac{- \frac{3}{4}}{{(\frac{- 1}{2})}^{2}}$

Step 7.2.2.5

Move the negative in front of the fraction.

$f' (\frac{15}{2}) = \frac{- \frac{3}{4}}{{(- \frac{1}{2})}^{2}}$

Step 7.2.2.6

Apply the product rule to $- \frac{1}{2}$ .

$f' (\frac{15}{2}) = \frac{- \frac{3}{4}}{{(- 1)}^{2} {(\frac{1}{2})}^{2}}$

Step 7.2.2.7

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

$f' (\frac{15}{2}) = \frac{- \frac{3}{4}}{1 {(\frac{1}{2})}^{2}}$

Step 7.2.2.8

Apply the product rule to $\frac{1}{2}$ .

$f' (\frac{15}{2}) = \frac{- \frac{3}{4}}{1 (\frac{1^{2}}{2^{2}})}$

Step 7.2.2.9

One to any power is one.

$f' (\frac{15}{2}) = \frac{- \frac{3}{4}}{1 (\frac{1}{2^{2}})}$

Step 7.2.2.10

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

$f' (\frac{15}{2}) = \frac{- \frac{3}{4}}{1 (\frac{1}{4})}$

Step 7.2.2.11

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

$f' (\frac{15}{2}) = \frac{- \frac{3}{4}}{\frac{1}{4}}$

Step 7.2.3

Multiply the numerator by the reciprocal of the denominator.

$f' (\frac{15}{2}) = - \frac{3}{4} \cdot 4$

Step 7.2.4

Cancel the common factor of $4$ .

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Step 7.2.4.1

Move the leading negative in $- \frac{3}{4}$ into the numerator.

$f' (\frac{15}{2}) = \frac{- 3}{4} \cdot 4$

Step 7.2.4.2

Cancel the common factor.

$f' (\frac{15}{2}) = \frac{- 3}{4} \cdot 4$

Step 7.2.4.3

Rewrite the expression.

$f' (\frac{15}{2}) = - 3$

Step 7.2.5

The final answer is $- 3$ .

$- 3$

Step 7.3

At $x = \frac{15}{2}$ the derivative is $- 3$ . Since this is negative, the function is decreasing on $(7, 8)$ .

Decreasing on $(7, 8)$ since $f' (x) < 0$

Step 8

Substitute a value from the interval $(8, 9)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $\frac{17}{2}$ in the expression.

$f' (\frac{17}{2}) = \frac{((\frac{17}{2}) - 9) ((\frac{17}{2}) - 7)}{{((\frac{17}{2}) - 8)}^{2}}$

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 $- 9$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (\frac{17}{2}) = \frac{(\frac{17}{2} - 9 \cdot \frac{2}{2}) (\frac{17}{2} - 7)}{{(\frac{17}{2} - 8)}^{2}}$

Step 8.2.1.2

Combine $- 9$ and $\frac{2}{2}$ .

$f' (\frac{17}{2}) = \frac{(\frac{17}{2} + \frac{- 9 \cdot 2}{2}) (\frac{17}{2} - 7)}{{(\frac{17}{2} - 8)}^{2}}$

Step 8.2.1.3

Combine the numerators over the common denominator.

$f' (\frac{17}{2}) = \frac{\frac{17 - 9 \cdot 2}{2} \cdot (\frac{17}{2} - 7)}{{(\frac{17}{2} - 8)}^{2}}$

Step 8.2.1.4

Simplify the numerator.

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Step 8.2.1.4.1

Multiply $- 9$ by $2$ .

$f' (\frac{17}{2}) = \frac{\frac{17 - 18}{2} \cdot (\frac{17}{2} - 7)}{{(\frac{17}{2} - 8)}^{2}}$

Step 8.2.1.4.2

Subtract $18$ from $17$ .

$f' (\frac{17}{2}) = \frac{\frac{- 1}{2} \cdot (\frac{17}{2} - 7)}{{(\frac{17}{2} - 8)}^{2}}$

Step 8.2.1.5

Move the negative in front of the fraction.

$f' (\frac{17}{2}) = \frac{- \frac{1}{2} \cdot (\frac{17}{2} - 7)}{{(\frac{17}{2} - 8)}^{2}}$

Step 8.2.1.6

To write $- 7$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (\frac{17}{2}) = \frac{- \frac{1}{2} \cdot (\frac{17}{2} - 7 \cdot \frac{2}{2})}{{(\frac{17}{2} - 8)}^{2}}$

Step 8.2.1.7

Combine $- 7$ and $\frac{2}{2}$ .

$f' (\frac{17}{2}) = \frac{- \frac{1}{2} \cdot (\frac{17}{2} + \frac{- 7 \cdot 2}{2})}{{(\frac{17}{2} - 8)}^{2}}$

Step 8.2.1.8

Combine the numerators over the common denominator.

$f' (\frac{17}{2}) = \frac{- \frac{1}{2} \cdot \frac{17 - 7 \cdot 2}{2}}{{(\frac{17}{2} - 8)}^{2}}$

Step 8.2.1.9

Simplify the numerator.

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Step 8.2.1.9.1

Multiply $- 7$ by $2$ .

$f' (\frac{17}{2}) = \frac{- \frac{1}{2} \cdot \frac{17 - 14}{2}}{{(\frac{17}{2} - 8)}^{2}}$

Step 8.2.1.9.2

Subtract $14$ from $17$ .

$f' (\frac{17}{2}) = \frac{- \frac{1}{2} \cdot \frac{3}{2}}{{(\frac{17}{2} - 8)}^{2}}$

Step 8.2.1.10

Combine exponents.

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Step 8.2.1.10.1

Multiply $\frac{3}{2}$ by $\frac{1}{2}$ .

$f' (\frac{17}{2}) = \frac{- \frac{3}{2 \cdot 2}}{{(\frac{17}{2} - 8)}^{2}}$

Step 8.2.1.10.2

Multiply $2$ by $2$ .

$f' (\frac{17}{2}) = \frac{- \frac{3}{4}}{{(\frac{17}{2} - 8)}^{2}}$

Step 8.2.2

Simplify the denominator.

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Step 8.2.2.1

To write $- 8$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (\frac{17}{2}) = \frac{- \frac{3}{4}}{{(\frac{17}{2} - 8 \cdot \frac{2}{2})}^{2}}$

Step 8.2.2.2

Combine $- 8$ and $\frac{2}{2}$ .

$f' (\frac{17}{2}) = \frac{- \frac{3}{4}}{{(\frac{17}{2} + \frac{- 8 \cdot 2}{2})}^{2}}$

Step 8.2.2.3

Combine the numerators over the common denominator.

$f' (\frac{17}{2}) = \frac{- \frac{3}{4}}{{(\frac{17 - 8 \cdot 2}{2})}^{2}}$

Step 8.2.2.4

Simplify the numerator.

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Step 8.2.2.4.1

Multiply $- 8$ by $2$ .

$f' (\frac{17}{2}) = \frac{- \frac{3}{4}}{{(\frac{17 - 16}{2})}^{2}}$

Step 8.2.2.4.2

Subtract $16$ from $17$ .

$f' (\frac{17}{2}) = \frac{- \frac{3}{4}}{{(\frac{1}{2})}^{2}}$

Step 8.2.2.5

Apply the product rule to $\frac{1}{2}$ .

$f' (\frac{17}{2}) = \frac{- \frac{3}{4}}{\frac{1^{2}}{2^{2}}}$

Step 8.2.2.6

One to any power is one.

$f' (\frac{17}{2}) = \frac{- \frac{3}{4}}{\frac{1}{2^{2}}}$

Step 8.2.2.7

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

$f' (\frac{17}{2}) = \frac{- \frac{3}{4}}{\frac{1}{4}}$

Step 8.2.3

Multiply the numerator by the reciprocal of the denominator.

$f' (\frac{17}{2}) = - \frac{3}{4} \cdot 4$

Step 8.2.4

Cancel the common factor of $4$ .

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Step 8.2.4.1

Move the leading negative in $- \frac{3}{4}$ into the numerator.

$f' (\frac{17}{2}) = \frac{- 3}{4} \cdot 4$

Step 8.2.4.2

Cancel the common factor.

$f' (\frac{17}{2}) = \frac{- 3}{4} \cdot 4$

Step 8.2.4.3

Rewrite the expression.

$f' (\frac{17}{2}) = - 3$

Step 8.2.5

The final answer is $- 3$ .

$- 3$

Step 8.3

At $x = \frac{17}{2}$ the derivative is $- 3$ . Since this is negative, the function is decreasing on $(8, 9)$ .

Decreasing on $(8, 9)$ since $f' (x) < 0$

Step 9

Substitute a value from the interval $(9, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 9.1

Replace the variable $x$ with $10$ in the expression.

$f' (10) = \frac{((10) - 9) ((10) - 7)}{{((10) - 8)}^{2}}$

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

Subtract $9$ from $10$ .

$f' (10) = \frac{1 (10 - 7)}{{(10 - 8)}^{2}}$

Step 9.2.1.2

Multiply $10 - 7$ by $1$ .

$f' (10) = \frac{10 - 7}{{(10 - 8)}^{2}}$

Step 9.2.1.3

Subtract $7$ from $10$ .

$f' (10) = \frac{3}{{(10 - 8)}^{2}}$

Step 9.2.2

Simplify the denominator.

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Step 9.2.2.1

Subtract $8$ from $10$ .

$f' (10) = \frac{3}{2^{2}}$

Step 9.2.2.2

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

$f' (10) = \frac{3}{4}$

Step 9.2.3

The final answer is $\frac{3}{4}$ .

$\frac{3}{4}$

Step 9.3

At $x = 10$ the derivative is $\frac{3}{4}$ . Since this is positive, the function is increasing on $(9, \infty)$ .

Increasing on $(9, \infty)$ since $f' (x) > 0$

Step 10

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

Increasing on: $(- \infty, 7), (9, \infty)$

Decreasing on: $(7, 8), (8, 9)$

Step 11