Calculus Examples

$f (x) = \frac{x^{2} - 14 x + 49}{x - 10}$

Step 1

Find the first derivative of the function.

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Step 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 $f (x) = x^{2} - 14 x + 49$ and $g (x) = x - 10$ .

$\frac{(x - 10) \frac{d}{d x} [x^{2} - 14 x + 49] - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 1.2

Differentiate.

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Step 1.2.1

By the Sum Rule, the derivative of $x^{2} - 14 x + 49$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 14 x] + \frac{d}{d x} [49]$ .

$\frac{(x - 10) (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 14 x] + \frac{d}{d x} [49]) - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 1.2.2

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

$\frac{(x - 10) (2 x + \frac{d}{d x} [- 14 x] + \frac{d}{d x} [49]) - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 1.2.3

Since $- 14$ is constant with respect to $x$ , the derivative of $- 14 x$ with respect to $x$ is $- 14 \frac{d}{d x} [x]$ .

$\frac{(x - 10) (2 x - 14 \frac{d}{d x} [x] + \frac{d}{d x} [49]) - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 1.2.4

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

$\frac{(x - 10) (2 x - 14 \cdot 1 + \frac{d}{d x} [49]) - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 1.2.5

Multiply $- 14$ by $1$ .

$\frac{(x - 10) (2 x - 14 + \frac{d}{d x} [49]) - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 1.2.6

Since $49$ is constant with respect to $x$ , the derivative of $49$ with respect to $x$ is $0$ .

$\frac{(x - 10) (2 x - 14 + 0) - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 1.2.7

Add $2 x - 14$ and $0$ .

$\frac{(x - 10) (2 x - 14) - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 1.2.8

By the Sum Rule, the derivative of $x - 10$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 10]$ .

$\frac{(x - 10) (2 x - 14) - (x^{2} - 14 x + 49) (\frac{d}{d x} [x] + \frac{d}{d x} [- 10])}{{(x - 10)}^{2}}$

Step 1.2.9

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

$\frac{(x - 10) (2 x - 14) - (x^{2} - 14 x + 49) (1 + \frac{d}{d x} [- 10])}{{(x - 10)}^{2}}$

Step 1.2.10

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10$ with respect to $x$ is $0$ .

$\frac{(x - 10) (2 x - 14) - (x^{2} - 14 x + 49) (1 + 0)}{{(x - 10)}^{2}}$

Step 1.2.11

Simplify the expression.

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Step 1.2.11.1

Add $1$ and $0$ .

$\frac{(x - 10) (2 x - 14) - (x^{2} - 14 x + 49) \cdot 1}{{(x - 10)}^{2}}$

Step 1.2.11.2

Multiply $- 1$ by $1$ .

$\frac{(x - 10) (2 x - 14) - (x^{2} - 14 x + 49)}{{(x - 10)}^{2}}$

Step 1.3

Simplify.

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Step 1.3.1

Apply the distributive property.

$\frac{(x - 10) (2 x - 14) - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 1.3.2

Simplify the numerator.

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Step 1.3.2.1

Simplify each term.

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Step 1.3.2.1.1

Expand $(x - 10) (2 x - 14)$ using the FOIL Method.

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Step 1.3.2.1.1.1

Apply the distributive property.

$\frac{x (2 x - 14) - 10 (2 x - 14) - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 1.3.2.1.1.2

Apply the distributive property.

$\frac{x (2 x) + x \cdot - 14 - 10 (2 x - 14) - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 1.3.2.1.1.3

Apply the distributive property.

$\frac{x (2 x) + x \cdot - 14 - 10 (2 x) - 10 \cdot - 14 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 1.3.2.1.2

Simplify and combine like terms.

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Step 1.3.2.1.2.1

Simplify each term.

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Step 1.3.2.1.2.1.1

Rewrite using the commutative property of multiplication.

$\frac{2 x \cdot x + x \cdot - 14 - 10 (2 x) - 10 \cdot - 14 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 1.3.2.1.2.1.2

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

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Step 1.3.2.1.2.1.2.1

Move $x$ .

$\frac{2 (x \cdot x) + x \cdot - 14 - 10 (2 x) - 10 \cdot - 14 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 1.3.2.1.2.1.2.2

Multiply $x$ by $x$ .

$\frac{2 x^{2} + x \cdot - 14 - 10 (2 x) - 10 \cdot - 14 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 1.3.2.1.2.1.3

Move $- 14$ to the left of $x$ .

$\frac{2 x^{2} - 14 \cdot x - 10 (2 x) - 10 \cdot - 14 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 1.3.2.1.2.1.4

Multiply $2$ by $- 10$ .

$\frac{2 x^{2} - 14 x - 20 x - 10 \cdot - 14 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 1.3.2.1.2.1.5

Multiply $- 10$ by $- 14$ .

$\frac{2 x^{2} - 14 x - 20 x + 140 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 1.3.2.1.2.2

Subtract $20 x$ from $- 14 x$ .

$\frac{2 x^{2} - 34 x + 140 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 1.3.2.1.3

Multiply $- 14$ by $- 1$ .

$\frac{2 x^{2} - 34 x + 140 - x^{2} + 14 x - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 1.3.2.1.4

Multiply $- 1$ by $49$ .

$\frac{2 x^{2} - 34 x + 140 - x^{2} + 14 x - 49}{{(x - 10)}^{2}}$

Step 1.3.2.2

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

$\frac{x^{2} - 34 x + 140 + 14 x - 49}{{(x - 10)}^{2}}$

Step 1.3.2.3

Add $- 34 x$ and $14 x$ .

$\frac{x^{2} - 20 x + 140 - 49}{{(x - 10)}^{2}}$

Step 1.3.2.4

Subtract $49$ from $140$ .

$\frac{x^{2} - 20 x + 91}{{(x - 10)}^{2}}$

Step 1.3.3

Factor $x^{2} - 20 x + 91$ using the AC method.

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Step 1.3.3.1

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 $91$ and whose sum is $- 20$ .

$- 13, - 7$

Step 1.3.3.2

Write the factored form using these integers.

$\frac{(x - 13) (x - 7)}{{(x - 10)}^{2}}$

Step 2

Find the second derivative of the function.

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Step 2.1

$f'' (x) = \frac{{(x - 10)}^{2} \frac{d}{d x} ((x - 13) (x - 7)) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{({(x - 10)}^{2})}^{2}}$

Step 2.2

Multiply the exponents in ${({(x - 10)}^{2})}^{2}$ .

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Step 2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f'' (x) = \frac{{(x - 10)}^{2} \frac{d}{d x} ((x - 13) (x - 7)) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{2 \cdot 2}}$

Step 2.2.2

Multiply $2$ by $2$ .

$f'' (x) = \frac{{(x - 10)}^{2} \frac{d}{d x} ((x - 13) (x - 7)) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{4}}$

Step 2.3

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x - 13$ and $g (x) = x - 7$ .

$f'' (x) = \frac{{(x - 10)}^{2} ((x - 13) \frac{d}{d x} (x - 7) + (x - 7) \frac{d}{d x} (x - 13)) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{4}}$

Step 2.4

Differentiate.

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Step 2.4.1

By the Sum Rule, the derivative of $x - 7$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 7]$ .

$f'' (x) = \frac{{(x - 10)}^{2} ((x - 13) (\frac{d}{d x} (x) + \frac{d}{d x} (- 7)) + (x - 7) \frac{d}{d x} (x - 13)) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{4}}$

Step 2.4.2

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

$f'' (x) = \frac{{(x - 10)}^{2} ((x - 13) (1 + \frac{d}{d x} (- 7)) + (x - 7) \frac{d}{d x} (x - 13)) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{4}}$

Step 2.4.3

Since $- 7$ is constant with respect to $x$ , the derivative of $- 7$ with respect to $x$ is $0$ .

$f'' (x) = \frac{{(x - 10)}^{2} ((x - 13) (1 + 0) + (x - 7) \frac{d}{d x} (x - 13)) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{4}}$

Step 2.4.4

Simplify the expression.

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Step 2.4.4.1

Add $1$ and $0$ .

$f'' (x) = \frac{{(x - 10)}^{2} ((x - 13) \cdot 1 + (x - 7) \frac{d}{d x} (x - 13)) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{4}}$

Step 2.4.4.2

Multiply $x - 13$ by $1$ .

$f'' (x) = \frac{{(x - 10)}^{2} (x - 13 + (x - 7) \frac{d}{d x} (x - 13)) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{4}}$

Step 2.4.5

By the Sum Rule, the derivative of $x - 13$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 13]$ .

$f'' (x) = \frac{{(x - 10)}^{2} (x - 13 + (x - 7) (\frac{d}{d x} (x) + \frac{d}{d x} (- 13))) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{4}}$

Step 2.4.6

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

$f'' (x) = \frac{{(x - 10)}^{2} (x - 13 + (x - 7) (1 + \frac{d}{d x} (- 13))) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{4}}$

Step 2.4.7

Since $- 13$ is constant with respect to $x$ , the derivative of $- 13$ with respect to $x$ is $0$ .

$f'' (x) = \frac{{(x - 10)}^{2} (x - 13 + (x - 7) (1 + 0)) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{4}}$

Step 2.4.8

Simplify by adding terms.

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Step 2.4.8.1

Add $1$ and $0$ .

$f'' (x) = \frac{{(x - 10)}^{2} (x - 13 + (x - 7) \cdot 1) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{4}}$

Step 2.4.8.2

Multiply $x - 7$ by $1$ .

$f'' (x) = \frac{{(x - 10)}^{2} (x - 13 + x - 7) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{4}}$

Step 2.4.8.3

Add $x$ and $x$ .

$f'' (x) = \frac{{(x - 10)}^{2} (2 x - 13 - 7) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{4}}$

Step 2.4.8.4

Subtract $7$ from $- 13$ .

$f'' (x) = \frac{{(x - 10)}^{2} (2 x - 20) - (x - 13) (x - 7) \frac{d}{d x} {(x - 10)}^{2}}{{(x - 10)}^{4}}$

Step 2.5

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = x - 10$ .

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Step 2.5.1

To apply the Chain Rule, set $u$ as $x - 10$ .

$f'' (x) = \frac{{(x - 10)}^{2} (2 x - 20) - (x - 13) (x - 7) (\frac{d}{d u} (u^{2}) \frac{d}{d x} (x - 10))}{{(x - 10)}^{4}}$

Step 2.5.2

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

$f'' (x) = \frac{{(x - 10)}^{2} (2 x - 20) - (x - 13) (x - 7) (2 u \frac{d}{d x} (x - 10))}{{(x - 10)}^{4}}$

Step 2.5.3

Replace all occurrences of $u$ with $x - 10$ .

$f'' (x) = \frac{{(x - 10)}^{2} (2 x - 20) - (x - 13) (x - 7) (2 (x - 10) \frac{d}{d x} (x - 10))}{{(x - 10)}^{4}}$

Step 2.6

Simplify with factoring out.

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Step 2.6.1

Multiply $2$ by $- 1$ .

$f'' (x) = \frac{{(x - 10)}^{2} (2 x - 20) - 2 (x - 13) (x - 7) ((x - 10) \frac{d}{d x} (x - 10))}{{(x - 10)}^{4}}$

Step 2.6.2

Factor $x - 10$ out of ${(x - 10)}^{2} (2 x - 20) - 2 (x - 13) (x - 7) ((x - 10) \frac{d}{d x} [x - 10])$ .

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Step 2.6.2.1

Factor $x - 10$ out of ${(x - 10)}^{2} (2 x - 20)$ .

$f'' (x) = \frac{(x - 10) ((x - 10) (2 x - 20)) - 2 (x - 13) (x - 7) ((x - 10) \frac{d}{d x} (x - 10))}{{(x - 10)}^{4}}$

Step 2.6.2.2

Factor $x - 10$ out of $- 2 (x - 13) (x - 7) ((x - 10) \frac{d}{d x} [x - 10])$ .

$f'' (x) = \frac{(x - 10) ((x - 10) (2 x - 20)) + (x - 10) (- 2 (x - 13) (x - 7) (\frac{d}{d x} (x - 10)))}{{(x - 10)}^{4}}$

Step 2.6.2.3

Factor $x - 10$ out of $(x - 10) ((x - 10) (2 x - 20)) + (x - 10) (- 2 (x - 13) (x - 7) (\frac{d}{d x} [x - 10]))$ .

$f'' (x) = \frac{(x - 10) ((x - 10) (2 x - 20) - 2 (x - 13) (x - 7) (\frac{d}{d x} (x - 10)))}{{(x - 10)}^{4}}$

Step 2.7

Cancel the common factors.

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Step 2.7.1

Factor $x - 10$ out of ${(x - 10)}^{4}$ .

$f'' (x) = \frac{(x - 10) ((x - 10) (2 x - 20) - 2 (x - 13) (x - 7) (\frac{d}{d x} (x - 10)))}{(x - 10) {(x - 10)}^{3}}$

Step 2.7.2

Cancel the common factor.

$f'' (x) = \frac{(x - 10) ((x - 10) (2 x - 20) - 2 (x - 13) (x - 7) (\frac{d}{d x} (x - 10)))}{(x - 10) {(x - 10)}^{3}}$

Step 2.7.3

Rewrite the expression.

$f'' (x) = \frac{(x - 10) (2 x - 20) - 2 (x - 13) (x - 7) (\frac{d}{d x} (x - 10))}{{(x - 10)}^{3}}$

Step 2.8

By the Sum Rule, the derivative of $x - 10$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 10]$ .

$f'' (x) = \frac{(x - 10) (2 x - 20) - 2 (x - 13) (x - 7) (\frac{d}{d x} (x) + \frac{d}{d x} (- 10))}{{(x - 10)}^{3}}$

Step 2.9

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

$f'' (x) = \frac{(x - 10) (2 x - 20) - 2 (x - 13) (x - 7) (1 + \frac{d}{d x} (- 10))}{{(x - 10)}^{3}}$

Step 2.10

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10$ with respect to $x$ is $0$ .

$f'' (x) = \frac{(x - 10) (2 x - 20) - 2 (x - 13) (x - 7) (1 + 0)}{{(x - 10)}^{3}}$

Step 2.11

Simplify the expression.

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Step 2.11.1

Add $1$ and $0$ .

$f'' (x) = \frac{(x - 10) (2 x - 20) - 2 (x - 13) (x - 7) \cdot 1}{{(x - 10)}^{3}}$

Step 2.11.2

Multiply $- 2$ by $1$ .

$f'' (x) = \frac{(x - 10) (2 x - 20) - 2 (x - 13) (x - 7)}{{(x - 10)}^{3}}$

Step 2.12

Simplify.

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Step 2.12.1

Apply the distributive property.

$f'' (x) = \frac{(x - 10) (2 x - 20) + (- 2 x - 2 \cdot - 13) (x - 7)}{{(x - 10)}^{3}}$

Step 2.12.2

Simplify the numerator.

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Step 2.12.2.1

Simplify each term.

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Step 2.12.2.1.1

Expand $(x - 10) (2 x - 20)$ using the FOIL Method.

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Step 2.12.2.1.1.1

Apply the distributive property.

$f'' (x) = \frac{x (2 x - 20) - 10 (2 x - 20) + (- 2 x - 2 \cdot - 13) (x - 7)}{{(x - 10)}^{3}}$

Step 2.12.2.1.1.2

Apply the distributive property.

$f'' (x) = \frac{x (2 x) + x \cdot - 20 - 10 (2 x - 20) + (- 2 x - 2 \cdot - 13) (x - 7)}{{(x - 10)}^{3}}$

Step 2.12.2.1.1.3

Apply the distributive property.

$f'' (x) = \frac{x (2 x) + x \cdot - 20 - 10 (2 x) - 10 \cdot - 20 + (- 2 x - 2 \cdot - 13) (x - 7)}{{(x - 10)}^{3}}$

Step 2.12.2.1.2

Simplify and combine like terms.

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Step 2.12.2.1.2.1

Simplify each term.

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Step 2.12.2.1.2.1.1

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{2 x \cdot x + x \cdot - 20 - 10 (2 x) - 10 \cdot - 20 + (- 2 x - 2 \cdot - 13) (x - 7)}{{(x - 10)}^{3}}$

Step 2.12.2.1.2.1.2

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

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Step 2.12.2.1.2.1.2.1

Move $x$ .

$f'' (x) = \frac{2 (x \cdot x) + x \cdot - 20 - 10 (2 x) - 10 \cdot - 20 + (- 2 x - 2 \cdot - 13) (x - 7)}{{(x - 10)}^{3}}$

Step 2.12.2.1.2.1.2.2

Multiply $x$ by $x$ .

$f'' (x) = \frac{2 x^{2} + x \cdot - 20 - 10 (2 x) - 10 \cdot - 20 + (- 2 x - 2 \cdot - 13) (x - 7)}{{(x - 10)}^{3}}$

Step 2.12.2.1.2.1.3

Move $- 20$ to the left of $x$ .

$f'' (x) = \frac{2 x^{2} - 20 \cdot x - 10 (2 x) - 10 \cdot - 20 + (- 2 x - 2 \cdot - 13) (x - 7)}{{(x - 10)}^{3}}$

Step 2.12.2.1.2.1.4

Multiply $2$ by $- 10$ .

$f'' (x) = \frac{2 x^{2} - 20 x - 20 x - 10 \cdot - 20 + (- 2 x - 2 \cdot - 13) (x - 7)}{{(x - 10)}^{3}}$

Step 2.12.2.1.2.1.5

Multiply $- 10$ by $- 20$ .

$f'' (x) = \frac{2 x^{2} - 20 x - 20 x + 200 + (- 2 x - 2 \cdot - 13) (x - 7)}{{(x - 10)}^{3}}$

Step 2.12.2.1.2.2

Subtract $20 x$ from $- 20 x$ .

$f'' (x) = \frac{2 x^{2} - 40 x + 200 + (- 2 x - 2 \cdot - 13) (x - 7)}{{(x - 10)}^{3}}$

Step 2.12.2.1.3

Multiply $- 2$ by $- 13$ .

$f'' (x) = \frac{2 x^{2} - 40 x + 200 + (- 2 x + 26) (x - 7)}{{(x - 10)}^{3}}$

Step 2.12.2.1.4

Expand $(- 2 x + 26) (x - 7)$ using the FOIL Method.

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Step 2.12.2.1.4.1

Apply the distributive property.

$f'' (x) = \frac{2 x^{2} - 40 x + 200 - 2 x (x - 7) + 26 (x - 7)}{{(x - 10)}^{3}}$

Step 2.12.2.1.4.2

Apply the distributive property.

$f'' (x) = \frac{2 x^{2} - 40 x + 200 - 2 x \cdot x - 2 x \cdot - 7 + 26 (x - 7)}{{(x - 10)}^{3}}$

Step 2.12.2.1.4.3

Apply the distributive property.

$f'' (x) = \frac{2 x^{2} - 40 x + 200 - 2 x \cdot x - 2 x \cdot - 7 + 26 x + 26 \cdot - 7}{{(x - 10)}^{3}}$

Step 2.12.2.1.5

Simplify and combine like terms.

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Step 2.12.2.1.5.1

Simplify each term.

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Step 2.12.2.1.5.1.1

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

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Step 2.12.2.1.5.1.1.1

Move $x$ .

$f'' (x) = \frac{2 x^{2} - 40 x + 200 - 2 (x \cdot x) - 2 x \cdot - 7 + 26 x + 26 \cdot - 7}{{(x - 10)}^{3}}$

Step 2.12.2.1.5.1.1.2

Multiply $x$ by $x$ .

$f'' (x) = \frac{2 x^{2} - 40 x + 200 - 2 x^{2} - 2 x \cdot - 7 + 26 x + 26 \cdot - 7}{{(x - 10)}^{3}}$

Step 2.12.2.1.5.1.2

Multiply $- 7$ by $- 2$ .

$f'' (x) = \frac{2 x^{2} - 40 x + 200 - 2 x^{2} + 14 x + 26 x + 26 \cdot - 7}{{(x - 10)}^{3}}$

Step 2.12.2.1.5.1.3

Multiply $26$ by $- 7$ .

$f'' (x) = \frac{2 x^{2} - 40 x + 200 - 2 x^{2} + 14 x + 26 x - 182}{{(x - 10)}^{3}}$

Step 2.12.2.1.5.2

Add $14 x$ and $26 x$ .

$f'' (x) = \frac{2 x^{2} - 40 x + 200 - 2 x^{2} + 40 x - 182}{{(x - 10)}^{3}}$

Step 2.12.2.2

Combine the opposite terms in $2 x^{2} - 40 x + 200 - 2 x^{2} + 40 x - 182$ .

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Step 2.12.2.2.1

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

$f'' (x) = \frac{- 40 x + 200 + 0 + 40 x - 182}{{(x - 10)}^{3}}$

Step 2.12.2.2.2

Add $- 40 x + 200$ and $0$ .

$f'' (x) = \frac{- 40 x + 200 + 40 x - 182}{{(x - 10)}^{3}}$

Step 2.12.2.2.3

Add $- 40 x$ and $40 x$ .

$f'' (x) = \frac{0 + 200 - 182}{{(x - 10)}^{3}}$

Step 2.12.2.2.4

Add $0$ and $200$ .

$f'' (x) = \frac{200 - 182}{{(x - 10)}^{3}}$

Step 2.12.2.3

Subtract $182$ from $200$ .

$f'' (x) = \frac{18}{{(x - 10)}^{3}}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$\frac{(x - 13) (x - 7)}{{(x - 10)}^{2}} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

$\frac{(x - 10) \frac{d}{d x} [x^{2} - 14 x + 49] - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 4.1.2

Differentiate.

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Step 4.1.2.1

By the Sum Rule, the derivative of $x^{2} - 14 x + 49$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 14 x] + \frac{d}{d x} [49]$ .

$\frac{(x - 10) (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 14 x] + \frac{d}{d x} [49]) - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 4.1.2.2

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

$\frac{(x - 10) (2 x + \frac{d}{d x} [- 14 x] + \frac{d}{d x} [49]) - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 4.1.2.3

Since $- 14$ is constant with respect to $x$ , the derivative of $- 14 x$ with respect to $x$ is $- 14 \frac{d}{d x} [x]$ .

$\frac{(x - 10) (2 x - 14 \frac{d}{d x} [x] + \frac{d}{d x} [49]) - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 4.1.2.4

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

$\frac{(x - 10) (2 x - 14 \cdot 1 + \frac{d}{d x} [49]) - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 4.1.2.5

Multiply $- 14$ by $1$ .

$\frac{(x - 10) (2 x - 14 + \frac{d}{d x} [49]) - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 4.1.2.6

Since $49$ is constant with respect to $x$ , the derivative of $49$ with respect to $x$ is $0$ .

$\frac{(x - 10) (2 x - 14 + 0) - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 4.1.2.7

Add $2 x - 14$ and $0$ .

$\frac{(x - 10) (2 x - 14) - (x^{2} - 14 x + 49) \frac{d}{d x} [x - 10]}{{(x - 10)}^{2}}$

Step 4.1.2.8

By the Sum Rule, the derivative of $x - 10$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 10]$ .

$\frac{(x - 10) (2 x - 14) - (x^{2} - 14 x + 49) (\frac{d}{d x} [x] + \frac{d}{d x} [- 10])}{{(x - 10)}^{2}}$

Step 4.1.2.9

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

$\frac{(x - 10) (2 x - 14) - (x^{2} - 14 x + 49) (1 + \frac{d}{d x} [- 10])}{{(x - 10)}^{2}}$

Step 4.1.2.10

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10$ with respect to $x$ is $0$ .

$\frac{(x - 10) (2 x - 14) - (x^{2} - 14 x + 49) (1 + 0)}{{(x - 10)}^{2}}$

Step 4.1.2.11

Simplify the expression.

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Step 4.1.2.11.1

Add $1$ and $0$ .

$\frac{(x - 10) (2 x - 14) - (x^{2} - 14 x + 49) \cdot 1}{{(x - 10)}^{2}}$

Step 4.1.2.11.2

Multiply $- 1$ by $1$ .

$\frac{(x - 10) (2 x - 14) - (x^{2} - 14 x + 49)}{{(x - 10)}^{2}}$

Step 4.1.3

Simplify.

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Step 4.1.3.1

Apply the distributive property.

$\frac{(x - 10) (2 x - 14) - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 4.1.3.2

Simplify the numerator.

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Step 4.1.3.2.1

Simplify each term.

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Step 4.1.3.2.1.1

Expand $(x - 10) (2 x - 14)$ using the FOIL Method.

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Step 4.1.3.2.1.1.1

Apply the distributive property.

$\frac{x (2 x - 14) - 10 (2 x - 14) - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 4.1.3.2.1.1.2

Apply the distributive property.

$\frac{x (2 x) + x \cdot - 14 - 10 (2 x - 14) - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 4.1.3.2.1.1.3

Apply the distributive property.

$\frac{x (2 x) + x \cdot - 14 - 10 (2 x) - 10 \cdot - 14 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 4.1.3.2.1.2

Simplify and combine like terms.

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Step 4.1.3.2.1.2.1

Simplify each term.

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Step 4.1.3.2.1.2.1.1

Rewrite using the commutative property of multiplication.

$\frac{2 x \cdot x + x \cdot - 14 - 10 (2 x) - 10 \cdot - 14 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 4.1.3.2.1.2.1.2

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

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Step 4.1.3.2.1.2.1.2.1

Move $x$ .

$\frac{2 (x \cdot x) + x \cdot - 14 - 10 (2 x) - 10 \cdot - 14 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 4.1.3.2.1.2.1.2.2

Multiply $x$ by $x$ .

$\frac{2 x^{2} + x \cdot - 14 - 10 (2 x) - 10 \cdot - 14 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 4.1.3.2.1.2.1.3

Move $- 14$ to the left of $x$ .

$\frac{2 x^{2} - 14 \cdot x - 10 (2 x) - 10 \cdot - 14 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 4.1.3.2.1.2.1.4

Multiply $2$ by $- 10$ .

$\frac{2 x^{2} - 14 x - 20 x - 10 \cdot - 14 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 4.1.3.2.1.2.1.5

Multiply $- 10$ by $- 14$ .

$\frac{2 x^{2} - 14 x - 20 x + 140 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 4.1.3.2.1.2.2

Subtract $20 x$ from $- 14 x$ .

$\frac{2 x^{2} - 34 x + 140 - x^{2} - (- 14 x) - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 4.1.3.2.1.3

Multiply $- 14$ by $- 1$ .

$\frac{2 x^{2} - 34 x + 140 - x^{2} + 14 x - 1 \cdot 49}{{(x - 10)}^{2}}$

Step 4.1.3.2.1.4

Multiply $- 1$ by $49$ .

$\frac{2 x^{2} - 34 x + 140 - x^{2} + 14 x - 49}{{(x - 10)}^{2}}$

Step 4.1.3.2.2

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

$\frac{x^{2} - 34 x + 140 + 14 x - 49}{{(x - 10)}^{2}}$

Step 4.1.3.2.3

Add $- 34 x$ and $14 x$ .

$\frac{x^{2} - 20 x + 140 - 49}{{(x - 10)}^{2}}$

Step 4.1.3.2.4

Subtract $49$ from $140$ .

$\frac{x^{2} - 20 x + 91}{{(x - 10)}^{2}}$

Step 4.1.3.3

Factor $x^{2} - 20 x + 91$ using the AC method.

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Step 4.1.3.3.1

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 $91$ and whose sum is $- 20$ .

$- 13, - 7$

Step 4.1.3.3.2

Write the factored form using these integers.

$f' (x) = \frac{(x - 13) (x - 7)}{{(x - 10)}^{2}}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{(x - 13) (x - 7)}{{(x - 10)}^{2}}$ .

$\frac{(x - 13) (x - 7)}{{(x - 10)}^{2}}$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{(x - 13) (x - 7)}{{(x - 10)}^{2}} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$\frac{(x - 13) (x - 7)}{{(x - 10)}^{2}} = 0$

Step 5.2

Set the numerator equal to zero.

$(x - 13) (x - 7) = 0$

Step 5.3

Solve the equation for $x$ .

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Step 5.3.1

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

$x - 13 = 0$

$x - 7 = 0$

Step 5.3.2

Set $x - 13$ equal to $0$ and solve for $x$ .

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Step 5.3.2.1

Set $x - 13$ equal to $0$ .

$x - 13 = 0$

Step 5.3.2.2

Add $13$ to both sides of the equation.

$x = 13$

Step 5.3.3

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

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Step 5.3.3.1

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

$x - 7 = 0$

Step 5.3.3.2

Add $7$ to both sides of the equation.

$x = 7$

Step 5.3.4

The final solution is all the values that make $(x - 13) (x - 7) = 0$ true.

$x = 13, 7$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{(x - 13) (x - 7)}{{(x - 10)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(x - 10)}^{2} = 0$

Step 6.2

Solve for $x$ .

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Step 6.2.1

Set the $x - 10$ equal to $0$ .

$x - 10 = 0$

Step 6.2.2

Add $10$ to both sides of the equation.

$x = 10$

Step 7

Critical points to evaluate.

$x = 13, 7$

Step 8

Evaluate the second derivative at $x = 13$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{18}{{((13) - 10)}^{3}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the denominator.

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Step 9.1.1

Subtract $10$ from $13$ .

$\frac{18}{3^{3}}$

Step 9.1.2

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

$\frac{18}{27}$

Step 9.2

Cancel the common factor of $18$ and $27$ .

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Step 9.2.1

Factor $9$ out of $18$ .

$\frac{9 (2)}{27}$

Step 9.2.2

Cancel the common factors.

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Step 9.2.2.1

Factor $9$ out of $27$ .

$\frac{9 \cdot 2}{9 \cdot 3}$

Step 9.2.2.2

Cancel the common factor.

$\frac{9 \cdot 2}{9 \cdot 3}$

Step 9.2.2.3

Rewrite the expression.

$\frac{2}{3}$

Step 10

$x = 13$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 13$ is a local minimum

Step 11

Find the y-value when $x = 13$ .

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Step 11.1

Replace the variable $x$ with $13$ in the expression.

$f (13) = \frac{{(13)}^{2} - 14 \cdot 13 + 49}{(13) - 10}$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify the numerator.

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Step 11.2.1.1

Raise $13$ to the power of $2$ .

$f (13) = \frac{169 - 14 \cdot 13 + 49}{13 - 10}$

Step 11.2.1.2

Multiply $- 14$ by $13$ .

$f (13) = \frac{169 - 182 + 49}{13 - 10}$

Step 11.2.1.3

Subtract $182$ from $169$ .

$f (13) = \frac{- 13 + 49}{13 - 10}$

Step 11.2.1.4

Add $- 13$ and $49$ .

$f (13) = \frac{36}{13 - 10}$

Step 11.2.2

Simplify the expression.

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Step 11.2.2.1

Subtract $10$ from $13$ .

$f (13) = \frac{36}{3}$

Step 11.2.2.2

Divide $36$ by $3$ .

$f (13) = 12$

Step 11.2.3

The final answer is $12$ .

$y = 12$

Step 12

Evaluate the second derivative at $x = 7$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{18}{{((7) - 10)}^{3}}$

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify the denominator.

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Step 13.1.1

Subtract $10$ from $7$ .

$\frac{18}{{(- 3)}^{3}}$

Step 13.1.2

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

$\frac{18}{- 27}$

Step 13.2

Reduce the expression by cancelling the common factors.

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Step 13.2.1

Cancel the common factor of $18$ and $- 27$ .

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Step 13.2.1.1

Factor $9$ out of $18$ .

$\frac{9 (2)}{- 27}$

Step 13.2.1.2

Cancel the common factors.

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Step 13.2.1.2.1

Factor $9$ out of $- 27$ .

$\frac{9 \cdot 2}{9 \cdot - 3}$

Step 13.2.1.2.2

Cancel the common factor.

$\frac{9 \cdot 2}{9 \cdot - 3}$

Step 13.2.1.2.3

Rewrite the expression.

$\frac{2}{- 3}$

Step 13.2.2

Move the negative in front of the fraction.

$- \frac{2}{3}$

Step 14

$x = 7$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 7$ is a local maximum

Step 15

Find the y-value when $x = 7$ .

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Step 15.1

Replace the variable $x$ with $7$ in the expression.

$f (7) = \frac{{(7)}^{2} - 14 \cdot 7 + 49}{(7) - 10}$

Step 15.2

Simplify the result.

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Step 15.2.1

Simplify the numerator.

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Step 15.2.1.1

Raise $7$ to the power of $2$ .

$f (7) = \frac{49 - 14 \cdot 7 + 49}{7 - 10}$

Step 15.2.1.2

Multiply $- 14$ by $7$ .

$f (7) = \frac{49 - 98 + 49}{7 - 10}$

Step 15.2.1.3

Subtract $98$ from $49$ .

$f (7) = \frac{- 49 + 49}{7 - 10}$

Step 15.2.1.4

Add $- 49$ and $49$ .

$f (7) = \frac{0}{7 - 10}$

Step 15.2.2

Simplify the expression.

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Step 15.2.2.1

Subtract $10$ from $7$ .

$f (7) = \frac{0}{- 3}$

Step 15.2.2.2

Divide $0$ by $- 3$ .

$f (7) = 0$

Step 15.2.3

The final answer is $0$ .

$y = 0$

Step 16

These are the local extrema for $f (x) = \frac{x^{2} - 14 x + 49}{x - 10}$ .

$(13, 12)$ is a local minima

$(7, 0)$ is a local maxima

Step 17