Calculus Examples

$f (x) = \frac{x^{2} - 12}{x - 4}$

Step 1

Find the $x$ values where the second derivative is equal to $0$ .

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Step 1.1

Find the second derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

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 $f (x) = x^{2} - 12$ and $g (x) = x - 4$ .

$f' (x) = \frac{(x - 4) \frac{d}{d x} (x^{2} - 12) - (x^{2} - 12) \frac{d}{d x} (x - 4)}{{(x - 4)}^{2}}$

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

By the Sum Rule, the derivative of $x^{2} - 12$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 12]$ .

$f' (x) = \frac{(x - 4) (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 12)) - (x^{2} - 12) \frac{d}{d x} (x - 4)}{{(x - 4)}^{2}}$

Step 1.1.1.2.2

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

$f' (x) = \frac{(x - 4) (2 x + \frac{d}{d x} (- 12)) - (x^{2} - 12) \frac{d}{d x} (x - 4)}{{(x - 4)}^{2}}$

Step 1.1.1.2.3

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12$ with respect to $x$ is $0$ .

$f' (x) = \frac{(x - 4) (2 x + 0) - (x^{2} - 12) \frac{d}{d x} (x - 4)}{{(x - 4)}^{2}}$

Step 1.1.1.2.4

Simplify the expression.

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Step 1.1.1.2.4.1

Add $2 x$ and $0$ .

$f' (x) = \frac{(x - 4) (2 x) - (x^{2} - 12) \frac{d}{d x} (x - 4)}{{(x - 4)}^{2}}$

Step 1.1.1.2.4.2

Move $2$ to the left of $x - 4$ .

$f' (x) = \frac{2 \cdot ((x - 4) x) - (x^{2} - 12) \frac{d}{d x} (x - 4)}{{(x - 4)}^{2}}$

Step 1.1.1.2.5

By the Sum Rule, the derivative of $x - 4$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 4]$ .

$f' (x) = \frac{2 (x - 4) x - (x^{2} - 12) (\frac{d}{d x} (x) + \frac{d}{d x} (- 4))}{{(x - 4)}^{2}}$

Step 1.1.1.2.6

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

$f' (x) = \frac{2 (x - 4) x - (x^{2} - 12) (1 + \frac{d}{d x} (- 4))}{{(x - 4)}^{2}}$

Step 1.1.1.2.7

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$f' (x) = \frac{2 (x - 4) x - (x^{2} - 12) (1 + 0)}{{(x - 4)}^{2}}$

Step 1.1.1.2.8

Simplify the expression.

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Step 1.1.1.2.8.1

Add $1$ and $0$ .

$f' (x) = \frac{2 (x - 4) x - (x^{2} - 12) \cdot 1}{{(x - 4)}^{2}}$

Step 1.1.1.2.8.2

Multiply $- 1$ by $1$ .

$f' (x) = \frac{2 (x - 4) x - (x^{2} - 12)}{{(x - 4)}^{2}}$

Step 1.1.1.3

Simplify.

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Step 1.1.1.3.1

Apply the distributive property.

$f' (x) = \frac{(2 x + 2 \cdot - 4) x - (x^{2} - 12)}{{(x - 4)}^{2}}$

Step 1.1.1.3.2

Apply the distributive property.

$f' (x) = \frac{2 x \cdot x + 2 \cdot (- 4 x) - (x^{2} - 12)}{{(x - 4)}^{2}}$

Step 1.1.1.3.3

Apply the distributive property.

$f' (x) = \frac{2 x \cdot x + 2 \cdot (- 4 x) - x^{2} + 12}{{(x - 4)}^{2}}$

Step 1.1.1.3.4

Simplify the numerator.

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Step 1.1.1.3.4.1

Simplify each term.

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Step 1.1.1.3.4.1.1

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

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Step 1.1.1.3.4.1.1.1

Move $x$ .

$f' (x) = \frac{2 (x \cdot x) + 2 \cdot (- 4 x) - x^{2} + 12}{{(x - 4)}^{2}}$

Step 1.1.1.3.4.1.1.2

Multiply $x$ by $x$ .

$f' (x) = \frac{2 x^{2} + 2 \cdot (- 4 x) - x^{2} + 12}{{(x - 4)}^{2}}$

Step 1.1.1.3.4.1.2

Multiply $2$ by $- 4$ .

$f' (x) = \frac{2 x^{2} - 8 x - x^{2} + 12}{{(x - 4)}^{2}}$

Step 1.1.1.3.4.1.3

Multiply $- 1$ by $- 12$ .

$f' (x) = \frac{2 x^{2} - 8 x - x^{2} + 12}{{(x - 4)}^{2}}$

Step 1.1.1.3.4.2

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

$f' (x) = \frac{x^{2} - 8 x + 12}{{(x - 4)}^{2}}$

Step 1.1.1.3.5

Factor $x^{2} - 8 x + 12$ using the AC method.

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Step 1.1.1.3.5.1

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 $12$ and whose sum is $- 8$ .

$- 6, - 2$

Step 1.1.1.3.5.2

Write the factored form using these integers.

$f' (x) = \frac{(x - 6) (x - 2)}{{(x - 4)}^{2}}$

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

$f'' (x) = \frac{{(x - 4)}^{2} \frac{d}{d x} ((x - 6) (x - 2)) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{({(x - 4)}^{2})}^{2}}$

Step 1.1.2.2

Multiply the exponents in ${({(x - 4)}^{2})}^{2}$ .

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Step 1.1.2.2.1

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

$f'' (x) = \frac{{(x - 4)}^{2} \frac{d}{d x} ((x - 6) (x - 2)) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{2 \cdot 2}}$

Step 1.1.2.2.2

Multiply $2$ by $2$ .

$f'' (x) = \frac{{(x - 4)}^{2} \frac{d}{d x} ((x - 6) (x - 2)) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{4}}$

Step 1.1.2.3

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 $f (x) = x - 6$ and $g (x) = x - 2$ .

$f'' (x) = \frac{{(x - 4)}^{2} ((x - 6) \frac{d}{d x} (x - 2) + (x - 2) \frac{d}{d x} (x - 6)) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{4}}$

Step 1.1.2.4

Differentiate.

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Step 1.1.2.4.1

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

$f'' (x) = \frac{{(x - 4)}^{2} ((x - 6) (\frac{d}{d x} (x) + \frac{d}{d x} (- 2)) + (x - 2) \frac{d}{d x} (x - 6)) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{4}}$

Step 1.1.2.4.2

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

$f'' (x) = \frac{{(x - 4)}^{2} ((x - 6) (1 + \frac{d}{d x} (- 2)) + (x - 2) \frac{d}{d x} (x - 6)) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{4}}$

Step 1.1.2.4.3

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$f'' (x) = \frac{{(x - 4)}^{2} ((x - 6) (1 + 0) + (x - 2) \frac{d}{d x} (x - 6)) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{4}}$

Step 1.1.2.4.4

Simplify the expression.

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Step 1.1.2.4.4.1

Add $1$ and $0$ .

$f'' (x) = \frac{{(x - 4)}^{2} ((x - 6) \cdot 1 + (x - 2) \frac{d}{d x} (x - 6)) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{4}}$

Step 1.1.2.4.4.2

Multiply $x - 6$ by $1$ .

$f'' (x) = \frac{{(x - 4)}^{2} (x - 6 + (x - 2) \frac{d}{d x} (x - 6)) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{4}}$

Step 1.1.2.4.5

By the Sum Rule, the derivative of $x - 6$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 6]$ .

$f'' (x) = \frac{{(x - 4)}^{2} (x - 6 + (x - 2) (\frac{d}{d x} (x) + \frac{d}{d x} (- 6))) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{4}}$

Step 1.1.2.4.6

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

$f'' (x) = \frac{{(x - 4)}^{2} (x - 6 + (x - 2) (1 + \frac{d}{d x} (- 6))) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{4}}$

Step 1.1.2.4.7

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6$ with respect to $x$ is $0$ .

$f'' (x) = \frac{{(x - 4)}^{2} (x - 6 + (x - 2) (1 + 0)) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{4}}$

Step 1.1.2.4.8

Simplify by adding terms.

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Step 1.1.2.4.8.1

Add $1$ and $0$ .

$f'' (x) = \frac{{(x - 4)}^{2} (x - 6 + (x - 2) \cdot 1) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{4}}$

Step 1.1.2.4.8.2

Multiply $x - 2$ by $1$ .

$f'' (x) = \frac{{(x - 4)}^{2} (x - 6 + x - 2) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{4}}$

Step 1.1.2.4.8.3

Add $x$ and $x$ .

$f'' (x) = \frac{{(x - 4)}^{2} (2 x - 6 - 2) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{4}}$

Step 1.1.2.4.8.4

Subtract $2$ from $- 6$ .

$f'' (x) = \frac{{(x - 4)}^{2} (2 x - 8) - (x - 6) (x - 2) \frac{d}{d x} {(x - 4)}^{2}}{{(x - 4)}^{4}}$

Step 1.1.2.5

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 $g (x) = x - 4$ .

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Step 1.1.2.5.1

To apply the Chain Rule, set $u$ as $x - 4$ .

$f'' (x) = \frac{{(x - 4)}^{2} (2 x - 8) - (x - 6) (x - 2) (\frac{d}{d u} (u^{2}) \frac{d}{d x} (x - 4))}{{(x - 4)}^{4}}$

Step 1.1.2.5.2

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

$f'' (x) = \frac{{(x - 4)}^{2} (2 x - 8) - (x - 6) (x - 2) (2 u \frac{d}{d x} (x - 4))}{{(x - 4)}^{4}}$

Step 1.1.2.5.3

Replace all occurrences of $u$ with $x - 4$ .

$f'' (x) = \frac{{(x - 4)}^{2} (2 x - 8) - (x - 6) (x - 2) (2 (x - 4) \frac{d}{d x} (x - 4))}{{(x - 4)}^{4}}$

Step 1.1.2.6

Simplify with factoring out.

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Step 1.1.2.6.1

Multiply $2$ by $- 1$ .

$f'' (x) = \frac{{(x - 4)}^{2} (2 x - 8) - 2 (x - 6) (x - 2) ((x - 4) \frac{d}{d x} (x - 4))}{{(x - 4)}^{4}}$

Step 1.1.2.6.2

Factor $x - 4$ out of ${(x - 4)}^{2} (2 x - 8) - 2 (x - 6) (x - 2) ((x - 4) \frac{d}{d x} [x - 4])$ .

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Step 1.1.2.6.2.1

Factor $x - 4$ out of ${(x - 4)}^{2} (2 x - 8)$ .

$f'' (x) = \frac{(x - 4) ((x - 4) (2 x - 8)) - 2 (x - 6) (x - 2) ((x - 4) \frac{d}{d x} (x - 4))}{{(x - 4)}^{4}}$

Step 1.1.2.6.2.2

Factor $x - 4$ out of $- 2 (x - 6) (x - 2) ((x - 4) \frac{d}{d x} [x - 4])$ .

$f'' (x) = \frac{(x - 4) ((x - 4) (2 x - 8)) + (x - 4) (- 2 (x - 6) (x - 2) (\frac{d}{d x} (x - 4)))}{{(x - 4)}^{4}}$

Step 1.1.2.6.2.3

Factor $x - 4$ out of $(x - 4) ((x - 4) (2 x - 8)) + (x - 4) (- 2 (x - 6) (x - 2) (\frac{d}{d x} [x - 4]))$ .

$f'' (x) = \frac{(x - 4) ((x - 4) (2 x - 8) - 2 (x - 6) (x - 2) (\frac{d}{d x} (x - 4)))}{{(x - 4)}^{4}}$

Step 1.1.2.7

Cancel the common factors.

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Step 1.1.2.7.1

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

$f'' (x) = \frac{(x - 4) ((x - 4) (2 x - 8) - 2 (x - 6) (x - 2) (\frac{d}{d x} (x - 4)))}{(x - 4) {(x - 4)}^{3}}$

Step 1.1.2.7.2

Cancel the common factor.

$f'' (x) = \frac{(x - 4) ((x - 4) (2 x - 8) - 2 (x - 6) (x - 2) (\frac{d}{d x} (x - 4)))}{(x - 4) {(x - 4)}^{3}}$

Step 1.1.2.7.3

Rewrite the expression.

$f'' (x) = \frac{(x - 4) (2 x - 8) - 2 (x - 6) (x - 2) (\frac{d}{d x} (x - 4))}{{(x - 4)}^{3}}$

Step 1.1.2.8

By the Sum Rule, the derivative of $x - 4$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 4]$ .

$f'' (x) = \frac{(x - 4) (2 x - 8) - 2 (x - 6) (x - 2) (\frac{d}{d x} (x) + \frac{d}{d x} (- 4))}{{(x - 4)}^{3}}$

Step 1.1.2.9

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

$f'' (x) = \frac{(x - 4) (2 x - 8) - 2 (x - 6) (x - 2) (1 + \frac{d}{d x} (- 4))}{{(x - 4)}^{3}}$

Step 1.1.2.10

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$f'' (x) = \frac{(x - 4) (2 x - 8) - 2 (x - 6) (x - 2) (1 + 0)}{{(x - 4)}^{3}}$

Step 1.1.2.11

Simplify the expression.

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Step 1.1.2.11.1

Add $1$ and $0$ .

$f'' (x) = \frac{(x - 4) (2 x - 8) - 2 (x - 6) (x - 2) \cdot 1}{{(x - 4)}^{3}}$

Step 1.1.2.11.2

Multiply $- 2$ by $1$ .

$f'' (x) = \frac{(x - 4) (2 x - 8) - 2 (x - 6) (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12

Simplify.

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Step 1.1.2.12.1

Apply the distributive property.

$f'' (x) = \frac{(x - 4) (2 x - 8) + (- 2 x - 2 \cdot - 6) (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12.2

Simplify the numerator.

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Step 1.1.2.12.2.1

Simplify each term.

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Step 1.1.2.12.2.1.1

Expand $(x - 4) (2 x - 8)$ using the FOIL Method.

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Step 1.1.2.12.2.1.1.1

Apply the distributive property.

$f'' (x) = \frac{x (2 x - 8) - 4 (2 x - 8) + (- 2 x - 2 \cdot - 6) (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.1.2

Apply the distributive property.

$f'' (x) = \frac{x (2 x) + x \cdot - 8 - 4 (2 x - 8) + (- 2 x - 2 \cdot - 6) (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.1.3

Apply the distributive property.

$f'' (x) = \frac{x (2 x) + x \cdot - 8 - 4 (2 x) - 4 \cdot - 8 + (- 2 x - 2 \cdot - 6) (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.2

Simplify and combine like terms.

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Step 1.1.2.12.2.1.2.1

Simplify each term.

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Step 1.1.2.12.2.1.2.1.1

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{2 x \cdot x + x \cdot - 8 - 4 (2 x) - 4 \cdot - 8 + (- 2 x - 2 \cdot - 6) (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.2.1.2

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

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Step 1.1.2.12.2.1.2.1.2.1

Move $x$ .

$f'' (x) = \frac{2 (x \cdot x) + x \cdot - 8 - 4 (2 x) - 4 \cdot - 8 + (- 2 x - 2 \cdot - 6) (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.2.1.2.2

Multiply $x$ by $x$ .

$f'' (x) = \frac{2 x^{2} + x \cdot - 8 - 4 (2 x) - 4 \cdot - 8 + (- 2 x - 2 \cdot - 6) (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.2.1.3

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

$f'' (x) = \frac{2 x^{2} - 8 \cdot x - 4 (2 x) - 4 \cdot - 8 + (- 2 x - 2 \cdot - 6) (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.2.1.4

Multiply $2$ by $- 4$ .

$f'' (x) = \frac{2 x^{2} - 8 x - 8 x - 4 \cdot - 8 + (- 2 x - 2 \cdot - 6) (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.2.1.5

Multiply $- 4$ by $- 8$ .

$f'' (x) = \frac{2 x^{2} - 8 x - 8 x + 32 + (- 2 x - 2 \cdot - 6) (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.2.2

Subtract $8 x$ from $- 8 x$ .

$f'' (x) = \frac{2 x^{2} - 16 x + 32 + (- 2 x - 2 \cdot - 6) (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.3

Multiply $- 2$ by $- 6$ .

$f'' (x) = \frac{2 x^{2} - 16 x + 32 + (- 2 x + 12) (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.4

Expand $(- 2 x + 12) (x - 2)$ using the FOIL Method.

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Step 1.1.2.12.2.1.4.1

Apply the distributive property.

$f'' (x) = \frac{2 x^{2} - 16 x + 32 - 2 x (x - 2) + 12 (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.4.2

Apply the distributive property.

$f'' (x) = \frac{2 x^{2} - 16 x + 32 - 2 x \cdot x - 2 x \cdot - 2 + 12 (x - 2)}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.4.3

Apply the distributive property.

$f'' (x) = \frac{2 x^{2} - 16 x + 32 - 2 x \cdot x - 2 x \cdot - 2 + 12 x + 12 \cdot - 2}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.5

Simplify and combine like terms.

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Step 1.1.2.12.2.1.5.1

Simplify each term.

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Step 1.1.2.12.2.1.5.1.1

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

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Step 1.1.2.12.2.1.5.1.1.1

Move $x$ .

$f'' (x) = \frac{2 x^{2} - 16 x + 32 - 2 (x \cdot x) - 2 x \cdot - 2 + 12 x + 12 \cdot - 2}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.5.1.1.2

Multiply $x$ by $x$ .

$f'' (x) = \frac{2 x^{2} - 16 x + 32 - 2 x^{2} - 2 x \cdot - 2 + 12 x + 12 \cdot - 2}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.5.1.2

Multiply $- 2$ by $- 2$ .

$f'' (x) = \frac{2 x^{2} - 16 x + 32 - 2 x^{2} + 4 x + 12 x + 12 \cdot - 2}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.5.1.3

Multiply $12$ by $- 2$ .

$f'' (x) = \frac{2 x^{2} - 16 x + 32 - 2 x^{2} + 4 x + 12 x - 24}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.1.5.2

Add $4 x$ and $12 x$ .

$f'' (x) = \frac{2 x^{2} - 16 x + 32 - 2 x^{2} + 16 x - 24}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.2

Combine the opposite terms in $2 x^{2} - 16 x + 32 - 2 x^{2} + 16 x - 24$ .

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Step 1.1.2.12.2.2.1

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

$f'' (x) = \frac{- 16 x + 32 + 0 + 16 x - 24}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.2.2

Add $- 16 x + 32$ and $0$ .

$f'' (x) = \frac{- 16 x + 32 + 16 x - 24}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.2.3

Add $- 16 x$ and $16 x$ .

$f'' (x) = \frac{0 + 32 - 24}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.2.4

Add $0$ and $32$ .

$f'' (x) = \frac{32 - 24}{{(x - 4)}^{3}}$

Step 1.1.2.12.2.3

Subtract $24$ from $32$ .

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

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{8}{{(x - 4)}^{3}}$ .

$\frac{8}{{(x - 4)}^{3}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $\frac{8}{{(x - 4)}^{3}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$\frac{8}{{(x - 4)}^{3}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$8 = 0$

Step 1.2.3

Since $8 \neq 0$ , there are no solutions.

No solution

Step 2

Find the domain of $f (x) = \frac{x^{2} - 12}{x - 4}$ .

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Step 2.1

Set the denominator in $\frac{x^{2} - 12}{x - 4}$ equal to $0$ to find where the expression is undefined.

$x - 4 = 0$

Step 2.2

Add $4$ to both sides of the equation.

$x = 4$

Step 2.3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, 4) \cup (4, \infty)$

Set-Builder Notation:

${x | x \neq 4}$

Interval Notation:

$(- \infty, 4) \cup (4, \infty)$

Set-Builder Notation:

${x | x \neq 4}$

Step 3

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- \infty, 4) \cup (4, \infty)$

Step 4

Substitute any number from the interval $(- \infty, 4)$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

Replace the variable $x$ with $0$ in the expression.

$f'' (0) = \frac{8}{{((0) - 4)}^{3}}$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify the denominator.

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Step 4.2.1.1

Subtract $4$ from $0$ .

$f'' (0) = \frac{8}{{(- 4)}^{3}}$

Step 4.2.1.2

Raise $- 4$ to the power of $3$ .

$f'' (0) = \frac{8}{- 64}$

Step 4.2.2

Reduce the expression by cancelling the common factors.

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Step 4.2.2.1

Cancel the common factor of $8$ and $- 64$ .

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Step 4.2.2.1.1

Factor $8$ out of $8$ .

$f'' (0) = \frac{8 (1)}{- 64}$

Step 4.2.2.1.2

Cancel the common factors.

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Step 4.2.2.1.2.1

Factor $8$ out of $- 64$ .

$f'' (0) = \frac{8 \cdot 1}{8 \cdot - 8}$

Step 4.2.2.1.2.2

Cancel the common factor.

$f'' (0) = \frac{8 \cdot 1}{8 \cdot - 8}$

Step 4.2.2.1.2.3

Rewrite the expression.

$f'' (0) = \frac{1}{- 8}$

Step 4.2.2.2

Move the negative in front of the fraction.

$f'' (0) = - \frac{1}{8}$

Step 4.2.3

The final answer is $- \frac{1}{8}$ .

$- \frac{1}{8}$

Step 4.3

The graph is concave down on the interval $(- \infty, 4)$ because $f'' (0)$ is negative.

Concave down on $(- \infty, 4)$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(4, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

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

$f'' (6) = \frac{8}{{((6) - 4)}^{3}}$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify the denominator.

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Step 5.2.1.1

Subtract $4$ from $6$ .

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

Step 5.2.1.2

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

$f'' (6) = \frac{8}{8}$

Step 5.2.2

Divide $8$ by $8$ .

$f'' (6) = 1$

Step 5.2.3

The final answer is $1$ .

$1$

Step 5.3

The graph is concave up on the interval $(4, \infty)$ because $f'' (6)$ is positive.

Concave up on $(4, \infty)$ since $f'' (x)$ is positive

Step 6

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave down on $(- \infty, 4)$ since $f'' (x)$ is negative

Concave up on $(4, \infty)$ since $f'' (x)$ is positive

Step 7