Calculus Examples

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

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 $f (x) = x$ and $g (x) = x^{2} - 9$ .

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

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

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

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

Step 1.1.1.2.2

Multiply $x^{2} - 9$ by $1$ .

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

Step 1.1.1.2.3

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

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

Step 1.1.1.2.4

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

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

Step 1.1.1.2.5

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

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

Step 1.1.1.2.6

Simplify the expression.

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Step 1.1.1.2.6.1

Add $2 x$ and $0$ .

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

Step 1.1.1.2.6.2

Multiply $2$ by $- 1$ .

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

Step 1.1.1.3

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

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

Step 1.1.1.4

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

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

Step 1.1.1.5

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

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

Step 1.1.1.6

Add $1$ and $1$ .

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

Step 1.1.1.7

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

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

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

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

Step 1.1.2.2

Differentiate.

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Step 1.1.2.2.1

Multiply the exponents in ${({(x^{2} - 9)}^{2})}^{2}$ .

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Step 1.1.2.2.1.1

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

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

Step 1.1.2.2.1.2

Multiply $2$ by $2$ .

$f'' (x) = \frac{{(x^{2} - 9)}^{2} \frac{d}{d x} (- x^{2} - 9) - (- x^{2} - 9) \frac{d}{d x} {(x^{2} - 9)}^{2}}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.2.2

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

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (\frac{d}{d x} (- x^{2}) + \frac{d}{d x} (- 9)) - (- x^{2} - 9) \frac{d}{d x} {(x^{2} - 9)}^{2}}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.2.3

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

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (- \frac{d}{d x} x^{2} + \frac{d}{d x} (- 9)) - (- x^{2} - 9) \frac{d}{d x} {(x^{2} - 9)}^{2}}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.2.4

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

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (- (2 x) + \frac{d}{d x} (- 9)) - (- x^{2} - 9) \frac{d}{d x} {(x^{2} - 9)}^{2}}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.2.5

Multiply $2$ by $- 1$ .

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (- 2 x + \frac{d}{d x} (- 9)) - (- x^{2} - 9) \frac{d}{d x} {(x^{2} - 9)}^{2}}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.2.6

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

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (- 2 x + 0) - (- x^{2} - 9) \frac{d}{d x} {(x^{2} - 9)}^{2}}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.2.7

Add $- 2 x$ and $0$ .

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (- 2 x) - (- x^{2} - 9) \frac{d}{d x} {(x^{2} - 9)}^{2}}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.3

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 $g (x) = x^{2} - 9$ .

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Step 1.1.2.3.1

To apply the Chain Rule, set $u$ as $x^{2} - 9$ .

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (- 2 x) - (- x^{2} - 9) (\frac{d}{d u} (u^{2}) \frac{d}{d x} (x^{2} - 9))}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.3.2

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

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (- 2 x) - (- x^{2} - 9) (2 u \frac{d}{d x} (x^{2} - 9))}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.3.3

Replace all occurrences of $u$ with $x^{2} - 9$ .

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (- 2 x) - (- x^{2} - 9) (2 (x^{2} - 9) \frac{d}{d x} (x^{2} - 9))}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.4

Differentiate.

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Step 1.1.2.4.1

Multiply $2$ by $- 1$ .

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (- 2 x) - 2 (- x^{2} - 9) ((x^{2} - 9) \frac{d}{d x} (x^{2} - 9))}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.4.2

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

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (- 2 x) - 2 (- x^{2} - 9) ((x^{2} - 9) (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 9)))}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.4.3

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

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (- 2 x) - 2 (- x^{2} - 9) ((x^{2} - 9) (2 x + \frac{d}{d x} (- 9)))}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.4.4

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

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (- 2 x) - 2 (- x^{2} - 9) ((x^{2} - 9) (2 x + 0))}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.4.5

Simplify the expression.

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Step 1.1.2.4.5.1

Add $2 x$ and $0$ .

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

Step 1.1.2.4.5.2

Move $2$ to the left of $x^{2} - 9$ .

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

Step 1.1.2.4.5.3

Multiply $2$ by $- 2$ .

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

Step 1.1.2.5

Simplify.

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Step 1.1.2.5.1

Apply the distributive property.

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (- 2 x) + (- 4 (- x^{2}) - 4 \cdot - 9) ((x^{2} - 9) x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.2

Apply the distributive property.

$f'' (x) = \frac{{(x^{2} - 9)}^{2} (- 2 x) + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3

Simplify the numerator.

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Step 1.1.2.5.3.1

Simplify each term.

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Step 1.1.2.5.3.1.1

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{- 2 {(x^{2} - 9)}^{2} x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.2

Rewrite ${(x^{2} - 9)}^{2}$ as $(x^{2} - 9) (x^{2} - 9)$ .

$f'' (x) = \frac{- 2 ((x^{2} - 9) (x^{2} - 9)) x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.3

Expand $(x^{2} - 9) (x^{2} - 9)$ using the FOIL Method.

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Step 1.1.2.5.3.1.3.1

Apply the distributive property.

$f'' (x) = \frac{- 2 (x^{2} (x^{2} - 9) - 9 (x^{2} - 9)) x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.3.2

Apply the distributive property.

$f'' (x) = \frac{- 2 (x^{2} x^{2} + x^{2} \cdot - 9 - 9 (x^{2} - 9)) x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.3.3

Apply the distributive property.

$f'' (x) = \frac{- 2 (x^{2} x^{2} + x^{2} \cdot - 9 - 9 x^{2} - 9 \cdot - 9) x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.4

Simplify and combine like terms.

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Step 1.1.2.5.3.1.4.1

Simplify each term.

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Step 1.1.2.5.3.1.4.1.1

Multiply $x^{2}$ by $x^{2}$ by adding the exponents.

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Step 1.1.2.5.3.1.4.1.1.1

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

$f'' (x) = \frac{- 2 (x^{2 + 2} + x^{2} \cdot - 9 - 9 x^{2} - 9 \cdot - 9) x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.4.1.1.2

Add $2$ and $2$ .

$f'' (x) = \frac{- 2 (x^{4} + x^{2} \cdot - 9 - 9 x^{2} - 9 \cdot - 9) x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.4.1.2

Move $- 9$ to the left of $x^{2}$ .

$f'' (x) = \frac{- 2 (x^{4} - 9 \cdot x^{2} - 9 x^{2} - 9 \cdot - 9) x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.4.1.3

Multiply $- 9$ by $- 9$ .

$f'' (x) = \frac{- 2 (x^{4} - 9 x^{2} - 9 x^{2} + 81) x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.4.2

Subtract $9 x^{2}$ from $- 9 x^{2}$ .

$f'' (x) = \frac{- 2 (x^{4} - 18 x^{2} + 81) x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.5

Apply the distributive property.

$f'' (x) = \frac{(- 2 x^{4} - 2 (- 18 x^{2}) - 2 \cdot 81) x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.6

Simplify.

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Step 1.1.2.5.3.1.6.1

Multiply $- 18$ by $- 2$ .

$f'' (x) = \frac{(- 2 x^{4} + 36 x^{2} - 2 \cdot 81) x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.6.2

Multiply $- 2$ by $81$ .

$f'' (x) = \frac{(- 2 x^{4} + 36 x^{2} - 162) x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.7

Apply the distributive property.

$f'' (x) = \frac{- 2 x^{4} x + 36 x^{2} x - 162 x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.8

Simplify.

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Step 1.1.2.5.3.1.8.1

Multiply $x^{4}$ by $x$ by adding the exponents.

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Step 1.1.2.5.3.1.8.1.1

Move $x$ .

$f'' (x) = \frac{- 2 (x \cdot x^{4}) + 36 x^{2} x - 162 x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.8.1.2

Multiply $x$ by $x^{4}$ .

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Step 1.1.2.5.3.1.8.1.2.1

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

$f'' (x) = \frac{- 2 (x \cdot x^{4}) + 36 x^{2} x - 162 x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.8.1.2.2

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

$f'' (x) = \frac{- 2 x^{1 + 4} + 36 x^{2} x - 162 x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.8.1.3

Add $1$ and $4$ .

$f'' (x) = \frac{- 2 x^{5} + 36 x^{2} x - 162 x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.8.2

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 1.1.2.5.3.1.8.2.1

Move $x$ .

$f'' (x) = \frac{- 2 x^{5} + 36 (x \cdot x^{2}) - 162 x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.8.2.2

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

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Step 1.1.2.5.3.1.8.2.2.1

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

$f'' (x) = \frac{- 2 x^{5} + 36 (x \cdot x^{2}) - 162 x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.8.2.2.2

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

$f'' (x) = \frac{- 2 x^{5} + 36 x^{1 + 2} - 162 x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.8.2.3

Add $1$ and $2$ .

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + (- 4 (- x^{2}) - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.9

Simplify each term.

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Step 1.1.2.5.3.1.9.1

Multiply $- 1$ by $- 4$ .

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + (4 x^{2} - 4 \cdot - 9) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.9.2

Multiply $- 4$ by $- 9$ .

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + (4 x^{2} + 36) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.10

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 1.1.2.5.3.1.10.1

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

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Step 1.1.2.5.3.1.10.1.1

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

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + (4 x^{2} + 36) (x^{2} x - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.10.1.2

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

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + (4 x^{2} + 36) (x^{2 + 1} - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.10.2

Add $2$ and $1$ .

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + (4 x^{2} + 36) (x^{3} - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.11

Expand $(4 x^{2} + 36) (x^{3} - 9 x)$ using the FOIL Method.

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Step 1.1.2.5.3.1.11.1

Apply the distributive property.

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 x^{2} (x^{3} - 9 x) + 36 (x^{3} - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.11.2

Apply the distributive property.

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 x^{2} x^{3} + 4 x^{2} (- 9 x) + 36 (x^{3} - 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.11.3

Apply the distributive property.

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 x^{2} x^{3} + 4 x^{2} (- 9 x) + 36 x^{3} + 36 (- 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.12

Simplify and combine like terms.

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Step 1.1.2.5.3.1.12.1

Simplify each term.

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Step 1.1.2.5.3.1.12.1.1

Multiply $x^{2}$ by $x^{3}$ by adding the exponents.

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Step 1.1.2.5.3.1.12.1.1.1

Move $x^{3}$ .

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 (x^{3} x^{2}) + 4 x^{2} (- 9 x) + 36 x^{3} + 36 (- 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.12.1.1.2

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

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 x^{3 + 2} + 4 x^{2} (- 9 x) + 36 x^{3} + 36 (- 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.12.1.1.3

Add $3$ and $2$ .

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 x^{5} + 4 x^{2} (- 9 x) + 36 x^{3} + 36 (- 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.12.1.2

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 x^{5} + 4 \cdot (- 9 x^{2} x) + 36 x^{3} + 36 (- 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.12.1.3

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 1.1.2.5.3.1.12.1.3.1

Move $x$ .

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 x^{5} + 4 \cdot (- 9 (x \cdot x^{2})) + 36 x^{3} + 36 (- 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.12.1.3.2

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

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Step 1.1.2.5.3.1.12.1.3.2.1

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

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 x^{5} + 4 \cdot (- 9 (x \cdot x^{2})) + 36 x^{3} + 36 (- 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.12.1.3.2.2

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

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 x^{5} + 4 \cdot (- 9 x^{1 + 2}) + 36 x^{3} + 36 (- 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.12.1.3.3

Add $1$ and $2$ .

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 x^{5} + 4 \cdot (- 9 x^{3}) + 36 x^{3} + 36 (- 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.12.1.4

Multiply $4$ by $- 9$ .

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 x^{5} - 36 x^{3} + 36 x^{3} + 36 (- 9 x)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.12.1.5

Multiply $- 9$ by $36$ .

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 x^{5} - 36 x^{3} + 36 x^{3} - 324 x}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.12.2

Add $- 36 x^{3}$ and $36 x^{3}$ .

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 x^{5} + 0 - 324 x}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.1.12.3

Add $4 x^{5}$ and $0$ .

$f'' (x) = \frac{- 2 x^{5} + 36 x^{3} - 162 x + 4 x^{5} - 324 x}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.2

Add $- 2 x^{5}$ and $4 x^{5}$ .

$f'' (x) = \frac{2 x^{5} + 36 x^{3} - 162 x - 324 x}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.3.3

Subtract $324 x$ from $- 162 x$ .

$f'' (x) = \frac{2 x^{5} + 36 x^{3} - 486 x}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.4

Simplify the numerator.

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Step 1.1.2.5.4.1

Factor $2 x$ out of $2 x^{5} + 36 x^{3} - 486 x$ .

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Step 1.1.2.5.4.1.1

Factor $2 x$ out of $2 x^{5}$ .

$f'' (x) = \frac{2 x (x^{4}) + 36 x^{3} - 486 x}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.4.1.2

Factor $2 x$ out of $36 x^{3}$ .

$f'' (x) = \frac{2 x (x^{4}) + 2 x (18 x^{2}) - 486 x}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.4.1.3

Factor $2 x$ out of $- 486 x$ .

$f'' (x) = \frac{2 x (x^{4}) + 2 x (18 x^{2}) + 2 x (- 243)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.4.1.4

Factor $2 x$ out of $2 x (x^{4}) + 2 x (18 x^{2})$ .

$f'' (x) = \frac{2 x (x^{4} + 18 x^{2}) + 2 x (- 243)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.4.1.5

Factor $2 x$ out of $2 x (x^{4} + 18 x^{2}) + 2 x (- 243)$ .

$f'' (x) = \frac{2 x (x^{4} + 18 x^{2} - 243)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.4.2

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$f'' (x) = \frac{2 x ({(x^{2})}^{2} + 18 x^{2} - 243)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.4.3

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$f'' (x) = \frac{2 x (u^{2} + 18 u - 243)}{{(x^{2} - 9)}^{4}}$

Step 1.1.2.5.4.4

Factor $u^{2} + 18 u - 243$ using the AC method.

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Step 1.1.2.5.4.4.1

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 $- 243$ and whose sum is $18$ .

$- 9, 27$

Step 1.1.2.5.4.4.2

Write the factored form using these integers.

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

Step 1.1.2.5.4.5

Replace all occurrences of $u$ with $x^{2}$ .

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

Step 1.1.2.5.4.6

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

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

Step 1.1.2.5.4.7

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 $b = 3$ .

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

Step 1.1.2.5.5

Simplify the denominator.

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Step 1.1.2.5.5.1

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

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

Step 1.1.2.5.5.2

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 $b = 3$ .

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

Step 1.1.2.5.5.3

Apply the product rule to $(x + 3) (x - 3)$ .

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

Step 1.1.2.5.6

Cancel the common factor of $x + 3$ and ${(x + 3)}^{4}$ .

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Step 1.1.2.5.6.1

Factor $x + 3$ out of $2 x (x + 3) (x - 3) (x^{2} + 27)$ .

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

Step 1.1.2.5.6.2

Cancel the common factors.

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Step 1.1.2.5.6.2.1

Factor $x + 3$ out of ${(x + 3)}^{4} {(x - 3)}^{4}$ .

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

Step 1.1.2.5.6.2.2

Cancel the common factor.

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

Step 1.1.2.5.6.2.3

Rewrite the expression.

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

Step 1.1.2.5.7

Cancel the common factor of $x - 3$ and ${(x - 3)}^{4}$ .

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Step 1.1.2.5.7.1

Factor $x - 3$ out of $(2 x (x - 3)) (x^{2} + 27)$ .

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

Step 1.1.2.5.7.2

Cancel the common factors.

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Step 1.1.2.5.7.2.1

Factor $x - 3$ out of ${(x + 3)}^{3} {(x - 3)}^{4}$ .

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

Step 1.1.2.5.7.2.2

Cancel the common factor.

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

Step 1.1.2.5.7.2.3

Rewrite the expression.

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

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

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{2 x (x^{2} + 27)}{{(x + 3)}^{3} {(x - 3)}^{3}}$ .

$\frac{2 x (x^{2} + 27)}{{(x + 3)}^{3} {(x - 3)}^{3}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $\frac{2 x (x^{2} + 27)}{{(x + 3)}^{3} {(x - 3)}^{3}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$\frac{2 x (x^{2} + 27)}{{(x + 3)}^{3} {(x - 3)}^{3}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$2 x (x^{2} + 27) = 0$

Step 1.2.3

Solve the equation for $x$ .

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Step 1.2.3.1

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

$x = 0$

$x^{2} + 27 = 0$

Step 1.2.3.2

Set $x$ equal to $0$ .

$x = 0$

Step 1.2.3.3

Set $x^{2} + 27$ equal to $0$ and solve for $x$ .

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Step 1.2.3.3.1

Set $x^{2} + 27$ equal to $0$ .

$x^{2} + 27 = 0$

Step 1.2.3.3.2

Solve $x^{2} + 27 = 0$ for $x$ .

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Step 1.2.3.3.2.1

Subtract $27$ from both sides of the equation.

$x^{2} = - 27$

Step 1.2.3.3.2.2

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

$x = \pm \sqrt{- 27}$

Step 1.2.3.3.2.3

Simplify $\pm \sqrt{- 27}$ .

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Step 1.2.3.3.2.3.1

Rewrite $- 27$ as $- 1 (27)$ .

$x = \pm \sqrt{- 1 (27)}$

Step 1.2.3.3.2.3.2

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{27}$

Step 1.2.3.3.2.3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{27}$

Step 1.2.3.3.2.3.4

Rewrite $27$ as $3^{2} \cdot 3$ .

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Step 1.2.3.3.2.3.4.1

Factor $9$ out of $27$ .

$x = \pm i \cdot \sqrt{9 (3)}$

Step 1.2.3.3.2.3.4.2

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

$x = \pm i \cdot \sqrt{3^{2} \cdot 3}$

Step 1.2.3.3.2.3.5

Pull terms out from under the radical.

$x = \pm i \cdot (3 \sqrt{3})$

Step 1.2.3.3.2.3.6

Move $3$ to the left of $i$ .

$x = \pm 3 i \sqrt{3}$

Step 1.2.3.3.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.2.3.3.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 3 i \sqrt{3}$

Step 1.2.3.3.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3 i \sqrt{3}$

Step 1.2.3.3.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3 i \sqrt{3}, - 3 i \sqrt{3}$

Step 1.2.3.4

The final solution is all the values that make $2 x (x^{2} + 27) = 0$ true.

$x = 0, 3 i \sqrt{3}, - 3 i \sqrt{3}$

Step 2

Find the domain of $f (x) = \frac{x}{x^{2} - 9}$ .

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Step 2.1

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

$x^{2} - 9 = 0$

Step 2.2

Solve for $x$ .

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Step 2.2.1

Add $9$ to both sides of the equation.

$x^{2} = 9$

Step 2.2.2

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

$x = \pm \sqrt{9}$

Step 2.2.3

Simplify $\pm \sqrt{9}$ .

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Step 2.2.3.1

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

$x = \pm \sqrt{3^{2}}$

Step 2.2.3.2

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

$x = \pm 3$

Step 2.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 3$

Step 2.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3$

Step 2.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3, - 3$

Step 2.3

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

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 3, - 3}$

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 3, - 3}$

Step 3

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

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

Step 4

Substitute any number from the interval $(- \infty, - 3)$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

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

$f'' (- 6) = \frac{2 (- 6) ({(- 6)}^{2} + 27)}{{((- 6) + 3)}^{3} {((- 6) - 3)}^{3}}$

Step 4.2

Simplify the result.

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Step 4.2.1

Multiply $2$ by $- 6$ .

$f'' (- 6) = \frac{- 12 ({(- 6)}^{2} + 27)}{{(- 6 + 3)}^{3} {(- 6 - 3)}^{3}}$

Step 4.2.2

Simplify the denominator.

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Step 4.2.2.1

Add $- 6$ and $3$ .

$f'' (- 6) = \frac{- 12 ({(- 6)}^{2} + 27)}{{(- 3)}^{3} {(- 6 - 3)}^{3}}$

Step 4.2.2.2

Subtract $3$ from $- 6$ .

$f'' (- 6) = \frac{- 12 ({(- 6)}^{2} + 27)}{{(- 3)}^{3} {(- 9)}^{3}}$

Step 4.2.2.3

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

$f'' (- 6) = \frac{- 12 ({(- 6)}^{2} + 27)}{- 27 {(- 9)}^{3}}$

Step 4.2.2.4

Raise $- 9$ to the power of $3$ .

$f'' (- 6) = \frac{- 12 ({(- 6)}^{2} + 27)}{- 27 \cdot - 729}$

Step 4.2.3

Simplify the numerator.

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Step 4.2.3.1

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

$f'' (- 6) = \frac{- 12 (36 + 27)}{- 27 \cdot - 729}$

Step 4.2.3.2

Add $36$ and $27$ .

$f'' (- 6) = \frac{- 12 \cdot 63}{- 27 \cdot - 729}$

Step 4.2.4

Reduce the expression by cancelling the common factors.

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Step 4.2.4.1

Multiply $- 27$ by $- 729$ .

$f'' (- 6) = \frac{- 12 \cdot 63}{19683}$

Step 4.2.4.2

Multiply $- 12$ by $63$ .

$f'' (- 6) = \frac{- 756}{19683}$

Step 4.2.4.3

Cancel the common factor of $- 756$ and $19683$ .

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Step 4.2.4.3.1

Factor $27$ out of $- 756$ .

$f'' (- 6) = \frac{27 (- 28)}{19683}$

Step 4.2.4.3.2

Cancel the common factors.

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Step 4.2.4.3.2.1

Factor $27$ out of $19683$ .

$f'' (- 6) = \frac{27 \cdot - 28}{27 \cdot 729}$

Step 4.2.4.3.2.2

Cancel the common factor.

$f'' (- 6) = \frac{27 \cdot - 28}{27 \cdot 729}$

Step 4.2.4.3.2.3

Rewrite the expression.

$f'' (- 6) = \frac{- 28}{729}$

Step 4.2.4.4

Move the negative in front of the fraction.

$f'' (- 6) = - \frac{28}{729}$

Step 4.2.5

The final answer is $- \frac{28}{729}$ .

$- \frac{28}{729}$

Step 4.3

The graph is concave down on the interval $(- \infty, - 3)$ because $f'' (- 6)$ is negative.

Concave down on $(- \infty, - 3)$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(- 3, 0)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

Replace the variable $x$ with $- 2$ in the expression.

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

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $2$ by $- 2$ .

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

Step 5.2.2

Simplify the denominator.

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Step 5.2.2.1

Add $- 2$ and $3$ .

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

Step 5.2.2.2

Subtract $3$ from $- 2$ .

$f'' (- 2) = \frac{- 4 ({(- 2)}^{2} + 27)}{1^{3} {(- 5)}^{3}}$

Step 5.2.2.3

One to any power is one.

$f'' (- 2) = \frac{- 4 ({(- 2)}^{2} + 27)}{1 {(- 5)}^{3}}$

Step 5.2.2.4

Raise $- 5$ to the power of $3$ .

$f'' (- 2) = \frac{- 4 ({(- 2)}^{2} + 27)}{1 \cdot - 125}$

Step 5.2.2.5

Multiply $- 125$ by $1$ .

$f'' (- 2) = \frac{- 4 ({(- 2)}^{2} + 27)}{- 125}$

Step 5.2.3

Simplify the numerator.

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Step 5.2.3.1

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

$f'' (- 2) = \frac{- 4 (4 + 27)}{- 125}$

Step 5.2.3.2

Add $4$ and $27$ .

$f'' (- 2) = \frac{- 4 \cdot 31}{- 125}$

Step 5.2.4

Reduce the expression by cancelling the common factors.

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Step 5.2.4.1

Multiply $- 4$ by $31$ .

$f'' (- 2) = \frac{- 124}{- 125}$

Step 5.2.4.2

Dividing two negative values results in a positive value.

$f'' (- 2) = \frac{124}{125}$

Step 5.2.5

The final answer is $\frac{124}{125}$ .

$\frac{124}{125}$

Step 5.3

The graph is concave up on the interval $(- 3, 0)$ because $f'' (- 2)$ is positive.

Concave up on $(- 3, 0)$ since $f'' (x)$ is positive

Step 6

Substitute any number from the interval $(0, 3)$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

Replace the variable $x$ with $2$ in the expression.

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

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $2$ by $2$ .

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

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

Add $2$ and $3$ .

$f'' (2) = \frac{4 (2^{2} + 27)}{5^{3} {(2 - 3)}^{3}}$

Step 6.2.2.2

Subtract $3$ from $2$ .

$f'' (2) = \frac{4 (2^{2} + 27)}{5^{3} {(- 1)}^{3}}$

Step 6.2.2.3

Raise $5$ to the power of $3$ .

$f'' (2) = \frac{4 (2^{2} + 27)}{125 {(- 1)}^{3}}$

Step 6.2.2.4

Raise $- 1$ to the power of $3$ .

$f'' (2) = \frac{4 (2^{2} + 27)}{125 \cdot - 1}$

Step 6.2.3

Simplify the numerator.

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Step 6.2.3.1

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

$f'' (2) = \frac{4 (4 + 27)}{125 \cdot - 1}$

Step 6.2.3.2

Add $4$ and $27$ .

$f'' (2) = \frac{4 \cdot 31}{125 \cdot - 1}$

Step 6.2.4

Simplify the expression.

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Step 6.2.4.1

Multiply $125$ by $- 1$ .

$f'' (2) = \frac{4 \cdot 31}{- 125}$

Step 6.2.4.2

Multiply $4$ by $31$ .

$f'' (2) = \frac{124}{- 125}$

Step 6.2.4.3

Move the negative in front of the fraction.

$f'' (2) = - \frac{124}{125}$

Step 6.2.5

The final answer is $- \frac{124}{125}$ .

$- \frac{124}{125}$

Step 6.3

The graph is concave down on the interval $(0, 3)$ because $f'' (2)$ is negative.

Concave down on $(0, 3)$ since $f'' (x)$ is negative

Step 7

Substitute any number from the interval $(3, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 7.1

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

$f'' (6) = \frac{2 (6) ({(6)}^{2} + 27)}{{((6) + 3)}^{3} {((6) - 3)}^{3}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $2$ by $6$ .

$f'' (6) = \frac{12 (6^{2} + 27)}{{(6 + 3)}^{3} {(6 - 3)}^{3}}$

Step 7.2.2

Simplify the denominator.

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Step 7.2.2.1

Add $6$ and $3$ .

$f'' (6) = \frac{12 (6^{2} + 27)}{9^{3} {(6 - 3)}^{3}}$

Step 7.2.2.2

Subtract $3$ from $6$ .

$f'' (6) = \frac{12 (6^{2} + 27)}{9^{3} \cdot 3^{3}}$

Step 7.2.2.3

Raise $9$ to the power of $3$ .

$f'' (6) = \frac{12 (6^{2} + 27)}{729 \cdot 3^{3}}$

Step 7.2.2.4

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

$f'' (6) = \frac{12 (6^{2} + 27)}{729 \cdot 27}$

Step 7.2.3

Simplify the numerator.

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Step 7.2.3.1

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

$f'' (6) = \frac{12 (36 + 27)}{729 \cdot 27}$

Step 7.2.3.2

Add $36$ and $27$ .

$f'' (6) = \frac{12 \cdot 63}{729 \cdot 27}$

Step 7.2.4

Reduce the expression by cancelling the common factors.

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Step 7.2.4.1

Multiply $729$ by $27$ .

$f'' (6) = \frac{12 \cdot 63}{19683}$

Step 7.2.4.2

Multiply $12$ by $63$ .

$f'' (6) = \frac{756}{19683}$

Step 7.2.4.3

Cancel the common factor of $756$ and $19683$ .

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Step 7.2.4.3.1

Factor $27$ out of $756$ .

$f'' (6) = \frac{27 (28)}{19683}$

Step 7.2.4.3.2

Cancel the common factors.

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Step 7.2.4.3.2.1

Factor $27$ out of $19683$ .

$f'' (6) = \frac{27 \cdot 28}{27 \cdot 729}$

Step 7.2.4.3.2.2

Cancel the common factor.

$f'' (6) = \frac{27 \cdot 28}{27 \cdot 729}$

Step 7.2.4.3.2.3

Rewrite the expression.

$f'' (6) = \frac{28}{729}$

Step 7.2.5

The final answer is $\frac{28}{729}$ .

$\frac{28}{729}$

Step 7.3

The graph is concave up on the interval $(3, \infty)$ because $f'' (6)$ is positive.

Concave up on $(3, \infty)$ since $f'' (x)$ is positive

Step 8

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

Concave down on $(- \infty, - 3)$ since $f'' (x)$ is negative

Concave up on $(- 3, 0)$ since $f'' (x)$ is positive

Concave down on $(0, 3)$ since $f'' (x)$ is negative

Concave up on $(3, \infty)$ since $f'' (x)$ is positive

Step 9