Calculus Examples

$f (x) = \frac{x^{2}}{x^{2} - 16}$

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 $f (x) = x^{2}$ and $g (x) = x^{2} - 16$ .

$\frac{(x^{2} - 16) \frac{d}{d x} [x^{2}] - x^{2} \frac{d}{d x} [x^{2} - 16]}{{(x^{2} - 16)}^{2}}$

Step 1.2

Differentiate.

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Step 1.2.1

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

$\frac{(x^{2} - 16) (2 x) - x^{2} \frac{d}{d x} [x^{2} - 16]}{{(x^{2} - 16)}^{2}}$

Step 1.2.2

Move $2$ to the left of $x^{2} - 16$ .

$\frac{2 \cdot (x^{2} - 16) x - x^{2} \frac{d}{d x} [x^{2} - 16]}{{(x^{2} - 16)}^{2}}$

Step 1.2.3

By the Sum Rule, the derivative of $x^{2} - 16$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16]$ .

$\frac{2 (x^{2} - 16) x - x^{2} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16])}{{(x^{2} - 16)}^{2}}$

Step 1.2.4

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

$\frac{2 (x^{2} - 16) x - x^{2} (2 x + \frac{d}{d x} [- 16])}{{(x^{2} - 16)}^{2}}$

Step 1.2.5

Since $- 16$ is constant with respect to $x$ , the derivative of $- 16$ with respect to $x$ is $0$ .

$\frac{2 (x^{2} - 16) x - x^{2} (2 x + 0)}{{(x^{2} - 16)}^{2}}$

Step 1.2.6

Simplify the expression.

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Step 1.2.6.1

Add $2 x$ and $0$ .

$\frac{2 (x^{2} - 16) x - x^{2} (2 x)}{{(x^{2} - 16)}^{2}}$

Step 1.2.6.2

Multiply $2$ by $- 1$ .

$\frac{2 (x^{2} - 16) x - 2 x^{2} x}{{(x^{2} - 16)}^{2}}$

Step 1.3

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

$\frac{2 (x^{2} - 16) x - 2 (x^{1} x^{2})}{{(x^{2} - 16)}^{2}}$

Step 1.4

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

$\frac{2 (x^{2} - 16) x - 2 x^{1 + 2}}{{(x^{2} - 16)}^{2}}$

Step 1.5

Add $1$ and $2$ .

$\frac{2 (x^{2} - 16) x - 2 x^{3}}{{(x^{2} - 16)}^{2}}$

Step 1.6

Simplify.

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Step 1.6.1

Apply the distributive property.

$\frac{(2 x^{2} + 2 \cdot - 16) x - 2 x^{3}}{{(x^{2} - 16)}^{2}}$

Step 1.6.2

Apply the distributive property.

$\frac{2 x^{2} x + 2 \cdot - 16 x - 2 x^{3}}{{(x^{2} - 16)}^{2}}$

Step 1.6.3

Simplify the numerator.

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Step 1.6.3.1

Simplify each term.

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Step 1.6.3.1.1

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

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Step 1.6.3.1.1.1

Move $x$ .

$\frac{2 (x \cdot x^{2}) + 2 \cdot - 16 x - 2 x^{3}}{{(x^{2} - 16)}^{2}}$

Step 1.6.3.1.1.2

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

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Step 1.6.3.1.1.2.1

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

$\frac{2 (x^{1} x^{2}) + 2 \cdot - 16 x - 2 x^{3}}{{(x^{2} - 16)}^{2}}$

Step 1.6.3.1.1.2.2

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

$\frac{2 x^{1 + 2} + 2 \cdot - 16 x - 2 x^{3}}{{(x^{2} - 16)}^{2}}$

Step 1.6.3.1.1.3

Add $1$ and $2$ .

$\frac{2 x^{3} + 2 \cdot - 16 x - 2 x^{3}}{{(x^{2} - 16)}^{2}}$

Step 1.6.3.1.2

Multiply $2$ by $- 16$ .

$\frac{2 x^{3} - 32 x - 2 x^{3}}{{(x^{2} - 16)}^{2}}$

Step 1.6.3.2

Combine the opposite terms in $2 x^{3} - 32 x - 2 x^{3}$ .

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Step 1.6.3.2.1

Subtract $2 x^{3}$ from $2 x^{3}$ .

$\frac{- 32 x + 0}{{(x^{2} - 16)}^{2}}$

Step 1.6.3.2.2

Add $- 32 x$ and $0$ .

$\frac{- 32 x}{{(x^{2} - 16)}^{2}}$

Step 1.6.4

Move the negative in front of the fraction.

$- \frac{32 x}{{(x^{2} - 16)}^{2}}$

Step 1.6.5

Simplify the denominator.

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Step 1.6.5.1

Rewrite $16$ as $4^{2}$ .

$- \frac{32 x}{{(x^{2} - 4^{2})}^{2}}$

Step 1.6.5.2

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 $b = 4$ .

$- \frac{32 x}{{((x + 4) (x - 4))}^{2}}$

Step 1.6.5.3

Apply the product rule to $(x + 4) (x - 4)$ .

$- \frac{32 x}{{(x + 4)}^{2} {(x - 4)}^{2}}$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $- 32$ is constant with respect to $x$ , the derivative of $- \frac{32 x}{{(x + 4)}^{2} {(x - 4)}^{2}}$ with respect to $x$ is $- 32 \frac{d}{d x} [\frac{x}{{(x + 4)}^{2} {(x - 4)}^{2}}]$ .

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

Step 2.2

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

Step 2.3

Differentiate using the Power Rule.

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Step 2.3.1

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

$f'' (x) = - 32 \frac{{(x + 4)}^{2} {(x - 4)}^{2} \cdot 1 - x \frac{d}{d x} ({(x + 4)}^{2} {(x - 4)}^{2})}{{({(x + 4)}^{2} {(x - 4)}^{2})}^{2}}$

Step 2.3.2

Multiply ${(x + 4)}^{2}$ by $1$ .

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

Step 2.4

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

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

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 $g (x) = x - 4$ .

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Step 2.5.1

To apply the Chain Rule, set $u_{1}$ as $x - 4$ .

$f'' (x) = - 32 \frac{{(x + 4)}^{2} {(x - 4)}^{2} - x ({(x + 4)}^{2} (\frac{d}{d u (1)} ({u_{1}}^{2}) \frac{d}{d x} (x - 4)) + {(x - 4)}^{2} \frac{d}{d x} ({(x + 4)}^{2}))}{{({(x + 4)}^{2} {(x - 4)}^{2})}^{2}}$

Step 2.5.2

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

$f'' (x) = - 32 \frac{{(x + 4)}^{2} {(x - 4)}^{2} - x ({(x + 4)}^{2} (2 u_{1} \frac{d}{d x} (x - 4)) + {(x - 4)}^{2} \frac{d}{d x} ({(x + 4)}^{2}))}{{({(x + 4)}^{2} {(x - 4)}^{2})}^{2}}$

Step 2.5.3

Replace all occurrences of $u_{1}$ with $x - 4$ .

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

Step 2.6

Differentiate.

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Step 2.6.1

Move $2$ to the left of ${(x + 4)}^{2}$ .

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

Step 2.6.2

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

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

Step 2.6.3

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

$f'' (x) = - 32 \frac{{(x + 4)}^{2} {(x - 4)}^{2} - x (2 {(x + 4)}^{2} (x - 4) (1 + \frac{d}{d x} (- 4)) + {(x - 4)}^{2} \frac{d}{d x} ({(x + 4)}^{2}))}{{({(x + 4)}^{2} {(x - 4)}^{2})}^{2}}$

Step 2.6.4

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

$f'' (x) = - 32 \frac{{(x + 4)}^{2} {(x - 4)}^{2} - x (2 {(x + 4)}^{2} (x - 4) (1 + 0) + {(x - 4)}^{2} \frac{d}{d x} ({(x + 4)}^{2}))}{{({(x + 4)}^{2} {(x - 4)}^{2})}^{2}}$

Step 2.6.5

Simplify the expression.

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Step 2.6.5.1

Add $1$ and $0$ .

$f'' (x) = - 32 \frac{{(x + 4)}^{2} {(x - 4)}^{2} - x (2 {(x + 4)}^{2} (x - 4) \cdot 1 + {(x - 4)}^{2} \frac{d}{d x} ({(x + 4)}^{2}))}{{({(x + 4)}^{2} {(x - 4)}^{2})}^{2}}$

Step 2.6.5.2

Multiply $2$ by $1$ .

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

Step 2.7

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 2.7.1

To apply the Chain Rule, set $u_{2}$ as $x + 4$ .

$f'' (x) = - 32 \frac{{(x + 4)}^{2} {(x - 4)}^{2} - x (2 {(x + 4)}^{2} (x - 4) + {(x - 4)}^{2} (\frac{d}{d u (2)} ({u_{2}}^{2}) \frac{d}{d x} (x + 4)))}{{({(x + 4)}^{2} {(x - 4)}^{2})}^{2}}$

Step 2.7.2

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

$f'' (x) = - 32 \frac{{(x + 4)}^{2} {(x - 4)}^{2} - x (2 {(x + 4)}^{2} (x - 4) + {(x - 4)}^{2} (2 u_{2} \frac{d}{d x} (x + 4)))}{{({(x + 4)}^{2} {(x - 4)}^{2})}^{2}}$

Step 2.7.3

Replace all occurrences of $u_{2}$ with $x + 4$ .

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

Step 2.8

Differentiate.

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Step 2.8.1

Move $2$ to the left of ${(x - 4)}^{2}$ .

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

Step 2.8.2

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

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

Step 2.8.3

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

$f'' (x) = - 32 \frac{{(x + 4)}^{2} {(x - 4)}^{2} - x (2 {(x + 4)}^{2} (x - 4) + 2 {(x - 4)}^{2} (x + 4) (1 + \frac{d}{d x} (4)))}{{({(x + 4)}^{2} {(x - 4)}^{2})}^{2}}$

Step 2.8.4

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

$f'' (x) = - 32 \frac{{(x + 4)}^{2} {(x - 4)}^{2} - x (2 {(x + 4)}^{2} (x - 4) + 2 {(x - 4)}^{2} (x + 4) (1 + 0))}{{({(x + 4)}^{2} {(x - 4)}^{2})}^{2}}$

Step 2.8.5

Combine fractions.

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Step 2.8.5.1

Add $1$ and $0$ .

$f'' (x) = - 32 \frac{{(x + 4)}^{2} {(x - 4)}^{2} - x (2 {(x + 4)}^{2} (x - 4) + 2 {(x - 4)}^{2} (x + 4) \cdot 1)}{{({(x + 4)}^{2} {(x - 4)}^{2})}^{2}}$

Step 2.8.5.2

Multiply $2$ by $1$ .

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

Step 2.8.5.3

Combine $- 32$ and $\frac{{(x + 4)}^{2} {(x - 4)}^{2} - x (2 {(x + 4)}^{2} (x - 4) + 2 {(x - 4)}^{2} (x + 4))}{{({(x + 4)}^{2} {(x - 4)}^{2})}^{2}}$ .

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

Step 2.8.5.4

Move the negative in front of the fraction.

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

Step 2.9

Simplify.

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Step 2.9.1

Apply the product rule to ${(x + 4)}^{2} {(x - 4)}^{2}$ .

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

Step 2.9.2

Apply the distributive property.

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

Step 2.9.3

Apply the distributive property.

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

Step 2.9.4

Simplify the numerator.

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Step 2.9.4.1

Factor $32 (x + 4) (x - 4)$ out of $32 {(x + 4)}^{2} {(x - 4)}^{2} + 32 (- x (2 {(x + 4)}^{2} (x - 4))) + 32 (- x (2 {(x - 4)}^{2} (x + 4)))$ .

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Step 2.9.4.1.1

Factor $32 (x + 4) (x - 4)$ out of $32 {(x + 4)}^{2} {(x - 4)}^{2}$ .

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

Step 2.9.4.1.2

Factor $32 (x + 4) (x - 4)$ out of $32 (- x (2 {(x + 4)}^{2} (x - 4)))$ .

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

Step 2.9.4.1.3

Factor $32 (x + 4) (x - 4)$ out of $32 (- x (2 {(x - 4)}^{2} (x + 4)))$ .

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

Step 2.9.4.1.4

Factor $32 (x + 4) (x - 4)$ out of $32 (x + 4) (x - 4) ((x + 4) (x - 4)) + 32 (x + 4) (x - 4) (- x (2 (x + 4)))$ .

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

Step 2.9.4.1.5

Factor $32 (x + 4) (x - 4)$ out of $32 (x + 4) (x - 4) ((x + 4) (x - 4) - x (2 (x + 4))) + 32 (x + 4) (x - 4) (- x (2 (x - 4)))$ .

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

Step 2.9.4.2

Combine exponents.

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Step 2.9.4.2.1

Multiply $2$ by $- 1$ .

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

Step 2.9.4.2.2

Multiply $2$ by $- 1$ .

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

Step 2.9.4.3

Simplify each term.

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Step 2.9.4.3.1

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

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Step 2.9.4.3.1.1

Apply the distributive property.

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

Step 2.9.4.3.1.2

Apply the distributive property.

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

Step 2.9.4.3.1.3

Apply the distributive property.

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

Step 2.9.4.3.2

Combine the opposite terms in $x \cdot x + x \cdot - 4 + 4 x + 4 \cdot - 4$ .

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Step 2.9.4.3.2.1

Reorder the factors in the terms $x \cdot - 4$ and $4 x$ .

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

Step 2.9.4.3.2.2

Add $- 4 x$ and $4 x$ .

$f'' (x) = - \frac{32 (x + 4) (x - 4) (x \cdot x + 0 + 4 \cdot - 4 - 2 x (x + 4) - 2 x (x - 4))}{{({(x + 4)}^{2})}^{2} {({(x - 4)}^{2})}^{2}}$

Step 2.9.4.3.2.3

Add $x \cdot x$ and $0$ .

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

Step 2.9.4.3.3

Simplify each term.

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Step 2.9.4.3.3.1

Multiply $x$ by $x$ .

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

Step 2.9.4.3.3.2

Multiply $4$ by $- 4$ .

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

Step 2.9.4.3.4

Apply the distributive property.

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

Step 2.9.4.3.5

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

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Step 2.9.4.3.5.1

Move $x$ .

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

Step 2.9.4.3.5.2

Multiply $x$ by $x$ .

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

Step 2.9.4.3.6

Multiply $4$ by $- 2$ .

$f'' (x) = - \frac{32 (x + 4) (x - 4) (x^{2} - 16 - 2 x^{2} - 8 x - 2 x (x - 4))}{{({(x + 4)}^{2})}^{2} {({(x - 4)}^{2})}^{2}}$

Step 2.9.4.3.7

Apply the distributive property.

$f'' (x) = - \frac{32 (x + 4) (x - 4) (x^{2} - 16 - 2 x^{2} - 8 x - 2 x \cdot x - 2 x \cdot - 4)}{{({(x + 4)}^{2})}^{2} {({(x - 4)}^{2})}^{2}}$

Step 2.9.4.3.8

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

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Step 2.9.4.3.8.1

Move $x$ .

$f'' (x) = - \frac{32 (x + 4) (x - 4) (x^{2} - 16 - 2 x^{2} - 8 x - 2 (x \cdot x) - 2 x \cdot - 4)}{{({(x + 4)}^{2})}^{2} {({(x - 4)}^{2})}^{2}}$

Step 2.9.4.3.8.2

Multiply $x$ by $x$ .

$f'' (x) = - \frac{32 (x + 4) (x - 4) (x^{2} - 16 - 2 x^{2} - 8 x - 2 x^{2} - 2 x \cdot - 4)}{{({(x + 4)}^{2})}^{2} {({(x - 4)}^{2})}^{2}}$

Step 2.9.4.3.9

Multiply $- 4$ by $- 2$ .

$f'' (x) = - \frac{32 (x + 4) (x - 4) (x^{2} - 16 - 2 x^{2} - 8 x - 2 x^{2} + 8 x)}{{({(x + 4)}^{2})}^{2} {({(x - 4)}^{2})}^{2}}$

Step 2.9.4.4

Combine the opposite terms in $x^{2} - 16 - 2 x^{2} - 8 x - 2 x^{2} + 8 x$ .

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Step 2.9.4.4.1

Add $- 8 x$ and $8 x$ .

$f'' (x) = - \frac{32 (x + 4) (x - 4) (x^{2} - 16 - 2 x^{2} - 2 x^{2} + 0)}{{({(x + 4)}^{2})}^{2} {({(x - 4)}^{2})}^{2}}$

Step 2.9.4.4.2

Add $x^{2} - 16 - 2 x^{2} - 2 x^{2}$ and $0$ .

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

Step 2.9.4.5

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

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

Step 2.9.4.6

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

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

Step 2.9.5

Combine terms.

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Step 2.9.5.1

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

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Step 2.9.5.1.1

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

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

Step 2.9.5.1.2

Multiply $2$ by $2$ .

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

Step 2.9.5.2

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

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Step 2.9.5.2.1

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

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

Step 2.9.5.2.2

Multiply $2$ by $2$ .

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

Step 2.9.5.3

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

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Step 2.9.5.3.1

Factor $x + 4$ out of $32 (x + 4) (x - 4) (- 3 x^{2} - 16)$ .

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

Step 2.9.5.3.2

Cancel the common factors.

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Step 2.9.5.3.2.1

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

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

Step 2.9.5.3.2.2

Cancel the common factor.

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

Step 2.9.5.3.2.3

Rewrite the expression.

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

Step 2.9.5.4

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

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Step 2.9.5.4.1

Factor $x - 4$ out of $32 (x - 4) (- 3 x^{2} - 16)$ .

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

Step 2.9.5.4.2

Cancel the common factors.

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Step 2.9.5.4.2.1

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

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

Step 2.9.5.4.2.2

Cancel the common factor.

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

Step 2.9.5.4.2.3

Rewrite the expression.

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

Step 2.9.6

Factor $- 1$ out of $- 3 x^{2}$ .

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

Step 2.9.7

Rewrite $- 16$ as $- 1 (16)$ .

$f'' (x) = - \frac{32 (- (3 x^{2}) - 1 \cdot 16)}{{(x + 4)}^{3} {(x - 4)}^{3}}$

Step 2.9.8

Factor $- 1$ out of $- (3 x^{2}) - 1 (16)$ .

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

Step 2.9.9

Rewrite $- (3 x^{2} + 16)$ as $- 1 (3 x^{2} + 16)$ .

$f'' (x) = - \frac{32 (- 1 (3 x^{2} + 16))}{{(x + 4)}^{3} {(x - 4)}^{3}}$

Step 2.9.10

Move the negative in front of the fraction.

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

Step 2.9.11

Multiply $- 1$ by $- 1$ .

$f'' (x) = 1 (\frac{32 (3 x^{2} + 16)}{{(x + 4)}^{3} {(x - 4)}^{3}})$

Step 2.9.12

Multiply $\frac{32 (3 x^{2} + 16)}{{(x + 4)}^{3} {(x - 4)}^{3}}$ by $1$ .

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

Step 3

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

$- \frac{32 x}{{(x + 4)}^{2} {(x - 4)}^{2}} = 0$

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

$f' (x) = \frac{(x^{2} - 16) \frac{d}{d x} (x^{2}) - x^{2} \frac{d}{d x} (x^{2} - 16)}{{(x^{2} - 16)}^{2}}$

Step 4.1.2

Differentiate.

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Step 4.1.2.1

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

$f' (x) = \frac{(x^{2} - 16) (2 x) - x^{2} \frac{d}{d x} (x^{2} - 16)}{{(x^{2} - 16)}^{2}}$

Step 4.1.2.2

Move $2$ to the left of $x^{2} - 16$ .

$f' (x) = \frac{2 \cdot ((x^{2} - 16) x) - x^{2} \frac{d}{d x} (x^{2} - 16)}{{(x^{2} - 16)}^{2}}$

Step 4.1.2.3

By the Sum Rule, the derivative of $x^{2} - 16$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16]$ .

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

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 $n = 2$ .

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

Step 4.1.2.5

Since $- 16$ is constant with respect to $x$ , the derivative of $- 16$ with respect to $x$ is $0$ .

$f' (x) = \frac{2 (x^{2} - 16) x - x^{2} (2 x + 0)}{{(x^{2} - 16)}^{2}}$

Step 4.1.2.6

Simplify the expression.

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Step 4.1.2.6.1

Add $2 x$ and $0$ .

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

Step 4.1.2.6.2

Multiply $2$ by $- 1$ .

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

Step 4.1.3

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

$f' (x) = \frac{2 (x^{2} - 16) x - 2 (x \cdot x^{2})}{{(x^{2} - 16)}^{2}}$

Step 4.1.4

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

$f' (x) = \frac{2 (x^{2} - 16) x - 2 x^{1 + 2}}{{(x^{2} - 16)}^{2}}$

Step 4.1.5

Add $1$ and $2$ .

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

Step 4.1.6

Simplify.

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Step 4.1.6.1

Apply the distributive property.

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

Step 4.1.6.2

Apply the distributive property.

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

Step 4.1.6.3

Simplify the numerator.

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Step 4.1.6.3.1

Simplify each term.

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Step 4.1.6.3.1.1

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

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Step 4.1.6.3.1.1.1

Move $x$ .

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

Step 4.1.6.3.1.1.2

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

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Step 4.1.6.3.1.1.2.1

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

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

Step 4.1.6.3.1.1.2.2

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

$f' (x) = \frac{2 x^{1 + 2} + 2 \cdot (- 16 x) - 2 x^{3}}{{(x^{2} - 16)}^{2}}$

Step 4.1.6.3.1.1.3

Add $1$ and $2$ .

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

Step 4.1.6.3.1.2

Multiply $2$ by $- 16$ .

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

Step 4.1.6.3.2

Combine the opposite terms in $2 x^{3} - 32 x - 2 x^{3}$ .

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Step 4.1.6.3.2.1

Subtract $2 x^{3}$ from $2 x^{3}$ .

$f' (x) = \frac{- 32 x + 0}{{(x^{2} - 16)}^{2}}$

Step 4.1.6.3.2.2

Add $- 32 x$ and $0$ .

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

Step 4.1.6.4

Move the negative in front of the fraction.

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

Step 4.1.6.5

Simplify the denominator.

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Step 4.1.6.5.1

Rewrite $16$ as $4^{2}$ .

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

Step 4.1.6.5.2

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 $b = 4$ .

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

Step 4.1.6.5.3

Apply the product rule to $(x + 4) (x - 4)$ .

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

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{32 x}{{(x + 4)}^{2} {(x - 4)}^{2}}$ .

$- \frac{32 x}{{(x + 4)}^{2} {(x - 4)}^{2}}$

Step 5

Set the first derivative equal to $0$ then solve the equation $- \frac{32 x}{{(x + 4)}^{2} {(x - 4)}^{2}} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$- \frac{32 x}{{(x + 4)}^{2} {(x - 4)}^{2}} = 0$

Step 5.2

Set the numerator equal to zero.

$32 x = 0$

Step 5.3

Divide each term in $32 x = 0$ by $32$ and simplify.

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Step 5.3.1

Divide each term in $32 x = 0$ by $32$ .

$\frac{32 x}{32} = \frac{0}{32}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $32$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{32 x}{32} = \frac{0}{32}$

Step 5.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{32}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Divide $0$ by $32$ .

$x = 0$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{32 x}{{(x + 4)}^{2} {(x - 4)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(x + 4)}^{2} {(x - 4)}^{2} = 0$

Step 6.2

Solve for $x$ .

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Step 6.2.1

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

${(x + 4)}^{2} = 0$

${(x - 4)}^{2} = 0$

Step 6.2.2

Set ${(x + 4)}^{2}$ equal to $0$ and solve for $x$ .

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Step 6.2.2.1

Set ${(x + 4)}^{2}$ equal to $0$ .

${(x + 4)}^{2} = 0$

Step 6.2.2.2

Solve ${(x + 4)}^{2} = 0$ for $x$ .

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Step 6.2.2.2.1

Set the $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 6.2.2.2.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 6.2.3

Set ${(x - 4)}^{2}$ equal to $0$ and solve for $x$ .

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Step 6.2.3.1

Set ${(x - 4)}^{2}$ equal to $0$ .

${(x - 4)}^{2} = 0$

Step 6.2.3.2

Solve ${(x - 4)}^{2} = 0$ for $x$ .

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Step 6.2.3.2.1

Set the $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 6.2.3.2.2

Add $4$ to both sides of the equation.

$x = 4$

Step 6.2.4

The final solution is all the values that make ${(x + 4)}^{2} {(x - 4)}^{2} = 0$ true.

$x = - 4, 4$

Step 6.3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 4, x = 4$

Step 7

Critical points to evaluate.

$x = 0$

Step 8

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{32 (3 {(0)}^{2} + 16)}{{((0) + 4)}^{3} {((0) - 4)}^{3}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the numerator.

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Step 9.1.1

Raising $0$ to any positive power yields $0$ .

$\frac{32 (3 \cdot 0 + 16)}{{(0 + 4)}^{3} {(0 - 4)}^{3}}$

Step 9.1.2

Multiply $3$ by $0$ .

$\frac{32 (0 + 16)}{{(0 + 4)}^{3} {(0 - 4)}^{3}}$

Step 9.1.3

Add $0$ and $16$ .

$\frac{32 \cdot 16}{{(0 + 4)}^{3} {(0 - 4)}^{3}}$

Step 9.2

Simplify the denominator.

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Step 9.2.1

Rewrite $0$ as $- 1 (0)$ .

$\frac{32 \cdot 16}{{(- 1 (0) + 4)}^{3} {(0 - 4)}^{3}}$

Step 9.2.2

Rewrite $4$ as $- 1 (- 4)$ .

$\frac{32 \cdot 16}{{(- 1 \cdot 0 - 1 \cdot - 4)}^{3} {(0 - 4)}^{3}}$

Step 9.2.3

Factor $- 1$ out of $- 1 \cdot 0 - 1 \cdot - 4$ .

$\frac{32 \cdot 16}{{(- 1 \cdot (0 - 4))}^{3} {(0 - 4)}^{3}}$

Step 9.2.4

Apply the product rule to $- 1 \cdot (0 - 4)$ .

$\frac{32 \cdot 16}{{(- 1)}^{3} \cdot {(0 - 4)}^{3} {(0 - 4)}^{3}}$

Step 9.2.5

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

$\frac{32 \cdot 16}{- 1 \cdot {(0 - 4)}^{3} {(0 - 4)}^{3}}$

Step 9.2.6

Multiply ${(0 - 4)}^{3}$ by ${(0 - 4)}^{3}$ by adding the exponents.

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Step 9.2.6.1

Move ${(0 - 4)}^{3}$ .

$\frac{32 \cdot 16}{- 1 \cdot ({(0 - 4)}^{3} {(0 - 4)}^{3})}$

Step 9.2.6.2

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

$\frac{32 \cdot 16}{- 1 \cdot {(0 - 4)}^{3 + 3}}$

Step 9.2.6.3

Add $3$ and $3$ .

$\frac{32 \cdot 16}{- 1 \cdot {(0 - 4)}^{6}}$

Step 9.3

Multiply $32$ by $16$ .

$\frac{512}{- 1 \cdot {(0 - 4)}^{6}}$

Step 9.4

Simplify the denominator.

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Step 9.4.1

Subtract $4$ from $0$ .

$\frac{512}{- 1 {(- 4)}^{6}}$

Step 9.4.2

Raise $- 4$ to the power of $6$ .

$\frac{512}{- 1 \cdot 4096}$

Step 9.5

Reduce the expression by cancelling the common factors.

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Step 9.5.1

Multiply $- 1$ by $4096$ .

$\frac{512}{- 4096}$

Step 9.5.2

Cancel the common factor of $512$ and $- 4096$ .

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Step 9.5.2.1

Factor $512$ out of $512$ .

$\frac{512 (1)}{- 4096}$

Step 9.5.2.2

Cancel the common factors.

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Step 9.5.2.2.1

Factor $512$ out of $- 4096$ .

$\frac{512 \cdot 1}{512 \cdot - 8}$

Step 9.5.2.2.2

Cancel the common factor.

$\frac{512 \cdot 1}{512 \cdot - 8}$

Step 9.5.2.2.3

Rewrite the expression.

$\frac{1}{- 8}$

Step 9.5.3

Move the negative in front of the fraction.

$- \frac{1}{8}$

Step 10

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Step 11

Find the y-value when $x = 0$ .

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Step 11.1

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

$f (0) = \frac{{(0)}^{2}}{{(0)}^{2} - 16}$

Step 11.2

Simplify the result.

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Step 11.2.1

Raising $0$ to any positive power yields $0$ .

$f (0) = \frac{0}{0^{2} - 16}$

Step 11.2.2

Simplify the denominator.

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Step 11.2.2.1

Raising $0$ to any positive power yields $0$ .

$f (0) = \frac{0}{0 - 16}$

Step 11.2.2.2

Subtract $16$ from $0$ .

$f (0) = \frac{0}{- 16}$

Step 11.2.3

Divide $0$ by $- 16$ .

$f (0) = 0$

Step 11.2.4

The final answer is $0$ .

$y = 0$

Step 12

These are the local extrema for $f (x) = \frac{x^{2}}{x^{2} - 16}$ .

$(0, 0)$ is a local maxima

Step 13