Calculus Examples

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

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

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

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

By the Sum Rule, the derivative of $x^{2} + 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4]$ .

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

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 $n = 2$ .

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

Step 1.1.1.2.3

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

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

Step 1.1.1.2.4

Simplify the expression.

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Step 1.1.1.2.4.1

Add $2 x$ and $0$ .

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

Step 1.1.1.2.4.2

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

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

Step 1.1.1.2.5

By the Sum Rule, the derivative of $4 - x^{2}$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [- x^{2}]$ .

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

Step 1.1.1.2.6

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

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

Step 1.1.1.2.7

Add $0$ and $\frac{d}{d x} [- x^{2}]$ .

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

Step 1.1.1.2.8

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

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

Step 1.1.1.2.9

Multiply.

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Step 1.1.1.2.9.1

Multiply $- 1$ by $- 1$ .

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

Step 1.1.1.2.9.2

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

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

Step 1.1.1.2.10

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

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

Step 1.1.1.2.11

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

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

Step 1.1.1.3

Simplify.

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Step 1.1.1.3.1

Apply the distributive property.

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

Step 1.1.1.3.2

Apply the distributive property.

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

Step 1.1.1.3.3

Apply the distributive property.

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

Step 1.1.1.3.4

Apply the distributive property.

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

Step 1.1.1.3.5

Simplify the numerator.

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Step 1.1.1.3.5.1

Simplify each term.

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Step 1.1.1.3.5.1.1

Multiply $2$ by $4$ .

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

Step 1.1.1.3.5.1.2

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

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Step 1.1.1.3.5.1.2.1

Move $x$ .

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

Step 1.1.1.3.5.1.2.2

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

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Step 1.1.1.3.5.1.2.2.1

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

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

Step 1.1.1.3.5.1.2.2.2

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

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

Step 1.1.1.3.5.1.2.3

Add $1$ and $2$ .

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

Step 1.1.1.3.5.1.3

Multiply $- 1$ by $2$ .

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

Step 1.1.1.3.5.1.4

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

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Step 1.1.1.3.5.1.4.1

Move $x$ .

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

Step 1.1.1.3.5.1.4.2

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

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Step 1.1.1.3.5.1.4.2.1

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

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

Step 1.1.1.3.5.1.4.2.2

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

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

Step 1.1.1.3.5.1.4.3

Add $1$ and $2$ .

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

Step 1.1.1.3.5.1.5

Multiply $2$ by $4$ .

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

Step 1.1.1.3.5.2

Combine the opposite terms in $8 x - 2 x^{3} + 2 x^{3} + 8 x$ .

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Step 1.1.1.3.5.2.1

Add $- 2 x^{3}$ and $2 x^{3}$ .

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

Step 1.1.1.3.5.2.2

Add $8 x$ and $0$ .

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

Step 1.1.1.3.5.3

Add $8 x$ and $8 x$ .

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

Step 1.1.1.3.6

Reorder terms.

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

Step 1.1.1.3.7

Simplify the denominator.

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Step 1.1.1.3.7.1

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

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

Step 1.1.1.3.7.2

Reorder $- x^{2}$ and $2^{2}$ .

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

Step 1.1.1.3.7.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = x$ .

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

Step 1.1.1.3.7.4

Apply the product rule to $(2 + x) (2 - x)$ .

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

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

Since $16$ is constant with respect to $x$ , the derivative of $\frac{16 x}{{(2 + x)}^{2} {(2 - x)}^{2}}$ with respect to $x$ is $16 \frac{d}{d x} [\frac{x}{{(2 + x)}^{2} {(2 - x)}^{2}}]$ .

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

Step 1.1.2.2

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

Step 1.1.2.3

Differentiate using the Power Rule.

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Step 1.1.2.3.1

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

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

Step 1.1.2.3.2

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

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

Step 1.1.2.4

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

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

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 $g (x) = 2 - x$ .

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Step 1.1.2.5.1

To apply the Chain Rule, set $u_{1}$ as $2 - x$ .

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

Step 1.1.2.5.2

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

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

Step 1.1.2.5.3

Replace all occurrences of $u_{1}$ with $2 - x$ .

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

Step 1.1.2.6

Differentiate.

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Step 1.1.2.6.1

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

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

Step 1.1.2.6.2

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

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

Step 1.1.2.6.3

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

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

Step 1.1.2.6.4

Add $0$ and $\frac{d}{d x} [- x]$ .

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

Step 1.1.2.6.5

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

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

Step 1.1.2.6.6

Multiply $- 1$ by $2$ .

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

Step 1.1.2.6.7

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

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

Step 1.1.2.6.8

Multiply $- 2$ by $1$ .

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

Step 1.1.2.7

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 $g (x) = 2 + x$ .

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Step 1.1.2.7.1

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

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

Step 1.1.2.7.2

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

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

Step 1.1.2.7.3

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

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

Step 1.1.2.8

Differentiate.

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Step 1.1.2.8.1

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

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

Step 1.1.2.8.2

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

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

Step 1.1.2.8.3

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

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

Step 1.1.2.8.4

Add $0$ and $\frac{d}{d x} [x]$ .

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

Step 1.1.2.8.5

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

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

Step 1.1.2.8.6

Combine fractions.

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Step 1.1.2.8.6.1

Multiply $2$ by $1$ .

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

Step 1.1.2.8.6.2

Combine $16$ and $\frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) + 2 {(2 - x)}^{2} (2 + x))}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$ .

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

Step 1.1.2.9

Simplify.

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Step 1.1.2.9.1

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

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

Step 1.1.2.9.2

Apply the distributive property.

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

Step 1.1.2.9.3

Apply the distributive property.

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

Step 1.1.2.9.4

Simplify the numerator.

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Step 1.1.2.9.4.1

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

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Step 1.1.2.9.4.1.1

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

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

Step 1.1.2.9.4.1.2

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

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

Step 1.1.2.9.4.1.3

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

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

Step 1.1.2.9.4.1.4

Factor $16 (2 + x) (2 - x)$ out of $16 (2 + x) (2 - x) ((2 + x) (2 - x)) + 16 (2 + x) (2 - x) (- x (- 2 (2 + x)))$ .

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

Step 1.1.2.9.4.1.5

Factor $16 (2 + x) (2 - x)$ out of $16 (2 + x) (2 - x) ((2 + x) (2 - x) - x (- 2 (2 + x))) + 16 (2 + x) (2 - x) (- x (2 (2 - x)))$ .

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

Step 1.1.2.9.4.2

Combine exponents.

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Step 1.1.2.9.4.2.1

Multiply $- 2$ by $- 1$ .

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

Step 1.1.2.9.4.2.2

Multiply $2$ by $- 1$ .

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

Step 1.1.2.9.4.3

Simplify each term.

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Step 1.1.2.9.4.3.1

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

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Step 1.1.2.9.4.3.1.1

Apply the distributive property.

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

Step 1.1.2.9.4.3.1.2

Apply the distributive property.

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

Step 1.1.2.9.4.3.1.3

Apply the distributive property.

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

Step 1.1.2.9.4.3.2

Simplify and combine like terms.

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Step 1.1.2.9.4.3.2.1

Simplify each term.

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Step 1.1.2.9.4.3.2.1.1

Multiply $2$ by $2$ .

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

Step 1.1.2.9.4.3.2.1.2

Multiply $- 1$ by $2$ .

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

Step 1.1.2.9.4.3.2.1.3

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

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

Step 1.1.2.9.4.3.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 1.1.2.9.4.3.2.1.5

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

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Step 1.1.2.9.4.3.2.1.5.1

Move $x$ .

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

Step 1.1.2.9.4.3.2.1.5.2

Multiply $x$ by $x$ .

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

Step 1.1.2.9.4.3.2.2

Add $- 2 x$ and $2 x$ .

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

Step 1.1.2.9.4.3.2.3

Add $4$ and $0$ .

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

Step 1.1.2.9.4.3.3

Apply the distributive property.

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

Step 1.1.2.9.4.3.4

Multiply $2$ by $2$ .

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

Step 1.1.2.9.4.3.5

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

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Step 1.1.2.9.4.3.5.1

Move $x$ .

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

Step 1.1.2.9.4.3.5.2

Multiply $x$ by $x$ .

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

Step 1.1.2.9.4.3.6

Apply the distributive property.

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

Step 1.1.2.9.4.3.7

Multiply $2$ by $- 2$ .

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

Step 1.1.2.9.4.3.8

Rewrite using the commutative property of multiplication.

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

Step 1.1.2.9.4.3.9

Simplify each term.

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Step 1.1.2.9.4.3.9.1

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

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Step 1.1.2.9.4.3.9.1.1

Move $x$ .

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

Step 1.1.2.9.4.3.9.1.2

Multiply $x$ by $x$ .

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

Step 1.1.2.9.4.3.9.2

Multiply $- 2$ by $- 1$ .

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

Step 1.1.2.9.4.4

Combine the opposite terms in $4 - x^{2} + 4 x + 2 x^{2} - 4 x + 2 x^{2}$ .

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Step 1.1.2.9.4.4.1

Subtract $4 x$ from $4 x$ .

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

Step 1.1.2.9.4.4.2

Add $4 - x^{2} + 2 x^{2}$ and $0$ .

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

Step 1.1.2.9.4.5

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

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

Step 1.1.2.9.4.6

Add $x^{2}$ and $2 x^{2}$ .

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

Step 1.1.2.9.5

Combine terms.

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Step 1.1.2.9.5.1

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

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Step 1.1.2.9.5.1.1

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

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

Step 1.1.2.9.5.1.2

Multiply $2$ by $2$ .

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

Step 1.1.2.9.5.2

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

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Step 1.1.2.9.5.2.1

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

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

Step 1.1.2.9.5.2.2

Multiply $2$ by $2$ .

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

Step 1.1.2.9.5.3

Cancel the common factor of $2 + x$ and ${(2 + x)}^{4}$ .

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Step 1.1.2.9.5.3.1

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

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

Step 1.1.2.9.5.3.2

Cancel the common factors.

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Step 1.1.2.9.5.3.2.1

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

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

Step 1.1.2.9.5.3.2.2

Cancel the common factor.

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

Step 1.1.2.9.5.3.2.3

Rewrite the expression.

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

Step 1.1.2.9.5.4

Cancel the common factor of $2 - x$ and ${(2 - x)}^{4}$ .

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Step 1.1.2.9.5.4.1

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

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

Step 1.1.2.9.5.4.2

Cancel the common factors.

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Step 1.1.2.9.5.4.2.1

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

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

Step 1.1.2.9.5.4.2.2

Cancel the common factor.

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

Step 1.1.2.9.5.4.2.3

Rewrite the expression.

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

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{16 (4 + 3 x^{2})}{{(2 + x)}^{3} {(2 - x)}^{3}}$ .

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

Step 1.2

Set the second derivative equal to $0$ then solve the equation $\frac{16 (4 + 3 x^{2})}{{(2 + x)}^{3} {(2 - x)}^{3}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

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

Step 1.2.2

Set the numerator equal to zero.

$16 (4 + 3 x^{2}) = 0$

Step 1.2.3

Solve the equation for $x$ .

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Step 1.2.3.1

Divide each term in $16 (4 + 3 x^{2}) = 0$ by $16$ and simplify.

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Step 1.2.3.1.1

Divide each term in $16 (4 + 3 x^{2}) = 0$ by $16$ .

$\frac{16 (4 + 3 x^{2})}{16} = \frac{0}{16}$

Step 1.2.3.1.2

Simplify the left side.

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Step 1.2.3.1.2.1

Cancel the common factor of $16$ .

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Step 1.2.3.1.2.1.1

Cancel the common factor.

$\frac{16 (4 + 3 x^{2})}{16} = \frac{0}{16}$

Step 1.2.3.1.2.1.2

Divide $4 + 3 x^{2}$ by $1$ .

$4 + 3 x^{2} = \frac{0}{16}$

Step 1.2.3.1.3

Simplify the right side.

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Step 1.2.3.1.3.1

Divide $0$ by $16$ .

$4 + 3 x^{2} = 0$

Step 1.2.3.2

Subtract $4$ from both sides of the equation.

$3 x^{2} = - 4$

Step 1.2.3.3

Divide each term in $3 x^{2} = - 4$ by $3$ and simplify.

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Step 1.2.3.3.1

Divide each term in $3 x^{2} = - 4$ by $3$ .

$\frac{3 x^{2}}{3} = \frac{- 4}{3}$

Step 1.2.3.3.2

Simplify the left side.

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Step 1.2.3.3.2.1

Cancel the common factor of $3$ .

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Step 1.2.3.3.2.1.1

Cancel the common factor.

$\frac{3 x^{2}}{3} = \frac{- 4}{3}$

Step 1.2.3.3.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 4}{3}$

Step 1.2.3.3.3

Simplify the right side.

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Step 1.2.3.3.3.1

Move the negative in front of the fraction.

$x^{2} = - \frac{4}{3}$

Step 1.2.3.4

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

$x = \pm \sqrt{- \frac{4}{3}}$

Step 1.2.3.5

Simplify $\pm \sqrt{- \frac{4}{3}}$ .

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Step 1.2.3.5.1

Rewrite $- \frac{4}{3}$ as $i^{2} \frac{2^{2}}{3}$ .

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Step 1.2.3.5.1.1

Rewrite $- 1$ as $i^{2}$ .

$x = \pm \sqrt{i^{2} \frac{4}{3}}$

Step 1.2.3.5.1.2

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

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

Step 1.2.3.5.2

Pull terms out from under the radical.

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

Step 1.2.3.5.3

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

$x = \pm i \sqrt{\frac{4}{3}}$

Step 1.2.3.5.4

Rewrite $\sqrt{\frac{4}{3}}$ as $\frac{\sqrt{4}}{\sqrt{3}}$ .

$x = \pm i \frac{\sqrt{4}}{\sqrt{3}}$

Step 1.2.3.5.5

Simplify the numerator.

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Step 1.2.3.5.5.1

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

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

Step 1.2.3.5.5.2

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

$x = \pm i \frac{2}{\sqrt{3}}$

Step 1.2.3.5.6

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

$x = \pm i (\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}})$

Step 1.2.3.5.7

Combine and simplify the denominator.

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Step 1.2.3.5.7.1

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

$x = \pm i \frac{2 \sqrt{3}}{\sqrt{3} \sqrt{3}}$

Step 1.2.3.5.7.2

Raise $\sqrt{3}$ to the power of $1$ .

$x = \pm i \frac{2 \sqrt{3}}{{\sqrt{3}}^{1} \sqrt{3}}$

Step 1.2.3.5.7.3

Raise $\sqrt{3}$ to the power of $1$ .

$x = \pm i \frac{2 \sqrt{3}}{{\sqrt{3}}^{1} {\sqrt{3}}^{1}}$

Step 1.2.3.5.7.4

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

$x = \pm i \frac{2 \sqrt{3}}{{\sqrt{3}}^{1 + 1}}$

Step 1.2.3.5.7.5

Add $1$ and $1$ .

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

Step 1.2.3.5.7.6

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 1.2.3.5.7.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$x = \pm i \frac{2 \sqrt{3}}{{(3^{\frac{1}{2}})}^{2}}$

Step 1.2.3.5.7.6.2

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

$x = \pm i \frac{2 \sqrt{3}}{3^{\frac{1}{2} \cdot 2}}$

Step 1.2.3.5.7.6.3

Combine $\frac{1}{2}$ and $2$ .

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

Step 1.2.3.5.7.6.4

Cancel the common factor of $2$ .

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Step 1.2.3.5.7.6.4.1

Cancel the common factor.

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

Step 1.2.3.5.7.6.4.2

Rewrite the expression.

$x = \pm i \frac{2 \sqrt{3}}{3^{1}}$

Step 1.2.3.5.7.6.5

Evaluate the exponent.

$x = \pm i \frac{2 \sqrt{3}}{3}$

Step 1.2.3.5.8

Combine $i$ and $\frac{2 \sqrt{3}}{3}$ .

$x = \pm \frac{i (2 \sqrt{3})}{3}$

Step 1.2.3.5.9

Move $2$ to the left of $i$ .

$x = \pm \frac{2 i \sqrt{3}}{3}$

Step 1.2.3.6

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

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Step 1.2.3.6.1

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

$x = \frac{2 i \sqrt{3}}{3}$

Step 1.2.3.6.2

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

$x = - \frac{2 i \sqrt{3}}{3}$

Step 1.2.3.6.3

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

$x = \frac{2 i \sqrt{3}}{3}, - \frac{2 i \sqrt{3}}{3}$

Step 2

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

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Step 2.1

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

$4 - x^{2} = 0$

Step 2.2

Solve for $x$ .

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Step 2.2.1

Subtract $4$ from both sides of the equation.

$- x^{2} = - 4$

Step 2.2.2

Divide each term in $- x^{2} = - 4$ by $- 1$ and simplify.

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Step 2.2.2.1

Divide each term in $- x^{2} = - 4$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 4}{- 1}$

Step 2.2.2.2

Simplify the left side.

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Step 2.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{- 4}{- 1}$

Step 2.2.2.2.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 4}{- 1}$

Step 2.2.2.3

Simplify the right side.

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Step 2.2.2.3.1

Divide $- 4$ by $- 1$ .

$x^{2} = 4$

Step 2.2.3

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

$x = \pm \sqrt{4}$

Step 2.2.4

Simplify $\pm \sqrt{4}$ .

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Step 2.2.4.1

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

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

Step 2.2.4.2

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

$x = \pm 2$

Step 2.2.5

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

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Step 2.2.5.1

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

$x = 2$

Step 2.2.5.2

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

$x = - 2$

Step 2.2.5.3

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

$x = 2, - 2$

Step 2.3

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

Interval Notation:

$(- \infty, - 2) \cup (- 2, 2) \cup (2, \infty)$

Set-Builder Notation:

${x | x \neq 2, - 2}$

Interval Notation:

$(- \infty, - 2) \cup (- 2, 2) \cup (2, \infty)$

Set-Builder Notation:

${x | x \neq 2, - 2}$

Step 3

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

$(- \infty, - 2) \cup (- 2, 2) \cup (2, \infty)$

Step 4

Substitute any number from the interval $(- \infty, - 2)$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

Replace the variable $x$ with $- 4$ in the expression.

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

Step 4.2

Simplify the result.

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Step 4.2.1

Remove parentheses.

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

Step 4.2.2

Simplify the numerator.

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Step 4.2.2.1

Raise $- 4$ to the power of $2$ .

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

Step 4.2.2.2

Multiply $3$ by $16$ .

$f'' (- 4) = \frac{16 (4 + 48)}{{(2 - 4)}^{3} {(2 - (- 4))}^{3}}$

Step 4.2.2.3

Add $4$ and $48$ .

$f'' (- 4) = \frac{16 \cdot 52}{{(2 - 4)}^{3} {(2 - (- 4))}^{3}}$

Step 4.2.3

Simplify the denominator.

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Step 4.2.3.1

Subtract $4$ from $2$ .

$f'' (- 4) = \frac{16 \cdot 52}{{(- 2)}^{3} {(2 - (- 4))}^{3}}$

Step 4.2.3.2

Multiply $- 1$ by $- 4$ .

$f'' (- 4) = \frac{16 \cdot 52}{{(- 2)}^{3} {(2 + 4)}^{3}}$

Step 4.2.3.3

Add $2$ and $4$ .

$f'' (- 4) = \frac{16 \cdot 52}{{(- 2)}^{3} \cdot 6^{3}}$

Step 4.2.3.4

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

$f'' (- 4) = \frac{16 \cdot 52}{- 8 \cdot 6^{3}}$

Step 4.2.3.5

Raise $6$ to the power of $3$ .

$f'' (- 4) = \frac{16 \cdot 52}{- 8 \cdot 216}$

Step 4.2.4

Reduce the expression by cancelling the common factors.

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Step 4.2.4.1

Multiply $16$ by $52$ .

$f'' (- 4) = \frac{832}{- 8 \cdot 216}$

Step 4.2.4.2

Multiply $- 8$ by $216$ .

$f'' (- 4) = \frac{832}{- 1728}$

Step 4.2.4.3

Cancel the common factor of $832$ and $- 1728$ .

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Step 4.2.4.3.1

Factor $64$ out of $832$ .

$f'' (- 4) = \frac{64 (13)}{- 1728}$

Step 4.2.4.3.2

Cancel the common factors.

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Step 4.2.4.3.2.1

Factor $64$ out of $- 1728$ .

$f'' (- 4) = \frac{64 \cdot 13}{64 \cdot - 27}$

Step 4.2.4.3.2.2

Cancel the common factor.

$f'' (- 4) = \frac{64 \cdot 13}{64 \cdot - 27}$

Step 4.2.4.3.2.3

Rewrite the expression.

$f'' (- 4) = \frac{13}{- 27}$

Step 4.2.4.4

Move the negative in front of the fraction.

$f'' (- 4) = - \frac{13}{27}$

Step 4.2.5

The final answer is $- \frac{13}{27}$ .

$- \frac{13}{27}$

Step 4.3

The graph is concave down on the interval $(- \infty, - 2)$ because $f'' (- 4)$ is negative.

Concave down on $(- \infty, - 2)$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(- 2, 2)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

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

$f'' (0) = \frac{16 (4 + 3 {(0)}^{2})}{{(2 + 0)}^{3} {(2 - (0))}^{3}}$

Step 5.2

Simplify the result.

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Step 5.2.1

Remove parentheses.

$f'' (0) = \frac{16 (4 + 3 {(0)}^{2})}{{(2 + 0)}^{3} {(2 - (0))}^{3}}$

Step 5.2.2

Simplify the numerator.

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Step 5.2.2.1

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

$f'' (0) = \frac{16 (4 + 3 \cdot 0)}{{(2 + 0)}^{3} {(2 - (0))}^{3}}$

Step 5.2.2.2

Multiply $3$ by $0$ .

$f'' (0) = \frac{16 (4 + 0)}{{(2 + 0)}^{3} {(2 - (0))}^{3}}$

Step 5.2.2.3

Add $4$ and $0$ .

$f'' (0) = \frac{16 \cdot 4}{{(2 + 0)}^{3} {(2 - (0))}^{3}}$

Step 5.2.3

Simplify the denominator.

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Step 5.2.3.1

Add $2$ and $0$ .

$f'' (0) = \frac{16 \cdot 4}{2^{3} {(2 - (0))}^{3}}$

Step 5.2.3.2

Multiply $- 1$ by $0$ .

$f'' (0) = \frac{16 \cdot 4}{2^{3} {(2 + 0)}^{3}}$

Step 5.2.3.3

Add $2$ and $0$ .

$f'' (0) = \frac{16 \cdot 4}{2^{3} \cdot 2^{3}}$

Step 5.2.3.4

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

$f'' (0) = \frac{16 \cdot 4}{8 \cdot 2^{3}}$

Step 5.2.3.5

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

$f'' (0) = \frac{16 \cdot 4}{8 \cdot 8}$

Step 5.2.4

Simplify the expression.

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Step 5.2.4.1

Multiply $16$ by $4$ .

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

Step 5.2.4.2

Multiply $8$ by $8$ .

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

Step 5.2.4.3

Divide $64$ by $64$ .

$f'' (0) = 1$

Step 5.2.5

The final answer is $1$ .

$1$

Step 5.3

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

Concave up on $(- 2, 2)$ since $f'' (x)$ is positive

Step 6

Substitute any number from the interval $(2, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

Replace the variable $x$ with $4$ in the expression.

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

Step 6.2

Simplify the result.

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Step 6.2.1

Remove parentheses.

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

Step 6.2.2

Simplify the numerator.

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Step 6.2.2.1

Raise $4$ to the power of $2$ .

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

Step 6.2.2.2

Multiply $3$ by $16$ .

$f'' (4) = \frac{16 (4 + 48)}{{(2 + 4)}^{3} {(2 - (4))}^{3}}$

Step 6.2.2.3

Add $4$ and $48$ .

$f'' (4) = \frac{16 \cdot 52}{{(2 + 4)}^{3} {(2 - (4))}^{3}}$

Step 6.2.3

Simplify the denominator.

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Step 6.2.3.1

Add $2$ and $4$ .

$f'' (4) = \frac{16 \cdot 52}{6^{3} {(2 - (4))}^{3}}$

Step 6.2.3.2

Multiply $- 1$ by $4$ .

$f'' (4) = \frac{16 \cdot 52}{6^{3} {(2 - 4)}^{3}}$

Step 6.2.3.3

Subtract $4$ from $2$ .

$f'' (4) = \frac{16 \cdot 52}{6^{3} {(- 2)}^{3}}$

Step 6.2.3.4

Raise $6$ to the power of $3$ .

$f'' (4) = \frac{16 \cdot 52}{216 {(- 2)}^{3}}$

Step 6.2.3.5

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

$f'' (4) = \frac{16 \cdot 52}{216 \cdot - 8}$

Step 6.2.4

Reduce the expression by cancelling the common factors.

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Step 6.2.4.1

Multiply $16$ by $52$ .

$f'' (4) = \frac{832}{216 \cdot - 8}$

Step 6.2.4.2

Multiply $216$ by $- 8$ .

$f'' (4) = \frac{832}{- 1728}$

Step 6.2.4.3

Cancel the common factor of $832$ and $- 1728$ .

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Step 6.2.4.3.1

Factor $64$ out of $832$ .

$f'' (4) = \frac{64 (13)}{- 1728}$

Step 6.2.4.3.2

Cancel the common factors.

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Step 6.2.4.3.2.1

Factor $64$ out of $- 1728$ .

$f'' (4) = \frac{64 \cdot 13}{64 \cdot - 27}$

Step 6.2.4.3.2.2

Cancel the common factor.

$f'' (4) = \frac{64 \cdot 13}{64 \cdot - 27}$

Step 6.2.4.3.2.3

Rewrite the expression.

$f'' (4) = \frac{13}{- 27}$

Step 6.2.4.4

Move the negative in front of the fraction.

$f'' (4) = - \frac{13}{27}$

Step 6.2.5

The final answer is $- \frac{13}{27}$ .

$- \frac{13}{27}$

Step 6.3

The graph is concave down on the interval $(2, \infty)$ because $f'' (4)$ is negative.

Concave down on $(2, \infty)$ since $f'' (x)$ is negative

Step 7

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

Concave down on $(- \infty, - 2)$ since $f'' (x)$ is negative

Concave up on $(- 2, 2)$ since $f'' (x)$ is positive

Concave down on $(2, \infty)$ since $f'' (x)$ is negative

Step 8