Calculus Examples

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

Step 1

Find the second derivative.

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Step 1.1

Find the first derivative.

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Step 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 $g (x) = 4 x^{2} - 1$ .

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

Step 1.1.2

Differentiate.

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Step 1.1.2.1

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

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

Step 1.1.2.2

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

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

Step 1.1.2.3

By the Sum Rule, the derivative of $4 x^{2} - 1$ with respect to $x$ is $\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 1]$ .

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

Step 1.1.2.4

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

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

Step 1.1.2.5

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

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

Step 1.1.2.6

Multiply $2$ by $4$ .

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

Step 1.1.2.7

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

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

Step 1.1.2.8

Simplify the expression.

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Step 1.1.2.8.1

Add $8 x$ and $0$ .

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

Step 1.1.2.8.2

Multiply $8$ by $- 1$ .

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

Step 1.1.3

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

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

Step 1.1.4

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

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

Step 1.1.5

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

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

Step 1.1.6

Add $1$ and $1$ .

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

Step 1.1.7

Subtract $8 x^{2}$ from $4 x^{2}$ .

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

Step 1.2

Find the second derivative.

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Step 1.2.1

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

Step 1.2.2

Differentiate.

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Step 1.2.2.1

Multiply the exponents in ${({(4 x^{2} - 1)}^{2})}^{2}$ .

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Step 1.2.2.1.1

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

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

Step 1.2.2.1.2

Multiply $2$ by $2$ .

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

Step 1.2.2.2

By the Sum Rule, the derivative of $- 4 x^{2} - 1$ with respect to $x$ is $\frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [- 1]$ .

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

Step 1.2.2.3

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

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

Step 1.2.2.4

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

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

Step 1.2.2.5

Multiply $2$ by $- 4$ .

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

Step 1.2.2.6

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

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

Step 1.2.2.7

Add $- 8 x$ and $0$ .

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

Step 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 $g (x) = 4 x^{2} - 1$ .

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Step 1.2.3.1

To apply the Chain Rule, set $u$ as $4 x^{2} - 1$ .

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

Step 1.2.3.2

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

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

Step 1.2.3.3

Replace all occurrences of $u$ with $4 x^{2} - 1$ .

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

Step 1.2.4

Differentiate.

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Step 1.2.4.1

Multiply $2$ by $- 1$ .

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

Step 1.2.4.2

By the Sum Rule, the derivative of $4 x^{2} - 1$ with respect to $x$ is $\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 1]$ .

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

Step 1.2.4.3

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

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

Step 1.2.4.4

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

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

Step 1.2.4.5

Multiply $2$ by $4$ .

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

Step 1.2.4.6

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

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

Step 1.2.4.7

Simplify the expression.

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Step 1.2.4.7.1

Add $8 x$ and $0$ .

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

Step 1.2.4.7.2

Move $8$ to the left of $4 x^{2} - 1$ .

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

Step 1.2.4.7.3

Multiply $8$ by $- 2$ .

$f'' (x) = \frac{{(4 x^{2} - 1)}^{2} (- 8 x) - 16 (- 4 x^{2} - 1) ((4 x^{2} - 1) x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5

Simplify.

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Step 1.2.5.1

Apply the distributive property.

$f'' (x) = \frac{{(4 x^{2} - 1)}^{2} (- 8 x) + (- 16 (- 4 x^{2}) - 16 \cdot - 1) ((4 x^{2} - 1) x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.2

Apply the distributive property.

$f'' (x) = \frac{{(4 x^{2} - 1)}^{2} (- 8 x) + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3

Simplify the numerator.

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Step 1.2.5.3.1

Simplify each term.

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Step 1.2.5.3.1.1

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{- 8 {(4 x^{2} - 1)}^{2} x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.2

Rewrite ${(4 x^{2} - 1)}^{2}$ as $(4 x^{2} - 1) (4 x^{2} - 1)$ .

$f'' (x) = \frac{- 8 ((4 x^{2} - 1) (4 x^{2} - 1)) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.3

Expand $(4 x^{2} - 1) (4 x^{2} - 1)$ using the FOIL Method.

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Step 1.2.5.3.1.3.1

Apply the distributive property.

$f'' (x) = \frac{- 8 (4 x^{2} (4 x^{2} - 1) - 1 (4 x^{2} - 1)) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.3.2

Apply the distributive property.

$f'' (x) = \frac{- 8 (4 x^{2} (4 x^{2}) + 4 x^{2} \cdot - 1 - 1 (4 x^{2} - 1)) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.3.3

Apply the distributive property.

$f'' (x) = \frac{- 8 (4 x^{2} (4 x^{2}) + 4 x^{2} \cdot - 1 - 1 (4 x^{2}) - 1 \cdot - 1) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.4

Simplify and combine like terms.

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Step 1.2.5.3.1.4.1

Simplify each term.

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Step 1.2.5.3.1.4.1.1

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{- 8 (4 \cdot (4 x^{2} x^{2}) + 4 x^{2} \cdot - 1 - 1 (4 x^{2}) - 1 \cdot - 1) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.4.1.2

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

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Step 1.2.5.3.1.4.1.2.1

Move $x^{2}$ .

$f'' (x) = \frac{- 8 (4 \cdot (4 (x^{2} x^{2})) + 4 x^{2} \cdot - 1 - 1 (4 x^{2}) - 1 \cdot - 1) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.4.1.2.2

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

$f'' (x) = \frac{- 8 (4 \cdot (4 x^{2 + 2}) + 4 x^{2} \cdot - 1 - 1 (4 x^{2}) - 1 \cdot - 1) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.4.1.2.3

Add $2$ and $2$ .

$f'' (x) = \frac{- 8 (4 \cdot (4 x^{4}) + 4 x^{2} \cdot - 1 - 1 (4 x^{2}) - 1 \cdot - 1) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.4.1.3

Multiply $4$ by $4$ .

$f'' (x) = \frac{- 8 (16 x^{4} + 4 x^{2} \cdot - 1 - 1 (4 x^{2}) - 1 \cdot - 1) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.4.1.4

Multiply $- 1$ by $4$ .

$f'' (x) = \frac{- 8 (16 x^{4} - 4 x^{2} - 1 (4 x^{2}) - 1 \cdot - 1) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.4.1.5

Multiply $4$ by $- 1$ .

$f'' (x) = \frac{- 8 (16 x^{4} - 4 x^{2} - 4 x^{2} - 1 \cdot - 1) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.4.1.6

Multiply $- 1$ by $- 1$ .

$f'' (x) = \frac{- 8 (16 x^{4} - 4 x^{2} - 4 x^{2} + 1) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.4.2

Subtract $4 x^{2}$ from $- 4 x^{2}$ .

$f'' (x) = \frac{- 8 (16 x^{4} - 8 x^{2} + 1) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.5

Apply the distributive property.

$f'' (x) = \frac{(- 8 (16 x^{4}) - 8 (- 8 x^{2}) - 8 \cdot 1) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.6

Simplify.

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Step 1.2.5.3.1.6.1

Multiply $16$ by $- 8$ .

$f'' (x) = \frac{(- 128 x^{4} - 8 (- 8 x^{2}) - 8 \cdot 1) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.6.2

Multiply $- 8$ by $- 8$ .

$f'' (x) = \frac{(- 128 x^{4} + 64 x^{2} - 8 \cdot 1) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.6.3

Multiply $- 8$ by $1$ .

$f'' (x) = \frac{(- 128 x^{4} + 64 x^{2} - 8) x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.7

Apply the distributive property.

$f'' (x) = \frac{- 128 x^{4} x + 64 x^{2} x - 8 x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.8

Simplify.

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Step 1.2.5.3.1.8.1

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

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Step 1.2.5.3.1.8.1.1

Move $x$ .

$f'' (x) = \frac{- 128 (x \cdot x^{4}) + 64 x^{2} x - 8 x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.8.1.2

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

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Step 1.2.5.3.1.8.1.2.1

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

$f'' (x) = \frac{- 128 (x \cdot x^{4}) + 64 x^{2} x - 8 x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.8.1.2.2

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

$f'' (x) = \frac{- 128 x^{1 + 4} + 64 x^{2} x - 8 x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.8.1.3

Add $1$ and $4$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{2} x - 8 x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.8.2

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

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Step 1.2.5.3.1.8.2.1

Move $x$ .

$f'' (x) = \frac{- 128 x^{5} + 64 (x \cdot x^{2}) - 8 x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.8.2.2

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

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Step 1.2.5.3.1.8.2.2.1

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

$f'' (x) = \frac{- 128 x^{5} + 64 (x \cdot x^{2}) - 8 x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.8.2.2.2

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

$f'' (x) = \frac{- 128 x^{5} + 64 x^{1 + 2} - 8 x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.8.2.3

Add $1$ and $2$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + (- 16 (- 4 x^{2}) - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.9

Simplify each term.

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Step 1.2.5.3.1.9.1

Multiply $- 4$ by $- 16$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + (64 x^{2} - 16 \cdot - 1) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.9.2

Multiply $- 16$ by $- 1$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + (64 x^{2} + 16) (4 x^{2} x - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.10

Simplify each term.

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Step 1.2.5.3.1.10.1

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

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Step 1.2.5.3.1.10.1.1

Move $x$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + (64 x^{2} + 16) (4 (x \cdot x^{2}) - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.10.1.2

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

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Step 1.2.5.3.1.10.1.2.1

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

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + (64 x^{2} + 16) (4 (x \cdot x^{2}) - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.10.1.2.2

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

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + (64 x^{2} + 16) (4 x^{1 + 2} - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.10.1.3

Add $1$ and $2$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + (64 x^{2} + 16) (4 x^{3} - 1 x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.10.2

Rewrite $- 1 x$ as $- x$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + (64 x^{2} + 16) (4 x^{3} - x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.11

Expand $(64 x^{2} + 16) (4 x^{3} - x)$ using the FOIL Method.

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Step 1.2.5.3.1.11.1

Apply the distributive property.

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 64 x^{2} (4 x^{3} - x) + 16 (4 x^{3} - x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.11.2

Apply the distributive property.

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 64 x^{2} (4 x^{3}) + 64 x^{2} (- x) + 16 (4 x^{3} - x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.11.3

Apply the distributive property.

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 64 x^{2} (4 x^{3}) + 64 x^{2} (- x) + 16 (4 x^{3}) + 16 (- x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12

Simplify and combine like terms.

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Step 1.2.5.3.1.12.1

Simplify each term.

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Step 1.2.5.3.1.12.1.1

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 64 \cdot (4 x^{2} x^{3}) + 64 x^{2} (- x) + 16 (4 x^{3}) + 16 (- x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12.1.2

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

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Step 1.2.5.3.1.12.1.2.1

Move $x^{3}$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 64 \cdot (4 (x^{3} x^{2})) + 64 x^{2} (- x) + 16 (4 x^{3}) + 16 (- x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12.1.2.2

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

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 64 \cdot (4 x^{3 + 2}) + 64 x^{2} (- x) + 16 (4 x^{3}) + 16 (- x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12.1.2.3

Add $3$ and $2$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 64 \cdot (4 x^{5}) + 64 x^{2} (- x) + 16 (4 x^{3}) + 16 (- x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12.1.3

Multiply $64$ by $4$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 256 x^{5} + 64 x^{2} (- x) + 16 (4 x^{3}) + 16 (- x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12.1.4

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 256 x^{5} + 64 \cdot (- 1 x^{2} x) + 16 (4 x^{3}) + 16 (- x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12.1.5

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

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Step 1.2.5.3.1.12.1.5.1

Move $x$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 256 x^{5} + 64 \cdot (- 1 (x \cdot x^{2})) + 16 (4 x^{3}) + 16 (- x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12.1.5.2

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

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Step 1.2.5.3.1.12.1.5.2.1

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

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 256 x^{5} + 64 \cdot (- 1 (x \cdot x^{2})) + 16 (4 x^{3}) + 16 (- x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12.1.5.2.2

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

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 256 x^{5} + 64 \cdot (- 1 x^{1 + 2}) + 16 (4 x^{3}) + 16 (- x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12.1.5.3

Add $1$ and $2$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 256 x^{5} + 64 \cdot (- 1 x^{3}) + 16 (4 x^{3}) + 16 (- x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12.1.6

Multiply $64$ by $- 1$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 256 x^{5} - 64 x^{3} + 16 (4 x^{3}) + 16 (- x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12.1.7

Multiply $4$ by $16$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 256 x^{5} - 64 x^{3} + 64 x^{3} + 16 (- x)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12.1.8

Multiply $- 1$ by $16$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 256 x^{5} - 64 x^{3} + 64 x^{3} - 16 x}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12.2

Add $- 64 x^{3}$ and $64 x^{3}$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 256 x^{5} + 0 - 16 x}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.1.12.3

Add $256 x^{5}$ and $0$ .

$f'' (x) = \frac{- 128 x^{5} + 64 x^{3} - 8 x + 256 x^{5} - 16 x}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.2

Add $- 128 x^{5}$ and $256 x^{5}$ .

$f'' (x) = \frac{128 x^{5} + 64 x^{3} - 8 x - 16 x}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.3.3

Subtract $16 x$ from $- 8 x$ .

$f'' (x) = \frac{128 x^{5} + 64 x^{3} - 24 x}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.4

Simplify the numerator.

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Step 1.2.5.4.1

Factor $8 x$ out of $128 x^{5} + 64 x^{3} - 24 x$ .

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Step 1.2.5.4.1.1

Factor $8 x$ out of $128 x^{5}$ .

$f'' (x) = \frac{8 x (16 x^{4}) + 64 x^{3} - 24 x}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.4.1.2

Factor $8 x$ out of $64 x^{3}$ .

$f'' (x) = \frac{8 x (16 x^{4}) + 8 x (8 x^{2}) - 24 x}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.4.1.3

Factor $8 x$ out of $- 24 x$ .

$f'' (x) = \frac{8 x (16 x^{4}) + 8 x (8 x^{2}) + 8 x (- 3)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.4.1.4

Factor $8 x$ out of $8 x (16 x^{4}) + 8 x (8 x^{2})$ .

$f'' (x) = \frac{8 x (16 x^{4} + 8 x^{2}) + 8 x (- 3)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.4.1.5

Factor $8 x$ out of $8 x (16 x^{4} + 8 x^{2}) + 8 x (- 3)$ .

$f'' (x) = \frac{8 x (16 x^{4} + 8 x^{2} - 3)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.4.2

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

$f'' (x) = \frac{8 x (16 {(x^{2})}^{2} + 8 x^{2} - 3)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.4.3

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

$f'' (x) = \frac{8 x (16 u^{2} + 8 u - 3)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.4.4

Factor by grouping.

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Step 1.2.5.4.4.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 16 \cdot - 3 = - 48$ and whose sum is $b = 8$ .

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Step 1.2.5.4.4.1.1

Factor $8$ out of $8 u$ .

$f'' (x) = \frac{8 x (16 u^{2} + 8 (u) - 3)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.4.4.1.2

Rewrite $8$ as $- 4$ plus $12$

$f'' (x) = \frac{8 x (16 u^{2} + (- 4 + 12) u - 3)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.4.4.1.3

Apply the distributive property.

$f'' (x) = \frac{8 x (16 u^{2} - 4 u + 12 u - 3)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.4.4.2

Factor out the greatest common factor from each group.

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Step 1.2.5.4.4.2.1

Group the first two terms and the last two terms.

$f'' (x) = \frac{8 x ((16 u^{2} - 4 u) + 12 u - 3)}{{(4 x^{2} - 1)}^{4}}$

Step 1.2.5.4.4.2.2

Factor out the greatest common factor (GCF) from each group.

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

Step 1.2.5.4.4.3

Factor the polynomial by factoring out the greatest common factor, $4 u - 1$ .

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

Step 1.2.5.4.5

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

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

Step 1.2.5.4.6

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

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

Step 1.2.5.4.7

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

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

Step 1.2.5.4.8

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 1$ .

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

Step 1.2.5.5

Simplify the denominator.

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Step 1.2.5.5.1

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

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

Step 1.2.5.5.2

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

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

Step 1.2.5.5.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 1$ .

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

Step 1.2.5.5.4

Apply the product rule to $(2 x + 1) (2 x - 1)$ .

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

Step 1.2.5.6

Cancel the common factor of $2 x + 1$ and ${(2 x + 1)}^{4}$ .

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Step 1.2.5.6.1

Factor $2 x + 1$ out of $8 x (2 x + 1) (2 x - 1) (4 x^{2} + 3)$ .

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

Step 1.2.5.6.2

Cancel the common factors.

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Step 1.2.5.6.2.1

Factor $2 x + 1$ out of ${(2 x + 1)}^{4} {(2 x - 1)}^{4}$ .

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

Step 1.2.5.6.2.2

Cancel the common factor.

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

Step 1.2.5.6.2.3

Rewrite the expression.

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

Step 1.2.5.7

Cancel the common factor of $2 x - 1$ and ${(2 x - 1)}^{4}$ .

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Step 1.2.5.7.1

Factor $2 x - 1$ out of $(8 x (2 x - 1)) (4 x^{2} + 3)$ .

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

Step 1.2.5.7.2

Cancel the common factors.

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Step 1.2.5.7.2.1

Factor $2 x - 1$ out of ${(2 x + 1)}^{3} {(2 x - 1)}^{4}$ .

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

Step 1.2.5.7.2.2

Cancel the common factor.

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

Step 1.2.5.7.2.3

Rewrite the expression.

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

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

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{8 x (4 x^{2} + 3)}{{(2 x + 1)}^{3} {(2 x - 1)}^{3}}$ .

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

Step 2

Set the second derivative equal to $0$ then solve the equation $\frac{8 x (4 x^{2} + 3)}{{(2 x + 1)}^{3} {(2 x - 1)}^{3}} = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

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

Step 2.2

Set the numerator equal to zero.

$8 x (4 x^{2} + 3) = 0$

Step 2.3

Solve the equation for $x$ .

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Step 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 $0$ .

$x = 0$

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

Step 2.3.2

Set $x$ equal to $0$ .

$x = 0$

Step 2.3.3

Set $4 x^{2} + 3$ equal to $0$ and solve for $x$ .

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Step 2.3.3.1

Set $4 x^{2} + 3$ equal to $0$ .

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

Step 2.3.3.2

Solve $4 x^{2} + 3 = 0$ for $x$ .

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Step 2.3.3.2.1

Subtract $3$ from both sides of the equation.

$4 x^{2} = - 3$

Step 2.3.3.2.2

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

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Step 2.3.3.2.2.1

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

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

Step 2.3.3.2.2.2

Simplify the left side.

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Step 2.3.3.2.2.2.1

Cancel the common factor of $4$ .

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Step 2.3.3.2.2.2.1.1

Cancel the common factor.

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

Step 2.3.3.2.2.2.1.2

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

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

Step 2.3.3.2.2.3

Simplify the right side.

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Step 2.3.3.2.2.3.1

Move the negative in front of the fraction.

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

Step 2.3.3.2.3

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

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

Step 2.3.3.2.4

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

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Step 2.3.3.2.4.1

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

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Step 2.3.3.2.4.1.1

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

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

Step 2.3.3.2.4.1.2

Factor the perfect power $1^{2}$ out of $3$ .

$x = \pm \sqrt{i^{2} \frac{1^{2} \cdot 3}{4}}$

Step 2.3.3.2.4.1.3

Factor the perfect power $2^{2}$ out of $4$ .

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

Step 2.3.3.2.4.1.4

Rearrange the fraction $\frac{1^{2} \cdot 3}{2^{2} \cdot 1}$ .

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

Step 2.3.3.2.4.1.5

Rewrite $i^{2} {(\frac{1}{2})}^{2}$ as ${(\frac{i}{2})}^{2}$ .

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

Step 2.3.3.2.4.2

Pull terms out from under the radical.

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

Step 2.3.3.2.4.3

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

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

Step 2.3.3.2.5

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

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Step 2.3.3.2.5.1

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

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

Step 2.3.3.2.5.2

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

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

Step 2.3.3.2.5.3

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

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

Step 2.3.4

The final solution is all the values that make $8 x (4 x^{2} + 3) = 0$ true.

$x = 0, \frac{i \sqrt{3}}{2}, - \frac{i \sqrt{3}}{2}$

Step 3

Find the points where the second derivative is $0$ .

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Step 3.1

Substitute $0$ in $f (x) = \frac{x}{4 x^{2} - 1}$ to find the value of $y$ .

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Step 3.1.1

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

$f (0) = \frac{0}{4 {(0)}^{2} - 1}$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Simplify the denominator.

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Step 3.1.2.1.1

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

$f (0) = \frac{0}{4 \cdot 0 - 1}$

Step 3.1.2.1.2

Multiply $4$ by $0$ .

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

Step 3.1.2.1.3

Subtract $1$ from $0$ .

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

Step 3.1.2.2

Divide $0$ by $- 1$ .

$f (0) = 0$

Step 3.1.2.3

The final answer is $0$ .

$0$

Step 3.2

The point found by substituting $0$ in $f (x) = \frac{x}{4 x^{2} - 1}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Step 4

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

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

Step 5

Substitute a value from the interval $(- \infty, 0)$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 0.1$ in the expression.

$f'' (- 0.1) = \frac{8 (- 0.1) (4 {(- 0.1)}^{2} + 3)}{{(2 (- 0.1) + 1)}^{3} {(2 (- 0.1) - 1)}^{3}}$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify the numerator.

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Step 5.2.1.1

Multiply $8$ by $- 0.1$ .

$f'' (- 0.1) = \frac{- 0.8 (4 {(- 0.1)}^{2} + 3)}{{(2 (- 0.1) + 1)}^{3} {(2 (- 0.1) - 1)}^{3}}$

Step 5.2.1.2

Multiply $- 0.8$ by $4 {(- 0.1)}^{2} + 3$ .

$f'' (- 0.1) = \frac{- 2.432}{{(2 (- 0.1) + 1)}^{3} {(2 (- 0.1) - 1)}^{3}}$

Step 5.2.2

Simplify the denominator.

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Step 5.2.2.1

Multiply $2$ by $- 0.1$ .

$f'' (- 0.1) = \frac{- 2.432}{{(- 0.2 + 1)}^{3} {(2 (- 0.1) - 1)}^{3}}$

Step 5.2.2.2

Add $- 0.2$ and $1$ .

$f'' (- 0.1) = \frac{- 2.432}{{0.8}^{3} {(2 (- 0.1) - 1)}^{3}}$

Step 5.2.2.3

Multiply $2$ by $- 0.1$ .

$f'' (- 0.1) = \frac{- 2.432}{{0.8}^{3} {(- 0.2 - 1)}^{3}}$

Step 5.2.2.4

Subtract $1$ from $- 0.2$ .

$f'' (- 0.1) = \frac{- 2.432}{{0.8}^{3} {(- 1.2)}^{3}}$

Step 5.2.2.5

Raise $0.8$ to the power of $3$ .

$f'' (- 0.1) = \frac{- 2.432}{0.512 {(- 1.2)}^{3}}$

Step 5.2.2.6

Raise $- 1.2$ to the power of $3$ .

$f'' (- 0.1) = \frac{- 2.432}{0.512 \cdot - 1.728}$

Step 5.2.3

Simplify the expression.

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Step 5.2.3.1

Multiply $0.512$ by $- 1.728$ .

$f'' (- 0.1) = \frac{- 2.432}{- 0.884736}$

Step 5.2.3.2

Divide $- 2.432$ by $- 0.884736$ .

$f'' (- 0.1) = 2.74884259$

Step 5.2.4

The final answer is $2.74884259$ .

$2.74884259$

Step 5.3

At $- 0.1$ , the second derivative is $2.74884259$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, 0)$ .

Increasing on $(- \infty, 0)$ since $f'' (x) > 0$

Step 6

Substitute a value from the interval $(0, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $0.1$ in the expression.

$f'' (0.1) = \frac{8 (0.1) (4 {(0.1)}^{2} + 3)}{{(2 (0.1) + 1)}^{3} {(2 (0.1) - 1)}^{3}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify the numerator.

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Step 6.2.1.1

Multiply $8$ by $0.1$ .

$f'' (0.1) = \frac{0.8 (4 \cdot {0.1}^{2} + 3)}{{(2 (0.1) + 1)}^{3} {(2 (0.1) - 1)}^{3}}$

Step 6.2.1.2

Multiply $0.8$ by $4 \cdot {0.1}^{2} + 3$ .

$f'' (0.1) = \frac{2.432}{{(2 (0.1) + 1)}^{3} {(2 (0.1) - 1)}^{3}}$

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

Multiply $2$ by $0.1$ .

$f'' (0.1) = \frac{2.432}{{(0.2 + 1)}^{3} {(2 (0.1) - 1)}^{3}}$

Step 6.2.2.2

Add $0.2$ and $1$ .

$f'' (0.1) = \frac{2.432}{{1.2}^{3} {(2 (0.1) - 1)}^{3}}$

Step 6.2.2.3

Multiply $2$ by $0.1$ .

$f'' (0.1) = \frac{2.432}{{1.2}^{3} {(0.2 - 1)}^{3}}$

Step 6.2.2.4

Subtract $1$ from $0.2$ .

$f'' (0.1) = \frac{2.432}{{1.2}^{3} {(- 0.8)}^{3}}$

Step 6.2.2.5

Raise $1.2$ to the power of $3$ .

$f'' (0.1) = \frac{2.432}{1.728 {(- 0.8)}^{3}}$

Step 6.2.2.6

Raise $- 0.8$ to the power of $3$ .

$f'' (0.1) = \frac{2.432}{1.728 \cdot - 0.512}$

Step 6.2.3

Simplify the expression.

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Step 6.2.3.1

Multiply $1.728$ by $- 0.512$ .

$f'' (0.1) = \frac{2.432}{- 0.884736}$

Step 6.2.3.2

Divide $2.432$ by $- 0.884736$ .

$f'' (0.1) = - 2.74884259$

Step 6.2.4

The final answer is $- 2.74884259$ .

$- 2.74884259$

Step 6.3

At $0.1$ , the second derivative is $- 2.74884259$ . Since this is negative, the second derivative is decreasing on the interval $(0, \infty)$

Decreasing on $(0, \infty)$ since $f'' (x) < 0$

Step 7

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(0, 0)$ .

$(0, 0)$

Step 8