Calculus Examples

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

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 $f (x) = x^{2} - 1$ and $g (x) = 2 x - 1$ .

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

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

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

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

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 $n = 2$ .

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

Step 1.1.1.2.3

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

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

Step 1.1.1.2.4

Simplify the expression.

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Step 1.1.1.2.4.1

Add $2 x$ and $0$ .

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

Step 1.1.1.2.4.2

Move $2$ to the left of $2 x - 1$ .

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

Step 1.1.1.2.5

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

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

Step 1.1.1.2.6

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

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

Step 1.1.1.2.7

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

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

Step 1.1.1.2.8

Multiply $2$ by $1$ .

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

Step 1.1.1.2.9

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

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

Step 1.1.1.2.10

Simplify the expression.

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Step 1.1.1.2.10.1

Add $2$ and $0$ .

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

Step 1.1.1.2.10.2

Multiply $2$ by $- 1$ .

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

Step 1.1.1.3

Simplify.

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Step 1.1.1.3.1

Apply the distributive property.

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

Step 1.1.1.3.2

Apply the distributive property.

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

Step 1.1.1.3.3

Apply the distributive property.

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

Step 1.1.1.3.4

Simplify the numerator.

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Step 1.1.1.3.4.1

Simplify each term.

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Step 1.1.1.3.4.1.1

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

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Step 1.1.1.3.4.1.1.1

Move $x$ .

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

Step 1.1.1.3.4.1.1.2

Multiply $x$ by $x$ .

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

Step 1.1.1.3.4.1.2

Multiply $2$ by $2$ .

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

Step 1.1.1.3.4.1.3

Multiply $2$ by $- 1$ .

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

Step 1.1.1.3.4.1.4

Multiply $- 2$ by $- 1$ .

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

Step 1.1.1.3.4.2

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

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

Step 1.1.1.3.5

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

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Step 1.1.1.3.5.1

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

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

Step 1.1.1.3.5.2

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

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

Step 1.1.1.3.5.3

Factor $2$ out of $2$ .

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

Step 1.1.1.3.5.4

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

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

Step 1.1.1.3.5.5

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

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

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

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

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

Step 1.1.2.2

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

Step 1.1.2.3

Differentiate.

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Step 1.1.2.3.1

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

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Step 1.1.2.3.1.1

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

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

Step 1.1.2.3.1.2

Multiply $2$ by $2$ .

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

Step 1.1.2.3.2

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

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

Step 1.1.2.3.3

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

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

Step 1.1.2.3.4

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

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

Step 1.1.2.3.5

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

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

Step 1.1.2.3.6

Multiply $- 1$ by $1$ .

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

Step 1.1.2.3.7

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

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

Step 1.1.2.3.8

Add $2 x - 1$ and $0$ .

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

Step 1.1.2.4

Multiply ${(2 x - 1)}^{2}$ by $2 x - 1$ by adding the exponents.

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Step 1.1.2.4.1

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

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Step 1.1.2.4.1.1

Raise $2 x - 1$ to the power of $1$ .

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

Step 1.1.2.4.1.2

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

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

Step 1.1.2.4.2

Add $2$ and $1$ .

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

Step 1.1.2.5

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = 2 x - 1$ .

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Step 1.1.2.5.1

To apply the Chain Rule, set $u$ as $2 x - 1$ .

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

Step 1.1.2.5.2

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

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

Step 1.1.2.5.3

Replace all occurrences of $u$ with $2 x - 1$ .

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

Step 1.1.2.6

Simplify with factoring out.

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Step 1.1.2.6.1

Multiply $2$ by $- 1$ .

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

Step 1.1.2.6.2

Factor $2 x - 1$ out of ${(2 x - 1)}^{3} - 2 (x^{2} - x + 1) ((2 x - 1) \frac{d}{d x} [2 x - 1])$ .

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Step 1.1.2.6.2.1

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

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

Step 1.1.2.6.2.2

Factor $2 x - 1$ out of $- 2 (x^{2} - x + 1) ((2 x - 1) \frac{d}{d x} [2 x - 1])$ .

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

Step 1.1.2.6.2.3

Factor $2 x - 1$ out of $(2 x - 1) {(2 x - 1)}^{2} + (2 x - 1) (- 2 (x^{2} - x + 1) (\frac{d}{d x} [2 x - 1]))$ .

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

Step 1.1.2.7

Cancel the common factors.

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Step 1.1.2.7.1

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

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

Step 1.1.2.7.2

Cancel the common factor.

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

Step 1.1.2.7.3

Rewrite the expression.

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

Step 1.1.2.8

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

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

Step 1.1.2.9

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

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

Step 1.1.2.10

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

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

Step 1.1.2.11

Multiply $2$ by $1$ .

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

Step 1.1.2.12

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

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

Step 1.1.2.13

Combine fractions.

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Step 1.1.2.13.1

Add $2$ and $0$ .

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

Step 1.1.2.13.2

Multiply $2$ by $- 2$ .

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

Step 1.1.2.13.3

Combine $2$ and $\frac{{(2 x - 1)}^{2} - 4 (x^{2} - x + 1)}{{(2 x - 1)}^{3}}$ .

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

Step 1.1.2.14

Simplify.

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Step 1.1.2.14.1

Apply the distributive property.

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

Step 1.1.2.14.2

Apply the distributive property.

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

Step 1.1.2.14.3

Simplify the numerator.

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Step 1.1.2.14.3.1

Simplify each term.

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Step 1.1.2.14.3.1.1

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

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

Step 1.1.2.14.3.1.2

Expand $(2 x - 1) (2 x - 1)$ using the FOIL Method.

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Step 1.1.2.14.3.1.2.1

Apply the distributive property.

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

Step 1.1.2.14.3.1.2.2

Apply the distributive property.

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

Step 1.1.2.14.3.1.2.3

Apply the distributive property.

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

Step 1.1.2.14.3.1.3

Simplify and combine like terms.

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Step 1.1.2.14.3.1.3.1

Simplify each term.

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Step 1.1.2.14.3.1.3.1.1

Rewrite using the commutative property of multiplication.

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

Step 1.1.2.14.3.1.3.1.2

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

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Step 1.1.2.14.3.1.3.1.2.1

Move $x$ .

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

Step 1.1.2.14.3.1.3.1.2.2

Multiply $x$ by $x$ .

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

Step 1.1.2.14.3.1.3.1.3

Multiply $2$ by $2$ .

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

Step 1.1.2.14.3.1.3.1.4

Multiply $- 1$ by $2$ .

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

Step 1.1.2.14.3.1.3.1.5

Multiply $2$ by $- 1$ .

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

Step 1.1.2.14.3.1.3.1.6

Multiply $- 1$ by $- 1$ .

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

Step 1.1.2.14.3.1.3.2

Subtract $2 x$ from $- 2 x$ .

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

Step 1.1.2.14.3.1.4

Apply the distributive property.

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

Step 1.1.2.14.3.1.5

Simplify.

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Step 1.1.2.14.3.1.5.1

Multiply $4$ by $2$ .

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

Step 1.1.2.14.3.1.5.2

Multiply $- 4$ by $2$ .

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

Step 1.1.2.14.3.1.5.3

Multiply $2$ by $1$ .

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

Step 1.1.2.14.3.1.6

Multiply $- 4$ by $2$ .

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

Step 1.1.2.14.3.1.7

Multiply $- 1$ by $- 4$ .

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

Step 1.1.2.14.3.1.8

Multiply $4$ by $2$ .

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

Step 1.1.2.14.3.1.9

Multiply $2 (- 4 \cdot 1)$ .

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Step 1.1.2.14.3.1.9.1

Multiply $- 4$ by $1$ .

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

Step 1.1.2.14.3.1.9.2

Multiply $2$ by $- 4$ .

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

Step 1.1.2.14.3.2

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

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Step 1.1.2.14.3.2.1

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

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

Step 1.1.2.14.3.2.2

Add $- 8 x + 2$ and $0$ .

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

Step 1.1.2.14.3.2.3

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

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

Step 1.1.2.14.3.2.4

Add $0$ and $2$ .

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

Step 1.1.2.14.3.3

Subtract $8$ from $2$ .

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

Step 1.1.2.14.4

Move the negative in front of the fraction.

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

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $- \frac{6}{{(2 x - 1)}^{3}}$ .

$- \frac{6}{{(2 x - 1)}^{3}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $- \frac{6}{{(2 x - 1)}^{3}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$- \frac{6}{{(2 x - 1)}^{3}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$6 = 0$

Step 1.2.3

Since $6 \neq 0$ , there are no solutions.

No solution

Step 2

Find the domain of $f (x) = \frac{x^{2} - 1}{2 x - 1}$ .

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Step 2.1

Set the denominator in $\frac{x^{2} - 1}{2 x - 1}$ equal to $0$ to find where the expression is undefined.

$2 x - 1 = 0$

Step 2.2

Solve for $x$ .

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Step 2.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 2.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 2.2.2.1

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Step 2.2.2.2

Simplify the left side.

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Step 2.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Step 2.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 2.3

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

Interval Notation:

$(- \infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

Set-Builder Notation:

${x | x \neq \frac{1}{2}}$

Interval Notation:

$(- \infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

Set-Builder Notation:

${x | x \neq \frac{1}{2}}$

Step 3

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

$(- \infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

Step 4

Substitute any number from the interval $(- \infty, \frac{1}{2})$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

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

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

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify the denominator.

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Step 4.2.1.1

Multiply $2$ by $0$ .

$f'' (0) = - \frac{6}{{(0 - 1)}^{3}}$

Step 4.2.1.2

Subtract $1$ from $0$ .

$f'' (0) = - \frac{6}{{(- 1)}^{3}}$

Step 4.2.1.3

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

$f'' (0) = - \frac{6}{- 1}$

Step 4.2.2

Simplify the expression.

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Step 4.2.2.1

Divide $6$ by $- 1$ .

$f'' (0) = 6$

Step 4.2.2.2

Multiply $- 1$ by $- 6$ .

$f'' (0) = 6$

Step 4.2.3

The final answer is $6$ .

$6$

Step 4.3

The graph is concave up on the interval $(- \infty, \frac{1}{2})$ because $f'' (0)$ is positive.

Concave up on $(- \infty, \frac{1}{2})$ since $f'' (x)$ is positive

Step 5

Substitute any number from the interval $(\frac{1}{2}, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

Replace the variable $x$ with $3$ in the expression.

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

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify the denominator.

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Step 5.2.1.1

Multiply $2$ by $3$ .

$f'' (3) = - \frac{6}{{(6 - 1)}^{3}}$

Step 5.2.1.2

Subtract $1$ from $6$ .

$f'' (3) = - \frac{6}{5^{3}}$

Step 5.2.1.3

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

$f'' (3) = - \frac{6}{125}$

Step 5.2.2

The final answer is $- \frac{6}{125}$ .

$- \frac{6}{125}$

Step 5.3

The graph is concave down on the interval $(\frac{1}{2}, \infty)$ because $f'' (3)$ is negative.

Concave down on $(\frac{1}{2}, \infty)$ since $f'' (x)$ is negative

Step 6

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

Concave up on $(- \infty, \frac{1}{2})$ since $f'' (x)$ is positive

Concave down on $(\frac{1}{2}, \infty)$ since $f'' (x)$ is negative

Step 7