Finite Math Examples

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

Step 1

Set the argument in $\ln (\frac{x^{2} - 1}{x} - 1)$ greater than $0$ to find where the expression is defined.

$\frac{x^{2} - 1}{x} - 1 > 0$

Step 2

Solve for $x$ .

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Step 2.1

Simplify $\frac{x^{2} - 1}{x} - 1$ .

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Step 2.1.1

Simplify the numerator.

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Step 2.1.1.1

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

$\frac{x^{2} - 1^{2}}{x} - 1 > 0$

Step 2.1.1.2

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

$\frac{(x + 1) (x - 1)}{x} - 1 > 0$

Step 2.1.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$\frac{(x + 1) (x - 1)}{x} - 1 \cdot \frac{x}{x} > 0$

Step 2.1.3

Simplify terms.

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Step 2.1.3.1

Combine $- 1$ and $\frac{x}{x}$ .

$\frac{(x + 1) (x - 1)}{x} + \frac{- x}{x} > 0$

Step 2.1.3.2

Combine the numerators over the common denominator.

$\frac{(x + 1) (x - 1) - x}{x} > 0$

Step 2.1.4

Simplify the numerator.

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Step 2.1.4.1

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

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Step 2.1.4.1.1

Apply the distributive property.

$\frac{x (x - 1) + 1 (x - 1) - x}{x} > 0$

Step 2.1.4.1.2

Apply the distributive property.

$\frac{x \cdot x + x \cdot - 1 + 1 (x - 1) - x}{x} > 0$

Step 2.1.4.1.3

Apply the distributive property.

$\frac{x \cdot x + x \cdot - 1 + 1 x + 1 \cdot - 1 - x}{x} > 0$

Step 2.1.4.2

Simplify and combine like terms.

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Step 2.1.4.2.1

Simplify each term.

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Step 2.1.4.2.1.1

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot - 1 + 1 x + 1 \cdot - 1 - x}{x} > 0$

Step 2.1.4.2.1.2

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

$\frac{x^{2} - 1 \cdot x + 1 x + 1 \cdot - 1 - x}{x} > 0$

Step 2.1.4.2.1.3

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

$\frac{x^{2} - x + 1 x + 1 \cdot - 1 - x}{x} > 0$

Step 2.1.4.2.1.4

Multiply $x$ by $1$ .

$\frac{x^{2} - x + x + 1 \cdot - 1 - x}{x} > 0$

Step 2.1.4.2.1.5

Multiply $- 1$ by $1$ .

$\frac{x^{2} - x + x - 1 - x}{x} > 0$

Step 2.1.4.2.2

Add $- x$ and $x$ .

$\frac{x^{2} + 0 - 1 - x}{x} > 0$

Step 2.1.4.2.3

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

$\frac{x^{2} - 1 - x}{x} > 0$

Step 2.1.4.3

Reorder terms.

$\frac{x^{2} - x - 1}{x} > 0$

Step 2.2

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x = 0$

$x^{2} - x - 1 = 0$

Step 2.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.4

Substitute the values $a = 1$ , $b = - 1$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Step 2.5

Simplify.

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Step 2.5.1

Simplify the numerator.

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Step 2.5.1.1

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

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 2.5.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 2.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot - 1}}{2 \cdot 1}$

Step 2.5.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{1 \pm \sqrt{1 + 4}}{2 \cdot 1}$

Step 2.5.1.3

Add $1$ and $4$ .

$x = \frac{1 \pm \sqrt{5}}{2 \cdot 1}$

Step 2.5.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm \sqrt{5}}{2}$

Step 2.6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 2.6.1

Simplify the numerator.

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Step 2.6.1.1

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

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 2.6.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 2.6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot - 1}}{2 \cdot 1}$

Step 2.6.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{1 \pm \sqrt{1 + 4}}{2 \cdot 1}$

Step 2.6.1.3

Add $1$ and $4$ .

$x = \frac{1 \pm \sqrt{5}}{2 \cdot 1}$

Step 2.6.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm \sqrt{5}}{2}$

Step 2.6.3

Change the $\pm$ to $+$ .

$x = \frac{1 + \sqrt{5}}{2}$

Step 2.7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 2.7.1

Simplify the numerator.

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Step 2.7.1.1

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

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 2.7.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 2.7.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot - 1}}{2 \cdot 1}$

Step 2.7.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{1 \pm \sqrt{1 + 4}}{2 \cdot 1}$

Step 2.7.1.3

Add $1$ and $4$ .

$x = \frac{1 \pm \sqrt{5}}{2 \cdot 1}$

Step 2.7.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm \sqrt{5}}{2}$

Step 2.7.3

Change the $\pm$ to $-$ .

$x = \frac{1 - \sqrt{5}}{2}$

Step 2.8

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}$

Step 2.9

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 0$

$x = \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}$

Step 2.10

Consolidate the solutions.

$x = 0, \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}$

Step 2.11

Find the domain of $\frac{x^{2} - 1}{x} - 1$ .

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Step 2.11.1

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

$x = 0$

Step 2.11.2

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

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

Step 2.12

Use each root to create test intervals.

$x < \frac{1 - \sqrt{5}}{2}$

$\frac{1 - \sqrt{5}}{2} < x < 0$

$0 < x < \frac{1 + \sqrt{5}}{2}$

$x > \frac{1 + \sqrt{5}}{2}$

Step 2.13

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 2.13.1

Test a value on the interval $x < \frac{1 - \sqrt{5}}{2}$ to see if it makes the inequality true.

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Step 2.13.1.1

Choose a value on the interval $x < \frac{1 - \sqrt{5}}{2}$ and see if this value makes the original inequality true.

$x = - 3$

Step 2.13.1.2

Replace $x$ with $- 3$ in the original inequality.

$\frac{{(- 3)}^{2} - 1}{- 3} - 1 > 0$

Step 2.13.1.3

The left side $- 3.\overline{6}$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 2.13.2

Test a value on the interval $\frac{1 - \sqrt{5}}{2} < x < 0$ to see if it makes the inequality true.

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Step 2.13.2.1

Choose a value on the interval $\frac{1 - \sqrt{5}}{2} < x < 0$ and see if this value makes the original inequality true.

$x = - 0.31$

Step 2.13.2.2

Replace $x$ with $- 0.31$ in the original inequality.

$\frac{{(- 0.31)}^{2} - 1}{- 0.31} - 1 > 0$

Step 2.13.2.3

The left side $1.91580645$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.13.3

Test a value on the interval $0 < x < \frac{1 + \sqrt{5}}{2}$ to see if it makes the inequality true.

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Step 2.13.3.1

Choose a value on the interval $0 < x < \frac{1 + \sqrt{5}}{2}$ and see if this value makes the original inequality true.

$x = 1$

Step 2.13.3.2

Replace $x$ with $1$ in the original inequality.

$\frac{{(1)}^{2} - 1}{1} - 1 > 0$

Step 2.13.3.3

The left side $- 1$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 2.13.4

Test a value on the interval $x > \frac{1 + \sqrt{5}}{2}$ to see if it makes the inequality true.

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Step 2.13.4.1

Choose a value on the interval $x > \frac{1 + \sqrt{5}}{2}$ and see if this value makes the original inequality true.

$x = 4$

Step 2.13.4.2

Replace $x$ with $4$ in the original inequality.

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

Step 2.13.4.3

The left side $2.75$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.13.5

Compare the intervals to determine which ones satisfy the original inequality.

$x < \frac{1 - \sqrt{5}}{2}$ False

$\frac{1 - \sqrt{5}}{2} < x < 0$ True

$0 < x < \frac{1 + \sqrt{5}}{2}$ False

$x > \frac{1 + \sqrt{5}}{2}$ True

$x < \frac{1 - \sqrt{5}}{2}$ False

$\frac{1 - \sqrt{5}}{2} < x < 0$ True

$0 < x < \frac{1 + \sqrt{5}}{2}$ False

$x > \frac{1 + \sqrt{5}}{2}$ True

Step 2.14

The solution consists of all of the true intervals.

$\frac{1 - \sqrt{5}}{2} < x < 0$ or $x > \frac{1 + \sqrt{5}}{2}$

Step 3

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

$x = 0$

Step 4

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

Interval Notation:

$(\frac{1 - \sqrt{5}}{2}, 0) \cup (\frac{1 + \sqrt{5}}{2}, \infty)$

Set-Builder Notation:

${x | \frac{1 - \sqrt{5}}{2} < x < 0, x > \frac{1 + \sqrt{5}}{2}}$

Step 5