Calculus Examples

$| 3 - \frac{1}{x} | < \frac{1}{2}$

Step 1

Write $| 3 - \frac{1}{x} | < \frac{1}{2}$ as a piecewise.

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Step 1.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$3 - \frac{1}{x} \geq 0$

Step 1.2

Solve the inequality.

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Step 1.2.1

Subtract $3$ from both sides of the inequality.

$- \frac{1}{x} \geq - 3$

Step 1.2.2

Multiply both sides by $x$ .

$- \frac{1}{x} x = - 3 x$

Step 1.2.3

Simplify the left side.

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Step 1.2.3.1

Cancel the common factor of $x$ .

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Step 1.2.3.1.1

Move the leading negative in $- \frac{1}{x}$ into the numerator.

$\frac{- 1}{x} x = - 3 x$

Step 1.2.3.1.2

Cancel the common factor.

$\frac{- 1}{x} x = - 3 x$

Step 1.2.3.1.3

Rewrite the expression.

$- 1 = - 3 x$

Step 1.2.4

Solve for $x$ .

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Step 1.2.4.1

Rewrite the equation as $- 3 x = - 1$ .

$- 3 x = - 1$

Step 1.2.4.2

Divide each term in $- 3 x = - 1$ by $- 3$ and simplify.

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Step 1.2.4.2.1

Divide each term in $- 3 x = - 1$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{- 1}{- 3}$

Step 1.2.4.2.2

Simplify the left side.

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Step 1.2.4.2.2.1

Cancel the common factor of $- 3$ .

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Step 1.2.4.2.2.1.1

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{- 1}{- 3}$

Step 1.2.4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 3}$

Step 1.2.4.2.3

Simplify the right side.

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Step 1.2.4.2.3.1

Dividing two negative values results in a positive value.

$x = \frac{1}{3}$

Step 1.2.5

Find the domain of $\frac{2 | 3 - \frac{1}{x} | - 1}{2}$ .

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Step 1.2.5.1

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

$x = 0$

Step 1.2.5.2

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

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

Step 1.2.6

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{1}{3}$

$x > \frac{1}{3}$

Step 1.2.7

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 1.2.7.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 1.2.7.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 1.2.7.1.2

Replace $x$ with $- 2$ in the original inequality.

$3 - \frac{1}{- 2} \geq 0$

Step 1.2.7.1.3

The left side $3.5$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.7.2

Test a value on the interval $0 < x < \frac{1}{3}$ to see if it makes the inequality true.

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Step 1.2.7.2.1

Choose a value on the interval $0 < x < \frac{1}{3}$ and see if this value makes the original inequality true.

$x = 0.17$

Step 1.2.7.2.2

Replace $x$ with $0.17$ in the original inequality.

$3 - \frac{1}{0.17} \geq 0$

Step 1.2.7.2.3

The left side $- 2.88235294$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.7.3

Test a value on the interval $x > \frac{1}{3}$ to see if it makes the inequality true.

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Step 1.2.7.3.1

Choose a value on the interval $x > \frac{1}{3}$ and see if this value makes the original inequality true.

$x = 3$

Step 1.2.7.3.2

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

$3 - \frac{1}{3} \geq 0$

Step 1.2.7.3.3

The left side $2.\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.7.4

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

$x < 0$ True

$0 < x < \frac{1}{3}$ False

$x > \frac{1}{3}$ True

$x < 0$ True

$0 < x < \frac{1}{3}$ False

$x > \frac{1}{3}$ True

Step 1.2.8

The solution consists of all of the true intervals.

$x < 0$ or $x \geq \frac{1}{3}$

Step 1.3

In the piece where $3 - \frac{1}{x}$ is non-negative, remove the absolute value.

$3 - \frac{1}{x} < \frac{1}{2}$

Step 1.4

Find the domain of $3 - \frac{1}{x} < \frac{1}{2}$ and find the intersection with $x < 0 or x \geq \frac{1}{3}$ .

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Step 1.4.1

Find the domain of $3 - \frac{1}{x} < \frac{1}{2}$ .

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Step 1.4.1.1

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

$x = 0$

Step 1.4.1.2

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

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

Step 1.4.2

Find the intersection of $x < 0 or x \geq \frac{1}{3}$ and $(- \infty, 0) \cup (0, \infty)$ .

$x < 0 or x \geq \frac{1}{3}$

Step 1.5

To find the interval for the second piece, find where the inside of the absolute value is negative.

$3 - \frac{1}{x} < 0$

Step 1.6

Solve the inequality.

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Step 1.6.1

Subtract $3$ from both sides of the inequality.

$- \frac{1}{x} < - 3$

Step 1.6.2

Multiply both sides by $x$ .

$- \frac{1}{x} x = - 3 x$

Step 1.6.3

Simplify the left side.

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Step 1.6.3.1

Cancel the common factor of $x$ .

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Step 1.6.3.1.1

Move the leading negative in $- \frac{1}{x}$ into the numerator.

$\frac{- 1}{x} x = - 3 x$

Step 1.6.3.1.2

Cancel the common factor.

$\frac{- 1}{x} x = - 3 x$

Step 1.6.3.1.3

Rewrite the expression.

$- 1 = - 3 x$

Step 1.6.4

Solve for $x$ .

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Step 1.6.4.1

Rewrite the equation as $- 3 x = - 1$ .

$- 3 x = - 1$

Step 1.6.4.2

Divide each term in $- 3 x = - 1$ by $- 3$ and simplify.

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Step 1.6.4.2.1

Divide each term in $- 3 x = - 1$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{- 1}{- 3}$

Step 1.6.4.2.2

Simplify the left side.

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Step 1.6.4.2.2.1

Cancel the common factor of $- 3$ .

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Step 1.6.4.2.2.1.1

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{- 1}{- 3}$

Step 1.6.4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 3}$

Step 1.6.4.2.3

Simplify the right side.

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Step 1.6.4.2.3.1

Dividing two negative values results in a positive value.

$x = \frac{1}{3}$

Step 1.6.5

Find the domain of $\frac{2 | 3 - \frac{1}{x} | - 1}{2}$ .

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Step 1.6.5.1

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

$x = 0$

Step 1.6.5.2

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

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

Step 1.6.6

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{1}{3}$

$x > \frac{1}{3}$

Step 1.6.7

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 1.6.7.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 1.6.7.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 1.6.7.1.2

Replace $x$ with $- 2$ in the original inequality.

$3 - \frac{1}{- 2} < 0$

Step 1.6.7.1.3

The left side $3.5$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 1.6.7.2

Test a value on the interval $0 < x < \frac{1}{3}$ to see if it makes the inequality true.

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Step 1.6.7.2.1

Choose a value on the interval $0 < x < \frac{1}{3}$ and see if this value makes the original inequality true.

$x = 0.17$

Step 1.6.7.2.2

Replace $x$ with $0.17$ in the original inequality.

$3 - \frac{1}{0.17} < 0$

Step 1.6.7.2.3

The left side $- 2.88235294$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 1.6.7.3

Test a value on the interval $x > \frac{1}{3}$ to see if it makes the inequality true.

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Step 1.6.7.3.1

Choose a value on the interval $x > \frac{1}{3}$ and see if this value makes the original inequality true.

$x = 3$

Step 1.6.7.3.2

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

$3 - \frac{1}{3} < 0$

Step 1.6.7.3.3

The left side $2.\overline{6}$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 1.6.7.4

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

$x < 0$ False

$0 < x < \frac{1}{3}$ True

$x > \frac{1}{3}$ False

$x < 0$ False

$0 < x < \frac{1}{3}$ True

$x > \frac{1}{3}$ False

Step 1.6.8

The solution consists of all of the true intervals.

$0 < x < \frac{1}{3}$

Step 1.7

In the piece where $3 - \frac{1}{x}$ is negative, remove the absolute value and multiply by $- 1$ .

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

Step 1.8

Find the domain of $- (3 - \frac{1}{x}) < \frac{1}{2}$ and find the intersection with $0 < x < \frac{1}{3}$ .

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Step 1.8.1

Find the domain of $- (3 - \frac{1}{x}) < \frac{1}{2}$ .

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Step 1.8.1.1

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

$x = 0$

Step 1.8.1.2

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

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

Step 1.8.2

Find the intersection of $0 < x < \frac{1}{3}$ and $(- \infty, 0) \cup (0, \infty)$ .

$0 < x < \frac{1}{3}$

Step 1.9

Write as a piecewise.

${\begin{cases} 3 - \frac{1}{x} < \frac{1}{2} & x < 0 or x \geq \frac{1}{3} \\ - (3 - \frac{1}{x}) < \frac{1}{2} & 0 < x < \frac{1}{3} \end{cases}$

Step 1.10

Simplify $- (3 - \frac{1}{x}) < \frac{1}{2}$ .

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Step 1.10.1

Apply the distributive property.

${\begin{cases} 3 - \frac{1}{x} < \frac{1}{2} & x < 0 or x \geq \frac{1}{3} \\ - 1 \cdot 3 - - \frac{1}{x} < \frac{1}{2} & 0 < x < \frac{1}{3} \end{cases}$

Step 1.10.2

Multiply $- 1$ by $3$ .

${\begin{cases} 3 - \frac{1}{x} < \frac{1}{2} & x < 0 or x \geq \frac{1}{3} \\ - 3 - - \frac{1}{x} < \frac{1}{2} & 0 < x < \frac{1}{3} \end{cases}$

Step 1.10.3

Multiply $- - \frac{1}{x}$ .

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Step 1.10.3.1

Multiply $- 1$ by $- 1$ .

${\begin{cases} 3 - \frac{1}{x} < \frac{1}{2} & x < 0 or x \geq \frac{1}{3} \\ - 3 + 1 \frac{1}{x} < \frac{1}{2} & 0 < x < \frac{1}{3} \end{cases}$

Step 1.10.3.2

Multiply $\frac{1}{x}$ by $1$ .

${\begin{cases} 3 - \frac{1}{x} < \frac{1}{2} & x < 0 or x \geq \frac{1}{3} \\ - 3 + \frac{1}{x} < \frac{1}{2} & 0 < x < \frac{1}{3} \end{cases}$

Step 2

Solve $3 - \frac{1}{x} < \frac{1}{2}$ when $x < 0 or x \geq \frac{1}{3}$ .

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Step 2.1

Solve $3 - \frac{1}{x} < \frac{1}{2}$ for $x$ .

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Step 2.1.1

Move all terms not containing $x$ to the right side of the inequality.

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Step 2.1.1.1

Subtract $3$ from both sides of the inequality.

$- \frac{1}{x} < \frac{1}{2} - 3$

Step 2.1.1.2

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- \frac{1}{x} < \frac{1}{2} - 3 \cdot \frac{2}{2}$

Step 2.1.1.3

Combine $- 3$ and $\frac{2}{2}$ .

$- \frac{1}{x} < \frac{1}{2} + \frac{- 3 \cdot 2}{2}$

Step 2.1.1.4

Combine the numerators over the common denominator.

$- \frac{1}{x} < \frac{1 - 3 \cdot 2}{2}$

Step 2.1.1.5

Simplify the numerator.

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Step 2.1.1.5.1

Multiply $- 3$ by $2$ .

$- \frac{1}{x} < \frac{1 - 6}{2}$

Step 2.1.1.5.2

Subtract $6$ from $1$ .

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

Step 2.1.1.6

Move the negative in front of the fraction.

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

Step 2.1.2

Multiply both sides by $x$ .

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

Step 2.1.3

Simplify.

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Step 2.1.3.1

Simplify the left side.

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Step 2.1.3.1.1

Cancel the common factor of $x$ .

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Step 2.1.3.1.1.1

Move the leading negative in $- \frac{1}{x}$ into the numerator.

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

Step 2.1.3.1.1.2

Cancel the common factor.

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

Step 2.1.3.1.1.3

Rewrite the expression.

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

Step 2.1.3.2

Simplify the right side.

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Step 2.1.3.2.1

Simplify $- \frac{5}{2} x$ .

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Step 2.1.3.2.1.1

Combine $x$ and $\frac{5}{2}$ .

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

Step 2.1.3.2.1.2

Move $5$ to the left of $x$ .

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

Step 2.1.4

Solve for $x$ .

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Step 2.1.4.1

Rewrite the equation as $- \frac{5 x}{2} = - 1$ .

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

Step 2.1.4.2

Multiply both sides of the equation by $- \frac{2}{5}$ .

$- \frac{2}{5} (- \frac{5 x}{2}) = - \frac{2}{5} \cdot - 1$

Step 2.1.4.3

Simplify both sides of the equation.

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Step 2.1.4.3.1

Simplify the left side.

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Step 2.1.4.3.1.1

Simplify $- \frac{2}{5} (- \frac{5 x}{2})$ .

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Step 2.1.4.3.1.1.1

Cancel the common factor of $2$ .

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Step 2.1.4.3.1.1.1.1

Move the leading negative in $- \frac{2}{5}$ into the numerator.

$\frac{- 2}{5} (- \frac{5 x}{2}) = - \frac{2}{5} \cdot - 1$

Step 2.1.4.3.1.1.1.2

Move the leading negative in $- \frac{5 x}{2}$ into the numerator.

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

Step 2.1.4.3.1.1.1.3

Factor $2$ out of $- 2$ .

$\frac{2 (- 1)}{5} \cdot \frac{- 5 x}{2} = - \frac{2}{5} \cdot - 1$

Step 2.1.4.3.1.1.1.4

Cancel the common factor.

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

Step 2.1.4.3.1.1.1.5

Rewrite the expression.

$\frac{- 1}{5} (- 5 x) = - \frac{2}{5} \cdot - 1$

Step 2.1.4.3.1.1.2

Cancel the common factor of $5$ .

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Step 2.1.4.3.1.1.2.1

Factor $5$ out of $- 5 x$ .

$\frac{- 1}{5} (5 (- x)) = - \frac{2}{5} \cdot - 1$

Step 2.1.4.3.1.1.2.2

Cancel the common factor.

$\frac{- 1}{5} (5 (- x)) = - \frac{2}{5} \cdot - 1$

Step 2.1.4.3.1.1.2.3

Rewrite the expression.

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

Step 2.1.4.3.1.1.3

Multiply.

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Step 2.1.4.3.1.1.3.1

Multiply $- 1$ by $- 1$ .

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

Step 2.1.4.3.1.1.3.2

Multiply $x$ by $1$ .

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

Step 2.1.4.3.2

Simplify the right side.

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Step 2.1.4.3.2.1

Multiply $- \frac{2}{5} \cdot - 1$ .

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Step 2.1.4.3.2.1.1

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{2}{5})$

Step 2.1.4.3.2.1.2

Multiply $\frac{2}{5}$ by $1$ .

$x = \frac{2}{5}$

Step 2.1.5

Find the domain of $- \frac{1}{x} + \frac{5}{2}$ .

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Step 2.1.5.1

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

$x = 0$

Step 2.1.5.2

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

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

Step 2.1.6

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{2}{5}$

$x > \frac{2}{5}$

Step 2.1.7

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 2.1.7.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 2.1.7.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 2.1.7.1.2

Replace $x$ with $- 2$ in the original inequality.

$3 - \frac{1}{- 2} < \frac{1}{2}$

Step 2.1.7.1.3

The left side $3.5$ is not less than the right side $0.5$ , which means that the given statement is false.

False

Step 2.1.7.2

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

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Step 2.1.7.2.1

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

$x = 0.2$

Step 2.1.7.2.2

Replace $x$ with $0.2$ in the original inequality.

$3 - \frac{1}{0.2} < \frac{1}{2}$

Step 2.1.7.2.3

The left side $- 2$ is less than the right side $0.5$ , which means that the given statement is always true.

True

Step 2.1.7.3

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

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Step 2.1.7.3.1

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

$x = 3$

Step 2.1.7.3.2

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

$3 - \frac{1}{3} < \frac{1}{2}$

Step 2.1.7.3.3

The left side $2.\overline{6}$ is not less than the right side $0.5$ , which means that the given statement is false.

False

Step 2.1.7.4

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

$x < 0$ False

$0 < x < \frac{2}{5}$ True

$x > \frac{2}{5}$ False

$x < 0$ False

$0 < x < \frac{2}{5}$ True

$x > \frac{2}{5}$ False

Step 2.1.8

The solution consists of all of the true intervals.

$0 < x < \frac{2}{5}$

Step 2.2

Find the intersection of $0 < x < \frac{2}{5}$ and $x < 0 or x \geq \frac{1}{3}$ .

$\frac{1}{3} \leq x < \frac{2}{5}$

Step 3

Solve $- 3 + \frac{1}{x} < \frac{1}{2}$ when $0 < x < \frac{1}{3}$ .

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Step 3.1

Solve $- 3 + \frac{1}{x} < \frac{1}{2}$ for $x$ .

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Step 3.1.1

Move all terms not containing $x$ to the right side of the inequality.

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Step 3.1.1.1

Add $3$ to both sides of the inequality.

$\frac{1}{x} < \frac{1}{2} + 3$

Step 3.1.1.2

To write $3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{1}{x} < \frac{1}{2} + 3 \cdot \frac{2}{2}$

Step 3.1.1.3

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

$\frac{1}{x} < \frac{1}{2} + \frac{3 \cdot 2}{2}$

Step 3.1.1.4

Combine the numerators over the common denominator.

$\frac{1}{x} < \frac{1 + 3 \cdot 2}{2}$

Step 3.1.1.5

Simplify the numerator.

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Step 3.1.1.5.1

Multiply $3$ by $2$ .

$\frac{1}{x} < \frac{1 + 6}{2}$

Step 3.1.1.5.2

Add $1$ and $6$ .

$\frac{1}{x} < \frac{7}{2}$

Step 3.1.2

Multiply both sides by $x$ .

$\frac{1}{x} x = \frac{7}{2} x$

Step 3.1.3

Simplify.

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Step 3.1.3.1

Simplify the left side.

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Step 3.1.3.1.1

Cancel the common factor of $x$ .

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Step 3.1.3.1.1.1

Cancel the common factor.

$\frac{1}{x} x = \frac{7}{2} x$

Step 3.1.3.1.1.2

Rewrite the expression.

$1 = \frac{7}{2} x$

Step 3.1.3.2

Simplify the right side.

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Step 3.1.3.2.1

Combine $\frac{7}{2}$ and $x$ .

$1 = \frac{7 x}{2}$

Step 3.1.4

Solve for $x$ .

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Step 3.1.4.1

Rewrite the equation as $\frac{7 x}{2} = 1$ .

$\frac{7 x}{2} = 1$

Step 3.1.4.2

Multiply both sides of the equation by $\frac{2}{7}$ .

$\frac{2}{7} \cdot \frac{7 x}{2} = \frac{2}{7} \cdot 1$

Step 3.1.4.3

Simplify both sides of the equation.

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Step 3.1.4.3.1

Simplify the left side.

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Step 3.1.4.3.1.1

Simplify $\frac{2}{7} \cdot \frac{7 x}{2}$ .

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Step 3.1.4.3.1.1.1

Cancel the common factor of $2$ .

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Step 3.1.4.3.1.1.1.1

Cancel the common factor.

$\frac{2}{7} \cdot \frac{7 x}{2} = \frac{2}{7} \cdot 1$

Step 3.1.4.3.1.1.1.2

Rewrite the expression.

$\frac{1}{7} (7 x) = \frac{2}{7} \cdot 1$

Step 3.1.4.3.1.1.2

Cancel the common factor of $7$ .

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Step 3.1.4.3.1.1.2.1

Factor $7$ out of $7 x$ .

$\frac{1}{7} (7 (x)) = \frac{2}{7} \cdot 1$

Step 3.1.4.3.1.1.2.2

Cancel the common factor.

$\frac{1}{7} (7 x) = \frac{2}{7} \cdot 1$

Step 3.1.4.3.1.1.2.3

Rewrite the expression.

$x = \frac{2}{7} \cdot 1$

Step 3.1.4.3.2

Simplify the right side.

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Step 3.1.4.3.2.1

Multiply $\frac{2}{7}$ by $1$ .

$x = \frac{2}{7}$

Step 3.1.5

Find the domain of $\frac{1}{x} - \frac{7}{2}$ .

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Step 3.1.5.1

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

$x = 0$

Step 3.1.5.2

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

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

Step 3.1.6

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{2}{7}$

$x > \frac{2}{7}$

Step 3.1.7

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 3.1.7.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 3.1.7.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 3.1.7.1.2

Replace $x$ with $- 2$ in the original inequality.

$- 3 + \frac{1}{- 2} < \frac{1}{2}$

Step 3.1.7.1.3

The left side $- 3.5$ is less than the right side $0.5$ , which means that the given statement is always true.

True

Step 3.1.7.2

Test a value on the interval $0 < x < \frac{2}{7}$ to see if it makes the inequality true.

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Step 3.1.7.2.1

Choose a value on the interval $0 < x < \frac{2}{7}$ and see if this value makes the original inequality true.

$x = 0.14$

Step 3.1.7.2.2

Replace $x$ with $0.14$ in the original inequality.

$- 3 + \frac{1}{0.14} < \frac{1}{2}$

Step 3.1.7.2.3

The left side $4.\overline{142857}$ is not less than the right side $0.5$ , which means that the given statement is false.

False

Step 3.1.7.3

Test a value on the interval $x > \frac{2}{7}$ to see if it makes the inequality true.

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Step 3.1.7.3.1

Choose a value on the interval $x > \frac{2}{7}$ and see if this value makes the original inequality true.

$x = 3$

Step 3.1.7.3.2

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

$- 3 + \frac{1}{3} < \frac{1}{2}$

Step 3.1.7.3.3

The left side $- 2.\overline{6}$ is less than the right side $0.5$ , which means that the given statement is always true.

True

Step 3.1.7.4

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

$x < 0$ True

$0 < x < \frac{2}{7}$ False

$x > \frac{2}{7}$ True

$x < 0$ True

$0 < x < \frac{2}{7}$ False

$x > \frac{2}{7}$ True

Step 3.1.8

The solution consists of all of the true intervals.

$x < 0$ or $x > \frac{2}{7}$

Step 3.2

Find the intersection of $x < 0 or x > \frac{2}{7}$ and $0 < x < \frac{1}{3}$ .

$\frac{2}{7} < x < \frac{1}{3}$

Step 4

Find the union of the solutions.

$\frac{2}{7} < x < \frac{2}{5}$

Step 5

Convert the inequality to interval notation.

$(\frac{2}{7}, \frac{2}{5})$

Step 6