Precalculus Examples

$\sqrt{\frac{3 - x}{5 - x}}$

Step 1

Set the radicand in $\sqrt{\frac{3 - x}{5 - x}}$ greater than or equal to $0$ to find where the expression is defined.

$\frac{3 - x}{5 - x} \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

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

$3 - x = 0$

$5 - x = 0$

Step 2.2

Subtract $3$ from both sides of the equation.

$- x = - 3$

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Dividing two negative values results in a positive value.

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

Step 2.3.2.2

Divide $x$ by $1$ .

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

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $- 3$ by $- 1$ .

$x = 3$

Step 2.4

Subtract $5$ from both sides of the equation.

$- x = - 5$

Step 2.5

Divide each term in $- x = - 5$ by $- 1$ and simplify.

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Step 2.5.1

Divide each term in $- x = - 5$ by $- 1$ .

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

Step 2.5.2

Simplify the left side.

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Step 2.5.2.1

Dividing two negative values results in a positive value.

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

Step 2.5.2.2

Divide $x$ by $1$ .

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

Step 2.5.3

Simplify the right side.

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Step 2.5.3.1

Divide $- 5$ by $- 1$ .

$x = 5$

Step 2.6

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

$x = 3$

$x = 5$

Step 2.7

Consolidate the solutions.

$x = 3, 5$

Step 2.8

Find the domain of $\frac{3 - x}{5 - x}$ .

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Step 2.8.1

Set the denominator in $\frac{3 - x}{5 - x}$ equal to $0$ to find where the expression is undefined.

$5 - x = 0$

Step 2.8.2

Solve for $x$ .

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Step 2.8.2.1

Subtract $5$ from both sides of the equation.

$- x = - 5$

Step 2.8.2.2

Divide each term in $- x = - 5$ by $- 1$ and simplify.

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Step 2.8.2.2.1

Divide each term in $- x = - 5$ by $- 1$ .

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

Step 2.8.2.2.2

Simplify the left side.

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Step 2.8.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 2.8.2.2.2.2

Divide $x$ by $1$ .

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

Step 2.8.2.2.3

Simplify the right side.

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Step 2.8.2.2.3.1

Divide $- 5$ by $- 1$ .

$x = 5$

Step 2.8.3

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

$(- \infty, 5) \cup (5, \infty)$

Step 2.9

Use each root to create test intervals.

$x < 3$

$3 < x < 5$

$x > 5$

Step 2.10

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 2.10.1

Test a value on the interval $x < 3$ to see if it makes the inequality true.

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Step 2.10.1.1

Choose a value on the interval $x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 2.10.1.2

Replace $x$ with $0$ in the original inequality.

$\frac{3 - (0)}{5 - (0)} \geq 0$

Step 2.10.1.3

The left side $0.6$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.10.2

Test a value on the interval $3 < x < 5$ to see if it makes the inequality true.

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Step 2.10.2.1

Choose a value on the interval $3 < x < 5$ and see if this value makes the original inequality true.

$x = 4$

Step 2.10.2.2

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

$\frac{3 - (4)}{5 - (4)} \geq 0$

Step 2.10.2.3

The left side $- 1$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.10.3

Test a value on the interval $x > 5$ to see if it makes the inequality true.

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Step 2.10.3.1

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Step 2.10.3.2

Replace $x$ with $8$ in the original inequality.

$\frac{3 - (8)}{5 - (8)} \geq 0$

Step 2.10.3.3

The left side $1.\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.10.4

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

$x < 3$ True

$3 < x < 5$ False

$x > 5$ True

$x < 3$ True

$3 < x < 5$ False

$x > 5$ True

Step 2.11

The solution consists of all of the true intervals.

$x \leq 3$ or $x > 5$

Step 3

Set the denominator in $\frac{3 - x}{5 - x}$ equal to $0$ to find where the expression is undefined.

$5 - x = 0$

Step 4

Solve for $x$ .

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Step 4.1

Subtract $5$ from both sides of the equation.

$- x = - 5$

Step 4.2

Divide each term in $- x = - 5$ by $- 1$ and simplify.

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Step 4.2.1

Divide each term in $- x = - 5$ by $- 1$ .

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

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Dividing two negative values results in a positive value.

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

Step 4.2.2.2

Divide $x$ by $1$ .

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

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Divide $- 5$ by $- 1$ .

$x = 5$

Step 5

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

Interval Notation:

$(- \infty, 3] \cup (5, \infty)$

Set-Builder Notation:

${x | x \leq 3, x > 5}$

Step 6