Find the Domain square root of (2x)/(x+9)-1

$\sqrt{\frac{2 x}{x + 9} - 1}$

Step 1

Set the radicand in $\sqrt{\frac{2 x}{x + 9} - 1}$ greater than or equal to $0$ to find where the expression is defined.

$\frac{2 x}{x + 9} - 1 \geq 0$

Step 2

Solve for $x$ .

Simplify $\frac{2 x}{x + 9} - 1$ .

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{x + 9}{x + 9}$ .

$\frac{2 x}{x + 9} - 1 \cdot \frac{x + 9}{x + 9} \geq 0$

Simplify terms.

Combine $- 1$ and $\frac{x + 9}{x + 9}$ .

$\frac{2 x}{x + 9} + \frac{- (x + 9)}{x + 9} \geq 0$

Combine the numerators over the common denominator.

$\frac{2 x - (x + 9)}{x + 9} \geq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{2 x - x - 1 \cdot 9}{x + 9} \geq 0$

Multiply $- 1$ by $9$ .

$\frac{2 x - x - 9}{x + 9} \geq 0$

Subtract $x$ from $2 x$ .

$\frac{x - 9}{x + 9} \geq 0$

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

$x - 9 = 0$

$x + 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

Subtract $9$ from both sides of the equation.

$x = - 9$

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

$x = 9$

$x = - 9$

Consolidate the solutions.

$x = 9, - 9$

Find the domain of $\frac{2 x}{x + 9} - 1$ .

Set the denominator in $\frac{2 x}{x + 9}$ equal to $0$ to find where the expression is undefined.

$x + 9 = 0$

Subtract $9$ from both sides of the equation.

$x = - 9$

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

$(- \infty, - 9) \cup (- 9, \infty)$

Use each root to create test intervals.

$x < - 9$

$- 9 < x < 9$

$x > 9$

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

Test a value on the interval $x < - 9$ to see if it makes the inequality true.

Choose a value on the interval $x < - 9$ and see if this value makes the original inequality true.

$x = - 12$

Replace $x$ with $- 12$ in the original inequality.

$\frac{2 (- 12)}{(- 12) + 9} - 1 \geq 0$

The left side $7$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 9 < x < 9$ to see if it makes the inequality true.

Choose a value on the interval $- 9 < x < 9$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{2 (0)}{(0) + 9} - 1 \geq 0$

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

False

Test a value on the interval $x > 9$ to see if it makes the inequality true.

Choose a value on the interval $x > 9$ and see if this value makes the original inequality true.

$x = 12$

Replace $x$ with $12$ in the original inequality.

$\frac{2 (12)}{(12) + 9} - 1 \geq 0$

The left side $0.\overline{142857}$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 9$ True

$- 9 < x < 9$ False

$x > 9$ True

$x < - 9$ True

$- 9 < x < 9$ False

$x > 9$ True

The solution consists of all of the true intervals.

$x < - 9$ or $x \geq 9$

Step 3

Set the denominator in $\frac{2 x}{x + 9}$ equal to $0$ to find where the expression is undefined.

$x + 9 = 0$

Step 4

Subtract $9$ from both sides of the equation.

$x = - 9$

Step 5

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

Interval Notation:

$(- \infty, - 9) \cup [9, \infty)$

Set-Builder Notation:

${x | x < - 9, x \geq 9}$

Step 6

Find the Domain square root of (2x)/(x+9)-1

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