Precalculus Examples

$\sqrt{\frac{x^{2}}{x + 1}}$

Step 1

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

$\frac{x^{2}}{x + 1} \geq 0$

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.

$x^{2} = 0$

$x + 1 = 0$

Step 2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Step 2.3

Simplify $\pm \sqrt{0}$ .

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Step 2.3.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 2.3.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Step 2.3.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 2.4

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.5

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

$x = 0$

$x = - 1$

Step 2.6

Consolidate the solutions.

$x = 0, - 1$

Step 2.7

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

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Step 2.7.1

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

$x + 1 = 0$

Step 2.7.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.7.3

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

$(- \infty, - 1) \cup (- 1, \infty)$

Step 2.8

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 0$

$x > 0$

Step 2.9

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 2.9.1

Test a value on the interval $x < - 1$ to see if it makes the inequality true.

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Step 2.9.1.1

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Step 2.9.1.2

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

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

Step 2.9.1.3

The left side $- 5.\overline{3}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.9.2

Test a value on the interval $- 1 < x < 0$ to see if it makes the inequality true.

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Step 2.9.2.1

Choose a value on the interval $- 1 < x < 0$ and see if this value makes the original inequality true.

$x = - 0.5$

Step 2.9.2.2

Replace $x$ with $- 0.5$ in the original inequality.

$\frac{{(- 0.5)}^{2}}{(- 0.5) + 1} \geq 0$

Step 2.9.2.3

The left side $0.5$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.9.3

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

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Step 2.9.3.1

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

$x = 2$

Step 2.9.3.2

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

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

Step 2.9.3.3

The left side $1.\overline{3}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.9.4

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

$x < - 1$ False

$- 1 < x < 0$ True

$x > 0$ True

$x < - 1$ False

$- 1 < x < 0$ True

$x > 0$ True

Step 2.10

The solution consists of all of the true intervals.

$- 1 < x \leq 0$ or $x \geq 0$

Step 2.11

Combine the intervals.

$x > - 1$

Step 3

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

$x + 1 = 0$

Step 4

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

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

Interval Notation:

$(- 1, \infty)$

Set-Builder Notation:

${x | x > - 1}$

Step 6