Precalculus Examples

$\sqrt{\sqrt{x + 1} + 1}$

Step 1

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

$x + 1 \geq 0$

Step 2

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

Step 3

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

$\sqrt{x + 1} + 1 \geq 0$

Step 4

Solve for $x$ .

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Step 4.1

Subtract $1$ from both sides of the inequality.

$\sqrt{x + 1} \geq - 1$

Step 4.2

To remove the radical on the left side of the inequality, square both sides of the inequality.

${\sqrt{x + 1}}^{2} \geq {(- 1)}^{2}$

Step 4.3

Simplify each side of the inequality.

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Step 4.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 1}$ as ${(x + 1)}^{\frac{1}{2}}$ .

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

Step 4.3.2

Simplify the left side.

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Step 4.3.2.1

Simplify ${({(x + 1)}^{\frac{1}{2}})}^{2}$ .

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Step 4.3.2.1.1

Multiply the exponents in ${({(x + 1)}^{\frac{1}{2}})}^{2}$ .

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Step 4.3.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x + 1)}^{\frac{1}{2} \cdot 2} \geq {(- 1)}^{2}$

Step 4.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 4.3.2.1.1.2.1

Cancel the common factor.

${(x + 1)}^{\frac{1}{2} \cdot 2} \geq {(- 1)}^{2}$

Step 4.3.2.1.1.2.2

Rewrite the expression.

${(x + 1)}^{1} \geq {(- 1)}^{2}$

Step 4.3.2.1.2

Simplify.

$x + 1 \geq {(- 1)}^{2}$

Step 4.3.3

Simplify the right side.

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Step 4.3.3.1

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

$x + 1 \geq 1$

Step 4.4

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

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Step 4.4.1

Subtract $1$ from both sides of the inequality.

$x \geq 1 - 1$

Step 4.4.2

Subtract $1$ from $1$ .

$x \geq 0$

Step 4.5

Find the domain of $\sqrt{x + 1} + 1$ .

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Step 4.5.1

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

$x + 1 \geq 0$

Step 4.5.2

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

Step 4.5.3

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

$[- 1, \infty)$

Step 4.6

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 0$

$x > 0$

Step 4.7

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 4.7.1

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

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Step 4.7.1.1

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

$x = - 4$

Step 4.7.1.2

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

$\sqrt{(- 4) + 1} + 1 \geq 0$

Step 4.7.1.3

The left side is not equal to the right side, which means that the given statement is false.

False

Step 4.7.2

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

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Step 4.7.2.1

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

$x = - 0.5$

Step 4.7.2.2

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

$\sqrt{(- 0.5) + 1} + 1 \geq 0$

Step 4.7.2.3

The left side $1.70710678$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 4.7.3

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

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Step 4.7.3.1

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

$x = 2$

Step 4.7.3.2

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

$\sqrt{(2) + 1} + 1 \geq 0$

Step 4.7.3.3

The left side $2.7320508$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 4.7.4

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 4.8

The solution consists of all of the true intervals.

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

Step 4.9

Combine the intervals.

$x \geq - 1$

Step 5

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

Interval Notation:

$[- 1, \infty)$

Set-Builder Notation:

${x | x \geq - 1}$

Step 6