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Examples

\sqrt{3 x + 9}

$\sqrt{3 x + 9}$

Step 1

Set the radicand in

\sqrt{3 x + 9}

$\sqrt{3 x + 9}$ greater than or equal to

0

$0$ to find where the expression is defined.

3 x + 9 \geq 0

$3 x + 9 \geq 0$

Step 2

Solve for

x

$x$ .

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Step 2.1

Subtract

9

$9$ from both sides of the inequality.

3 x \geq - 9

$3 x \geq - 9$

Step 2.2

Divide each term in

3 x \geq - 9

$3 x \geq - 9$ by

3

$3$ and simplify.

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Step 2.2.1

Divide each term in

3 x \geq - 9

$3 x \geq - 9$ by

3

$3$ .

\frac{3 x}{3} \geq \frac{- 9}{3}

$\frac{3 x}{3} \geq \frac{- 9}{3}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of

3

$3$ .

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Step 2.2.2.1.1

Cancel the common factor.

\frac{3 x}{3} \geq \frac{- 9}{3}

$\frac{3 x}{3} \geq \frac{- 9}{3}$

Step 2.2.2.1.2

Divide

x

$x$ by

1

$1$ .

x \geq \frac{- 9}{3}

$x \geq \frac{- 9}{3}$

x \geq \frac{- 9}{3}

$x \geq \frac{- 9}{3}$

x \geq \frac{- 9}{3}

$x \geq \frac{- 9}{3}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide

- 9

$- 9$ by

3

$3$ .

x \geq - 3

$x \geq - 3$

x \geq - 3

$x \geq - 3$

x \geq - 3

$x \geq - 3$

x \geq - 3

$x \geq - 3$

Step 3

The domain is all values of

x

$x$ that make the expression defined.

Interval Notation:

[- 3, \infty)

$[- 3, \infty)$

Set-Builder Notation:

{x | x \geq - 3}

${x | x \geq - 3}$

Step 4

Enter YOUR Problem

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[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$