Precalculus Examples

f (x) = \sqrt{9 - x^{2}}

$f (x) = \sqrt{9 - x^{2}}$

Step 1

Set the radicand in

\sqrt{9 - x^{2}}

$\sqrt{9 - x^{2}}$ greater than or equal to

0

$0$ to find where the expression is defined.

9 - x^{2} \geq 0

$9 - x^{2} \geq 0$

Step 2

Solve for

x

$x$ .

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Step 2.1

Subtract

9

$9$ from both sides of the inequality.

- x^{2} \geq - 9

$- x^{2} \geq - 9$

Step 2.2

Divide each term in

- x^{2} \geq - 9

$- x^{2} \geq - 9$ by

- 1

$- 1$ and simplify.

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Step 2.2.1

Divide each term in

- x^{2} \geq - 9

$- x^{2} \geq - 9$ by

- 1

$- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

\frac{- x^{2}}{- 1} \leq \frac{- 9}{- 1}

$\frac{- x^{2}}{- 1} \leq \frac{- 9}{- 1}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Dividing two negative values results in a positive value.

\frac{x^{2}}{1} \leq \frac{- 9}{- 1}

$\frac{x^{2}}{1} \leq \frac{- 9}{- 1}$

Step 2.2.2.2

Divide

x^{2}

$x^{2}$ by

1

$1$ .

x^{2} \leq \frac{- 9}{- 1}

$x^{2} \leq \frac{- 9}{- 1}$

x^{2} \leq \frac{- 9}{- 1}

$x^{2} \leq \frac{- 9}{- 1}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide

- 9

$- 9$ by

- 1

$- 1$ .

x^{2} \leq 9

$x^{2} \leq 9$

x^{2} \leq 9

$x^{2} \leq 9$

x^{2} \leq 9

$x^{2} \leq 9$

Step 2.3

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

\sqrt{x^{2}} \leq \sqrt{9}

$\sqrt{x^{2}} \leq \sqrt{9}$

Step 2.4

Simplify the equation.

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Step 2.4.1

Simplify the left side.

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Step 2.4.1.1

Pull terms out from under the radical.

| x | \leq \sqrt{9}

$| x | \leq \sqrt{9}$

| x | \leq \sqrt{9}

$| x | \leq \sqrt{9}$

Step 2.4.2

Simplify the right side.

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Step 2.4.2.1

Simplify

\sqrt{9}

$\sqrt{9}$ .

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Step 2.4.2.1.1

Rewrite

9

$9$ as

3^{2}

$3^{2}$ .

| x | \leq \sqrt{3^{2}}

$| x | \leq \sqrt{3^{2}}$

Step 2.4.2.1.2

Pull terms out from under the radical.

| x | \leq | 3 |

$| x | \leq | 3 |$

Step 2.4.2.1.3

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

3

$3$ is

3

$3$ .

| x | \leq 3

$| x | \leq 3$

| x | \leq 3

$| x | \leq 3$

| x | \leq 3

$| x | \leq 3$

| x | \leq 3

$| x | \leq 3$

Step 2.5

Write

| x | \leq 3

$| x | \leq 3$ as a piecewise.

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Step 2.5.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

x \geq 0

$x \geq 0$

Step 2.5.2

In the piece where

x

$x$ is non-negative, remove the absolute value.

x \leq 3

$x \leq 3$

Step 2.5.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

x < 0

$x < 0$

Step 2.5.4

In the piece where

x

$x$ is negative, remove the absolute value and multiply by

- 1

$- 1$ .

- x \leq 3

$- x \leq 3$

Step 2.5.5

Write as a piecewise.

{\begin{matrix} x \leq 3 & x \geq 0 - x \leq 3 & x < 0 \end{matrix}

${\begin{cases} x \leq 3 & x \geq 0 \\ - x \leq 3 & x < 0 \end{cases}$

{\begin{matrix} x \leq 3 & x \geq 0 - x \leq 3 & x < 0 \end{matrix}

${\begin{cases} x \leq 3 & x \geq 0 \\ - x \leq 3 & x < 0 \end{cases}$

Step 2.6

Find the intersection of

$x \leq 3$ and

$x \geq 0$ .

$0 \leq x \leq 3$

Step 2.7

Solve

$- x \leq 3$ when

$x < 0$ .

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Step 2.7.1

Divide each term in

$- x \leq 3$ by

$- 1$ and simplify.

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Step 2.7.1.1

Divide each term in

$- x \leq 3$ by

$- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Step 2.7.1.2

Simplify the left side.

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Step 2.7.1.2.1

Dividing two negative values results in a positive value.

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

Step 2.7.1.2.2

Divide

$x$ by

$1$ .

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

Step 2.7.1.3

Simplify the right side.

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Step 2.7.1.3.1

Divide

$3$ by

$- 1$ .

$x \geq - 3$

Step 2.7.2

Find the intersection of

$x \geq - 3$ and

$x < 0$ .

$- 3 \leq x < 0$

Step 2.8

Find the union of the solutions.

$- 3 \leq x \leq 3$

Step 3

The domain is all values of

$x$ that make the expression defined.

Interval Notation:

$[- 3, 3]$

Set-Builder Notation:

${x | - 3 \leq x \leq 3}$

Step 4