Precalculus Examples

$f (x) = \sqrt{25 - x^{2}}$

Step 1

Set the radicand in $\sqrt{25 - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$25 - x^{2} \geq 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Subtract $25$ from both sides of the inequality.

$- x^{2} \geq - 25$

Step 2.2

Divide each term in $- x^{2} \geq - 25$ by $- 1$ and simplify.

Tap for more steps...

Step 2.2.1

Divide each term in $- x^{2} \geq - 25$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 25}{- 1}$

Step 2.2.2

Simplify the left side.

Tap for more steps...

Step 2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 25}{- 1}$

Step 2.2.2.2

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{- 25}{- 1}$

Step 2.2.3

Simplify the right side.

Tap for more steps...

Step 2.2.3.1

Divide $- 25$ by $- 1$ .

$x^{2} \leq 25$

Step 2.3

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

$\sqrt{x^{2}} \leq \sqrt{25}$

Step 2.4

Simplify the equation.

Tap for more steps...

Step 2.4.1

Simplify the left side.

Tap for more steps...

Step 2.4.1.1

Pull terms out from under the radical.

$| x | \leq \sqrt{25}$

Step 2.4.2

Simplify the right side.

Tap for more steps...

Step 2.4.2.1

Simplify $\sqrt{25}$ .

Tap for more steps...

Step 2.4.2.1.1

Rewrite $25$ as $5^{2}$ .

$| x | \leq \sqrt{5^{2}}$

Step 2.4.2.1.2

Pull terms out from under the radical.

$| x | \leq | 5 |$

Step 2.4.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $5$ is $5$ .

$| x | \leq 5$

Step 2.5

Write $| x | \leq 5$ as a piecewise.

Tap for more steps...

Step 2.5.1

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

$x \geq 0$

Step 2.5.2

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

$x \leq 5$

Step 2.5.3

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

$x < 0$

Step 2.5.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 5$

Step 2.5.5

Write as a piecewise.

${\begin{cases} x \leq 5 & x \geq 0 \\ - x \leq 5 & x < 0 \end{cases}$

Step 2.6

Find the intersection of $x \leq 5$ and $x \geq 0$ .

$0 \leq x \leq 5$

Step 2.7

Solve $- x \leq 5$ when $x < 0$ .

Tap for more steps...

Step 2.7.1

Divide each term in $- x \leq 5$ by $- 1$ and simplify.

Tap for more steps...

Step 2.7.1.1

Divide each term in $- x \leq 5$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{5}{- 1}$

Step 2.7.1.2

Simplify the left side.

Tap for more steps...

Step 2.7.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{5}{- 1}$

Step 2.7.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{5}{- 1}$

Step 2.7.1.3

Simplify the right side.

Tap for more steps...

Step 2.7.1.3.1

Divide $5$ by $- 1$ .

$x \geq - 5$

Step 2.7.2

Find the intersection of $x \geq - 5$ and $x < 0$ .

$- 5 \leq x < 0$

Step 2.8

Find the union of the solutions.

$- 5 \leq x \leq 5$

Step 3

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

Interval Notation:

$[- 5, 5]$

Set-Builder Notation:

${x | - 5 \leq x \leq 5}$

Step 4