Precalculus Examples

$f (x) = \frac{3 x - 2}{x - 3} + \frac{\sqrt{16 - x^{2}} - 4}{2 - | x - 3 |}$

Step 1

Set the radicand in $\sqrt{16 - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$16 - x^{2} \geq 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Subtract $16$ from both sides of the inequality.

$- x^{2} \geq - 16$

Step 2.2

Divide each term in $- x^{2} \geq - 16$ by $- 1$ and simplify.

Tap for more steps...

Step 2.2.1

Divide each term in $- x^{2} \geq - 16$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 16}{- 1}$

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.

$\frac{x^{2}}{1} \leq \frac{- 16}{- 1}$

Step 2.2.2.2

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

$x^{2} \leq \frac{- 16}{- 1}$

Step 2.2.3

Simplify the right side.

Tap for more steps...

Step 2.2.3.1

Divide $- 16$ by $- 1$ .

$x^{2} \leq 16$

Step 2.3

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

$\sqrt{x^{2}} \leq \sqrt{16}$

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.

$| x | \leq \sqrt{16}$

Step 2.4.2

Simplify the right side.

Tap for more steps...

Step 2.4.2.1

Simplify $\sqrt{16}$ .

Tap for more steps...

Step 2.4.2.1.1

Rewrite $16$ as $4^{2}$ .

$| x | \leq \sqrt{4^{2}}$

Step 2.4.2.1.2

Pull terms out from under the radical.

$| x | \leq | 4 |$

Step 2.4.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$| x | \leq 4$

Step 2.5

Write $| x | \leq 4$ 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.

$x \leq 4$

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 $- 1$ .

$- x \leq 4$

Step 2.5.5

Write as a piecewise.

${\begin{cases} x \leq 4 & x \geq 0 \\ - x \leq 4 & x < 0 \end{cases}$

Step 2.6

Find the intersection of $x \leq 4$ and $x \geq 0$ .

$0 \leq x \leq 4$

Step 2.7

Solve $- x \leq 4$ when $x < 0$ .

Tap for more steps...

Step 2.7.1

Divide each term in $- x \leq 4$ by $- 1$ and simplify.

Tap for more steps...

Step 2.7.1.1

Divide each term in $- x \leq 4$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{4}{- 1}$

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.

$\frac{x}{1} \geq \frac{4}{- 1}$

Step 2.7.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{4}{- 1}$

Step 2.7.1.3

Simplify the right side.

Tap for more steps...

Step 2.7.1.3.1

Divide $4$ by $- 1$ .

$x \geq - 4$

Step 2.7.2

Find the intersection of $x \geq - 4$ and $x < 0$ .

$- 4 \leq x < 0$

Step 2.8

Find the union of the solutions.

$- 4 \leq x \leq 4$

Step 3

Set the denominator in $\frac{3 x - 2}{x - 3}$ equal to $0$ to find where the expression is undefined.

$x - 3 = 0$

Step 4

Add $3$ to both sides of the equation.

$x = 3$

Step 5

Set the denominator in $\frac{\sqrt{16 - x^{2}} - 4}{2 - | x - 3 |}$ equal to $0$ to find where the expression is undefined.

$2 - | x - 3 | = 0$

Step 6

Solve for $x$ .

Tap for more steps...

Step 6.1

Subtract $2$ from both sides of the equation.

$- | x - 3 | = - 2$

Step 6.2

Divide each term in $- | x - 3 | = - 2$ by $- 1$ and simplify.

Tap for more steps...

Step 6.2.1

Divide each term in $- | x - 3 | = - 2$ by $- 1$ .

$\frac{- | x - 3 |}{- 1} = \frac{- 2}{- 1}$

Step 6.2.2

Simplify the left side.

Tap for more steps...

Step 6.2.2.1

Dividing two negative values results in a positive value.

$\frac{| x - 3 |}{1} = \frac{- 2}{- 1}$

Step 6.2.2.2

Divide $| x - 3 |$ by $1$ .

$| x - 3 | = \frac{- 2}{- 1}$

Step 6.2.3

Simplify the right side.

Tap for more steps...

Step 6.2.3.1

Divide $- 2$ by $- 1$ .

$| x - 3 | = 2$

Step 6.3

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x - 3 = \pm 2$

Step 6.4

The complete solution is the result of both the positive and negative portions of the solution.

Tap for more steps...

Step 6.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x - 3 = 2$

Step 6.4.2

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

Tap for more steps...

Step 6.4.2.1

Add $3$ to both sides of the equation.

$x = 2 + 3$

Step 6.4.2.2

Add $2$ and $3$ .

$x = 5$

Step 6.4.3

Next, use the negative value of the $\pm$ to find the second solution.

$x - 3 = - 2$

Step 6.4.4

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

Tap for more steps...

Step 6.4.4.1

Add $3$ to both sides of the equation.

$x = - 2 + 3$

Step 6.4.4.2

Add $- 2$ and $3$ .

$x = 1$

Step 6.4.5

The complete solution is the result of both the positive and negative portions of the solution.

$x = 5, 1$

Step 7

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

Interval Notation:

$[- 4, 1) \cup (1, 3) \cup (3, 4]$

Set-Builder Notation:

${x | - 4 \leq x \leq 4, x \neq 1, 3}$

Step 8