Precalculus Examples

$\frac{2 x}{16 - x^{2}} < 0$

Step 1

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$2 x = 0$

$16 - x^{2} = 0$

Step 2

Divide each term in $2 x = 0$ by $2$ and simplify.

Tap for more steps...

Step 2.1

Divide each term in $2 x = 0$ by $2$ .

$\frac{2 x}{2} = \frac{0}{2}$

Step 2.2

Simplify the left side.

Tap for more steps...

Step 2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Step 2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{2}$

Step 2.3

Simplify the right side.

Tap for more steps...

Step 2.3.1

Divide $0$ by $2$ .

$x = 0$

Step 3

Subtract $16$ from both sides of the equation.

$- x^{2} = - 16$

Step 4

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

Tap for more steps...

Step 4.1

Divide each term in $- x^{2} = - 16$ by $- 1$ .

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

Step 4.2

Simplify the left side.

Tap for more steps...

Step 4.2.1

Dividing two negative values results in a positive value.

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

Step 4.2.2

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

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

Step 4.3

Simplify the right side.

Tap for more steps...

Step 4.3.1

Divide $- 16$ by $- 1$ .

$x^{2} = 16$

Step 5

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

$x = \pm \sqrt{16}$

Step 6

Simplify $\pm \sqrt{16}$ .

Tap for more steps...

Step 6.1

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

$x = \pm \sqrt{4^{2}}$

Step 6.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 4$

Step 7

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

Tap for more steps...

Step 7.1

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

$x = 4$

Step 7.2

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

$x = - 4$

Step 7.3

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

$x = 4, - 4$

Step 8

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 0$

$x = 4, - 4$

Step 9

Consolidate the solutions.

$x = 0, 4, - 4$

Step 10

Find the domain of $\frac{2 x}{16 - x^{2}}$ .

Tap for more steps...

Step 10.1

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

$16 - x^{2} = 0$

Step 10.2

Solve for $x$ .

Tap for more steps...

Step 10.2.1

Subtract $16$ from both sides of the equation.

$- x^{2} = - 16$

Step 10.2.2

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

Tap for more steps...

Step 10.2.2.1

Divide each term in $- x^{2} = - 16$ by $- 1$ .

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

Step 10.2.2.2

Simplify the left side.

Tap for more steps...

Step 10.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 10.2.2.2.2

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

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

Step 10.2.2.3

Simplify the right side.

Tap for more steps...

Step 10.2.2.3.1

Divide $- 16$ by $- 1$ .

$x^{2} = 16$

Step 10.2.3

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

$x = \pm \sqrt{16}$

Step 10.2.4

Simplify $\pm \sqrt{16}$ .

Tap for more steps...

Step 10.2.4.1

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

$x = \pm \sqrt{4^{2}}$

Step 10.2.4.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 4$

Step 10.2.5

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

Tap for more steps...

Step 10.2.5.1

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

$x = 4$

Step 10.2.5.2

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

$x = - 4$

Step 10.2.5.3

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

$x = 4, - 4$

Step 10.3

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

$(- \infty, - 4) \cup (- 4, 4) \cup (4, \infty)$

Step 11

Use each root to create test intervals.

$x < - 4$

$- 4 < x < 0$

$0 < x < 4$

$x > 4$

Step 12

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Tap for more steps...

Step 12.1

Test a value on the interval $x < - 4$ to see if it makes the inequality true.

Tap for more steps...

Step 12.1.1

Choose a value on the interval $x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

Step 12.1.2

Replace $x$ with $- 6$ in the original inequality.

$\frac{2 (- 6)}{16 - {(- 6)}^{2}} < 0$

Step 12.1.3

The left side $0.6$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 12.2

Test a value on the interval $- 4 < x < 0$ to see if it makes the inequality true.

Tap for more steps...

Step 12.2.1

Choose a value on the interval $- 4 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 12.2.2

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

$\frac{2 (- 2)}{16 - {(- 2)}^{2}} < 0$

Step 12.2.3

The left side $- 0.\overline{3}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 12.3

Test a value on the interval $0 < x < 4$ to see if it makes the inequality true.

Tap for more steps...

Step 12.3.1

Choose a value on the interval $0 < x < 4$ and see if this value makes the original inequality true.

$x = 2$

Step 12.3.2

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

$\frac{2 (2)}{16 - {(2)}^{2}} < 0$

Step 12.3.3

The left side $0.\overline{3}$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 12.4

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Tap for more steps...

Step 12.4.1

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

Step 12.4.2

Replace $x$ with $6$ in the original inequality.

$\frac{2 (6)}{16 - {(6)}^{2}} < 0$

Step 12.4.3

The left side $- 0.6$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 12.5

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 4$ False

$- 4 < x < 0$ True

$0 < x < 4$ False

$x > 4$ True

$x < - 4$ False

$- 4 < x < 0$ True

$0 < x < 4$ False

$x > 4$ True

Step 13

The solution consists of all of the true intervals.

$- 4 < x < 0$ or $x > 4$

Step 14

Convert the inequality to interval notation.

$(- 4, 0) \cup (4, \infty)$

Step 15