Finite Math Examples

$\ln (\frac{x^{2}}{{(x + 1)}^{3}})$

Step 1

Set the denominator in $\frac{x^{2}}{{(x + 1)}^{3}}$ equal to $0$ to find where the expression is undefined.

${(x + 1)}^{3} = 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3

Set the argument in $\ln (\frac{x^{2}}{{(x + 1)}^{3}})$ less than or equal to $0$ to find where the expression is undefined.

$\frac{x^{2}}{{(x + 1)}^{3}} \leq 0$

Step 4

Solve for $x$ .

Tap for more steps...

Step 4.1

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

$x^{2} = 0$

${(x + 1)}^{3} = 0$

Step 4.2

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

$x = \pm \sqrt{0}$

Step 4.3

Simplify $\pm \sqrt{0}$ .

Tap for more steps...

Step 4.3.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 4.3.2

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

$x = \pm 0$

Step 4.3.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 4.4

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 4.5

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4.6

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

$x = 0$

$x = - 1$

Step 4.7

Consolidate the solutions.

$x = 0, - 1$

Step 4.8

Find the domain of $\frac{x^{2}}{{(x + 1)}^{3}}$ .

Tap for more steps...

Step 4.8.1

Set the denominator in $\frac{x^{2}}{{(x + 1)}^{3}}$ equal to $0$ to find where the expression is undefined.

${(x + 1)}^{3} = 0$

Step 4.8.2

Solve for $x$ .

Tap for more steps...

Step 4.8.2.1

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 4.8.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4.8.3

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

$(- \infty, - 1) \cup (- 1, \infty)$

Step 4.9

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 0$

$x > 0$

Step 4.10

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 4.10.1

Test a value on the interval $x < - 1$ to see if it makes the inequality true.

Tap for more steps...

Step 4.10.1.1

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Step 4.10.1.2

Replace $x$ with $- 4$ in the original inequality.

$\frac{{(- 4)}^{2}}{{((- 4) + 1)}^{3}} \leq 0$

Step 4.10.1.3

The left side $- 0.\overline{592}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 4.10.2

Test a value on the interval $- 1 < x < 0$ to see if it makes the inequality true.

Tap for more steps...

Step 4.10.2.1

Choose a value on the interval $- 1 < x < 0$ and see if this value makes the original inequality true.

$x = - 0.5$

Step 4.10.2.2

Replace $x$ with $- 0.5$ in the original inequality.

$\frac{{(- 0.5)}^{2}}{{((- 0.5) + 1)}^{3}} \leq 0$

Step 4.10.2.3

The left side $2$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 4.10.3

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

Tap for more steps...

Step 4.10.3.1

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

$x = 2$

Step 4.10.3.2

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

$\frac{{(2)}^{2}}{{((2) + 1)}^{3}} \leq 0$

Step 4.10.3.3

The left side $0.\overline{148}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 4.10.4

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

$x < - 1$ True

$- 1 < x < 0$ False

$x > 0$ False

$x < - 1$ True

$- 1 < x < 0$ False

$x > 0$ False

Step 4.11

The solution consists of all of the true intervals.

$x < - 1$ or $x = 0$

Step 5

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 1, x = 0$

Step 6