Solve the Inequality for x (x+1)/(x+2)<=0

$\frac{x + 1}{x + 2} \leq 0$

Step 1

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

$x + 1 = 0$

$x + 2 = 0$

Step 2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 4

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

$x = - 1$

$x = - 2$

Step 5

Consolidate the solutions.

$x = - 1, - 2$

Step 6

Find the domain of $\frac{x + 1}{x + 2}$ .

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

$(- \infty, - 2) \cup (- 2, \infty)$

Step 7

Use each root to create test intervals.

$x < - 2$

$- 2 < x < - 1$

$x > - 1$

Step 8

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

Test a value on the interval $x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

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

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

The left side $1.5$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 2 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < - 1$ and see if this value makes the original inequality true.

$x = - 1.5$

Replace $x$ with $- 1.5$ in the original inequality.

$\frac{(- 1.5) + 1}{(- 1.5) + 2} \leq 0$

The left side $- 1$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 0$

Replace $x$ with $0$ in the original inequality.

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

The left side $0.5$ is greater than the right side $0$ , which means that the given statement is false.

False

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

$x < - 2$ False

$- 2 < x < - 1$ True

$x > - 1$ False

$x < - 2$ False

$- 2 < x < - 1$ True

$x > - 1$ False

Step 9

The solution consists of all of the true intervals.

$- 2 < x \leq - 1$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$- 2 < x \leq - 1$

Interval Notation:

$(- 2, - 1]$

Step 11

Solve the Inequality for x (x+1)/(x+2)<=0

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