Solve the Inequality for x Solve the inequality. (x-1)/(x+1)<=0

Solve the inequality. $\frac{x - 1}{x + 1} \leq 0$

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

$x - 1 = 0$

$x + 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

$x = 1$

$x = - 1$

Consolidate the solutions.

$x = 1, - 1$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$x > 1$

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 $x < - 1$ 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.

$x = - 4$

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

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

The left side $1.\overline{6}$ is greater than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 0$

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

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

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.

$x = 4$

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

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

The left side $0.6$ 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.

$x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

$x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

The solution consists of all of the true intervals.

$- 1 < x \leq 1$

The result can be shown in multiple forms.

Inequality Form:

$- 1 < x \leq 1$

Interval Notation:

$(- 1, 1]$

Solve the Inequality for x Solve the inequality. (x-1)/(x+1)<=0

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