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

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

Step 1

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

$x^{2} - 1 = 0$

$x^{2} + 1 = 0$

Step 2

Add $1$ to both sides of the equation.

$x^{2} = 1$

Step 3

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

$x = \pm \sqrt{1}$

Step 4

Any root of $1$ is $1$ .

$x = \pm 1$

Step 5

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

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

$x = 1$

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

$x = - 1$

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

$x = 1, - 1$

Step 6

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 7

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

$x = \pm \sqrt{- 1}$

Step 8

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

Step 9

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

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

$x = i$

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

$x = - i$

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

$x = i, - i$

Step 10

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

$x = 1, - 1$

$x = i, - i$

Step 11

Consolidate the solutions.

$x = 1, - 1$

Step 12

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$x > 1$

Step 13

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.

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

The left side $0.88235294$ 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.

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

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.

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

The left side $0.88235294$ 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

Step 14

The solution consists of all of the true intervals.

$- 1 \leq x \leq 1$

Step 15

The result can be shown in multiple forms.

Inequality Form:

$- 1 \leq x \leq 1$

Interval Notation:

$[- 1, 1]$

Step 16

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

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