Solve for x (x^3-x)/(x^2+1)>0

$\frac{x^{3} - x}{x^{2} + 1} > 0$

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

$x^{3} - x = 0$

$x^{2} + 1 = 0$

Factor the left side of the equation.

Factor $x$ out of $x^{3} - x$ .

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} - x = 0$

Factor $x$ out of $- x$ .

$x \cdot x^{2} + x \cdot - 1 = 0$

Factor $x$ out of $x \cdot x^{2} + x \cdot - 1$ .

$x (x^{2} - 1) = 0$

Rewrite $1$ as $1^{2}$ .

$x (x^{2} - 1^{2}) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$x ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$x (x + 1) (x - 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 1 = 0$

$x - 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 1$ equal to $0$ and solve for $x$ .

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $x (x + 1) (x - 1) = 0$ true.

$x = 0, - 1, 1$

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

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

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

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

$x = \pm i$

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.

$x = i, - i$

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

$x = 0, - 1, 1$

$x = i, - i$

Consolidate the solutions.

$x = 0, - 1, 1$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 0$

$0 < x < 1$

$x > 1$

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.

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

The left side $- 3.52941176$ is not greater than the right side $0$ , which means that the given statement is false.

False

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

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

$x = - 0.5$

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

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

The left side $0.3$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 0.5$

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

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

The left side $- 0.3$ is not greater than the right side $0$ , which means that the given statement is false.

False

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.

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

The left side $3.52941176$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 1$ False

$- 1 < x < 0$ True

$0 < x < 1$ False

$x > 1$ True

$x < - 1$ False

$- 1 < x < 0$ True

$0 < x < 1$ False

$x > 1$ True

The solution consists of all of the true intervals.

$- 1 < x < 0$ or $x > 1$

The result can be shown in multiple forms.

Inequality Form:

$- 1 < x < 0 or x > 1$

Interval Notation:

$(- 1, 0) \cup (1, \infty)$

Solve for x (x^3-x)/(x^2+1)>0

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