Precalculus Examples

$\arccos (\frac{2 x}{1 - x^{2}})$

Step 1

Set the argument in $\arccos (\frac{2 x}{1 - x^{2}})$ greater than or equal to $- 1$ to find where the expression is defined.

$\frac{2 x}{1 - x^{2}} \geq - 1$

Step 2

Solve for $x$ .

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Step 2.1

Multiply both sides by $1 - x^{2}$ .

$\frac{2 x}{1 - x^{2}} (1 - x^{2}) = - (1 - x^{2})$

Step 2.2

Simplify.

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Step 2.2.1

Simplify the left side.

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Step 2.2.1.1

Cancel the common factor of $1 - x^{2}$ .

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Step 2.2.1.1.1

Cancel the common factor.

$\frac{2 x}{1 - x^{2}} (1 - x^{2}) = - (1 - x^{2})$

Step 2.2.1.1.2

Rewrite the expression.

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

Step 2.2.2

Simplify the right side.

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Step 2.2.2.1

Simplify $- (1 - x^{2})$ .

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Step 2.2.2.1.1

Apply the distributive property.

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

Step 2.2.2.1.2

Multiply $- 1$ by $1$ .

$2 x = - 1 - - x^{2}$

Step 2.2.2.1.3

Multiply $- - x^{2}$ .

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Step 2.2.2.1.3.1

Multiply $- 1$ by $- 1$ .

$2 x = - 1 + 1 x^{2}$

Step 2.2.2.1.3.2

Multiply $x^{2}$ by $1$ .

$2 x = - 1 + x^{2}$

Step 2.2.2.1.4

Reorder $- 1$ and $x^{2}$ .

$2 x = x^{2} - 1$

Step 2.3

Solve for $x$ .

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Step 2.3.1

Subtract $x^{2}$ from both sides of the equation.

$2 x - x^{2} = - 1$

Step 2.3.2

Add $1$ to both sides of the equation.

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

Step 2.3.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.3.4

Substitute the values $a = - 1$ , $b = 2$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (- 1 \cdot 1)}}{2 \cdot - 1}$

Step 2.3.5

Simplify.

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Step 2.3.5.1

Simplify the numerator.

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Step 2.3.5.1.1

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1 \cdot 1}}{2 \cdot - 1}$

Step 2.3.5.1.2

Multiply $- 4 \cdot - 1 \cdot 1$ .

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Step 2.3.5.1.2.1

Multiply $- 4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4 \cdot 1}}{2 \cdot - 1}$

Step 2.3.5.1.2.2

Multiply $4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4}}{2 \cdot - 1}$

Step 2.3.5.1.3

Add $4$ and $4$ .

$x = \frac{- 2 \pm \sqrt{8}}{2 \cdot - 1}$

Step 2.3.5.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 2.3.5.1.4.1

Factor $4$ out of $8$ .

$x = \frac{- 2 \pm \sqrt{4 (2)}}{2 \cdot - 1}$

Step 2.3.5.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot - 1}$

Step 2.3.5.1.5

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{2}}{2 \cdot - 1}$

Step 2.3.5.2

Multiply $2$ by $- 1$ .

$x = \frac{- 2 \pm 2 \sqrt{2}}{- 2}$

Step 2.3.5.3

Simplify $\frac{- 2 \pm 2 \sqrt{2}}{- 2}$ .

$x = 1 \pm \sqrt{2}$

Step 2.3.6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 2.3.6.1

Simplify the numerator.

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Step 2.3.6.1.1

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1 \cdot 1}}{2 \cdot - 1}$

Step 2.3.6.1.2

Multiply $- 4 \cdot - 1 \cdot 1$ .

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Step 2.3.6.1.2.1

Multiply $- 4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4 \cdot 1}}{2 \cdot - 1}$

Step 2.3.6.1.2.2

Multiply $4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4}}{2 \cdot - 1}$

Step 2.3.6.1.3

Add $4$ and $4$ .

$x = \frac{- 2 \pm \sqrt{8}}{2 \cdot - 1}$

Step 2.3.6.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 2.3.6.1.4.1

Factor $4$ out of $8$ .

$x = \frac{- 2 \pm \sqrt{4 (2)}}{2 \cdot - 1}$

Step 2.3.6.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot - 1}$

Step 2.3.6.1.5

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{2}}{2 \cdot - 1}$

Step 2.3.6.2

Multiply $2$ by $- 1$ .

$x = \frac{- 2 \pm 2 \sqrt{2}}{- 2}$

Step 2.3.6.3

Simplify $\frac{- 2 \pm 2 \sqrt{2}}{- 2}$ .

$x = 1 \pm \sqrt{2}$

Step 2.3.6.4

Change the $\pm$ to $+$ .

$x = 1 + \sqrt{2}$

Step 2.3.7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 2.3.7.1

Simplify the numerator.

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Step 2.3.7.1.1

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1 \cdot 1}}{2 \cdot - 1}$

Step 2.3.7.1.2

Multiply $- 4 \cdot - 1 \cdot 1$ .

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Step 2.3.7.1.2.1

Multiply $- 4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4 \cdot 1}}{2 \cdot - 1}$

Step 2.3.7.1.2.2

Multiply $4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4}}{2 \cdot - 1}$

Step 2.3.7.1.3

Add $4$ and $4$ .

$x = \frac{- 2 \pm \sqrt{8}}{2 \cdot - 1}$

Step 2.3.7.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 2.3.7.1.4.1

Factor $4$ out of $8$ .

$x = \frac{- 2 \pm \sqrt{4 (2)}}{2 \cdot - 1}$

Step 2.3.7.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot - 1}$

Step 2.3.7.1.5

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{2}}{2 \cdot - 1}$

Step 2.3.7.2

Multiply $2$ by $- 1$ .

$x = \frac{- 2 \pm 2 \sqrt{2}}{- 2}$

Step 2.3.7.3

Simplify $\frac{- 2 \pm 2 \sqrt{2}}{- 2}$ .

$x = 1 \pm \sqrt{2}$

Step 2.3.7.4

Change the $\pm$ to $-$ .

$x = 1 - \sqrt{2}$

Step 2.3.8

The final answer is the combination of both solutions.

$x = 1 + \sqrt{2}, 1 - \sqrt{2}$

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

Solve for $x$ .

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Step 2.4.2.1

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

$1 + x = 0$

$1 - x = 0$

Step 2.4.2.2

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

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Step 2.4.2.2.1

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

$1 + x = 0$

Step 2.4.2.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.4.2.3

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

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Step 2.4.2.3.1

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

$1 - x = 0$

Step 2.4.2.3.2

Solve $1 - x = 0$ for $x$ .

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Step 2.4.2.3.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 2.4.2.3.2.2

Divide each term in $- x = - 1$ by $- 1$ and simplify.

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Step 2.4.2.3.2.2.1

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Step 2.4.2.3.2.2.2

Simplify the left side.

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Step 2.4.2.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Step 2.4.2.3.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Step 2.4.2.3.2.2.3

Simplify the right side.

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Step 2.4.2.3.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 2.4.2.4

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

$x = - 1, 1$

Step 2.4.3

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

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

Step 2.5

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1 - \sqrt{2}$

$1 - \sqrt{2} < x < 1$

$1 < x < 1 + \sqrt{2}$

$x > 1 + \sqrt{2}$

Step 2.6

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

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Step 2.6.1

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

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Step 2.6.1.1

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

$x = - 4$

Step 2.6.1.2

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

$\frac{2 (- 4)}{1 - {(- 4)}^{2}} \geq - 1$

Step 2.6.1.3

The left side $0.5\overline{3}$ is greater than the right side $- 1$ , which means that the given statement is always true.

True

Step 2.6.2

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

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Step 2.6.2.1

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

$x = - 0.71$

Step 2.6.2.2

Replace $x$ with $- 0.71$ in the original inequality.

$\frac{2 (- 0.71)}{1 - {(- 0.71)}^{2}} \geq - 1$

Step 2.6.2.3

The left side $- 2.86348054$ is less than the right side $- 1$ , which means that the given statement is false.

False

Step 2.6.3

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

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Step 2.6.3.1

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

$x = 0$

Step 2.6.3.2

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

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

Step 2.6.3.3

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

True

Step 2.6.4

Test a value on the interval $1 < x < 1 + \sqrt{2}$ to see if it makes the inequality true.

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Step 2.6.4.1

Choose a value on the interval $1 < x < 1 + \sqrt{2}$ and see if this value makes the original inequality true.

$x = 2$

Step 2.6.4.2

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

$\frac{2 (2)}{1 - {(2)}^{2}} \geq - 1$

Step 2.6.4.3

The left side $- 1.\overline{3}$ is less than the right side $- 1$ , which means that the given statement is false.

False

Step 2.6.5

Test a value on the interval $x > 1 + \sqrt{2}$ to see if it makes the inequality true.

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Step 2.6.5.1

Choose a value on the interval $x > 1 + \sqrt{2}$ and see if this value makes the original inequality true.

$x = 5$

Step 2.6.5.2

Replace $x$ with $5$ in the original inequality.

$\frac{2 (5)}{1 - {(5)}^{2}} \geq - 1$

Step 2.6.5.3

The left side $- 0.41\overline{6}$ is greater than the right side $- 1$ , which means that the given statement is always true.

True

Step 2.6.6

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

$x < - 1$ True

$- 1 < x < 1 - \sqrt{2}$ False

$1 - \sqrt{2} < x < 1$ True

$1 < x < 1 + \sqrt{2}$ False

$x > 1 + \sqrt{2}$ True

$x < - 1$ True

$- 1 < x < 1 - \sqrt{2}$ False

$1 - \sqrt{2} < x < 1$ True

$1 < x < 1 + \sqrt{2}$ False

$x > 1 + \sqrt{2}$ True

Step 2.7

The solution consists of all of the true intervals.

$x < - 1$ or $1 - \sqrt{2} \leq x < 1$ or $x \geq 1 + \sqrt{2}$

Step 3

Set the argument in $\arccos (\frac{2 x}{1 - x^{2}})$ less than or equal to $1$ to find where the expression is defined.

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

Step 4

Solve for $x$ .

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Step 4.1

Multiply both sides by $1 - x^{2}$ .

$\frac{2 x}{1 - x^{2}} (1 - x^{2}) = 1 (1 - x^{2})$

Step 4.2

Simplify.

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Step 4.2.1

Simplify the left side.

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Step 4.2.1.1

Cancel the common factor of $1 - x^{2}$ .

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Step 4.2.1.1.1

Cancel the common factor.

$\frac{2 x}{1 - x^{2}} (1 - x^{2}) = 1 (1 - x^{2})$

Step 4.2.1.1.2

Rewrite the expression.

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

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Simplify $1 (1 - x^{2})$ .

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Step 4.2.2.1.1

Multiply $1 - x^{2}$ by $1$ .

$2 x = 1 - x^{2}$

Step 4.2.2.1.2

Reorder $1$ and $- x^{2}$ .

$2 x = - x^{2} + 1$

Step 4.3

Solve for $x$ .

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Step 4.3.1

Add $x^{2}$ to both sides of the equation.

$2 x + x^{2} = 1$

Step 4.3.2

Subtract $1$ from both sides of the equation.

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

Step 4.3.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4.3.4

Substitute the values $a = 1$ , $b = 2$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Step 4.3.5

Simplify.

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Step 4.3.5.1

Simplify the numerator.

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Step 4.3.5.1.1

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 4.3.5.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 4.3.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1}}{2 \cdot 1}$

Step 4.3.5.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Step 4.3.5.1.3

Add $4$ and $4$ .

$x = \frac{- 2 \pm \sqrt{8}}{2 \cdot 1}$

Step 4.3.5.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 4.3.5.1.4.1

Factor $4$ out of $8$ .

$x = \frac{- 2 \pm \sqrt{4 (2)}}{2 \cdot 1}$

Step 4.3.5.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Step 4.3.5.1.5

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{2}}{2 \cdot 1}$

Step 4.3.5.2

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 \sqrt{2}}{2}$

Step 4.3.5.3

Simplify $\frac{- 2 \pm 2 \sqrt{2}}{2}$ .

$x = - 1 \pm \sqrt{2}$

Step 4.3.6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 4.3.6.1

Simplify the numerator.

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Step 4.3.6.1.1

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 4.3.6.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 4.3.6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1}}{2 \cdot 1}$

Step 4.3.6.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Step 4.3.6.1.3

Add $4$ and $4$ .

$x = \frac{- 2 \pm \sqrt{8}}{2 \cdot 1}$

Step 4.3.6.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 4.3.6.1.4.1

Factor $4$ out of $8$ .

$x = \frac{- 2 \pm \sqrt{4 (2)}}{2 \cdot 1}$

Step 4.3.6.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Step 4.3.6.1.5

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{2}}{2 \cdot 1}$

Step 4.3.6.2

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 \sqrt{2}}{2}$

Step 4.3.6.3

Simplify $\frac{- 2 \pm 2 \sqrt{2}}{2}$ .

$x = - 1 \pm \sqrt{2}$

Step 4.3.6.4

Change the $\pm$ to $+$ .

$x = - 1 + \sqrt{2}$

Step 4.3.7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 4.3.7.1

Simplify the numerator.

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Step 4.3.7.1.1

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 4.3.7.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 4.3.7.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1}}{2 \cdot 1}$

Step 4.3.7.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Step 4.3.7.1.3

Add $4$ and $4$ .

$x = \frac{- 2 \pm \sqrt{8}}{2 \cdot 1}$

Step 4.3.7.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 4.3.7.1.4.1

Factor $4$ out of $8$ .

$x = \frac{- 2 \pm \sqrt{4 (2)}}{2 \cdot 1}$

Step 4.3.7.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Step 4.3.7.1.5

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{2}}{2 \cdot 1}$

Step 4.3.7.2

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 \sqrt{2}}{2}$

Step 4.3.7.3

Simplify $\frac{- 2 \pm 2 \sqrt{2}}{2}$ .

$x = - 1 \pm \sqrt{2}$

Step 4.3.7.4

Change the $\pm$ to $-$ .

$x = - 1 - \sqrt{2}$

Step 4.3.8

The final answer is the combination of both solutions.

$x = - 1 + \sqrt{2}, - 1 - \sqrt{2}$

Step 4.4

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

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Step 4.4.1

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

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

Step 4.4.2

Solve for $x$ .

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Step 4.4.2.1

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

$1 + x = 0$

$1 - x = 0$

Step 4.4.2.2

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

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Step 4.4.2.2.1

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

$1 + x = 0$

Step 4.4.2.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4.4.2.3

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

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Step 4.4.2.3.1

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

$1 - x = 0$

Step 4.4.2.3.2

Solve $1 - x = 0$ for $x$ .

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Step 4.4.2.3.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 4.4.2.3.2.2

Divide each term in $- x = - 1$ by $- 1$ and simplify.

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Step 4.4.2.3.2.2.1

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Step 4.4.2.3.2.2.2

Simplify the left side.

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Step 4.4.2.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Step 4.4.2.3.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Step 4.4.2.3.2.2.3

Simplify the right side.

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Step 4.4.2.3.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 4.4.2.4

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

$x = - 1, 1$

Step 4.4.3

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

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

Step 4.5

Use each root to create test intervals.

$x < - 1 - \sqrt{2}$

$- 1 - \sqrt{2} < x < - 1$

$- 1 < x < - 1 + \sqrt{2}$

$- 1 + \sqrt{2} < x < 1$

$x > 1$

Step 4.6

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

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Step 4.6.1

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

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Step 4.6.1.1

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

$x = - 5$

Step 4.6.1.2

Replace $x$ with $- 5$ in the original inequality.

$\frac{2 (- 5)}{1 - {(- 5)}^{2}} \leq 1$

Step 4.6.1.3

The left side $0.41\overline{6}$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 4.6.2

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

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Step 4.6.2.1

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

$x = - 2$

Step 4.6.2.2

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

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

Step 4.6.2.3

The left side $1.\overline{3}$ is greater than the right side $1$ , which means that the given statement is false.

False

Step 4.6.3

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

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Step 4.6.3.1

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

$x = 0$

Step 4.6.3.2

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

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

Step 4.6.3.3

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

True

Step 4.6.4

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

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Step 4.6.4.1

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

$x = 0.71$

Step 4.6.4.2

Replace $x$ with $0.71$ in the original inequality.

$\frac{2 (0.71)}{1 - {(0.71)}^{2}} \leq 1$

Step 4.6.4.3

The left side $2.86348054$ is greater than the right side $1$ , which means that the given statement is false.

False

Step 4.6.5

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

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Step 4.6.5.1

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

$x = 4$

Step 4.6.5.2

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

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

Step 4.6.5.3

The left side $- 0.5\overline{3}$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 4.6.6

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

$x < - 1 - \sqrt{2}$ True

$- 1 - \sqrt{2} < x < - 1$ False

$- 1 < x < - 1 + \sqrt{2}$ True

$- 1 + \sqrt{2} < x < 1$ False

$x > 1$ True

$x < - 1 - \sqrt{2}$ True

$- 1 - \sqrt{2} < x < - 1$ False

$- 1 < x < - 1 + \sqrt{2}$ True

$- 1 + \sqrt{2} < x < 1$ False

$x > 1$ True

Step 4.7

The solution consists of all of the true intervals.

$x \leq - 1 - \sqrt{2}$ or $- 1 < x \leq - 1 + \sqrt{2}$ or $x > 1$

Step 5

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

$1 - x^{2} = 0$

Step 6

Solve for $x$ .

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Step 6.1

Subtract $1$ from both sides of the equation.

$- x^{2} = - 1$

Step 6.2

Divide each term in $- x^{2} = - 1$ by $- 1$ and simplify.

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Step 6.2.1

Divide each term in $- x^{2} = - 1$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 1}{- 1}$

Step 6.2.2

Simplify the left side.

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Step 6.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{- 1}{- 1}$

Step 6.2.2.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 1}{- 1}$

Step 6.2.3

Simplify the right side.

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Step 6.2.3.1

Divide $- 1$ by $- 1$ .

$x^{2} = 1$

Step 6.3

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

$x = \pm \sqrt{1}$

Step 6.4

Any root of $1$ is $1$ .

$x = \pm 1$

Step 6.5

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

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Step 6.5.1

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

$x = 1$

Step 6.5.2

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

$x = - 1$

Step 6.5.3

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

$x = 1, - 1$

Step 7

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

Interval Notation:

$(- \infty, - 1 - \sqrt{2}] \cup [1 - \sqrt{2}, - 1 + \sqrt{2}] \cup [1 + \sqrt{2}, \infty)$

Set-Builder Notation:

${x | x \leq - 1 - \sqrt{2}, 1 - \sqrt{2} \leq x \leq - 1 + \sqrt{2}, x \geq 1 + \sqrt{2}}$

Step 8