Precalculus Examples

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

Step 1

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

$x + 1 = 0$

Step 2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3

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

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

Step 4

Solve for $x$ .

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Step 4.1

Factor using the perfect square rule.

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Step 4.1.1

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

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

Step 4.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 4.1.3

Rewrite the polynomial.

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

Step 4.1.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

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

Step 4.2

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

$x + 1 = 0$

Step 4.3

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

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

$x^{2} - 1 = 0$

Step 6

Solve for $x$ .

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Step 6.1

Add $1$ to both sides of the equation.

$x^{2} = 1$

Step 6.2

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

$x = \pm \sqrt{1}$

Step 6.3

Any root of $1$ is $1$ .

$x = \pm 1$

Step 6.4

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

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Step 6.4.1

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

$x = 1$

Step 6.4.2

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

$x = - 1$

Step 6.4.3

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:

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

Set-Builder Notation:

${x | x \neq 1, - 1}$

Step 8