Pre-Algebra Examples

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

Step 1

Find the domain for $f (x) = \frac{(x + 1) (x - 1)}{| x - 1 |}$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the absolute value function.

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Step 1.1

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

$| x - 1 | = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x - 1 = \pm 0$

Step 1.2.2

Plus or minus $0$ is $0$ .

$x - 1 = 0$

Step 1.2.3

Add $1$ to both sides of the equation.

$x = 1$

Step 1.3

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

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 1}$

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 1}$

Step 2

For each $x$ value, there is one $y$ value. Select a few $x$ values from the domain. It would be more useful to select the values so that they are around the $x$ value of the absolute value vertex.

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Step 2.1

Substitute the $x$ value $- 2$ into $f (x) = \frac{(x + 1) (x - 1)}{| x - 1 |}$ . In this case, the point is $(- 2, 1)$ .

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Step 2.1.1

Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = \frac{((- 2) + 1) ((- 2) - 1)}{| (- 2) - 1 |}$

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Simplify the numerator.

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Step 2.1.2.1.1

Add $- 2$ and $1$ .

$f (- 2) = \frac{- 1 (- 2 - 1)}{| - 2 - 1 |}$

Step 2.1.2.1.2

Subtract $1$ from $- 2$ .

$f (- 2) = \frac{- 1 \cdot - 3}{| - 2 - 1 |}$

Step 2.1.2.2

Simplify the denominator.

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Step 2.1.2.2.1

Subtract $1$ from $- 2$ .

$f (- 2) = \frac{- 1 \cdot - 3}{| - 3 |}$

Step 2.1.2.2.2

The absolute value is the distance between a number and zero. The distance between $- 3$ and $0$ is $3$ .

$f (- 2) = \frac{- 1 \cdot - 3}{3}$

Step 2.1.2.3

Simplify the expression.

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Step 2.1.2.3.1

Multiply $- 1$ by $- 3$ .

$f (- 2) = \frac{3}{3}$

Step 2.1.2.3.2

Divide $3$ by $3$ .

$f (- 2) = 1$

Step 2.1.2.4

The final answer is $1$ .

$y = 1$

Step 2.2

Substitute the $x$ value $- 1$ into $f (x) = \frac{(x + 1) (x - 1)}{| x - 1 |}$ . In this case, the point is $(- 1, 0)$ .

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Step 2.2.1

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = \frac{((- 1) + 1) ((- 1) - 1)}{| (- 1) - 1 |}$

Step 2.2.2

Simplify the result.

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Step 2.2.2.1

Simplify the numerator.

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Step 2.2.2.1.1

Add $- 1$ and $1$ .

$f (- 1) = \frac{0 (- 1 - 1)}{| - 1 - 1 |}$

Step 2.2.2.1.2

Subtract $1$ from $- 1$ .

$f (- 1) = \frac{0 \cdot - 2}{| - 1 - 1 |}$

Step 2.2.2.2

Simplify the denominator.

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Step 2.2.2.2.1

Subtract $1$ from $- 1$ .

$f (- 1) = \frac{0 \cdot - 2}{| - 2 |}$

Step 2.2.2.2.2

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$f (- 1) = \frac{0 \cdot - 2}{2}$

Step 2.2.2.3

Simplify the expression.

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Step 2.2.2.3.1

Multiply $0$ by $- 2$ .

$f (- 1) = \frac{0}{2}$

Step 2.2.2.3.2

Divide $0$ by $2$ .

$f (- 1) = 0$

Step 2.2.2.4

The final answer is $0$ .

$y = 0$

Step 2.3

Substitute the $x$ value $0$ into $f (x) = \frac{(x + 1) (x - 1)}{| x - 1 |}$ . In this case, the point is $(0, - 1)$ .

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Step 2.3.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = \frac{((0) + 1) ((0) - 1)}{| (0) - 1 |}$

Step 2.3.2

Simplify the result.

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Step 2.3.2.1

Simplify the numerator.

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Step 2.3.2.1.1

Rewrite $0$ as $- 1 (0)$ .

$f (0) = \frac{(- 1 \cdot 0 + 1) (0 - 1)}{| 0 - 1 |}$

Step 2.3.2.1.2

Rewrite $1$ as $- 1 (- 1)$ .

$f (0) = \frac{(- 1 \cdot 0 - 1 \cdot - 1) (0 - 1)}{| 0 - 1 |}$

Step 2.3.2.1.3

Factor $- 1$ out of $- 1 \cdot 0 - 1 \cdot - 1$ .

$f (0) = \frac{- 1 \cdot ((0 - 1) (0 - 1))}{| 0 - 1 |}$

Step 2.3.2.1.4

Raise $0 - 1$ to the power of $1$ .

$f (0) = \frac{- 1 \cdot ((0 - 1) (0 - 1))}{| 0 - 1 |}$

Step 2.3.2.1.5

Raise $0 - 1$ to the power of $1$ .

$f (0) = \frac{- 1 \cdot ((0 - 1) (0 - 1))}{| 0 - 1 |}$

Step 2.3.2.1.6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (0) = \frac{- 1 \cdot {(0 - 1)}^{1 + 1}}{| 0 - 1 |}$

Step 2.3.2.1.7

Add $1$ and $1$ .

$f (0) = \frac{- 1 \cdot {(0 - 1)}^{2}}{| 0 - 1 |}$

Step 2.3.2.2

Simplify the denominator.

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Step 2.3.2.2.1

Subtract $1$ from $0$ .

$f (0) = \frac{- 1 \cdot {(0 - 1)}^{2}}{| - 1 |}$

Step 2.3.2.2.2

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

$f (0) = \frac{- 1 \cdot {(0 - 1)}^{2}}{1}$

Step 2.3.2.3

Simplify the numerator.

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Step 2.3.2.3.1

Subtract $1$ from $0$ .

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

Step 2.3.2.3.2

Raise $- 1$ to the power of $2$ .

$f (0) = \frac{- 1 \cdot 1}{1}$

Step 2.3.2.4

Simplify the expression.

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Step 2.3.2.4.1

Multiply $- 1$ by $1$ .

$f (0) = \frac{- 1}{1}$

Step 2.3.2.4.2

Divide $- 1$ by $1$ .

$f (0) = - 1$

Step 2.3.2.5

The final answer is $- 1$ .

$y = - 1$

Step 2.4

Substitute the $x$ value $2$ into $f (x) = \frac{(x + 1) (x - 1)}{| x - 1 |}$ . In this case, the point is $(2, 3)$ .

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Step 2.4.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = \frac{((2) + 1) ((2) - 1)}{| (2) - 1 |}$

Step 2.4.2

Simplify the result.

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Step 2.4.2.1

Simplify the numerator.

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Step 2.4.2.1.1

Add $2$ and $1$ .

$f (2) = \frac{3 (2 - 1)}{| 2 - 1 |}$

Step 2.4.2.1.2

Subtract $1$ from $2$ .

$f (2) = \frac{3 \cdot 1}{| 2 - 1 |}$

Step 2.4.2.2

Simplify the denominator.

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Step 2.4.2.2.1

Subtract $1$ from $2$ .

$f (2) = \frac{3 \cdot 1}{| 1 |}$

Step 2.4.2.2.2

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$f (2) = \frac{3 \cdot 1}{1}$

Step 2.4.2.3

Simplify the expression.

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Step 2.4.2.3.1

Multiply $3$ by $1$ .

$f (2) = \frac{3}{1}$

Step 2.4.2.3.2

Divide $3$ by $1$ .

$f (2) = 3$

Step 2.4.2.4

The final answer is $3$ .

$y = 3$

Step 2.5

The absolute value can be graphed using the points around the vertex $(- 2, 1), (- 1, 0), (0, - 1), (2, 3)$

$\begin{array}{c} x & y \\ - 2 & 1 \\ - 1 & 0 \\ 0 & - 1 \\ 2 & 3 \end{array}$

Step 3