Pre-Algebra Examples

$x + \frac{x}{| x |}$

Step 1

Find the domain for $y = x + \frac{x}{| x |}$ 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 $\frac{x}{| x |}$ equal to $0$ to find where the expression is undefined.

$| x | = 0$

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 $| x | = \pm x$ .

$x = \pm 0$

Step 1.2.2

Plus or minus $0$ is $0$ .

$x = 0$

Step 1.3

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

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 0}$

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 0}$

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

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Step 2.1.1

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

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

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Simplify each term.

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Step 2.1.2.1.1

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

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

Step 2.1.2.1.2

Divide $- 2$ by $2$ .

$f (- 2) = - 2 - 1$

Step 2.1.2.2

Subtract $1$ from $- 2$ .

$f (- 2) = - 3$

Step 2.1.2.3

The final answer is $- 3$ .

$y = - 3$

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Simplify the result.

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

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

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

Step 2.2.2.1.2

Divide $- 1$ by $1$ .

$f (- 1) = - 1 - 1$

Step 2.2.2.2

Subtract $1$ from $- 1$ .

$f (- 1) = - 2$

Step 2.2.2.3

The final answer is $- 2$ .

$y = - 2$

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Simplify the result.

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Step 2.3.2.1

Simplify each term.

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Step 2.3.2.1.1

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

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

Step 2.3.2.1.2

Divide $1$ by $1$ .

$f (1) = 1 + 1$

Step 2.3.2.2

Add $1$ and $1$ .

$f (1) = 2$

Step 2.3.2.3

The final answer is $2$ .

$y = 2$

Step 2.4

Substitute the $x$ value $2$ into $f (x) = x + \frac{x}{| x |}$ . 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.

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

Step 2.4.2

Simplify the result.

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Step 2.4.2.1

Simplify each term.

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Step 2.4.2.1.1

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

$f (2) = 2 + \frac{2}{2}$

Step 2.4.2.1.2

Divide $2$ by $2$ .

$f (2) = 2 + 1$

Step 2.4.2.2

Add $2$ and $1$ .

$f (2) = 3$

Step 2.4.2.3

The final answer is $3$ .

$y = 3$

Step 2.5

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

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

Step 3