Graph f(x)=x/(|x|)+(x-1)/(|x-1|)

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

Step 1

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

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

$| x | = 0$

Solve for $x$ .

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

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

$| x - 1 | = 0$

Solve for $x$ .

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

$x - 1 = \pm 0$

Plus or minus $0$ is $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 0, 1}$

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 0, 1}$

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.

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

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

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

Simplify the result.

Subtract $1$ from $- 2$ .

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

Simplify each term.

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

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

Divide $- 2$ by $2$ .

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

Simplify the denominator.

Subtract $1$ from $- 2$ .

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

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

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

Divide $- 3$ by $3$ .

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

Subtract $1$ from $- 1$ .

$f (- 2) = - 2$

The final answer is $- 2$ .

$y = - 2$

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

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

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

Simplify the result.

Subtract $1$ from $2$ .

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

Simplify each term.

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

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

Divide $2$ by $2$ .

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

Simplify the denominator.

Subtract $1$ from $2$ .

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

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

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

Divide $1$ by $1$ .

$f (2) = 1 + 1$

Add $1$ and $1$ .

$f (2) = 2$

The final answer is $2$ .

$y = 2$

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

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

Step 3

Graph f(x)=x/(|x|)+(x-1)/(|x-1|)

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