Graph y=1/(|x|)

$y = \frac{1}{| x |}$

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

Set the denominator in $\frac{1}{| x |}$ 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 $0$ .

$x = 0$

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:

${x | x \neq 0}$

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

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

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

Simplify the result.

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

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

The final answer is $\frac{1}{2}$ .

$y = \frac{1}{2}$

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

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

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

Simplify the result.

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

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

Divide $1$ by $1$ .

$f (- 1) = 1$

The final answer is $1$ .

$y = 1$

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

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

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

Simplify the result.

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

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

The final answer is $\frac{1}{2}$ .

$y = \frac{1}{2}$

The absolute value can be graphed using the points around the vertex $(- 2, 0.5), (- 1, 1), (1, 1), (2, 0.5)$

$\begin{array}{c} x & y \\ - 2 & 0.5 \\ - 1 & 1 \\ 1 & 1 \\ 2 & 0.5 \end{array}$

Graph y=1/(|x|)

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