Calculus Examples

$\frac{| x |}{x + 1}$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = \frac{| x |}{x + 1}$ is $(0, 0)$ .

Tap for more steps...

Step 1.1

To find the $x$ coordinate of the vertex, set the inside of the absolute value $x$ equal to $0$ . In this case, $x = 0$ .

$x = 0$

Step 1.2

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

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

Step 1.3

Simplify $\frac{| 0 |}{(0) + 1}$ .

Tap for more steps...

Step 1.3.1

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

$y = \frac{0}{0 + 1}$

Step 1.3.2

Add $0$ and $1$ .

$y = \frac{0}{1}$

Step 1.3.3

Divide $0$ by $1$ .

$y = 0$

Step 1.4

The absolute value vertex is $(0, 0)$ .

$(0, 0)$

Step 2

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

Tap for more steps...

Step 2.1

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

$x + 1 = 0$

Step 2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.3

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 3

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.

Tap for more steps...

Step 3.1

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

Tap for more steps...

Step 3.1.1

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

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

Step 3.1.2

Simplify the result.

Tap for more steps...

Step 3.1.2.1

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

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

Step 3.1.2.2

Add $- 2$ and $1$ .

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

Step 3.1.2.3

Divide $2$ by $- 1$ .

$f (- 2) = - 2$

Step 3.1.2.4

The final answer is $- 2$ .

$y = - 2$

Step 3.2

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

Tap for more steps...

Step 3.2.1

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

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

Step 3.2.2

Simplify the result.

Tap for more steps...

Step 3.2.2.1

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

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

Step 3.2.2.2

Add $1$ and $1$ .

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

Step 3.2.2.3

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

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

Step 3.3

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

Tap for more steps...

Step 3.3.1

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

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

Step 3.3.2

Simplify the result.

Tap for more steps...

Step 3.3.2.1

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

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

Step 3.3.2.2

Add $2$ and $1$ .

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

Step 3.3.2.3

The final answer is $\frac{2}{3}$ .

$y = \frac{2}{3}$

Step 3.4

The absolute value can be graphed using the points around the vertex $(0, 0), (- 2, - 2), (1, 0.5), (2, 0.67)$

$\begin{array}{c} x & y \\ - 2 & - 2 \\ 0 & 0 \\ 1 & 0.5 \\ 2 & 0.67 \end{array}$

Step 4