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# Absolute Value Function

Absolute value is a term used in mathematics to indicate the distance of a number from the zero of a number line. Absolute value of a number is always positive. The absolute value of x is denoted as |x|, gives the distance between x and 0. Absolute value of 5, |5|, gives the distance between 0 and 5.

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## Absolute Value Function Definition

Suppose x is a real number. Then the absolute value of x is defined as follows:

It is also called piece-wise function.

Definition of absolute value function is read as:

If x is greater than or equal to zero, then use |x| = x
That is, if a number is non negative, then its absolute value is itself.

For example:

| 3 | = 3, | 6 | = 6.

If x is less than zero, then use |x| = - x
That is, if a number is negative, then its absolute value is its opposite.

For example:

| - 3 | = 3, | - 6 | = 6.

Use the below widget to evaluate absolute equation.

## Absolute Value Functions and Graphs

The function f(x) = |x| is called the absolute value function.

The domain of the absolute value function is the set of real numbers and the range is the set of positive real numbers.
If x is positive then f(x) = x and if x is negative then f(x) = - x so that f(x) is always positive. Graph of absolute value function is always "V" shaped.

Graph of f(x) = |x| or y = |x|

## Graphing Absolute Value Functions

A function of the form f(x) = |ax + b| + c, where a$\neq$0 is an absolute value function. Graphs of absolute value functions are not look like the graphs of linear functions. Because of behavior of absolute values, it is important to include negative inputs when graphing absolute value functions.

Step for graphing an absolute value function:

Step 1:
Write two equations using the definition of absolute value.

Step 2: Solve both equations for y.

Step 3: Shade the lines in proper "V".

### Solved Examples

Question 1: Graph f(x) = |x + 3|

Solution:

Step 1: Put y = |x + 3|

Step 2: When x ≥ 0

 x 0 1 2 3 y = x + 3 3 4 5 6

Step 3: When x < 0

 x -1 -3 -4 -5 -6 y = - (x + 3) 2 0 1 2 3

Step 4: Graph the values

Question 2: Graph f(x) = |x| + 2

Solution:

Step 1: Put y = |x| + 2

Step 2: When x ≥ 0

 x 0 1 2 y = x + 2 2 3 4

Step 3: When x < 0

 x -1 -2 -3 y = - x + 2 3 4 5

Step 4: Graph the values

## Derivative of Absolute Value Function

Derivatives of absolute values can be done by using the formula:

$\frac{\mathrm{d} }{\mathrm{d} x}$ |f(x)| = $\frac{f(x)}{|f(x)|}$.f '(x)

where f(x) $\neq$ 0

### Solved Examples

Question 1: Find the derivative of |x + 9|

Solution:

Let f(x) = x + 9

derivative of |f(x)| using formula

$\frac{\mathrm{d} }{\mathrm{d} x}$ |f(x)| = $\frac{f(x).f^'(x)}{|f(x)|}$

f '(x) = $\frac{\mathrm{d} }{\mathrm{d} x}$(x + 9) = 1

$\frac{\mathrm{d} }{\mathrm{d} x}$ |x + 9| = $\frac{(x + 9). 1}{|x + 9|}$

= $\frac{x + 9}{|x + 9|}$

if x > 0 then |x + 9| = x + 9

$\frac{\mathrm{d} }{\mathrm{d} x}$ |x + 9| = $\frac{x + 9}{|x + 9|}$

=
$\frac{x + 9}{x + 9}$

= 1

if x < 0 then |x + 9| = -(x + 9)

$\frac{\mathrm{d} }{\mathrm{d} x}$ |x + 9| = $\frac{x + 9}{|x + 9|}$

=
$\frac{x + 9}{-(x + 9)}$

= -1

and f '(x) does not exist at x = - 9.

Question 2: Find the derivative of |2x + 1|

Solution:

Let f(x) = 2x + 1

derivative of |f(x)| using formula

$\frac{\mathrm{d} }{\mathrm{d} x}$ |f(x)| =  $\frac{f(x).f^'(x)}{|f(x)|}$

f '(x) = $\frac{\mathrm{d} }{\mathrm{d} x}$(2x + 1) = 2

$\frac{\mathrm{d} }{\mathrm{d} x}$|2x + 1| = $\frac{2x + 1}{|2x + 1|}$ * 2

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

if x > 0 then |2x + 1| = 2x + 1

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

= 2

if x < 0 then |2x + 1| = -(2x + 1)

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

= - 2

and f '(x) does not exist at x = $\frac{-1}{2}$.

Question 3: Find the derivative of  x +  |x +7|

Solution:

Let g(x) = x + |f(x)|, where f(x) = x + 7

g'(x) = 1 + $\frac{\mathrm{d} }{\mathrm{d} x}$ |f(x)|

derivative of |f(x)| using formula

$\frac{\mathrm{d} }{\mathrm{d} x}$ |f(x)| = $\frac{f(x).f^'(x)}{|f(x)|}$

f '(x) = $\frac{\mathrm{d} }{\mathrm{d} x}$(x + 7) = 1

and f '(x) does not exist at x = - 7

$\frac{\mathrm{d} }{\mathrm{d} x}$ |x + 7| = $\frac{(x + 7). 1}{|x + 7|}$

= $\frac{x + 7}{|x + 7|}$

if x > 0 then |x + 7| = x + 7

$\frac{x + 7}{|x + 7|}$ = $\frac{x + 7}{x + 7}$

= 1

$\frac{\mathrm{d} }{\mathrm{d} x}$ g(x) = 1 + 1 = 2

if x < 0 then |x + 7| = -(x + 7)

$\frac{x + 7}{|x + 7|}$ = $\frac{x + 7}{-(x + 7)}$

= - 1

$\frac{\mathrm{d} }{\mathrm{d} x}$ g(x) = 1 - 1 = 0

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