In mathematics, the absolute value of any real number is the non-negative value of that number without regard to its sign. It is similar to the distance of any number from zero in a number line. For example, if we have to find the value of |12| and |-12|, then it is 12. Because absolute value of -12 is 12, and the absolute value of 12 is also 12. This function is one of the most used functions in mathematics. Absolute value is used in various fields of mathematics like in complex numbers, ordered rings, quaternions and vector spaces.
Absolute value function (also known as mod function) is defined as:
f(x) = |x| =  $\left\{\begin{matrix} x &;x \geq 0 \\ -x&; x <0 \end{matrix}\right.$

This function is closely related to the notation of distance, norm and magnitude in various physics and mathematical contexts.

Notaion of absolute value function : f(x) = |x|
It is also sometimes written as : abs (x)
The graph of |X| makes a right angle at origin.

Graphically this can be shown as follows:

The domain of Absolute value function is R (real numbers)
The range of the same is All Positive Real Numbers [0,+$\infty$).

The absolute value parent function is:  y = |x|

## Graphing

Graph gives the pictorial representation of the function. We can draw a graph for absolute functions as similar we do for other algebraic functions. Let us understand this concept with the help of an example. For the graphing of any absolute value function, follow the following example,
Let y = |x - 1|
Then,
Pick any values of x (positive and negative both)
As,
When x= -2, y = |-2-1| = |-3| = 3
When x= -1  y = |-1-1| = |-2| = 2
When x= 0   y = |0-1| = |-1| = 1
When x= 1   y = |1-1| = |0| = 0
When x= 2   y = |2-1| = |1| = 1
When x= 3   y = |3-1| = |2| = 2
When x= 4   y = |4-1|= |3|= 3
Now using above points plot a graph and we will get a graph of absolute value function.

## Vertex

For computing ‘vertex’ of any absolute value function firstly write down what is inside the mod and put it equal to zero. Then, solve for x.
Now, using value of x find y from original equation.
And we will get ordered pair which is the ‘vertex’ of an absolute function.

For Example:
Let y = |x+3|-1    …………….(1)

Putting x+3 = 0

We get x = -3

Now using value of x in (1) we get,
y = |-3+3|-1

y = |0|-1

y = -1

The vertex for y = |x+3|-1 is (-3,-1).

## Derivative

It is very easy to differentiate absolute funciton. The real absolute value function has a derivative for every z $\neq$ 0. Its derivative for z $\neq$ 0 is given by the step function.

$\frac{d|z|}{dz}$ = $\frac{z}{|z|}$ = $\left\{\begin{matrix} 1&;x > 0 \\ -1&; x <0 \end{matrix}\right.$

It is is not differentiable at x = 0.

Let |z| be an absolute value function of which we have to find the derivative,

Recall that |z| = $\sqrt{Z^2}$

By using this we can find the derivative of |z|,

$\frac{d}{dz}$ |Z| = $\frac{d}{dz}$ $\sqrt{Z^2}$

=$\frac{d}{dz}$ $(Z^2)^{\frac{1}{2}}$

=$\frac{1}{2}$ $(Z^2)^{\frac{-1}{2}}$.2Z

=$\frac{Z}{\sqrt{Z^2}}$

=$\frac{z}{|Z|}$

Therefore we have the following formula:

$\frac{d}{dz}$ |Z| = $\frac{z}{|Z|}$

## Inverse

Inverse function is one that takes the output and returns the input of the given function. We can better understand this concept with the help of example.
Consider an example, Let f(y) = |y - 3| + 2 for x $\geq$ 3

To find the inverse of absolute function, let us first see the graph of the function:

We can see that due to the restriction in domain the graph divides into two halves. But for x $\geq$ 3 we have the right half of the absolute value function.

Therefore, for finding then inverse of the absolute value function f(x) = |x-3|+2 for x=3 we have the right half, i.e. f(x) = (x-3)+2 for x $\geq$ 3. Also we have the range of the function y$\geq$ 2 which will be the domain of inverse function.
Now, let's solve the inverse of function

F(x) = (x-3)+2

Replace f(x) by y

y = (x-3)+2

now interchange the places of x and y and solve for y to get the inverse

x = (y-3)+2

x-2 = y-3

y = x+1

$f^{(-1)}(x)$=x+1

Since the range of the original function is y $\geq$ 2, therefore the domain of the inverse function will be x$\geq$ 2.

$f^{(-1)} (x)$=x+1for x $\geq$ 2

## Integration

Indefinite Integration:

$\int|x|dx$ = $\frac{x|x|}{2}$ + C ; C is content of integration

To find the definite integral of an absolute value function we use the following property

$\int_{a}^{b}f(x)dx$ = $\int_{a}^{c}f(x)dx$ + $\int_{a}^{b}f(x)dx$

Where,a< b< cGENERALIZATION: The above property can be generalized in following form

$\int_{a}^{b}f(x)dx$ = $\int_{a}^{c_1}f(x)dx$ + $\int_{a}^{c_2}f(x)dx$ + ... +$\int_{c_n}^{b}f(x)dx$

Where, $a< c_1< c_2<....c_n<b$
Let an absolute value function |5x-3|

We have to evaluate $\int_{0}^{1}|5x-3|dx$

Solution:

|5x-3|= $\left\{\begin{matrix} -(5x-3)\ when\ 5x - 3 < 0\ i.e., x < \frac{3}{5}\\ 5x -3\ when\ 5x -3 \geq 0\ i.e., x \geq \frac{3}{5} \end{matrix}\right.$

$\int_{0}^{1}|5x-3|dx$

= $\int_{0}^{\frac{3}{5}}|5x-3|dx$ + $\int_{\frac{3}{5}}^{0}|5x-3|dx$

= $\int_{0}^{\frac{3}{5}}-(5x-3)dx$ + $\int_{\frac{3}{5}}^{1}(5x-3)dx$

= $[3x-\frac{5x^2}{2}]_{0}^{\frac{3}{5}}+[\frac{5x^{2}}{2} -3x]_{\frac{3}{5}}^{1}$

= ($\frac{9}{5}$-$\frac{9}{10}$)+(-$\frac{1}{2}$ + $\frac{9}{10}$) = $\frac{13}{10}$

## Examples

Example 1: Find the derivative of |x - 10|

Solution: Let f(x) = |x - 10|

Derivative of the absolute function can be found by using the below formula:

$\frac{d|(f(x)|}{dx}$ = $\frac{f(x)}{|f(x)|}$

$\frac{d|x-10|}{dx}$ = $\frac{x-10}{|x-10|}$  (using above formula)

For x < 0

$\frac{x-10}{|x-10|}$ = $\frac{x-10}{-(x-10)}$ = -1

For x > 0

$\frac{x-10}{|x-10|}$ = $\frac{x-10}{x-10}$ = 1

Example 2:
Evaluate $\int_{0}^{1}|2x-1|dx$?

Solution:
|2x-1|= $\left\{\begin{matrix} -(2x-1)\ when\ 2x - 1 < 0\ i.e., x < \frac{1}{2}\\ 2x -1\ when\ 2x -1 \geq 0\ i.e., x \geq \frac{1}{2} \end{matrix}\right.$

$\int_{0}^{1}|2x-1|dx$

= $\int_{0}^{\frac{1}{2}}|2x-1|dx$ + $\int_{\frac{1}{2}}^{0}|2x-1|dx$

= $\int_{0}^{\frac{1}{2}}-(2x-1)dx$ + $\int_{\frac{1}{2}}^{1}(2x-1)dx$

= $[x-\frac{2x^2}{2}]_{0}^{\frac{1}{2}}+[\frac{2x^{2}}{2} -x]_{\frac{1}{2}}^{1}$

= $[x-x^2]_0^{\frac{1}{2}} + [x^2-x]_{\frac{1}{2}}^1$

= $\frac{1}{4}$ + $\frac{3}{4}$
= 1