For each non-negative value of x, f(x) is equal to x. But for negative values of x, the value of f(x) is the negative of the value of x

$f(x) = |x|= \begin{cases} x & \text{ if } x \geq 0 \\ -x & \text{ if } x < 0 \end{cases} $

It is also called the

Example

|-3| =3

|-2|=2

|2|=2

We know that all the real number can be plotted on the real number line. The absolute value function is used to measure the distance between two numbers on the number line.. Thus, the distance between x and 0 is |x - 0| = |x|. So |x| represent the distance of the number from the origin.Similarly the distance between x and y is |x - y|. Thus, the distance from -3 to -1 is |-3 - (-1)| = |-3 + 1| = |2| = 2, and the distance from -3 to 5 is |-3 - 5| = |-8| = 8

For $f : R \rightarrow R , y = f(x) = |x|$ for each $x \in R$

Domain = R

Range = $R^+ ={x \in R : x \geq 0}$

The graph of the modulus function is shown below. It coincide with the graph of identity function y=x for $x \geq 0$ and for x< 0, it is a graph of linear function y=-x

Here at x=0 ,the graph is turning upward. So x=0 is the turning point in this graph

1. For any real number x , we have

$\sqrt {x^2} =|x|$

2. ||x||= |x|

3. if a and b are positive real numbers

a. $x^2 \leq a^2 \Leftrightarrow |x| \leq a \Leftrightarrow -a \leq x \leq a$

b. $x^2 \geq a^2 \Leftrightarrow |x| \geq a \Leftrightarrow x \leq -a \; or x \geq a$

c. $x^2 < a^2 \Leftrightarrow |x| < a \Leftrightarrow -a < x < a$

d. $x^2 > a^2 \Leftrightarrow |x| > a \Leftrightarrow x < -a \; or x > a$

e. $a^2 \leq x^2 \leq b^2 \Leftrightarrow a \leq |x| \leq b \Leftrightarrow x \in [-b,-a] \cup [a,b]$

f $a^2 < x^2 < b^2 \Leftrightarrow a < |x| < b \Leftrightarrow x \in (-b,-a) \cup (a,b)$

4. if a is negative

a. |x| \geq a , x \in R$

b. |x| \leq a , x =\phi$

3. For real number x and y

a. |xy|=|x||y|

b. $ |\frac {x}{y}| =\frac {|x|}{|y|} ,y \neq 0$

c. $|x +y | \leq |x| + |y|$

$|x -y| \leq |x| + |y| $

for $x,y \geq 0 \; or \; x,y < 0 , |x+y| =|x| + |y|$

d. $|x + y | \geq |x| - |y|$

$|x -y| \geq |x| - |y| $

for $x,y \geq 0 \; and \; |x| \geq |y| y or \; x,y < 0 \; and \; |x| \geq |y| , |x- y| =|x| - |y|$

a. $y =|1-x|$

b. $y = 2 -|1-x|$

c. $y= \frac {2}{\sqrt {x -|x|}} $

a. $y =|1-x|$

Clearly This is defined for $x \in R$. So Domain is R

Now $|1 -x| \geq 0 $ for all $x \in R$

So Range is $[0,\infty )$

b. $y = 2 -|1-x|$

Clearly This is defined for $x \in R$. So Domain is R

Now $|1-x| \geq 0$ for all $x \in R$

or

$-|1-x| \leq 0$

$2- |1-x| \leq 2$ for all $x \in R$

So Range is $(-\infty,2]$

c.$y= \frac {2}{\sqrt {x -|x|}} $

Now

$x -|x| = \begin{cases} x-x=0 & \text{ if } x \geq 0 \\ x+x=2x & \text{ if } x \leq 0 \end{cases}$

Therefore, $\frac {2}{\sqrt {x -|x|}}$ is undefined for all $x \in R$

So domain of the function is $\phi$

2. Draw the graph of the below function?

a. y=x + |x|

b. y = x -|x|

c. y =|x-1|

We need to convert this function into simple form before drawing the graph

a. y=x + |x|

This can be written as

$y= x +|x| = \begin{cases} x+x=2x & \text{ if } x \geq 0 \\ x-x=0 & \text{ if } x < 0 \end{cases}$

So, This is a linear function y =2x for $x \geq 0$ and linear function y =0 for $x < 0$

So graph will be

b.y = x -|x|

This can be written as

$y= x -|x| = \begin{cases} x-x=0 & \text{ if } x \geq 0 \\ x+x=2x & \text{ if } x < 0 \end{cases}$

So, This is a linear function y =0 for $x \geq 0$ and linear function y =2x for $x < 0$

So graph will be

c.y =|x-1|

This can be written as

$y= |x -1| = \begin{cases} x-1 & \text{ if } x \geq 1 \\ 1 -x & \text{ if } x < 1 \end{cases}$

So, This is a linear function y =x-1 for $x \geq 1$ and linear function y =1-x for $x < 1$

So graph will be

Important thing to note here. Turning point of the graph is x=1 here

- Cartesian Products
- |
- What is relations?
- |
- What is Function
- |
- Domain of Function
- |
- Range of Function
- |
- Identity Function
- |
- Constant Function
- |
- Linear Function
- |
- Modules Function
- |
- Greatest Integer Function
- |
- Polynomial Function
- |
- Algebra of Real Function

Thanks for visiting our website. From feedback of our visitors we came to know that sometimes you are not able to see the answers given under "Answers" tab below questions. This might happen sometimes as we use javascript there. So you can view answers where they are available by reloding the page and letting it reload properly by waiting few more seconds before clicking the button.

We really do hope that this resolve the issue. If you still hare facing problems then feel free to contact us using feedback button or contact us directly by sending is an email at **[email protected]**

We are aware that our users want answers to all the questions in the website. Since ours is more or less a one man army we are working towards providing answers to questions available at our website.

Class 11 Maths Class 11 Physics