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

Class 11 Maths Class 11 Physics

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