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Modulus Function

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Modulus Function Definition

Modulus function is an interesting topic in Math and an important one from a competitive examinations’ point of view. The modulus function only gives a positive value of any variable or a number as the output. It is also known as the absolute value function because it gives a non-negative value for any independent variable, no matter if it is positive or negative. 


In other words, a modulus function gives the magnitude of a number. Therefore, this function is also considered as the distance of a number from the origin or zero on the graph. 


It is commonly represented as y = |x|, where x represents a real number, and y = f(x), representing all positive real numbers, including 0, and f:R→R and x ∈ R.


The expression in which a modulus can be defined is given below:


 f(x)   = \[\left\{\begin{matrix} x & if x \geq 0\\ -x & if x < 0 \end{matrix}\right.\]


Here, x represents any non-negative number, and the function generates a positive equivalent of x. For a negative number, x<0, the function generates (−x) where −(−x) = positive value of x. 


However, there are different cases for a modular function and can mean differently for various contexts. 

Case 1:

For y = |x|, where x is a real number, i.e., x > 0, since variables can have real values only. Here, the modulus function of the real variable stays the positive value of the real number. For example,


For x = 2, 


y = |2|, i.e., = 2.

Case 2: 

For y = |f(x)|, here we use f(x) instead of |x|, and therefore, the modulus changes the function value and properties, modifying the overall function. A few examples are given below: 


|f(x)| = a ;    a > 0    => f(x) = ± a


|f(x)| = a;    a = 0    => f(x) = 0


|f(x)|= a;    a < 0    => There is no solution to this equation because a modulus function can never result in a negative outcome. 

Modulus Function Graph 

In  modulus function, every time |x| = 4, the value of x = ±4.


For plotting the graph, we need to take certain values first,


When x = −5 then y = f(x) = |x|


y = |−5| = 5


Similarly, for x = −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, the respective values of y will be 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5. 


(Image will be Uploaded Soon)


Here for x > 0, the graph represents a line where y = x. Similarly for x < 0, the graph is a line where y = −x. Also, the vertex of the modulus graph  y = |x| is given by (0,0).


Thus, from the graph, we can conclude that the values of the modulus function are always positive for all the values of x.


As the modulus function is understood as a non-negative value, therefore, it can be said that the modulus of a variable is similar to that of the square root of the square of the variable. Here's how:


|x| = \[\sqrt{x^{2}}\]

Domain and Range of Modulus Function

In the graph, both the lines hold true the definition of modulus functions and help define the domain and range of modulus function, i.e., 


the domain = R (or Real Numbers)


Range = [0,∞); where the range of modulus function is the upper half of the real numbers (R+), i.e., all the positive real numbers, including 0.

Properties of Modulus Function

Since the modulus function can be effective to find inequality between the numbers, here are the following properties of the modulus function:

Here are some other non-negative expressions that can explain the non-negative value of the modulus function:

  1. The even exponent of an expression or variable can be defined as x2n , where n ∈ Z.

  2. The even root of a variable can be defined as x1/2n, where n ∈ Z.

  3. The value of y can be defined as y = 1−sinx or y = 1−cosx, (since sinx ≤ 1 and cosx ≤1) 

  • When a > 0, 

Here, x lies between −a and a, not considering the endpoints of the interval, i.e.,

|x| < a; a > 0 ⇒ −a < x < a


|x| > a; a > 0  ⇒ x < − a or x > a ⇒ x ∈ (−∞, −a) ∪ (a, ∞)

Since the inequalities can be useful to express intervals in the compact form, here's an example of the cosec trigonometric function range that is defined as x ∈ (−∞, −1] ∪ [1,∞}, represented as:

|x| ≥ 1

|f(x)| < a; a > 0, ⇒ −a < f(x) < a

  • For x, y as real variables:

  1. |x − y| = 0, ⇔ x = y

  2. |x + y| ≤ |x| + |y|

  3. |x − y| ≥ ||x| − |y||

  4. |xy| = |x| * |y|

  5. |x/y| = |x| / |y|, where |y| ≠ 0.

  • For p and q as positive real numbers:

  1. x2 ≤ p2 ⇔ |x| ≤ p ⇔ −p ≤ x ≤ p

  2. x2 ≥ p2 ⇔ |x| ≥ p ⇔ x ≤ −p , x ≥ p

  3. x2 < p2 ⇔ |x| < p ⇔ −p < x < p

  4. x2 > p2 ⇔ |x| > p ⇔ x < −p, x > p 

  5. p2 ≤ x2 ≤ q2 ⇔ p ≤ |x| ≤ q ⇔ x ∈ [−q,−p] ∪ [p,q]

  6. p2 < x2 <q2⇔ p < |x| < q ⇔ x ∈ (−q, −p) ∪ (p, q)

Modulus Function Questions (Solved)

Example 1: A function f is defined on R as: 

f(x) = \[\left\{\begin{matrix} \frac{|x|}{x}, & x\neq 0\\ 0, & x = 0 \end{matrix}\right.\]

Plot the graph.

Solution:

When x is a positive integer, the function can be defined as: 


f (x) =  \[\frac{|x|}{x}\] = \[\frac{x}{x}\] = 1


When x is a negative integer, the function can be defined as:


f(x) =  \[\frac{|x|}{x}\] = \[\frac{-x}{x}\] = -1


Therefore, the f can be redefined as: 


f(x) =  \[\left\{\begin{matrix} 1, & &x>0 \\ 0 & & x=0\\ -1& & x<0 \end{matrix}\right.\]


The filled dot at (0,0), and the hollow dots at (0,1), (0,−1), represent that f(0) has the value as 0, instead of 1 or −1. Such a function is also known as the Signum function.

Example 2: Solve |x + 2| = 6 using modulus function.

Solution: We know that the modulus function always gives a non-negative output, therefore we have two cases:

If x + 2 > 0, then |x + 2| = x + 2 and

If x + 2 < 0, then |x + 2| = −(x + 2)


Case 1: If x + 2 > 0, we have

|x + 2| = x + 2

⇒ x + 2 = 6

⇒ x = 6 − 2 = 4


Case 2: If x + 2 < 0, we have

|x + 2| = − (x + 2)

⇒ − (x + 2) = 6

⇒ − x − 2 = 6

⇒ x = −2 − 6 = −8

Hence, the solution for x is −8 and 4.

We can say −8 < x < 4.

Practice Questions!

  1. Plot the graph for the modulus functions: 

  • y = 3x + 4

  • Y = x2 + 8

  1. Obtain the domain and range of the modulus functions:

  • f(x) = 7 + |3 − x|

  • f(x) = 2 − |x − 8|

FAQs on Modulus Function

1. What is the absolute value of a modulus function?

The absolute value of a modulus function needs to be non-negative. Since the absolute value of a modulus function generally defines the distance between two points, therefore it can be expressed as:


For f(x), where x represents 0, therefore,


|f(0)| = 0, which isn't a positive, but a non-negative value where x ≥ 0.


Similarly,


For a negative integer like −6,


f(x) = f(−6)


|f(−6)| = −(−6)


= |−6|


= 6.


Therefore, |x| = x, where x is ≤ 0, or a non-positive number.


It can be inferred that the absolute value of any modulus function needs to be non-negative always, not necessarily meaning positive.

2. Is Modulus Function Onto?

For f: R → R, the


f(x) = |x|, where x, x ≥ 0 and −x, if x < 0.


Therefore, f(−1) = |−1| = 1.


and f(1) = |1| = 1.


But since, −1 not equal to 1, and f(−1) = f(1),


It can be said that f cannot be considered one-one. 


Similarly when −1 ∈ R, R represents the real number set.


f(x) = |x|, the value that stays non-negative. Therefore, there isn't any value x present in the domain R where f(x) = |x| = −1. 


Therefore, f isn't onto. 


Hence, it is proved that the modulus function is neither one-one nor onto.