The modulus function, also called the absolute value function, is a special type of function in mathematics that always gives the positive value (or magnitude) of a number or expression. It tells us how far a number is from zero on the number line, without considering its sign.

For example:

• |5| = 5
• |−5| = 5
• |0| = 0

So, no matter if the input is positive or negative, the output is always zero or positive.

The modulus function is written as f(x) = |x|. It is defined for all real numbers, which means you can put any number into the function. The domain is all real numbers (−∞, ∞), and the range is all non-negative numbers [0, ∞).

What is a Modulus Function?

The modulus function, also called the absolute value function, is a function that gives the positive value (or magnitude) of any number or variable, no matter if the number is negative or positive. It simply removes the sign and shows how far the number is from zero.

In math, the modulus of a number x is written as |x|. This means:

• If x is positive or zero, the value remains the same.
• If x is negative, the function turns it into a positive value.

So, the result of the modulus function is always zero or greater. It never gives a negative output.

The modulus also represents the distance of the number from zero on the number line. That’s why it is always a non-negative value.

The function \(f\left(x\right)=\left|x\right|\) is called a modulus function. Here |x| denotes the modulus of x, where x is a real number. If x is non-negative then f(x) will be of the equivalent value x. However, if x is negative, then f(x) will denote the magnitude of x, that is, f(x) = -x if x is negative.

Modulus function formula is \(\begin{Bmatrix}f\left(x\right)&=x&x\ge0\\
f\left(x\right)&=-x &x<0\end{Bmatrix}\)

This indicates if the value of x is higher than or equivalent to 0, then the modulus function catches the actual value, but if in case x is smaller than 0 then the function catches minus of the actual value or original.

Domain and Range of Modulus Function

We can use the modulus function for every real number. The range of the modulus function is defined as the collection of non-negative real quantities and is expressed as [0,∞) whereas the domain of the function is R, where R relates to the collection of all positive real numbers.

Hence the domain of |x| is R and its range is [0, ∞).

If f: P → Q is a function, then the set P is named as the domain of the function f and set Q is designated as the co-domain of the function f.

If f: P → Q is a function, then the range of f consists of those components of Q which are connected with at least one element of P. It is expressed by f(P).

Thus, f(P) = {y : y = f(x) for some x ∈ P}

Graph of Modulus Function

Line Graph of the modulus function extends in the first and the second quadrants as the coordinates of the points on the graph are of the pattern (x, y), (-x, y). The function f: R →R defined by f(x) = |x| for each x ∈R is called the modulus function. This implies that for every non-negative value of x, f(x) is equivalent to x. Although for negative conditions of x, the value of f(x) is negative concerning the value of x.

Now let us understand how to plot the mod x graph or the graph for a modulus function. Consider x to be a variable, taking values from -6 to 6. While determining modulus for the positive values of ‘x’, the line outlined in the graph is ‘y = x’ and for the negative values of ‘x’, the line sketched in the graph is ‘y = -x’.

Properties of Modulus Function

So far we have seen the definition, the formula and the graph of the mod function, let us now examine the various properties of the mod function:

Property 1: For any real number x, we have:

\(\sqrt{x^2}=\left|x\right|\)

\(\left|\left|x\right|\right|=\left|x\right|\)

Property 2: If a and b denote positive real numbers, then:

\(\left(a\right)x^2\le a^2\leftrightarrow|x|\le a\leftrightarrow−a\le x\le a\)

\(\left(b\right)\ x^2\ge a^2\leftrightarrow|x|\ge a\leftrightarrow x\le−a\text{ or}\ x\ge a\)

\(\left(c\right)x^2<a^2\leftrightarrow|x|<a\leftrightarrow−a<x<a\)

\(\left(d\right)\ x^2>a^2\leftrightarrow|x|>a\leftrightarrow x<−a\text{ or}\ x>a\)

\(\left(e\right)\ a^2\le x^2\le b^2\leftrightarrow a\le|x|\le b\leftrightarrow x∈[−b,−a]∪[a,b]\)

\(\left(f\right)\ a^2<x^2<b^2\leftrightarrow a<|x|<b\leftrightarrow x∈(−b,−a)∪(a,b)\)

Property 3: If a is negative, then:

\(\left(a\right)\ |x|\ge a,\text{ or }|x|\ne a\ x\in R\)

\(\left(b\right)\ |x|\le a,x=ϕ\)

Property 4: For any real number x and y:

\(\left(a\right)\ |xy|=|x||y|\)

\(\left(b\right)\ \left|\frac{x}{y}\right|=\left|\frac{x}{y}\right|\text{ for }y\ne0\)

\(\left(c\right)\ |x+y|\le|x|+|y|\)

\(\left(d\right)\ |x−y|\le|x|+|y|\)

\(\left(e\right)\text{ For }x,y\ge0\text{ or }x,y<0,|x+y|=|x|+|y|\)

\(\left(f\right)|x+y|\ge|x|−|y|\)

\(\left(g\right)|x−y|\ge|x|−|y|\)

\(\left(h\right)\text{ For }x,y\ge0\text{ and }|x|\ge|y|\text{ or }x,y<0\text{ and }|x|\ge|y|,|x−y|=|x|−|y|\)

Property 5: The absolute value has the coming four primary properties (a and b are real numbers):

\(a]\ ∣a∣\ge0\text{ Non−negativity}\)

\(b]\ ∣a∣=0⟺a=0\text{ Positive−definiteness}\)

\(c]\ ∣ab∣=∣a∣∣b∣\text{ Multiplicativity}\)

\(d]\ ∣a+b∣\le∣a∣+∣b∣\text{ Subadditivity, specifically the triangle inequality}\)

Derivative of Modulus Function

Having a good idea regarding the various properties of modulus functions lets us learn about the differentiation of mod x or derivative of mod x. Until now we are familiar that a mod function f(x) = |x| is equal to x if x > 0 and -x if x < 0.

\(\begin{Bmatrix}f\left(x\right)&=x&x\ge0\\
f\left(x\right)&=-x &x<0\end{Bmatrix}\)

Hence, the derivative of modulus function is equal to 1 if x > 0 and -1 if x < 0. The point to consider is that the derivative of the mod function is not defined for x = 0.

\(\begin{matrix}\frac{d\left\{f\left(x\right)\right\}}{dx}&=1&x>0\\
\frac{d\left\{f\left(x\right)\right\}}{dx}&=-1&x<0\end{matrix}\)

Therefore the derivative of mod function can be formulated as \(\frac{d\left(\left|x\right|\right)}{dx}=\frac{x}{\left|x\right|}\)

For all values of x and x which are not equal to 0.

Learn about Integral Calculus

Integration of Modulus Function

Considering the formula of the modulus function and integration formulas, the integral of the mod function is given as follows:

If f(x) = |x| is equal to x if x > 0 and -x if x < 0 then the integration for the function is:

\(\begin{Bmatrix}f\left(x\right)&=x&x\ge0\\
f\left(x\right)&=-x &x<0\end{Bmatrix}\)

\(\begin{matrix}\int_{ }^{ }f\left(x\right)&=\left(\frac{1}{2}\right)x^2+C&x\ge0\\
\int_{ }^{ }f\left(x\right)&=-\left(\frac{1}{2}\right)x^2+C&x<0\end{matrix}\)

Hence the integration of the modulus function can be stated as:

\(\begin{matrix}\int_{ }^{ }\left|x\right|&=\left(\frac{1}{2}\right)x^2+C&\text{if }x\ge0\\
\int_{ }^{ }\left|x\right|&=-\left(\frac{1}{2}\right)x^2+C&\text{if }x<0\end{matrix}\)

Learn about Applications of Derivatives

Application of Modulus Function (|x|)

Now that we know how the modulus function works, let’s look at how it is applied through some simple examples. The modulus function gives the distance of a number from zero, and it always gives a non-negative result.

Let’s take the function f(x) = |x|, and check how it behaves for different values of x:

• If x = –3, then f(–3) = |–3| = 3. Since –3 is less than zero, we take the opposite sign.
• If x = 3, then f(3) = |3| = 3. Since 3 is already positive, we keep it as it is.
• If x = 0, then f(0) = |0| = 0. The distance from zero is just zero.

So, whether we use 3 or –3, the answer is the same:
|3| = |–3| = 3

This shows that the modulus function always gives the positive value of any number, no matter its sign.

Important Notes on the Modulus Function

• The modulus function is also known as the absolute value function. It gives the positive value of any number, no matter whether the number is negative or positive.
  It is written as f(x) = |x|.
• The domain of the modulus function includes all real numbers. That means you can use any value of x, whether it’s negative, zero, or positive.
• The range of the modulus function is all values greater than or equal to 0. This is because the output is never negative.
• The graph of the basic modulus function y = |x| has a V-shape and its lowest point (called the vertex) is at (0, 0).
• If the function is written as y = a |x − h| + k, then the vertex of the graph is at the point (h, k).

Solved Examples of Modulus Functions

With the knowledge of modulus sign, definition and the various properties let’s now learn how to solve modulus questions with some solved examples:

Example 1: Obtain the domain and range for the function y = |4−x|.

Solution: Given function y = |4−x|.

Clearly, we can say that this equation is defined for x ∈ R. Hence the domain is R.

Now |4−x| ≥ 0 for all x ∈ R

Therefore the range is [0, ∞)

Example 2: Obtain the domain and range of the given function y = 3 − |2 − x|.

Solution: y = 3 − |2 − x|

Clearly, we can say that this function is defined for x ∈ R. So the domain of the given function is R.

Now as we know |2−x|≥0 for all x∈R

or can say:

−|2−x|≤0

3−|2−x|≤3 for all x∈R

Therefore the range is (−∞,3]

Example 3: Obtain the domain and range of the given function.

\(y=\frac{2}{\sqrt{x−|x|}}\)

Solution: Here,

\(x−\left|x\right|=\begin{matrix}x−x&=0&\text{ if }x\ge0\\
x+x&=2x&\text{ if }x\le0\end{matrix}\)

Therefore,\(y=\frac{2}{\sqrt{x−|x|}}\) is undefined for all x ∈ R
Hence the domain of the function is ϕ.

Example 4: Find the value of the mod function |x| for x = -12 and x = 11

Solution: If x = -12, then |x| = |-12| = 12

If x = 11, then |x| = |11| = 11

This implies that for |x| =12 for x = -12 and |x| = 11 for x = 11.

This satisfies the mod function property which says that for a negative number or positive number the output of |x| is a positive number only.

Example 5: Solve |x + 4| = 9 using mod function definition.

Solution: We know that the modulus function output is always non-negative, Hence we will have two cases:

If x + 4 > 0, that is the value of |x + 4| = x + 4 and if x + 4 < 0, that is the value of |x + 4| = -(x + 4).

Case 1: If x + 4 > 0, we have

|x + 4| = x + 4

⇒ x + 4 = 9

⇒ x = 9- 4 = 5

Case 2: If x + 4 < 0, we have

|x + 4| = -(x + 4)

⇒ -(x + 4) = 9

⇒ -x – 4 = 9

⇒ x = -4 – 9 = -13

Answer: Therefore, the solution is x = 5, -13.

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