But what is the reasoning behind this? This happens because it represents the distance from zero to a Number x on the Number Line.

The distance from zero to 2 is 2, and the distance from zero to -2 is also 2, therefore $f\left(2\right)=\left|2\right|=2$, and$f(-2)=|-2|=2$

This is why $\left|x\right|$ represents the value of a number x disregarding its sign.

If you have an expression inside the modulus function, calculate the value inside, then find the positive version of the result.

If you have the function $f\left(x\right)=|x-3|+1$ find$f(-2)$

$f(-2)=|-2-3|+1$

$f(-2)=|-5|+1=5+1=6$

## Equation of modulus functions

The **equation of a modulus function** is denoted as follows:

The **domain** of a modulus function is the set of all Real Numbers, and the **range** is the set of all Real Numbers greater than or equal to zero. From the equation, we can say that if the number inside the modulus function is already positive, you leave it like that, but if the number is negative, then the result will be the positive version of that number (as if you were multiplying the negative number by -1).

### Properties of modulus functions

The properties of modulus functions are:

The modulus or absolute value of a number will always give a positive result.

$\left|4\right|=4,|-5|=5$

The modulus of a number x will give the same result as the modulus of -x.

$\left|x\right|=|-x|=x$

$\left|4\right|=|-4|=4$

The

**modulus of the product of two values**a and b can be split into the product of two separate modulus values.

$|axb|=\left|a\right|x\left|b\right|$

$|2(-3\left)\right|=\left|2\right|\times |-3|\phantom{\rule{0ex}{0ex}}|-6|=2x3\phantom{\rule{0ex}{0ex}}6=6$

The

**modulus of the division of two values**a and b can be split into the division of two separate modulus values.

$\left|\frac{a}{b}\right|=\frac{\left|a\right|}{\left|b\right|}$

$\left|\frac{-9}{3}\right|=\frac{|-9|}{\left|3\right|}\phantom{\rule{0ex}{0ex}}|-3|=\frac{9}{3}\phantom{\rule{0ex}{0ex}}3=3$

The

**modulus of the sum or subtraction of two values,**a and b,**cannot be split**into the sum or subtraction of two separate modulus values.

$|a\pm b|\ne \left|a\right|\pm \left|b\right|$

**Sum:**

**Subtraction:**

When

**solving Equations**, modulus functions involve an extra step:

Keeping in mind that the value of x inside a modulus function could be positive or negative, you need to solve the equation considering both cases, therefore you will get two solutions.

For the equation $|3x-2|=4$, we can obtain 2 possible solutions as follows:

1) **Solution 1:**

2) **Solution 2:**

## How do you graph modulus functions?

To draw a graph of a modulus function you need to substitute values of x into $f\left(x\right)=\left|x\right|$, to obtain the corresponding values of y, as $y=f\left(x\right)$. You will get a table of x and y values that you will need to plot on the coordinate plane. We will substitute the values of x from -2 to 2.

x | y |

-2 | 2 |

-1 | 1 |

0 | 0 |

1 | 1 |

2 | 2 |

To sketch the graph of the modulus function $y=|ax+b|$, you need to sketch $y=ax+b$, and reflect the portion of the line that goes below the x-axis into the x-axis.

Sketch the graph for $y=|x-1|$ showing the points where they cross the coordinate axes.

Ignoring the modulus, you need to sketch the graph of $y=x-1$

When $y=0$, $x=1$

The line crosses the x-axis at (1, 0)

- When $x=0$, $y=-1$

The line crosses the y-axis at (0, -1)

Sketch the graph for $y=x-1$:

- For the negative values of y, reflect in the x-axis. In this case, (0, -1) becomes (0, 1)

## Solving equations involving modulus functions

When you have an equation like $|3x-1|=5$, you can use its graph to help you find its solution by following these steps:

Sketch the Graphs of both sides of the equation separately. In this case, $y=|3x-1|$ and$y=5$

Identify the intersection points of the two Graphs. In this case,

**A**corresponds to the intersection point between y = 5 and the original section of the graph of $|3x-1|=3x-1$, and**B**represents the intersection between $y=5$ and the reflected section of the graph of $|3x-1|=-(3x-1)$.Find both solutions:

**A:** $3x-1=5$

$3x=5+1$

$3x=6$

$x=\frac{6}{3}$

$x=2$

**B:** $-(3x-1)=5$

$-3x+1=5$

$-3x=5-1$

$-3x=4$

$x=\frac{-4}{3}$

### Solving inequalities involving modulus functions

Based on the previous example, now we are going to solve the inequality $|3x-1|>5$. You need to proceed in the same way as before to find the values of x at intersection points A and B, which are $x=2$ and $x=\frac{-4}{3}$.

After you have the points of intersection, you can look at the graph to identify the values of x that satisfy the inequality $|3x-1|>5$.

The inequality is true when the graph of $y=|3x-1|$ is above the graph of $y=5$, this happens when $x<\frac{-4}{3}$ or $x>2$. In set notation:$\{x:x<\frac{-4}{3}\}\cup \{x:x>2\}$

### Inverse of a modulus function

The **inverse of a modulus function** is **not a function** unless you restrict its domain so that it can be a one-to-one function. To achieve this, we need to restrict its domain to only one half of the graph. You can choose either half if it is not specified in the question.

Find the inverse of the function$f\left(x\right)=|x+1|$

We will restrict the domain of the function to only the reflected section of the graph (left of x = -1), which can be denoted as $f\left(x\right)=-(x+1)$ for $x\le -1$. Now we can find the inverse, because this section of the graph is a one-to-one function.

Following the steps to find the inverse of a function:

- Replace f (x) by y

$f\left(x\right)=-(x+1)$

$y=-(x+1)$

- Swap x and y, and solve for y

$x=-y-1$

$y=-x-1$

${f}^{-1}\left(x\right)=-x-1$ This is the inverse function of $f\left(x\right)=|x+1|$

The domain of the inverse function is the range of the original function, which is $y\ge 0$. Therefore, the domain of the inverse function ${f}^{-1}\left(x\right)=-x-1$ is $x\ge 0$.

## How do you differentiate a modulus function?

To find the derivative of the modulus function, we need to look at the equation of a modulus function again:

$f\left(x\right)=\left|x\right|=\left\{\begin{array}{l}xifx\ge 0\\ -xifx<0\end{array}\right.$

We know that $\frac{d}{dx}x=1$, therefore we can say the following:

$\frac{d}{dx}\left(\right|x\left|\right)=\left\{\begin{array}{ll}1& ifx>0\\ -1& ifx<0\\ Notdefined& ifx=0\end{array}\right.$

In general, $\frac{d}{dx}\left(\right|x\left|\right)=\frac{x}{\left|x\right|}$for all values of x apart from$x=0$

If we substitute some values of x in the previous equation, then we can see that the statements in the piecewise function above are true:

$\frac{d}{dx}\left(\right|-1\left|\right)=\frac{-1}{|-1|}=\frac{-1}{1}=-1$

$\frac{d}{dx}\left(\right|0\left|\right)=\frac{0}{\left|0\right|}=\frac{0}{0}=undefined$

$\frac{d}{dx}\left(\right|1\left|\right)=\frac{1}{\left|1\right|}=\frac{1}{1}=1$

## How do you integrate a modulus function?

To find the integral of a modulus function, we can proceed as follows:

We know that the modulus function is defined like this,

$f\left(x\right)=\left|x\right|=\left\{\begin{array}{l}xifx\ge 0\\ -xifx<0\end{array}\right.$

Therefore, we need to calculate the integral for x and -x.

Remember that x has an exponent of 1 ($x={x}^{1}$)

Using the Integration formula: $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+c$

$\int \left|x\right|dx=\left\{\begin{array}{ll}\frac{1}{2}{x}^{2}+c& ifx\ge 0\\ -\left(\frac{1}{2}{x}^{2}\right)+c& ifx<0\end{array}\right.$

## Modulus Functions - Key takeaways

- The modulus of a number x will be the same number, but positive.
- The modulus of a number x represents the distance from zero to that number x on the number line.
- To sketch the graph of the modulus function $y=|ax+b|$, you need to sketch $y=ax+b$, and reflect the portion of the line that goes below the x-axis into the x-axis.
- Sketching the graphs of equations or inequalities involving modulus functions can help to solve them, by finding the x-coordinates of the intersection points of the two graphs.
- The inverse of a modulus function is not a function, unless you restrict its domain to only one half of the graph, so that it can be a one-to-one function.
- When finding the derivative and integral of a modulus function, there will be two possible solutions, considering when $f\left(x\right)=\left|x\right|=x$ (for $x\ge 0$) and when $f\left(x\right)=\left|x\right|=-x$ (for $x<0$).

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##### Frequently Asked Questions about Modulus Functions

How do you integrate a modulus function?

∫|x| dx = ½x² + c if x ≥ 0, -(½x²) + c if x < 0

Define modulus function and describe its graph.

Modulus functions, also known as absolute value functions, are generically represented as f(x) = |x|. The modulus of a number x will be the same number, but positive. The graph of a modulus function has a characteristic v-shape.

How to draw a graph of a modulus function?

To sketch the graph of the modulus function y = |ax+b|, sketch y = ax+b, and reflect the portion of the line that goes below the x-axis into the x-axis.

How to differentiate a modulus function?

d/dx (|x|) = 1 if x > 0, -1 if x < 0, not defined if x = 0

d/dx (|x|) = x/|x| for all values of x apart from x = 0

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