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Definition, properties and graphing of absolute value

absolute value

Absolute value represents the distance from one point on the number line to the point of origin.

We’ll be needing a number line and a few points, for example: \ A (4), \ B(2), C(-2). We’ll mark them on the number line and observe their distance from the number zero. That distance is also called a module of those numbers.

From the picture we can conclude that the distance between points A and O equals 4, from point B to point O is 2 and from point C to point O is 2. Notice how the absolute value of positive numbers is exactly that number, and the absolute value of negative numbers is the positive version of that number.

This will always be true, whether we’re working with real, rational, irrational, integers or natural numbers.

Properties of absolute value

An absolute value is marked by two vertical lines placed to the left and right of a number, and it is defined as:

mark absolute value

This means that it doesn’t matter whether the x is positive or negative, its absolute value will always be positive.

\mid x \mid = 0 if and only if x= 0

\mid x \mid \ge 0 for every x \in \mathbb{R}

An absolute value also has some properties:

  1. \mid a \cdot b \mid = \mid a \mid \cdot \mid b \mid for every a, b \in \mathbb{R}
  2. \mid \frac{a}{b} \mid = \frac{\mid a \mid}{\mid b \mid}, b \not= 0 for every a \in \mathbb{R}, b \in \mathbb{R} \setminus \{ 0 \}
  3. \mid a + b \mid \le \mid a \mid + \mid b \mid, for every a, b \in \mathbb{R}

 

Are those equalities true?

  1. It doesn’t matter whether A or B are negative or positive, their product will always be the same. The sign in front of the product of two numbers depends on whether are both positive, both negative or one positive and other negative. Because the absolute value will always give positive value, this doesn’t really matter.
  2. Similarly, a quotient is positive if both the divisor and dividend have the same sign in front of them (be it a “+” or “-“), so it doesn’t matter if you divide two positive numbers, or if you can remove the minus sign from the result.
  3. Why is there an inequality? Well, the logic we used for multiplication and division cannot be applied here, because it’s not the same if you add two positive numbers, and if you have subtraction and then you make it positive. For example:

\ |2|- |4| = 2 - 4 = - 2 and \ |2 - 4| = |-2| = 2
 

Example 1. Calculate the value of the given expression if:

\ a = 2, \ b = -5 ,

\ |(|a|-|b|)| = ?.

First, we’ll include the values of a and b into the task:

\ |(|2|-|-5|)| = ?

When you have more than one absolute value, you treat them just like parenthesis; first solve inside the absolute value and then outside.

\ |(|2|-|-5|)| = |2-5| = |-3| = 3

Example 2. Calculate the value of given expression if:

a = -\frac{-1}{2} , b = \frac{4}{3}. \frac{2|a - 3b|}{|b|} = ?

\frac{2 \mid -\frac{1}{2} - 4 \mid}{\mid \frac{4}{3} \mid}

=\frac{2 \mid -\frac{9}{2} \mid}{\frac{4}{3}}

= \frac{9}{\frac{4}{3}}

= \frac{27}{4}


Graf of absolute value

Example 1.

We all know how to draw the graph of \ f(x) = x. But what if you start wondering how you would draw a graph of \ f(x)= |x|?

Let’s see what happens in the range from -3 to 3.

number fo making graph

You can notice that this resulted with a same value for opposite numbers. This means that the function will be symmetrical.symmetrical functions of absolute value

Now let’s graphically review the difference between \ |a+b| and \ |a| + |b|

Example 2.

We’ll observe two functions: \ f(x) = |x - 2| and \ g(x) = |x| - 2

absolute function

Parts of two different functions overlapped. Why? Well, when you have a constant inside your absolute value, the whole graph will be moved to the right if that constant is negative, and if it’s positive, to the left. But if you have a constant outside of absolute value (e.g. the “-2” in the function g(x)), the graph will move upward if that value is positive, and downward if it’s negative. The x coordinate of the vertex (the point in which the two lines connect, the tip of the graph) is exactly the value by which you shifted the graph left or right, and the y coordinate is the value by which you shifted the graph up or down.

overlapping functions

You have only two more options for a simple graph with absolute value, and that is if a constant greater than 0 multiplies your variable, or if a constant lesser than zero does it.

Example 3.

We’ll draw functions : \ f(x) = 2 |x|, g(x) = \frac{1}{2}|x| , \ z(x) = -2 |x|

three absolute value functions

From here, you can conclude that if you have smaller coefficients multiplying an absolute value, its graph will become more widespread, and if you have larger coefficients, it will become narrower. And if you multiply the function with a negative coefficient, the graph will move to the negative side of y – axis.

graph of three different absolute value functions


Simple absolute equalities and inequalities

Example 1.
Solve: \ |x| + 2 = 4 .

These equalities are solved in a similar way as are normal equalities. That means, by moving the unknowns (it doesn’t matter if they are bound with module, or not) to the left side, and the constants to the right. Remember to change the sign of the constant due to the side change and this is what we get:

\ |x| = 2. Now, here is the different part. This equation has two solutions: 2 and -2. That’s because \ |2| = |-2| = 2

Example 2.

\mid -0.2x \mid + \mid -\frac{3}{4} \mid = 2 -> \mid -\frac{1}{5}x \mid + \frac{3}{4} = 2

\mid -\frac{1}{5}x \mid = \frac{5}{4} here we can use property number 1 (\mid a * b \mid = \mid a \mid * \mid b \mid)

\mid -\frac{1}{5} \mid \mid x \mid = \frac{5}{4}

\frac{1}{5} \mid x \mid = \frac{5}{4} \setminus 5

\mid x \mid = \frac{25}{4} \rightarrow x_1 = \frac{25}{4}, x_2 = -\frac{25}{4}

 

Example 3.

\ |-2x - 2| = 2

First we have to know for what values of x will the value of the whole module become negative, and for what positive. That is done by “searching zeroes” (equating the expression with 0, and then solving that equation) from the module:

\ -2x - 2 = 0 
\ -2x = 2 
\ x = -1 
This value is very important. When the value of x crosses this point, it immediately changes its sign. How do we know on which side of this point will the values be positive and on which negative?
Choose some random number from the interval \ <-\infty, -1 > and then see the value of the module in that case.

For example we’ll take \ -2: -2(-2) - 2 = 2 which is greater than zero. And that means that for the interval <-∞, -1 > everything inside module will remain the same, and

1. Case : \ x < -1 -> everything inside the module remains the same: \ - 2x - 2 = 2 -> \ x = - 2

2. Case : \ x > -1 -> everything inside the module changes sign  \ 2x + 2 = 2 -> \ x = 0

To check our solution, we can simply replace the x with an appropriate value:

For -2: \ |-2(-2)-2| = 2 , \ |2| = 2

For 0: \ |0-2| = 2 , \ |2| = 2

Also, you should always check if your x is from the interval you are looking for. If it’s not, the x  is not included in the set of solutions. -2 < -1, 0 > -1. Since – 2 and 0 are parts of our interval, they are included in our set of solutions.

Example 4.

|\frac{x-1}{x-2}|=\frac{1}{2}

Solution:

We use properties: |a|=|b| \Leftrightarrow a=b or a=-b

\frac{|x-1|}{|x-2|}=\frac{1}{2} /\cdot 2|x-2|, x \neq 2

2|x-1|=|x-2|

|2x-2|=|x-2|

Now, we solve this like before:

1. 2x-2=x-2

2x-2=x-2   /+(-x+2)

2x-2-x+2=x-2-x+2

x=0

2. 2x-2=-x+2

2x-2=-x+2 /+(x-2)

2x-2+x-2=-x+2+x-2

x=\frac{4}{3}

Example 5.

\ |x| > 2

Now, variable x can again be either positive or negative. So we’ll break it down to cases:

Case 1: for \ x > 0 -> \ x > 2
Case 2: for \ x > 0 -> \ - x > 2 -> \ x < - 2
And the final solution is the union of those two intervals: \ <-\infty, -2 > U < 2, \infty >.
The solution represented graphically:

Example 6.

\ |x + 3| > 4 (the same as the equations)

\ x + 3 = 0 -> \ x = - 3

For x = 0 we have \ 0 + 3 = 3 > 0.

1. Case: \ x > 3 -> everything inside module remains the same -> \ x + 3 > 4 , \ x > 1

2. Case: \ x < 3 -> everything inside module changes sign  -> \ - x - 3 > 4 , \ - x > 7 , \ x > - 7

Here we have to look out for the conditions:
In the first case, we are looking only at numbers greater than three. Since we got a solution x > 1, we have to look at their intersection, which is \ x > 3

In the second case, we are only looking at numbers lesser than 3, and we got the solution x > -7. Once again, we are looking for the intersection. If you’re having trouble imagining their intersection, you can always help yourself by drawing it on the number line:

So our solution for the second case is \ <-7, 3 >

And the solution to the whole task is: $\ <-7, 3 > .

 

Absolute value worksheets

  Solve and graph functions - Integers (1.2 MiB, 438 hits)

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  Equations - Integers (509.2 KiB, 392 hits)

  Equations - Decimals (542.2 KiB, 364 hits)

  Equations - Fractions (784.8 KiB, 351 hits)

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