Real Numbers:Absolute Value

Absolute Values

Absolute Values represented using two vertical bars, $\vert$, are common in Algebra. They are meant to signify the number's distance from 0 on a number line. If the number is negative, it becomes positive. And if the number was positive, it remains positive:

$\left\vert 4\right\vert =4\,$
$\left\vert -4\right\vert =4\,$

For a formal definition:

$|x|={\begin{cases}x,&{\text{if }}x\geq 0\\-x,&{\text{if }}x<0\end{cases}}$

This can be read aloud as the following:

If $x\geq 0$, then $|x|=x$
If $x<0$, then $|x|=-x$

The formal definition is simply a declaration of what the function represents at certain restrictions of the $x$-value. For any $x<0$, the output of the graph of the function on the $xy$ plane is that of the linear function $y=-x$. If $x\geq 0$, then the output is that of the linear function $y=x$.

For our purposes, it does not technically matter whether $x\geq 0{\text{ and }}x<0$ or $x>0{\text{ and }}x\leq 0$. As long as you pick one and are consistent with it, it does not matter how this is defined. By convention, it is usually defined as in the beginning formal definition.

Please note that the opposite (the negative, -) of a negative number is a positive. For example, the opposite of $-1$ is $1$. Usually, some books and teachers would refer to opposite number as the negative of the given magnitude. For convenience, this may be used, so always keep in mind this shortcut in language.

Properties of the Absolute Value Function

We will define the properties of the absolute value function. This will be important to know when taking the CLEP exam since it can drastically speed up the process of solving absolute value equations. Finally, the practice problems in this section will test you on your knowledge on absolute value equations. We recommend you learn these concepts to the best of your abilities. However, this will not be explicitly necessary by the time one takes the exam.

Domain and Range

Let $f(x)=|x|$ whose mapping is $f:\mathbb {R} \to \mathbb {R}$. By definition,

$|x|={\begin{cases}-x&{\text{if}}&x<0\\x&{\text{if}}&x\geq 0\end{cases}}$.

Because it can only be the case that $y=-x{\text{ if }}x<0$ and $y=x{\text{ if }}x\geq 0$, it is not possible for $|x|<0$. However, since $x$ has no restriction, the domain, $A$, has no restriction. Thus, if $B$ represents the range of the function, then $A=\{x\in \mathbb {R} \}$ and $B=\{y\geq 0|y\in \mathbb {R} \}$. Template:Calculus/Def By the above definition, there exists an absolute minimum to the parent function, and it exists at the origin, $O(0,0)$

Even or odd?

Recall the definition of an even and an odd function. Let there be a function $f:A\to B$

If $x\in A$ and $f(-x)=f(x)$, then $f$ is even.

If $x\in A$ and $f(-x)=-f(x)$, then $f$ is odd.

Template:ExampleRobox Let $f:\mathbb {R} \to \mathbb {R} :x\mapsto |x|$. By definition,

$f(x)=|x|={\begin{cases}-x&{\text{if}}&x<0\\x&{\text{if}}&x\geq 0\end{cases}}$.

Suppose $x\in A$. Let $x>0\Rightarrow -x<0$.

$f(x)=x$

$f(-x)=-(-x)=x$

$\Rightarrow f(-x)=f(x)\blacksquare$

Template:Robox/Close Because $f(x)$ is even, it is also the case that it is symmetrical. A review of this can be found here (Graphs and Their Properties).

One-to-one and onto?

Recall the definitions of injective and surjective.

If $u,v\in A$, and $f(u)=f(v)\Rightarrow u=v$, then $f(x)$ is injective.

If for all $b\in B$ there is an $a\in A$ such that $f(a)=b$, then $f(x)$ is surjective.

Template:ExampleRobox Suppose $u,v\in \mathbb {R}$ and $f(u)=f(v)$. By the previous proof, we showed $f(x)$ is even. As such, we can use the value $v=-u$ to make the following statement:

$f(u)=f(v)\Rightarrow u\neq v$

Therefore, $f(x)$ is non-injective. Template:Robox/Close Because we have not established how to prove these statements through algebraic manipulation, we will be deriving properties as we go to gain a further understanding of these new functions. Establishing whether a function is surjective is simply through checking the definition (negating if otherwise to establish it as non-surjective). Template:ExampleRobox Suppose $b\in \mathbb {R}$. There exists an element $b=-1\in \mathbb {R}$, for which $f(x)=|x|\neq -1$ for all $x\in \mathbb {R}$.$\blacksquare$ Template:Robox/Close A review of the definitions can be found here (Definition and Interpretations of Functions).

Intercepts and Inflections of the Parent Function

Figure 1: $f(x)=|x|$ graphed on the first and second quadrant (above the $x$ axis), showing only the positive $y$ values.

With all the information provided from the previous sections, we can derive the graph of the parent function $f(x)=|x|$. It is even, and therefore, symmetrical about the $y$-axis since there is an $x$-intercept at $x=0$. Finally, because we know the domain and range, we know the minimum of the function is at $O(0,0)$, and we know the definition of the function, we can easily show that the graph of $f(x)=|x|$ is the following image to the right (Figure 1).

A summary of what you should see from the graph is this:

• Domain: $\{x\in \mathbb {R} \}$.
• Range: $\{y\geq 0|y\in \mathbb {R} \}$.
• There is an absolute minimum at $O(0,0)$.
• There is one $x$-intercept at $x=0$.
• There is one $y$-intercept at $y=0$.
• The graph is even and symmetrical about the $y$-axis.
• The graph is non-injective and non-surjective.
• The graph has no inflection point.

Transformations of the Parent Function

Many times, one will not be working with the parent function. Many real life applications of this function involve at least some manipulation to either the input or the output: vertical stretching/contraction, horizontal stretching/contraction, reflection about the $x$-axis, reflection about the $y$-axis, and vertical/horizontal shifting. Luckily, not much changes when it comes to the manipulation of these functions. The exceptions will be talked about in more detail:

The properties not listed above are exceptions to the general rule about functions found in the chapter Algebra of Functions. The exceptions are not anything substantial. The only difference with what we found generally versus what we have provided above are simply a result of what we found in the previous section.

• There is no reflection about the $y$-axis because the function is even and symmetrical.
• There is no horizontal expansion and contraction because it gives the same result as vertical expansion and contraction (this will be proven later).

We now have all the information we will need to know about absolute value functions now.

Graphing Absolute Value Functions

This subsection is absolutely not optional. You will be asked these questions very explicitly, so it is a good idea to understand this section. If you didn't read the previous subsection, you are not going to understand how any of this makes sense.

Fortunately, the idea behind graphing any arbitrary function is mostly dependent on what you know about the function. Therefore, we can easily be able to graph functions. These examples should hopefully be further confirmation of what you learned in Algebra of Functions.

Method 1: Follow procedure from Algebra of Functions

This method will work for any arbitrary function. However, it will not always be the quickest method for absolute value functions. We follow the following steps. Let $f(x)$ be the parent function and $g(x)=Af(ax+b)+c$.

1. Factor $ax+b$ so that $ax+b=a\left(x+{\frac {b}{a}}\right)$.
2. Horizontally shift $f(x)$ to the left/right by ${\frac {b}{a}}$.
3. Horizontally contract/expand $f\left(x-{\frac {b}{a}}\right)$ by $a$.
4. Vertically expand/contract/flip $f\left(a\left(x-{\frac {b}{a}}\right)\right)$ by $A$.
5. Vertically shift $Af\left(a\left(x-{\frac {b}{a}}\right)\right)$ upward/downward by $c$.

Since $f(x)={\frac {1}{2}}|2x+6|-5$ has $A={\frac {1}{2}}$, $a=2$, $b=6$, and $c=-5$, we may apply these steps as given to get to our desired result. As this should be review, we will not be meticulously graphing each step. As such, only the final function (and the parent function in red) will be shown.

Method 3: Find absolute minimum or maximum, graph one half, reflect.

While method 1 will always work for any arbitrary, continuous function, method 3 is fastest for the absolute value function that composes a linear function.

First, we should try to find the vertex. We know from Algebra of Functions that the only thing that will affect the location of the vertex in even functions is the $x-a$ term on the inner composed linear function and the vertical shift of the entire function, $c$.

Rewriting the absolute value equation as shown below will allow us to find the vertex of the function.

$f(x)={\frac {1}{2}}|2x+6|-5={\frac {1}{2}}|2(x+3)|-5$

This then tells us the vertex is at $(-3,-5)$.

This method then tells us to graph the slopes. However, how should that work? Recall the formal definition of an arbitrary absolute value function:

$|g(x)|={\begin{cases}-g(x)&{\text{if}}&x<x_{0}\\g(x)&{\text{if}}&x\geq x_{0}\end{cases}}$

In the above definition of a general absolute value function, $g(x_{0})=-g(x_{0})=0$. This means that where the $x$-value implies a vertex on the function, that is how we restrict absolute value function.

In our instance, $|g(x)|=|2x+6|$, for which $2(-3)+6=0$, so $x_{0}=-3$. We can say, thusly, that

 :$|2x+6|={\begin{cases}-2x-6&{\text{if}}&x<3\\2x+6&{\text{if}}&x\geq 3\end{cases}}$

To be continued.