A Graphical Analysis of the Sums of Absolute Value Functions

A GRAPHICAL ANALYSIS OF THE SUMS OF ABSOLUTE VALUEFUNCTIONS

ANSHUMAN DAS

1. Introduction

The American high school mathematics curriculum has startlingly little to say aboutabsolute value functions containing more than one absolute value bracket. We learn, forexample, how to graph functions such as f(x) = 2|x− 3|. But almost never do we encountermore complicated functions such as f(x) = 2|x− 3|+ 4|x+ 8|+ 7|x− 5|, which exhibit muchmore interesting graphical behavior. The basic graphical pattern, for example, is no longerthat of a “V” shape.

The main problem that I wish to solve is how we can learn about these sums and use themto our advantage. My aim in studying these functions is to fill this hole in our curriculum.One sign of how little attention we pay to such topics is that it is impossible to find a neatpresentation of results on this subject, either in a textbook or online. In studying thesefunctions, I found some interesting behavior that later allowed me to add more and more tomy findings.

Below, I study the behavior of sums of absolute value functions and their applications. Ihandle this case by looking at a general parametric family of functions:

f(x) =k∑

i=1

mi|x− ai|

where the numbers represented by the variables m and a are real constants.My approach into learning about the sums was to generally experiment with the behav-

ior. I plotted several equations into a grapher and held numbers like the constants a andcoefficients m as my independent variables, where as the shape of the graph and its behaviorbecame my dependent variable. From this behavior, I generated hypotheses of how eachindividual behavior shows the interaction of the function with the graph and then proved itwith algebra.

In the end, I was able to prove that the sums of absolute value functions were actuallycontinuous piecewise affine functions

After presenting concrete results on this basic case, I generalize by characterizing thequalitative behavior of more general sums. Through my investigation, I was able to generatea theory of how these functions operate.

2. Examples of Sums of Absolute Value Functions

2.1. Two Absolute Value Terms. To investigate these functions, I will take examples,and then generalize the example to all scenarios. I begin by studying the simplest case of a

Date: August 24, 2014.1

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sum of two absolute value functions. Let us look at the function:

f(x) = 3|x− 2|+ 2|x + 1|

Here is a graph of the function shown below:

Figure 1. Sum of two absolute value functions with positive end slope

Just by looking at this graph, we can see two very interesting things. The first is that thecorners of the function, or the non-differentiable points on the function, have x -coordinatesof −1 and +2, respectively, from their positions on the graph. Now if we look at the actualfunction, we see that the constant inside the absolute value brackets are −2 and +1, exactlythe opposite of the x -coordinates. Keep this idea stored for later, because I will explain thislater.

The second aspect of the function is the end behavior slope. The magnitude of the slopeof both the rays is equal and appears to be 5. Now look at the coefficient of each absolutevalue bracket: +3 and +2, respectively on the function. The slope magnitude appears to bethe sum of the two coefficients, but that is just a claim for now: we will justify this later.

Now let us look at another example:

Figure 2. Sum of two absolute value functions with zero end slope

A GRAPHICAL ANALYSIS OF THE SUMS OF ABSOLUTE VALUE FUNCTIONS 3

Here the corners are situated at x = (−2) and x = 3. Like before, the corners seem to bepositioned the negative of the constant added, or in other terms, follow the pattern x − a,where a is the x -coordinate of the corner.

Also, the end behavior slopes are equal to zero. If we look at the coefficients of eachabsolute value bracket, their sum is equal to zero: 4 + (−4) = 0.

2.2. Three Absolute Value Functions. This behavior holds to be true for two functionswith two absolute value brackets, so now we move on to functions with three brackets. Ourfirst example is the function:

f(x) = 2|x− 3|+ |x− 1|+ 5|x + 4|

After graphing the function, we obtain:

Figure 3. Sum of three absolute value functions

This function appears to have no more than 3 corners. The corners, like before, are alllocated at the negative of the actual constant inside of the absolute value brackets, or inother words, x = {3, 1,−4}.

The magnitudes of the end behavior slopes appear to be 8, as predicted by the coefficientsof the absolute value brackets, respectively 2, 1, and 5. Their sum, as stated before, seemsto give us the slope magnitude, which is 2 + 1 + 5 = 8. How does this hold true for bothdouble and triple absolute value brackets?

2.3. N Absolute Value Functions. In this last example, we provide a complicated casewith a function containing 8 absolute value brackets:

f(x) = −|x− 6| − |x− 4| − |x− 2| − |x|+ |x + 2|+ |x + 4|+ |x + 6|+ |x + 8|

The graph of this function appears below:

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Figure 4. Sum of eight absolute value functions

If you look closely, you will notice that this is not a smooth curve, but rather a connectionof segments. There are precisely 8 corners, each of whose x -coordinate is located at thedesignated constant inside the absolute value bracket: x = {6, 4, 2, 0,−2,−4,−6,−8}. Inaddition, the end behavior slope magnitude is equal to 0. Coincidentally, the sum of all thecoefficients of the absolute value terms adds up to 0: −1+(−1)+(−1)+(−1)+1+1+1+1 = 0.

By now, you may have also seen another interesting behavior of these functions: thenumber of absolute value brackets seems to equal the number of the corners. Why does thiswork? How do the sums of absolute value functions lead to these new functions?

3. Functional Corners

One of the key characteristics of a corner is its non-differentiability. We can use this factto prove that the non-differentiable points appear at an x-coordinate where of the absolutevalue bracket can be reduced to zero and that there are the same number of non-differentiablepoints as there are irreducible absolute value terms. First, we start by explaining what howto calculate the derivative of an absolute value function. If we rewrite the function as thesquare root of the square of the function, then it becomes easier to differentiate it using thechain rule:

d

dx

k∑i=1

mi|x− ai| =d

dx(m1|x− a1|+ . . . + mk|x− ak|)

= m1x− a1|x− a1|

+ . . . + mkx− ak|x− ak|

=k∑

i=1

mix− ai|x− ai|

Since we already know that all the parts of this function are linear and we also know thatthe graph is continuous, then this shows that the points are corners and non-differentiable.Now let’s try to find the slope at the point x = ai. As we can see, this forces the ith absolutevalue derivative term to blow up to infinity. Thus, the slope is not calculable at ai. Thisholds true for all ai: if we have, for example, 137 irreducible absolute value terms, then all137 terms have a value of x such that the slope is not calculable.


This result also shows that the non-differentiable points appear at the x-coordinate of ai,or at the constant inside each absolute value term.

4. General Behavior

It must be shown that the behavior of sum of absolute value functions is not unique tospecific functions. To show this, we must look at the behavior of singular absolute valuefunctions. Since these functions are essentially piecewise functions, we can rewrite, forexample, |x| as:

f(x) =

{−x when x < 0x when x ≥ 0

So at the point of non-differentiability, the function is broken up into subsections. Since thenon-differentiable points are at the x-coordinates of the functional corner, then it becomeseasier to break up sums of absolute value functions. Let’s say that our function is of thegeneral form:

f(x) =n∑

i=1

mi|x− ai|

Now let us define ai such that a1 < a2 < a3 < . . . < an. From this definition, we canbreak up the single function into a piecewise function with separate domains of x ≤ a1,a1 < x ≤ a2, . . . , an < x. From the first piece of the function, if x is less than a1, then thevalues inside every absolute value are non-positive, so the function becomes:

f(x) =n∑

i=1

mi|ai − x|

For the next piece of the function, a1 < x ≤ a2, the value of the absolute value will benon-negative for values of x > a1, but non-positive for any x ≤ a2:

f(x) = m1|x− a1|+n∑

i=2

mi|ai − x|

The next piece, a2 < x ≤ a3, has a similar result: the absolute value is non-negative forany a2 < x and non-positive for any x ≤ a3:

f(x) = m1|x− a1|+ m2|x− a2|+n∑

i=3

mi|ai − x|

=2∑

i=1

mi|ai − x|+n∑

i=3

mi|ai − x|

If we generalize this to a point anywhere on the function such that aj < x and x > aj+1,then we can rewrite the sums as:

f(x) =

j∑i=1

mi|ai − x|+n∑

i=j+1

mi|ai − x|

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By expanding out the complete sum, and rearranging the variables, we obtain the function:

f(x) =

(j∑

i=1

mi −n∑

i=j+1

mi

)x +

(n∑

i=j+1

miai −j∑

i=1

miai

)From this fact, we actually see that it takes the form of f(x) = mx + b, which makes it

evident that every piece of the function is affine, but each piece takes on a different affinefunction.

Proposition 1. The function f(x) =∑

mi|x−ai| is a continuous piecewise linear function.

Proof. The explanation is clearly stated above, so to restate would be trivial. �

The slopes of each line segment would be given by the expanded affine function:

m =

j∑i=1

mi −n∑

i=j+1

mi

Through this equation, we can actually see that the end slopes are of opposite sign. Thiswill be a very important concept to note later when disproving the converse statement.

Remark 1. There are some characteristic behaviors that are significant about the sums ofabsolute value functions. Through the proposition, there are some interesting generalizationswe can get out of the sums:

(1) The slope at any differentiable point on the function is equal to the total sum, whichincludes both negative and positive parts of the coefficients. The sum is dependent onthe location of the differentiable point with respect to local non-differentiable points.

(2) The number of non-differentiable points of the function is equal to the number ofabsolute value brackets.

(3) The non-differentiable points appear at an x-coordinate equal to the negated constantinside each absolute value term.

So now that we know that sums of absolute value functions are actually continuous piece-wise affine functions, does this mean that the converse is true? This is, in fact, not thecase. For the converse to be true, all conditions must be satisfied: oppositely signed endslopes and correctly placed non-differentiable points. However, there are numerous piecewisecounterexamples that involve end slopes of completely different magnitude. Of course, thisjust means that there is the option of either correct end behavior or correct inner behavior.The conditions that need to be satisfied are opposite end slopes and correctly placed non-differentiable points. Depending on the situation, you have the option between either of thetwo, but not both.

5. Generation of a Function from the Graph

This is not to say, however, that if given a graph that has a predetermined function, thatthere is no possible way to write the function. We assume the general series expansion of theequation to a limited number of absolute value brackets, depending on the number of non-differentiable points. From this, based on the x-coordinate of each non-differentiable point,we can figure out what each individual an term will be. Now the last piece of information


is the coefficient of each absolute value term. To solve for this, we need a set of equationsto solve for the unknowns, and then the last thing to do is to solve the system for eachindividual coefficient.

The first equation you will need is the sum of the coefficients. By looking at the graph,we can determine the RHS ray slope, and from there, our first equation will be:

mr =k∑

i=1

mi

where k is the number of absolute value terms.The following equations will all be equations based on substitutions of the individual non-

differentiable points on the function’s graph. This should give you k equations, but onlyk − 1 equations will be sufficient. So in total, you will have k − 1 equations plus the firstequation for a total of k equations and k variables. This should give you the coefficients ofeach absolute value term, and consequently, the equation of the function.

Consider the following example:Suppose we have the graph that looks like:

Figure 5. Example problem

Let’s note some characteristic behavior of the graph:

• End behavior slopes are equal to zero• Three non-differentiable points located at (−3, 12), (2,−8), and (4,−12), respectively

From this, we can write a general equation of the function:

y = f(x) = m1|x + 3|+ m2|x− 2|+ m3|x− 4|

If we know that the coordinates of the non-differentiable points, we can substitute the val-ues of x and y into the function and obtain three equations (any one of which is unnecessary).So the three equations are:

12 = 5m2 + 7m3(1)

−8 = 5m1 + 2m3(2)

−12 = 7m1 + 2m2(3)

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One of these equations is redundant, so we can arbitrarily choose one of them, and get ridof it. Next we need an equation for the end behavior slope: because the RHS and LHS rayslopes are 0, we can say the sum of the coefficients is equal to 0:

0 = m1 + m2 + m3(4)

Now let’s eliminate the third equation from the first set of three equations obtained fromthe substitution of the non-differentiable points. After solving the system using whatevermethod, the variables come out to: m1 = −2, m2 = 1, and m3 = 1. Therefore, the finalequation of the function is:

f(x) = −2|x + 3|+ |x− 2|+ |x− 4|

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A Graphical Analysis of the Sums of Absolute Value Functions