Looking at the Picture
Absolute Value Equations and Inequalities
April 19, 2009
1
Graphing Absolute Value Functions
One way to define absolute values is to introduce the function
(
−x x < 0
|x| =
x
x≥0
(1)
The graph of this function is easy to draw: from (1) we see that for −∞ < x < 0,
its graph is a straight line with slope −1, while for 0 ≤ x < ∞, it is a straight
line with slope 1, both lines joined at (0, 0):
In all our problems we are only taking on absolute values of linear expression,
like, as an example, |1 − 3x|. The graph of this function can be obtained from
the one for |x|, via transformations like “shrinking” or “shifting”. With a little
thought, you may convince yourself, in any case, that the graph will still be a
wedge, with the vertex at the point where 1 − 3x = 0, that is x = 31 :
1
2
Equations and Inequalities, From The Viewpoint of Functions
Here are a few equations, and a few inequalities:
5x − 1 = 0
(2)
2x2 − 3x + 1 = 2 − x
(3)
8x − 1 < 2
(4)
6x + 1 ≥ 2 + x
(5)
x2 − 1 < 0
(6)
We can “read” them as follows. We introduce functions, connected to each line
above:
(2) f (x) = 5x − 1
(3) g(x) = 2x2 − 3x + 1, h(x) = 2 − x
(4) k(x) = 8x − 1
(5) m(x) = 6x + 1, n(x) = 2 + x
(6) p(x) = x2 − 1
The problems can now be written as: “find the values of the variablex, such
that, respectively,”
(2) f (x) = 0
(3) g(x) = h(x)
2
(4) k(x) < 2
(5) m(x) ≥ n(x)
(6) p(x) < 0
If we have the graphs of the functions, we can answer the questions by looking
at the graphs, and see which values of x do the job. For example, the solution
to (5) would be given by all values of x, for which the graph of m lies above the
graph of n.
Don’t jump to conclusions: this method of graphical solution is worthless in
terms of getting hard solutions (hand graphs cannot give you but very poorly
approximate values, and using a graphing calculator or graphing utility is cheating: the software uses algebra and, more generally, hard numerical methods to
display the graph – and you still have to eyeball numbers from the picture,
which will give you, again, a very rough approximate solution). The method is
however extremely useful as a supplement to the hard work of solving problems,
because it gives you a visual impression of the problem.
For instance, it makes it easy to understand why a problem has as many
solutions as it does. In problem (3), as an example, solutions will be points
where the two graphs cross. Looking at a picture, shows that there will be two
such points:
3
Apply To Absolute Value Problems
Strange as it sounds, it turns out that absolute value inequalities cut a lot more
slack than equations, when the solutions are “trivial” (that is “no solution” or “all
real numbers”). If you follow the rules, and read the results carefully, inequalities
will lead you to the correct answer regardless, whereas equations will not. Let’s
see how this can happen.
3
3.1
Simple Absolute Value Equations
There is a reason why, when confronted with a problem like
|2x − 1| = −1
(7)
you should “stop right away”, and declare there is no solution. If you tried to
solve this problem “blindly”, you could do something like
2x − 1 = −1 or 2x − 1 = 1
(8)
x = 0 or x = 1
which are both wrong, as the problem has no solution. What is going on here?
First look at a problem that does have a solution. Consider
|2x − 1| = 1
(9)
.Look at the graph. We graph q(x) = |2x − 1|, and look whether the graph ever
takes the value 1 (of course, it does):
Fig.1
Now, let’s solve the equation algebraically:
2x − 1 = 1 or 2x − 1 = −1
x = 1 or x = 0
If we solve these two equations graphically, we see that we are looking now at
the (different) function s(x) = 2x − 1, and matching it with the values y = −1
and y = 1:
4
Fig.2
This “disentangling” of the absolute value graph works nicely.
Let’s look now at the first problem (|2x − 1| = −1). We graph q(x) =
|2x − 1|, and look whether the graph ever takes the value −1 (of course, it
doesn’t):
Now let’s draw the graphs for the equations (8) that we wrote while trying
to solve this: it turns out we ended up solving (9), not (7)!
3.2
Simple Absolute Value Inequalities
In an inequality, you look at the values of x that have the graph stay below or
over a reference line. For example, if we look again at
|2x − 1| < 1
we can go back to our graph in Fig.1, and see that the inequality holds when
0 < x < 1. The algebraic solution goes like
2x − 1 < 1 and 2x − 1 > −1
5
and we get the same result if we look at the corresponding graph, Fig.2
Now, let’s look at the “no solution” problem
|2x − 1| < −1
If we try to solve this problem automatically, this results in
2x − 1 < −1 and 2x − 1 > 1
x < 0 and x > 2
That’s the outer bands in Fig.2, and since it is an AND situation, the result is
∅, because a number cannot be, at the same time, less than 0 and greater than
1. This “failsafe” operation of inequalities extends to much more complicated
situations.
3.3
More Complicated Situations
If the variable x appears elsewhere, besides the absolute value term, things
become, apparently, more complex. In real terms, though, they are exactly as
involved as in the “simple” case. The difference is that it is far less obvious than
in the “simple” case whether the problem has no solutions or not.
Still, not much different than in the previous sections, the “equation” case
does not have the automatic safeguards that the “inequality” case has.
3.3.1
Inequalities
There are different issues that can be at play. We may have the unknown appear
outside the absolute value, or, additionally, we may have more than one absolute
value to handle. Still, algebra will see you through. Let’s try two examples.
1 Let’s solve
x − |3x − 1| ≤ 1 − 3x
Algebraically, this works like this:
− |3x − 1| ≤ 1 − 3x − x = 1 − 4x
|3x − 1| ≥ 4x − 1
which splits into
3x − 1 ≥ 4x − 1
−x ≥ 0
x≤0
OR
3x − 1 ≤ − (4x − 1)
3x − 1 ≤ −4x + 1
6
7x ≤ 2
2
x≤
7
Given the “OR” join, the final answer is
{x : −∞ < x ≤ 2}
Now, let’s look at the picture. We graph both f (x) = x − |3x − 1|, and 1 − 3x,
and look how they compare:
Fig. 3
As we can see, we do, indeed, have one solution only, and it looks like the
graph of x − |3x − 1| is below the graph of 1 − 3x for x less than a number that
seems reasonably close to 27 ≈ .29
3.3.2
Equations
If we try to solve the equation
x − |3x − 1| = 1 − 3x
graphically, we, obviously, end up with the same picture as Fig. 3. This leads
us to conclude that the only solution is x = 72 . If we work algebraically, we find
− |3x − 1| = 1 − 3x − x
|3x − 1| = 4x − 1
which splits in
3x − 1 = 4x − 1
−x = 0
OR
3x − 1 = 1 − 4x
7x = 2
7
2
7
and we just saw how only the second result is legitimate. If we look at the
graphs, we see what has been going on. We would be comparing the graph of
3x − 1 with those of 4x − 1, and of 1 − 4x:
x=
Again, the problem we had before us has been “subtly” changed when moving
to the without-the-absolute-value form, but the inequality version saved us from
mistakes, thanks to its handling of the logical join of the solutions we found.
3.3.3
More Than One Absolute Value
We didn’t really take on inequalities involving more than one absolute value
– only equations. As suggested by the previous examples, though, we have to
expect to be confronted by bogus solutions. Let’s look at a couple of examples,
with the graphs helping us grasp the situation better
Example Consider
|3 − 2x| + 1 = |x − 1|
The algebraic procedure goes as follows. We start by concentrating on one
of the two, for example, the second:
|x − 1| = 1 + |3 − 2x|
which splits into
x − 1 = 1 + |3 − 2x|
|3 − 2x| = x − 2
(10)
or
x − 1 = −1 − |3 − 2x|
− |3 − 2x| = x
|3 − 2x| = −x
8
(11)
Equation (10) leads to
3 − 2x = x − 2
−3x = −5
3
x=−
5
or
3 − 2x = 2 − x
−x = −1
x=1
Equation (11) leads to
3 − 2x = −x
−x = −3
x=3
or
3 − 2x = x
−3x = −3
x=1
So, the “candidate” solutions are − 53 , 1, 3. Plugging them into the original
equation will decide which are legitimate:
x = − 53 Plugging into |3 − 2x| + 1 = |x − 1| results into a left hand side equal
to
3 − 2 − 3 + 1 = 3 + 6 + 1 = 21 + 1 = 26
5
5
5
5
and a right hand side equal to
3
− − 1 = − 8 = 8
5
5 5
so this is a spurious solution.
x = 1 The left hand side is
|3 − 2| + 1 = 2
and the right hand side is
|1 − 1| = 0
so this is also a spurious solution.
x = 3 The right hand side is
|3 − 2 · 3| + 1 = |−3| + 1 = 3 + 1 = 4
and the left hand side is
|3 − 1| = 2
So, it seems we have no solution at all.
9
Going to the graphs, we find
e
The picture shows clearly that the problem does, indeed, have no solution.
You may wonder what do those spurious solutions correspond to. The answer
is that the equation we get by “splitting” substitute a straight line in place of
the wedge, and, really, split the target into two. However, when there is no
solution, as here, “straightening” the wedges will produce crossing that there
were not there at the start.
A second type of problem with more than one absolute value is one with
nested absolute values. For example,
|2x + |1 − 3x|| = x + 1
(12)
The obvious way to handle this is to start with the outer absolute value. We
have a “split” of
2x + |1 − 3x| = x + 1
|1 − 3x| = −x + 1
(13)
or
2x + |1 − 3x| = −1 − x
|1 − 3x| = −1 − 3x
(13) splits, in turn, into
1 − 3x = −x + 1
−2x = 0
x=0
or
1 − 3x = x − 1
−4x = −2
1
x=
2
10
(14)
while (14) splits into
1 − 3x = −1 − 3x
1 = −1
that is “no solution”, or
1 − 3x = 1 + 3x
−6x = −2
1
x=
3
Hence, our “candidates” are 0, 13 , 12 .Plugging then into (12) results in
x = 0 Easily, 1 = 1, so this is a solution
x=
1
3
We have on the left hand side
1 2 + 1 − 3 1 = 2 = 2
3 3 3 3
and on the right hand side
1
4
+1=
3
3
showing a spurious solution.
x=
1
2
On the left hand side we find
1 2 + 1 − 3 1 = 1 + − 1 = 3
2 2
2 2
while the right hand side is
1
3
+1=
2
2
and we have a second legitimate solution.
Let’s look at the graph of the original problem:
11
Note how getting rid of the outer absolute value on the left hand side, and
“splitting” the right hand side, does not create a fictitious solution:
Equation (13) has the following graph:
indicating it has no solution. However, splitting it does “create” a pseudosolution (the blue line is the graph of 1 − 3x):
12