Looking at the Picture Absolute Value Equations and Inequalities April 19, 2009

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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

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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)

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(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:

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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.

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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:

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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:

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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):

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