James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A8
appendix A Numbers, Inequalities, and Absolute Values
2 2
3 5 7
FIGURE 9
Example 7 Solve | x 2 5 | , 2.
SOLUTION 1 By Property 5 of (6), | x 2 5 |
Therefore, adding 5 to each side, we have
, 2 is equivalent to
22 , x 2 5 , 2
3 , x , 7
and the solution set is the open interval s3, 7d.
SOLUTION 2 Geometrically the solution set consists of all numbers x whose distance
from 5 is less than 2. From Figure 9 we see that this is the interval s3, 7d.
Example 8 Solve | 3x 1 2 | > 4.
SOLUTIOn By Properties 4 and 6 of (6), | 3x 1 2 |
> 4 is equivalent to
3x 1 2 > 4 or 3x 1 2 < 24
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In the first case 3x > 2, which gives x > 2 3 . In the second case 3x < 26, which gives
x < 22. So the solution set is
hx | x < 22 or x > 2 3j − s2`, 22g ø f 2 3 , `)
Another important property of absolute value, called the Triangle Inequality, is used
frequently not only in calculus but throughout mathematics in general.
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7 The Triangle Inequality If a and b are any real numbers, then
| a 1 b | < | a | 1 | b |
Observe that if the numbers a and b are both positive or both negative, then the two
sides in the Triangle Inequality are actually equal. But if a and b have opposite signs,
the left side involves a subtraction and the right side does not. This makes the Tri angle
Inequality seem reasonable, but we can prove it as follows.
Notice that
2| a | < a < | a |
is always true because a equals either | a | or 2 | a | . The corresponding statement for b is
Adding these inequalities, we get
2| b | < b < | b |
2s| a | 1 | b | d < a 1 b < | a | 1 | b |
If we now apply Properties 4 and 5 (with x replaced by a 1 b and a by | a | 1 | b |), we
obtain
| a 1 b | < | a | 1 | b |
which is what we wanted to show.
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