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

appendix A Numbers, Inequalities, and Absolute Values A5
First we subtract 1 from each side of the inequality (using Rule 1 with c − 21):
x , 7x 1 4
Then we subtract 7x from both sides (Rule 1 with c − 27x):
26x , 4
Now we divide both sides by 26 (Rule 4 with c − 2 1 6):
x . 2 4 6 − 22 3
These steps can all be reversed, so the solution set consists of all numbers greater than
2 2 3 . In other words, the solution of the inequality is the interval (22 3 , `). n
Example 2 Solve the inequalities 4 < 3x 2 2 , 13.
SOLUTIOn Here the solution set consists of all values of x that satisfy both inequalities.
Using the rules given in (2), we see that the following inequalities are equivalent:
4 < 3x 2 2 , 13
6 < 3x , 15 (add 2)
2 < x , 5 (divide by 3)
Therefore the solution set is f2, 5d.
n
Example 3 Solve the inequality x 2 2 5x 1 6 < 0.
SOLUTIOn First we factor the left side:
sx 2 2dsx 2 3d < 0
We know that the corresponding equation sx 2 2dsx 2 3d − 0 has the solutions 2
and 3. The numbers 2 and 3 divide the real line into three intervals:
A visual method for solving Exam ple 3
is to use a graphing device to graph the
parabola y − x 2 2 5x 1 6 (as in Figure
4) and observe that the curve lies on
or below the x-axis when 2 < x < 3.
y
y=≈-5x+6
s2`, 2d s2, 3d s3, `d
On each of these intervals we determine the signs of the factors. For instance,
x [ s2`, 2d ? x , 2 ? x 2 2 , 0
Then we record these signs in the following chart:
Interval x 2 2 x 2 3 sx 2 2d sx 2 3d
x , 2 2 2 1
2 , x , 3 1 2 2
x . 3 1 1 1
0
FIGURE 4
1 2 3 4
x
Another method for obtaining the information in the chart is to use test values. For
instance, if we use the test value x − 1 for the interval s2`, 2d, then substitution in
x 2 2 5x 1 6 gives
1 2 2 5s1d 1 6 − 2
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