Section 3.8 Solving Polynomial Inequalities
Example.
Find all
x
such that
x
2
+
x

2
<
0.
Solution.
x
2
+
x

2=(
x
+2)(
x

1). The expression (
x
x

1) changes sign at
x
=

2and
x
=1.
If
x<

2, then (
x
x

1)
>
0.
If

2
<x<
1, then (
x
x

1)
<
0.
If 1
<x
,then(
x
x

1)
>
0.
Hence,
x
2
+
x

2
<
0 on the interval (

2
,
1).
Example.
Solve graphically:
x
4
≤
x
+1.
Solution. From the graph below,
x
=

.
70
,y
=
.
24 and
x
=1
.
21
=2
.
16 are approximate points
where the graphs of
x
4
and
x
+1 intersect. Hence the approximate solution is the interval [

.
70
,
1
.
21].
–5
0
5
–2
–1
1
2
y
=
x
4
,
y
=
x
+1
Example.
Solve
f
(
x
)=
x

1
x

2
>
0.
Graphical solution. The graph below indicates the solution consists of
x
in the intervals (
∞
,
1)
and (2
,
∞
).
–10
–5
0
5
10
1234
y
=
x

1
x

2
Analytical solution. The strategic values of
x
where
f
(
x
) may change sign are
x
,
2.
If
∞
1, then
f
(
x
)
>
0
If 1
2, then
f
(
x
)
<
0
If 2
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '11
 Kutter
 Algebra, Inequalities, Negative and nonnegative numbers, Example. Solve

Click to edit the document details