This preview shows pages 1–2. Sign up to view the full content.
MATH 137
Assignment 4  Solutions
1. (a) If
f
(
x
) =
x
3
+
bx
2
+
cx
+
d
and
x
6
= 0, then
f
(
x
) =
x
3
(1 +
b
x
+
c
x
2
+
d
x
3
).
Suppose that

x

is chosen large enough so that
±
±
±
±
b
x
±
±
±
±
,
±
±
±
c
x
2
±
±
±
,
±
±
±
±
d
x
3
±
±
±
±
are all less than
1
4
. Then
±
±
±
±
b
x
+
c
x
2
+
d
x
3
±
±
±
±
≤
±
±
±
±
b
x
±
±
±
±
+
±
±
±
c
x
2
±
±
±
+
±
±
±
±
d
x
3
±
±
±
±
<
3
4
<
1.
Let
b
x
+
c
x
2
+
d
x
3
=
A
. Now

A
≤ 
A

, so
A
≥ 
A

and thus 1 +
A
≥
1
 
A

>
0.
Therefore when

x

is large enough, the sign of
f
(
x
) is the same as the sign of
x
3
which is
the same as the sign of
x
.
(b) From (a) there exist numbers
u
and
v
such that
f
(
u
)
<
0 and
f
(
v
)
>
0. A polynomial
function is continuous, so by the Intermediate Value Theorem here is a number w such
that
f
(
w
) = 0.
(c) “Yes”. One could repeat the argument to show that if
f
(
x
) =
x
m
+
b
m

1
x
m

1
+
···
+
b
0
then when

x

is large enough
1 +
b
m

1
x
+
···
+
b
0
x
m
>
0
.
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 '08
 SPEZIALE
 Math

Click to edit the document details