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
.
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 Fall '08
 SPEZIALE
 Math, lim, Continuous function, 3a2

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