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Problems graded: 18.4, 18.9, 19.4 (10 points each); the others were worth
5 points for completion (19.1 counted as two problems). To convert from 60
points to 100, we multiplied everyone’s score by 1.5 (rounding up) and then
added ten points to everyone’s score.
18.4. We deFne
f
:
S
→
R
as follows:
f
(
x
) =
1
x

x
0
. This is continuous
by Theorem 17.4(iii) (the denominator is nonzero in
S
as
x
n
=
x
0
); however,
as
x
n
→
x
0
,

f
(
x
)
 → ∞
(by the limit laws from Section 9); since
{
x
n
}
is a
sequence in
S
,
f
is unbounded.
18.8.
Because the product of zero and any other number is zero while
f
(
a
)
f
(
b
)
<
0, either
f
(
a
)
>
0 (in which case
f
(
b
)
<
0 so
f
(
b
)
<
0
< f
(
a
))
or
f
(
a
)
<
0 (in which case
f
(
b
)
>
0 so
f
(
a
)
<
0
< f
(
b
)); in any case, 0 is
between
f
(
a
) and
f
(
b
) so the Intermediate Value Theorem states that there
exists
x
∈
(
a, b
) with
f
(
x
) = 0.
18.9. We write
f
(
x
) =
a
0
+
a
1
x
+
. . .
+
a
n
x
n
, where
n
is odd and
a
n
n
= 0.
By dividing by
a
n
(which does not change the location of the roots) we may
suppose
a
n
was originally 1.
Choose
M >

a
0

+
. . .
+

a
n

1

so that
M
is also greater than 1 and note
that the magnitude of the
x
j
term in
f
(
M
) for
j < n
is

a
j

M
j
≤ 
a
j

M
n

1
so,
by the triangle inequality,
f
(
M
)
≥
M
j
+1
−
Σ
n

1
j
=0

a
j

M
j
=
M
j
(
M
−
Σ
n

1
j
=0

a
j

)
>
0
.
Similiarly,
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 Spring '09
 WEISBART

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