Math 3110 Homework 11 Solutions
Exercise 12.1.1
Let
f
be a continuous function whose values are always rational numbers.
Assume for
contradiction that
f
is not a constant function. Then there exist
x
1
, x
2
with
x
1
< x
2
such
that
f
(
x
1
) =
f
(
x
2
). Without loss of generality, assume
f
(
x
1
)
< f
(
x
2
). Then by Theorem
2.5, we can find an irrational number
k
such that
f
(
x
1
)
< k < f
(
x
2
). Since
f
is continuous,
then by the Intermediate Value Theorem (Corollary 12.1), there exists a number
c
with
x
1
< c < x
2
such that
k
=
f
(
c
).
But this contradicts that the values of
f
are always
rational. So
f
must be constant.
Exercise 12.1.3
Let
f
(
x
) =
x
2
(
x

1), which has roots
x
= 0 and
x
= 1. Bolzano’s Theorem cannot find the
root
x
= 0 because
f
(
x
)
<
0 immediately to both sides of the root, so
f
(
x
) does not change
sign on a small neighborhood of
x
= 0. Newton’s method can still find this root, however, if
we start close enough to it.
Exercise 12.2.3
The positive roots of
f
(
x
) =
x

tan(
x
) are exactly the
x
coordinates of the intersection
points of the functions
y
=
x
and
y
= tan(
x
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 '08
 RAMAKRISHNA
 Math, Topology, Intermediate Value Theorem, Continuous function, Bolzano’s Theorem

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