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Math 125A, Fall 2009
Solutions for Homework 7
(Prepared by Matt Low)
1. It follows from Theorem 29.4 that
f
0
is a constant function. In other words,
f
0
(
x
) =
C
for some
constant
C
. De±ne a new function
g
(
x
) =
f
(
x
)
±
Cx
. Note that
g
0
(
x
) = 0, so Theorem 29.4
again implies that
g
(
x
) =
D
for some constant
D
. Thus
f
(
x
) =
Cx
+
D
for all
x
2
R
.
2. (a) True. Consider any interval of the form [
n;n
+ 1] for some integer
n
. By the mean value
theorem (Theorem 29.3), there is a number
c
2
(
n;n
+ 1) such that:
f
0
(
c
) =
f
(
n
+ 1)
±
f
(
n
)
(
n
+ 1)
±
n
=
0
±
0
1
= 0
Because there are in±nitely many of these intervals [
n;n
+ 1], there are also in±nitely many
of these points
c
at which
f
0
(
c
) = 0.
(b) False. For example, consider the function
f
de±ned in Problem 5. In Problem 5(b), you
showed that
T
(
x
) = 0 for all
x
. But
f
(
x
)
6
= 0 when
x
6
= 0.
(c) True. This is an immediate consequence of Corollary 31.4.
3. By Theorem 29.8, the derivative of any function satis±es intermediate value theorem. But
g
(
x
)
does not. For example,
g
(
±
1) =
±
2 and
g
(1) = 2, but there is no number
c
at which
g
(
c
) = 0.
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This note was uploaded on 03/18/2012 for the course MATH 67 taught by Professor Schilling during the Fall '07 term at UC Davis.
 Fall '07
 Schilling

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