4.12
(a) False. One counterexample is
f
(
x
) =
e

x
.
(b) False. Take
f
(
x
) = 0. (Any constant function will work.)
(c) False. Consider
f
(
x
) =
x
if 0
6
=
x
6
= 1
,
1
if
x
= 0
,
0
if
x
= 1
.
(d) True. If

f
(
x
)
 ≤
M
for every
x
∈
R
, then
M
+ 1 is not in the image of
f
.
(e) False. Take
f
(
x
) =
x
2
.
4.13
We assume without loss of generality that
n >
rev(
n
) so that
a > c
. Then
x
=
n

rev(
n
) = (100
a
+ 10
b
+
c
)

(100
c
+ 10
b
+
a
)
= (100
a
+ 10(
b

1) + (10 +
c
))

(100
c
+ 10
b
+
a
)
= (100(
a

1) + 10(10 +
b

1) + (10 +
c
))

(100
c
+ 10
b
+
a
)
= 100((
a

1)

c
) + 10((9 +
b
)

b
) + ((10 +
c
)

a
)
= 100((
a

c
)

1) + 10
·
9 + (10

(
a

c
))
.
(Note that these algebraic manipulations are motivated by the notion of “borrowing”
as in the standard algorithm for subtraction from grade school.) Since 0
≤
c < a
≤
9,
we have 1
≤
a

c
≤
9, which implies that 0
≤
(
a

c
)

1
≤
8 and 1
≤
10

(
a

c
)
≤
9.
Hence the digits of
x
are (
a

c
)

1
,
9, and 10

(
a

c
). This yields
x
+ rev(
x
) = [100((
a

c
)

1) + 10
·
9 + (10

(
a

c
))]
+ [100(10

(
a

c
)) + 10
·
9 + ((
a

c
)

1)]
= 100
·
9 + 2
·
10
·
9 + 9 = 1089
.
4.22
We first show
f
is surjective.
We have
f
(1
/
2) = 0, so it remains to show that
if
z
∈
R
, z
6
= 0, then there is some
y
∈
(0
,
1) such that
f
(
x
) =
z
.
We claim that
y
= (
z

1 +
√
z
2
+ 1)
/
2
z
will work.
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 Spring '10
 Math, Inverse function, grade school, Surjection, constant function

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