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Math 23b Homework 2 Solutions
1. In each part, the rules are given by welldeﬁned functions on certain domains whose
union is
R
, so we just need to check that these functions agree wherever their corre
sponding domains overlap.
(a) For 2
< x <
4,
x > x

1
>
0, so

x

1

=
x

1 =

x
 
1, so the rules deﬁne a
function on
R
.
(b) Note that

1
<
0
<
2 so 0 lies in both domains. Since

0

1

= 1
6
=

1 =

0

1,
this implies the rules do not deﬁne a function.
(c) There is no overlap in the domains, so the rules deﬁne a function on
R
.
(d) Expanding the numerator of ((
x
+3)
2

9)
/x
gives (
x
+3)
2

9 =
x
2
+6
x
+9

9 =
x
(
x
+ 6). Hence, for
x
6
= 0, we have ((
x
+ 3)
2

9)
/x
=
x
+ 6, so the functions
agee on the intersection of their domains, so the rules deﬁne a function on
R
.
(e) The ﬁrst two domains both contain the interval [2
,
4], but for
x
≥
2
>
0, the
square root theorem gives
√
x
2
=
x
, so the rules deﬁne a function on
R
.
2. Let
f
and
g
be realvalued functions on
R
.
(a)
Claim:
If
f
and
g
are bounded, then
f
+
g
is also bounded.
Proof.
By assumption, there exist constants
M
and
N
such that

f
(
x
)
 ≤
M
and

g
(
x
)
 ≤
N
for all
x
∈
R
, so by the triangle inequality, we have for each
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 Spring '09
 BongLian
 Math

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