Math 131A Analysis Summer Session A
Homework 2 Solutions
1. Find all solutions to the following:
(a)

x

3
  
x
+ 6

= 5
(b)

x

3
 ≥ 
x
+ 6

(c)

x
2

8

= 7
x

20
Solution.
(a) We have 4 cases to check.
Case 1:
x

3
≥
0
and
x
+ 6
≥
0
. This case is equivalent to
x
≥
3
.
5 =

x

3
  
x
+ 6

=
x

3

(
x
+ 6) =

9
A contradiction. Therefore no solution exists in this case.
Case 2:
x

3
<
0
and
x
+ 6
≥
0
. This case is equivalent to

6
≤
x <
3
.
5 =

x

3
  
x
+ 6

= 3

x

(
x
+ 6) =

2
x

3
Thus
x
=

4
is a solution.
Case 3:
x

3
<
0
and
x
+ 6
<
0
. This case is equivalent to
x <

6
.
5 =

x

3
  
x
+ 6

= 3

x
+ (
x
+ 6) = 9
A contradiction. Therefore no solution exists in this case.
Case 4:
x

3
≥
0
and
x
+ 6
<
0
This case is impossible.
The solution set is
{
4
}
.
(b) We have the same cases to check:
Case 1:
x

3
≥
0
and
x
+ 6
≥
0
. This case is equivalent to
x
≥
3
.
x

3 =

x

3
 ≥ 
x
+ 6

=
x
+ 6
which implies

3
≥
6
A contradiction. Therefore no solution exists in this case.
Case 2:
x

3
<
0
and
x
+ 6
≥
0
. This case is equivalent to

6
≤
x <
3
.
3

x
=

x

3
 ≥ 
x
+ 6

=
x
+ 6
So
2
x
≤ 
3
or
x
≤ 
3
2
. Since in this case, we also have

6
≤
x <
3
, the values

6
≤
x
≤ 
3
2
all represent solutions.
1
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Case 3:
x

3
<
0
and
x
+ 6
<
0
. This case is equivalent to
x <

6
.
3

x
=

x

3
 ≥ 
x
+ 6

=

x

6
So
3
≥ 
6
, which is true for every
x
in this case. Thus the values
x <

6
all
represent solutions.
The solution set is
{
x
∈
R
:
x
≤ 
3
2
}
.
(c) We have 2 cases to check.
Case 1:
x
2

8
≥
0
x
2

8 =

x
2

8

= 7
x

20
Thus
0 =
x
2

7
x
+ 12 = (
x

3)(
x

4)
. So
x
= 3
and
x
= 4
are both solutions
in this case since
3
2

8 = 9

8 = 1
>
0
and
4
2

8 = 16

8 = 8
>
0
.
Case 2:
x
2

8
<
0
8

x
2
=

x
2

8

= 7
x

20
Thus
0 =
x
2
+ 7
x

28
. The roots are

7
±
√
7
2

4(

28)
2
=

7
±
√
161
2
But

7

√
161
2
2
>
(

7

12
2
)
2
>
8
and

7+
√
161
2
2
>
(
5
.
68
2
)
2
= 8
.
0656
>
8
so
these roots cannot be solutions.
The solution set is
{
3
,
4
}
.
2. Let
A
=
{
3
x
2

2

:

1
≤
x
≤
1
}
, a bounded set.
Use Calculus to find inf
A
and sup
A
.
Solution.
Note

3
x
2

2

= 0
precisely when
x
=
±
2
3
.
Since

3
x
2

2
 ≥
0
for all
x
and
±
2
3
∈
[

1
,
1]
,
inf
A
= min
A
= 0
.
Also,
3
x
2

2
≥
0
precisely when
x <

2
3
or
x >
2
3
. Thus

3
x
2

2

= 2

3
x
2
when

2
3
≤
x
≤
2
3
.
The function
f
(
x
) = 2

3
x
2
is differentiable on
(

2
3
,
2
3
)
with derivative
f
(
x
) =

6
x
. Thus
f
is decreasing on
(

2
3
,
0)
,
increasing on
(0
,
2
3
)
and has a local max
2
at
x
= 0
. On
(

1
,

2
3
)
and on
(
2
3
,
1)
,
f
(
x
) = 3
x
2

2
has derivative
f
(
x
) = 6
x
.
Thus
f
is decreasing on
(

1
,

2
3
)
, increasing on
(
2
3
,
1)
, and

3(1)
2

2

=

3((

1)
2
)

2

= 1
. Thus
sup
A
= max
A
= 2
.
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