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Homework # 1
Due: 05/23/06
#1
Find the domain of the following functions
a)
f
(
x
) =
3
x
2

2
x
+ 1
x
2

4
x

21
The domains of 3
x
2

2
x
+ 1 and
x
2

4
x

21 are all real numbers. So the domain of
f
is all real numbers except where
x
2

4
x

21 = 0. So
x
2

4
x

21 = 0
(
x

7)(
x
+ 3) = 0
x
= 7
,

3
So the domain is all real numbers except
x
=

3 and
x
= 7.
b)
f
(
x
) =
√
x
+ 3

√
x

2 +
3
√
x
2

1
You can only take the square root of a positive number, so the domain of
√
x
+ 3 is where
x
+ 3
≥
0, or
x
≥ 
3.
Similarly, the domain of
√
x

2 is
x

2
≥
0 or
x
≥
2.
You can take the cube root of any number, so the domain of
3
√
x
2

1 is all real numbers.
The domain of
f
is the intersection of all of these. So it’s all numbers that are bigger
than both 3 and 2. This means the domain is
x
≥
2 or [2
,
∞
)
c)
f
(
x
) =
1
√
x
2

1
The domain of
f
is the domain of
√
x
2

1 minus the points where it’s zero. The domain
of
√
x
2

1 is where
x
2

1
≥
0. To ±nd where this is positive, ±rst ±nd where it’s zero and
then test values.
x
2

1 = 0
(
x
+ 1)(
x

1) = 0
x
=

1
,
1
So now we look at the number line:

1
1
1
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View Full DocumentSo we want to test values less than 1, between 1 and 1, and greater than 1. Plugging in
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 Spring '08
 varies
 Real Numbers

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