Homework 1 - PreCalc Review-solutions - Version 028...

Version 028 – Homework 1 - PreCalc Review – Helleloid – (58250) 1
This print-out should have 19 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
Welcome to Quest. This homework over-
laps #2 and #3. Use it as a precalculus review
and diagnostic. Good luck with M408K.
001 10.0 points
Rewrite the expression
radicalBig
a +
√
b .
1. ( a + b
1 / 2
)
1 / 2
correct
2. a
1 / 2
+ b
1 / 4
3. a + b
4. a + b
1 / 2
5. a + b
1 / 4
Explanation:
Since
√
x = x
1 / 2
, we see that
radicalBig
a +
√
b = ( a + b
1 / 2
)
1 / 2
.
002 10.0 points
Rationalize the numerator of
√
x + 4 −
√
x − 3
x
.
1.
7
x (
√
x + 4 +
√
x − 3)
correct
2.
1
x (
√
x + 4 +
√
x − 3)
3.
x
√
x + 4 +
√
x − 3
4.
7
x (
√
x + 4 −
√
x − 3)
5.
7 x
√
x + 4 −
√
x − 3
Explanation:
By the difference of squares,
(
√
x + 4 −
√
x − 3)(
√
x + 4 +
√
x − 3)
= (
√
x + 4)
2
− (
√
x − 3)
2
= 7 .
Thus, after multiplying both the numerator
and the denominator in the given expression
by
√
x + 4 +
√
x − 3 ,
we obtain
7
x (
√
x + 4 +
√
x − 3)
.
003 10.0 points
Simplify the expression
parenleftBig
xy
− 3
√
z
parenrightBig
6
÷
parenleftBig
y
2
x
1 / 3
z
− 3
parenrightBig
12
as much as possible, leaving no negative ex-
ponents and no radicals.
1.
x
10
y 42 z 39
correct
2.
x
2
y 6 z 33
3.
x
10
y
6
z
33
4.
x
10
y
42
z 39
5.
x
10
z
39
y 42
Explanation:
By the Laws of Exponents
parenleftBig
xy
− 3
√
z
parenrightBig
6
=
x
6
y 18 z 3
,
while
parenleftBig
y
2
x 1 / 3 z − 3
parenrightBig
12
=
y
24
z
36
x
4
.

Version 028 – Homework 1 - PreCalc Review – Helleloid – (58250) 2
Consequently, the given expression can be
rewritten as
x
6
y
18
z
3
×
x
4
y
24
z
36
=
x
10
y
42
z
39
.
004 10.0 points
Simplify the expression
f ( x ) =
5 +
15
x − 4
4 + 60
parenleftBig
x
x 2 − 16
parenrightBig
as much as possible.
1. f ( x ) =
x − 4
x − 16
2. f ( x ) =
5
4
parenleftBig
x + 4
x + 16
parenrightBig
correct
3. f ( x ) =
5
4
parenleftBig
x + 4
2 x + 16
parenrightBig
4. f ( x ) =
x + 4
x − 16
5. f ( x ) =
5
4
parenleftBig
x − 4
x + 16
parenrightBig
Explanation:
After bringing the numerator to a common
denominator it becomes
5 x − 20 + 15
x − 4
=
5 x − 5
x − 4
.
Similarly, after bringing the denominator to
a common denominator and factoring it be-
comes
4 x
2
− 64 + 60 x
x
2
− 16
=
4( x − 1)( x + 16)
x
2
− 16
.
Consequently,
f ( x ) =
5 +
15
x − 4
4 + 60
parenleftBig
x
x 2 − 16
parenrightBig
=
5 x − 5
4( x − 1)( x + 16)
parenleftBig
x
2
− 16
x − 4
parenrightBig
.
On the other hand,
x
2
− 16 = ( x + 4)( x − 4) .
Thus, finally, we see that
f ( x ) =
5
4
parenleftbigg
x + 4
x + 16
parenrightbigg
.
005 10.0 points
Find the solution set of the inequality
x − 2
x − 4
≤
x
x − 1
,
expressing your answer in interval notation.
1. ( −∞ , − 2 ) ∪ (1 , 4)
2. ( −∞ , − 2 ] ∪ (1 , 4) correct
3. ( − 2 , 1) ∪ (4 , ∞ )
4. [ − 2 , 1) ∪ (4 , ∞ )
5. ( −∞ , − 2 ] ∪ (4 , ∞ )
Explanation:
To solve the inequality
x − 2
x − 4
≤
x
x − 1
we first bring all the terms to one side:
x − 2
x − 4
−
x
x − 1
≤ 0 ,
and then bring the left hand side to a common