ExamView - PreCalc 12 Chp 4 Rev Pack 2015 v1.tst

ID: A
PreCalc 12 Chapter 4 Review Pack v1
Answer Section
MULTIPLE CHOICE
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14. ANS:
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C
PTS: 0
DIF:
4.2 Combining Functions Algebraically
Relations and Functions
KEY:
C
PTS: 0
DIF:
4.2 Combining Functions Algebraically
Relations and Functions
KEY:
C
PTS: 0
DIF:
4.1 Combining Functions Graphically
Relations and Functions
KEY:
C
PTS: 0
DIF:
4.2 Combining Functions Algebraically
Relations and Functions
KEY:
A
PTS: 0
DIF:
4.2 Combining Functions Algebraically
Relations and Functions
KEY:
D
PTS: 0
DIF:
4.2 Combining Functions Algebraically
Relations and Functions
KEY:
D
PTS: 0
DIF:
4.2 Combining Functions Algebraically
Relations and Functions
KEY:
A
PTS: 0
DIF:
4.2 Combining Functions Algebraically
Relations and Functions
KEY:
B
PTS: 0
DIF:
4.3 Introduction to Composite Functions
Relations and Functions
KEY:
B
PTS: 0
DIF:
4.3 Introduction to Composite Functions
Relations and Functions
KEY:
C
PTS: 0
DIF:
4.3 Introduction to Composite Functions
Relations and Functions
KEY:
C
PTS: 0
DIF:
4.3 Introduction to Composite Functions
Relations and Functions
KEY:
B
PTS: 0
DIF:
4.3 Introduction to Composite Functions
Relations and Functions
KEY:
A
PTS: 0
DIF:
4.3 Introduction to Composite Functions
Relations and Functions
KEY:
1
Easy
LOC: 12.RF1
Conceptual Understanding | Procedural Knowledge
Easy
LOC: 12.RF1
Conceptual Understanding | Procedural Knowledge
Moderate
LOC: 12.RF1
Conceptual Understanding | Procedural Knowledge
Easy
LOC: 12.RF1
Conceptual Understanding | Procedural Knowledge
Easy
LOC: 12.RF1
Conceptual Understanding | Procedural Knowledge
Moderate
LOC: 12.RF1
Conceptual Understanding | Procedural Knowledge
Moderate
LOC: 12.RF1
Conceptual Understanding | Procedural Knowledge
Moderate
LOC: 12.RF1
Conceptual Understanding | Procedural Knowledge
Easy
LOC: 12.RF1
Conceptual Understanding | Procedural Knowledge
Easy
LOC: 12.RF1
Conceptual Understanding | Procedural Knowledge
Easy
LOC: 12.RF1
Conceptual Understanding | Procedural Knowledge
Easy
LOC: 12.RF1
Conceptual Understanding | Procedural Knowledge
Moderate
LOC: 12.RF1
Conceptual Understanding | Procedural Knowledge
Moderate
LOC: 12.RF1
Conceptual Understanding | Procedural Knowledge
ID: A
15. ANS:
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TOP:
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22. ANS:
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23. ANS:
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TOP:
A
PTS: 0
DIF: Easy
4.3 Introduction to Composite Functions
LOC: 12.RF1
Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
A
PTS: 0
DIF: Easy
4.3 Introduction to Composite Functions
LOC: 12.RF1
Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
A
PTS: 0
DIF: Easy
4.4 Determining Restrictions on Composite Functions
LOC: 12.RF1
Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
B
PTS: 0
DIF: Easy
4.4 Determining Restrictions on Composite Functions
LOC: 12.RF1
Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
A
PTS: 0
DIF: Easy
4.4 Determining Restrictions on Composite Functions
LOC: 12.RF1
Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
D
PTS: 0
DIF: Moderate
4.4 Determining Restrictions on Composite Functions
LOC: 12.RF1
Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
D
PTS: 0
DIF: Moderate
4.4 Determining Restrictions on Composite Functions
LOC: 12.RF1
Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
C
PTS: 0
DIF: Moderate
4.4 Determining Restrictions on Composite Functions
LOC: 12.RF1
Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
B
PTS: 0
DIF: Moderate
4.4 Determining Restrictions on Composite Functions
LOC: 12.RF1
Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
2
ID: A
SHORT ANSWER
24. ANS:
The domain is: x  ò
The range is approximately: y  2.3
PTS: 0
DIF: Moderate
REF: 4.1 Combining Functions Graphically
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge | Communication
3
ID: A
25. ANS:
The domain is: x  ò
The range is: y  ò
PTS: 0
DIF: Moderate
REF: 4.1 Combining Functions Graphically
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge | Communication
26. ANS:
x2
(x  1) 2
Domain: x  2 , x  1
q(x) 
PTS: 0
DIF: Moderate
REF: 4.2 Combining Functions Algebraically
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
27. ANS:
h(x)  x  4  4x  3
Domain: x  4
PTS: 0
DIF: Easy
REF: 4.2 Combining Functions Algebraically
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
4
ID: A
28. ANS:
q(x) 
3x
3x  4
4
Domain:   x  3
3
PTS: 0
DIF: Moderate
REF: 4.2 Combining Functions Algebraically
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
29. ANS:
x 2  4x  36
g ÊÁË f(x) ˆ˜¯ 
8
Domain: x  ò; Range: y  4
PTS: 0
DIF: Moderate
REF: 4.3 Introduction to Composite Functions
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
30. ANS:
Forms of the equation may vary.
g ÁÊË f(x) ˜ˆ¯  8x 3  24x 2  24x  12
Domain: x  ò; Range: y  ò
PTS: 0
DIF: Moderate
REF: 4.3 Introduction to Composite Functions
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
31. ANS:
a) f ÊÁË g(1) ˆ˜¯  4
b) g ÊÁË f(3) ˆ˜¯  4
PTS: 0
DIF: Easy
REF: 4.3 Introduction to Composite Functions
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
32. ANS:
Since f ÊÁË g(x) ˆ˜¯  g ÊÁË f(x) ˆ˜¯  x , the functions are inverses.
PTS: 0
DIF: Moderate
REF: 4.3 Introduction to Composite Functions
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
5
ID: A
33. ANS:
g ÊÁË f(x) ˆ˜¯ 
3x  8
Domain: x  
8
3
PTS: 0
DIF: Easy
REF: 4.4 Determining Restrictions on Composite Functions
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
34. ANS:
1
g ÊÁË f(x) ˆ˜¯ 
x6 5
Domain: x  6 and x  19
PTS: 0
DIF: Moderate
REF: 4.4 Determining Restrictions on Composite Functions
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge
PROBLEM
35. ANS:
Equations may vary.
The domain of q(x)  x 2  1 is x  ò .
So, the domains of f(x) and g(x) must also be x  ò , and g(x) must never equal 0.
Choose a function g(x) that is never equal to 0 and has domain x  ò ,
such as g(x)  x 2  1 .
Multiply g(x) by (x 2  1 ) to get f(x) .
f(x)  (x 2  1)(x 2  1)
f(x)  x 4  x 2  x 2  1
f(x)  x 4  1
x4  1
So, q(x)  2
x 1
PTS: 0
DIF: Moderate
REF: 4.2 Combining Functions Algebraically
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge | Communication
6
ID: A
36. ANS:
Equations may vary.
a) k(x)  3x 2  11x  4
k(x)  (3x 2 )  (11x)  (4)
So, f(x)  3x 2 , g(x)  11x , and h(x)  4
b) k(x)  3x 2  11x  4
k(x)  (3x 2 )  (11x  4)
So, f(x)  3x 2 and g(x)  11x  4
c) k(x)  3x 2  11x  4
Factor: k(x)  3x 2  11x  4
k(x)  (3x  1)(x  4)
So, f(x)  3x  1 and g(x)  x  4
PTS: 0
DIF: Moderate
REF: 4.2 Combining Functions Algebraically
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge | Communication
7
ID: A
37. ANS:
a) From the graph: domain: x  ò ; range: y  0
b) From the graph: domain: x  0 ; range: y  0
c) The domain of h(x) is the same as the domain of g(x) , because x cannot be negative; that is x  0 .
The range of h(x) is the same as the ranges of f(x) and g(x) ; that is, all real numbers greater than or equal
to 0, or y  0 .
d) From the graphs, approximate values are:
x
0
1
2
3
4
f(x)
0
1
2
3
4
g(x)
0
3.5
4.9
6.0
6.9
f(x)  g(x)
0
4.5
6.9
9.0
10.9
Plot points at: (0, 0), (1, 4.5), (2, 6.9), (3, 9.0), (4, 10.9)
Join the points with a smooth curve.
The graph verifies the domain and range of y  h(x) .
PTS: 0
DIF: Moderate
REF: 4.1 Combining Functions Graphically
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge | Communication
8
ID: A
38. ANS:
a) Since the denominator cannot be 0, the domain of f(x) is: x  1
b) Sample response:
Use synthetic division to determine (x 2  4x  2)  (x  1).
The function can be written as: f(x)  x  5 
So, g(x)  x  5 ; h(x)  7; and k(x)  x  1
7
x1
PTS: 0
DIF: Difficult
REF: 4.2 Combining Functions Algebraically
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge | Communication
39. ANS:
The function g(x) is a horizontal line with y-intercept 3.
a) When g(x) is added to f(x) ,
the graph of y  f(x) will be translated 3 units down.
b) When g(x) is subtracted from f(x) ,
the graph of y  f(x) will be translated 3 units up.
c) When f(x) is multiplied by g(x) ,
the graph of y  f(x) will be stretched vertically by a factor of 3 and
reflected in the x-axis.
d) When f(x) is divided by g(x) ,
1
the graph of y  f(x) will be compressed vertically by a factor of and
3
reflected in the x-axis.
PTS: 0
DIF: Moderate
REF: 4.2 Combining Functions Algebraically
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Communication
9
ID: A
40. ANS:
Determine an explicit equation for f ÊÁË g (x) ˆ˜¯ .
In f(x)  ax  b , replace x with cx  d .
f ÊÁË g(x) ˆ˜¯  a(cx  d)  b
f ÊÁË g(x) ˆ˜¯  acx  ad  b
f ÁÊË g(x) ˜ˆ¯  acx  (ad  b)
Determine an explicit equation for g ÊÁË f(x) ˆ˜¯ .
In g(x)  cx  d , replace x with ax  b .
g ÊÁË f(x) ˆ˜¯  c(ax  b)  d
g ÁÊË f(x) ˜ˆ¯  cax  cb  d
g ÊÁË f(x) ˆ˜¯  acx  (bc  d)
Both composite functions are linear functions.
For example, use the linear functions f(x)  2x  3 and g(x)  x  5 .
f ÊÁË g(x) ˆ˜¯  2(x  5)  3
f ÁÊË g(x) ˜ˆ¯  2x  10  3
f ÊÁË g(x) ˆ˜¯  2x  7
g ÊÁË f(x) ˆ˜¯  (2x  3)  5
g ÊÁË f(x) ˆ˜¯  2x  3  5
g ÊÁË f(x) ˆ˜¯  2x  8
f ÊÁË g(x) ˆ˜¯  2x  7 and g ÊÁË f(x) ˆ˜¯  2x  8 are both linear functions.
PTS: 0
DIF: Difficult
REF: 4.3 Introduction to Composite Functions
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge | Communication
10
ID: A
41. ANS:
1
a) For g ÊÁË f(x) ˆ˜¯ , replace x in g(x)  x 2  5x with
.
x4
ÊÁ 1 ˆ˜ 2
Ê
ˆ
˜˜  5 ÁÁÁ 1 ˜˜˜
Ê
ˆ
g ÁË f(x) ˜¯  ÁÁÁÁ
˜
Á
ÁË x  4 ˜˜¯
Ë x  4 ˜¯
The domain of f(x) is x  4 .
The domain of g(x) is x  ò .
So, the domain of g ÊÁË f(x) ˆ˜¯ is x  4 .
1
with x 2  5x .
b) For f ÊÁË g(x) ˆ˜¯ , replace x in f(x) 
x4
1
f ÊÁË g(x) ˆ˜¯  2
x  5x  4
The domain of g(x) is x  ò .
The domain of f(x) is x  4 .
So, the value of g(x) cannot be 4. That is,
x 2  5x  4
2
x  5x  4  0
(x  1)(x  4)  0
So, x  1 or x  4
So, the domain of f ÊÁË g(x) ˆ˜¯ is x  1 and x  4 .
PTS: 0
DIF: Moderate
REF: 4.4 Determining Restrictions on Composite Functions
LOC: 12.RF1
TOP: Relations and Functions
KEY: Conceptual Understanding | Procedural Knowledge | Communication
42. ANS:
Complete the square:
9x 2  24x  12  9x 2  24x  12  4  4
 (3x  4) 2  4
So, y 
(3x  4) 2  4
Let f ÊÁÁ g ÊÁË h(x) ˆ˜¯ ˆ˜˜  (3x  4) 2  4 .
¯
Ë
Replace (3x  4) 2 with x.
Then, g ÊÁË h(x) ˆ˜¯  (3x  4) 2 and f(x) 
Use g ÊÁË h(x) ˆ˜¯  (3x  4) 2 .
x4
Replace 3x  4 with x.
Then, h(x)  3x  4 and g(x)  x 2
Possible functions are: f(x) 
x  4 , g(x)  x 2 , and h(x)  3x  4
PTS: 0
DIF: Difficult
REF: 4.4 Determining Restrictions on Composite Functions
LOC: 12.RF1
TOP: Relations and Functions
KEY: Procedural Knowledge | Communication | Problem-Solving Skills
11

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