Practice Test - Chapter 1
Determine whether the given relation
represents y as a function of x.
4. PARKING The cost of parking a car downtown is
$0.75 per 30 minutes for a maximum of $4.50.
Parking is charged per second.
a. Write a function for c(x), the cost of parking a car
for x hours.
b. Find c(2.5).
c. What is the domain for c(x)? Explain your
reasoning.
1. y 3 – x = 5
SOLUTION: When x = 1, y = ±
. Therefore, the relation is not
one-to-one and not a function.
SOLUTION: a. A cost of $0.75 per 30 minutes is equal to a cost
of $1.50 per hour. The charge maximizes at $4.50 so
will remain unchanged when the car is parked for
longer than 3 hours.
c(x) =
2. b. 2.5 · 1.5 = 3.75
c. D = [0, 3]; Sample answer: The number of hours
must be greater than or equal to 0.
SOLUTION: The graph passes the Vertical Line Test, so the
relation is a function.
State the domain and range of each function.
3. y =
SOLUTION: For every x-value, there is only one corresponding yvalue. Therefore, the relation is one-to-one and a
function.
Also, the graph of the relation passes the Vertical
Line Test.
5. SOLUTION: The graph continues for all values of x and has a
minimum y-value of −3. D = (− , ), R = [−3, )
4. PARKING The cost of parking a car downtown is
$0.75 per 30 minutes for a maximum of $4.50.
Parking is charged per second.
a. Write a function for c(x), the cost of parking a car
for x hours.
b. Find c(2.5).
c. What is the domain for c(x)? Explain your
reasoning.
SOLUTION: a. A cost of $0.75 per 30 minutes is equal to a cost
of $1.50 per hour. The charge maximizes at $4.50 so
will remain unchanged when the car is parked for
longer than 3 hours.
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c(x) =
6. SOLUTION: The graph continues for all values of x such that x ≤ 5 and has a minimum y-value of 0. D = (− , 5], R =
[0, )
Find the y-intercept(s) and zeros for each
function.
7. f (x) = 4x2 – 8x − 12
SOLUTION: Page 1
SOLUTION: (x, −y)
(0.25, −1)
(1, −2)
(−2.25, 3)
(−4, 4)
(0.25, 1)
(1, 2)
(2.25, 3)
(4, 4)
The graph continues for all values of x such that x ≤ 5 and has
a minimum
y-value1of 0. D = (− , 5], R =
Practice
Test
- Chapter
[0, )
Find the y-intercept(s) and zeros for each
function.
7. f (x) = 4x2 – 8x − 12
Determine whether each function is continuous
at x = 3. If discontinuous, identify the type of
discontinuity as infinite, jump , or removable.
10. SOLUTION: SOLUTION: f(3) = 6, ,
The function is continuous at x = 3.
11. f (x) =
SOLUTION: 3
2
8. f (x) = x + 4x + 3x
SOLUTION: not continuous; removable discontinuity at x = 3
Find the average rate of change for each
function on the interval [−2, 6].
12. f (x) = −x4 + 3x
SOLUTION: 9. MULTIPLE CHOICE Which relation is
symmetric about the x-axis?
A −x2 − yx = 2
3
B x y = −8
C y = |x|
D −y 2 = −4x
SOLUTION: In order for a relation to be symmetric with respect
to the x-axis, (x, y) must correspond with (x, −y).
This occurs for choice D only.
(x, y)
(x, −y)
(0.25, 1)
(0.25, −1)
(1, 2)
(1, −2)
(2.25, 3)
(−2.25, 3)
(4, 4)
(−4, 4)
13. f (x) =
SOLUTION: Determine whether each function is continuous
at x = 3. If discontinuous, identify the type of
discontinuity as infinite, jump , or removable.
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SOLUTION: f(3) = 6, ,
Use the graph of each function to estimate
SOLUTION: Practice Test - Chapter 1
f is decreasing on (− , −1.5), increasing on (−1.5,
0), decreasing on (0, 1.5), and increasing on (1.5,
).
16. Which function is shown in the graph?
13. f (x) =
SOLUTION: F f (x) = |x − 4| − 3
G f (x) = |x − 4| + 3
H f (x) = |x + 4| − 3
J f (x) = |x + 4| + 3
SOLUTION: Use the graph of each function to estimate
intervals to the nearest 0.5 unit on which the
function is increasing or decreasing.
The parent function g(x) = | x | is shifted down 3
units. Therefore, F and H are possible choices. The
graph is also shifted 4 units left, which corresponds
to choice H.
Identify the parent function f (x) of g(x). Then
sketch the graph of g(x).
17. g(x) = −(x + 3)3
SOLUTION: f(x) = x
14. 3
SOLUTION: f is increasing on (−
).
, 2.5) and decreasing on (2.5,
18. g(x) = |x2 − 4|
SOLUTION: 15. SOLUTION: f(x) = x
2
f is decreasing on (− , −1.5), increasing on (−1.5,
0), decreasing on (0, 1.5), and increasing on (1.5,
).
16. Which function is shown in the graph?
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Given f (x) = x − 6 and g(x) = x 2 − 36, find each
function and its domain.
Practice Test - Chapter 1
There are no restrictions on the domain.
Given f (x) = x − 6 and g(x) = x 2 − 36, find each
function and its domain.
19. 21. TEMPERATURE In most countries, temperature
is measured in degrees Celsius. The equation that
relates degrees Fahrenheit with degrees Celsius is F
=
SOLUTION: C + 32.
a. Write C as a function of F.
b. Find two functions f and g such that C = [f
(F).
g]
SOLUTION: a.
The value of x cannot equal −6 or 6. In the original
expression, if x = 6 or −6, a division by 0 would
occur, which is undefined.
20. [g f ](x)
b. When making a composition, look for the
operation that must be done first. This will represent
the innermost function in the composition, in this
case, g(x). Sample answer: (x) =
SOLUTION: x; g(x) = x − 32
Determine whether f has an inverse function. If
it does, find the inverse function and state any
restrictions on its domain.
22. f (x) = (x – 2)3
SOLUTION: Graph f (x).
There are no restrictions on the domain.
21. TEMPERATURE In most countries, temperature
is measured in degrees Celsius. The equation that
relates degrees Fahrenheit with degrees Celsius is F
=
C + 32.
a. Write C as a function of F.
b. Find two functions f and g such that C = [f
(F).
g]
The graph passes the Horizontal Line Test.
Therefore, an inverse exists.
SOLUTION: a.
There is no domain restriction.
b. When making a composition, look for the
This will represent
the innermost function in the composition, in this
eSolutions
Manual that
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operation
mustbybeCognero
done first.
case, g(x). Sample answer: (x) =
x; g(x) = x − 32
23. f (x) =
SOLUTION: Page 4
Practice
- Chapter
1
There isTest
no domain
restriction.
The domain of f
−1
(x) is x ≠ 1.
24. f (x) =
23. f (x) =
SOLUTION: SOLUTION: Graph f (x).
Graph f (x).
The graph passes the Horizontal Line Test.
Therefore, an inverse exists.
The graph passes the Horizontal Line Test.
Therefore, an inverse exists.
Graph the inverse.
Graph the inverse.
The range of f (x) = [0, ∞), so the domain of f
must be limited to [0, ∞).
−1
(x)
25. f (x) = x2 – 16
The domain of f
−1
(x) is x ≠ 1.
24. f (x) =
SOLUTION: Graph f (x).
SOLUTION: Graph f (x).
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The graph does not pass the Horizontal Line Test.
Page 5
Therefore, an inverse does not exist.
The range
of f -(x) = [0, ∞), so the domain of f
Practice
Test
Chapter 1
−1
(x)
must be limited to [0, ∞).
25. f (x) = x2 – 16
SOLUTION: Graph f (x).
The graph does not pass the Horizontal Line Test.
Therefore, an inverse does not exist.
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