Download 3.    SOLUTION:  

This notation can be read as all values between
negative 31 and 64 inclusive. The bracket
indicates that the 64 is included, while the
parenthesis indicates that the negative 31 is not
included.
1-1 Functions
Write each set of numbers in set-builder and
interval notation, if possible.
3. x ≤ −4
9. {− 0.25, 0, 0.25, 0.50, …}
SOLUTION: The set includes multiples of 0.25, starting with 0.25
(–1) or –0.25..
SOLUTION: The set includes all real numbers less than or equal
to –4.
In set-builder notation the set can be described as
{ x | 0.25n = x, n ≥ −1, n }. This notation can be
read as x such that x is 0.25 times n for all n
greater than or equal to −1, with n as an element
of the integers.
In set-builder notation this set can be described as {x
| x ≤ −4, x }. This notation can be read as x
such that x is less than or equal to -4, with x as
an element of the real numbers.
In interval notation it can be described as (− , −4].
This notation can be read as all values between
negative infinity and negative 4 inclusive. The
bracket indicates that the -4 is included.
The set cannot be described using one or more
inequalities, and therefore cannot be written in
interval notation.
12. all multiples of 8
SOLUTION: 6. −31 < x ≤ 64
This set includes all integers that are multiples of 8. SOLUTION: In set-builder notation the set can be described as {x
| x = 8n, n }. This notation can be read as x
such that x is 8 times n, with n as an element of
the integers.
The set includes all real numbers greater than –31
and less than or equal to 64.
In set-builder notation this set can be described as {x
| −31 < x ≤ 64, x }. This notation can be read as
x such that x is greater than -31 and x is less
than or equal to -64, with x as an element of the
real numbers.
In interval notation it can be described as (–31, 64].
This notation can be read as all values between
negative 31 and 64 inclusive. The bracket
indicates that the 64 is included, while the
parenthesis indicates that the negative 31 is not
included.
The set cannot be described using one or more
inequalities, and therefore cannot be written in
interval notation.
Determine whether each relation represents y
as a function of x.
15. The input value x is a bank account number and the
output value y is the account balance.
SOLUTION: Each value of x cannot be assigned to more than one
y-value because a bank account can only have one
balance at a given time. Therefore, the sentence
describes y as a function of x.
9. {− 0.25, 0, 0.25, 0.50, …}
SOLUTION: The set includes multiples of 0.25, starting with 0.25
(–1) or –0.25..
In set-builder notation the set can be described as
{ x | 0.25n = x, n ≥ −1, n }. This notation can be
read as x such that x is 0.25 times n for all n
greater than or equal to −1, with n as an element
of the integers.
The set cannot be described using one or more
inequalities, and therefore cannot be written in
interval notation.
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12. all multiples of 8
18. SOLUTION: Each x-value is not assigned to exactly one y-value.
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When x = 0.04 there are two corresponding yvalues, y = 449 and y = 451. Therefore, the table
does not represent y as a function of x.
1-1
Each value of x cannot be assigned to more than one
y-value because a bank account can only have one
balance at a given time. Therefore, the sentence
Functions
describes
y as a function of x.
24. =y – 6
SOLUTION: To determine whether this equation represents y as a
function of x, solve the equation for y.
18. SOLUTION: Each x-value is not assigned to exactly one y-value.
When x = 0.04 there are two corresponding yvalues, y = 449 and y = 451. Therefore, the table
does not represent y as a function of x.
The graph of this equation is quadratic. All quadratic
equations in which x is the squared term will open up
or open down and will pass the vertical line test. All
quadratic equations in which y is the squared term
will open left or open right and will not pass the
vertical line test. This graph does not pass the
vertical line test, so this equation is not a function.
21. 3y + 4x = 11
SOLUTION: To determine whether this equation represents y as a
function of x, solve the equation for y.
For every x-value there will be two corresponding yvalues.
All linear functions that can be written in slopeintercept form are non-vertical lines when graphed.
Thus, they will pass the vertical line test and are
functions. This equation represents y as a function of
x, because for every x-value there is exactly one
corresponding y-value.
27. SOLUTION: 24. =y – 6
SOLUTION: To determine whether this equation represents y as a
function of x, solve the equation for y.
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The graph of this equation is quadratic. All quadratic
A vertical line at x = –2 intersects the graph at more
than one point. Therefore, the graph does not
represent y as a function of x.
Find each function value.
30. g(x) = 2x2 + 18x – 14
a. g(9)
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A vertical line at x = –2 intersects the graph at more
than one point. Therefore, the graph does not
represent y as a function of x.
1-1 Functions
Find each function value.
30. g(x) = 2x2 + 18x – 14
a. g(9)
b. g(3x)
c. g(1 + 5m)
SOLUTION: 27. 2
To find each value, replace x in g(x) = 2x + 18x –
14.
SOLUTION: a.
b.
A vertical line at x = –2 intersects the graph at more
than one point. Therefore, the graph does not
represent y as a function of x.
Find each function value.
30. g(x) = 2x2 + 18x – 14
a. g(9)
b. g(3x)
c. g(1 + 5m)
c.
SOLUTION: 2
To find each value, replace x in g(x) = 2x + 18x –
14.
a.
b.
33. a. g(−2)
b. g(5x)
c. g(8 – 4b)
SOLUTION: c.
To find each value, replace x in
.
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SOLUTION: To find each value, replace x in
1-1 Functions
.
36. a. g(−2)
b. g(3m)
c. g(4m – 2)
SOLUTION: To find each value, replace m in
.
State the domain of each function.
39. SOLUTION: When the denominator of
36. is zero, the expression is a. g(−2)
b. g(3m)
c. g(4m – 2)
undefined.
SOLUTION: To find each value, replace m in
.
Therefore, the domain of this function is all real
numbers except x = –4 and x = –1, which can be
written as (− , −4) (−4, −1) (−1, ).
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Page 4
42. SOLUTION: This function is defined only when x ≠ 0 and x + 1 ≠ 0. Therefore, the function is defined for all real
numbers except x = 0 and x = –1. The domain of f
(x) is (– , –1)
(–1, 0) (0, ).
1-1 Functions
State the domain of each function.
Find f (−5) and f (12) for each piecewise function.
39. 48. SOLUTION: When the denominator of
SOLUTION: is zero, the expression is To find f (–5), use f (x) = –4x + 3.
undefined.
Therefore, the domain of this function is all real
numbers except x = –4 and x = –1, which can be
written as (− , −4) (−4, −1) (−1, ).
2
To find f (12), use f (x) = 3x + 1.
42. SOLUTION: The square root of a negative number cannot be a
2
real number, so 6 − x ≥ 0. 2
If 6 − x ≥ 0, then .
If x were greater than
or less than –
, the
2
expression 6 – x would be negative, and thus would
not be a real number. The domain of h(x) is [–
45. f (x) =
,
].
+ SOLUTION: This function is defined only when x ≠ 0 and x + 1 ≠ 0. Therefore, the function is defined for all real
numbers except x = 0 and x = –1. The domain of f
(x) is (– , –1)
(–1, 0) (0, ).
Find f (−5) and f (12) for each piecewise function.
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48. Page 5