Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
324
Relations and Functions
What makes a function a function?
Note: Problems 15.1–15.2 refer to relation p defined below.
p : {(1,9), (2,6), (5,1), (8,6)}
15.1 Evaluate p(1) and p(8).
A relation defines a relationship between two sets of numbers. Relation p is
defined as a list of coordinate pair (a,b), such that each input value a is paired
with an output value b.
To evaluate p(1), locate the ordered pair with abscissa 1 and report the
ordinate. Relation p contains the ordered pair (1,9), so p(1) = 9. Note
that p(1) ≠ 5. Although the point (5,1) belongs to the relation, 1 is the
ordinate of that coordinate pair, not the abscissa.
To evaluate p(8), identify the ordered pair with abscissa 8: (8,6). The ordinate
of that pair is 6, so p(8) = 6.
Note: Problems 15.1–15.2 refer to relation p defined below.
p : {(1,9), (2,6), (5,1), (8,6)}
15.2 Is p a function? Why or why not?
A relation is a function if and only if every abscissa value is paired with exactly
one ordinate value. Relation p states that p(1) = 9, p(2) = 5, p(5) = 1, and
p(8) = 6. Each input value (1, 2, 5, and 8) is paired with only one output value,
so p is a function.
Note: Problems 15.3–15.4 refer to relation q defined below.
q : {(–4, –1), (–2,3), (0,6), (–2,9), (–6, –13)}
15.3 Evaluate q(0) and q(–1).
To evaluate q(0), identify the coordinate pair with an abscissa of 0: (0,6).
Relation q defines q(0) as the ordinate of that pair: q(0) = 6.
Because none of the ordered pair constituting relation q have an abscissa of –1,
q(–1) is undefined.
Note: Problems 15.3–15.4 refer to relation q defined below.
q : {(–4, –1), (–2,3), (0,6), (–2,9), (–6, –13)}
15.4 Is q a function? Why or why not?
To be a function, each of a relation’s input values must be paired with only one
corresponding output value. Notice that two coordinate pairs in relation q con-
tain the same abscissa: (–2,3) and (–2,9). Therefore, q(–2) = 3 and q(–2) = 9.
Relation q pairs abscissa value –2 with two different ordinate values, so q is not
a function.
The left
number in
each ordered
pair is the input:
1, 2, 5, 8; p is not
dened for any
other values, so p(3),
for example, is
undened.
“Abscissa”
is the fancy
word for x-value
of a point, and
“ordinate” means
the y-value. The
abscissa of (1,9)
is 1 and the
ordinate is 9.
None of the
x-values in p are
repeated—each only
appears once. That
means each x-value
is paired with only
one y-value, so it’s
a function.
Q does con-
tain the point (–4, –1),
but –1 is the output
(ordinate) of that pair,
not the input.
Every input of
a function can only
produce one output. In
this case, the input –2
produces two outputs,
3 and 9.
Chapter Fifteen — Functions
The Humongous Book of Algebra Problems
325
Note: Problems 15.5–15.6 refer to the relation .
15.5 Evaluate r(–3) and r(4).
To evaluate r(–3) and r(4), substitute x = –3 and x = 4 into the relation.
Note: Problems 15.5–15.6 refer to the relation .
15.6 Assuming x is a real number, is r(x) a function? Why or why not?
Relation r(x) is a function, as every real number has exactly one absolute value.
The absolute value of a real number greater than or equal to zero equals the
number itself, and the absolute value of a real number less than zero is defined
as the opposite of that number.
Note: Problems 15.7–15.8 refer to the relation s(x) = 5 ± x.
15.7 Evaluate s(0) and s(–1).
To evaluate s(0), substitute x = 0 into s(x) = 5 ± x.
Evaluate s(–1).
The expression contains a ± symbol, so consider both possibilities as you
evaluate s(1).
s(0) = 5 ± 0
had a ± sign in it,
too, but adding 0 to
5 and subtracting 0
from 5 both produce
the same result: 5.
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