230 CHAPTER 3 Linear Equations and Inequalities in Two Variables; Functions
OBJECTIVE 2 Understand the definition of a function.We now investigate
an important type of relation called a function.
Determining Whether Relations Are FunctionsDetermine whether each relation is a function.
(a)
Each first component appears once and only once. The relation is a function.
(b)
The first component 9 appears in two ordered pairs and corresponds to two dif-
ferent second components. Therefore, this relation is not a function. NOW TRY
519 , 3 2 , 19 , - 32 , 1 4, 2 26
51 - 2, 4 2 , 1 - 1, 1 2 , 1 0, 0 2 , 1 1, 1 2 , 1 2, 4 26
EXAMPLE 2
Most functions have an infinite number of ordered pairs and are usually defined
with equations that tell how to get the second components (outputs), given the first
components (inputs). Here are some everyday examples of functions.
1.The cost yin dollars charged by an express mail company is a function of the
weight xin pounds determined by the equation
2.In Cedar Rapids, Iowa, the sales tax is 7% of the price of an item. The tax yon
a particular item is a function of the price x, because
3.The distance dtraveled by a car moving at a constant speed of 45 mph is a
function of the time t. Thus,
The function concept can be illustrated by an input-output “machine,” as seen in
FIGURE 42. The express mail company equation provides an out-
put (the cost yin dollars) for a given input (the weight xin pounds).
y=1.5 1 x- 12 + 9
d= 45 t.
y=0.07x.
y=1.5 1 x- 12 + 9.
x = 3
(pounds)
Domain
valuey = 12
(dollars)
Range
valuey = 1.5( 3 – 1) + 9An input-output (function) machine
for y = 1.5(x – 1) + 9
FIGURE 42FunctionA functionis a set of ordered pairs in which each first component corresponds
to exactly one second component.
The relation in Example 1(a), is a function. The
relation in Example 1(b)is nota function, because
the same first component, 3, corresponds to more than one second component. If the
ordered pairs in Example 1(b)were interchanged, giving the relation
the result wouldbe a function. In that case, each domain element ( f irst component)
corresponds to exactly one range element (second component).
51 5, 3 2 , 1 6, 3 2 , 1 7, 3 2 , 1 8, 3 26 ,
51 3, 5 2 , 1 3, 6 2 , 1 3, 7 2 , 1 3, 8 26
51 0, 1 2 , 1 2, 5 2 , 1 3, 8 2 , 1 4, 2 26 ,
NOW TRY
EXERCISE 2
Determine whether each
relation is a function.
(a)
(b)
1 3, 1 2 , 1 8, 1 26
51 - 1, - 32 , 1 0, 2 2 ,
1 4, 3 2 , 1 4, 5 26
51 - 1, 2 2 , 1 0, 1 2 , 1 1, 0 2 ,
NOW TRY ANSWERS
- (a)not a function (b)function
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