Section 8.2 Exponential Functions
771
Version: Fall 2007
8.2 Exponential Functions
Let’s suppose that the current population of the city of Pleasantville is 10 000 and that
the population is growing at a rate of 2% per year. In order to analyze the population
growth over a period of years, we’ll try to develop a formula for the population as a
function of time, and then graph the result.
First, note that at the end of one year, the population increase is 2% of 10 000,
or 200 people. We would now have 10 200 people in Pleasantville. At the end of the
second year, take another 2% of 10 200, which is an increase of 204 people, for a total of
10 404. Because the increase each year is not constant, the graph of population versus
time cannot be a line. Hence, our eventual population function will not be linear.
To develop our population formula, we start by letting the function
P
(
t
) represent
the population of Pleasantville at time
t
, where we measure
t
in years. We will start
time at
t
= 0 when the initial population of Pleasantville is 10000. In other words,
P
(0) = 10000. The key to understanding this example is the fact that the population
increases by 2% each year. We are making an assumption here that this overall growth
accounts for births, deaths, and people coming into and leaving Pleasantville. That is,
at the end of the ﬁrst year, the population of Pleasantville will be 102% of the initial
population. Thus,
P
(1) = 1
.
02
P
(0) = 1
.
02(10000)
.
(1)
We could multiply out the right side of this equation, but it will actually be more useful
to leave it in its current form.
Now each year the population increases by 2%. Therefore, at the end of the second
year, the population will be 102% of the population at the end of the ﬁrst year. In
other words,
P
(2) = 1
.
02
P
(1)
.
(2)
If we replace
P
(1) in
equation (2)
with the result found in
equation (1)
, then
P
(2) = (1
.
02)(1
.
02)(10000) = (1
.
02)
2
(10000)
.
(3)
Let’s iterate one more year. At the end of the third year, the population will be
102% of the population at the end of the second year, so
P
(3) = 1
.
02
P
(2)
.
(4)
However, if we replace
P
(2) in
equation (4)
with the result found in
equation (3)
,
we obtain
P
(3) = (1
.
02)(1
.
02)
2
(10000) = (1
.
02)
3
(10000)
.
(5)
The pattern should now be clear. The population at the end of
t
years is given by
the function
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