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Arkansas Tech University
MATH 2243: Business Calculus
Dr. Marcel B. Finan
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Exponential Functions
Recall that linear functions are functions that change at a constant rate. For
example, if f (x) = mx + b then f (x + 1) = m(x + 1) + b = f (x) + m. So when x
increases by 1, the y value changes by m. In contrast, an exponential function
is a function that changes by a constant positive factor.
If we denote the constant factor by b then an exponential function is a function of the form
f (t) = a · bt , b > 0, b 6= 1.
Since b = f (0), we call a the initial value. We call b the base or growth
factor of f (t).
Remark 5.1
Why b is restricted to b > 0 and b 6= 1? Since t is allowed
√ to have any value, a
negative b will create meaningless expressions such as b (if t = 12 ). Also, for
b = 1 the function f (t) = a is called a constant function and its graph is a
horizontal line.
Example 5.1
Identify the initial value and the growth factor of the exponential function f (t) =
0.75(0.2)t .
Solution.
The initial value is 0.75 and the growth factor is b = 0.2
Note that is, if f (t) = a · bt then f (t + 1) = bf (t). That is, f (t) changes by
a constant factor every time the input is increased by 1.
Writing b = 1 + r we obtain f (t + 1) = f (t) + rf (t). This implies that
r=
f (t + 1) − f (t)
.
f (t)
The right-hand side is a percent rate. Thus, an exponential function is a function
that changes at a constant percent rate.
Example 5.2
Suppose you are offered a job at a starting salary of $40,000 per year. To
strengthen the offer, the company promises annual raises of 6% per year for the
first 10 years. Let P (t) be your salary after t years. Find a formula for P (t)
and then compute your projected salary after 4 years from now.
Solution.
A formula for f (t) is f (t) = 40, 000(1.06)t . After four years, the projected salary
is f (4) = 40, 000(1.06)4 ≈ $50, 499.08.
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Example 5.3
The amount in milligrams of a drug in the body t hours after taking a pill is
given by A(t) = 25(0.85)t .
(a) What is the initial dose given?
(b) What percent of the drug leaves the body each hour?
(c) What is the amount of drug left after 10 hours?
Solution.
(a) Initial dose given is A(0) = 25 mg.
(b) r = a − 1 = 0.85 − 1 = −.15 so that 15% of the drug leaves the body each
hour.
(c) A(10) = 25(0.85)10 ≈ 4.92 mg.
Exponential functions are used to model increasing quantities such as population growth problems.
Example 5.4
Suppose that you are observing the behavior of cell duplication in a lab. In
one experiment, you started with one cell and the cells doubled every minute.
That is, the population cell is increasing at the constant rate of 100%. Write
an equation to determine the number (population) of cells after t hours.
Solution.
Table 1 below shows the number of cells for the first 5 minutes. Let P (t) be the
number of cells after t minutes.
t
P(t)
0
1
1 2 3
2 4 8
Table 1
4
16
5
32
At time 0, i.e t=0, the number of cells is 1 or 20 = 1. After 1 minute, when
t = 1, there are two cells or 21 = 2. After 2 minutes, when t = 2, there are 4
cells or 22 = 4.
Therefore, one formula to estimate the number of cells (size of population) after
t minutes is the equation (model)
f (t) = 2t .
The graph of P (t) is given in Figure 5.1
Figure 5.1
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Exponential functions can also model decreasing quantities known as decay
models.
Example 5.5
If you start a biology experiment with 5,000,000 cells and 45% of the cells are
dying every minute, write an equation to determine the number of cells that are
alive after t minutes.
Solution.
In this case, we have a = 5, 000, 000 and b = 0.55 so that P (t) = 5, 000, 000(0.55)t .
The graph of P (t) is given in Figure 5.2
Figure 5.2
From the previous two examples we observe the following:
• The graph of f (t) = a · bt is increasing for b > 1 and decreasing for 0 < b < 1.
• The graph opens up.
• The vertical intercept is a.
• The domain consists of all real numbers.
• The range consists of all positive numbers (so no horizontal intercepts).
Recognizing an Exponential Function Defined by Data
Suppose that f is a function defined by a table of values. If f is an exponential
abt+n
function then f can be written in the form f (t) = abt . Thus, f (t+n)
f (t) = abtn =
bn . This says that the ratios of y values are constant for equally spaced t values.
Example 5.6
Decide if the function is linear or exponential?Find a formula for each case.
t
f (t)
g(t)
0
12.5
0
1
13.75
2
2
15.125
4
3
16.638
6
4
18.301
8
Solution.
15.125
16.638
18.301
Since 13.75
12.5 ≈ 13.75 ≈ 15.125 ≈ 16.638 ≈ 1.1 then f (t) is an exponential function.
To find a formula for f (t) = abt we use the first two points obtaining 12.5 =
f (0) = a and 13.75 = f (1) = ab = 12.5b. Hence, b = 13.75
12.5 ≈ 1.1 so that
f (t) = 12.5(1.1)t .
On the other hand, equal increments in x corresponds to equal increments in
the g-values so that g(t) is linear, say g(t) = mt + b. Since g(0) = 0 then b = 0.
Also, 2 = g(1) = m so that g(t) = 2t
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The Effect of the Parameters a and b
Recall that an exponential function with base b and initial value a is a function
of the form f (t) = a · bt . In what follows, we assume that a > 0. Since a = f (0),
the point (0, a) is the vertical intercept of f (t).
We know that the slope of a linear function measures the steepness of the graph.
Similarly, the parameter b measures the steepness of the graph of an exponential function. The greater the value of b, the more rapidly the graph rises or
equivalently, the smaller the value of b, the more rapidly the graph falls.
• General Observations
(i) For b > 1, as t decreases, the function values get closer and closer to 0.
Symbolically, as t → −∞, y → 0. For 0 < b < 1, as t increases, the function
values gets closer and closer to the horizontal axis. That is, as t → ∞, y → 0.
We call the horizontal axis, a horizontal asymptote.
(ii) The domain of an exponential function consists of the set of all real numbers
whereas the range consists of the set of all positive real numbers.
(iii) The graph of f (t) = abt with a > 0 is always concave up.
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