Families.
*See Exercise 23, Section 1.3.
a.Find the value of kfor 2, and 3, using the data for the 4-ft pendulum.
b.Use a graphing calculator to plot the data in the table and to graph the function for the three values of k(and their corresponding values of n) found in part a. Which function best fits the data?
L5kTn n51,
2.4 Exponential Functions 79
Later in this section, in Examples 5 and 6, we will see that the answers to these questions depend on exponential functions.
In earlier sections we discussed functions involving expressions such as
or where the variable or variable expression is the base of an exponential expression, and the exponent is a constant. In an exponential function, the variable is in the exponent and the base is a constant.
x21, x2,12x1123,
Exponential Functions
How much interest will an investment earn? What is the oxygen consumption of yearling salmon?
2.4
APPLY IT
Exponential Function
An exponential functionwith base ais defined as
f1x2 5ax, where a+0 and au1.
(If the function is the constant function
Exponential functions may be the single most important type of functions used in prac- tical applications. They are used to describe growth and decay, which are important ideas in management, social science, and biology.
Figure 46 shows a graph of the exponential function defined by You could
plot such a curve by hand by noting that and
and then drawing a smooth curve through the points (0, 1), (1, 2), and (2, 4). This graph is typical of the graphs of exponential functions of the form where The y-intercept is (0, 1). Notice that as xgets larger and larger, the function also gets larger. As xgets more and more negative, the function becomes smaller and smaller, approaching but never reaching 0. Therefore, the x-axis is a horizontal asymp- tote, but the function only approaches the left side of the asymptote. In contrast, rational functions approach both the left and right sides of the asymptote. The graph suggests that the domain is the set of all real numbers and the range is the set of all positive numbers.
Graphing an Exponential Function Graph
SOLUTION The graph, shown in Figure 47, is the horizontal reflection of the graph of given in Figure 46. Since this graph is typical of the graphs of exponential functions of the form where The domain includes all real numbers and the range includes all positive numbers. The y-intercept is (0, 1). Notice that this function, with is decreasing over its domain.
In the definition of an exponential function, notice that the base ais restricted to posi- tive values, with negative or zero bases not allowed. For example, the function
could not include such numbers as or in the domain because the y-values would not be real numbers. The resulting graph would be at best a series of separate points having little practical use.
Graphing an Exponential Function Graph
SOLUTION The graph of is the vertical reflection of the graph of so this is a decreasing function. (Notice that is not the same as In we raise 2 to the xpower and then take the negative.) The 3 indicates that the graph should be translated
22x, 1222x.
22x
y52x, y5 22x
f1x2 5 22x13.
x51/4
x51/2
y5 1242x 11/22x,
22x5 f1x2 5
0,a,1.
y5ax
22x51/2x5 11/22x, f1x2 52x
f1x2 522x. a.1.
y5ax,
121, 1/22,
122, 1/42,
2254,
2152, 2051,
22151/2,
22251/4,
f1x2 52x. f1x2 51.)
a51,
FOR REVIEW
To review the properties of expo- nents used in this section, see Section R.6.
–4 –2 0 2 4 x
y
2 4 6 8
y = 2x
FIGURE 46
FOR REVIEW
Recall from Section 2.2 that the graph of is the reflection of the graph of about the y-axis.
f1x2 f12x2
EXAMPLE 1
EXAMPLE 2 FIGURE 47
–4 –2 0 2 4 x
y
2 4 6 8
y = 2–x
vertically 3 units, as compared to the graph of Since would have y- intercept this function has y-intercept which is up 3 units. For negative values of x, the graph approaches the line which is a horizontal asymptote. The graph is shown in Figure 48.
Exponential Equations In Figures 46 and 47, which are typical graphs of expo- nential functions, a given value of xleads to exactly one value of Because of this, an equation with a variable in the exponent, called an exponential equation, often can be solved using the following property.
ax. y513,0, 22,
10, 212, y5 22. y5 22
f(x)
x y = 3 8
–5 –2
f(x) = –2 + 3x
0 3
FIGURE 48
If and a.0,a21, ax5ay,then x5y.
The value is excluded, since for example, even though To solve using this property, work as follows.
Solving Exponential Equations (a)Solve
SOLUTION First rewrite both sides of the equation so the bases are the same. Since and
Multiply exponents.
(b)Solve
SOLUTION Since the bases must be the same, write 32 as and 128 as giving
Multiply exponents.
Now use the property from above to get
Verify this solution in the original equation. TRY YOUR TURN 1
Compound Interest The calculation of compound interest is an important applica- tion of exponential functions. The cost of borrowing money or the return on an investment is called interest. The amount borrowed or invested is the principal,P. The rate of inter- estris given as a percent per year, and tis the time,measured in years.
x5 26 3 . 3x526 10x2557x121 210x25527x121.
12522x215 1272x13 322x215128x13
27, 25
322x215128x13.
x5 3 2. 2x53 32x533 1322x533 9x527 27533,
9532 9x527.
x5 7 3 3x57 23x527
23x527 223.
12513, a51
EXAMPLE 3
YOUR TURN 1 Solve
25x/25125x13.
FOR REVIEW
Recall from Section R.6 that 1am2n5amn.
2.4 Exponential Functions 81
Simple Interest
The product of the principal P, rate r, and time tgives simple interest,I:
. I5Prt
With compound interest, interest is charged (or paid) on interest as well as on the prin- cipal. To find a formula for compound interest, first suppose that Pdollars, the principal, is deposited at a rate of interest rper year. The interest earned during the first year is found using the formula for simple interest.
At the end of one year, the amount on deposit will be the sum of the original principal and the interest earned, or
(1)
If the deposit earns compound interest, the interest earned during the second year is found from the total amount on deposit at the end of the first year. Thus, the interest earned during the second year (again found by the formula for simple interest), is
(2)
so the total amount on deposit at the end of the second year is the sum of amounts from (1) and (2) above, or
In the same way, the total amount on deposit at the end of three years is
After tyears, the total amount on deposit, called the compound amount,is
When interest is compounded more than once a year, the compound interest formula is adjusted. For example, if interest is to be paid quarterly (four times a year), of the inter- est rate is used each time interest is calculated, so the rate becomes and the number of compounding periods in tyears becomes 4t. Generalizing from this idea gives the follow- ing formula.
r/4,
1/4
P111r2t. P111r23.
P111r21P111r2r5P111r2 111r2 5P111r22. 3P111r2 4 1r2 112 5P111r2r,
P1Pr5P111r2.
First-year interest5P.r.15Pr.
Compound Amount
If Pdollars is invested at a yearly rate of interest rper year, compounded mtimes per year for tyears, the compound amountis
A5Pa11 r
mbtm dollars.
Compound Interest
Inga Moffitt invests a bonus of $9000 at 6% annual interest compounded semiannually for 4 years. How much interest will she earn?
SOLUTION Use the formula for compound interest with and
Use a calculator.
The investment plus the interest is $11,400.93. The interest amounts to
$90005$2400.93. $11,400.932
<11,400.93 5900011.0328 59000a11 0.06
2 b4122 A5Pa11 r
mbtm
t54. P59000,r50.06,m52,
EXAMPLE 4
YOUR TURN 2 Find the interest earned on $4400 at 3.25% interest compounded quarterly for 5 years.
NOTE When using a calculator to compute the compound interest, store each partial result in the calculator and avoid rounding off until the final answer.
The Number e Perhaps the single most useful base for an exponential function is the num- ber e, an irrational number that occurs often in practical applications. The famous Swiss math- ematician Leonhard Euler (pronounced “oiler”) (1707–1783) was the first person known to have referred to this number as e, and the notation has continued to this day. To see how the number eoccurs in an application, begin with the formula for compound interest,
Suppose that a lucky investment produces annual interest of 100%, so that Suppose also that you can deposit only $1 at this rate, and for only one year. Then and Substituting these values into the formula for compound interest gives
As interest is compounded more and more often, mgets larger and the value of this expres- sion will increase. For example, if (interest is compounded annually),
so that your $1 becomes $2 in one year. Using a graphing calculator, we produced Figure 49 (where mis represented by X and by to see what happens as mbecomes larger and larger. A spreadsheet can also be used to produce this table.
The table suggests that as mincreases, the value of gets closer and closer to a fixed number, called e. As we shall see in the next chapter, this is an example of a limit.
1111/m2m Y1)
1111/m2m
a11 1
mbm5 a11 1
1b152152, m51
Pa11 r
mbt1m251a11 1
mb11m25 a11 1 mbm.
t51. r51.00P551.1
Pa11 r mbtm.
Definition of e
As mbecomes larger and larger, becomes closer and closer to the number e, whose approximate value is 2.718281828.
a11 1
mbm
1 8 50100 1000 10000
2 2.5658 2.6916 2.7048 2.7169 2.7181 2.7183 X5100000
X Y1
100000
FIGURE 49
The value of eis approximated here to 9 decimal places. Euler approximated eto 18 decimal places. Many calculators give values of usually with a key labeled Some require two keys, either INV LN or 2nd LN. (We will define in the next section.) In Figure 50, the functions and are graphed for comparison. Notice that is between and because eis between 2 and 3. For the graphs show that All three functions have y-intercept It is difficult to see from the
graph, but when
The number eis often used as the base in an exponential equation because it provides a good model for many natural, as well as economic, phenomena. In the exercises for this section, we will look at several examples of such applications.
Continuous Compounding In economics, the formula for continuous com- poundingis a good example of an exponential growth function. Recall the formula for compound amount
A5Pa11 r mbtm, x,0.
3x,ex,2x 10, 12.
3x.eexx.2x. 2x 3x, x.0,
y53x y5ex,
y52x, ex, ln x ex.
y
x y = 2 8
2 4 6
–1 0 1 2 3 y = e y = 3x
x x
FIGURE 50
2.4 Exponential Functions 83 where mis the number of times annually that interest is compounded. As mbecomes larger and larger, the compound amount also becomes larger but not without bound. Recall that as mbecomes larger and larger, becomes closer and closer to e. Similarly,
becomes closer and closer to e. Let us rearrange the formula for compound amount to take advantage of this fact.
This last expression becomes closer and closer to as m becomes larger and larger, which describes what happens when interest is compounded continuously. Essentially, the number of times annually that interest is compounded becomes infinitely large. We thus have the following formula for the compound amount when interest is compounded continuously.
Pert m
r .rt5tm 5Pc a11 1
1m/r2 b
m/r
drt 5Pa11 1
1m/r2btm A5Pa11 r
mbtm
a11 1
1m/r2 b
m/r
1111/m2m
Continuous Compounding
If a deposit of Pdollars is invested at a rate of interest rcompounded continuously for tyears, the compound amount is
dollars.
A5Pert
Continuous Compound Interest
If $600 is invested in an account earning 2.75% compounded continuously, how much would be in the account after 5 years?
SOLUTION In the formula for continuous compounding, let and to get
or $688.44. TRY YOUR TURN 3
In situations that involve growth or decay of a population, the size of the population at a given time toften is determined by an exponential function of t. The next example illus- trates a typical application of this kind.
Oxygen Consumption
Biologists studying salmon have found that the oxygen consumption of yearling salmon (in appropriate units) increases exponentially with the speed of swimming according to the function defined by
where xis the speed in feet per second. Find the following.
(a) The oxygen consumption when the fish are still f1x2 5100e0.6x, A5600e 510.02752<688.44,
0.0275 P5600, t55, r5
EXAMPLE 5
YOUR TURN 3 Find the amount after 4 years if $800 is invested in an account earning 3.15% compounded continuously.
EXAMPLE 6 APPLY IT
APPLY IT
SOLUTION When the fish are still, their speed is 0. Substitute 0 for x:
When the fish are still, their oxygen consumption is 100 units.
(b)The oxygen consumption at a speed of 2 ft per second SOLUTION Find as follows.
At a speed of 2 ft per second, oxygen consumption is about 332 units rounded to the nearest integer. Because the function is only an approximation of the real situation, further accu- racy is not realistic.
Food Surplus
A magazine article argued that the cause of the obesity epidemic in the United States is the decreasing cost of food (in real terms) due to the increasing surplus of food. Source:The New York Times Magazine.As one piece of evidence, the following table was provided, which we have updated, showing U.S. corn production (in billions of bushels) for selected years.
f122 5100e10.621225100e1.2<332
f122
e051 5100.15100.
f102 5100e10.621025100e0
FOR REVIEW
Refer to the discussion on linear regression in Section 1.3. A simi- lar process is used to fit data points to other types of functions.
Many of the functions in this chapter’s applications were deter- mined in this way, including that given in Example 6.
EXAMPLE 7
1930 1.757
1940 2.207
1950 2.764
1960 3.907
1970 4.152
1980 6.639
1990 7.934
2000 9.968
2005 11.112
Production Year (billions of bushels)
(a)Plot the data. Does the production appear to grow linearly or exponentially?
SOLUTION Figure 51 shows a graphing calculator plot of the data, which suggests that corn production is growing exponentially.
(b)Find an exponential function in the form of that models this data, where xis the year and is the production of corn. Use the data for 1930 and 2005.
SOLUTION Since we have Using we have
Divide by
Take the root.
Thus Figure 52 shows that this function fits the data well.
(c)Determine the expected annual percentage increase in corn production during this time period.
SOLUTION Since ais 1.0249, the production of corn each year is 1.0249 times its value the previous year, for a rate of increase of 0.024952.49%per year.
p1x2 51.75711.02492x21930.
<1.0249.
75th
a5 a11.112 1.757b1/75
1.757.
a755 11.112 1.757 p12005251.757a20052193051.757a75511.112
x52005, p051.757.
p1193025p0a05p0,
p1x2 p1x25p0ax21930
1920 2010
12
0
FIGURE 51
1920 2010
12
0
y 5 1.757(1.0249)x21930
FIGURE 52
2.4 Exponential Functions 85
(d)Graph pand estimate the year when corn production will be double what it was in 2005.
SOLUTION Figure 53 shows the graphs of and on the same coordinate axes. (Note that the scale in Figure 53 is different than the scale in Fig- ures 51 and 52 so that larger values of xand are visible.) Their graphs intersect at approximately 2033, which is thus the year when corn production will be double its 2005 level. In the next section, we will see another way to solve such problems that does not require the use of a graphing calculator.
Another way to check whether an exponential function fits the data is to see if points whose x-coordinates are equally spaced have y-coordinates with a constant ratio. This must
be true for an exponential function because if then and
so
This last expression is constant if is constant, that is, if the x-coordinates are equally spaced.
In the previous example, all data points but the last have x-coordinates 10 years apart, so we can compare the ratios of corn production for any of these first pairs of years. Here are the ratios for 1930–1940 and for 1990–2000:
These ratios are identical to 3 decimal places, so an exponential function fits the data very well. Not all ratios are this close; using the values at 1970 and 1980, we have
From Figure 52, we can see that this is because the 1970 value is below the exponential curve and the 1980 value is above the curve.
Another way to find an exponential function that fits a set of data is to use a graphing calculator or computer program with an exponential regression feature. This fits an exponential function through a set of points using the least squares method, introduced in Section 1.3 for fitting a line through a set of points. On a TI-84 Plus, for example, enter the year into the list L1and the corn production into L2. For simplicity, subtract 1930 from each year, so that 1930 corresponds to Selecting ExpRegfrom the STAT CALCmenu yields which is close to the function we found in Example 7(b).
1.72811.02542x, y 5
x50.
1.599. 6.639/4.1525
9.968
7.93451.256 2.207
1.75751.256 x22x1 f1x22
f1x12 5 a.bx2
a.bx15bx22x1. a.bx2,
f1x22 5 f1x2 5a.bx, f1x125a.bx1 p1x2
y52.11.112522.224 p1x2
1920 2050
24
0 Intersection
X2033.1738 Y22.224 y 1.757(1.0249)x1930 y 22.224
FIGURE 53
TECHNOLOGY NOTE
2.4 EXERCISES
A ream of 20-lb paper contains 500 sheets and is about 2 in.
high. Suppose you take one sheet, fold it in half, then fold it in half again, continuing in this way as long as possible. Source:
The AMATYC Review.
1.Complete the table.
19. 20.
21. 22.
23. 24.
25. 26.
27. 28.
Graph each of the following.
29. 30.
31. 32.
33. In our definition of exponential function, we ruled out negative values ofa. The author of a textbook on mathematical econom- ics, however, obtained a “graph” of by plotting the following points and drawing a smooth curve through them.
y5 1222x y54e2x/221 y5 23e22x12
y5 22ex23 y55ex12
ex215x1651 27x59x21x
8x252x14 5x21x51
2x224x5a1 16bx24 520x05 1
25
20x058 e2x5 1e42x13
1e3222x5e2x15 16x135642x25
Number of Folds 1 2 3 4 5 10 50
Layers of Paper
*
*
2.After folding 50 times (if this were possible), what would be the height (in miles) of the folded paper?
For Exercises 3–11, match the correct graph A–F to the func- tion without using your calculator. Notice that there are more functions than graphs; some of the functions are equivalent.
After you have answered all of them, use a graphing calculator to check your answers. Each graph in this group is plotted on the window by
3. 4.
5. 6.
7. 8.
9. 10.
11.y53x21
y5 22132x y52232x
y5 a1 3bx y53132x
y53x11 y5a1
3b12x
y532x y53x
324, 44. 322, 24
22 2
4
24
22 2
4
24
22 2
4
24
22 2
4
24
2 2
4
4
22 2
4
24
12.In Exercises 3 –11, there were more formulas for functions than there were graphs. Explain how this is possible.
Solve each equation.
13. 14.
15. 16.
17.4x58x11 18.25x5125x12 ex5 1
e5 3x5 1
81
4x564 2x532
x 0 1 2 3
y 1/16 21/8 1/4 21/2 1 22 4 28 21
22 23 24
The graph oscillates very neatly from positive to negative val- ues ofy. Comment on this approach. (This exercise shows the dangers of point plotting when drawing graphs.)
34. Explain why the exponential equation cannot be solved using the method described in Example 3.
35. Explain why when but
when
36.A friend claims that as xbecomes large, the expression gets closer and closer to 1, and 1 raised to any power is still 1.
Therefore, gets closer and closer to 1 as x gets larger. Use a graphing calculator to graph f on How might you use this graph to explain to the friend why does not approach 1 as x becomes large?
What does it approach?
APPLICATIONS
B u s i n e s s a n d E c o n o m i c s