Section 8.2 Eponential Functions 771 8.2 Eponential Functions Let s suppose that the current population of the cit of Pleasantville is 10 000 and that the population is growing at a rate of 2% per ear. In order to analze the population growth over a period of ears, we ll tr 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 ear, 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 ear, take another 2% of 10 200, which is an increase of 204 people, for a total of 10 404. Because the increase each ear 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 b letting the function P (t) represent the population of Pleasantville at time t, where we measure t in ears. We will start time at t = 0 when the initial population of Pleasantville is 10 000. In other words, P (0) = 10 000. The ke to understanding this eample is the fact that the population increases b 2% each ear. 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 ear, the population of Pleasantville will be 102% of the initial population. Thus, P (1) = 1.02P (0) = 1.02(10 000). (1) We could multipl out the right side of this equation, but it will actuall be more useful to leave it in its current form. Now each ear the population increases b 2%. Therefore, at the end of the second ear, the population will be 102% of the population at the end of the first ear. In other words, P (2) = 1.02P (1). (2) If we replace P (1) in equation (2) with the result found in equation (1), then P (2) = (1.02)(1.02)(10 000) = (1.02) 2 (10 000). (3) Let s iterate one more ear. At the end of the third ear, the population will be 102% of the population at the end of the second ear, so P (3) = 1.02P (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 (10 000) = (1.02) 3 (10 000). (5) The pattern should now be clear. The population at the end of t ears is given b the function 1 Coprighted material. See: http://msenu.redwoods.edu/intalgtet/
772 Chapter 8 Eponential and Logarithmic Functions P (t) = (1.02) t (10 000). It is traditional in mathematics and science to place the initial population in front in this formula, writing instead P (t) = 10 000(1.02) t. (6) Our function P (t) is defined b equation (6) for all positive integers {1, 2, 3,...}, and P (0) = 10 000, the initial population. Figure 1 shows a plot of our function. Although points are plotted onl at integer values of t from 0 to 40, that s enough to show the trend of the population over time. The population starts at 10 000, increases over time, and the earl increase (the difference in population from one ear to the net) also gets larger as time passes. 23000 P (population) P (t)=10 000(1.02) t 10000 0 40 Figure 1. Graph of population P (t) of Pleasantville for t = 0, 1, 2, 3,... t (ears) Eample 7. We can now use the function P (t) to predict the population in later ears. Assuming that the growth rate of 2% continues, what will the population of Pleasantville be after 40 ears? What will it be after 100 ears? Substitute t = 40 and t = 100 into equation (6). The population in 40 ears will be and the population in 100 ears will be P (40) = 10 000(1.02) 40 22 080, P (100) = 10 000(1.02) 100 72 446. What would be different if we had started with a population of 12 000? B tracing over our previous steps, it should be eas to see that the new formula would be P (t) = 12 000(1.02) t. Similarl, if the growth rate had been 3% per ear instead of 2%, then we would have ended up with the formula P (t) = 10 000(1.03) t.
Section 8.2 Eponential Functions 773 Thus, b letting P 0 represent the initial population, and r represent the growth rate (in decimal form), we can generalize the formula to P (t) = P 0 (1 + r) t. (8) Note that our formula for the function P (t) is different from the previous functions that we ve studied so far, in that the input variable t is part of the eponent in the formula. Thus, this is a new tpe of function. Now let s contrast the situation in Pleasantville with the population dnamics of Ghosttown. Ghosttown also starts with a population of 10 000, but several factories have closed, so some people are leaving for better opportunities. In this case, the population of Ghosttown is decreasing at a rate of 2% per ear. We ll again develop a formula for the population as a function of time, and then graph the result. First, note that at the end of one ear, the population decrease is 2% of 10 000, or 200 people. We would now have 9 800 people left in Ghosttown. At the end of the second ear, take another 2% of 9 800, which is a decrease of 196 people, for a total of 9 604. As before, because the decrease each ear is not constant, the graph of population versus time cannot be a line, so our eventual population function will not be linear. Now let the function P (t) represent the population of Ghosttown at time t, where we measure t in ears. The initial population of Ghosttown at t = 0 is 10 000, so P (0) = 10 000. Since the population decreases b 2% each ear, at the end of the first ear the population of Ghosttown will be 98% of the initial population. Thus, P (1) = 0.98P (0) = 0.98(10 000). (9) Each ear the population deccreases b 2%. Therefore, at the end of the second ear, the population will be 98% of the population at the end of the first ear. In other words, P (2) = 0.98P (1). (10) If we replace P (1) in equation (10) with the result found in equation (9), then P (2) = (0.98)(0.98)(10 000) = (0.98) 2 (10 000). (11) Let s iterate one more ear. At the end of the third ear, the population will be 98% of the population at the end of the second ear, so P (3) = 0.98P (2). (12) However, if we replace P (2) in equation (12) with the result found in equation (11), we obtain P (3) = (0.98)(0.98) 2 (10 000) = (0.98) 3 (10 000). (13) The pattern should now be clear. The population at the end of t ears is given b the function
774 Chapter 8 Eponential and Logarithmic Functions or equivalentl, P (t) = (0.98) t (10 000), P (t) = 10 000(0.98) t. (14) Our function P (t) is defined b equation (14) for all positive integers {1, 2, 3,...}, and P (0) = 10 000, the initial population. Figure 2 shows a plot of our function. Although points are plotted onl at integer values of t from 0 to 40, that s enough to show the trend of the population over time. The population starts at 10 000, decreases over time, and the earl decrease (the difference in population from one ear to the net) also gets smaller as time passes. 10000 P (population) 5000 P (t)=10 000(0.98) t 0 0 40 Figure 2. Graph of population P (t) of Ghosttown for t = 0, 1, 2, 3,... t (ears) Eample 15. Assuming that the rate of decrease continues at 2%, predict the population of Ghosttown after 40 ears and after 100 ears. Substitute t = 40 and t = 100 into equation (14). The population in 40 ears will be P (40) = 10 000(0.98) 40 4457, and the population in 100 ears will be P (100) = 10 000(0.98) 100 1326. Note that if we had instead started with a population of 9 000, for eample, then the new formula would be P (t) = 9 000(0.98) t. Similarl, if the rate of decrease had been 5% per ear instead of 2%, then we would have ended up with the formula P (t) = 10 000(0.95) t. Thus, b letting P 0 represent the initial population, and r represent the growth rate (in decimal form), we can generalize the formula to P (t) = P 0 (1 r) t. (16)
Section 8.2 Eponential Functions 775 Definition As noted before, our functions P (t) in our Pleasantville and Ghosttown eamples are a new tpe of function, because the input variable t is part of the eponent in the formula. Definition 17. An eponential function is a function of the form f(t) = b t, where b > 0 and b 1. b is called the base of the eponential function. More generall, a function of the form f(t) = Ab t, where b > 0, b 1, and A 0, is also referred to as an eponential function. In this case, the value of the function when t = 0 is f(0) = A, so A is the initial amount. In applications, ou will almost alwas encounter eponential functions in the more general form Ab t. In fact, note that in the previous population eamples, the function P (t) has this form P (t) = Ab t, with A = P 0, b = 1 + r in Pleasantville, and b = 1 r in Ghosttown. In particular, A = P 0 is the initial population. Since eponential functions are often used to model processes that var with time, we usuall use the input variable t (although of course an variable can be used). Also, ou ma be curious wh the definition sas b 1, since 1 t just equals 1. We ll eplain this curiosit at the end of this section. Graphs of Eponential Functions We ll develop the properties for the basic eponential function b t first, and then note the minor changes for the more general form Ab t. For a working eample, let s use base b = 2, and let s compute some values of f(t) = 2 t and plot the result (see Figure 3). 16 f(t)=2 t t f(t) = 2 t 1 2 2 4 3 8 4 16 (a) t 4 Figure 3. Plotting points (t, f(t)) defined b the function f(t) = 2 t, with t = 1, 2, 3, 4,.... (b)
776 Chapter 8 Eponential and Logarithmic Functions Recall from the previous section that 2 t is also defined for negative eponents t and the 0 eponent. Thus, the eponential function f(t) = 2 t is defined for all integers. Figure 4 shows a new table and plot with points added at 0 and negative integer values. t f(t) = 2 t 4 1/16 3 1/8 2 1/4 1 1/2 0 0 1 2 2 4 3 8 4 16 16 f(t)=2 t t 4 (a) Figure 4. Plotting points (t, f(t)) defined b the function f(t) = 2 t, with t =..., 3, 2, 1, 0, 1, 2, 3.... However, the previous section showed that 2 t is also defined for rational and irrational eponents. Therefore, the domain of the eponential function f(t) = 2 t is the set of all real numbers. When we add in the values of the function at all rational and irrational values of t, we obtain a final continuous curve as shown in Figure 5. 16 f(t)=2 t (b) Figure 5. t 4 Note several properties of the graph in Figure 5: a) Moving from left to right, the curve rises, which means that the function increases as t increases. In fact, the function increases rapidl for positive t. b) The graph lies above the t-ais, so the values of the function are alwas positive. Therefore, the range of the function is (0, ). c) The graph has a horizontal asmptote = 0 (the t-ais) on the left side. This means that the function almost dies out (the values get closer and closer to 0) as t approaches.
Section 8.2 Eponential Functions 777 What about the graphs of other eponential functions with different bases? We ll use the calculator to eplore several of these. First, use our calculator to compare 1 () = 2 and 2 () = 3. As can be seen in Figure 6(a), the graph of 3 rises faster that 2 for > 0, and dies out faster for < 0. 10 2 10 3 2 1 1 2 2 Figure 6. 3 () = 4 (a) (b) Comparing functions 1 () = 2, 2 () = 3, and Net, add in 3 () = 4. The result is shown in Figure 6(b). Again, increasing the size of the base to b = 4 results in a function which rises even faster on the right and likewise dies out faster on the left. If ou continue to increase the size of the base b, ou ll see that this trend continues. That s not terribl surprising because, if we compute the value of these functions at a fied positive, for eample at = 2, then the values increase: 2 2 < 3 2 < 4 2 <.... Similarl, at = 2, the values decrease: 2 2 > 3 2 > 4 2 >.... All of the functions in our eperiments so far share the properties listed in (a) (c) above: the function increases, the range is (0, ), and the graph has a horizontal asmptote = 0 on the left side. Now let s tr smaller values of the base b. First use the calculator to plot the graph of 1 () = (1/2) (see Figure 7(a)). 2 3 10 1 2 (a) Graph of 1 () = (1/2) Figure 7. (b) Comparing functions 1 () = (1/2), 2 () = (1/3), and 3 () = (1/4)
778 Chapter 8 Eponential and Logarithmic Functions This graph is much different. It rises rapidl to the left, and almost dies out on the right. Compare this with 2 () = (1/3) and 3 () = (1/4) (see Figure 7(b)). As the base gets smaller, the graph rises faster on the left, and dies out faster on the right. Using reflection properties, it s eas to understand the appearance of these last three graphs. Note that ( ) 1 = (2 1 ) = 2, (18) 2 so it follows that the graph of ( ) 1 2 is just a reflection in the -ais of the graph of 2 (see Figure 8). g 5 f 2 Figure 8. Comparing functions f() = 2 and g() = (1/2) = 2 Thus, we seem to have two different tpes of graphs, and therefore two tpes of eponential functions: one tpe is increasing, and the other decreasing. Our eperiments above, along with a little more eperimentation, should convince ou that b is increasing for b > 1, and decreasing for 0 < b < 1. The first tpe of functions are called eponential growth functions, and the second tpe are eponential deca functions. Properties of Eponential Growth Functions: f() = b with b > 1 The domain is the set of all real numbers. Moving from left to right, the graph rises, which means that the function increases as increases. The function increases rapidl for positive. The graph lies above the -ais, so the values of the function are alwas positive. Therefore, the range is (0, ). The graph has a horizontal asmptote = 0 (the -ais) on the left side. This means that the function almost dies out (the values get closer and closer to 0) as approaches. The second propert above deserves some additional eplanation. Looking at Figure 6(b), it appears that 2 and 3 increase rapidl as increases, but 1 appears to increase slowl. However, this is due to the fact that the graph of 1 () = 2
Section 8.2 Eponential Functions 779 is onl shown on the interval [ 2, 2]. In Figure 5, the same function is graphed on the interval [ 4, 4], and it certainl appears to increase rapidl in that graph. The point here is that eponential growth functions eventuall increase rapidl as increases. If ou graph the function on a large enough interval, the function will eventuall become ver steep on the right side of the graph. This is an important propert of the eponential growth functions, and will be eplored further in the eercises. Properties of Eponential Deca Functions: f() = b with 0 < b < 1 The domain is the set of all real numbers. Moving from left to right, the graph falls, which means that the function decreases as increases. The function decreases rapidl for negative. The graph lies above the -ais, so the values of the function are alwas positive. Therefore, the range is (0, ). The graph has a horizontal asmptote = 0 (the -ais) on the right side. This means that the function almost dies out (the values get closer and closer to 0) as approaches. Wh do we refrain from using the base b = 1? After all, 1 is certainl defined: it has the value 1 for all. But that means that f() = 1 is just a constant linear function its graph is a horizontal line. Therefore, this function doesn t share the same properties as the other eponential functions, and we ve alread classified it as a linear function. Thus, 1 is not considered to be an eponential function. Eample 19. Plot the graph of the function f() = (1.5). Identif the range of the function and the horizontal asmptote. Since the base 1.5 is larger than 1, this is an eponential growth function. Therefore, its graph will have a shape similar to the graphs in Figure 6. The graph rises, there will be a horizontal asmptote = 0 on the left side, and the range of the function is (0, ). The graph can then be plotted b hand b using this knowledge along with approimate values at = 2, 1, 0, 1, 2. See Figure 9. f() = (1.5) 5 2 0.44 1 0.67 0 1 1 1.5 2 2.25 3 f (a) Figure 9. (b) Graph of f() = (1.5)
780 Chapter 8 Eponential and Logarithmic Functions Eample 20. Plot the graph of the function g() = (0.2). Identif the range of the function and the horizontal asmptote. Since the base 0.2 is smaller than 1, this is an eponential deca function. Therefore, its graph will have a shape similar to the graphs in Figure 7. The graph falls, there will be a horizontal asmptote = 0 on the right side, and the range of the function is (0, ). The graph can then be plotted b hand b using this knowledge along with approimate values at = 2, 1, 0, 1, 2. See Figure 10. g() = (0.2) 2 25 1 5 0 1 1 0.2 2 0.04 (a) Figure 10. g 28 (b) Graph of g() = (0.2) Eample 21. Plot the graph of the function h() = 2 1. Identif the range of the function and the horizontal asmptote. The graph of h can be obtained from the graph of f() = 2 (see Figure 5) b a vertical shift down 1 unit. Therefore, the horizontal asmptote = 0 of the graph of f will also be shifted down 1 unit, so the graph of h has a horizontal asmptote = 1. Similarl, the range of f will be shifted down to ( 1, ) = Range(h). The graph can then be plotted b hand b using this knowledge along with approimate values at = 2, 1, 0, 1, 2. See Figure 11. h() = 2 1 2 0.75 1 0.5 0 0 8 2 1 1 2 3 3 = 1 (a) (b) Figure 11. Graph of h() = 2 1 In later sections of this chapter, we will also see more general eponential functions of the form f() = Ab (in fact, the Pleasantville and Ghosttown functions at the h
Section 8.2 Eponential Functions 781 beginning of this section are of this form). If A is positive, then the graphs of these functions can be obtained from the basic eponential graphs b vertical scaling, so the graphs will have the same general shape as either the eponential growth curves (if b > 1) or the eponential deca curves (if 0 < b < 1) we plotted earlier.