Exponential FunctionsLet’s suppose that the current population of the city of Pleasantville is 10 000 and thatthe population is growing at a rate of 2% per year. In order to analyze the populationgrowth over a period of years, we’ll try to develop a formula for the population as afunction 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 thesecond year, take another 2% of 10 200, which is an increase of 204 people, for a total of10 404. Because the increase each year is not constant, the graph of population versustime 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) representthe population of Pleasantville at time t, where we measure t in years. We will starttime at t = 0 when the initial population of Pleasantville is 10 000. In other words,P (0) = 10 000. The key to understanding this example is the fact that the populationincreases by 2% each year. We are making an assumption here that this overall growthaccounts 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 initialpopulation. Thus,

P (1) = 1.02P (0) = 1.02(10 000). (1)

We could multiply out the right side of this equation, but it will actually be more usefulto leave it in its current form.

Now each year the population increases by 2%. Therefore, at the end of the secondyear, the population will be 102% of the population at the end of the first year. Inother 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 year. At the end of the third year, the population will be102% of the population at the end of the second year, 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 years is given bythe function

Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/5

336

P (t) = (1.02)t(10 000).

It is traditional in mathematics and science to place the initial population in front inthis formula, writing instead

P (t) = 10 000(1.02)t. (6)

Our function P (t) is defined by 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 only at integer values of t from 0 to 40, that’s enough toshow the trend of the population over time. The population starts at 10 000, increasesover time, and the yearly increase (the difference in population from one year to thenext) also gets larger as time passes.

t (years)

P (population)P (t)=10 000(1.02)t

40010000

23000

Figure 1. Graph of population P (t)of Pleasantville for t = 0, 1, 2, 3, . . .

Example 7. We can now use the function P (t) to predict the population in lateryears. Assuming that the growth rate of 2% continues, what will the population ofPleasantville be after 40 years? What will it be after 100 years?

Substitute t = 40 and t = 100 into equation (6). The population in 40 years will be

P (40) = 10 000(1.02)40 ≈ 22 080,

and the population in 100 years will be

P (100) = 10 000(1.02)100 ≈ 72 446.

What would be different if we had started with a population of 12 000? By tracingover our previous steps, it should be easy to see that the new formula would be

P (t) = 12 000(1.02)t.

Similarly, if the growth rate had been 3% per year instead of 2%, then we would haveended up with the formula

P (t) = 10 000(1.03)t.

337

Thus, by letting P0 represent the initial population, and r represent the growth rate(in decimal form), we can generalize the formula to

P (t) = P0(1 + r)t. (8)

Note that our formula for the function P (t) is different from the previous functionsthat we’ve studied so far, in that the input variable t is part of the exponent in theformula. Thus, this is a new type of function.

Now let’s contrast the situation in Pleasantville with the population dynamics ofGhosttown. Ghosttown also starts with a population of 10 000, but several factorieshave closed, so some people are leaving for better opportunities. In this case, thepopulation of Ghosttown is decreasing at a rate of 2% per year. We’ll again develop aformula for the population as a function of time, and then graph the result.

First, note that at the end of one year, the population decrease is 2% of 10 000,or 200 people. We would now have 9 800 people left in Ghosttown. At the end ofthe second year, take another 2% of 9 800, which is a decrease of 196 people, for atotal of 9 604. As before, because the decrease each year is not constant, the graph ofpopulation versus time cannot be a line, so our eventual population function will notbe linear.

Now let the function P (t) represent the population of Ghosttown at time t, wherewe measure t in years. The initial population of Ghosttown at t = 0 is 10 000, soP (0) = 10 000. Since the population decreases by 2% each year, at the end of the firstyear the population of Ghosttown will be 98% of the initial population. Thus,

P (1) = 0.98P (0) = 0.98(10 000). (9)

Each year the population deccreases by 2%. Therefore, at the end of the second year,the population will be 98% of the population at the end of the first year. In otherwords,

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 year. At the end of the third year, the population will be98% of the population at the end of the second year, 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 years is given bythe function

338

P (t) = (0.98)t(10 000),

or equivalently,

P (t) = 10 000(0.98)t. (14)

Our function P (t) is defined by 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 only at integer values of t from 0 to 40, that’s enough toshow the trend of the population over time. The population starts at 10 000, decreasesover time, and the yearly decrease (the difference in population from one year to thenext) also gets smaller as time passes.

t (years)

P (population)

P (t)=10 000(0.98)t

400

10000

5000

0

Figure 2. Graph of population P (t)of Ghosttown for t = 0, 1, 2, 3, . . .

Example 15. Assuming that the rate of decrease continues at 2%, predict thepopulation of Ghosttown after 40 years and after 100 years.

Substitute t = 40 and t = 100 into equation (14). The population in 40 years will be

P (40) = 10 000(0.98)40 ≈ 4457,

and the population in 100 years will be

P (100) = 10 000(0.98)100 ≈ 1326.

Note that if we had instead started with a population of 9 000, for example, thenthe new formula would be

P (t) = 9 000(0.98)t.

Similarly, if the rate of decrease had been 5% per year instead of 2%, then we wouldhave ended up with the formula

P (t) = 10 000(0.95)t.

Thus, by letting P0 represent the initial population, and r represent the growth rate(in decimal form), we can generalize the formula to

P (t) = P0(1− r)t. (16)

339

DefinitionAs noted before, our functions P (t) in our Pleasantville and Ghosttown examples area new type of function, because the input variable t is part of the exponent in theformula.

Definition 17. An exponential function is a function of the form

f(t) = bt,

where b > 0 and b �= 1. b is called the base of the exponential function.More generally, a function of the form

f(t) = Abt,

where b > 0, b �= 1, and A �= 0, is also referred to as an exponential function. Inthis case, the value of the function when t = 0 is f(0) = A, so A is the initialamount.

In applications, you will almost always encounter exponential functions in the moregeneral form Abt. In fact, note that in the previous population examples, the functionP (t) has this form P (t) = Abt, with A = P0, b = 1 + r in Pleasantville, and b = 1− rin Ghosttown. In particular, A = P0 is the initial population.

Since exponential functions are often used to model processes that vary with time,we usually use the input variable t (although of course any variable can be used). Also,you may be curious why the definition says b �= 1, since 1t just equals 1. We’ll explainthis curiosity at the end of this section.

Graphs of Exponential FunctionsWe’ll develop the properties for the basic exponential function bt first, and then notethe minor changes for the more general form Abt. For a working example, let’s use baseb = 2, and let’s compute some values of f(t) = 2t and plot the result (see Figure 3).

t f(t) = 2t

1 22 43 84 16

t4

y16 f(t)=2t

(a) (b)

Figure 3. Plotting points (t, f(t)) defined by the function f(t) =2t, with t = 1, 2, 3, 4, . . ..

340

Recall from the previous section that 2t is also defined for negative exponents tand the 0 exponent. Thus, the exponential function f(t) = 2t is defined for all integers.Figure 4 shows a new table and plot with points added at 0 and negative integervalues.

t f(t) = 2t

−4 1/16−3 1/8−2 1/4−1 1/20 01 22 43 84 16

t4

y16 f(t)=2t

(a) (b)

Figure 4. Plotting points (t, f(t)) defined by the function f(t) =2t, with t = . . . ,−3,−2,−1, 0, 1, 2, 3 . . ..

However, the previous section showed that 2t is also defined for rational and irra-tional exponents. Therefore, the domain of the exponential function f(t) = 2t is theset of all real numbers. When we add in the values of the function at all rational andirrational values of t, we obtain a final continuous curve as shown in Figure 5.

t4

y16 f(t)=2t

Figure 5.

Note several properties of the graph in Figure 5:

a) Moving from left to right, the curve rises, which means that the function increasesas t increases. In fact, the function increases rapidly for positive t.

b) The graph lies above the t-axis, so the values of the function are always positive.Therefore, the range of the function is (0,∞).

c) The graph has a horizontal asymptote y = 0 (the t-axis) on the left side. Thismeans that the function almost “dies out” (the values get closer and closer to 0) ast approaches −∞.

341

What about the graphs of other exponential functions with different bases? We’lluse the calculator to explore several of these.

First, use your calculator to compare y1(x) = 2x and y2(x) = 3x. As can be seenin Figure 6(a), the graph of 3x rises faster that 2x for x > 0, and dies out faster forx < 0.

x2

y10

y1

y2

x2

y10

y1

y2

y3

(a) (b)

Figure 6. Comparing functions y1(x) = 2x, y2(x) = 3x, andy3(x) = 4x

Next, add in y3(x) = 4x. The result is shown in Figure 6(b). Again, increasingthe size of the base to b = 4 results in a function which rises even faster on the rightand likewise dies out faster on the left. If you continue to increase the size of the baseb, you’ll see that this trend continues. That’s not terribly surprising because, if wecompute the value of these functions at a fixed positive x, for example at x = 2, thenthe values increase: 22 < 32 < 42 < . . .. Similarly, at x = −2, the values decrease:2−2 > 3−2 > 4−2 > . . ..

All of the functions in our experiments so far share the properties listed in (a)–(c) above: the function increases, the range is (0,∞), and the graph has a horizontalasymptote y = 0 on the left side. Now let’s try smaller values of the base b. First usethe calculator to plot the graph of y1(x) = (1/2)x (see Figure 7(a)).

x2

y10

y1

y2y3

(a) Graph ofy1(x) = (1/2)x

(b) Comparing functionsy1(x) = (1/2)x,y2(x) = (1/3)x,

and y3(x) = (1/4)x

Figure 7.

342

This graph is much different. It rises rapidly to the left, and almost dies out on theright. Compare this with y2(x) = (1/3)x and y3(x) = (1/4)x (see Figure 7(b)). Asthe base gets smaller, the graph rises faster on the left, and dies out faster on the right.

Using reflection properties, it’s easy to understand the appearance of these lastthree graphs. Note that (

12

)x= (2−1)x = 2−x, (18)

so it follows that the graph of(1

2)x is just a reflection in the y-axis of the graph of 2x

(see Figure 8).

x2

y5

g f

Figure 8. Comparing functionsf(x) = 2x and g(x) = (1/2)x = 2−x

Thus, we seem to have two different types of graphs, and therefore two types of expo-nential functions: one type is increasing, and the other decreasing. Our experimentsabove, along with a little more experimentation, should convince you that bx is increas-ing for b > 1, and decreasing for 0 < b < 1. The first type of functions are calledexponential growth functions, and the second type are exponential decay functions.

Properties of Exponential Growth Functions: f(x) = bx with b > 1

• The domain is the set of all real numbers.

• Moving from left to right, the graph rises, which means that the functionincreases as x increases. The function increases rapidly for positive x.

• The graph lies above the x-axis, so the values of the function are always positive.Therefore, the range is (0,∞).

• The graph has a horizontal asymptote y = 0 (the x-axis) on the left side. Thismeans that the function almost “dies out” (the values get closer and closer to0) as x approaches −∞.

The second property above deserves some additional explanation. Looking atFigure 6(b), it appears that y2 and y3 increase rapidly as x increases, but y1 ap-pears to increase slowly. However, this is due to the fact that the graph of y1(x) = 2x

343

is only shown on the interval [−2, 2]. In Figure 5, the same function is graphed on theinterval [−4, 4], and it certainly appears to increase rapidly in that graph. The pointhere is that exponential growth functions eventually increase rapidly as x increases.If you graph the function on a large enough interval, the function will eventually be-come very steep on the right side of the graph. This is an important property of theexponential growth functions, and will be explored further in the exercises.

Properties of Exponential Decay Functions: f(x) = bx 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 de-creases as x increases. The function decreases rapidly for negative x.

• The graph lies above the x-axis, so the values of the function are always positive.Therefore, the range is (0,∞).

• The graph has a horizontal asymptote y = 0 (the x-axis) on the right side. Thismeans that the function almost “dies out” (the values get closer and closer to0) as x approaches ∞.

Why do we refrain from using the base b = 1? After all, 1x is certainly defined:it has the value 1 for all x. But that means that f(x) = 1x is just a constant linearfunction – its graph is a horizontal line. Therefore, this function doesn’t share the sameproperties as the other exponential functions, and we’ve already classified it as a linearfunction. Thus, 1x is not considered to be an exponential function.

Example 19. Plot the graph of the function f(x) = (1.5)x. Identify the range ofthe function and the horizontal asymptote.

Since the base 1.5 is larger than 1, this is an exponential growth function. Therefore,its graph will have a shape similar to the graphs in Figure 6. The graph rises, therewill be a horizontal asymptote y = 0 on the left side, and the range of the function is(0,∞). The graph can then be plotted by hand by using this knowledge along withapproximate values at x = −2,−1, 0, 1, 2. See Figure 9.

x f(x) = (1.5)x

−2 0.44−1 0.670 11 1.52 2.25 x3

y5

f

(a) (b)

Figure 9. Graph of f(x) = (1.5)x

344

Example 20. Plot the graph of the function g(x) = (0.2)x. Identify the range ofthe function and the horizontal asymptote.

Since the base 0.2 is smaller than 1, this is an exponential decay function. Therefore,its graph will have a shape similar to the graphs in Figure 7. The graph falls, therewill be a horizontal asymptote y = 0 on the right side, and the range of the functionis (0,∞). The graph can then be plotted by hand by using this knowledge along withapproximate values at x = −2,−1, 0, 1, 2. See Figure 10.

x g(x) = (0.2)x

−2 25−1 50 11 0.22 0.04

x2

y28g

(a) (b)

Figure 10. Graph of g(x) = (0.2)x

Example 21. Plot the graph of the function h(x) = 2x − 1. Identify the range ofthe function and the horizontal asymptote.

The graph of h can be obtained from the graph of f(x) = 2x (see Figure 5) by avertical shift down 1 unit. Therefore, the horizontal asymptote y = 0 of the graph of fwill also be shifted down 1 unit, so the graph of h has a horizontal asymptote y = −1.Similarly, the range of f will be shifted down to (−1,∞) = Range(h). The graph canthen be plotted by hand by using this knowledge along with approximate values atx = −2,−1, 0, 1, 2. See Figure 11.

x h(x) = 2x − 1−2 −0.75−1 −0.50 01 12 3 x3

y8

h

y=−1

(a) (b)

Figure 11. Graph of h(x) = 2x − 1

In later sections of this chapter, we will also see more general exponential functionsof the form f(x) = Abx (in fact, the Pleasantville and Ghosttown functions at the

345

beginning of this section are of this form). If A is positive, then the graphs of thesefunctions can be obtained from the basic exponential graphs by vertical scaling, so thegraphs will have the same general shape as either the exponential growth curves (ifb > 1) or the exponential decay curves (if 0 < b < 1) we plotted earlier.

346

Exercises

1. The current population of Fortunais 10,000 hearty souls. It is known thatthe population is growing at a rate of4% per year. Assuming this rate remainsconstant, perform each of the followingtasks.

a. Set up an equation that models thepopulation P (t) as a function of timet.

b. Use the model in the previous part topredict the population 40 years fromnow.

c. Use your calculator to sketch the graphof the population over the next 40years.

2. The population of the town of Imag-ination currently numbers 12,000 people.It is known that the population is grow-ing at a rate of 6% per year. Assumingthis rate remains constant, perform eachof the following tasks.

a. Set up an equation that models thepopulation P (t) as a function of timet.

b. Use the model in the previous part topredict the population 30 years fromnow.

c. Use your calculator to sketch the graphof the population over the next 30years.

3. The population of the town of De-spairia currently numbers 15,000 individ-uals. It is known that the population isdecaying at a rate of 5% per year. As-suming this rate remains constant, per-form each of the following tasks.

Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/6

a. Set up an equation that models thepopulation P (t) as a function of timet.

b. Use the model in the previous part topredict the population 50 years fromnow.

c. Use your calculator to sketch the graphof the population over the next 50years.

4. The population of the town of Hope-less currently numbers 25,000 individu-als. It is known that the population isdecaying at a rate of 6% per year. As-suming this rate remains constant, per-form each of the following tasks.

a. Set up an equation that models thepopulation P (t) as a function of timet.

b. Use the model in the previous part topredict the population 40 years fromnow.

c. Use your calculator to sketch the graphof the population over the next 40years.

In Exercises 5-12, perform each of thefollowing tasks for the given function.

a. Find the y-intercept of the graph ofthe function. Also, use your calcula-tor to find two points on the graph tothe right of the y-axis, and two pointsto the left.

b. Using your five points from (a) as aguide, set up a coordinate system ongraph paper. Choose and label ap-propriate scales for each axis. Plotthe five points, and any additionalpoints you feel are necessary to dis-

347

cern the shape of the graph.c. Draw the horizontal asymptote with

a dashed line, and label it with itsequation.

d. Sketch the graph of the function.e. Use interval notation to describe both

the domain and range of the function.

5. f(x) = (2.5)x

6. f(x) = (0.1)x

7. f(x) = (0.75)x

8. f(x) = (1.1)x

9. f(x) = 3x + 1

10. f(x) = 4x − 5

11. f(x) = 2x − 3

12. f(x) = 5x + 2

In Exercises 13-20, the graph of an ex-ponential function of the form f(x) =bx + c is shown. The dashed red line isa horizontal asymptote. Determine therange of the function. Express your an-swer in interval notation.

13.

x

y

5

5

14.

x

y

5

5

15.

x

y

5

5

16.

x

y

5

5

17.

x

y

5

5

348

18.

x

y

5

5

19.

x

y

5

5

20.

x

y

5

5

In Exercises 21-32, compute f(p) at thegiven value p.

21. f(x) = (1/3)x; p = −4

22. f(x) = (3/4)x; p = 1

23. f(x) = 5x; p = 5

24. f(x) = (1/3)x; p = 4

25. f(x) = 4x; p = −4

26. f(x) = 5x; p = −3

27. f(x) = (5/2)x; p = −3

28. f(x) = 9x; p = 3

29. f(x) = 5x; p = −4

30. f(x) = 9x; p = 0

31. f(x) = (6/5)x; p = −4

32. f(x) = (3/5)x; p = 0

In Exercises 33-40, use your calcula-tor to evaluate the function at the givenvalue p. Round your answer to the near-est hundredth.

33. f(x) = 10x; p = −0.7.

34. f(x) = 10x; p = −1.60.

35. f(x) = (2/5)x; p = 3.67.

36. f(x) = 2x; p = −3/4.

37. f(x) = 10x; p = 2.07.

38. f(x) = 7x; p = 4/3.

39. f(x) = 10x; p = −1/5.

40. f(x) = (4/3)x; p = 1.15.

41. This exercise explores the propertythat exponential growth functions even-tually increase rapidly as x increases. Letf(x) = 1.05x. Use your graphing calcu-lator to graph f on the intervals

(a) [0, 10] and (b) [0, 100].For (a), use Ymin = 0 and Ymax = 10.For (b), use Ymin = 0 and Ymax = 100.Make accurate copies of the images inyour viewing window on your homeworkpaper. What do you observe when youcompare the two graphs?

349

Answers

1.

a) P (t) = 10 000(1.04)t

b) P (40) ≈ 48 101

c)

3.

a) P (t) = 15 000(0.95)t

b) P (50) ≈ 1 154

c)

5.

a) The y-intercept is (0, 1). Evaluatethe function at x = 1, 2,−1,−2 toobtain the points (1, 2.5), (2, 6.25),(−1, 0.4), (−2, 0.16) (other answers arepossible).

b) See the graph in part (d).

c) The horizontal asymptote is y = 0.See the graph in part (d).

d)

x3

y10

y=0

f(x)=(2.5)x

e) Domain = (−∞,∞), Range = (0,∞)

350

7.

a) The y-intercept is (0, 1). Evaluatethe function at x = 1, 2,−1,−2 toobtain the points (1, 0.75), (2, 0.56),(−1, 1.34), (−2, 1.78) (other answersare possible).

b) See the graph in part (d).

c) The horizontal asymptote is y = 0.See the graph in part (d).

d)

x5

y5

y=0

f(x)=(0.75)x

e) Domain = (−∞,∞), Range = (0,∞)

9.

a) The y-intercept is (0, 2). Evaluatethe function at x = 1, 2,−1,−2 to ob-tain the points (1, 4), (2, 10), (−1, 1.34),(−2, 1.11) (other answers are possi-ble).

b) See the graph in part (d).

c) The horizontal asymptote is y = 1.See the graph in part (d).

d)

x3

y20

y=1

f(x)=3x+1

e) Domain = (−∞,∞), Range = (1,∞)

11.

a) The y-intercept is (0,−2). Evalu-ate the function at x = 1, 2,−1,−2to obtain the points (1,−1), (2, 1),(−1,−2.5), (−2,−2.75) (other answersare possible).

b) See the graph in part (d).

c) The horizontal asymptote is y = −3.See the graph in part (d).

d)

x5

y5

y=−3

f(x)=2x−3

e) Domain = (−∞,∞), Range = (−3,∞)

13. (−1,∞)

15. (2,∞)

17. (2,∞)

351

19. (−2,∞)

21. 81

23. 3125

25.1

256

27.8

125

29.1

625

31.6251296

33. 0.20

35. 0.03

37. 117.49

39. 0.63

41.

a) The graph on the interval [0, 10] in-creases very slowly. In fact, the graphlooks almost linear.

b) The graph on the interval [0, 100] in-creases slowly at first, but then in-creases very rapidly on the second halfof the interval.

352