Copyright © Cengage Learning. All rights reserved. 9 Nonlinear Functions and Models.

Copyright © Cengage Learning. All rights reserved.

9 Nonlinear Functionsand Models

Copyright © Cengage Learning. All rights reserved.

9.2 Exponential Functions and Models

3

Exponential Functions and Models

The Laws of Exponents

If b and c are positive and x and y are any real numbers, then the following laws hold:

Law Quick Examples

4


Law Quick Examples

5


Exponential Function

An exponential function has the form

f (x) = Abx,

where A and b are constants with A ≠ 0 and b positive and not equal to 1. We call b the base of the exponential function.

Quick Example

f (x) = 2x

f (1) = 21 = 2

f (–3) = 2–3 =

f (0) = 20 = 1

Technology: A*b^x

A = 1, b = 2;Technology: 2^x

2^1

2^(–3)

2^0

6

Exponential Functions from the Numerical and Graphical Points of View

7


The following table shows values of f (x) = 3(2x) for some values of x (A = 3, b = 2):

Its graph is shown in Figure 10.

Notice that the y-intercept is A = 3 (obtained by setting x = 0).

Figure 10

8


In general:

In the graph of f (x) = Abx, A is the y-intercept, or the value of y when x = 0.

What about b? Notice from the table that the value of y is multiplied by b = 2 for every increase of 1 in x. If we decrease x by 1, the y coordinate gets divided by b = 2.

The value of y is multiplied by b for every one-unit increase of x.

9


On the graph, if we move one unit to the right from any

point on the curve, the y coordinate doubles. Thus, the

curve becomes dramatically steeper as the value of x

increases. This phenomenon is called exponential

growth.

10


Exponential Function Numerically and Graphically

For the exponential function f (x) = Abx :

Role of A

f (0) = A, so A is the y-intercept of the graph of f.

Role of b

If x increases by 1, f (x) is multiplied by b.

If x increases by 2, f (x) is multiplied by b2.

If x increases by x, f (x) is multiplied by bx.

...

If x increases by 1, y is multiplied by b.

11


Quick Example

Technology: 2^x; 2^(-x)

12


When x increases by 1, f2(x) is multiplied by . The

function f1(x) = 2x illustrates exponential growth, while

illustrates the opposite phenomenon:

exponential decay.

13

Example 1 – Recognizing Exponential Data Numerically and Graphically

Some of the values of two functions, f and g, are given in the following table:

One of these functions is linear, and the other is exponential. Which is which?

14

Example 1 – Solution

Remember that a linear function increases (or decreases)

by the same amount every time x increases by 1.

The values of f behave this way: Every time x increases by

1, the value of f (x) increases by 4. Therefore, f is a linear

function with a slope of 4. Because f (0) = 1, we see that

f (x) = 4x + 1

is a linear formula that fits the data.

15


On the other hand, every time x increases by 1, the value of g(x) is multiplied by 3.

Because g(0) = 2, we find that

g(x) = 2(3x)

is an exponential function fitting the data.

We can visualize the two functions f and g by plotting the data points (Figure 11).

cont’d

Figure 11

16


The data points for f (x) clearly lie along a straight line, whereas the points for g(x) lie along a curve.

The y coordinate of each point for g(x) is 3 times the y coordinate of the preceding point, demonstrating that the curve is an exponential one.

cont’d

17

Applications

18

Applications

Recall some terminology we mentioned earlier: A quantity y

experiences exponential growth if y = Abt with b 1. (Here

we use t for the independent variable, thinking of time.)

It experiences exponential decay if y = Abt with 0 b 1.

19

Example 3(a) – Exponential Growth and Decay

Compound Interest

If $2,000 is invested in a mutual fund with an annual yield of 12.6% and the earnings are reinvested each month, then the future value after t years is

which can be written as 2,000(1.010512)t, so A = 2,000 and b = 1.010512.

This is an example of exponential growth, because b 1.

20

Example 3(b) – Exponential Growth and Decay

Carbon DecayThe amount of carbon 14 remaining in a sample that originally contained A grams is approximately

C(t) = A(0.999879)t.

This is an instance of exponential decay, because b 1.

cont’d

21

The Number e and More Applications

22


Suppose we invest $1 in the bank for 1 year at 100% interest, compounded m times per year.

If m = 1, then 100% interest is added every year, and so our money doubles at the end of the year. In general, the accumulated capital at the end of the year is

(1+1/m)^m

23


Now, we are interested in what A becomes for large values of m. Below is an Excel sheet showing the quantity for larger and larger values of m.

24


Something interesting does seem to be happening! Thenumbers appear to be getting closer and closer to a specific value.

In mathematical terminology, we say that the numbers converge to a fixed number, 2.71828 . . . , called the limiting value of the quantities .

This number, called e, is one of the most important in mathematics.

The number e is irrational, just as the more familiar number is, so we cannot write down its exact numerical value. To 20 decimal places,

e = 2.71828182845904523536. . . .

25


We now say that, if $1 is invested for 1 year at 100% interest compounded continuously, the accumulated money at the end of that year will amount to $e = $2.72 (to the nearest cent).

The Number e and Continuous Compounding

The number e is the limiting value of the quantities as m gets larger and larger, and has the value2.71828182845904523536 . . .

If $P is invested at an annual interest rate r compounded continuously, the accumulated amount after t years is

A(t) = Per t.

26


Quick Example

If $100 is invested in an account that bears 15% interest compounded continuously, at the end of 10 years the investment will be worth

A(10) = 100e(0.15)(10) = $448.17.

27

Example 5 – Continuous Compounding

a. You invest $10,000 at Fastrack Savings & Loan, which pays 6% compounded continuously. Express the balance in your account as a function of the number of years t and calculate the amount of money you will have after 5 years.

b. Your friend has just invested $20,000 in Constant Growth Funds, whose stocks are continuously declining at a rate of 6% per year. How much will her investment be worth in 5 years?

28

Example 5 – Continuous Compounding

c. During which year will the value of your investment first exceed that of your friend?

29

Example 5(a) – Solution

We use the continuous growth formula with P = 10,000,

r = 0.06, and t variable, getting

A(t) = Per t = 10,000e0.06t.

In five years,

A(5) = 10,000e0.06(5)

= 10,000e0.3

≈ $13,498.59.

30

Example 5(b) – Solution

Because the investment is depreciating, we use a negative

value for r and take P = 20,000, r = –0.06, and t = 5, getting

A(t) = Per t = 20,000e–0.06t

A(5) = 20,000e–0.06(5)

= 20,000e–0.3

≈ $14,816.36.

cont’d

31

Example 5(c) – Solution

We can answer the question now using a graphing calculator, a spreadsheet, or the Function Evaluator and Grapher tool at the Website.

Just enter the exponential models of parts (a) and (b) and create tables to compute he values at the end of several years:

cont’d

32

Example 5(c) – Solution cont’d

33

Example 5(c) – Solution

From the table, we see that the value of your investment overtakes that of your friend after t = 5 (the end of year 5) and before t = 6 (the end of year 6).

Thus your investment first exceeds that of your friend sometime during year 6.

cont’d

34


Exponential Functions: Alternative Form

We can write any exponential function in the following alternative form:

f (x) = Aerx

where A and r are constants. If r is positive, f modelsexponential growth; if r is negative, f models exponential decay.

35


Quick Examples

1. f (x) = 100e0.15x

2. f (t) = Ae–0.000 121 01t

Exponential growth A = 100, r = 0.15

Exponential decay of carbon 14;r = –0.000 121 01

36

Exponential Regression

37

Example 6 – Exponential Regression: Health Expenditures

The following table shows annual expenditure on health in the United States from 1980 through 2009 (t = 0 represents 1980).

a. Find the exponential regression model

C(t) = Abt

for the annual expenditure.

b. Use the regression model to estimate the expenditure in 2002 (t = 22; the actual expenditure was approximately $1,640 billion).

38


a. We use technology to obtain the exponential regression curve (see Figure 13):

C(t) ≈ 296(1.08)t

b. Using the model

C(t) ≈ 296(1.08)t we find that

C(22) ≈ 296(1.08)22

≈ $1,609 billion

which is close to the actual number of around $1,640 billion.

Coefficients rounded

Figure 13

Copyright © Cengage Learning. All rights reserved. 9 Nonlinear Functions and Models.

Documents

Transcript of Copyright © Cengage Learning. All rights reserved. 9 Nonlinear Functions and Models.