Chapter 6

Exponential and Logarithmic Functions and Applications

Section 6.1

Section 6.1 Exponential Functions and Their Graphs

• Exponential Functions: Definition, Characteristics, and Graphs

• Transformations of Graphs of Exponential Functions

• Introduction to Exponential Growth and Decay

• Average Rate of Change

• Different Time Periods

• Increasing or Decreasing by a Constant Percent

• Comparing Linear and Exponential Growth

• Applications

Exponential Function

An exponential function is one of the form f(x) = bx , where b and x are real numbers. Constant b is the base of the function, b > 0, and b 1; variable x is the exponent.

The domain of f(x) = bx is the set of all real numbers.

The following are examples of exponential functions:

f(x) = 5x + 3 f(x) = –3x f(x) = 8–x

The following are not exponential functions:

f(x) = x5 + 3 f(x) = (–3)x f(x) = 0x f(x) = 1x

Note: f(x)= –3x is not the same as f(x) = (–3)x. The latter one is not exponential because the base is negative, b = –3. Observe that the first one, f(x) = –3x, is obtained by multiplying f(x)= 3x by –1, and it is exponential.

x6f(x)

Graphing Exponential Functions

For each function, complete the table and graph the function:

f(x) = 3x f(x) = (1/3)x

x y = 3x

–2

–1

0

1

2

x y = (1/3)x

–2

–1

0

1

2

x y = 3x

–2 3–2 = 1/9

–1 3–1 = 1/3

0 30 = 1

1 31 = 3

2 32 = 9

x y = (1/3)x

–2 (1/3) –2 = 9

–1 (1/3) –1 = 3

0 (1/3)0 = 1

1 (1/3)1 = 1/3

2 (1/3)2 = 1/9

Comparing Graphs

Both graphs have the same domain and range:Domain = (–∞, ∞); Range = (0, ∞)

The y-intercept is (0, 1).

The x-axis is a horizontal asymptote.

Both are one-to-one functions.

f(x) = 3x, with b > 1, is increasing.

f(x) = (1/3)x, with 0 < b < 1, is decreasing.

Graphs of Exponential Functions

For the exponential function f(x) = bx, where b and x are real numbers, b > 0, and b 1:

The domain is the set of all real numbers, or (–∞, ∞), and the range is the set of all positive real numbers, or (0, ∞).

The y-intercept is (0, 1).

The x-axis (that is, the line y = 0) is a horizontal asymptote.

The function is one-to-one.

(continued on the next slide)

Graphs of Exponential Functions (contd.)

The graph has one of the following shapes:

If b > 1, the function is increasing; the larger the value of b, the steeper the increase.

If 0 < b < 1, the function is decreasing; the smaller the value of b, the faster the decrease.

Transformations of Graphs of Exponential Functions (I)

Apply transformations to the graph of f(x) = 3x to construct the graph of each function. Determine the horizontal asymptote and the y-intercept for each function.

a. f(x) = 3x + 2 b. f(x) = 3x – 2

a. Vertical shift, up 2 units Horizontal asymptote: y = 2 y-intercept from graph: (0, 3) Verifying y-intercept:

f(0) = 30 + 2 = 1 + 2 = 3

b. Vertical shift, down 2 units Horizontal asymptote: y = –2

y-intercept from graph: (0, –1) Verifying y-intercept: f(0) = 30 – 2 = 1 – 2 = –1

 

 

Transformations of Graphs of Exponential Functions (III)

Apply transformations to the graph of f(x) = 3x to construct the graph of each function. Determine the horizontal asymptote and the y-intercept for each function.

a. f(x) = –3x b. f(x) = 3–x

a. Reflection about the x-axis Horizontal asymptote: y = 0

y-intercept from graph: (0, –1) Verifying y-intercept:

f(0) = –(30) = –1

b. Reflection about the y-axis Horizontal asymptote: y = 0 y-intercept from graph: (0, 1) Verifying y-intercept: f(0) = 3–(0) = 1

 

 

Effect of Multiplying f(x) = bx by a Constant c

If the exponential function of the form f(x) = bx is multiplied by a constant c, where |c| > 1, the graph will be stretched by a factor equivalent to the constant.

If the exponential function of the form f(x) = bx is multiplied by a constant c, where 0 < |c| < 1, the graph will be compressed by a factor equivalent to the constant.

The y-intercept of f(x) = cbx is (0, c).

 

 

 

Graph f(x) = 4(3x) and compare it to the graph of f(x) = 3x.

The graph of f(x) = 3x has been vertically stretched by a factor of 4.

Both graphs have the same horizontal asymptote: y = 0

The y-intercept of f(x) = 4(3x) is (0, 4); the y-coordinate corresponds to the value of the factor c = 4.

Verifying y-intercept: f(0) = 4(30) = 4(1) = 4

 

 

Graph f(x) = (1/4)(3x) and compare it to the graph of f(x) = 3x.

The graph of f(x) = 3x has been vertically compressed by a factor of 1/4.

Both graphs have the same horizontal asymptote: y = 0

The y-intercept of f(x) = (1/4)(3x) is (0, 1/4); the y-coordinate corresponds to the value of the factor c = 1/4.

Verifying y-intercept: f(0) = (1/4)(30) = (1/4)(1) = 1/4

 

 

Without graphing nor performing calculations, determine ifthe function f(x) = 8(1.7)x is increasing or decreasing, and find its vertical intercept.

The base of the exponential function is b = 1.7

Since b > 1, this function is increasing.

The y-intercept of an exponential function of the form f(x) = cbx is (0, c).

Since c = 8, the vertical intercept is (0, 8).

 

 

 

Find the exponential function whose graph is shown below.

We are looking for an increasing exponential function f(x) = cbx.

Since the y-intercept is (0, 1), c = 1.

The point (2, 7.84) lies on its graph, thus we know that f(2) = 7.84, and we look for a base b such that b2 = 7.84.

Since the base cannot be negative, b = 2.8.

The exponential function is f(x) = 2.8x.

8.284.7b

84.7b2

Find the exponential function whose graph is shown below.

We are looking for a decreasing exponential function f(x) = cbx.

Since the y-intercept is (0, 3), c = 3.

The point (3, 1.536) lies on its graph, thus we know that f(3) = 1.536, and we look for a base b such that 3b3 = 1.536.

The exponential function is f(x) = 3(0.8)x.

8.0512.0b

512.0b

536.1b3

3

3

3

Basic Exponential Growth and Decay

The basic exponential growth and decay function isgiven by: P(t) = P0bt

where:

t = time

P0 = P(0) = the initial amount, or value of P at time 0; P > 0

b = growth factor if b > 1; decay factor if 0 < b < 1

 

 

 

The following table displays values for an exponential function. Determine whether the function represents exponential growth or decay. Find the growth or decay factor; complete the table.  

As the t-values increase, the values of P(t) increase very fast. This is an example of an exponential growth model. From the table, we can see that P(0) = 2. So, we have P(t) = 2bt.

To find the value of the growth factor, b, we will choose any given point - other than the initial value - and replace t and P(t) accordingly, then solve for b.

(continued on the next slide)

 

 

 

t 0 1 2 3 4 5

P(t) 2 128 8162

(Contd.) 

Using the coordinates (2, 128), we can find the value of b in P(t) = 2bt. 2b2 = 128

b2 = 64

b = 8

The growth factor is b = 8, and the function is P(t) = 2(8)t.

The complete table is shown next.

 

 

 

t 0 1 2 3 4 5

P(t) 2 128 8,162

t 0 1 2 3 4 5

P(t) 2 16 128 1024 8,162

65,536

As of 2014, Facebook was considered the world’s largest social network. The number of Facebook active users in millions through 2010 can be modeled by the function P(t) = 1.53(2.82)t, where t is the number of years after 2004. Sources: www.facebook.com; www.crunchbase.com, www.bizma.com

a. Identify the value of P0 and the growth or decay factor.

P(t) = 1.53(2.82)t . Therefore, P0 = 1.53 and b = 2.82.

Since the base is greater than 1, b = 2.82 is the growth factor.

b. Without graphing, determine whether the function is increasing or decreasing. Explain your decision.

The function is increasing because b > 1.

(continued on the next slide)

(Contd.)As of 2014, Facebook was considered the world’s largest social network. The number of Facebook active users in millions through 2010 can be modeled by the function P(t) = 1.53(2.82)t, where t is the number of years after 2004. Sources: www.facebook.com; www.crunchbase.com, www.bizma.com

c. Graph the function with your calculator. Label the axes and show your viewing window. Identify and interpret the vertical intercept. Active users

Year [0, 10, 1] by [0, 550, 50]

The vertical intercept in P(t) = 1.53(2.82)t is (0,1.53), which approximates the number of active users (in millions) in 2004.

(continued on the next slide)

 

 

 

(Contd.)

d. Approximate the number of Facebook active users in 2008. Find the answer graphically, numerically, and algebraically. Round your answer to the nearest whole number

In 2008, there were approximately 97 million Facebook active users, as demonstrated next.

Graphically Numerically

Algebraically P(4) = 1.53(2.82)4 = 96.76 97

 

 

 

Average Rate of Change

Recall that if (x1, f(x1)) and (x2, f(x2)) represent two distinct points on the graph of a function f, the average rate of change of f between x1 and x2 is given by

Using the data from the previous problem, find the average rate of change in the number of Facebook active users between 2006 and 2008. Round to the nearest whole number.

For 2006, x = 2. So, P(2) = 1.53(2.82)2 = 12.17

For 2008, x = 4. So, P(4) = 1.53(2.82)4 = 96.76

The average rate of change in the number of Facebook active users between 2006-2008 was 42 million. 

 

 

 

12

12

xx)x(f)x(f

xy

423.4224

17.1276.96xy

Different Time Periods

If an exponential problem states that the growth or decay factor b occurs at a fixed time period, like every year, month, week, day, hour, etc., we can use P(t) = P0bt for time t.

If the problem refers to a different time period, such as every 3 days, every 5 months, every 2 years etc., then we need to adjust the time exponent accordingly.

Hint: Think of “every n amount of time” as t/n. For example, every 3 days would be reflected as t/3, every 5 months, t/5, etc.

Increasing or Decreasing by a Constant Percent

For an exponential growth or decay model P(t) = P0bt:

If b > 1, the growth rate satisfies b = 1 + r, where r is the percent increase.

Example: If a quantity grows by 2%, the growth rate is 0.02, therefore, b = 1 + 0.02 = 1.02.

If 0 < b < 1, the decay rate satisfies b = 1 – r, where r is the percent decrease.

Example: If a quantity decays by 2%, the decay rate is 0.02, therefore,b = 1 – 0.02 = 0.98.

Suppose that a particular bacteria culture initially contained 500 bacteria and they increased by 25% every 3 weeks. a. Write an exponential function N(t) that models this problem for t weeks.

N(t) = 500(1.25)t/3

b. Determine the number of bacteria after 6 weeks. Round your answer to the nearest whole number.

N(6) = 500(1.25)6/3 = 500(1.25)2 781 bacteria

c. Determine the number of bacteria after 5 months. Round to the nearest whole number. (Assume 4 weeks per month.)

For 5 months, t = 20. N(20) = 500(1.25)20/3 2213 bacteria

Caution: Remember to enclose the exponent “20/3” within a set of parentheses when performing calculations with the calculator.

Comparing Linear and Exponential Growth

Compare the linear function f(x) = 2 + 3x and the exponential function g(x) = 2(3)x for 0 x 5.

The output values for the exponential function grow much faster.

Increasing the x-values by one in the linear function has the effect of adding 3 more units to the value of f(x), whereas increasing the x-values by one in the exponential function has the effect of multiplying g(x) by the growth factor 3 each time.

The average rate of change for a linear function is constant, while the average rate of change of the exponential function is not.

x f(x) = 2 + 3x g(x) = 2(3)x

0 2 + 3(0) = 2 2(3)0 = 2

1 2 + 3(1) = 5 2(3)1 = 6

2 2 + 3(2) = 8 2(3)2 = 18

3 2 + 3(3) = 11 2(3)3 = 54

4 2 + 3(4) = 14 2(3)4 =162

A medical clinic has identified 2 cases of a new virus. Each subsequent day, a new patient arrives with the same virus, and four days later, the clinic has treated a total of 6 patientswho have contracted the virus.

a. Suppose that the number of patients with this virus grows linearly. Write a function L(x) that models this problem, where x represents days.

L(x) = 2 + x

b. Use your linear function to estimate how many patients will contract the virus in 12 days.

L(12) = 2 + 12 = 14 patients

(continued on the next slide)

(Contd.)c. Suppose that the number of patients with this virus grows exponentially. Write a function E(x) that models this problem, where x represents days.

The initial number of patients was 2. So, E(0) = 2, and we haveE(x) = 2bx.

During the 4th day, there were 6 cases, thus we can use the point (4, 6) to find the growth factor, b.

6 = 2b4

3 = b4

(3)(1/4) = b

1.3161= b

Therefore, our function is E(x) = 2(1.361)x.

(continued on the next slide)

(Contd.)d. Use your exponential function to estimate how many patients will contract the virus after 12 days.

E(12) = 2(1.361)12 = 54 patients e. Graph both functions with your calculator. Label the axes and show your viewing window. Compare both growths.

Virus patients 

Days [0, 30, 5] by [0, 100, 20]

Both graphs are increasing and they confirm that the exponential growth overruns the linear one.

Using your textbook, practice the problems assigned by your instructor to review the concepts from Section 6.1.