Slide 4.1- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 4.1- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES

Exponential Functions

Learn the definition of exponential function.

Learn to graph exponential functions.

Learn to solve exponential equations.

Learn to use transformations on exponential functions.

SECTION 4.1

1

2

3

4


EXPONENTIAL FUNCTION

A function f of the form

is called an exponential function with base a. The domain of the exponential function is (–∞, ∞).

f x ax , a 0 and a 1,


EXAMPLE 1 Evaluating Exponential Functions

a. Let f x 3x 2. Find f 4 .b. Let g x 210x. Find g 2 .

c. Let h x 1

9

x

. Find h 3

2

.

Solution

a. f 4 34 2 32 9

b. g 2 210 2 21

102 21

100 0.02

c. Let h 3

2

1

9

3

2 9 1

3

2 93

2 27


RULES OF EXPONENTS

Let a, b, x, and y be real numbers with a > 0 and b > 0. Then

ax ay axy ,

ax

ay ax y ,

ab x axbx ,

ax yaxy ,

a0 1,

a x 1

ax 1

a

x

.


EXAMPLE 2Graphing an Exponential Function with

Base a > 1

Graph the exponential function

Solution

Make a table of values.

f x 3x.

x –3 –2 –1 0 1 2 3

y = 3x 1/27 1/9 1/3 1 3 9 27

Plot the points and draw a smooth curve.



Base a > 1

Solution continued

This graph is typical for exponential functions when a > 1.



Base 0 < a < 1

Sketch the graph of

Solution

Make a table of values.

y 1

2

x

.

x –3 –2 –1 0 1 2 3

y = (1/2)x 8 4 2 1 1/2 1/4 1/8

Plot the points and draw a smooth curve.



Base 0 < a < 1

Solution continued

As x increases in the positive direction, y decreases towards 0.


PROPERTIES OF EXPONENTAIL FUNCTIONS

Let f (x) = ax, a > 0, a ≠ 1. Then

1. The domain of f (x) = ax is (–∞, ∞).

2. The range of f (x) = ax is (0, ∞); thus, the entire graph lies above the x-axis.

3. For a > 1,i. f is an increasing function; thus, the

graph is rising as we move from left to right.


ii. As x ∞, y = ax increases indefinitely and very rapidly.

4. For 0 < a < 1,i. f is a decreasing function; thus, the

graph is falling as we scan from left to right.

iii. As x –∞, the values of y = ax get closer and closer to 0.

ii. As x –∞, y = ax increases indefinitely and very rapidly.

iii. As x ∞, the values of y = ax get closer and closer to 0.


5. Each exponential function f is one-to-one. Thus,

ii. f has an inverse

6. The graph of f (x) = ax has no x-intercepts. In other words, the graph of f (x) = ax never crosses the x-axis. Put another way, there is no value of x that will cause f (x) = ax to equal 0.

ax1 ax2 ,i. if x1 = x2;


7. The graph of f (x) = ax has y-intercept 1. If we substitute x = 0 in the equation y = ax , we obtain y = a0 = 1, which yields 1 as the y-intercept.

8. The x-axis is a horizontal asymptote for every exponential function of the form f (x) = ax.


EXAMPLE 4Finding a Base Value for an Exponential Function

Find a if the graph of the exponential function f (x) = ax contains the point (2, 49). Solution

Write y = f (x) so that we have y = ax.

Solution set is {–7, 7}. The base for an exponential function must be positive, so a = 7 and the exponential function is f (x) = 7x.

49 a2

7 a

Since the point (2, 49) is on the graph, we have,


EXAMPLE 5 Solving an Exponential Equation

Solve for x: 52 x 1 25.

Solution

52 x 1 25

52 x 1 52

2x 1 2

2x 3

x 3

2

The solution set is3

2

.


EXAMPLE 6 Finding the First Coordinate, Given the Second

a. Let f x 3x 2.

so, 3x 2 1

273x 2 3 3

Find x so that f x 1

27,

b. Let g x 5x x 3 . Find x so that g x 1

25.

So the point 1,1

27

is on the graph of f.

2x 1 3

x 1

Solution

a. f x 3x 2 and f x 1

27


EXAMPLE 6 Finding the First Coordinate, Given the Second

Solution continued

b. g x 5x x 3 and g x 1

25

So there are two points

1,1

25

on the graph of g.

2,1

25

and

so, 5x x 3 5 2

x x 3 2

x2 3x 2

x2 3x 2 0

x 1 x 2 0

x 1 or x 2


TRANSFORMATIONS ON EXPONENTIAL FUNCTION f (x) = ax

Transformation Equation Effect on Equation

HorizontalShift

y = ax+b

= f (x + b)Shift the graph of y = ax, b units(i) left if b > 0.(ii) right if b < 0.

VerticalShift

y = ax + b = f (x) + b

Shift the graph of y = ax, b units(i) up if b > 0.(ii) down if b < 0.




Stretching or Compressing(Vertically)

y = cax

= c f (x)Multiply the y coordinates by c. The graph of y = ax is vertically(i) stretched if c > 1.(ii) compressed if

0 < c < 1.




Reflection y = –ax = – f (x) The graph of y = ax is reflected in the x-axis.

The graph of y = ax is reflected in the y-axis.

y = a–x = f (–x)


EXAMPLE 7 Sketching Graphs

Use transformations to sketch the graph of each function.

a. f x 3x 4

State the domain and range of each function and the horizontal asymptote of its graph.

b. f x 3x1

c. f x 3x d. f x 3x 2



Solution a.

Domain: (–∞, ∞)

Range: (–4, ∞)

Horizontal Asymptote: y = –4



Solution b.


Range: (0, ∞)

Horizontal Asymptote: y = 0



Solution c.


Range: (–∞, 0)




Solution d.


Range: (–∞, 2)



EXAMPLE 8 Comparing Exponential and Power Functions

Compare the graphs of f x x2

The graph of g(x) is higher than f (x) in the interval [0, 2). Thus, 2x > x2 for 0 < x < 2.

g x 2x

for x ≥ 0.and

Solution



The graph of f intersects the graph of g at x = 2.Thus,2x = x2 at x = 2; both = 4, (2, 4) is a point on each graph

Solution continued



The graph of f is higher than the graph of g in the interval (2, 4).Thus,x2 > 2x for 2 < x < 4.

Solution continued



The graph of f intersects the graph of g at x = 4.Thus,2x = x2 at x = 4; both = 16, (2, 16) is a point on each graph

Solution continued



The graph of g is higher than the graph of f for x > 4.Thus,2x > x2 for x > 4.

Solution continued


EXAMPLE 9 Bacterial Growth

A technician to the French microbiologist Louis Pasteur noticed that a certain culture of bacteria in milk doubles every hour. If the bacteria count B(t) is modeled by the equation

B t 20002t ,

a. the initial number of bacteria,b. the number of bacteria after 10 hours; andc. the time when the number of bacteria will be

32,000.

with t in hours, find


EXAMPLE 9 Bacterial Growth

B0 B 0 200020 20001 2000a. Initial size

b. B 10 2000210 2,048,000

32000 20002t

16 2t

c. Find t when B(t) = 32,000

24 2t

4 t4 hours after the starting time, the number of bacteria will be 32,000.

Solution

Slide 4.1- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

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