slide 4.1- 1 copyright © 2007 pearson education, inc. publishing as pearson addison-wesley
TRANSCRIPT
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
Slide 4.1- 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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,
Slide 4.1- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 4.1- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
.
Slide 4.1- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 4.1- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 2Graphing an Exponential Function with
Base a > 1
Solution continued
This graph is typical for exponential functions when a > 1.
Slide 4.1- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 3Graphing an Exponential Function with
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.
Slide 4.1- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 3Graphing an Exponential Function with
Base 0 < a < 1
Solution continued
As x increases in the positive direction, y decreases towards 0.
Slide 4.1- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 4.1- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 4.1- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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;
Slide 4.1- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 4.1- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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,
Slide 4.1- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
.
Slide 4.1- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 4.1- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 4.1- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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.
Slide 4.1- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
TRANSFORMATIONS ON EXPONENTIAL FUNCTION f (x) = ax
Transformation Equation Effect on Equation
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.
Slide 4.1- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
TRANSFORMATIONS ON EXPONENTIAL FUNCTION f (x) = ax
Transformation Equation Effect on Equation
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)
Slide 4.1- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 4.1- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 7 Sketching Graphs
Solution a.
Domain: (–∞, ∞)
Range: (–4, ∞)
Horizontal Asymptote: y = –4
Slide 4.1- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 7 Sketching Graphs
Solution b.
Domain: (–∞, ∞)
Range: (0, ∞)
Horizontal Asymptote: y = 0
Slide 4.1- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 7 Sketching Graphs
Solution c.
Domain: (–∞, ∞)
Range: (–∞, 0)
Horizontal Asymptote: y = 0
Slide 4.1- 25 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 7 Sketching Graphs
Solution d.
Domain: (–∞, ∞)
Range: (–∞, 2)
Horizontal Asymptote: y = 2
Slide 4.1- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 4.1- 27 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 8 Comparing Exponential and Power Functions
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
Slide 4.1- 28 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 8 Comparing Exponential and Power Functions
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
Slide 4.1- 29 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 8 Comparing Exponential and Power Functions
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
Slide 4.1- 30 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
EXAMPLE 8 Comparing Exponential and Power Functions
The graph of g is higher than the graph of f for x > 4.Thus,2x > x2 for x > 4.
Solution continued
Slide 4.1- 31 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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
Slide 4.1- 32 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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