Transcript 5.2 Exponential Functions

Chapter 5: Exponential and Logarithmic
Functions
5.1
5.2
5.3
5.4
5.5
Inverse Functions
Exponential Functions
Logarithms and Their Properties
Logarithmic Functions
Exponential and Logarithmic Equations and
Inequalities
5.6 Further Applications and Modeling with
Exponential and Logarithmic Functions
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Slide 5-2
5.2 Exponential Functions
If a > 0, a  1, then
f(x) = ax
is the exponential function with base a.
• Additional Properties of Exponents
–
–
–
–
ax is a unique real number.
ab = ac if and only if b = c.
If a > 1 and m < n, then am < an.
If 0 < a < 1 and m < n, then am > an.
Copyright © 2007 Pearson Education, Inc.
e.g. 4  4
2
3
e.g.  12    12 
2
3
Slide 5-3
5.2 Graphs of Exponential Functions
Example Graph f ( x)  2 x. Determine the domain
and range of f.
Solution
There is no x-intercept. Any number to the zero
power is 1, so the y-intercept is (0,1). The domain
is (–,), and the range is (0,).
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Slide 5-4
5.2 Graph of f (x) = ax, a > 1
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Slide 5-5
5.2 Graph of f (x) = ax, 0 < a < 1
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Slide 5-6
5.2 Comparing Graphs
Example Explain why the graph of g ( x )   12 x
is a reflection across the y-axis of the graph of
f ( x)  2 x.
Analytic Solution
Show that g(x) = f(–x).
x
1

g ( x)   
 2
1 x
 2 
 2 x  f ( x)
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Slide 5-7
5.2 Comparing Graphs
Graphical Solution
The graph below indicates that g(x) is a reflection
across the y-axis of f(x).
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Slide 5-8
5.2 Using Translations to Graph an
Exponential Function
x
Example Explain how the graph of y  2  3
x
is obtained from the graph of y  2 .
Solution
y  2 x
y  2 x  3
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reflect 2 x across the x - axis
 2 x is translated up 3 units
Slide 5-9
5.2 Example using Graphs to Evaluate
Exponential Expressions
Example
Use a graph to evaluate .5 2.
Solution
With x   2  1.414214 , we find
x
that y  2.6651441 from the graph of y = .5 .
Copyright © 2007 Pearson Education, Inc.
Slide 5-10
5.2 Exponential Equations (Type I)
Example
Solve 25 x  125.
Solution
25 x  125
5 
2 x
5
3
5 5
2x
3
2x  3
3
x
2
Write with the same base.
a 
m n
a
mn
Set exponents equal and solve.
Verify : 25 2  125. The solution set is 32.
3
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Slide 5-11
5.2 Using a Graph to Solve Exponential
Inequalities
Example
Solve the inequality 1.5
x 1

  0.
27 x
8
Solution
Using the graph below, the graph lies
above the x-axis for values of x less than .5.
The solution set for y > 0 is
(–,.5).
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Slide 5-12
5.2 The Natural Number e
•
•
Named after Swiss mathematician Leonhard
Euler
1 x
e involves the expression 1  x  :
as x  , 1  1x   e  2.718281828
x
•
•
e is an irrational number
Since e is an important
base, calculators are
programmed to find
powers of e.
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Slide 5-13
5.2 Compound Interest
•
Recall simple earned interest I  Prt , where
– P is the principal (or initial investment),
– r is the annual interest rate, and
– t is the number of years.
•
If A is the final balance at the end of each year,
then
1st year : A  P  Pr  P(1  r )
2nd year : A  P(1  r )  P(1  r )r  P(1  r )(1  r )  P(1  r ) 2

nth year : A  P(1  r ) n
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Slide 5-14
5.2 Compound Interest Formula
Suppose that a principal of P dollars is invested at an annual
interest rate r (in percent), compounded n times per year.
Then, the amount A accumulated after t years is given by the
formula
nt
r


A  P 1   .
 n
Example Suppose that $1000 is invested at an annual
rate of 8%, compounded quarterly. Find the total amount in
the account after 10 years if no withdrawals are made.
Solution
.08 

A  1000 1 

4



4 10
 2208.039664
The final balance is $2208.04.
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Slide 5-15
5.2 Continuous Compounding Formula
If P dollars is deposited at a rate of interest r compounded
continuously for t years, the final amount A in dollars on
deposit is
rt
A  Pe .
Example Suppose $5000 is deposited in an account
paying 8% compounded continuously for 5 years. Find the
total amount on deposit at the end of 5 years.
Solution
A  5000 e.08( 5)  5000 e.4  7459 .12
The final balance is $7459.12.
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Slide 5-16
5.2 Modeling the Risk of Alzheimer’s
Disease
Example The chances of dying of influenza or
pneumonia increase exponentially after age 55
according to the function defined by
f ( x)  r (1.082) x ,
where r is the risk (in decimal form) at age 55 and x
is the number of years greater than 55. What is the
risk at age 75?
20
Solution x = 75 – 55 = 20, so f ( x)  r (1.082)  4.84r
Thus, the risk is almost fives times as great at age 75
as at age 55.
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Slide 5-17