§ 9.1

Exponential Functions
Consider the function f ( x )

2 x
Does this function look different from other functions we have worked with before?
If you said, “yes” – you are right. It is different. The difference is that
x is in the exponent – not the base.
Functions like this one are called “exponential functions.”
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 9.1

Exponential Functions
You will need a calculator for evaluating exponential expressions.
Any scientific calculator will work for this purpose.
Many real-life situations, including population growth, growth of epidemics, radioactive decay, and other changes that involve rapid increase or decrease, can be described using exponential functions.
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 9.1

Exponential Functions
The exponential function f with base b is defined by f
 b x or y
 b x where b is a positive constant other than 1 ( b > 0 and ) and x is any real number.
Page 640
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 9.1

Exponential Functions
The exponential function f with base b is defined by f
 b x or y
 b x where b is a positive constant other than 1 ( b > 0 and ) and x is any real number.
Page 640
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 9.1

Exponential Functions
The exponential function f with base b is defined by f
 b x or y
 b x where b is a positive constant other than 1 ( b > 0 and ) and x is any real number.
x

1
1 f

2 x g(x)

10 x h ( x )

3 x

1 j ( x )

2 not
F
 x 2 G( x )

1 x H ( x )

(

1 ) x J ( x )
 x x
•
Variable is the base and not the exponent.
•
The base cannot be 1.
•
The base of an exponential function must be positive.
•
Variable is both the base and the exponent
Page 640
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 9.1

Exponential Functions
  b T x f
 
 x f
 
 b x consists of all positive real numbers.
f
 
 x
    y -intercept is 1.
3) If b
  x function. The greater the value of b , the steeper the increase.
4) If 0 < b
  x < 1, has a graph that goes down to the right and is a decreasing function. The smaller the value of b , the steeper the decrease.
  x 5) The graph of approaches, but does not touch, the x -axis. The x -axis, or y = 0, is a horizontal asymptote.
See graph on page 644.
Page 644
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 9.1

Exponential Functions
EXAMPLE
 
3 x and
 
3
 x Graph in the same rectangular coordinate system. How is the graph of g related to the graph of f ?
SOLUTION
We begin by setting up a table showing some of the coordinates for f and g , selecting integers from -2 to 2 for x . x
-2
-1
0
1
2 f f f f f f
 

3 x
 
 


3

3

1
2


1
1
/
/
3
9
 
 
 



3
0
3
1
3
2



3
1
9 g
  g

 
3


3

   x
3
2 g
 

3

  
3
1

9

3 g

3

0 
3
0 
1 g g
 


3

1
3

2


1 /
1 /
3
1
3
2


1 /
1 /
3
9
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 9.1

Exponential Functions
CONTINUED
We plot the points for each function and connect them with a smooth curve. Because of the scale on the y -axis, some points on each function are not shown. The graph of g is a reflection of the graph of f across the y -axis.
g
 

3

-2.00, 9.00
x
-2.00, 0.11
7
6
5
9
8
4
-1.00, 3.00
3
2
-1
1
-1.00, 0.33
0
0
0.00, 1.00
-2 1
2.00, 9.00
1.00, 3.00
f
 

3 x
2
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 9.1

Exponential Functions
Pages 645-646
Not on tests.
An irrational number, symbolized by the letter e , appears as the base in many applied exponential functions. This irrational number is approximately equal to 2.72 More accurately, e

2 .
71828 
The number e is called the natural base . The function f
 
 e x is called the natural exponential function .
Remember that e is not a variable. It’s just an irrational number.
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 9.1

Exponential Functions
EXAMPLES from homework
Do 2-10 (even) using calculator.
Do 12
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 9.1

Exponential Functions
EXAMPLES from homework
Do 12-16
•
Opposite causes graph to appear in quadrants 3 and 4 since f(x) will always be negative as in 14 and 16
•
-1 in number 13 causes graph to cross the y axis at (0, 0) rather than (0, 1)
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 9.1

Exponential Functions
EXAMPLES from homework
Do 18
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 9.1

Exponential Functions
EXAMPLES from homework
20
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 9.1

Exponential Functions
EXAMPLES from homework
26
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 9.1

Exponential Functions in Application
EXAMPLE
In college, we study large volumes of information – information that, unfortunately, we do not often retain for very long. The function f

80 e

0 .
5 x 
20 describes the percentage of information, f ( x ), that a particular person remembers x weeks after learning the information.
Find the percentage of information that is remembered after 4 weeks.
Page 646-647
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 9.1

Exponential Functions in Application
CONTINUED
SOLUTION
Because we want to know the percentage of information retained after 4 weeks, we replace x with 4.
f f f f f

80 e

0 .
5 x 
20
 
 
 
 




80 e

0 .
5

4
80 e

2
80
30 .

83

0 .
14

20

20
20
This is the given function.
Replace x with 4.
Multiply -0.5 and 4.
Evaluate the exponent.
Finish simplifying.
Therefore, four weeks after learning information, a certain person retains about 30.83% of that information.
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 9.1

Exponential Functions
After t years, the balance, A , in an account with principal
P and annual interest rate r (in decimal form) is given by the following formulas:
1) For n compoundings per year:
A

P
 

1 r n nt
2) For continuous compounding:
A

Pe rt .
Pages 647-648
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 9.1

Exponential Functions in Application
EXAMPLE
Find the accumulated value of an investment of $5000 for 10 years at an interest rate of 6.5% if the money is (a) compounded semiannually, (b) compounded monthly, (c) compounded continuously.
SOLUTION
We are trying to determine what the accumulated value of an investment is. Therefore, we are looking for A . We first determine the values for t , P , and r . Since the investment will accumulate for 10 years, t = 10. Since the initial investment is
$5000, P = 5000. And since the interest rate is 6.5%, r = 0.065.
Now we are ready to use the appropriate formulas to answer the questions.
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 9.1

Exponential Functions in Application
CONTINUED
(a) Since the investment is being compounded semiannually, n = 2. We now solve for A .
nt
A

P
 

1 r n
2

10
A

5000

0 .
065
2
A

A

5000

1

0 .
0325

20
5000

1 .
0325

20
A

5000

1 .
90
This is the equation to use.
Replace P with 5000, r with 0.065, n with 2 and t with 10.
Divide and multiply.
Add.
Evaluate the exponent.
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 9.1

Exponential Functions in Application
CONTINUED
A

9479 .
19 Multiply.
Therefore, the accumulated value of the investment is $9,479.19.
(b) Since the investment is being compounded monthly, n = 12.
We now solve for A .
nt
A

P
 

1 r n
12

10
0 .
065
A

5000
 12
This is the equation to use.
Replace n
P with 5000, with 12 and t r with 10.
with 0.065,
A

A

5000

1

0 .
0054

120
5000

1 .
0054

120
Divide and multiply.
Add.
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 9.1

Exponential Functions in Application
CONTINUED
A

5000

1 .
91
A

9541 .
92
Evaluate the exponent.
Multiply.
Therefore, the accumulated value of the investment is $9,541.92.
(c) Since the investment is being compounded continuously, there is no n value. We now solve for A .
A

Pe rt This is the equation to use.
A

5000 e
0 .
065

10
A

5000 e
0 .
65
A

5000

1 .
92
Replace P with 5000, r with 0.065, and t with 10.
Multiply.
Evaluate the exponent.
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 9.1

Exponential Functions in Application
CONTINUED
A

9600 Multiply.
Therefore, the accumulated value of the investment is $9,600.
You may wish to remember the compound interest formula. Almost everyone needs to either borrow or invest – so that’s a formula that is applicable for many!
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 9.1

DONE

Exponential Functions
EXAMPLE
The 1986 explosion at the Chernobyl nuclear power plant in the former Soviet Union sent about 1000 kilograms of radioactive
    x
30 describes the amount, f ( x ), in kilograms, of cesium-137 remaining in Chernobyl x years after 1986. If even 100 kilograms of cesium-137 remain in Chernobyl’s atmosphere, the area is considered unsafe for human habitation. Find f (80) and determine if Chernobyl will be safe for human habitation by
2066.
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 9.1

Exponential Functions
CONTINUED
SOLUTION
In finding f (80), we are finding how many kilograms of cesium-
137 are in Chernobyl 80 years after 1986, or in 2066.
f f f f f
 

1000
  x
30
 

1000
  80
30
 

1000
  8
3
 
 


1000
157

0 .
157

This is the given function.
Replace x with 80.
Divide.
Evaluate the exponent.
Multiply.
Chernobyl will not be safe for human habitation by 2066 with approximately 157 kilograms of cesium-137.
Blitzer, Intermediate Algebra, 5e – Slide #26 Section 9.1