Transcript Lecture 2 Economic Models, Functions, Logs, Exponents, e

Lecture 2
Economic Models, Functions,
Logs, Exponents, e
Variables, Constant, Parameters

Variables: magnitude can change
 Price, profit, revenue…
 Represented by symbols
 Can be ‘frozen’ by setting value
Try to setup models to obtain solutions to
variables
 Endogenous variable: determined by model
 Exogenous variable: determined outside
model

Variables, Constant, Parameters

Constants, do not change, but can be joined
with variables.
 Called coefficients
 Ex: 1000q,

1000 is constant, quantity is variable
When constants are not set?
 Ex: βq, where β stands for the coefficient and q
for quantity
 Now, β can change! A ‘variable’ constant is called
a parameter.
Equation vs. Identity

An identity, is a definition, like profit, revenue,
and cost
  R  C
where π is profit, R is revenue, C is cost
 Behavioral equations

Specify how variables interact
R  pq
C  wl  rk
 Conditional equations

Optimization and Equilibrium conditions are examples
MC  MR
Qd  Qs
Real Numbers

Rational numbers: (‘ratio’)
 Whole numbers:
+ Integers
 Fractions

Irrational numbers
 Can’t be expressed as a fraction
 Nonrepeating, nonterminating decimals

Ex:
Real Numbers
combine rational and irrational
2
 Non-imaginary, i=

1
Set Properties and Set Notation
Definition: A set is any collection of objects
specified in such a way that we can determine
whether a given object is or is not in the
collection.
 Notation:
eA
means “e is an element of A”, or “e belongs to
set A”.
The notation
eA
means “e is not an element of A”.

Null Set
Example. What are the real number solutions
of the equation
x2 + 1 = 0?
There is no answer to this… you can write this as
{ }, [], or .

Set Builder Notation

Sometimes it is convenient to represent sets using
set builder notation. For example, instead of
representing the set A (letters in the alphabet) by
the roster method, we can use
A = {x | x is a letter of the English alphabet}
which means the same as A = {a , b, c, d, e, …, z}
In statistics… {x|a} would be read x, given a.
Here, we will use the vertical line in set notation
 Example.
{x | x2 = 9} = {3, -3}
This is read as “the set of all x such that the square
of x equals 9.” This set consists of the two numbers
3 and -3.
Union of Sets
The union of two sets A and B is the set of all
elements formed by combining all the
elements of set A and all the elements of set B
into one set. It is written A  B.
A
B
In the Venn diagram on the
left, the union of A and B is
the entire region shaded.
Intersection of Sets
The intersection of two sets A and B is the set
of all elements that are common to both A
and B. It is written A  B.
A
B
In the Venn diagram on the
left, the intersection of A
and B is the shaded region.
The Complement of a Set
The complement of a set A is defined as the set
of elements that are contained in U, the
universal set, but not contained in set A. The
symbolism and notation for the complement of
set A are A '  {x U x  A}
In the Venn diagram on the left, the
rectangle represents the universe.
A is the shaded area outside the
set A.
Application
A marketing survey of 1,000 commuters found that 600
answered listen to the news, 500 listen to music, and 300
listen to both. Let N = set of commuters in the sample who
listen to news and M = set of commuters in the sample who
listen to music. Find the number of commuters in the set
N M '
The number of elements in a set A is denoted by n(A), so in
this case we are looking for
n( N  M ')
Solution
The study is based on 1000 commuters, so n(U)=1000.
The number of elements in the four sections in the Venn diagram
need to add up to 1000.
The red part represents the commuters who listen to both news
and music. It has 300 elements.
The set N (news listeners) consists of
a green part and a red part. N has 600
elements, the red part has 300, so the
green part must also be 300.
Continue in this fashion.
Solution
(continued)
U
200 people listen to
neither news nor music
N
300 listen
to news
but not
music.
M
200
listen to
music
but not
news
300 listen to both
music and news
N M '
is the green part,
which contains 300
commuters.
Functions


The previous graph is the graph of a function. The
idea of a function is this: a correspondence between
two sets D and R such that to each element of the
first set, D, there corresponds one and only one
element of the second set, R.
The first set is called the domain, and the set of
corresponding elements in the second set is called
the range.
For example, the cost of a pizza (C) is related to the
size of the pizza. A 10 inch diameter pizza costs
$9.00, while a 16 inch diameter pizza costs $12.00.
Function Definition

You can visualize a function by the following diagram which
shows a correspondence between two sets: D, the domain of
the function, gives the diameter of pizzas, and R, the range
of the function gives the cost of the pizza.
10
12
16
9.00
10.00
12.00
domain D
or x
range R
or f(x)
Functions Specified by Equations

Consider the equation
y  x 2
2
-2
Input:
x = -2
Process:
square (–2),
then subtract 2
(-2,2) is an
ordered pair of
the function.
2
Output:
result is 2
Vertical Line Test for a Function
If you have the graph of an equation, there is an easy
way to determine if it is the graph of an function. It is
called the vertical line test which states that:
An equation defines a function if each vertical line in
the coordinate system passes through at most one
point on the graph of the equation. If any vertical line
passes through two or more points on the graph of
an equation, then the equation does not define a
function.
Vertical Line Test for a Function
(continued)
This graph is not the graph of a
function because you can draw a
vertical line which crosses it
twice.
This is the graph of a
function because any vertical
line crosses only once.
Function Notation

The following notation is used to describe functions.
The variable y will now be called f (x).

This is read as “ f of x” and simply means the y
coordinate of the function corresponding to a given x
2
y  x 2
value.
Our previous equation
f ( x)  x  2
2
can now be expressed as
Function Evaluation

Consider our function

What does f (-3) mean?
f ( x)  x 2  2
Function Evaluation


2
f
(
x
)

x
2
Consider our function
What does f (-3) mean?
Replace x with the value –3 and evaluate the
expression
2
f (3)  (3)  2

The result is 11 . This means that the point (-3,11)
is on the graph of the function.
Some Examples

1.
f (a)  3(a)  2
f (6  h)  3(6  h)  2  18  3h  2
 16  3h

KEEP THIS ‘h’ example in mind for derivatives
Domain of a Function

Consider
f ( x)  3x  2
f (0)  ?
f (0)  3(0)  2  2

which is not a real number.
Question: for what values of x is the function
defined?
Domain of a Function

Answer:
is defined only when the radicand (3x-2) is equal to or
greater than zero. This implies that
x
2
3
Domain of a Function
(continued)

Therefore, the domain of our function is the set of
real numbers that are greater than or equal to 2/3.

Example: Find the domain of the function
1
f ( x) 
x4
2
Domain of a Function
(continued)

Therefore, the domain of our function is the set of real
numbers that are greater than or equal to 2/3.

Example: Find the domain of the function
1
f ( x) 
x4
2

Answer:
 x x  8 , [8, )
Domain of a Function:
Another Example

Find the domain of
1
f ( x) 
3x  5
Domain of a Function:
Another Example

Find the domain of
1
f ( x) 
3x  5

In this case, the function is defined for all
values of x except where the denominator of
the fraction is zero. This means all real
numbers x except 5/3.
Mathematical Modeling
The price-demand function for a company is
given by
p( x)  1000  5 x,
0  x  100
where x (variable) represents the number of
items and p(x) represents the price of the item.
1000 and -5 are constants. Determine the
revenue function and find the revenue
generated if 50 items are sold.
Solution
Revenue = Price ∙ Quantity, so
R(x)= p(x) ∙ x = (1000 – 5x) ∙ x
When 50 items are sold, x = 50, so we will
evaluate the revenue function at x = 50:
R(50)  (1000  5(50)) 50  37,500
0  x  100
The domain of the function has already been
specified. What is the ‘range’ over this domain?
Solution
Example 2.4 (5) from Chiang
Production Problem

Either coal (C) or gas (G) can be used to
produce steel. If Pc=100 and Pg=500, draw an
isocost curve to limit expenditures to $10,000
Production Problem

If the price of gas declines by 20%? What
happens to the budget line? What if the price
of coal rises by 25%? Expenditures rise 50%?

Production Problem
All together?
Break-Even and Profit-Loss Analysis




Any manufacturing company has costs C and
revenues R.
The company will have a loss if R < C, will break even
if R = C, and will have a profit if R > C.
Costs include fixed costs such as plant overhead,
etc. and variable costs, which are dependent on the
number of items produced.
C = a + bx
(x is the number of items produced)
a and b are parameters
Break-Even and Profit-Loss Analysis
(continued)



Price-demand functions, usually determined by
financial departments, play an important role in
profit-loss analysis.
p = m – nx
(x is the number of items than can be sold at $p per
item.) [Note: m and n are PARAMETERS]
The revenue function is
R = (number of items sold) ∙ (price per item)
= xp = x(m - nx)
The profit function is
P = R - C = x(m - nx) - (a + bx)
Example of Profit-Loss Analysis
A company manufactures notebook computers. Its
marketing research department has determined that the
data is modeled by the price-demand function p(x) =
2,000 - 60x, when 1 < x < 25, (x is in thousands).
What is the company’s revenue function and what is its
domain?
Answer to Revenue Problem
Since Revenue = Price ∙ Quantity,
R( x)  x  p( x)  x  (2000  60 x)  2000 x  60 x 2
The domain of this function is the same as the domain
of the price-demand function, which is 1 ≤ x ≤ 25 (in
thousands.)
First, is the demand function,
and the plot of it. Below, you
can see that x*p(x) is our
revenue function…
But, firms don’t maximize
revenue, they maximize
profits, so we have to consider
our cost function
Profit Problem
The financial department for the company in the preceding
problem has established the following cost function for
producing and selling x thousand notebook computers:
C(x) = 4,000 + 500x
(x is in thousand dollars).
Write a profit function for producing and selling x thousand
notebook computers, and indicate the domain of this function.
Answer to Profit Problem
Since Profit = Revenue - Cost, and our revenue function
from the preceding problem was R(x) = 2000x - 60x2,
P(x) = R(x) - C(x) = 2000x - 60x2 - (4000 + 500x)
= -60x2 + 1500x – 4000.
The domain of this function is the same as the domain of
the original price-demand function, 1< x < 25 (in
thousands.)
Now, to get this profit function
First, is the cost function, and
the plot of it. Below, you can
see that the revenue function
plotted with our cost function
gives us a visual
representation of profit…
Where these two lines cross,
profit is ZERO, since R=C. If
But, firms don’t maximize
revenue, they maximize
profits, so we have to consider
our cost function. We want to
maximize the difference
between the two functions.
This, is R-C, or our
profit function
Below, I have solved
for this using
Mathematica’s
Maximize function
Types of functions

Polynomial functions
x
f ( x)  x
2
y
-3
-2
-1
0
1
2
3
Solution of Problem 1
x
y
-3
9
-2
4
-1
0
1
0
1
1
2
4
3
9
Polynomials or Quadratics can shift

Now, sketch the related
graph given by the
equation below and
explain, in words, how
it is related to the first
function you graphed.
f ( x  2)  ( x  2)
x
y
-3
-2
-1
2
0
1
2
3
Solution of Problem 2
x
y
-3 25
-2 16
-1 9
0 4
1 1
2 0
3 1
Cubic functions
x
y
30
-3
-27
-2
-8
-1
-1
0
0
1
1
2
8
3
27
20
10
0
-4
-3
-2
-1
0
-10
-20
-30
1
2
3
4
Problem 4

Now, graph the
following related
function:
f ( x)  5  x  5
3
x
y
-3
-2
-1
0
1
2
3
Solution to Problem 4
x
y
40
-3
-22
30
-2
-3
20
-1
4
10
0
5
0
-4
-3
-2
-1
0
1
6
-10
2
13
-20
3
32
-30
1
2
3
4
Problem 5
h( x )  x

Graph

What is the domain of this function?
Solution to Problem 5

The domain is all non-negative real
numbers. Here is the graph:
6
4
2
0
-5
0
-2
-4
-6
5
10
15
20
25
30
Problem 6
 h( x )   x

Graph:

Explain, in words, how it compares to
problem 5.
Solution to Problem 6
6
4
2
0
-5
0
5
10
15
20
25
30
-2
-4
-6
Conclusion:
The graph of the function –f (x) is a reflection of the graph of
f (x) across the x axis. That is, if the graphs of f (x) and –f (x)
were folded along the x axis, the two graphs would coincide.
Absolute Value Function


Graph the absolute value function. Be sure to
choose x values that are both positive and
negative, as well as zero.
Graph
a( x  1)  2 | x  1 | 2
Absolute Value Function
(continued)
Notice the
symmetry of
the graph.
10
8
6
a ( x)  | x |
4
2
0
-10
-5
0
-2
-4
5
10
Absolute Value Function
(continued)
a( x  1)  2  | x  1 | 2
Shifted left one
unit and down
two units.
8
7
6
5
4
3
2
1
0
-10
-5
-1 0
-2
-3
5
10
The purple line
uses the absolute
value function
(Abs) and the
second takes the
square root of the
square… to get
the absolute
value.
The bottom shifts
the axes.
Summary of
Graph Transformations




Vertical Translation: y = f (x) + k
 k > 0 Shift graph of y = f (x) up k units.
 k < 0 Shift graph of y = f (x) down |k| units.
Horizontal Translation: y = f (x+h)
 h > 0 Shift graph of y = f (x) left h units.
 h < 0 Shift graph of y = f (x) right |h| units.
Reflection: y = -f (x)
Reflect the graph of y = f (x) in the x axis.
Vertical Stretch and Shrink: y = Af (x)
 A > 1: Stretch graph of y = f (x) vertically by multiplying
each ordinate value by A.
 0<A<1: Shrink graph of y = f (x) vertically by multiplying
each ordinate value by A.
Piecewise-Defined Functions



Earlier we noted that the absolute value of a real
number x can be defined as
  x if x  0
| x|
 x if x  0
Notice that this function is defined by different
rules for different parts of its domain. Functions
whose definitions involve more than one rule are
called piecewise-defined functions.
Graphing one of these functions involves graphing
each rule over the appropriate portion of the
domain.
Example of a
Piecewise-Defined Function
Graph the function
2  2 x if x  2
f ( x)  
 x  2 if x  2
Example of a
Piecewise-Defined Function
Graph the function
2  2 x if x  2
f ( x)  
 x  2 if x  2
Notice that the point
(2,0) is included but
the point (2, -2) is
not.
Rational Functions
x 1
y 2
x  2x  4
Unit Elastic Curve (Rect. Hyperbola)
a
a
y  , or rather q  so that pq  a where a is a constant
x
p
No matter what the price is, the quantity changes such that
expenditures are always equal to a constant (a). Thus, the
percentage change in quantity is always equal (and
opposite) the percentage change in price!
To sketch a quadratic
You need to find the places it crosses the xaxis, and you need to find where it turns
around. When you don’t have Mathematica,
or know how to take a derivative this can be
tricky.
 Trust me, derivatives are easier… and make
more sense.

Vertex Form of the Quadratic Function
It is convenient to convert the general
form of a quadratic equation
f ( x)  ax  bx  c
2
to what is known as the vertex form:
f ( x)  a( x  h)  k
2
Note, if inside parentheses is x-2, h=2.
Completing the Square to Find
the Vertex of a Quadratic Function
The example below illustrates the
procedure: f ( x)  3x2  6 x 1
Consider
Complete the square to find the
vertex.
Completing the Square to Find
the Vertex of a Quadratic Function
The example below illustrates the
procedure: f ( x)  3x2  6 x 1
Consider
Complete the square to find the
vertex.
Solution:
 Factor the coefficient of x2 out of the first two terms:
f (x) = -3(x2 - 2x) -1
Completing the square (continued)
 Add 1 to complete the square inside the parentheses. Because
of the -3 outside the parentheses, we have actually added -3,
so we must add +3 to the outside. The trick is adding and
subtracting a number that makes parentheses sum of squares
f (x) = -3(x2 - 2x +1) -1+3
f (x) = -3(x - 1)2 + 2
The vertex is (1, 2)
 The quadratic function opens down since the
coefficient of the x2 term is -3, which is
negative.

There has to be an easier way…
There is… which will become obvious when we get to start
taking derivatives and knowing the quadratic function. This
takes a quadratic function which has many values of f(x), and
finds the values where f(x) EQUALS zero (a quadratic
equation). These are also called the ROOTS of the quadratic
function.
 b  b 2  4ac
where ax 2  bx  c  0
2a
If both roots are real… split the difference and find the values of x
and f(x)
Intercepts of a Quadratic Function

Find the x and y intercepts of
f ( x)  3x  6 x  1
2
Intercepts of a Quadratic Function

Find the x and y intercepts of
f ( x)  3x  6 x  1
2

x intercepts: Set f (x) = 0:

Use the quadratic formula:
x = b  b  4ac
2
2a
0  3x  6 x  1
2
6  62  4(3)(1) 6  24
=

 0.184,1.816
2(3)
6
Intercepts of a Quadratic Function
(continued)

y intercept: Let x = 0. If x = 0, then y = -1,
so (0, -1) is the y intercept.
f ( x)  3x  6 x  1
2
f (0)  3(0)  6(0)  1
2
Generalization
For

f ( x)  a( x  h)2  k
If a  0, then the graph of f is a parabola.
 If a > 0, the graph opens upward.
 If a < 0, the graph opens downward. Vertex is (h , k)




Axis of symmetry: x = h
f (h) = k is the minimum if a > 0, otherwise the
maximum
Domain = set of all real numbers  y y  k 
Range:  y y  k  if a < 0. If a > 0, the range is
Solving Quadratic Inequalities
Solve the quadratic inequality -3x2 + 6x -1 > 0
Solving Quadratic Inequalities
Solve the quadratic inequality -3x2 + 6x -1 > 0
Answer:
This inequality holds for those values of x for which
the graph of f (x) is at or above the x axis. This
happens for x between the two x intercepts, including
the intercepts. Thus, the solution set for the
quadratic inequality is
0.184 < x < 1.816.
Application of Quadratic Functions
A Macon, Georgia, peach orchard farmer
now has 20 trees per acre. Each tree
produces, on the average, 300 peaches.
For each additional tree that the farmer
plants, the number of peaches per tree is
reduced by 10. How many more trees
should the farmer plant to achieve the
maximum yield of peaches? What is the
maximum yield?
Solution:
Yield = (number of peaches per tree) x
(number of trees)
 Yield = 300 x 20 = 6000 (currently)
 Plant one more tree: Yield = ( 300 – 1(10)) * (
20 + 1) =
290 x 21 = 6090 peaches.
 Plant two more trees:
 Yield = ( 300 – 2(10)* ( 20 + 2) = 280 x 22 =
6160
Solution (continued)
Let x represent the number of additional
2

10
x
 100 x  6000
trees.
Then Yield =( 300 – 10x) (20 + x)=
 To find the maximum yield, note that the Y (x)
function is a quadratic function opening
downward. Hence, the vertex of the function
will be the maximum value of the yield. Graph
is below, with the y value in thousands.

Solution
(continued)

Complete the square to find the vertex of the
parabola:
2

10(
x
 10 x  25)  6000  250
 Y (x) =

We have to add 250 on the outside since we
multiplied –10 by 25 = -250.
Solution
(continued)


Y (x) = 10( x  5)2  6250
Thus, the vertex of the quadratic function is (5 ,
6250) . So, the farmer should plant 5 additional trees
and obtain a yield of 6250 peaches. We know this
yield is the maximum of the quadratic function since
the the value of a is -10. The function opens
downward, so the vertex must be the maximum.
Break-Even Analysis
The financial department of a company that produces digital
cameras has the revenue and cost functions for x million
cameras are as follows:
R(x) = x(94.8 - 5x)
C(x) = 156 + 19.7x. Both have domain 1 < x < 15
Break-even points are the production levels at which
R(x) = C(x). Find the break-even points algebraically to the
nearest thousand cameras.
Solution to Break-Even Problem
Set R(x) equal to C(x):
x(94.8 - 5x) = 156 + 19.7x
-5x2 + 75.1x - 156 = 0
75.1  75.12  4(6)(156)
x
2(5)
x = 2.490 or 12.530
The company
breaks even at x =
2.490 and 12.530
million cameras.
Solution to
Break-Even
Problem
(continued)
If we graph the cost and
revenue functions on a
graphing utility, we obtain the
following graphs, showing the
two intersection points:
We can also plot the profit
function to see where the
profit is maximized…
Quadratic Regression
A visual inspection of the plot of a data set might indicate
that a parabola would be a better model of the data than a
straight line. In that case, rather than using linear
regression to fit a linear model to the data, we would use
quadratic regression on a graphing calculator to find the
function of the form y = ax2 + bx + c that best fits the
data.
Example of Quadratic Regression
An automobile tire manufacturer collected the data in
the table relating tire pressure x (in pounds per square
inch) and mileage (in thousands of miles.)
x
Mileage
28
45
30
52
32
55
34
51
36
47
Polynomial Functions
A polynomial function is a function that can be written in
the form
an x  an 1 x
n
n 1
   a1 x  a0
for n a nonnegative integer, called the degree of the
polynomial. The domain of a polynomial function is the
set of all real numbers.
A polynomial of degree 0 is a constant. A polynomial of
degree 1 is a linear function. A polynomial of degree 2 is a
quadratic function.
Shapes of Polynomials
A polynomial is called odd if it only contains odd
powers of x
 It is called even if it only contains even powers of x
Let’s look at the shapes of some even and odd
polynomials
Look for some of the following properties:

 Symmetry
 Number of x axis intercepts
 Number of local maxima/minima
 What happens as x goes to +∞ or -∞?
Graph of Odd Polynomial
Example 1
f ( x)  x  5 x  4 x
5
3
15.0
10.0
5.0
0.0
-3
-2
-1
0
-5.0
-10.0
-15.0
1
2
3
Graph of Odd Polynomial
Example 2
f ( x)  x  27 x
3
60.0
40.0
20.0
0.0
-8
-6
-4
-2
0
-20.0
-40.0
-60.0
2
4
6
8
Graph of Even Polynomial
Example 1
4
2
f ( x)  x  6 x
30.0
25.0
20.0
15.0
10.0
5.0
0.0
-4
-3
-2
-1
-5.0
-10.0
-15.0
0
1
2
3
4
Graph of Even Polynomial
Example 2
f ( x)  3 x  1
2
30.0
25.0
20.0
15.0
10.0
5.0
0.0
-4
-3
-2
-1
0
-5.0
1
2
3
4
Observations
Odd Polynomials

For an odd polynomial,
 the graph is symmetric about the origin
 the graphs starts negative, ends positive, or vice
versa, depending on whether the leading
coefficient is positive or negative
 either way, a polynomial of degree n crosses the x
axis at least once, at most n times.
Observations
Even Polynomials

For an even polynomial,
 the graph is symmetric about the y axis
 the graphs starts negative, ends negative, or starts
and ends positive, depending on whether the
leading coefficient is positive or negative
 either way, a polynomial of degree n crosses the x
axis at most n times. It may or may not cross at all.
Characteristics of polynomials





Graphs of polynomials are continuous. One can sketch the
graph without lifting up the pencil.
Graphs of polynomials have no sharp corners.
Graphs of polynomials usually have turning points, which is a
point that separates an increasing portion of the graph from
a decreasing portion.
A polynomial of degree n can have at most n linear factors.
Therefore, the graph of a polynomial function of positive
degree n can intersect the x axis at most n times.
A polynomial of degree n may intersect the x axis fewer than
n times.
Rational Functions


A rational function is a quotient of two polynomials, P(x) and
Q(x), for all x such that Q(x) is not equal to zero.
Example: Let P(x) = x + 5 and Q(x) = x – 2, then
R(x) =
x5
x2
is a rational function whose domain is all real values of x with
the exception of 2 (Why?)
Vertical Asymptotes of
Rational Functions
x values at which the function is undefined represent vertical
asymptotes to the graph of the function. A vertical asymptote is a line
of the form x = k which the graph of the function approaches but does
not cross. In the figure below, which is the graph of
the line x = 2 is a
vertical asymptote.
x5
x2
Horizontal Asymptotes of
Rational Functions
A horizontal asymptote of a function is a line
of the form y = k which the graph of the
function approaches as x approaches 
For example, in the
graph of
x5
x2
the line y = 1 is a
horizontal asymptote.
Generalizations about
Asymptotes of Rational Functions



The number of vertical asymptotes of a rational
function
f (x) = n(x)/d(x) is at most equal to the degree of
d(x).
A rational function has at most one horizontal
asymptote.
The graph of a rational function approaches the
horizontal asymptote (when one exists) both as
x increases and decreases without bound.
Exponential functions

The equation

for b>0 defines the exponential function with
base b . The domain is the set of all real
numbers, while the range is the set of all
positive real numbers.
f ( x)  b
x
Riddle



Here is a problem related to exponential functions:
Suppose you received a penny on the first day of
December, two pennies on the second day of
December, four pennies on the third day, eight
pennies on the fourth day and so on. How many
pennies would you receive on December 31 if this
pattern continues?
Would you rather take this amount of money or
receive a lump sum payment of $10,000,000?
Solution
Complete the table:
Day
1
2
No. pennies
1
2
3
4
5
4
8
16
6
7
32
64
2^1
2^2
2^3
...
Solution
(continued)




Now, if this pattern continued, how many pennies
would you have on Dec. 31?
Your answer should be 230 (two raised to the
thirtieth power). The exponent on two is one less
than the day of the month. See the preceding slide.
What is 230?
1,073,741,824 pennies!!! Move the decimal point
two places to the left to find the amount in dollars.
You should get: $10,737,418.24
Solution
(continued)


The obvious answer to the question is to take the
number of pennies on December 31 and not a lump
sum payment of $10,000,000
(although I would not mind having either amount!)
This example shows how an exponential function
grows extremely rapidly. In this case, the
exponential function
f ( x)  2
is used to model this problem.
x
Graph of f ( x)  2

x
Use a table to graph the exponential function
above. Note: x is a real number and can be
replaced with numbers such as 2 as well as
other irrational numbers. We will use integer
values for x in the table:
Table of values
x
-4
-3
-2
y
2-4 = 1/24 = 1/16
2-3 = 1/8
2-2 = 1/4
-1
2-1 = 1/2
0
1
2
20 = 1
21 = 2
22 = 4
Graph of y =
f ( x)  2
x
Characteristics of the graphs of
x
f ( x)  b where b> 1
All graphs will approach the x axis as x
becomes unbounded and negative
 All graphs will pass through (0,1) (y
intercept)
 There are no x intercepts.
 Domain is all real numbers
 Range is all positive real numbers.
 The graph is always increasing on its domain.
 All graphs are continuous curves.

Graphs of







f ( x)  b
x
if 0 < b < 1
All graphs will approach the x axis as x gets
large.
All graphs will pass through (0,1) (y intercept)
There are no x intercepts.
Domain is all real numbers
Range is all positive real numbers.
The graph is always decreasing on its domain.
All graphs are continuous curves.
1
f ( x)  2  x
2
x
Graph of
Using a table of values, you will obtain the following graph.
The graphs of f ( x)  b x and
f ( x)  b  xwill be reflections
of each other about the y-axis, in general.
12
10
8
graph of y = 2^(-x)
6
approaches the positive x-axis as x gets large
4
2
passes through (0,1)
0
-4
-2
0
2
4
Base e Exponential Functions



Of all the possible bases b we can use for the
exponential function y = bx, probably the most useful
one is the exponential function with base e.
The base e is an irrational number, and, like π,
cannot be represented exactly by any finite decimal
fraction.
However, e can be approximated as closely as we
like by evaluating the expression as x gets infinitely
large
 1
1  
 x
x
Exponential Function With Base e
 1
1  
 x
x
The table to the left
illustrates what happens
to the expression
x
1
2
10
2.59374246
100
2.704813829
1000
10000
1000000
 1
1  
 x
x
as x gets increasingly
2.716923932 larger. As we can see from
2.718145927 the table, the values
2.718280469 approach a number
whose approximation is
2.718
Leonhard Euler
1707-1783


Leonhard Euler first demonstrated that
1

1



x

x
will approach a fixed constant we now call “e”.
So much of our mathematical notation is due to Euler that it will come as
no surprise to find that the notation e for this number is due to him. The
claim which has sometimes been made, however, that Euler used the
letter e because it was the first letter of his name is ridiculous. It is
probably not even the case that the e comes from "exponential", but it
may have just be the next vowel after "a" and Euler was already using the
notation "a" in his work. Whatever the reason, the notation e made its
first appearance in a letter Euler wrote to Goldbach in 1731.
http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/e.html#s19
Graph of f ( x)  e

x
Graph is similar to the
graphs of
f ( x)  2 x
and
f ( x )  3x
It has the same
characteristics as these
graphs
Relative Growth Rates




Functions of the form y = cekt, where c and k are constants and
the independent variable t represents time, are often used to
model population growth and radioactive decay.
Note that if t = 0, then y = c. So, the constant c represents the
initial population (or initial amount.)
The constant k is called the relative growth rate. If the relative
growth rate is k = 0.02, then at any time t, the population is
growing at a rate of 0.02y persons (2% of the population) per
year.
We say that population is growing continuously at relative
growth rate k to mean that the population y is given by the
model y = cekt.
Growth and Decay Applications:
Atmospheric Pressure

The atmospheric pressure p
decreases with increasing
height. The pressure is
related to the number of
kilometers h above the sea
level by the formula:
P(h)  760e
0.145 h
 Find the pressure at sea
level (h = 0)
 Find the pressure at a
height of 7 kilometers.
Solution
 Find the pressure at sea level (h = 0)
P(0)  760e0  760
 Find the pressure at a height of
7 kilometers
P(7)  760e0.145(7)  275.43
Depreciation of a Machine
A machine is initially worth V0
dollars but loses 10% of its
value each year. Its value after
t years is given by the formula
V (t )  V0 (0.9 )
t
Find the value after 8 years of
a machine whose initial value
is $30,000.
Depreciation of a Machine
A machine is initially worth V0
dollars but loses 10% of its
value each year. Its value after
t years is given by the formula
V (t )  V0 (0.9 )
t
Find the value after 8 years of
a machine whose initial value
is $30,000.

Solution:
V (t )  V0 (0.9t )
V (8)  30000(0.98 )  $12,914
Compound Interest

The compound interest formula is
 r
A  P 1  
 n

nt
Here, A is the future value of the investment,
P is the initial amount (principal), r is the
annual interest rate as a decimal, n
represents the number of compounding
periods per year, and t is the number of
years
Compound Interest Problem

Find the amount to which $1500 will grow if
deposited in a bank at 5.75% interest
compounded quarterly for 5 years.
Compound Interest Problem

Find the amount to which $1500 will grow if
deposited in a bank at 5.75% interest compounded
quarterly for 5 years.

Solution: Use the compound interest formula:
 r
A  P 1  
 n
nt
Substitute P = 1500, r = 0.0575, n = 4 and t = 5 to
obtain
(4)(5)
 0.0575 
A  1500  1 

4


=$1995.55
Logarithmic Functions

In this section, another type of function will
be studied called the logarithmic function.
There is a close connection between a
logarithmic function and an exponential
function. We will see that the logarithmic
function and exponential functions are
inverse functions. We will study the concept
of inverse functions as a prerequisite for our
study of logarithmic function.
One to One Functions
We wish to define an inverse of a function.
Before we do so, it is necessary to discuss the
topic of one to one functions.
First of all, only certain functions are one to
one.
Definition: A function is said to be one to one if
distinct inputs of a function correspond to
distinct outputs. That is, if
x1  x2 , f ( x1 )  f ( x2 )
Graph of One to One Function
This is the graph of a one to
one function. Notice that if we
choose two different x values,
the corresponding y values are
different. Here, we see that if
x = 0, then y = 1, and if x = 1,
then y is about 2.8.
Now, choose any other pair of
x values. Do you see that the
corresponding y values will
always be different?
5
4
3
2
1
0
-1
0
1
2
Horizontal Line Test

Recall that for an equation to be a function, its graph
must pass the vertical line test. That is, a vertical line
that sweeps across the graph of a function from left
to right will intersect the graph only once at each x
value.

There is a similar geometric test to determine if a
function is one to one. It is called the horizontal line
test. Any horizontal line drawn through the graph of
a one to one function will cross the graph only once.
If a horizontal line crosses a graph more than once,
then the function that is graphed is not one to one.
Which Functions Are One to One?
40
12
30
10
8
20
6
10
4
0
-4
-2
0
2
2
4
-10
-20
-30
0
-4
-2
0
2
4
Definition of Inverse Function
Given a one to one function, the inverse
function is found by interchanging the x and y
values of the original function. That is to say, if
an ordered pair (a,b) belongs to the original
function then the ordered pair (b,a) belongs
to the inverse function.
 Note: If a function is not one to one (fails the
horizontal line test) then the inverse of such
a function does not exist.

Logarithmic Functions

The logarithmic function with base two is defined to
be the inverse of the one to one exponential
function
9
y2

8
x
7
6
graph of y = 2^(x)
5
Notice that the exponential
function
x
4
approaches the negative x-axis as x gets
3
large
y2
2
passes through (0,1)
1
0
-4
is one to one and therefore has
an inverse.
-2
0
2
4
Inverse of an Exponential Function



Start with y  2
Now, interchange x and y coordinates:
x
x2
y
There are no algebraic techniques that can be used to solve
for y, so we simply call this function y the logarithmic function
with base 2. The definition of this new function is:
log 2 x  y
if and only if
x  2y
Graph, Domain, Range
of Logarithmic Functions



The domain of the logarithmic function y = log2x is
the same as the range of the exponential function y
= 2x. Why?
The range of the logarithmic function is the same as
the domain of the exponential function (Again,
why?)
Another fact: If one graphs any one to one function
and its inverse on the same grid, the two graphs will
always be symmetric with respect to the line y = x.
Logarithmic-Exponential
Conversions
Study the examples below. You should be able to convert a
logarithmic into an exponential expression and vice versa.
1. log 4 (16)  x  4 x  16  x  2
2.
1
1
3
log 3 ( )  log 3 ( 3 )  log 3 (3 )  3
27
3
3.
125  5
4.
3
 log5 125  3
1
2
1
81  9  81  9  log81  9  
2
Solving Equations
Using the definition of a logarithm, you can
solve equations involving logarithms.
Examples:
log b (1000)  3  b3  1000  b3  103  b  10
log6  x   5  6  x  7776  x
5
In each of the above, we converted from log form to
exponential form and solved the resulting equation.
Properties of Logarithms
These are the properties of logarithms. M and N are
positive real numbers, b not equal to 1, and p and x are
real numbers.
(For 4, we need x > 0).
5. log b MN  log b M  log b N
1.log b (1)  0
2.log b (b)  1
3.log b b x  x
4.b
log b x
x
M
6. log b
 log b M  log b N
N
7. log b M p  p log b M
8. log b M  log b N iff M  N
Solving Logarithmic Equations
log 4 ( x  6)  log 4 ( x  6)  3 
1.
Solve for x:
2.
Product rule
3.
Special product
4.
Definition of log
5.
2
100

x

x can be +10 only
Why?
10  x 
6.
log 4 ( x  6)( x  6)  3 
log 4  x 2  36   3 
43  x 2  36 
64  x  36 
2
x  10
Another Example
1. Solve:
Another Example
1. Solve:
2. Quotient rule
3. Simplify
(divide out common factor π)
4. rewrite
5 definition of logarithm
6. Property of exponentials
Common Logs and Natural Logs

Common log
log x  log10 x
If no base is indicated,
the logarithm is
assumed to be base 10.

Natural log
ln( x)  log e x
e  2.7181828
Solving a Logarithmic Equation
Solve for x. Obtain the exact
solution of this equation in terms
of e (2.71828…)
ln (x + 1) – ln x = 1
Solving a Logarithmic Equation
Solve for x. Obtain the exact
solution of this equation in terms
of e (2.71828…)
Quotient property of logs
Definition of (natural log)
Multiply both sides by x
Collect x terms on left side
Factor out common factor
Solve for x
ln (x + 1) – ln x = 1
x 1
ln
1
x
1
e
e
x 1
x
ex = x + 1
ex - x = 1
x(e - 1) = 1
1
x
e 1
Application
How long will it take money to double
if compounded monthly at 4 %
interest?
Application
How long will it take money to double
if compounded monthly at 4 %
interest?
1. Compound interest formula
2. Replace A by 2P (double the
amount)
3. Substitute values for r and m
4. Divide both sides by P
5. Take ln of both sides
6. Property of logarithms
7. Solve for t and evaluate expression
Solution:
mt
r

A  P 1  
 m
12 t
 0.04 
2 P  P 1 

12 

2  (1.003333...)12t
ln 2  ln  (1.003333...)12t 
ln 2  12t ln(1.00333...)
ln 2
 t  t  17.36
12 ln(1.00333...)