SPEA Math Camp - Oncourse

SPEA Math Camp
Course Materials
Professor Henry Wakhungu
August 2012

Table of Contents
Day One: Basic Algebra, Geometry, and Miscellaneous Topics .................................................................... 4
Day One Class Notes ................................................................................................................................. 4
Day One In-Class Exercises ...................................................................................................................... 24
Day One In-Class Exercises Solutions ...................................................................................................... 28
Day One Homework ................................................................................................................................ 34
Day One Homework Solutions ................................................................................................................ 36
Day Two: Solving and Manipulating Equations ........................................................................................... 38
Day Two Class Notes ............................................................................................................................... 38
Day Two In-Class Exercises...................................................................................................................... 56
Day Two In-Class Exercises Solutions ...................................................................................................... 58
Day Two Homework ................................................................................................................................ 65
Day Two Homework Solutions ................................................................................................................ 66
Day Three: Functions and their Graphs ...................................................................................................... 70
Logarithmic and Exponential Functions ...................................................................................................... 70
Day Three Class Notes ............................................................................................................................. 70
Day Three In-Class Exercises ................................................................................................................... 99
Day Three In-Class Exercises Solutions ................................................................................................. 102
Day Three Homework ........................................................................................................................... 115
Day Three Homework Solutions ........................................................................................................... 117
Day Four: Derivatives, Optimization, and Integration .............................................................................. 126
Day Four Class Notes............................................................................................................................. 126
Day Four In-Class Exercises ................................................................................................................... 138
Day Four In-Class Exercises Solutions ................................................................................................... 139
Day Four Homework ............................................................................................................................. 141
Day Four Homework Solutions ............................................................................................................. 143
Day Five: Word Problems and Applications .............................................................................................. 148
Day Five Class Notes ............................................................................................................................. 148
Day Five In-Class Exercises .................................................................................................................... 157
Day Five In-Class Exercises Solutions .................................................................................................... 159
Day Five Homework .............................................................................................................................. 167
Day Five Homework Solutions .............................................................................................................. 170
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Supplemental Problems ........................................................................................................................ 184
Solutions to Supplemental Problems .................................................................................................... 202
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MATH CAMP
School of Public and Environmental Affairs
Day One: Basic Algebra, Geometry, and Miscellaneous Topics
Common Mathematical Symbols
Day One Class Notes
= Equals
Approximately equal to (used especially for decimal representations of
irrational numbers)
Not equal to
> Greater than
< Less than
Greater than or equal to
Less than or equal to
Plus or minus
Minus or plus
Therefore, implies
If and only if
Set of real numbers
Sum
Infinity (positive)
Negative infinity
x Absolute value of x
f(x)
Types of Numbers
Function of x
Real Numbers: Basically all numbers
Integers (Whole Numbers): 0,
1, 2,
3...
Rational Numbers: Numbers that can be represented as a fraction of integers
Irrational Numbers: Real numbers that aren’t rational (they will have infinite decimals)
Positive Numbers: Numbers that are greater than 0
Negative Numbers: Numbers that are less than 0
Non-zero Numbers: Numbers that aren’t equal to zero, both positive and negative
Non-Negative Numbers: Numbers greater than or equal to zero
Complex Numbers: Contain the symbol i, includes all real and imaginary numbers
Math Camp 2011 Page 4

GREEK LETTERS
Letter Lower Case Upper Case
Alpha
Beta
Gamma
Delta
Epsilon
Zeta
Eta
Theta
Iota
Kappa
Lambda
Mu
Nu
Zi
Omicron
Pi
Rho
Sigma
Tau
Upsilon
Phi
Chi
Psi
Omega
Math Camp 2011 Page 5

A FEW BASIC MATHEMATICAL LAWS
Commutative Law of Addition
Commutative Law of Multiplication
Associative Law of Addition
Associative Law of Multiplication
Distributive Law
a b b a
ab ba
a ( b c)
( a b)
c
a ( bc)
( ab)
c
( a b)
c ac bc
NEGATIVE NUMBERS AND ZERO
The number that we must add to the number n to get 0 we call the opposite of n. The symbol for
the opposite of n is –n. Therefore, n + (the opposite of n) = 0. In mathematical symbols, we
have n + (-n) = 0.
The subtraction problem m - n = ( ) has the same answer as the addition problem m + (-n) = ( ).
Multiplying a number by -1 gives the opposite of that number.
1 n
n
The product of two negative numbers is positive.
3 5
15
The product of a negative number and a positive number is negative.
( 5)(8) 40
The product of any number and 0 is 0.
( 8)(0)
0
Division of anything by 0 is undefined.
ORDERS OF OPERATIONS
Mathematical expressions often contain many operators ( , ,
,
)
, and we need to follow a
certain order when evaluating expressions with multiple operators. If we don’t follow the
necessary steps, then we will get different answers.
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For instance, suppose we wanted to evaluate the expression, 3 25. If we add and then
multiply, we get 25. If we multiply and then add, we get 13 — two very different answers.
Thus, we need rules to help us figure out the correct order for performing these operations.
Rules for the Orders of Operations:
1. Start by evaluating any expressions that you find in parentheses ( ), brackets [ ], or braces
{ }. If there are parentheses (or brackets or braces) nested within each other (meaning
that one set of parentheses is completely within another parentheses), then evaluation the
expressions in the innermost parentheses first. Then work outward.
2. Evaluate all exponents that are on single numbers. If a parentheses has an exponent on it,
then you have to evaluate the expression inside the parentheses before you evaluate the
exponent on the parentheses.
3. Evaluate the multiplications and divisions. Start from the left and do them in the order
they appear as you move from left to right.
4. Evaluate the additions and subtractions. Start from the left and do them in the order that
they appear as you move from left to right.
Examples:
3 56
3
30 33 => do the multiplication before the addition
2
(3 8)
14
2(
5)
14
10
14
4 => evaluate the expression in the parentheses
first by subtracting 8 from 3, then multiply 2 by that quantity, then add the remaining
quantities together from left to right
10
85
2 310
83 310
24 3
34 31
3 => evaluate the
innermost parentheses first by subtracting 2 from 5, then evaluate all the expressions in
the outermost parentheses by multiplying 8 by 3 and then adding 10 to that quantity,
finally end by adding/subtracting the remaining quantities
Note: You have to be careful when using a calculator to evaluate these expressions. A scientific
calculator should be programmed to follow the order of operations. A regular calculator,
however, will evaluate the operators in the order that you enter them in the calculator. So you
have to remember to enter multiplications and divisions before additions and subtractions, etc.
ROOTS AND EXPONENTS
Introduction to Simple Exponents
Let b be a real number and n be a positive integer. The product of n of the b’s is (b) (b) (b)…(b)
{done n times}. This is called “b to the nth power,” or “b raised to the n,” or “b to the n.”
We write (5) (5) (5) as 5 3 . We read this as 5 to the 3 rd power. To calculate 5 3 we multiply 5 by
itself three times, so (5) (5) (5) = 125. In this particular case, the base is 5 and the exponent is 3.
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To calculate 4 3 , just multiply (4) (4) (4) = (16) (4) = 64.
To calculate 6 5 , just multiply (6) (6) (6) (6) (6) = 7,776.
To summarize, if x is the base, then
x x
1
x x x
2
x x x x
3
x x x x x
4
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Rules for Working with Exponents
The following fourteen rules will help guide you in manipulating expressions that have
exponents in them. Each of these rules will be discussed in more detail below with
accompanying examples.
1
Rule 1: b b
Rule 2: b 0 1, (where b 0)
0
Rule 3: 0 is undefined
Rule 4:
Rule 5:
Rule 6:
Rule 7:
b
b
b
1
2
b
1
n
n
2
m
n
b
b
n
b
1
n
b
, (where b 0)
, (where b 0)
, (where b 0)
n n
Rule 8: x x x m
Rule 9: b
1
m
r s rs
b b , (where 0
r s
Rule 10:
rs
b )
b b , (where b 0)
r r
a b a b , (where b 0)
Rule 11:
r
r
r
a a
Rule 12: , (where b 0)
r
b b
r
b rs
Rule 13: b , (where b 0)
s
b
a
Rule 14:
b
Square Roots
r
b
a
r
, (where b 0)
For any positive number b,
1
2
b
shall be that positive number whose square is b; that is,
( b
1
2
)
2
1
2
1
2
b b
b
This is called the square root of b. It can also be written as b .
Math Camp 2011 Page 9

1
2
So 9 = the square root of 9 = 9 3
1
2
25 = the square root of 25 = 25 5
There are a few things to keep in mind about square roots:
1. Negative numbers do not have square roots (at least in the real number system). For
example, 9 is not a real number.
2. The square root of 0 is 0 because 0 2 0 . So 0 0 .
3. The square root of a positive number is positive.
4. If 0
b then b b
2
. So 2 2 2.
2
5. And finally, we have Rule 5 from above, which says b b . Because
2
b is a
2
positive number, the square root of b must be a positive number (see #3 above), so
2
we need the absolute value to make sure that the square root of b is positive. For
example, 5 2 5 5
Other Fractional Exponents
.
If b is a number whose nth power (where n is a positive integer) equals the number x, then b is
called the nth root of x. In other words, if b n = x, then b is the nth root of x. This can be written
as n x b.
When we write expressions using the radical sign, , we say that the expression is
written in radical form.
Note that
b 2
b .
Negative Exponents
For any number b (other than 0), b -1 shall be the reciprocal of b:
1
b
1
b
For any number b (other than 0) and any positive integer n, b -n shall be the reciprocal of b n :
n
1
b
(Rule 7)
n
b
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If n is a positive integer and b is a number other than 0, then
Manipulating Exponents
n
1 1
n
b b
b
n
(Rule 7 and Rule 12)
The basic law for multiplying number with exponents is:
b
r s rs
b b
(Rule 9)
Basically this means that when we multiply the bases we can add the exponents together. So if
you want to multiply (2 3 )(2 4 ), then it would be:
2
3
2
4
(2
2
2)(2
2
2
2) 2
2
2
2
2
2
2 2
7
2
34
Note: This does not work if the bases are different. The base must be the same for this rule to
5 2
apply. For instance, we could not apply this rule to the product 3 7 because 3 and 7 are two
different bases.
Examples of Exponent Rules:
1
2
16
9
16
9
1
2
1
2
16
9
4
3
6 3
5 5
5
9
1
4
4
81 81 3
16
5
2
5
5
16 4 1024
7
7
10
8
7
108
7
2
This could also be written as
7
7
10
8
7
10
7
8
7
10(
8)
7
2
(3
)
4 2
3
4(
2)
3
8
6 6 6
3 7 3 7
2
5
16
1
2
1
2
16
2
5
16
5
1
2
1
2
2
4
5
1
4 5
20
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Dealing with Negative Bases
When the base of an exponent is negative the following rules apply. If the exponent is odd, then
the final answer will be negative. If the exponent is even, then the answer will be positive.
Examples:
(-2) 1 = -2
(-2) 2 = (-2)(-2) = 4
(-2) 3 = -8
(-2) 4 = 16
Scientific Notation
The population of the world is approximately 6,598,678,357 people (www.census.gov, May
2007). Big numbers like this are difficult to read and to say. Even if we rounded this number off
to 6,000,000,000, it is time-consuming to count the zeros. To deal with big numbers like this,
scientists have developed what is called scientific notation.
Before we explain exactly what scientific notation is, take a look at the following:
10 1 = 10 ten
10 2 = (10)(10) = 100 hundred
10 3 = (10)(10)(10) = 1,000 thousand
10 4 = (10)(10)(10)(10) = 10,000 ten thousand
10 5 = (10)(10)(10)(10)(10) = 100,000 hundred thousand
10 6 = (10)(10)(10)(10)(10)(10) = 1,000,000 million
Every number larger than 10 can be written as:
So,
500 = (5)(10 2 )
683 = (6.83)(10 2 ) = 6.83 x 10 2
1,784 = (1.784)(10 3 )
(b)(10 n ) or, equivalently b x 10 n
Because numbers in scientific notation both have a base of 10, we can easily multiply two
numbers is scientific notation. We just multiply the bases and then add the exponents on 10.
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3 8
5(10 ) 7(10 )
(5)(7)(10
(5)(7)(10
(5)(7)(10
35(10
11
)
(3.5)(10)(10
(3.5)(10
12
3
11
)
)(10
38
)
)
11
)
8
)
Scientific Notation for Small Numbers
n
Every positive number less than 1 can be written in the form b(10
) , where b is at least 1 but
less than 10 and n is a positive integer.
So 510
5 . 00005
Note: Sometimes you will see a different form for scientific notation if you use a calculator or
excel.
5.67 E-03 = 5.67 x 10 -3
3.89 E07 = 3.89 x 10 7
FRACTIONS
Defining Fractions
The fraction b
a represents a portion. It means that we are taking a parts out of b parts.
The top number in a fraction is called the numerator (in this case, a). The bottom number in a
fraction is called the denominator (in this case, b).
Fundamental Principle: The value of a fraction is not changed by multiplying or dividing the
numerator and denominator by the same nonzero expression.
For instance,
x 3
y 3
is still equal to y
x .
Adding Fractions
To add two fractions with the same denominator, just add the numerators and place them over
the denominator.
a b a b
c c c
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To add two fractions with different denominators, we have to replace at least one of the
denominators with a denominator that will allow us to add the two fractions as above. This is
called the least common denominator or lowest common denominator.
If the positive integers a and b are denominators of fractions, then the smallest positive integer
that both a and b divide evenly (that is, the remainder of the division is 0) is called the lowest
common denominator for the numbers a and b, or the lowest common denominator of the two
fractions.
A positive integer is called a prime if it is larger than 1 and cannot be written as the product of
smaller positive integers. If an integer is not prime, then it can be written as the product of two
or more primes.
To find a common denominator, you can break each number down into its prime components
and use those to find the lowest common denominator.
2 7 2 4 7 3 8
Example 1:
3 4 3 4 4 3 12
5 7 5 7 5 7 2
Example 2:
6 3 23
3 23
3 2
21
12
5
6
29
12
14
6
To add an integer and a fraction, you have to turn the integer into a fraction by dividing it by 1
and then follow the rules above.
Subtracting Fractions
4 5 4 5 3 4 1 15 4
5
3 1 3 1 3 3 1 3 3
Subtracting fractions is very similar to adding them. If the base is the same, then you can just
subtract the numerators, but if the bases are different, then you have to find a common
denominator before subtracting the two fractions.
19
3
19
6
Example 2:
9 6 3
Example 1:
8 8 8
3 6 3 8 6 5 24 30 6 3
5 8 5 8 8 5 40 40 40 20
Note: When working with fractions, you should express your final answer in the smallest terms
6 3
possible. That is why we simplify from to . This is called factoring or canceling out.
40 20
Math Camp 2011 Page 14

The cancellation rule tells us how to reduce a fraction.
nb n
mb m
Basically, all you need to do is to factor the numerator and denominator, and then cancel out all
the common factors.
3
x (2y
2)
5x(
y 1)
3
x (2)( y 1)
5x(
y 1)
2x
5
2
Multiplying Fractions
To multiply two fractions, we multiply their numerators to get the numerator of the answer and
then multiply their denominators to get the denominator of the answer. In mathematical
notation,
a
b
c
d
ac
bd
Example:
3
4
2 3
2
7 4
7
6
28
3
14
Multiplying an Integer and a Fraction
To multiply an integer and a fraction, just turn the integer into a fraction by dividing it by 1.
Example:
4 5 4
5
3 1 3
20
3
Dividing Fractions
The quotient of two fractions is written as
a
a c or as
b
b d c
d
To divide a fraction, you multiply the first fraction by the reciprocal of the second fraction. So in
order to divide by d
c you actually multiply by c
d . This is often called cross-multiplication.
a
b
c
d
a
b
c
d
a
b
d
c
Math Camp 2011 Page 15

To divide a fraction by an integer, just turn the integer into a fraction by dividing it by 1.
4 4 2 4 1 4 2
Example: 2
5 5 1 5 2 10 5
Note: When working with fractions, you should express your final answer in the smallest terms
4 2
possible. That is why we simplify from to . 10 5
WORKING WITH ALGEBRAIC EXPRESSIONS
Introducing Polynomials
A monomial is an expression that has the form kx n where k is a number, n is a non-negative
integer and x is a letter. The integer n is the degree of the monomial. The number k is called the
coefficient of the monomial. The letter x is called the variable.
The sum of any number of monomials is called a polynomial. It is customary to write a
polynomial so that the degrees of the monomials decrease from left to right. The following are
examples of polynomials.
17x
3x
5
9x
3
38x
5
13
3x
x
4
3x
4
5x
3
4x
5x
1
2
6x
6
The degree of a polynomial is the degree of the monomial (within the polynomial) that has the
highest degree.
3x 3 5x has degree 3
5 2
5x 4x
3x
5 has degree 5
5x 1 has degree 1
Simplifying Polynomials
When we have a polynomial (algebraic expression), we can simplify it by collecting all the like
terms. You can collect like terms by adding (or subtracting) the coefficients on monomials of
the same degree.
For instance,
3x 5x
(3 5) x 8x
Math Camp 2011 Page 16

To simplify the following expression, we will collect all the like terms.
15x
3
15x
(15 1)
x
14x
3x
5 2x
3
3
x
3
3
6x
2x
4x
(2 4) x
2
2
2
x
8x
5
2
3
2
6x
10
4x
3x
6x
x 5 10
(3 6 1)
x ( 5
10)
2
x
Multiplying Polynomials
One basic rule of algebra is the distributive rule. The distributive rule says:
a(
b c)
ab ac
The distributive rule is what allows us to collective like terms to simplify expressions. The
distributive rule can also help us to multiply out polynomials. We use the distributive rule
several times to multiply two polynomials.
The reminder we use to do this is called FOIL—first, outside, inside, last. This is how it works.
First, multiply the first term from each expression. Then multiply the terms on the outside of the
expression. Then proceed to the terms on the inside, and then take the terms at the end of each
expression.
2
ax
bcx
d acx adx bcx bd
Example:
(5x
3)(2x
1)
5x
2x
5x
(
1)
3
2x
(3)( 1)
10x
Basic Manipulation of Algebraic Expressions
2
5x
6x
3 10x
2
x 3
The same rules that we used in manipulating fractions apply to working with algebraic
expressions as well.
Before we learn how to add, subtract, multiply and divide fractions of polynomials, we need to
learn how to simplify fractions involving polynomials. Whenever you have a fraction that has
polynomials in it, you can factor both the numerator and the denominator and cancel out the
factors that appear in both the numerator and the denominator.
4
6x
3 x (3)(2x
1)
2
4
x
2
x (2x
1)
2
x (2x
1)
3x
Math Camp 2011 Page 17

Adding Polynomial Fractions
If the denominators of two algebraic expressions are the same, then you can simply add the
numerators together.
2
5x
3 x
x 1
2
5x
1
6x
x 1
2
5x
4
x 1
To add fractions of polynomials where the denominators are not equal, then we have to find a
lowest common denominator. One common denominator that will always work is just the
product of the two denominators.
To add the two fractions,
2
6x
4 2x
x 1
, the common denominator is x ( x
2 2)
.
2
x x 2
2
6x
4 2x
x 1
2
x x 2
2
2
6x
4 x 2 2x
x 1
x
2
2
x x 2 x 2 x
3
2
3 2
(6x
12x
4x
8) (2x
x x)
2
x(
x 2)
8x
3
5x
x(
x
2
2
11x
8
2)
Subtracting Polynomial Fractions
Again, to subtract two algebraic fractions, first find a common denominator, and then subtract.
2
5x
2 x 1
3
x x
2 2
5x
2 x x 1
2 3
x x x
3 2 2
(5x
2x
) ( x 1)
3
x
3 2
5x
x 1
3
x
Math Camp 2011 Page 18

Multiplying Polynomial Fractions
Again, we multiply algebraic fractions the same way we multiply numerical fractions:
a c ac
b d bd
Example:
x 1
x 3 ( x 1)(
x 3)
3x
2 x 2 (3x
2)( x 2)
x
3x
2
2
3x
x 3 x
6x
2x
4 3x
2
2
2x
3
8x
4
To multiply a simple polynomial by a fractional algebraic expression, use the same method as
when we multiplied a fraction by an integer. That is, turn the polynomial into a fraction by
giving it a denominator of 1.
2
2
x 2 x 4 x 2 x 2x
4x
8 x 6x
8
( x 4)
x 2 1 x 2 x 2 x 2
Dividing Polynomial Fractions
And, once again, to divide algebraic expressions, we follow the same rules as for regular
fractions—just cross multiplication.
3x
2
5
Evaluating Algebraic Expressions
3
3x
2
2
3x
2
5
2
3x
2 3x
2
To evaluate an algebraic expression, all you have to do is plug in a number into each variable in
the expression. This allows us to find the value of an expression in a particular case.
Example 1: Evaluate the following expression when x = 1.
3x
2
3(1)
3 9 5
7
9x 5
2
9 1
5
Example 2: Evaluate the following expression when a = 3 and b = -2.
a
2
(3)
3ab
5b
2
2
8a
4b
3
3(3)( 2)
5( 2)
9 18
20 24 8 3 16
2
8(3) 4( 2)
3
3
15
3
Math Camp 2011 Page 19

Remember: The expression
2
x is not the same thing as 2
need to evaluate it as x
2 x
x. If you have
x 2
2
x ( x)(
x)
.
Example:
Evaluate the following expressions for x = 3
x
2
( x )
5 3
2
2
5 ( 3)
PERCENT CHANGE
5 9
5 4
2
5 9 5 14
x
. If you are given
, then you need to evaluate it as
2
x , then you
We can figure out the percent change in a quantity by forming a ratio of the quantities that we
are interested in. When dealing with a percent change, the original price or quantity becomes the
denominator. This is because we want to make all comparison relative to this value.
Example 1: Suppose that Big Macs cost $3.15 last year and $3.65 this year. What is the percent
change?
3.65 3.15
Well, .16 16%
. The price of Big Macs increased by 16% this year.
3.15
Example 2: Suppose that your GPA was 3.9 last semester and now it is 3.5. What is the percent
change?
3.5 3.9
Well, .
1. Thus your GPA decreased by 10%.
3.9
WORKING WITH SUMMATION SIGNS
Defining the Summation Sign
Mathematics uses a lot of symbols in order to simplify presentation of material. One helpful
symbol is the summation sign, which is capital sigma, or . This sign can be used to tell us to
add several things together.
For instance, if we wanted to add several numbers, we could write the following:
x
x
1
x2
x x4
x5
Math Camp 2011 Page 20

If we wanted to represent this same sum using the summation symbol, we would write:
5
i1
The summation symbol itself tells us that we are going to add. The x
i
tells us that we are
going to add values of x. The letter i basically stands as an index. Under the summation sign,
we learn what our starting value for the index is—in this case it is i = 1. And the number on top
of the summation sign tells us the ending value for the index—in this case 5. So the above
symbol tells us to add all the values of x starting with x
1
and ending with x
5
(which includes all
the values in between the two).
One formula using the summation symbol that you will see again in your statistics class is the
following:
n
xi
i
x 1 , which is sometimes also written just as
n
x
i
x
x
n
When the summation is written without the index, it means to sum over all the possible or
available values.
The above formula is the formula for finding the mean (or average) of a set of numbers. It tells
us that if we have n numbers, we sum them all, and then divide by n.
Example: Find the mean of the following numbers: 1, 7, 3, 22, 64, 13
x
n
i1
n
x
i
x x
1
2
x
3
x
n
4
x
5
x
6
1
7 3 22 62 13
110
18.33
6
6
Rules for Summation Signs
The following rules are helpful when working with summation signs.
Rule 1:
Rule 2:
c nc (Sum of a Constant)
cxi
c
xi
where c is any constant
i i
xi
Rule 3: x
y
i i
xi
Rule 4: x
y
y
y
i
i
Math Camp 2011 Page 21

2
Rule 5: x
2
i
xi
Rule 6: x
i
y
i
x
i
y
i
2 2 2
xi
i
xi
yi
2
Rule 7: y
x y
i
A couple other sums that you will definitely encounter in statistical applications are x 2 and
2
x . For the first formula 2 x , we square each value of x and then sum those squares
x
together. For the second formula 2
i
, we sum the x values and then square the final number.
Example: Given the numbers—9, 17, 32, 16, 8, 2, 9, 7, 3, 18—find x , x 2 , and x 2
x 9 17
32 16
8 2 9 7 3 18
121
2 2 2 2 2 2 2 2 2 2
x 9 17
32 16
8 2 9 7 3 18
81
289 1024
64 4 81
49 9 324 1925
2
2 2
(9 17
32 16
8 2 9 7 3 18)
121 14,
641
x
Summation Application Problem
An important statistical concept is the variance. At this point it isn’t important for you to know
what the variance is or why it is important (you will learn that in your statistics class), but we can
use what we know about summation signs to help us calculate it. There are two different
formulas for calculating the variance, and we will use both.
Trout, Inc. feeds fingerling trout in special ponds and markets them when they attain a certain
weight. A sample of 10 trout were isolated in a pond and fed a special food mixture, designated
RT-10. At the end of the experimental period, the weights of the trout were (in grams): 124, 125,
125, 123, 120, 124, 127, 125, 126, and 121.
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th edition, McGraw-Hill: Boston. p. 108)
2
x
x
2
a. Calculate the variance using the deviation formula. s
n 1
x 124 125
125
123
120
124
127
125
126
121
x
n
10
2
1240
10
124
s
2
2
2
2
2
x
x 124
124 (125 124)
(125 124)
... 121124
n 1
10 1
2
42
4.6667
9
Math Camp 2011 Page 22

. Calculate the variance using the direct formula.
s
2
x
2
x
n
n 1
2
x 124 125
125
123
120
124
127
125
126
121
1240
s
2
x
2
2 2 2 2 2 2 2 2 2 2
124
125
125
123
120
124
127
125
126
121
153,
802
x
2
x
n
n 1
2
2
1240
153,802
10
4.6667
10 1
Math Camp 2011 Page 23

ORDERS OF OPERATIONS
MATH CAMP
School of Public and Environmental Affairs
Day One In-Class Exercises
1. Evaluate the following expressions using the order of operations:
a. 10 35 56
2 15
150 15 3 6 7
2
b. 21
ROOTS AND EXPONENTS
2. Calculate the following:
a. 2 3
b. 3 2
c. 6 5
d. 1 35
e. 10 2
f. 7 2
g. (-1) 3
h. (-1) 600
i. 3 4
j. 5 0
k. 19 1
l. 6 6
3. Compute:
a. 5 -1
b. 3 -3
c. 17 -2
2
1
d.
3
e. 2 -10
4. Evaluate the following:
1
2
a. 25
b.
1
25
1
2
2
3
c. 125
5. Simplify the following:
5 8
a. 3 3
2 3
b. ( 6) ( 6)
( 6)
c.
6
x
x
2
4
Math Camp 2011 Page 24

d.
e.
2
2
2
f.
5
x
6
x
1
9
7
9
2
g. 3
5
h.
2
7
1 1
5
2
3
i.
4
6. Write the following in the form 10 n :
a. 100
b. 1000
c. (10)(10)(10)(10)(10)
d. (10)(10)(10)(10)(10)(10)(10)
e. ten to the seventh power
f. 10
7. Write the following in scientific notation:
a. 983
b. 1,542
c. 10,000,000
d. 987,400
e. 732,000,000,000
f. .00035
g. .01
h. -.006
i. .0000000892
8. Multiply the following numbers and express the result in scientific notation.
5 3
a. 4(10 ) 5(10 )
FRACTIONS
b. 3(10
3 3
) 3(10 )
9. Compute the following:
a.
5 2
3 5
b.
6 3
7
c.
5 2
3 9
d.
7 2
13
Math Camp 2011 Page 25

e.
3 2
4 6
f.
3 4
8 5
g.
7 5
4
WORKING WITH ALGEBRAIC EXPRESSIONS
10. Simplify the following polynomials by collecting like terms:
a. 6x 3x
5x
2
2
3
b. 5x
2 3x
x 3x
12x
3x
5 x
c. 6( x 4) 3x
2 x
2 5x
11. Multiply the following polynomials:
a. 6x
42
x 1
b. x
3x
3
12. Compute the following:
5x 1
9x
1
a.
x x
2
6x
7 x 1
b.
x 1
x 2
6x 3 9x
4
c.
5x
2 x
8x
2 5x
7x
1
d.
x 8 x
3
13. Evaluate the following expression:
a. 5x 3 y 8x
9y
3, where x = 3 and y = 5
14. A radioactive substance decays in the following manner: In any second, half of it
disappears. If it weighs one ounce at the beginning, how much is left after
a. One second
b. Two seconds
c. Three seconds
d. Four seconds
PERCENT CHANGE
15. Suppose that you paid $5,498 in taxes last year and $6,743 in taxes this year. What is the
percentage increase in the amount you paid for taxes?
Math Camp 2011 Page 26

WORKING WITH SUMMATION SIGNS
x
16. Find x for the following:
n
a. -8, 34, -63, 25, 101, -90, 5, 76, 43, -5
x
17. For each set of values, find x , x 2 , and 2
a. 5, 12, 8, 3, 4
18. Calculate 2
x i
x
for the following. To do this you will first need to compute
x
x .
n
a. 5, 3, -11, 10, 17, 3, -21, 36
19. The annual report of Dennis Industries cited these primary earnings per common share
for the past five years: $2.68, $1.03, $2.26, $4.30, and $3.58.
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th edition, McGraw-Hill: Boston. p. 106)
x
a. What is the mean primary earnings per share of common stock? x
n
2
x
x
2
b. What is the variance? s
n 1
20. An insurance company wants to determine the strength of the relationship between the
number of hours a person works and the number of injuries or accidents that person has
over a period of one week. The data follow. Compute r.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 487)
Hours
worked, x
No. of
accidents, y
r
40 32 36 44 41
1 0 3 8 5
n
xy
x
y
2
2
2
x
x
n
y
n
y
2
Math Camp 2011 Page 27

ORDERS OF OPERATIONS
MATH CAMP
School of Public and Environmental Affairs
Day One In-Class Exercises Solutions
1. Evaluate the following expressions using the order of operations:
2
10
35 5
6 15
10
35 536
15
10
7 36
15
a.
10
252 15
242 15
227
150
b.
150
ROOTS AND EXPONENTS
15
36
7 2 21 150 15
36
14 21 150 15
3
20
15
60
21 150 75 21 2 21 23
2. Calculate the following:
a. 2 3 = 8
b. 3 2 = 9
c. 6 5 = 7776
d. 1 35 = 1
e. 10 2 = 100
f. 7 2 = 49
g. (-1) 3 = -1
h. (-1) 600 = 1
i. 3 4 = 81
j. 5 0 = 1
k. 19 1 = 19
l. 6 6 = 46,656
3. Compute:
a. 5 -1 1
= . 2
5
b. 3 -3 1 1
. 037
3
3 27
c. 17 -2 1 1
. 00346
2
17 289
2
2
1 3
3 9
d. 9
2
3
1 1 1
e. 2 -10 1 1
. 000977
10
2 1024
4. Evaluate the following:
a. 25 5
25 2 1
2
21
Math Camp 2011 Page 28

1
2
1
2
1 1 1 1
b. . 20
1
25
25 5
2
25
2
3
3
c. 125 125 5 25
5. Simplify the following:
5 8 58
13
a. 3 3
3 3
2 3 4 234
b. ( 6)
( 6)
( 6)
( 6)
( 6)
c.
d.
e.
f.
6 2 62
8
x x x x
1 7 17
8
9 9
9 9
7
2 72
5
2 2
2
2
x 5
x x
1
56
1
6
x
x
5
3 3 3
2 2 5 10
g.
h.
2
1
5
2
1
2
(25)
1
2
10
1
1
10
2
3 3 9
i.
2
4 4 16
6. Write the following in the form 10 n :
a. 100 = 10 2
b. 1000 = 10 3
c. (10)(10)(10)(10)(10) = 10 5
d. (10)(10)(10)(10)(10)(10)(10) = 10 7
e. ten to the seventh power = 10 7
f. 10 = 10 1
7. Write the following in scientific notation:
a. 983 = (9.83)(10 2 )
b. 1,542 = (1.542)(10 3 )
c. 10,000,000 = (10 7 )
d. 987,400 = (9.874)(10 5 )
e. 732,000,000,000 = (7.32)(10 11 )
f. .00035 = (3.5)(10 -4 )
g. .01 = 10 -2
h. -.006 = (-6)(10 -3 )
i. .0000000892 = (8.92)(10 -8 )
8. Multiply the following numbers and express the result in scientific notation.
5 3
53
8
9
a. 4(10 ) 5(10
) (45)(10
) (20)(10 ) (2.0)(10 )
3 3
33
6
b. 3(10 ) 3(10
) (3
3)(10 ) (9)(10 )
9
Math Camp 2011 Page 29

FRACTIONS
9. Compute the following:
a.
5 2 5 5 2 3 25 6 25 6 31
3 5 3 5 5 3 15 15 15 15
b.
6 6 3 6 3 7 6 21 6 21 27
3
7 7 1 7 1 7 7 7 7 7
c.
d.
5 2 5 3 2 15 2 13
3 9 3 3 9 9 9 9
7 7 2 7 2 14
2
13 13 1 131
13
e.
3 2 3
2 6 1
4 6 4 6 24 4
f.
3 4 3 5 35
15
8 5 8 4 8
4 32
g.
7 7 5 7 1 7 1
7
5
4 4 1 4 5 4 5
20
WORKING WITH ALGEBRAIC EXPRESSIONS
10. Simplify the following polynomials by collecting like terms:
a. 6x 3x
5x
8x
b.
5x
2 3x
x
3
9x
2
2
x 3x
12x
7
2
3x
5 x
3
2
2
6( x 4) 3x
2 x 5x
6x
24 3x
2 x 5x
c.
2
x 8x
22
11. Multiply the following polynomials:
2
2
a. 6x
42
x 1 12x
6x
8x
4 12x
2x
4
2
2
b. x
3x
3 x 3x
3x
9 x 9
12. Compute the following:
5x
1
9x
1
5x
1
9x
1
14x
a.
14
x x x x
Math Camp 2011 Page 30

.
2
2
6x
7 x 1
6x
7 x 2 x 1
x 1
x 2 x 1
x 2 x 2
2
3 2
6x
12x
7x
14
x x x 1
2
2
x x 2x
2 x x 2x
2
2
3 2
6x
12x
7x
14
x x x 1
2
x x 2x
2
3 2
x 7x
18x
15
2
x x 2
x
x
1
1
c.
3
3
6x
9x
4 6x
(9x
4)
5x
2 x x(5x
2)
2
6x
(9x
4)
5x
2
3
54x
24x
5x
2
2
2
2
3 2 2
8x
5x
7x
1
8x
5x
x 3 8x
24x
5x
15x
2
x 8 x 3 x 8 7x
1
7x
x 56x
8
d.
3 2
8x
29x
15x
2
7x
55x
8
13. Evaluate the following expression:
a. 5x 3 y 8x
9y
3, where x = 3 and y = 5
3
5(3) (5) 8(3) 9(5) 3 675 24 45 3 741
14. A radioactive substance decays in the following manner: In any second, half of it
disappears. If it weighs one ounce at the beginning, how much is left after
1 1
a. One second 1
ounce
2 2
1 1 1
b. Two seconds ounce
2 2 4
1 1 1
c. Three seconds ounce
4 2 8
1 1 1
d. Four seconds ounce
8 2 16
PERCENT CHANGE
15. Suppose that you paid $5,498 in taxes last year and $6,743 in taxes this year. What is the
percentage increase in the amount you paid for taxes?
6743 5498
.23 => the amount of taxes you paid increased by 23%
5498
WORKING WITH SUMMATION SIGNS
x
16. Find x for the following:
n
a. -8, 34, -63, 25, 101, -90, 5, 76, 43, -5 = 11.8
Math Camp 2011 Page 31

17. For each set of values, find x , x 2 , and x 2
a. 5, 12, 8, 3, 4
x 5 12
8 3 4 32
x
2
5
2
12
2
8
2
3
2
4
2
258
2
2 2
(5 12
8 3 4) 32 1024
x
18. Calculate 2
x i
x
x
x .
n
a. 5, 3, -11, 10, 17, 3, -21, 36
for the following. To do this you will first need to compute
x
x
n
5 3 1110
17
3 21
36
8
42
5.25
8
2
2
2
2
2
x i
x x1
x x2
x x3
x ... x8
x
2
2
2
2
2
5
5.25 3
5.25 11
5.25 10
5.25 17
5.25 3
5.25
2
2
21
5.25 36
5.25 2069. 5
19. The annual report of Dennis Industries cited these primary earnings per common share
for the past five years: $2.68, $1.03, $2.26, $4.30, and $3.58.
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th edition, McGraw-Hill: Boston. p. 106)
a. What is the mean primary earnings per share of common stock?
x 2.68 1.03
2.26 4.30 3.58 13.85
x
$2.77
n
5
5
b. What is the variance?
2
x
x
2
s
n 1
2
2
2
2
2.68
2.77 1.03
2.77 2.26
2.77 (4.30 2.77) 3.58
2.77
6.2928
$1.57
4
5 1
2
2
Math Camp 2011 Page 32

20. An insurance company wants to determine the strength of the relationship between the
number of hours a person works and the number of injuries or accidents that person has
over a period of one week. The data follow. Compute r.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 487)
Hours
worked, x
No. of
accidents, y
r
40 32 36 44 41
1 0 3 8 5
n
xy
x
y
2
2
2
x
x
n
y
n
y
x 193
y 17
xy 705
x 2
7, 537
y 2
99
r
n
xy
x
y
2
2
2
x
x
n
y
2
2
n
y
5(705) (193)(17)
2
2
5(7,537)
(193) 5(99)
(17)
.814
Math Camp 2011 Page 33

MATH CAMP
School of Public and Environmental Affairs
Day One Homework
1. Simplify the following:
a.
2
5
7
1
2
2
b. 3
.16
c.
2
4b
c
4
b c
1
2
1
4
1
2
2. Compute the following:
a. 7 x
2 5x
62
x 9
b.
2
3x 2 9x
x 12
x x
3. Suppose that your nonprofit organization had 135 clients last year and 156 clients this year.
By what percent did your clientele increase?
4. Suppose that you had 3 cousins in 1995 and 23 cousins in 2005. What is the percent change?
5. Show that x
x 0
i
. (This is something that you will see again in your statistics class.
These are called deviations from the mean.) (Hint: You will use the fact that
x
x )
n
Math Camp 2011 Page 34

6. Dave’s Automatic Door installs automatic garage door openers. Based on a sample, following
are the times, in minutes, required to install 10 doors: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42.
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th edition, McGraw-Hill: Boston. p. 108)
a. Calculate the variance using the deviation formula.
b. Calculate the variance using the direct formula. s
2
s
2
x x
n 1
2
x
2
x
n
n 1
2
Math Camp 2011 Page 35

MATH CAMP
School of Public and Environmental Affairs
Day One Homework Solutions
1. Simplify the following:
2
2
5 7 7
a.
2
7 5 5
b. 3.16
2
2
1
3.16 1
2
2
1
2 3.16 3. 16
c.
2
4b
c
4
b c
b
1
1
2
1
4
1 1 2
8 4 2
c
2
1
2
4
b c
2
4b
c
1 2
8 8
c
2b
1
4
1
2
1
8
1
2
c
2b
4
b
1
2
1
2bc
1
4
2
b
1
8
c
1
2
2
1 1
4 2
c
1 1
2 2
b
2
2b
c
1
1
8
c
1
4
b
c
2
2(
1)
1 1
8 4
2. Compute the following:
2
7x
5x
6 2x
9
14x
a. 3 2
14x
53x
33x
54
3
10x
2
12x
63x
2
45x
54
b.
2
2
2
3x 2 9x
x 12
3x
2 9x
x 12
9x
4x
10
x x
x
x
3. Suppose that your nonprofit organization had 135 clients last year and 156 clients this year.
By what percent did your clientele increase?
156 1.16 => your clientele is 116% of what is was last year, so it increased by
135
16%
4. Suppose that you had 3 cousins in 1995 and 23 cousins in 2005. What is the percent change?
23 3 20
6.67 => your number of cousins has increased by 667%
3 3
Math Camp 2011 Page 36

5. Show that x
x 0
These are called deviations from the mean.)
i
. (This is something that you will see again in your statistics class.
x x
x
i
( x x1
) ( x x2
) ... ( x xn
)
x
i
nx
x
i
n
xi
n
x
i
0
6. Dave’s Automatic Door, referred to in Exercise 3, installs automatic garage door openers.
Based on a sample, following are the times, in minutes, required to install 10 doors: 28, 32, 24,
46, 44, 40, 54, 38, 32, and 42.
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th edition, McGraw-Hill: Boston. p. 108)
x
x
n
a. Calculate the variance using the deviation formula.
2
x
x
2
s
n 1
28 32 24 46 44 40 54 38 32 42 380
38
10
10
s
2
2
2
x
x 28
38 ... 42
38
744
82.6667
n 1
10 1
9
b. Calculate the variance using the direct formula.
x
2
x
2
s
n
n 1
x 28 32 24 46 44 40 54 38 32 42 380
2
2
x
2
28
2
32
2
24
2
46
2
...
32
2
42
2
15,184
s
2
2
x 380
2
x
n
n 1
15,184
10
10 1
2
744
82.66667
9
Math Camp 2011 Page 37

MATH CAMP
School of Public and Environmental Affairs
Day Two: Solving and Manipulating Equations
Day Two Class Notes
FACTORING
If an algebraic expression is the product of other algebraic expressions, then each of the algebraic
expressions that were multiplied together is called a factor of the product. Suppose that a, b, and
c are algebraic expressions. If a b c then a and b are both factors of c.
2
2
For instance, because ( x 1)(
x 2) x 2x
x 2 x x 2 , we can say that (x-1) and
2
(x+2) are factors of x x 2 .
The process of finding factors of an algebraic expression is called factoring. Factoring is useful
because it helps us to solve equations or otherwise manipulate algebraic expressions. We want
to factor a polynomial until it is prime. A prime algebraic expression is one that doesn’t have
any monomial or polynomial factors with integer coefficients other than one and itself.
3 2
2
Example 1: x 3x
2x
6 ( x 3)( x 2)
Each of the two factors, (x-3) and x 2 2
are prime because the integer coefficients are
not divisible by anything other than 1 and themselves.
4 2
2 2
2
Example 2: 20x
48x
y 16y
(4x
8y)(5x
2y)
Even though we have factored this equation into two separate factors, we still aren’t done
because the factor 4x 2 8y
isn’t prime yet. We can still factor it further.
4 2
2 2
2
2
2
So 20x
48x
y 16y
(4x
8y)(5x
2y)
4( x 2y)(5x
2y)
3 2
Example 3: To factor the polynomial 10x 15x
5x
, we have to factor out the monomial 5x,
which is common to all three terms in the polynomial.
3 2
2
So 10x
15x
5x
5x(2x
3x
1)
Example 4: Even if the whole polynomial might not have a common monomial to factor out, we
can often group terms together and factor a monomial out of the group.
2x
2 5x
2xy
5y
x(2x
5) y(2x
5) ( x y)(2x
5)
Math Camp 2011 Page 38

Formulas for Factoring
The following formulas can be useful in helping us to factor various polynomials into their prime
factors.
2
Formula 1: x ( a b)
x ab ( x a)(
x b)
Formula 2:
x
2
2
2xy
y ( x y
2
2
Formula 3: acx ( ad bc)
xy bdy ( ax by)(
cx dy)
)
2 2
Formula 4: (Difference of Two Squares): x y ( x y)(
x y)
3 3
2
2
Formula 5: (Difference of Two Cubes): x y ( x y)(
x xy y )
3 3
2
2
Formula 6: (Sum of Two Cubes): x y ( x y)(
x xy y )
2
2
Formula 7: (A Variant on the Difference of Two Squares): x d ( x d )( x d )
Using these factoring formulas, we can follow this general procedure to factor polynomials into
their prime factors.
(1) If all the terms in the polynomial have a common factor (other than 1), then factor out the
common factor.
(2) If the polynomial has only two terms, then determine if it is the difference of two squares,
the difference of two cubes or the sum of two cubes. If the polynomial falls into one of
these categories, then you can use it is, then you can use Formulas 4, 5, 6 or 7 to factor
the equation.
(3) If the polynomial has three terms, then you will need to use Formulas 1, 2 or 3 to factor
it.
(4) If the polynomial has more than three terms, try to factor the polynomial by grouping
terms together that might have a common factor.
a. Arrange the four terms so that the first two terms have a common factor and the
second two terms have a common factor.
b. Use the distributive property to factor each group of two terms.
(5) Look at your polynomial again to see if the terms in any of the factors have a common
factor. If so, then factor it out.
(6) Check your work to make sure that you factored everything correctly.
These examples show how to use the factoring formulas to factor polynomials.
Example 1:
Example 2:
Example 3:
2
2
x 2x
15
x ( 3)
x 5x
( 3)(5)
by Formula 1
x(
x 3) 5( x 3) ( x 3)( x 5)
2
2
x 4x
4 ( x 2)( x 2) ( x 2) by Formula 2
9
2
2
a 24a
16
(3a
4) by Formula 2
2 2
Example 4: 25u
36v
(5u
6v)(5u
6v)
by Formula 4
Math Camp 2011 Page 39

2
2
Example 5: To factor the polynomial 6x
11xy
10y
into the product of two binomials
( ax by)(
cx dy)
, such as in Formula 3, we have to find a and c such that ac = 6 and b and d
such that bd = -10 and such that ad + bc = -11. The possibilities for b and d are: (1,-10); (-1, 10);
(2,-5) and (-2, 5). The possibilities for a and c are: (1,6) or (2,3). By testing these combinations,
2
2
we find that 6x
11xy
10y
(2x
5y)(3x
2y)
.
3
3 3
8a
27 (2a)
3 (2a
3) (2a)
Example 6:
2
(2a
3)(4a
6a
9)
2
(2a)(3)
3
2
Example 7: x 81
( x 81)( x 81) ( x 9)( x 9)
by Formula 7
Example 8:
64a
3
b
3
(4a)
(4a
b)16a
2
3
3
b
(4a
b)
(4a)
4ab
b
2
2
(4a)(
b)
b
2
2
by Formula 6
by Formula 5
2
2
Example 9: x 2xy
3xy
6y
xx
2y)
3y(
x 2y)
( x 3y ( x 2y)
by grouping terms
together and using the distributive property to factor out their common factor
QUICK POLYNOMIAL REVIEW
Remember that the degree of a polynomial is equal to the degree of the monomial with the
highest degree in the polynomial.
f ( x)
a
0
Constant function, degree = 0 (think of a x a 1
a )
f ( x)
a bx
Linear function, degree = 1
f ( x)
cx
2
a bx
Quadratic function, degree = 2
f ( x)
dx
…and so on
2 3
x bx cx Cubic function, degree = 3
SOLVING EQUATIONS
As mentioned earlier, one of the reasons that we need to know how to factor polynomials is
because we want to be able to find all the roots of the polynomial. A root of a polynomial is a
number that when substituted into a polynomial in place of the variable, the entire equation
equals 0. A root is also called a solution to the equation.
Solving Linear Equations
We can find these roots by solving for x in an equation. We’ll start with a simple linear equation
(a polynomial of degree 1). We want to find the value of x that will make the equation equal 0.
Math Camp 2011 Page 40

To solve for x, remember one very important rule—whatever you do to one side of the equation,
you must also do to the other side. So if you add 5 to one side, then you must add five to the
other side, and so forth.
3
Example: Solve the equation 5x x 4 for x.
2
3
5x
x 4
2
3 3 3
5x
x x x 4
2 2 2
3
5x
x 4
2
3
2 (5x)
2 x (2)( 4)
2
10x
3x
8
7x
8
7x
7
x
8
7
8
7
8
3
Therefore x is the root of the equation 5x
x 4 . We can check this answer by
7
2
8
substituting x into the equation in the place of x.
7
This is how to find a root for a polynomial of degree 1 (a linear function). Now we will find the
roots for a polynomial of degree 2 (a quadratic function). One thing to note: the number of roots
(solutions) to an equation will be equal to the degree of the equation. So our linear equation
above had only one solution. Quadratic equations will have two roots.
Finding Solutions to Quadratic Equations
A second degree algebraic expression with one variable is called a quadratic equation, and it
2
written as: ax bx c 0 . This is the standard form for a quadratic equation.
To find the solutions (roots) to a quadratic equation, follow these steps:
2
(1) Write the equation in the standard form with the x term coming first and having a
positive coefficient. If necessary, rewrite the equation so that all the terms are on the left
hand side of the equation and zero is on the right hand side of the equation.
(2) Factor the left hand side of the equation using the factoring formulas.
(3) Set each of these factors equal to zero (using the Zero-Factory Property) and solve for the
variable.
Math Camp 2011 Page 41

(4) Check each of the solutions in the original equation.
The Zero-Factor Property states that if a b 0 , then either a 0 or b 0 . This property helps
us to find the solutions to equations that have been factored. For instance, if we know that
( 2x 1)(
x 3) 0 then either 2x 1
0 or x 3 0.
Example: Solve the equation x 2 4 for x.
2
x 4
x
2
4 0
( x 2)( x 2) 0
Rewrite the equation in standard form
Factor using Formula 4
x 2 0
x 2
Set each of the factor equal to zero and solve
x 2 0
x 2
x
2
( 2)
(2)
4
2
2
4
4
Check each solution in the original equation
Therefore, the solutions to this equation are x 2 and x 2
.
2
Example: Find the solutions to the equation x 4x 21 0.
x
2
4x
21 0
( x 7)( x 3) 0
x 7 0 x 7
x 3 0 x 3
x
2
(7)
4x 21 0
2
4 7 21 0
49 28 21 0
0 0
x
2
( 3)
4x 21 0
2
4( 3)
21 0
9 12
21 0
0 0
2
Thus, x = 7 and x = -3 are the solutions to the equation x 4x 21 0.
Math Camp 2011 Page 42

Remember: The number of roots to an equation is equal to the degree of the equation. So a
quadratic equation always has two roots.
Example: Find the roots of the equation 4y 2 16y 20 0 .
4y
2
4( y
16y
20 0
2
4y
5) 0
4( y 5)( y 1)
0
y 5 0
y 5
y 1
0
y 1
Using the Quadratic Equation
In the event that a quadratic equation cannot be easily factored, we can also use the quadratic
equation to find the solutions to the equation. Remember that a second degree algebraic
2
expression with one variable is typically written as: ax bx c 0 . Anytime we have an
equation of this form, the roots to the equation can be found by substituting the coefficients a, b,
and c into the formula:
b
x
b
2 4ac
2a
Example: Find the roots to the equation 6x
2 7x 5 0 using the quadratic formula. In this
case a = 6, b = -7 and c = -5.
b
x
( 7)
7
7 169
12
7 13
12
2
b 4ac
2a
( 7)
2(6)
49 120
12
2
4(6)( 5)
So
7 13
20 5
x and
12 12 3
7 13
x
12
6 1
are the solutions to the equation.
12 2
Math Camp 2011 Page 43

Imaginary Numbers
Sometimes when determining the roots of an equation, you will encounter a square root of a
negative number, such as
to define i, where i 1
.
2
n
. In order to take the square root of a negative number, we need
Example: Using the quadratic equation, we can find the roots of the equation
25x
2 10x 13
0
b
x
10
10
10
1200
50
10
10
10
20i
50
1
2i
5
2
b 4ac
2a
(10)
100 1300
50
11200
50
1 1200
50
10
i 400 3
50
3
2(25)
3
4(25)(13)
When dealing with real world application, we usually won’t be dealing with imaginary roots. An
imaginary root might be a mathematical solution to a problem but not a practical solution to a
problem. However, you should be aware that imaginary roots do exist.
Solving Radical Equations
Radical equations are equations where the variable appears in a radical, such as a square root or
cube root. We often solve these equations by changing the form of the equation to make it linear
or quadratic. However, extraneous roots can arise in certain mathematical situations when we
change the form of an equation. For instance, when we square an equation to get rid of a square
root, we can introduce extraneous or false roots because squaring both a number and the opposite
of that number will give you the same result. In this case, one, both or neither solution might
Math Camp 2011 Page 44
2

work. In order to determine if extraneous roots are present, you must go back and check each of
the roots to see if it satisfies the original equation.
Example: Solve the following equation: x 10 x 10
x 10
x 10
x 10
10 x
2
x 10 10
x
x 10
(10 x)(10
x)
x 10
100 10x
10x
x
x
x
2
2
20x
100
x 10
0
21x
90 0
( x 15)(
x 6) 0
x 6,15
2
Now we have to check both of these solutions.
2
x 10
x 10
6 10
6 10
16 6 10
4 6 10
10 10
Because x = 6 solves the original equation, it is a solution.
x 10
x 10
15 10
15
10
25 15
10
5 15
10
20 10
Because x = 15 doesn’t solve the original equation, it is not a
solution.
Solving Cubic and Higher Polynomials
To find the solutions (roots) to a higher degree polynomial (usually a cubic polynomial), follow
these steps:
(1) Write the equation in standard form with the right hand side of the equation being zero
and the other terms being written in decreasing order on the left hand side of the
equation.
(2) Factor the left hand side of the equation using the factoring formulas.
Math Camp 2011 Page 45

a. If the polynomial cannot be factored using the formulas, then it would be
necessary to use a computer to find the solutions to the equation.
(3) Set each of these factors equal to zero (using the Zero-Factory Property) and solve for the
variable.
(4) Check each of the solutions in the original equation.
Example: Find the roots of the following equation: 27x 3 8
0 .
27x
3
8 0
(3x
2)(9x
2
6x
4) 0
Factor using Formula 5
3x
2 0
3x
2
x
2
3
Set each factor equal to zero and solve
9x 2 6x 4 0
Solve this equation using the quadratic formula.
x
x
x
x
b
6
2
b 4ac
2a
6
4(9)(4)
2(9)
6 36 144
18
6 i 108
18
2
SOLVING SYSTEMS OF EQUATIONS
Not only do we want to solve one equation with one variable for its roots, but we also want to be
able to solve two or more related equations for their roots as well. Two or more of these related
equations are called systems of equations, or simultaneous equations. A system of two equations
could look like this:
ax
by c
dx
ey f
Math Camp 2011 Page 46

Substitution Method
The substitution method for solving a system of equations involves solving one of the equations
for one variable in terms of the second variable, and then substituting that into the second
equation.
x
4y
6
2x
y 3
Start by solving the first equation for x in terms of y.
x 4y
6
x 6 4y
Then substitute this in for x in the second equation, and then solve that equation for y.
2x
y 3
2(6 4y)
y 3
12 8y
y 3
9y
3 12
9y
9
y 1
Now substitute y 1
into the first equation to get a value for x.
x 4y
6
x 4( 1)
6
x 4 6
x 6 4
x 2
So the solution set for this system of equations is (2,-1).
Addition-Subtraction Method
This method is based on the principle that when you multiply any equation by a constant, it is
still the same equation. As long as we are consistent about performing the same operation on
both sides of the equation, then we are still dealing with the same equation.
Math Camp 2011 Page 47

Example: Suppose we want to find the solution set for the following system of equations.
x
y 1
3x
4y
2
We want to find a multiple of one equation that when added to (or subtracted from) the other
equation, it will cancel out one of the variables. Since we have x in equation 1 and 3x in
equation 2, we could multiply equation 1 by the number -3 and then add the two equations
together in order to get rid of the x.
x
y 1
3x
4y
2
(
3)
x ( 3)
y ( 3)(1)
3x
4y
2
3x
3y
3
3x
4y
2
3x
3x
4y
3y
2 ( 3)
7 y 1
1
y
7
Now substitute the solution for y back into the first equation to get x.
x y 1
1
x 1
7
1
x 1
7
x 1
1
7
7
7
1
7
6
7
Therefore the solution set to this system of equations is
6 1
, .
7 7
Systems of Equations with No Solutions
In order for a system of equations to have one unique solution, the systems need to be
independent and consistent. We also need to have as many equations as we have variables, so if
we have two variables, then we need to have two equations in order to solve it. If we have three
variables, then we need to have three equations in order to find the solution.
Math Camp 2011 Page 48

What does it mean to be consistent and independent? Well, dependent equations (opposite of
independent) are equations that are a linear combination of each other. This means that you can
obtain one equation by multiplying the other equation by a non-zero constant.
Example: Find the solution set to the following system of equations.
3x
2y
4
6x
4y
8
Using the addition/subtraction method, we just multiply equation 1 by -2 and add that to the
second equation.
3x
2y
4
6x
4y
8
(
2)(3x)
( 2)(2y)
( 2)(4)
6x
4y
8
6x
4y
8
6x
4y
8
If we were to try to add these two equations together, then not only would x cancel out, but y
would as well. There is no unique solution because there are actually infinite solutions to this
system of equations. These equations actually describe the same exact line, so any point on that
line is a solution. This gives us an infinite solution set.
This is an example of a set of dependent equations.
Inconsistency basically means that there is no solution to the system of equations. There is no
point (x, y) that will simultaneously satisfy both equations. Consider the following system of
equations:
6x
3y
5
2x
y 4
If we were to graph these two lines, we would see that they are parallel and never intersect. If
we solve equation 1 in terms of y, we get:
6x
3y
5
3y
5 6x
5
y 2x
3
Math Camp 2011 Page 49

Now substitute this into equation 2:
2x
y 4
2x
5
2x
3
5
4
3
5
2x
4
3
2x
4
The final result is not true. Therefore this system of equations is inconsistent and we cannot find
a unique solution.
SOLVING INEQUALITIES
An inequality is any equation that uses the symbols
inequalities:
, ,
,
. So the following would all be
a b
ax b 0
5 3
ax
2
5 bx 8
When we solve an inequality, we are trying to find all the values of the variable that make the
statement true. Solving inequalities is very similar to solving other types of equations.
Remember the rule—whatever you do to one side of the equation, you also must do to the other
side of the equation. With inequalities, there are just a few other things to pay attention to.
Some Rules to Remember When Solving Inequalities:
(1) Just like other equations, you can add or subtract the same number from both sides of the
inequality.
(2) When multiplying or dividing an inequality by a negative number, you need to change
the direction of the inequality (from < to >, etc).
(3) When multiplying an inequality by a positive number, you do not need to change the
direction of the inequality.
(4) If you take the reciprocal of both sides of an inequality, then you need to change the
direction of the inequality.
(5) If you square both sides of an inequality, then you do not need to change the sign of the
inequality.
Math Camp 2011 Page 50

Example 1: Solve the following inequality for x.
2x
5 21
2x
21
5
2x
16
16
x 8
2
x 8
Example 2: Solve the following inequality for x.
x
9 1
3
x
1
9
3
x
8
3
x ( 8)(3)
x 24
x 24
Inequalities with No Solution or Infinitely Many Solutions
It is possible when solving an inequality that we cannot find a simple solution to the equation. In
this case there is either no solution or infinitely many solutions, depending on the result of the
problem.
Example 1: Solve this inequality 3(
x 5) 6x
3x
20 .
3( x 5) 6x
3x
20
3x
15
6x
3x
20
3x
15
3x
20
15 20
Because 15 < 20 is true, there are infinitely many solutions to this inequality.
Math Camp 2011 Page 51

Example 2: Solve the inequality
4(
x 1) x 23 3x
.
4( x 1)
x 23 3x
4x
4 x 23 3x
4x
4 4x
23
4 23
Because 4 > 23 is never true, there are no solutions to this inequality.
Solving Quadratic Inequalities
When solving quadratic inequalities, we have to be a little bit more careful. We will need to use
a number line to help us to determine which solutions are correct. Follow these steps when
solving quadratic inequalities:
(1) Write the inequality in standard form with zero on the right hand side.
(2) Factor the inequality on the left hand side.
(3) Set the factors equal to zero and solve for the roots.
(4) Plot these roots on a number line.
(5) For each segment of the number line, determine the signs of the two factors in that
interval.
(6) Find the segments on the number line that makes the original inequality true.
2
Example: Solve the inequality x x 12
.
x
x
2
2
x 12
x 12
0
x
2
x 12
0
( x 4)( x 3) 0
x 4
x 3
(-)(-) (-)(+) (+)(+)
(1)
x < -3
(2)
x = -3 x = 4
-3 < x < 4
(3)
x > 4
Math Camp 2011 Page 52

We plot the two solutions x 3
and x 4 on the number line. These two solutions
divide the number line into three intervals. In the interval on the left, interval (1), we have
x 3 . In this range, the first factor of the equation above, ( x 4)
, would be negative because
if we insert any number less than -3 into the equation, we would get a negative number. In
interval 1, the second factor ( x 3)
would also be negative because we would be adding three to
some number less than -3. So both factors would be negative in this interval. The product of
these two negative numbers would be positive, and that satisfies the original equation that the
product of these two factors be greater than 0. So x 3
is a solution of this equation.
In interval (2), where 3 x 4, we would get a negative factor and a positive factor.
The product of these two factors would be negative, so that does not satisfy the original equation.
Thus, 3 x 4 is not a solution to the inequality.
In interval (3), where x 4 , both factors would be positive. The product of two positive
factors is positive, so x 4 is also a solution to the inequality.
Solving Absolute Value Equations
The following formulas are helpful in working with equations and inequalities that use absolute
values.
Formula 1:
a
a
a
Formula 2:
a
a
Formula 3:
a b
b a
Formula 4:
ab
a b
Formula 5:
2
b
b
2
Formula 6:
a b
a b
Formula 7:
a b
a b
Formula 8:
Formula 9:
true for )
x a is equivalent to x = a or x = -a, where a is any positive number
x a is equivalent to a x a , where a is any positive number (The same is
Formula 10: x a is equivalent to x a
or x a , where a is any positive number (The same
is true for )
Math Camp 2011 Page 53

There are three main steps to follow when solving inequalities that involve an absolute value.
(1) First you will need to get rid of the absolute value sign by rewriting the original equation.
a. If the equation involves an equality (=), such as ax b c then write the original
equation as the two equations ax b c or ax b c
. (By Formula 8)
b. If the equation uses the greater than sign (>), such as ax b c then rewrite the
equation as ax b c or ax b c
. (by Formula 10)
c. If the equation uses a less than sign (

Now check both solutions
y 3 6 2
y 3 6 2
7 3 6 2
1
3 6 2
4 6 2
4 6 2
10 2
10 2
Neither y 7
nor y 1 work as solutions to this problem.
Example 3: 9x
5 4
9x
5 4
9x
4 5
9x
9
x 1
9x
5 4
9x
4
5
9x
1
x
1
9
Example 4: 9x
5 4
4 9x
5 4
4 5 9x
4 5
1 9x
9
1
x 1
9
Taking the Square Root in an Equation
When taking the square root of a variable that is a square, we need to remember to use Rule 5 for
working with exponents (see Day One Class Notes), which says
rules for working with absolute values to finish solving the equation.
2
Example: Solve the equation x 25 0 .
x
x
2
2
x
25 0
25
2
x 5
x 5
25
b
2
b . We can then use our
Math Camp 2011 Page 55

MATH CAMP
School of Public and Environmental Affairs
Day Two In-Class Exercises
FACTORING
1. Factor the given polynomial:
a. 5a 2 25ab
b.
3 3
x n
3x
c.
2
2
4x
29xy
25y
d. x
2 16
e.
3
64a b
SOLVING EQUATIONS
3
2. Find the solutions (roots) to the following equations either directly or by using the
factoring formulas if necessary:
a. 2t
2 11
0
b. 25x
2 85x 30 0
c. x 3 x 5
3. Find the solutions to the following equations by using the quadratic formula:
2
a. x 4x 3 0
2
b. 6y
6 y
4. Find the solutions to the following higher-order equation:
a. 64x
3 125
5. Find three consecutive positive integers, the sum of whose squares is 770.
SOLVING SYSTEMS OF EQUATIONS
6. Find the solution set to the following systems of equations using the substitution method:
a.
x
y 1
2x
y 3
b.
3y
x 5
x
2y
10
7. Find the solution set to the following systems of equations using the addition/subtraction
method. If the system of equations cannot be solved, then indicate if it is dependent or
inconsistent.
4x
y 5 0
a.
3y
12x
15
Math Camp 2011 Page 56

.
c.
2x
4y
5
4x
5y
6
x 3 y 5
7
2 3
x 4 2y
3
2
3 5
SOLVING INEQUALITIES
8. Solve the following inequalities for x:
2
a. x 9 0
b. 17x
15
5
c. 3x 2 9x
9. Solve the following absolute value equation:
a. 2x
7 5
Math Camp 2011 Page 57

MATH CAMP
School of Public and Environmental Affairs
Day Two In-Class Exercises Solutions
FACTORING
1. Factor the given polynomial:
a. 5a
2 25ab
5a(
a 5b)
3 3 3
b. x
n 3x
x ( x
n 3)
2
2
c. 4x
29xy
25y
(4x
25y)(
x y)
2
d. x 16
( x 4)( x 4)
3 3
2
2
e. 64a
b (4a
b)(16a
4ab
b )
SOLVING EQUATIONS
2. Find the solutions (roots) to the following equations either directly or by using the
factoring formulas if necessary:
2
2t
11
0
a.
2t
t
2
2
t
11
11
2
11
2
b.
25x
2
(25x
25x
10
25x
10
10
x
25
85x
30 0
10)( x 3) 0
0
2
5
x 3 0
x 3
Math Camp 2011 Page 58

c.
x 3 x 5
2
x 3 x
5
x 3 ( x 5)( x 5)
x 3 x
x
x
2
2
10x
x 25 3 0
11x
28 0
( x 7)( x 4) 0
x
4, 7
2
5x
5x
25
2
Because we squared the equations, we need to check the roots to see if they are
extraneous or not.
x 3 x 5
4 3 4 5
1 4
1 4
Accordingly, x = 4 is not a root of this equation.
x 3 x 5
7 3 7 5
4 2
2 2
Accordingly, x = 7 is a root of this equation.
3. Find the solutions to the following equations by using the quadratic formula:
2
a. x 4x 3 0
b
x
( 4)
x
4
x
2
b 4ac
2a
( 4)
2(1)
16 12
2
2
4(1)(3)
4
2
4 4 2
1,3
2
Math Camp 2011 Page 59

.
6y
6 y
y
2
2
6y
6 0
b
x
( 6)
x
6
x
2
b 4ac
2a
( 6)
36 24
2
2(1)
6 2 15
3 15
2
4. Find the solutions to the following higher-order equation:
3
64x
125
0
a.
2
(4x
5)(16x
20x
25) 0
4x
5 0
4x
5
x
5
4
2
4(1)( 6)
6 60
2
16x
2
20x
25 0
2
b b 4ac
x
2a
20
x
20
2(16)
4(16)(25)
20 400 1600
x
32
20 20i
3 5 5i
x
32
8
2
3
5. Find three consecutive positive integers, the sum of whose squares is 770.
Let x = the first positive integer
Let x +1 = the second consecutive positive integer
Let x + 2 = the third consecutive positive integer
Then we know that
2
2
x ( x 1)
( x 2)
x
x
2
2
3x
3x
( x 1)(
x 1)
( x 2)( x 2) 770
x
2
2
2
2x
1
x
6x
5 770 0
6x
765 0
2
2
770
4x
4 770
Math Camp 2011 Page 60

x
x
x
x
b
6
6
6
2
b 4ac
2a
(6)
2
2(3)
36 9180
6
9216
6
4(3)( 765)
6 96
15, 17
6
Thus the consecutive positive integers are 15, 16, and 17
SOLVING SYSTEMS OF EQUATIONS
6. Find the solution set to the following systems of equations using the substitution method:
x
y 1
a.
2x
y 3
x y 1
x y
1
Solve the first equation for x
2x
y 3
2( y
1)
y 3
2y
2 y 3
y 1
y 1
Substitute this value for x in the second equation
x y
1
x 11
x 2
Insert this value for y back into the first equation
b.
3y
x 5
x
2y
10
Math Camp 2011 Page 61

3y
x 5
x 5 3y
x 3y
5
Solve the first equation for x
x 2y
10
(3y
5) 2y
10
5y
5
y 1
Substitute this value of x into the second equation
x 3y
5
x 3( 1)
5
x 3
5
Substitute this value of y back into the first equation
x 8
7. Find the solution set to the following systems of equations using the addition/subtraction
method. If the system of equations cannot be solved, then indicate if it is dependent or
inconsistent.
4x
y 5 0
a.
3y
12x
15
This system of equations is dependent because the first equation is equal
to the second equation multiplied by 3.
b.
2x
4y
5
4x
5y
6
2x
4y
5
4x
5y
6
4x
8y
10
4x
5y
6
4x
4x
8y
5y
10
6
3y
4
y
4
3
2x
4y
5
4
2x
4 5
3
6x
16
15
6x
1
1
x
6
Math Camp 2011 Page 62

Math Camp 2011 Page 63
c.
2
5
3
2
3
4
7
3
5
2
3
y
x
y
x
5
70
14
1
69
6
6
5
9
1
6
5
69
6
9
1
6
5
23
2
3
30
9
6
20
5
42
10
2
9
3
30
3)
3(2
4)
5(
42
5)
2(
3)
3(
2
5
3
2
3
4
7
3
5
2
3
x
x
y
y
x
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
4
8
2
23
2
15
23
2
3(5)
23
2
3
y
y
y
y
y
x
SOLVING INEQUALITIES
8. Solve the following inequalities for x:
a.
3
3,
3
9
9
9
0
9
2
2
2
x
x
x
x
x
x
x
b.
17
10
10
17
15
5
17
5
15
17
x
x
x
x

c.
3x
3x
2
2
9x
9x
0
3x(
x 3) 0
3x
0
x 0
x 3 0
x 3
(-)(-) (+)(-) (+)(+)
x = 0 x = 3
Therefore, the solutions to this inequality are x 0 and x 3 .
9. Solve the following absolute value equation:
2x
7 5
2x
7 5
2x
5 7
2x
2
x 1
2x
7 5
2x
5
7
2x
12
x 6
Now we check both solutions.
2( 1)
7 5
2( 6)
7 5
2 7 5
12
7 5
5 5
5 5
5 5
5 5
Both of the solutions work, so we have x 1
and x 6
as solutions to this equation.
Math Camp 2011 Page 64

MATH CAMP
School of Public and Environmental Affairs
Day Two Homework
1. Factor the given polynomial:
3 6
a. 64x y
b.
u
3 3
3v
u 27v
2. Find the solutions to the following equations by using the factoring formulas and the
quadratic formula, if necessary:
a.
2 2
3x 24x
60x
b.
4
2
5x
405x
0
c. 64x
3 8
0
d.
3
a 125
0
3. Two hikers leave town A at the same time and walk by two different routes to town B.
The average speed of one hiker is 1 mile per hour more than the average speed of the
other hiker. The slower hiker reaches town B ½ hour before the faster hiker because the
route taken by the faster hiker is 15 miles long, while the routes taken by the slower hiker
is only 10 miles long. What is the average speed of each hiker? (Hint: distance =
rate*time)
4. Find the solution set to the following system of equations using the addition/subtraction
method. You might have to rearrange the equations to be in the correct format first. If
the system cannot be solved, then indicate whether it is dependent or inconsistent.
a. 3x
9y
18
0
x 3y
6
b.
x y 10
7 5 7
y 14
x
3 3
5. In the United States, normal household voltage is 115 volts. However, it is not
uncommon for actual voltage to differ from normal voltage by at most 5 volts. Express
this situation as an inequality involving an absolute value. Use x as the actual voltage
and solve for x. (Michael Sullivan, College Algebra, Upper Saddle Rive, NJ: Prentice Hall, 2005, p. 140)
Math Camp 2011 Page 65

MATH CAMP
School of Public and Environmental Affairs
Day Two Homework Solutions
1. Factor the given polynomial:
3 6
2 2 2 4
a. 64x
y (4x
y )(16x
4xy
y )
b.
u 3v
u
3
(3v
u)(9v
27v
2
3
(27v
3vu
u
2
3
u
3
) ( 3v
u)
) (3v
u)
(3v
u)
(9v
2
3vu
u
2
) 1
2. Find the solutions to the following equations by using the factoring formulas and the
quadratic formula, if necessary:
a.
3x
3x
2
2
27x
24x
24x
2
2
2
60x
60x
0
3x(9x
20) 0
3x
0
x 0
60x
0
9x
20 0
9x
20
x
20
9
b.
5x
5x
4
2
5x
2
405x
( x
x 0
2
0
2
0
81) 0
x
2
81 0
( x 9)( x 9) 0
x 9, 9
Math Camp 2011 Page 66

c.
64x
3
8 0
(4x
2)(16x
2
8x
4) 0
16x
2
8x
4 0
x
b
2
b 4ac
2a
4x
2 0
4x
2
x
1
2
x
x
x
x
8
8
2
4(16)(4)
2(16)
8 64 256
32
8 192
32
8 8i
3 1
i 3
32 4
d.
a
3
125
0
( a 5)( a
2
5a
25) 0
a
2
5a
25 0
a 5 0
a 5
2
b b 4ac
x
2a
5 25 4(1)(25)
x
2
5 75
x
2
5 5i
3
x
2
3. Two hikers leave town A at the same time and walk by two different routes to town B.
The average speed of one hiker is 1 mile per hour more than the average speed of the
other hiker. The slower hiker reaches town B ½ hour before the faster hiker because the
route taken by the faster hiker is 15 miles long, while the routes taken by the slower hiker
is only 10 miles long. What is the average speed of each hiker? (Hint: distance =
rate*time)
Math Camp 2011 Page 67

Let x be the number of miles per hour in average speed of the slower hiker
Let (x+1) be the number of miles per hour in the average speed of the faster hiker
Given that d rt , we know that the time it takes the faster hiker to reach his
15
destination is . And this should be equal to the time it takes the slower hiker
x 1
10 1
to reach the destination plus ½ hour .
x 2
15 10 1
So
x 1
x 2
Now we just need to solve this equation for x…first we need to clear the
denominators.
15 10 1
x 1
x 2
15
10 1
2x(
x 1)
2x(
x 1)
2x(
x 1)
x 1
x 2
2x(15)
20( x 1)
x(
x 1)
30x
20x
20 x
x
x
2
2
21x
30x
20 0
9x
20 0
( x 5)( x 4) 0
x 4
x 5
2
x
So the average speed of the slower hiker is either 4 or 5 miles per hour. This
means that the average speed of the faster hiker is either 5 or 6 miles per hour.
4. Find the solution set to the following system of equations using the addition/subtraction
method. You might have to rearrange the equations to be in the correct format first. If
the system cannot be solved, then indicate whether it is dependent or inconsistent.
a.
3x
9y
18
0
x
3y
6
This system of equations cannot be solved because the equations are dependent.
The first equation is a multiple of the second equation.
Math Camp 2011 Page 68

.
x y 10
7 5 7
y 14
x
3 3
x y 10
7 5 7
y 14
x
3 3
3x
y 14
5x
7 y 50
3(3) y 14
3x
y 14
y 14 9
5x
7 y 50
y 5
21x
7 y 98
5x
21x
7 y 7 y 50 98
16x
48
x 3
5. In the United States, normal household voltage is 115 volts. However, it is not
uncommon for actual voltage to differ from normal voltage by at most 5 volts. Express
this situation as an inequality involving an absolute value. Use x as the actual voltage and
solve for x. (Michael Sullivan, College Algebra, Upper Saddle Rive, NJ: Prentice Hall,
2005, p. 140)
AV NV 5
where AV = actual voltage and NV = normal voltage
x 115
5
5 x 115
5
5 115
x 5 115
110 x 120
Math Camp 2011 Page 69

MATH CAMP
School of Public and Environmental Affairs
Day Three: Functions and their Graphs
Logarithmic and Exponential Functions
Day Three Class Notes
FUNCTIONS
Definition of a Function
A function describes the relationship between two variables, x and y. A function maps all the
possible x values to the values of y in such a way that each x value is only mapped to one value
of y. We denote a function as y f (x), which means that y is a function of x. We say that x is
the independent variable and y is the dependent variable because the values of y are dependent on
x.
Domain and Range
The domain of a function is all the possible values that we can substitute in the function for x.
The range is all the possible values that we can get for y. Unless stated otherwise, we assume
that the domain of a function is all real numbers that produce a real value on the dependent
variable, y. The range will then be all the values of the dependent variable, y, that we get by
using all the values of the independent variable, x.
Example 1: The function f ( x)
x 2 2 has a domain that includes all real numbers because we
can insert any number in for x and get a real value of y. The range of this function is all positive
numbers greater than or equal to 2, meaning that the y values will range anywhere from 2 to
positive infinity.
Example 2: The function f ( x)
ln x (the natural log of x) has a domain that includes all positive
real numbers greater than 0. We can’t insert 0 in for x because ln 0 is not defined. So 0 is
definitely not in the domain of our function. The range will be all real numbers from negative
infinity to positive infinity.
1
Example 3: The function f ( x)
has a domain of all real numbers except 0, because division
x
by 0 is undefined. The range is all real numbers from negative infinity to positive infinity.
1
Example 4: The function f ( x)
is trickier. The domain of the function will be all
3 x
possible x values such that 3 x 0 because we won’t get real number values of y if we take the
Math Camp 2011 Page 70

square root of a negative number. Luckily, we can solve for the domain of the function as
follows:
3 x 0
x
3
x 3
1
Therefore the domain of f ( x)
is all values of x that are less than -3. The range will be
3 x
all positive numbers greater than 0.
Elementary Functions and their Graphs
Now that we understand the basic principles of functions, we can look at some of the most basic
and common functions that you will encounter. This is a list of these functions, followed by
their basic graphs:
f ( x)
a
Constant Function
f ( x)
x
Identity Function (a special case of the linear function)
2
f ( x)
x
Square Function
3
f ( x)
x
Cube Function
f ( x)
x Square Root Function
f ( x)
x
Absolute Value Function
f
x
( x)
e
Exponential Function
f
f ( x)
mx b
Linear Function
2
f ( x)
ax bx c , a 0
Quadratic Function
( x)
n
x
Nth root Function
n
f ( x)
a x a
a
f ( x)
n
p(
x)
an
x
q(
x)
b x
n1
n 1
x ...
a1x
0
Polynomial Function
n
a
b
m
m
x
( x)
a , 0,
a 1
n1
m1
x
x
n1
m1
... a x a
1
1
0
... b x b
0
Rational Function
f a Exponential Function
f ( x)
log x , b 0,
b 1
Logarithmic Function
b
The graphs of examples of these basic functions are below:
Math Camp 2011 Page 71

f ( x)
3
f ( x)
x
Constant Function
Identity Function
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
3
f ( x)
x
Cube Function
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
2
f ( x)
x
Square Function
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
f ( x)
x
f ( x)
x
Square Root Function
Absolute Value Function
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
Math Camp 2011 Page 72

3
f ( x)
x
f ( x)
2x
1
Nth Root Function
Linear Function
5
4
3
2
1
0
-5 -3 -1 1 3 5
-1
-2
-3
-4
-5
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
2 1 3
f ( x)
x x
2 2
3 2
f ( x)
x x x 1
Quadratic Function
Polynomial Function
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
f
Exponential Function
x
( x)
e
f ( x)
ln( x)
Logarithmic Function
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
Math Camp 2011 Page 73

1
f ( x)
x 3
Rational Function
x
2 2
f ( x)
x
Exponential Function
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
5
4
3
2
1
0
-5 -3 -1
-1
1 3 5
-2
-3
-4
-5
BASICS OF GRAPHING FUNCTIONS
Graphing Ordered Pairs
Graphing is a useful way of presenting and interpreting numbers. Many people find that the
visual presentation of a graph makes the numbers easier to understand. The first place to start
when discussing graphing is to graph ordered pairs. An ordered pair is a pair of numbers ( x,
y)
where the ordering of the numbers is important. In this case, we know that if we see the ordered
pair (3,5), that x = 3 and y = 5 because of the order in which the numbers were presented to us.
The following pair of lines to the right are
called the Cartesian coordinate system, and
we can use this to graph ordered pairs. The
horizontal line represents the values of x and
the vertical line represents the values of y.
So if we wanted to plot the point (2,3) on
the graph, we would go two spaces to the
right on the x (horizontal) axis, and then up
three spaces on the y (vertical) axis and
draw a point. Then we will have plotted the
point (2,3).
Math Camp 2011 Page 74

Y
The graph below shows the plots of the following points: (1,3), (0,8), (-3,0), (-2,-7),
(-1,5), (5,5), (8,-4), (10,0), (9,3), (-3,-9).
Plotted Points
10
8 0, 8
6
-1, 5 5, 5
4
1, 3
9, 3
2
-3, 0 0
10, 0
-10 -5 -2 0 5 10
-4
8, -4
-6
-2, -7
-8
-3, -9 -10
X
Given a scatter plot (as above) of data, we can identify certain types of relationships between x
and y.
1. We have a positive linear relationship when the points fall in an approximate line that
ascends from left to right. In this type of relationship, values of y increase as values
of x increase.
2. We have a negative linear relationship when the points fall in an approximate line that
descends from left to right. In this type of relationship, values of y decrease as values
of x increase.
3. We have a nonlinear relationship when the points fall in some identifiable pattern
than isn’t a straight line. This is often a curve or a horseshoe shape. The type of
curve describes the nature of the relationship.
4. We have no relationship between x and y when the dots don’t fall in any discernible
pattern. It would look like a bunch of points randomly scattered around.
Positive Relationship
8
6
4
2
0
-6 -4 -2 0
-2
2 4 6
-4
-6
Negative Relationship
6
4
2
0
-6 -4 -2 0
-2
2 4 6
-4
-6
-8
Math Camp 2011 Page 75

Nonlinear Relationship
No Relationship
3
2
1
0
-6 -4 -2 -1 0 2 4 6
-2
-3
-4
-5
-6
-7
1.5
1
0.5
0
-6 -4 -2 0 2 4 6
-0.5
-1
-1.5
Transformations of Basic Graphs
We can use a basic function to create a new function by performing a transformation on that
function. For instance, we can add or subtract any number from the function y f (x)
by
transforming the function to y f ( x)
c . We could also transform y f (x)
by adding or
subtracting any number from x within the function, such as y f ( x c)
. Performing these
operations has a predictable impact on the graph of y f (x). These rules are summarized
below:
Vertical Translation Rule 1:
Vertical Translation Rule 2:
y f ( x)
c => Shifts the graph up by c units
y f ( x)
c => Shifts the graph down by c units
Horizontal Translation Rule 1: y f ( x c)
=> Shifts the graph left by c units
Horizontal Translation Rule 1: y f ( x c)
=> Shifts the graph right by c units
Reflection Rule: y f (x)
=> Reflects the graph of y f (x)
over a horizontal line
Flattening Rule: y kf(x)
=> Flattens the graph (by multiplying each y value by k, when
k 1)
Expanding Rule: y kf(x)
=> Expands the graph (by multiplying each y value by k, when
0 k 1)
2
Suppose we have a quadratic function f ( x)
ax b. The graph of a quadratic function is
called a parabola. We will use this parabola to illustrate the graph transformation rules above.
Math Camp 2011 Page 76

2
f ( x)
x
f ( x)
x 2 5
Basic Parabola Vertical Transformation Rule 1
Entire graph shifts upward by 5 units
f ( x)
x 2 5
Vertical Transformation Rule 2
Entire graph shifts downward by 5 units
2
f ( x)
10x
Flattening Rule
Flattens the entire graph
Math Camp 2011 Page 77

1
f ( x)
x
10
2
f ( x)
x
2
Expansion Rule
Reflection Rule
Entire graph is expanded out larger Entire graph is reflected over the line y 0
2
f ( x)
( x 5)
2
f ( x)
( x 5)
Horizontal Transformation Rule 1 Horizontal Transformation Rule 2
Shifts entire graph left by 5 units Shifts entire graph right by 5 units
Math Camp 2011 Page 78

LINEAR FUNCTIONS AND THEIR GRAPHS
A linear function is a special type of function because it has the same slope over its entire
domain. The slope of a linear function is constant. However, in other functions, the slope varies
across the domain of the function.
Now that we can plot single points on a pair of axes, we can now draw the graph of a linear
function (or a line). An equation for a line is an equation in two variables where both variables
are of degree 1. Suppose that we had the following equation for a line:
3x
5y
15
3
y x 3
5
Before we actually graph this line, we should talk about some of its properties. First of all, this
equation tells us that y is a function of x. This function (or equation) tells us how y changes for
different values of x. Just as a reminder, x would be an independent variable and y is the
dependent variable.
In order to graph this line, we need to know the points that it passes through. We can make a
table of the values of x and y that define this line. Then we just plot those points and draw a line
through them.
x y
-10 9
-5 6
0 3
5 0
10 -3
-10, 9
-5, 6
10
8
6
4
2
0, 3
0
5, 0
-15 -10 -5 0 5 10 15
-2
10, -3
-4
Because a line is determined by two points, we actually only need to figure out two points on the
graph, and then we can draw the line connecting them. The easiest way to get these two points is
to substitute 0 in for x and then get the corresponding y value. Then substitute in 0 for y and get
the corresponding x value.
To plot the line 7x 3y
42 , we set x = 0 and find that y = 14 and then we set y = 0 and get x =
6. We can then plot these two points and draw a line through them.
Math Camp 2011 Page 79

20
15
0, 14
10
5
0
6, 0
-7 -5 -3 -1 1 3 5 7
-5
-10
The place where the line crosses the x-axis is called the x-intercept. The place where the line
crosses the y-axis is called the y-intercept.
Finding the Distance between Two Points
Suppose you have two points, called ( x
1,
y1)
and ( x
2,
y2
) . The distance between these two
points is given by the formula:
Distance between x , ) and , )
(
1
y1
(
2
y2
2
x = 2
x
2
x1
y2
y1
Example: Given the points (3,5) and (-7,0), find the distance between them.
2
2
2 2
7 3 0
5 ( 10)
( 5)
100 25 125 5 5
Finding the Midpoint of a Line Segment
This formula gives us the coordinates of the point that is exactly halfway between two other
points, or the midpoint.
Midpoint of line segment =
x x
2
y
,
2
1 2 1
y2
Example: Given the points (3,6) and (-2,4), the midpoint between these two points would be:
x x
2
y
,
2
1 2 1
y2
3 2 6 4 1
, ,5
2 2 2
Math Camp 2011 Page 80

Finding the Slope of a Line
The slope of a line is an important concept in mathematics. It tells us the nature of the
relationship between x and y — whether they are positive or negatively related or unrelated. It
also allows us to quantify this relationship by saying how much y will change when you change x
by one unit.
Given that we have two points called ( x
1,
y1)
and ( x
2,
y2
) , we can find the slope of the line
connecting these two points using the following formula.
m
rise
run
y
x
y
x
2
2
y1
x
1
Example: Given the points (3,5) and (-7,0), find the slope of the line connecting the two points.
rise
m
run
y
x
2
2
y
x
1
1
0 5 5
7 3 10
1
2
Two lines that have the same slope are said to be parallel. If you have two lines and the product
of their slopes is -1 or ( m
1m2
1)
, then the lines are perpendicular (meaning that they are at
right angles to each other). Two lines that have the same slope and y-intercept will be exactly on
top of each other are said to be collinear.
A positive slope means that the x and y are positively related to each other. This means that as x
increases, y will also increase (and vice versa). If the slope is negative, then it means that x and y
have a negative relationship. So as x decreases, y will increase (and vice versa). If the slope of
the line is 0, then there is no relationship between x and y.
Finding the Equation of a Line
Generally, the equation for a line is of the form ax by c , where and a and b are both not
equal to zero. However, there are two other ways to write the equation for a line that are more
helpful to us. The first is called the point-slope form, which follows this format:
y y m( x 1) where
1
x
m
rise
run
y
x
2
2
y1
x
1
The second is called the slope-intercept form of the line, which takes the form y mx b , where
m is the slope of the line and b is the y-intercept (or the point on the y-axis that the line passes
through).
Math Camp 2011 Page 81

Given any two points, we can not only determine the distance between the two points and the
slope of the line connecting the two points, but we can also determine the equation for the line
itself.
Example 1: Find the equation of the line through the two points (2,1) and (4,7). In order to do
this, we first compute the slope.
rise
m
run
y
x
2
2
y
x
1
1
7 1
6
3
4 2 2
Then the equation for the line would be:
y y
1
y 1
3( x 2)
y 3x
6 1
y 3x
5
m( x x ) 1
We can also use a line’s equation to give us information about the line itself.
Example 2: Suppose you are given the equation for a line 7x
21y 3 0 . Determine the
slope.
In order to determine the slope, we need to first put the equation in slope-intercept form.
7x
21y
3 0
21y
7x
3
1
y x
3
The slope of the line is the coefficient on x, which is
1
7
1
.
3
Economics Application Problem
(This application problem was taken in its entirety from Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences,
and Social Sciences, tenth edition, Upper Saddle River, NJ: Prentice Hall, pp. 46-47)
At the beginning of the twenty-first century, the world demand for crude oil was about 75
million barrels per day and the price of a barrel fluctuated between $20 and $40. Suppose that
the daily demand for crude oil is 76.1 million barrels when the price is $25.52 per barrel and this
demand drops to 74.9 million barrels when the price rises to $33.68. Assuming a linear
relationship between the demand x and the price p, find a linear function in the form p ax b
that models the price-demand relationship for crude oil. Use this model to predict the demand if
the price rises to $39.12 per barrel.
Math Camp 2011 Page 82

Find the equation of the line through (76.1, 25.52) and (74.9, 33.68). We first find the slope of
the line:
m
y
x
2
2
y
x
1
1
33.68 25.52 8.16
6.8
74.9 76.1 1.2
Use the point-slope form to find the equation of the line:
p p
1
m( x x ) 1
p 25.52 6.8(
x 76.1)
p 25.52 6.8x
517.48
p 6.8x
543
To find the demand when the price is $39.12 per barrel, we solve the equation p 39. 12 for x:
p 39.12
6.8x
543 39.12
6.8x
503.88
503.88
x 74.1 million barrels per day
6.8
The daily supply for crude oil also varies with the price. Suppose that the daily supply is 73.4
million barrels when the price is $23.84 and this supply rises to 77.4 million barrels when the
price rises to $34.24. Assuming a linear relationship between the supply x and the price p, find a
linear function in the form p ax b that models the price-supply relationship for crude oil.
Use this model to predict the supply if the price drops to $20.98 per barrel.
Find the equation of the line through (73.4, 23.84) and (77.4, 34.24). We first find the slope of
the line:
m
y
x
2
2
y
x
1
1
34.24 23.84 10.4
2.6
77.4 73.4 4
Use the point-slope form to find the equation of the line:
p p
1
m( x x ) 1
p 23.84 2.6( x 73.4)
p 23.84 2.6x
190.84
p 2.6x
167
Math Camp 2011 Page 83

To find the demand when the price drops to $20.98 per barrel, we solve the equation p 20. 98
for x.
p 20.98
2.6x
167
20.98
2.6x
187.98
187.98
x
2.6
x 72.4 million barrels per day
In a free competitive market, the price of a product is determined by the relationship between
supply and demand. The price tends to stabilize at the point of intersection of the demand and
supply functions. This is the equilibrium point.
The equilibrium point for the supply and demand functions is found as follows:
2.6x
167
6.8x
543
9.4x
710
710
x
9.4
x 75.532 million barrels (equilibrium quantity)
p 2.6(75.532)
167
$29.38 (equilibrium price)
QUADRATIC FUNCTIONS AND THEIR GRAPHS
2
Quadratic functions, such as y ax bx c , are also common functions that you will
encounter in economics and other classes. Knowing the properties of quadratic functions can
help us to solve many important problems. As mentioned earlier, the graph of a quadratic
function is called a parabola. The domain of a quadratic function is all real numbers. The range
of a quadratic function depends on the values of a, b, and c.
Just like with a line, it is helpful to know the x- and y-intercepts of a parabola. We can easily
find the y-intercept by substituting in 0 for x. We can find the x-intercepts by substituting 0 in
for y and solving for x. As we learned yesterday, a quadratic function will have two roots, so it
can potentially cross the x-axis in two places.
2
Example: Find the x- and y-intercepts for the following quadratic function y x 2x
15.
y x
y 0
2
2
y 15
2x
15
2 0 15
Math Camp 2011 Page 84

y x
0 x
2x
15
2x
15
0 ( x 5)( x 3)
x 3, 5
A graph of this parabola would look like this:
2
2
10
-1 0123456789
-9 -4 -3
-2
1 6
-4
-6
-5
-10
-9
-7
-8
-12
-11
-14
-13
-15
-17
-16
-18
Notice that the parabola crosses the y-axis at -15 and the x-axis at -5 and 3.
Generally, the graph of a quadratic function is a parabola. In most cases, the parabola will have
a line of symmetry that is parallel to the vertical axis. The vertex of a parabola is the lowest or
highest point on the parabola. The highest point on a parabola is called the maximum, and this
occurs in parabolas that open downward. The lowest point on a parabola is called the minimum
and this occurs on parabolas that open upward. The maximum and minimum always occur at the
vertex of the parabola. The line of symmetry that goes though the vertex and is parallel to the
vertical axis is called the axis of the parabola.
Finding the minimum or maximum value of a quadratic function can help us to solve many
important real-world problems. For instance, a company’s revenue function is a quadratic
equation. If we want to find the maximum revenue, all we have to do is find the maximum of the
parabola. Because we know that the maximum or minimum value will always occur at a
parabola’s vertex, all we have to do is find the vertex of the parabola. (When we learn about
derivatives, we will see another way to easily determine the maximum or minimum point of a
parabola.)
Math Camp 2011 Page 85

If we have a quadratic function of the form,
b
will be found at , f
2a
b
2a
y ax
2
bx c , then the vertex of the parabola
. We know that the vertex will be a minimum value if a 0
and it will yield a maximum value if a 0. The axis of symmetry will be the line
b
x .
2a
Example: Find the vertex of this parabola and determine whether it is a maximum or minimum
2
point. f ( x)
2x
12x
9
b
a
2
12
12
3
2
2 4
b
f
2a
f
2
3 23
12
3
9 18 36 9 9
The vertex for this parabola is the point ( 3, 9)
and because a 2 0, this is a minimum point
on the parabola.
POLYNOMIAL AND RATIONAL FUNCTIONS
Polynomial Functions
Remember that a polynomial function will be of the form:
n
f ( x)
a x a a x a
n
n1
n
1
x ...
1
0
We have already discussed some special cases of the polynomial function, such as straight lines
and parabolas. The domain of a polynomial function is the set of all real numbers.
When we graph higher-level polynomial functions, we will see that the degree of the polynomial
is related to the graph of the polynomial. Even-degree polynomials have similar shapes, and
odd-degree polynomials will also have similar graphs. Below are some graphs of polynomial
functions.
Math Camp 2011 Page 86

2
f ( x)
2x
1
f ( x)
x 3x
2. 5
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
f ( x)
x
3 4x
5
4
3
2
1
0
-5 -3 -1 -1 1 3 5
-2
-3
-4
-5
f ( x)
2x
4
4x
5
4
3
2
1
0
-2
-3
-4
-5
2
2x
1
2
-5 -3 -1 1 3 5
-1
5 3
6 4 2
f ( x)
x 5x
4x
1
f ( x)
x 7x
14x
x 5
5
5
4
3
2
1
0
-5 -3 -1 1 3 5
-1
-2
-3
-4
4
3
2
1
0
-5 -3 -1 1 3 5
-1
-2
-3
-4
-5
-5
Math Camp 2011 Page 87

As you look at these graphs, the graphs on the left hand side are of odd degree and the graphs on
the right hand side are of even degree. The graphs of the odd degree polynomials all begin
negative and end positive and they all cross the x-axis at least once. The graphs of the even
degree polynomials all begin positive and end positive and have at least one turn in the graph.
This is what is called a turning point. A turning point is a place on a continuous graph that
separates a section where the function is increasing from a section where the function is
decreasing.
The graph of a polynomial function of degree n will have at most n-1 turning points and can only
cross the x-axis at most n times.
Rational Functions
Remember that a rational function is of the form:
f ( x)
p(
x)
q(
x)
an
x
b x
m
n
m
a
b
n1
m1
x
x
n1
m1
... a x a
1
1
0
... b x b
0
The domain of a rational function is all real numbers except those that make the denominator
equal to zero.
We can find the x- and y-intercepts for a rational function in the same way that we find the
intercepts for other functions.
Many times a graph of a rational function will not be continuous, meaning that it has no holes or
breaks in the graph. Because of this, many graphs of rational functions can have vertical or
horizontal asymptotes. A vertical asymptote helps up to describe the behavior of the graph of the
function when it gets close to a certain number. The graph will never cross over an asymptote.
1
The graph below is a graph of the rational function f ( x)
. The graph has a vertical
x 3
5
asymptote at the line x 3. Notice how the graph
4
of the function gets closer and closer to the line
x 3 but never crosses it.
3
2
1
0
-5 -3 -1 1 3 5
-1
We can easily find the vertical asymptotes of a
rational function by solving for all the values of x
that make the denominator equal to zero.
-2
The graph above has a horizontal asymptote at the
-3
line y 0 . Horizontal asymptotes can be found
-4
by dividing each term of the numerator and
-5
denominator by the highest power of x in the
function. Then you look at what happens to the
graph as you increase and decrease the values of x.
Math Camp 2011 Page 88

1
For instance, in our example f ( x)
, we divide all the terms in the numerator and
x 3
1
denominator by x, because that is the highest power of x in the function. So we get f ( x)
x
3
1
x
As we increase the values of x toward positive infinity, the values of x
1 and x
3 will get closer
and closer to zero. This makes the values of the entire function get closer and closer to zero (or
zero divided by 1). As we decrease the values of x toward negative infinity, we see that the
function will once again approach zero. This indicates that we have a horizontal asymptote at
y 0 .
EXPONENTIAL FUNCTIONS
Exponential functions have a wide variety of real-world applications, so they are very useful to
learn. We can apply exponential functions to such things as the growth of money in your bank
account to radioactive decay to the growth of an endangered species to how fast someone can
learn to use a computer.
Basics of the Exponential Function
As you might guess, the exponential function is just a function of x where x is the exponent of
some number (or expression), as opposed to raising x to some exponent. So the exponential
function is given by:
f ( x)
for any b > 0 and b 1. We exclude b = 1 because 1 raised to any power is equal to 1, which is
just the constant function. We also want b 0 because we want to avoid imaginary numbers.
The domain of the exponential function is all real numbers, and the range of the exponential
function is the set of all positive numbers.
x
b
The graphs of the functions
f 2
x
( x)
and
f
x
( x)
2 are below.
f ( x)
2
x
Math Camp 2011 Page 89

10
9
8
7
6
5
4
3
2
1
0
-10 -5 0 5 10
f ( x)
1
x
2
2
x
10
9
8
7
6
5
4
3
2
1
0
-10 -5 0 5 10
Notice a couple of things about these graphs. First of all, these two graphs are reflections of each
x
other across the y-axis. Second, the graph of f ( x)
2 is the same thing as the graph of
x
1
f ( x)
. We can summarize the properties of these graphs as follows:
2
x
Basic Properties of the Exponential Function, f ( x)
b , where b 1
x
(1) The domain of f ( x)
b is all real numbers
x
(2) The range of f ( x)
b is the set of all positive numbers
Math Camp 2011 Page 90

x
(3) The graph of f ( x)
b will always pass through the point (0,1) because b 0 1 for any
base b. Thus, the y-intercept is 1.
x
(4) The graph of f ( x)
b is continuous, meaning that there are no breaks or holes in the
graph.
x
(5) The x-axis is a horizontal asymptote for the graph of f ( x)
b . Because the x-axis is a
horizontal asymptote, there are no x-intercepts.
x
(6) The graph of f ( x)
b increases as x increases.
x
The graph of f ( x)
b when 0 b 1
has the same properties as when b 1 except that
f
x
( x)
b decreases (rather than increases) as x increases.
The Base e
There is a special number that often comes up in mathematical applications. It is the number e,
which is an irrational number (so when written out in decimal form its decimals continue to
eternity). When written out to several decimal places, we have e 2. 7182818 . When we want
x
x
to write a function using e, we usually write exp( x)
e . The graph of exp( x)
e is below and
x
is very similar to the graph of f ( x)
b (because in this caseb e 2. 7182818.
10
9
8
7
6
5
4
3
2
1
0
-10 -5 0 5 10
Exponential Probability Application Problem
(This application problem was taken from Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, pp. 422-423)
Between 9:00pm and 10:00pm cars arrive at Burger King’s drive-thru at the rate of 12 cars per
hour (0.2 car per minute). The following formula from the field of probability can be used to
determine the probability that a car will arrive within t minutes of 9:00pm.
F(
t)
1
e
0.2t
(a) Determine the probability that a car will arrive within 5 minutes of 9:00pm (before
9:05pm).
(b) Determine the probability that a car will arrive within 30 minutes of 9:00pm (before
9:30pm).
(c) What value does F approach as t becomes unbounded in the positive direction?
(d) Within how many minutes of 9:00pm will the probability of a car arriving equal 50%?
Math Camp 2011 Page 91

Solutions:
a) The probability that a car will arrive within 5 minutes is found by evaluating F (t)
at
t 5
0.2(5)
F ( t)
1
e 0.63212
We conclude that there is a 63% probability that a car will arrive within 5 minutes.
b) The probability that a car will arrive within 30 minutes is found by evaluating F (t)
at
t 30
0.2(30)
F ( t)
1
e 0.9975
There is a 99.75% probability that a car will arrive within 30 minutes.
c) As time passes, the probability that a car will arrive increases. The value that F
approaches can be found by letting
as
t . Thus, F approaches 1 as t gets large.
.2t
t . Since e
0. 2t
0 1
e
0.2t
, it follows that e 0
d) Within 3.5 minutes of 9:00pm, the probability of a car arriving equals 50% .
Formulas for Common Applications of Exponential Functions
Exponential Growth and Decay
A population experiencing exponential growth (or growth compounded continuously) can be
modeled using the formula:
rt
A A0e
where A = the size of the population in the future
A
0
= the size of the population at time t 0
r = the rate of population growth (as a decimal)
t = time in years
Exponential decay can be modeled using a similar formula:
A
0
A e
rt
where A = amount in the future
A
0
= amount at time t 0
r = the rate of decay (as a decimal)
t = time in years
Compound Interest
Math Camp 2011 Page 92

A P
1
r
m
mt
where A = future value of the account
P = the principal or present value of the account
r = annual rate (as a decimal)
m = the number of times per year the interest is compounded (for instance, for interest
compounded quarterly m 4)
Continuous Compound Interest
rt
A Pe
where A = future value of the account in t years
P = the principal or present value of the account or amount invested
r = rate (as a decimal)
t = number of years
LOGARITHMIC FUNCTIONS
Basic Information about Logarithmic Functions
The logarithmic function is the inverse of the exponential function (with base b). This function
is denoted f ( x)
log
b
x , where b 0 andb
1. The domain of this function is the set of
positive numbers and the range is the set of real numbers.
6
5
4
3
2
1
0
-2
-1
0 2 4 6 8 10
4
-2
3
-3
2
-4
1
0
-2
-1
0 2 4 6 8 10
-2
The graph to the left is of the function
f ( x)
log x , where b 1.
b
Notice that the graph of f ( x)
log x is a
x
reflection of the graph f ( x)
b across the
line y x .
The graph to the left is of the function
f ( x)
log x , where 0 b 1.
b
b
-3
Math Camp 2011 Page 93
-4
-5
-6

Notice that this second graph (where 0 b 1) is a reflection of the original graph (where b 1)
across the x-axis.
Basic Properties of the Logarithmic Function, f ( x)
log x , where b 1
(1) The domain of f ( x)
log x is all real numbers
b
(2) The range of f ( x)
log x is the set of all positive numbers
b
(3) The graph of f ( x)
log
b
x will always pass through the point (1,0) because log
b
1 0
for any base b. Thus, the x-intercept is 1.
(4) The graph of f ( x)
log
b
x is continuous, meaning that there are no breaks or holes in
the graph.
(5) The y-axis is a vertical asymptote for the graph of f ( x)
log
b
x . Because the y-axis is a
horizontal asymptote, there are no y-intercepts.
(6) The graph of f ( x)
log x increases as x increases.
b
The graph of f ( x)
log x when 0 b 1
has the same properties as when b 1 except that
b
f ( x)
log x decreases (rather than increases) as x increases.
b
The number e is also special number when working with logarithms as well. The inverse of
x
f ( x)
e is the function f ( x)
log
e
x , which is commonly shortened to f ( x)
ln x . This is
known as the natural log of x. The natural log behaves the same as any other log function.
Relationship between Logarithms and Exponents
x
The following equations provide an illustration of the relationship between f ( x)
b and
f ( x)
log x . (Remember that these two functions are inverses of each other)
b
3 2 9 is equal to log 3
9 2
b
Math Camp 2011 Page 94

1 1
2
9
1
3
is equal to
log
1
3
1
9
1
1
10 3 is equal to log 10
3
1000
1000
5 0 1 is equal to log 5
1 0
1
2
Therefore, whenever we see a logarithmic function, we can express it as an exponential function
and vice versa.
Example 1: If we have log 9
81 2 , we know that 9 2 81.
Example 2: If we have 2 10 1024 , we can also express it as log 2
1024 10 .
In addition to this, whenever we have the logarithmic function f ( x)
log
b
x we can also solve
for either x or b, which can be helpful in solving many equations.
Example 1: Solve log
b
49 2
for b.
b
2
b
49
1
2
2
1 1
b
2
49
1
b
49
1
b
7
49
1
2
Example 2: Solve log 6
x 4for x.
x
6 4
x 1296
Rules for Manipulating Logarithms
Just as we had several rules for working with exponents, we have several rules that can help use
to manipulate equations that contain logarithms.
Rule 1:
Rule 2:
Rule 3:
log
b
b y
b
log b x
log
b
y
x
xy log
b
x log
b
y
Math Camp 2011 Page 95

Rule 4:
Rule 5:
log
b
x
y
log
n
log x nlog
b
b
x log
b
x
b
y
Rule 6: log
b
1 0 for all bases b
Rule 7: log
b
b 1 for all bases b
Rule 8:
Rule 9:
1
log
b
log b
n
n
log
n
b
x log
b
x
1
n
1
log
n
b
x
Rule 10:
log
log
b
x
log
n
n
x
b
2 3
5
Example 1: log 32 log 48
log 2 2 log 2 5 by Rule 1
2 2
2
2
Example 2: log
3
243 log
3
9
27 log
3
9 log
3
27 2 3 5 by Rule 3
25
Example 3: log
5
log
5
25 log
5
125 2 3 1
by Rule 4
125
1
This is equivalent to log
5
log
5
1
log
5
5 0 1
1
5
Example 4: Given that log 10
2 0. 3010 and log 10
3 0. 4771, evaluate the following.
log
10
6 log
10
23
log
10
2 log
10
3 0.3010 0.4771 0.7781
5
Example 5: log
10
32 log
10
2 5log
10
2 5(0.3010)
1.
505 by Rule 5
1 1 3
log
2
2
2
3
by Rule 5
2 2 2
Example 6: 8 log 8 log
8
1
2
Logs can also be helpful when we are working with very large and very small numbers that can
be expressed in scientific notation.
Example:
log
10
log
623 log
10
10
6.23 2log
6.2310
10
2
10 log
log
10
10
(6.23) log
10
10
6.23 2 0.7945 2 2.7945
2
Math Camp 2011 Page 96

Alcohol and Driving Application Problem
(This application problem was taken from Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, pp. 435-436)
The concentration of alcohol in a person’s blood is measurable. Recent medical research
suggests that the risk R (given as a percent) of having an accident while driving a car can be
modeled by the equation
kx
R 6e
where x is the variable concentration of alcohol in the blood and k is a constant.
(a) Suppose that a concentration of alcohol in the blood of 0.04 results in a 10% risk ( R 10
) of an accident. Find the concentration k in the equation.
(b) Using this value of k, what is the risk if the concentration is 0.17?
(c) Using the same value of k, what concentration of alcohol corresponds to a risk of 100%?
(d) If the law asserts that anyone with a risk of having an accident of 20% or more should not
have driving privileges, at what concentration of alcohol in the blood should a driver be
arrested and charged with a DUI (Driving While Under the Influence)?
Solutions:
(a) For a concentration of alcohol in the blood of 0.04 and a risk of 10%, we let x 0. 04and
R 10 in the equation and solve for k.
kx
R 6e
10 6e
k (0.04)
10 k (0.04)
e
6
10
0.04k
ln 0.510826
6
k 12.77
(b) Using k = 12.77 and x = 0.17 in the equation, we find the risk R to be
rx (12.77)(0.17)
R 6e
6e
52.6
For a concentration of alcohol in the blood of 0.17, the risk of an accident is about 52.6%.
(c) Using k = 12.77 and R = 100 in the equation, we find the concentration x of alcohol in the
blood to be
kx
R 6e
100 6e
12.77x
100 12.77x
e
6
100
12.77x
ln 2.8134
6
x 0.22
For a concentration of alcohol in the blood of 0.22, the risk of an accident is 100%.
Math Camp 2011 Page 97

(d) Using k = 12.77 and R = 20 in the equation, we find the concentration x of alcohol in the
blood to be
kx
R 6e
20 6e
12.77x
20 12.77x
e
6
20
12.77x
ln 1.204
6
x 0.094
A driver with a concentration of alcohol in the blood of 0.094 or more (9.4%) should be
arrested and charged with a DUI.
Math Camp 2011 Page 98

MATH CAMP
School of Public and Environmental Affairs
Day Three In-Class Exercises
FUNCTIONS
1. Find the domain and range of the following functions:
2
a. f ( x)
x 3x
5
b. f ( x)
3x
12
BASICS OF GRAPHING FUNCTIONS
2. Plot the points (1,9), (-2,-6), (-4, 10), (0, 7), (9,1), (-5,0), (5,-7) and (10,3) on a graph.
3. Write the function whose graph is the graph of
a. Shifted to the right by 8 units
b. Shifted to the left by 3 units
c. Shifted up by 4 units
d. Shifted down by 10 units
e. Reflected about the x-axis
f. Flattened by a factor of 3
g. Expanded by a factor of .5
4
y x , but is:
LINEAR FUNCTIONS AND THEIR GRAPHS
4. Graph the following lines:
a. 4x
6y 9 0
b. 5x
2y
0
5. Draw a graph of the line connecting these two points and find the length of the line.
a. (0,4), (0,-4)
6. What is the midpoint of the two points?
a. (0,3), (5,1)
b. (-3,-7), (-2,9)
7. Draw a graph of the line connecting these two points and find the slope of the line.
a. (4,-1), (-2,3)
b. (-5,3), (1,-3)
8. Find an equation of the line through the two given points. Graph the line.
a. ( 0,0),(6,2)
Math Camp 2011 Page 99

9. Find an equation of the line through the given point and with the given slope. Graph the
line.
1
a. ( 0,6), m
4
QUADRATIC FUNCTIONS AND THEIR GRAPHS
10. Graph the following equations:
a. f ( x)
( x 6)
2 9
b. f ( x)
5x
5
11. Find the x- and y-intercepts of the following functions:
2
a. f ( x)
x 5x
14
12. Suppose that you are given the following equation for a demand curve for widgets. In
this case, Q = the quantity of widgets demanded and P = the price of widgets.
Q 8.5
.
05P
a. Solve this equation for P.
b. Graph the equation for the line that you found in Part A.
c. Suppose that you know that revenue is the price multiplied by the quantity, or
R P Q
. Using the equation for P that you found in Part A, determine the
revenue function.
d. Graph the revenue function from Part C. (Revenue will go on the vertical axis)
e. If the quantity sold was 2 units, what would the price be? What would the
revenue be?
f. Just by looking at this graph, at approximately what quantity will revenue be the
greatest?
POLYNOMIAL AND RATIONAL FUNCTIONS
13. Find the domain and range of the following function:
x 3
a. f ( x)
x 6
14. Graph the following equations:
a. f ( x)
x 3 3
b. f ( x)
x 4
1
c. f ( x)
x 5
15. Find the x- and y-intercepts of the following functions:
a. f ( x)
x 3 8
Math Camp 2011 Page 100

EXPONENTIAL FUNCTIONS
16. Graph the following functions:
a.
b.
1
f ( x)
4
x
f ( x)
4
x
0.06t
17. In a certain culture, if A is the number of bacteria present at t minutes, then A ke
where k is a constant. If there are 1000 bacteria present initially, how many bacteria will
be present after 1 hour has elapsed?
LOGARITHMIC FUNCTIONS
18. Find the value of the following logarithms:
a. log 5
25
b. log 2
16
19. Solve the following equations for x or b, respectively.
a. log 5
x 2
b.
2
log 27
x
3
c.
1
log
b
5
3
d. log
10
x log
10(
x 15)
2
20. Simplify these logarithmic expressions:
a.
2
4log
b
x log
b
y log
b
z
3
b.
3 4 2
log
b
x log
b
y log
b
z
5 5 5
21. Find the following logarithms given that log 10
2 0. 3010 , log 10
3 0. 4771, and
log 10
7 0.8451:
a. log 7,000
b. log .00007
c. log 1400
d. log .0003
22. Between 12:00pm and 1:00pm, cars arrive at Citibank’s drive-thru at the rate of 6 cars
per hour (0.1 car per minute). The following formula from statistics can be used to
determine the probability that a car will arrive within t minutes of 12:00pm.
0.1t
F(
t)
1
e
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 440)
a. Determine how many minutes are needed for the probability to reach 50%
b. Determine how many minutes are needed for the probability to reach 80%
Math Camp 2011 Page 101

y
MATH CAMP
School of Public and Environmental Affairs
Day Three In-Class Exercises Solutions
FUNCTIONS
1. Find the domain and range of the following functions:
2
a. f ( x)
x 3x
5 Domain = all real numbers, Range = all real numbers
b. f ( x)
3x
12
Domain = x 4, Range = y 0
BASICS OF GRAPHING FUNCTIONS
2. Plot the points (1,9), (-2,-6), (-4, 10), (0, 7), (9,1), (-5,0), (5,-7) and (10,3) on a graph.
Answers to #1
12
-4, 10 10
1, 9
8
0, 7
6
4
10, 3
2
9, 1
-5, 0
0
-6 -4 -2 -2 0 2 4 6 8 10 12
-4
-2, -6 -6
-8
5, -7
x
3. Write the function whose graph is the graph of
a. Shifted to the right by 8 units
b. Shifted to the left by 3 units
4
y x , but is:
f ( x)
( x 8)
f ( x)
( x 3)
c. Shifted up by 4 units f ( x)
x 4 4
d. Shifted down by 10 units f ( x)
x 4 10
e. Reflected about the x-axis
f. Flattened by a factor of 3
g. Expanded by a factor of .5
LINEAR FUNCTIONS AND THEIR GRAPHS
f ( x)
x
Math Camp 2011 Page 102
4
4
f ( x)
3x
1
f ( x)
x
2
4
4
4

4. Graph the following lines:
a. 4x
6y 9 0
10
8
6
4
2
0
-15 -10 -5 0 5 10 15
-2
b. 5x
2y
0
-4
-6
30
20
10
0
-15 -10 -5 0 5 10 15
-10
-20
-30
5. Draw a graph of the line connecting these two points and find the length of the line.
a. (0,4), (0,-4)
5
4
3
2
1
0
-1 -0.5 -1 0 0.5 1
-2
-3
-4
-5
6. What is the midpoint of the two points?
2
2
2 2
0
0
4 4 (0) ( 8)
64 8
Math Camp 2011 Page 103

a. (0,3), (5,1)
x x
2
y
,
2
1 2 1
y2
b. (-3,-7), (-2,9)
x x
2
y
,
2
1 2 1
y2
0 5 3 1
5
, ,2
2 2 2
3 2 7 9
,
2 2
5
,1
2
7. Draw a graph of the line connecting these two points and find the slope of the line.
a. (4,-1), (-2,3)
3.5
3
2.5
2
1.5
1
0.5
0
-3 -2 -1 -0.5 0 1 2 3 4 5
-1
-1.5
m
rise
run
y
x
2
2
y
x
1
1
3 ( 1)
2 4
4
6
2
3
Math Camp 2011 Page 104

. (-5,3), (1,-3)
4
3
2
1
0
-6 -5 -4 -3 -2 -1 0
-1
1 2
-2
-3
-4
rise
m
run
y
x
2
2
y
x
1
1
3 3
1
( 5)
6
1
6
8. Find an equation of the line through the two given points. Graph the line.
a. ( 0,0),(6,2)
m
rise
run
y
x
2
2
y
x
1
1
2 0
6 0
2
6
1
3
y y
1
y 0
1
y
3
m( x x1)
x
1
(
3
x 0)
4
3
2
1
0
-15 -10 -5 0
-1
5 10 15
-2
-3
-4
Math Camp 2011 Page 105

9. Find an equation of the line through the given point and with the given slope. Graph the
line.
1
a. ( 0,6), m
4
y y m( x x1)
1
1
y 6 ( x 0)
4
1
y x 6
4
9
8
7
6
5
4
3
2
1
0
-15 -10 -5 0 5 10 15
QUADRATIC FUNCTIONS AND THEIR GRAPHS
10. Graph the following equations:
a. f ( x)
( x 6)
2 9
300
250
200
150
100
50
0
-20 -10 0 10 20 30
Math Camp 2011 Page 106

. f ( x)
5x
5
50
40
30
20
10
0
-15 -10 -5 -10 0 5 10 15
-20
-30
-40
-50
11. Find the x- and y-intercepts of the following functions:
2
a. f ( x)
x 5x
14
x-intercepts => x 7, 2
y-intercept => -14
12. Suppose that you are given the following equation for a demand curve for widgets. In
this case, Q = the quantity of widgets demanded and P = the price of widgets.
Q 8.5
.
05P
a. Solve this equation for P.
Q 8.5 .05P
.05P
Q 8.5
Q 8.5
P
.05 .05
P 20Q
170
b. Graph the equation for the line that you found in Part A.
400
300
200
100
0
-15 -10 -5 0 5 10 15 20
-100
-200
Math Camp 2011 Page 107

c. Suppose that you know that revenue is the price multiplied by the quantity, or
R P Q
. Using the equation for P that you found in Part A, determine the
revenue function.
R P Q
R ( 20Q
170)
Q
R 20Q
2
170Q
d. Graph the revenue function from Part C. (Revenue will go on the vertical axis)
1000
0
-15 -10 -5 0 5 10 15 20 25
-1000
-2000
-3000
-4000
-5000
-6000
e. If the quantity sold was 2 units, what would the market clearing price be? What
would the revenue be?
P 20Q
170
P 20(2)
170
P 40
170
P 130
R 20Q
R 20(2)
R 20(4)
340
R 80
340
R 260
2
170Q
2
170(2)
f. Just by looking at this graph, at approximately what quantity will revenue be the
greatest?
Somewhere around 4 widgets
Math Camp 2011 Page 108

POLYNOMIAL AND RATIONAL FUNCTIONS
13. Find the domain and range of the following function:
x 3
a. f ( x)
Domain = all real numbers except x 6 (vertical asymptote),
x 6
range = all real numbers except y 1 (horizontal asymptote)
14. Graph the following equations:
a. f ( x)
x 3 3
50
40
30
20
10
0
-15 -10 -5 -10 0 5 10 15
-20
-30
-40
-50
b. f ( x)
x 4
20
15
10
5
0
-20 -10 0 10 20
-5
Math Camp 2011 Page 109

c.
1
f ( t)
t 5
5
4
3
2
1
0
-10 -5 -1 0 5
-2
-3
-4
-5
15. Find the x- and y-intercepts of the following functions:
a. f ( x)
x 3 8
x-intercepts => x 2
y-intercept => 8
EXPONENTIAL FUNCTIONS
16. Graph the following functions:
1
a. f ( x)
4
x
10
9
8
7
6
5
4
3
2
1
0
-10 -5 0 5 10
Math Camp 2011 Page 110

.
f ( x)
4
x
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
-10 -5 0 5 10
0.06t
17. In a certain culture, if A is the number of bacteria present at t minutes, then A ke
where k is a constant. If there are 1000 bacteria present initially, how many bacteria will
be present after 1 hour has elapsed?
At t = 0, we have 1,000 bacteria. We can substitute this into our equation to solve for k.
A ke
0.06t
1000 ke
1000 ke
1000 k
0.060
0
We want to find the number of bacteria present at t = 60 minutes (which is one hour).
A ke
0.06t
A 1000e
A 1000e
3.6
A 1000(36.598)
A 36,598
0.0660
LOGARITHMIC FUNCTIONS
18. Find the value of the following logarithms:
a. log 5
25 2
b. log 2
16 4
Math Camp 2011 Page 111

19. Solve the following equations for x or b, respectively.
a. log x 2
5
2
5
x
x 25
b.
log
27
2
3
27
x 9
x
x
2
3
c.
d.
log
b
1
3
b
5
5
3
1
3
1
3
b
5
b 125
log
log
10
100 x
x
2
2
10
10
( x 20)( x 5) 0
x 20
x 5
3
x log
x ( x 15)
2
x ( x 15)
2
10
15x
15x
100
0
( x 15)
2
20. Simplify these logarithmic expressions:
a.
4log
b
log
log
b
b
2
x log
3
x
x
4
y
4
log
z
2
3
b
b
y log
y
2
3
b
log
b
z
z
Math Camp 2011 Page 112

.
3
log
5
log
log
log
log
b
b
b
b
4
x log
5
x
5
b
3
5
3
5
x y
2
5
z
3
x y
2
z
log
4
5
4
3
x y
2
z
4
b
1
5
b
2
y log
5
y
4
5
log
b
b
z
z
2
5
21. Find the following logarithms given that log 10
2 0. 3010 , log 10
3 0. 4771, and
log 10
7 0.8451:
3
3
a. log 7,000 log 7 10
log 7 log 10 (.8451) 3 3. 8451
10 10 10
5
5
b. log.00007 log7 10
log7 log10 (.8451) 5 4.
1549
2
2
log1400 log 7 210
log 7 log 2 log10
c.
.8451
.3010 2 3.1461
4
4
d. log.0003 log3 10
log3 log10 .4771
4 3.
5229
22. Between 12:00pm and 1:00pm, cars arrive at Citibank’s drive-thru at the rate of 6 cars
per hour (0.1 car per minute). The following formula from statistics can be used to
determine the probability that a car will arrive within t minutes of 12:00pm.
0.1t
F(
t)
1
e
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 440)
a. Determine how many minutes are needed for the probability to reach 50%
0.1t
F(
t)
1
e
.50 1
e
.5 1
e
.5 e
.5 e
0.1t
ln .5 ln e
0.1t
0.1t
0.1t
0.1t
ln .5 0.1t
ln .5 .693
t 6.93
0.1 0.1
It will take 6.93 minutes for the probability to reach 50%.
Math Camp 2011 Page 113

. Determine how many minutes are needed for the probability to reach 80%
F(
t)
1
e
.80 1
e
.8 1
e
.2 e
.2 e
0.1t
ln .2 ln e
0.1t
0.1t
0.1t
0.1t
0.1t
ln .2 0.1t
ln .2 1.61
t 16.09
0.1 0.1
It will take 16.09 minutes for the probability to reach 80%
Math Camp 2011 Page 114

MATH CAMP
School of Public and Environmental Affairs
Day Three Homework
1. Find an equation of the line through the given point with the given slope. Graph the line.
1
( 2, 5),
m
2
2. Suppose that you are given the following equation for a demand curve for toy elephants.
In this case, Q = the quantity of elephants demanded and P = the price of toy elephants.
Q
P 425
.8
a. Solve this equation for P.
b. Suppose that you know that revenue is the price multiplied by the quantity, or
R P Q
. Using the equation for P that you found in Part A, determine the
revenue function.
c. Graph the revenue function from Part B.
d. If the quantity sold was 5 units, what would the price be? What would the
revenue be?
e. Just by looking at the graph, at approximately what quantity will revenue be the
greatest?
f. Now suppose that the cost function of producing toy elephants is given by
C 50 5Q . When you make approximately 4 toy elephants, what is the cost?
g. We also know that profit is the revenue minus the cost, or R C . Using the
revenue function from Part B and the cost function from Part F, find the profit
function.
h. Graph the profit function from Part G.
i. By looking at the graph, at approximately what quantity will you have the highest
profit?
j. If you sell 3 toy elephants, what will be your profit?
3. A rare species of insect was discovered in the Amazon Rain Forest. To protect the
species, environmentalists declare the insect endangered and transplant the insects into a
protected area. The population of the insect t years after being transplanted is given by P.
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, pp. 329)
50(1 0.5t)
P(
t)
(2 0.01 t)
a. How many insects were discovered? In other words, what was the population
when t 0 ?
b. What will the population be after 5 years?
c. Determine the horizontal asymptote of P (t)
. What is the largest population that
the protected area can sustain?
Math Camp 2011 Page 115

4. The research department in a company that manufactures AM/FM clock radios
established the following price-demand, cost, and revenue functions:
p( x)
50 1.
25x
Price-demand function
C( x)
160 10x
Cost function
R( x)
xp(
x)
x(50
1.25x)
Revenue function
Where x is in thousands of units, and C(x)
and R (x)
are in thousands of dollars. All
three functions have domain 1 x 40 . (Barnett, Ziegler and Byleen, Finite Mathematics for Business,
Economics, Life Sciences, and Social Sciences, tenth edition, Upper Saddle River, NJ: Prentice Hall, pp. 74)
a. Graph the cost function and the revenue function simultaneously in the same
coordinate system.
b. Determine algebraically when R C . Then, with the aid of part (A), determine
when R C and R C to the nearest unit.
c. Determine algebraically the maximum revenue (to the nearest thousand dollars)
and the output (to the nearest unit) that produces the maximum revenue. What is
the wholesale price of the radio (to the nearest dollar) at this output?
0.3t
5. If Y grams of a radioactive substance are present after t seconds, then Y be where b
is a constant. If 100 grams of the substance is present initially, how much is present after
5 seconds?
6. It took from the dawn of humanity to 1830 for the population to grow to the first billion
people, just 100 more years (by 1930) for the second billion, and 3 billion more were
added in only 60 more years (by 1990). In 2002, the estimated world population was 6.2
billion with an annual growth rate of 1.25% compounded continuously.
(Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences, tenth edition, Upper
Saddle River, NJ: Prentice Hall, pp. 108-109)
a. Write an equation that models the world population growth, letting 2002 be year
0.
b. Based on the model, what is the expected world population (to the nearest
hundred million) in 2010? In 2030?
7. Solve the following for x or b, respectively.
a. log 0.1 1
b
b. log
2(
x 1)
log
2(3x
5)
log
2(5x
3)
2
kt
8. Psychologists sometimes use the function L(
t)
A(1
e ) to measure the amount L
learned at time t. The number A represents the amount to be learned, and the number k
measures the rate of learning. Suppose that a student has an amount A of 200 vocabulary
works to learn. A psychologist determines that the student learned 20 vocabulary words
after 5 minutes.
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 440)
a. Determine the rate of learning k.
b. Approximately how many words will the student have learned after 10 minutes?
c. After 15 minutes?
d. How long does it take for the student to learn 180 words?
Math Camp 2011 Page 116

MATH CAMP
School of Public and Environmental Affairs
Day Three Homework Solutions
1. Find an equation of the line through the given point with the given slope. Graph the line.
1
( 2, 5),
m
2
y y
1
y ( 5)
1
y
2
1
y x 6
2
m( x x ) 1
1
(
2
x 1
5
x 2)
0
-15 -10 -5 0 5 10 15
-2
-4
-6
-8
-10
-12
2. Suppose that you are given the following equation for a demand curve for toy elephants.
In this case, Q = the quantity of elephants demanded and P = the price of toy elephants.
Q
P 425
.8
a. Solve this equation for P.
P
Q 425
.8
.8Q
P
(.8)(425)
.8Q
P
340
P .8Q
340
Math Camp 2011 Page 117

. Suppose that you know that revenue is the price multiplied by the quantity, or
R P Q
. Using the equation for P that you found in Part A, determine the
revenue function.
R P Q
R ( .8Q
340) Q
R .8Q
2
340Q
c. Graph the revenue function from Part B.
200000
0
-1500 -1000 -500 0
-200000
500 1000 1500 2000
-400000
-600000
-800000
-1000000
-1200000
-1400000
d. If the quantity sold was 5 units, what would the price be? What would the
revenue be?
P .8Q
340
P .8(5)
340 336
R .8Q
R .8(5)
R .8(25)
1700
R 20
1700
R 1680
2
340Q
2
340(5)
e. Just by looking at the graph, at approximately what quantity will revenue be the
greatest?
Somewhere around 250 or so
f. Now suppose that the cost function of producing toy elephants is given by
C 50 5Q . When you make approximately 4 toy elephants, what is the cost?
Math Camp 2011 Page 118

C 50 5Q
C 50 5(4)
C 50 20
C 70
g. We also know that profit is the revenue minus the cost, or R C . Using the
revenue function from Part B and the cost function from Part F, find the profit
function.
R C
.8Q
340Q
(50 5Q)
2
.8Q
335Q
50
h. Graph the profit function from Part G.
2
200000
0
-1500 -1000 -500 0
-200000
500 1000 1500 2000
-400000
-600000
-800000
-1000000
-1200000
-1400000
i. By looking at the graph, at approximately what quantity will you have the highest
profit?
Somewhere around 200 or 250, definitely between 0 and 500
j. If you sell 3 toy elephants, what will be your profit?
2
.8Q
335Q
50
.8(3)
947.8
2
335(3) 50
3. A rare species of insect was discovered in the Amazon Rain Forest. To protect the
species, environmentalists declare the insect endangered and transplant the insects into a
protected area. The population of the insect t years after being transplanted is given by P.
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, pp. 329)
50(1 0.5t)
P(
t)
(2 0.01 t)
Math Camp 2011 Page 119

a. How many insects were discovered? In other words, what was the population
when t 0 ?
50(1 0.5t)
P(
t)
(2 0.01 t)
50(1 0.5 0)
P(0)
(2 0.01
0)
50
25
2
b. What will the population be after 5 years?
50(1 0.5t)
P(
t)
(2 0.01 t)
50 1
(0.5)(5)
P(5)
2 (0.01)(5)
50(1 2.5)
2 .05
175
2.05
85.37
After 5 years, there will be approximately 85 insects.
c. Determine the horizontal asymptote of P (t)
. What is the largest population that
the protected area can sustain?
To find the horizontal asymptote of P (t)
we need to divide every term in the
numerator and denominator by t.
50(1 0.5t)
P(
t)
(2 0.01 t)
50 25t
2 0.01t
50
25
t
2
0.01
t
50 2
As t , 0 and 0 .
t t
50
25
This means that the expression t 25
will draw closer to 2500
2
0.01
0.01
t
as t . The same is true as t . Therefore, the horizontal
asymptote is P ( t)
2500 . This asymptote tells us that the protected area
cannot sustain more insects than 2500.
4. The research department in a company that manufactures AM/FM clock radios
established the following price-demand, cost, and revenue functions:
Math Camp 2011 Page 120

p( x)
50 1.
25x
Price-demand function
C( x)
160 10x
Cost function
R( x)
xp(
x)
x(50
1.25x)
Revenue function
Where x is in thousands of units, and C(x)
and R (x)
are in thousands of dollars. All
three functions have domain 1 x 40 . (Barnett, Ziegler and Byleen, Finite Mathematics for Business,
Economics, Life Sciences, and Social Sciences, tenth edition, Upper Saddle River, NJ: Prentice Hall, pp. 74)
a. Graph the cost function and the revenue function simultaneously in the same
coordinate system.
600
500
400
300
200
100
0
0 10 20 30 40
b. Determine algebraically when R C . Then, with the aid of part (A), determine
when R C and R C to the nearest unit.
R C
x(50
1.25x)
160 10x
50x
1.25x
1.25x
1.25x
2
2
2
160 10x
50x
10x
160
0
40x
160
0
b
x
2
b 4ac
2a
40 1600 800
2.5
40
(40)
4( 1.25)(
160)
2( 1.25)
40 28.28 4.688
2.5 27.312
2
As seen in the graph, R C when x 4. 688 or when x 27. 312 . Also,
R C when 4.688
x 27. 312
c. Determine algebraically the maximum revenue (to the nearest thousand dollars)
and the output (to the nearest unit) that produces the maximum revenue. What is
the wholesale price of the radio (to the nearest dollar) at this output?
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To find the maximum revenue, we need to find the vertex of the revenue curve,
which will occur at:
b 50
x 20
2a
2( 1.25)
The output when we hit the maximum revenue is 20 units.
So the maximum revenue will be:
b
2
R R (20) 1.25(20)
50(20) 500
1000
$500
2a
The price at this output will be:
p ( 20) 50 1.25(20)
50 25 $25
0.3t
5. If Y grams of a radioactive substance are present after t seconds, then Y be where b
is a constant. If 100 grams of the substance is present initially, how much is present after
5 seconds?
We know that there are 100 grams of substance initially, so there are 100 grams of the
substance at t = 0. We can plug this into our formula to get b.
Y = be -0.3t
100 = be (-0.3)(0)
100 = be 0
b = 100
Now we can solve for Y when t = 5, given that b = 100.
Y = 100e -0.3t
Y = 100e (-0.3)(5) = 100e -1.5 = 100(0.223) = 22.3 grams
Math Camp 2011 Page 122

6. It took from the dawn of humanity to 1830 for the population to grow to the first billion
people, just 100 more years (by 1930) for the second billion, and 3 billion more were
added in only 60 more years (by 1990). In 2002, the estimated world population was 6.2
billion with an annual growth rate of 1.25% compounded continuously.
(Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences, tenth edition, Upper
Saddle River, NJ: Prentice Hall, pp. 108-109)
a. Write an equation that models the world population growth, letting 2002 be year
0.
P =6.2e (0.0125)t
b. Based on the model, what is the expected world population (to the nearest
hundred million) in 2010? In 2030?
7. Solve the following for x or b, respectively.
a. log 0.1 1
b
b. log
2(
x 1)
log
2(3x
5) log
2(5x
3) 2
Math Camp 2011 Page 123

kt
8. Psychologists sometimes use the function L(
t)
A(1
e ) to measure the amount L
learned at time t. The number A represents the amount to be learned, and the number k
measures the rate of learning. Suppose that a student has an amount A of 200 vocabulary
works to learn. A psychologist determines that the student learned 20 vocabulary words
after 5 minutes.
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 440)
a. Determine the rate of learning k.
b. Approximately how many words will the student have learned after 10 minutes?
c. After 15 minutes?
Math Camp 2011 Page 124

d. How long does it take for the student to learn 180 words?
It would take approximately 115 minutes to learn 180 words.
Math Camp 2011 Page 125

MATH CAMP
School of Public and Environmental Affairs
DERIVATIVES
What are Derivatives?
Day Four: Derivatives, Optimization, and Integration
Day Four Class Notes
Slopes can help us to understand the relationships between variables. When we have an equation
for a line, finding the slope is really straightforward and easy to do. When we have functions
that take other shapes, however, finding the slope at any particular point on the function becomes
a little harder. In addition, in many functions the slope is different at different points on the
curve. In order to do this, we are going to have to learn a few basic principles of calculus,
specifically taking the derivative of a function.
Because finding the slope of a line is easy, we can exploit the properties of a line in order to find
the slope of another function at a given point.
A tangent line is a straight line that intersects a curve at a single point and has the same slope as
the curve at the point of intersection. Thus, if we can find the slope of the tangent line, we can
find the slope of the curve at the point of tangency. The graph below shows a parabola with a
tangent line.
60
40
20
0
-15 -10 -5 0 5 10 15
-20
-40
-60
We want to know this slope so that we can know the rate of change of the function (curve) at that
particular point. This rate of change has many important real world applications.
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Given a curve and a tangent line, if we know the equation for the line, we can easily determine
the slope of the curve at the point where the tangent line and the curve meet. The problem is that
we don’t always know the equation for a line tangent to the curve at a particular point, so we
need another way to find the slope at that point. Taking the derivative of the curve at that point
can do this for us.
The derivative is the slope of the line tangent to the curve at a given point. This slope can give
us important information about the curve that we are interested in. For one thing, the derivative
of a curve tells us the instantaneous rate of change of the curve. A really curvy function can
change at different rates in different places—it can be increasing or decreasing, shallow or deep,
etc. (A line has the same slope at every point, but other curves have a different slope at different
points along the curve.)
As mentioned earlier, the instantaneous rate of change has many real world applications. For
instance, velocity is the derivative of the position function. Marginal cost is the derivative of the
cost function, and marginal revenue is the derivative of the revenue function.
In math camp, we are only going to learn some simple derivatives—derivatives for polynomials.
These will be the most common kinds of derivatives that you will encounter in your SPEA
classes. But we can apply the rules of derivatives to almost any functional equation. The
process of finding the slope of the tangent line is referred to as taking the derivative of (or
differentiating) the functional equation.
Let y denote the dependent variable and let x denote the independent variable. We can write y as
a function of x as so: y f (x)
where f (x)
is some unspecified functional relationship between
dy
x and y. The derivative of this function is represented by the notation , which is basically the
dx
derivative of y with respect to x.
Rules of Differentiation
d
Rule 1: ( c)
0
dx
for any constant c.
Rule 2:
d
dx
( cx)
c for any cx where c is a constant and x is a variable.
Rule 3: A power function of the form
d d n
n1
( y)
( cx ) n cx .
dx dx
n
y cx , where c and n are constants, has the derivative
Rule 4: (Sum Rule) For any function y f ( x)
g(
x)
where f (x)
and g(x)
are both functions of
dy df dg
x. Then we have .
dx dx dx
Math Camp 2011 Page 127

dy df dg
Rule 5: (Product Rule) For any function y f ( x)
g(
x)
, g f
dx dx dx
Rule 6: (Quotient Rule) Given the function
f ( x)
y ,
g(
x)
dy
dx
df dg
g
f
dx dx
2
g
Rule 7: (Chain Rule) This is used for functions nested within functions, such as y f ( g(
x))
. It
dy df dg
states ( g(
x)) . Basically, this means that we take the derivative of the outer function
dx dx dx
while leaving the inner function alone inside of it, and then multiply by the derivative of the
inner function.
Rule 8:
Rule 9:
d (ln x)
1
dx x
d
dx
x
( e ) e
x
Examples of the Rules of Differentiation
dy
Example 1: Suppose the y f (x)
is y = 10, where 10 is a constant. Then 0 . This makes
dx
sense because the function y = 10 is a horizontal line which has a slope of 0.
dy dy
Example 2: Suppose that y f (x)
is y 14x. Then ( y)
(14x)
14 . This also makes
dx dx
sense because the equation y 14x
is the equation for a line with slope 14 and a y-intercept of 0.
3
Example 3: Given the function y 4x , the derivative would be
dy dy 3
31
2
( y)
(4x
) 3
4x
12x
.
dx dx
Math Camp 2011 Page 128

2
2
Example 4: Suppose that we have y 5 x x . Then f ( x)
5x
and g( x)
x . Using the
df 2
21
dy
rules above, (5x
) 25x
10x
and ( x)
1. Using our Sum Rule, we get
dx
dx
dy df dg
dy 2
10x
1. So (5x
x)
10x
1.
dx dx dx
dx
2 4
Example 5: Let y (3x
)(5x
).
Example 6: Let
2
4x
x
dy
dx
df
dx
g
dy df
(3x
dx
dx
dy
dx
dy
(6x)(5x
dx
df
(2x
dx
dg
dx
2x
2
y . Then x 1
df dg
g
dy
dx dx
2
dx g
dy
dx
dy
dx
2
)
(5x
21
4
41
2
3x
(5
x ) 4
5x
4
f
4
) (20x
dg
)
(5x
dx
3
)(3x
2
)
4
(3x
)
(3x
21
2
2
2
2x
(
x 1)
(1)(2x
) (4
x)(
x 1)
2x
2
2
4x
2x
2x
2
dg
) (
x 1)
( x 1
2x
dx
2
( x 1)
( x 1)
2
f
2
2
2x
4x
2
x 2x
2
x
2
2
2
)
2
)
2x
2
Example 7: Let y 5x
3 3 . This is tricky because we have a square root function of an
algebraic expression. In this case, f(x) is the square root function and g ( x)
5x 3 3 . So
applying the Chain Rule, we get:
Math Camp 2011 Page 129

dy
dx
dy
dx
dy
dx
df
dx
1
2
( g(
x))
15x
3
2 5x
3
1 1
3 1
5 3 2
x 3
5x
0
2
3
dg
dx
Finding the Equation of a Tangent Line
Using our derivatives, we can now find the equation for a tangent line at any point on a curve.
2
Find the tangent line to the curve f ( x)
3x
4x
1at the point where x = 2. Well, when x = 2,
2
we know that y f (2) 3(2 ) 4(2) 1
12
8
1
5, so the point of tangency is (2,5). The
slope at this point should be the derivative of the curve at x = 2.
dy
dx
d
dx
(3x
2
4x
1)
23x
21
4 6x
4
We evaluate the first derivative at x = 2 and get 6(2)
4 12
4 8. Thus, the slope at this
point is 8.
Now that we have a point and a slope, we can use the point-slope form of a line to get the
equation for the tangent line.
y y
1
y 5 8( x 2)
y 8x
16
5
y 8x
11
m( x x ) 1
OPTIMIZATION AND APPLICATIONS OF DERIVATIVES
In many real-world applications, finding a maximum or minimum point of an equation is
important. The derivative can help us to these types of optimization problems.
Maximizing Revenue Application Problem
2
Suppose a firm has a revenue function of R 170Q
20Q
, where Q is the quantity of goods
sold. Find the point where revenue is maximized. Let’s graph the revenue function first. This
graph would be:
Math Camp 2011 Page 130

1000
500
0
-15 -10 -5 -500 0 5 10 15 20 25
-1000
-1500
-2000
-2500
-3000
-3500
-4000
-4500
Just by looking at the graph, we can see that revenue will be maximized at the very top of the
parabola (somewhere in the neighborhood of Q=4). The tangent line at the top of the parabola
will have a slope of zero. So if we can find a point where the slope of our curve is 0, then we
know where the maximum revenue is.
Let’s take the derivative of the revenue function to find the slope.
dR
dx
170 40Q
Now let’s set the derivative equal to zero and solve for Q.
170 40Q
0
40Q
170
170
Q
40
17
4
17
Therefore, when quantity equals , we will have the maximum revenue. We can substitute this
4
value of Q back into our equation to find out what this revenue will be.
Second Derivatives
R 170Q
20Q
17 17
R 170 20
4 4
2
2
$361.25
Not only can the derivative give us valuable information about a function, but the second
derivative of a function can also provide us with valuable information. Often, we are interested
Math Camp 2011 Page 131

in places on a curve where the first derivative is 0 (meaning that the slope is 0). This can happen
at minimum and maximum points. But from the first derivative, we can’t tell whether or not we
are at a maximum or a minimum point. The second derivative gives us an easy way to determine
if we have a maximum or a minimum.
We know that at a maximum point, the slope changes from positive to zero to negative. At a
minimum point, the slope changes from negative to zero to positive. Therefore, if the second
derivative is negative, then we know we are at a local maximum. If the second derivative is
positive, then we are at a local minimum.
We get the second derivative by taking the derivative of a function twice.
2
Example: Find the maximum or minimum of the following function: f ( x)
5x
10x
12
.
Use the second derivative to determine whether this point is indeed a maximum or a minimum.
2
df
The first derivative of f ( x)
5x
10x
12
is f ( x)
10x
10. We can set this
dx
derivative equal to zero to get the location of the maximum or minimum.
10x
10
0
10x
10
x 1
f (1) 5(1)
2
10(1)
12
5 10
12
7
The point of the curve that is our maximum or minimum point is ( 1,7)
.
Now we can take the second derivative of the function.
2
f ( x)
5x
10x
12
df
f ( x)
10x
10
dx
2
d f
f ( x)
10
2
dx
Because the second derivative is positive, the point is a minimum point.
INTEGRATION
Integration, also known as antidifferentiation, is the reverse of differentiation. The indefinite
integral
f ( x)
dx
is a function whose derivative is f(x). You will be given a function, f(x), and need to find a
function, F(x), such that F'
( x)
f ( x).
The function
f ( x)
dx F(
x)
C
Math Camp 2011 Page 132

expresses that all antiderivatives of f(x) are the function F(x) + C, where C is any constant.
Each time we differentiate a function, we also derive an integration formula. The derivative
d 2
(5x
) 10x
d(
x)
can be turned into the integration formula
2
10xdx
5x
C.
Notation
f ( x)
dx F(
x)
C where F'
( x)
f ( x)
Where
f ( x)
dx is the indefinite integral of f(x).
f(x)
is the integral sign.
is the integrand and is the quantity to be integrated.
dx
F(x)
C
indicates that x is the variable with respect to which the integration is to take
place.
is the antiderivative.
is the constant of integration.
The process for finding f ( x)
dx is called indefinite integration and f ( x)
dx is read as “the
indefinite integral” of f of x dx.
Math Camp 2011 Page 133

Helpful formulas
Differentiation Formula Corresponding Integration Formula
d r
r1
( x ) dx rx r1
r
rx dx x C or
dx
r1
r x
x dx C,
r 1
r 1
d ( e x ) e
x (1) x x
e dx e C
dx
d x
1
1
(ln )
dx x
dx ln x C
x
d
(sin x)
cos x
dx
cos xdx sin x C
d
dx
(cos x)
sin
x
d 2
sin xdx cos x C
(tan x)
sec x sec 2 xdx tan x C
dx
We will concentrate on two methods for evaluating indefinite integrals – integration by
substitution and integration by parts.
Integration by substitution
The following formula is called integration by substitution and is often used to transform a
complicated integral into a simpler one
f ( g(
x))
g'(
x)
dx F(
g(
x))
C.
(1)
Example 1:
2 3
( x 1)
2xdx
3
2
2 3
Set f ( x)
x , g(
x)
x 1, then f ( g(
x))
( x 1)
and g'
( x)
2x.
1 4
An antiderivative F(x) of f(x) is given by F(
x)
x , so that, by formula (1), we have
4
( x
2
3
1 2 4
1) 2xdx
F(
g(
x))
C ( x 1)
C.
4
Math Camp 2011 Page 134

Example 1 can be solved another way using equation 1. Using equation 1, g(x) is replaced by the
new variable u, and g '(
x)
dx is replaced by du. Replacing the variables from equation 1 with u
and du reduces the complex expression f(g(x)) into a simpler one, f(u):
f ( g(
x))
g'(
x)
dx f ( u)
du
Since u = g(x), we then obtain
f ( u)
du F(
u)
C.
f ( g(
x))
g'(
x)
dx F(
u)
C F(
g(
x))
C.
It is important to note that replacing g’(x)dx by du is only a correct mathematical statement
because doing so leads to the correct answers.
Evaluating Example 1 using the new method
Set u = x 2 + 1. Then
d
du ( x
2 1)
dx 2xdx
, and
dx
( x
2
1)
3
2xdx
1 4
u
4
1
( x
4
C
2
1)
4
u
3
du
C
(since u =x 2 +1).
Method for integration of function of the form f '(
g(
x))
g'(
x).
1. Define a new variable u = g(x).
2. Transform the integral with respect to x into an integral with respect to u by replacing
g(x) everywhere by u and g′(x)dx by du.
3. Integrate the resulting function of u.
4. Rewrite the answer in terms of x by replacing u by g(x).
Example 2:
(ln x)
2
x
dx
Math Camp 2011 Page 135

Let u = lnx and du = (1/x)
x 2
(ln )
1 2
dx
x
x
2
u du
dx (ln x)
3
u
C
3
3
(ln x)
3
C
(since u = ln x).
Integration by Parts
Integration by Parts is another strategy for simplifying integrals and is also known as the product
rule. Let f(x) and g(x) be any two functions and G(x) be an antiderivative of g(x). The product
rule states
d
[ f ( x)
G(
x)]
d(
x)
f ( x)
G'(
x)
f '( x)
G(
x)
f ( x)
g(
x)
f '( x)
G(
x)
[since G’(x) = g(x)]
Therefore,
This last formula can be rewritten as follows
f ( x)
G(
x)
f ( x)
g(
x)
dx f '( x)
G(
x)
dx.
f ( x)
g(
x)
dx f ( x)
G(
x)
f '( x)
G(
x)
dx
(2)
Equation 2 is one of the most important techniques of integration.
Example 1:
xe x dx
Set f(x) =x, g(x) =e x . Then f’(x) = 1, G(x) = e x , and equation (1) yields
x
x
x
x x
xe dx xe 1
e
dx xe e
1. The integrand (the integrated function) is the product of functions f(x) = x and g(x) =
e x .
2. To compute f’(x) and G(x), differentiate f(x) and integrate g(x).
C
Math Camp 2011 Page 136

3. Calculate f '(
x)
G(
x)
dx
The above principles may be applied to all problems involving integration by parts.
Example 2:
2
x sin xdx
Let f(x) =x 2 , g(x) = sinx, f’(x) = 2x, and G(x) = -cosx.
2
2
sin xdx x
cos x 2x
(
x cos x)
dx
x
2
cos x 2
xcos
xdx
(2)
You can used integration by parts on x cos xdx.
Let f(x) = x , g(x) = cosx, f’(x) =1, and G(x) =
sin x.
Combine (2) and (3)
xcos
xdx xsin
x 1sin
xdx
xsin x cos x C
(3)
2
2
x sin xdx x
cos x 2( xsin
x cos x)
C
x
2
cos x 2xsin
x 2cos x C.
The Integration section was adapted from L.J. Goldstein et al, Calculus & Its Applications 10 th ed. and B.L. Bleau.
Forgotten Calculus 2 nd ed. These are helpful sources for additional information on integration and definite integrals.
An in-depth lesson in integration will be given in the class Applied Math for Environmental Science for the Master of
Science Environmental Science. A basic understanding of integration is satisfactory for the Master of Public Affairs.
Math Camp 2011 Page 137

MATH CAMP
School of Public and Environmental Affairs
Day Four In-Class Exercises
DERIVATIVES
1. Find the following derivatives:
a.
7
f ( x)
7x
b. f ( x)
26
c. f ( x)
x
5
d. f ( x)
x 3x
1
5
e. f ( x)
11x
ln
x
f.
OPTIMIZATION
f
7 2
3x
5x
3
x)
17x
1
(
2
2
2. Suppose that your revenue function is R 150Q
10Q
. At what quantity is revenue
maximized? What is the revenue at this point?
INTEGRATION
3. ( 2 7 ) dx
X 3 2
2
3
4. (
x x 7) (2x
1)
dx
5. xsin xdx
6. x 2 ln xdx
Math Camp 2011 Page 138

MATH CAMP
School of Public and Environmental Affairs
Day Four In-Class Exercises Solutions
DERIVATIVES
1. Find the following derivatives:
7
71
a. f ( x)
7x
7 7x
49x
b. f ( x)
26 0
c.
1
2
1 1
1
2
2
f ( x)
x x x x
2 2
1
6
1
2
1
x
d.
f ( x)
1
(
2
x
5
x
5
3x
1
3x
1)
1
2
(5x
1
(
2
4
x
5
3)
3x
1)
2( x
5x
5
1
1
2
4
5x
3
4
3x
1)
1
2
3
5
4
5
e. f ( x)
11x
ln
x (55x
)(ln x)
(11x
)
7 2
3x
5x
3
f ( x)
2
17x
1
8
3 6
357x
170x
21x
10x
102x
f.
4 2
289x
34x
1
8 6
255x
21x
112x
4 2
289x
34x
1
1
x
6
2
7 2
21x
10x17x
1 34x3
x 5x
3
8
(17x
2
170x
3
1)
2
102x
OPTIMIZATION
2
2. Suppose that your revenue function is R 150Q
10Q
. At what quantity is revenue
maximized? What is the revenue at this point?
dR
dQ
150
150 20Q
20Q
0
20Q
150
150
Q
20
15
2
R 150Q
10Q
562.5
2
15 15
150 10
2 2
2
Math Camp 2011 Page 139

INTEGRATION
3. ( 2 7 ) dx
2
3
X 3 2
du
u 1
3
u
du
7 7
5
2
3
1
u 3 3
3
C u C (2 7x)
7 5 35 35
3
5
3
5
C
2
3
4. (
x x 7) (2x
1)
dx
3 du
u ( 2x
1)
(2x
1)
4
u C 4
u
3
du
5. xsin xdx
f ( x)
x,
f '( x)
1,
x
cos x
g(
x)
sin x
G(
x)
cos
x
xsin
xdx x
cos x
cos xdx
x
cos x sin x C
6. x 2 ln xdx
1
( cos
x)
dx
2
f ( x)
ln x,
g(
x)
x
1
3
f '( x)
,
x
G(
x)
x
3
3
3
2 x 1 x
x ln xdx ln x
3
x 3
3
x 1 2
ln x x dx
3 3
3
x 1 3
ln x x C
3 9
dx
MATH CAMP
Math Camp 2011 Page 140

School of Public and Environmental Affairs
Day Four Homework
DERIVATIVES
1. Find derivatives of the following:
a. f ( x)
ln( x 2 1)
b.
c.
3x
f ( x)
f
4
2
2x
7x
x 5
2
x 5x
3
x)
7x
x 7
(
2
2
d. f ( x)
4x
8x
13
e. f ( x)
x
2
x
2 1
f.
1 1
f ( x)
1 2
x
x
g.
f
3
( x)
( x 5x
3)
3
2. Find the derivatives of the following and evaluate the derivative at the given point:
2
a. f ( x)
5x
3x
7 , x = -1
b.
c.
10
f ( x)
, x = 25
x
2
f ( x)
x x , x = 1
3. Find all the maxima and minima of the following curve. (Hint: Find the first derivative
and set it equal to zero. Then solve for x. Then you can use the second derivative to
determine whether each point is a maximum or minimum.)
3
2x
2
f ( x)
x 4x
19
3
2 3
4. Suppose a firm assess its profit function as 10 48Q
15Q
Q .
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, p. 59)
a. Compute the firm’s profit for the following levels of output: Q 2 , Q 8, and
Q 14 .
Math Camp 2011 Page 141

5. Evaluate
b. Derive an expression for marginal profit (Hint: marginal profit is the first
derivative of the profit function).. Compute marginal profit at Q 2 , Q 8, and
Q 14 . Confirm that the profit is maximized at Q 8.
a. xe x 2
2 dx
2 3
b. 3x x 1dx
2 x
c.
dx
2
2x
8x
1
8
d. x(
x 5) dx
2
e. x ln xdx
Math Camp 2011 Page 142

MATH CAMP
School of Public and Environmental Affairs
Day Four Homework Solutions
DERIVATIVES
1. Find derivatives of the following:
2
1 2x
a. f ( x)
ln( x 1)
f '( x)
(2x)
2
2
x 1
x 1
b.
3x
f ( x)
4
2
2x
7x
x 5
c.
f
2
x 5x
3
x)
7x
x 7
(
2
2
d. f ( x)
4x
8x
13
Math Camp 2011 Page 143

e. ( 2 1
3
1
f x)
x f '( x)
2x
2x
2( x )
2
x
x
3
f.
1 1
f ( x)
1 2
x
x
1 2
1 2 1 1 1
f ( x)
( x
) 2
( x ) 1
2 3 2 3
x x x x x x
1
2
x
x
2 2
3
3
3
3
2 2
g. f ( x)
( x 5x
3) f '( x)
3( x 5x
3) (3x
5)
2. Find the derivatives of the following and evaluate the derivative at the given point:
2
a. f ( x)
5x
3x
7 , x = -1
f '( x)
10x
3
f '( 1)
10( 1)
3 10
3 7
b.
c.
10
f ( x)
, x = 25
x
10
f '( x)
x
2
5
f '( x)
3
2
3
2
x
5 5
f '(25)
3
3
2 25
25
2
f ( x)
x x , x = 1
f '( x)
1
2x
f '( x)
1
2x
f '( x)
1
2 1
5
125
1
25
3. Find all the maxima and minima of the following curve. (Hint: Find the first derivative
and set it equal to zero. Then solve for x. Then you can use the second derivative to
determine whether each point is a maximum or minimum.)
3
2x
2
f ( x)
x 4x
19
3
Math Camp 2011 Page 144

2 3
4. Suppose a firm assess its profit function as 10 48Q
15Q
Q .
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, p. 59)
a. Compute the firm’s profit for the following levels of output: Q 2 , Q 8, and
Q 14 .
b. Derive an expression for marginal profit (Hint: marginal profit is the first
derivative of the profit function).. Compute marginal profit at Q 2 , Q 8, and
Q 14 . Confirm that the profit is maximized at Q 8.
It appears that Q = 2 and Q = 8 are both roots to the marginal profit equation (or
derivative of the profit equation). Either point could be the maximum point. We need to
use the second derivative to see which point is the maximum point.
Math Camp 2011 Page 145

Because the second derivative is negative at the point Q = 8, this is the point that
maximizes profit.
5. Evaluate
a. xe x 2
2 dx
e
e
u
x
2
e
x
2
C
C
2xdx
2 3
b. 3x x 1dx
u x
3x
x
3
2
u
3
1
x
C
C
e
1dx
u
du
3
u x 1
d 3
2
du ( x 1)
dx 3x
dx
dx
2
(
3
2
3
3
2
3
1)
3
2
udu
2 x
c.
dx
2
2x
8x
1
1
1
(2
x)
dx
3
2x
8x
1
4
(
4)(2
2
2x
1
8x
1
x)
dx
1
4
1 1
du
u 4
u
1
2
du
Math Camp 2011 Page 146

Math Camp 2011 Page 147
C
x
x
C
u
C
u
2
1
2
2
1
2
1
1)
8
(2
2
1
2
1
2
4
1
d. dx
x
x
8
5)
(
1,
)
'(
,
)
(
x
f
x
x
f
9
8
5)
(
9
1
)
(
5)
(
)
(
x
x
G
x
x
g
C
x
x
x
C
x
x
x
dx
x
x
x
dx
x
x
x
dx
x
x
10
9
10
9
9
9
9
9
8
5)
(
90
1
5)
(
9
1
5)
(
10
1
9
1
5)
(
9
1
5)
(
9
1
5)
(
9
1
5)
(
9
1
1
5)
(
9
1
5)
(
e. xdx
x ln
2
,
1
)
'(
,
ln
)
(
x
x
f
x
x
f
3
2
3
)
(
)
(
x
x
G
x
x
g
C
x
x
x
dx
x
x
x
dx
x
x
x
x
xdx
x
3
3
2
3
3
3
2
9
1
ln
3
3
1
ln
3
3
1
ln
3
ln

MATH CAMP
School of Public and Environmental Affairs
Day Five: Word Problems and Applications
Day Five Class Notes
STRATEGIES FOR SOLVING WORD PROBLEMS
The following list of steps can help you to understand and solve word problems in mathematics.
1. Read through the problem a couple times.
2. List the values that are known and those that are unknown.
3. Identify what value you need to determine—what final answer you are looking for.
4. Assign a variable to each unknown quantity.
5. Draw a picture, if possible.
6. Use formulas and equations to describe the relationship between the knowns and the
unknowns. Basically you are trying to express the relationships in mathematical terms.
7. Use mathematical procedures to solve these equations for the unknowns.
8. Check your answers to see if they are plausible.
Example: A ladder that is 30 feet long is leaning against a building. The bottom of the ladder is
10 feet away from the building. How high on the building does the ladder reach?
Knowns
We know the length of the ladder is 30 feet.
The ladder is 10 feet from the building
Unknown
Where the ladder reaches on the building, and this is what we want to find out. We will
call this x.
Math Camp 2011 Page 148

Building
Ladder
30 feet
Find this
distance
= x feet
10 feet
The ladder makes a right triangle against the building, so we can use the properties of a right
triangle to relate our knowns and unknowns. In this case we know that the square of the
2 2 2
hypotenuse is equal to the sum of the squares of the sides, so a b c . In this case c = 30
feet and a = 10 feet. We just need to find b.
a
2
10
100 b
b
2
2
b
b
2
b
2
2
800
c
2
30
2
900
800 28.3
Therefore the ladder is 28.3 feet high against the building.
APPLICATIONS IN STATISTICS
We have already applied our mathematical knowledge to several problems in statistics, such as
calculating the mean and variance. We will now apply what we have learned this week to some
other important statistical concepts.
In your statistics class, you will learn about regression lines. We often want to find the equation
for a line that best fits our data. This gives us an additional method of describing the relationship
between x and y beyond such methods as a scatter plot. The formula for a regression line is
y a bx where a is the y-intercept and b is the slope. Notice that this is basically the slopeintercept
form a line. Given values for x and y, we can compute a and b using the following
formulas.
a
2
y
x
x
xy
2
n
x
x 2
Math Camp 2011 Page 149

n
xy
x
y
2
n
x
x 2
Example: These data were obtained from a survey of the number of years people smoked and
the percentage of lung damage they sustained. Draw a scatter plot of the data. Then calculate r.
Then calculate a and b and find the regression equation. Predict the percentage of lung damage
for a person who has smoked for 30 years.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. pp. 494-495)
Years, x 22 14 31 36 9 41 19
Damage, 20 14 54 63 17 71 23
y
80
70
60
50
40
30
20
10
0
0 10 20 30 40 50
r
n
xy
x
y
2
2
2
x
x
n
y
n
y
2
The easiest way to calculate r would be to calculate the component parts of r. These components
are shown in the following table.
Years, Damage,
xy x^2 y^2
x y
22 20 440 484 400
14 14 196 196 196
31 54 1674 961 2916
36 63 2268 1296 3969
9 17 153 81 289
41 71 2911 1681 5041
19 23 437 361 529
Sum 172 262 8079 5060 13340
Math Camp 2011 Page 150

So we have
x 172
y 262
xy 8, 079
x
2 5,060
y 2
13, 340
r
n
xy
x
y
2
2
2
x
x
n
y
2
n
y
7(8079) (172)(262)
2
2
7(5060)
(172) 7(13340) (262)
.96
a
2
y
x
x
xy
2
2
n
x
x
(262)(5060) (172)(8079)
10.944
2
7(5060) (172)
n
b
xy
x
y
2
2
n
x
x
7(8079) (172)(262)
1.969
2
7(5060) (172)
Therefore, the equation for the regression line is y 1.969x
10.
944. The scatter plot with this
line imposed on it is shown below.
Math Camp 2011 Page 151

80
70
y = 1.9686x - 10.944
60
50
40
30
20
10
0
0 10 20 30 40 50
Now if we want to predict the amount of damage done when the person has smoked for 30 years,
we just substitute x = 30 into this equation to get y.
y 1.969x 10.944
1.969(30)
10.944
48.126
APPLICATIONS IN ECONOMICS
Another important idea in economics is elasticity. Elasticity measures the responsiveness of the
sales of a particular good to changes in the price of that good. The price elasticity of demand is
the ratio of the percent change in quantity and the percent change in price.
E P
Q
Q
P
P
( Q1
Q0
)
( P P )
1
0
Q
P
0
0
Well, given that we know that the first derivative is the instantaneous rate of change, we can
rewrite this equation in terms of the first derivative. Remember that the elasticity is measuring
the change in quantity in relation to the change in price. So,
Q
Q ( Q1
Q0
) Q
E P
P
( P1
P0
) P0
P
0
dQ
Q
dP
P
dQ
P
dP
Q
Example: Suppose we had a demand function like the following:
Q 850 50P
74Pop
.
6A
where P is the price, Pop is the local population in thousands and A is the advertising
expenditure.
Math Camp 2011 Page 152

If P = $2.50 and A = $400 and Pop = 50, what is the quantity?
Q 850 50P
74Pop
.6A
Q 850 50(2.50) 74(50) .6(400)
4665
If we wanted to find the price elasticity, we just take the derivative of the demand function with
respect to P. We treat all other variables as if they were just constants (or numbers).
Q 850 50P
74Pop
.
6A
dQ
Q 50
dP
E P
dQ
P
dP
Q
2.50
3540
50 .
0035
APPLICATIONS IN FINANCE
One important concept in public finance is the idea of discounting future benefits. When doing a
cost-benefit analysis of a project, it is important to discount the future benefits of the project in
order to get at true view of the value of the project. We need to discount future benefits because
of positive interest rates. The existence of positive interest rates implies that a dollar of benefits
in the future will be less than a dollar of benefits right now. This is because we could invest less
than a dollar right now and because of positive interest rates that amount will grow to be equal to
a dollar at some point in the future.
For instance, if the interest rate is 10% per year, then we need only invest $90.91 dollars today to
have that money grow to be $100 in one year. This means that the present value of receiving
$100 on year from now is $90.91. We call this the present value of receiving $100 in one year.
The formula for calculating present value is:
PV
X
( 1
r)
n
where X is the dollar amount to be received in the future, n is the number of years, and r is called
the social discount rate, or interest rate.
If the benefits of a project will accrue over several years, then we can add up their present values
in this fashion:
PV
X
n
i
i
i1 (1 r)
Math Camp 2011 Page 153

Example: Suppose you are going to build a new park that will yield the following benefits
Year 1 $0
Year 2 $0
Year 3 $500
Year 4 $1000
Year 5 $5000
Assuming a social discount rate of 5%, what is the present value of the project? Assuming a
social discount rate of 20%, what is the present value of the project?
PV
PV
PV
PV
PV
PV
PV
n
X
i
(1 r)
i1
X
1
(1 r)
0
(1 .05)
0
(1.05)
1
1
0 0 500 1000
1.05 1.1025 1.157625 1.21550625
0 0 431.92 822.70 3917.62
$5,172.25
X
2
(1 r)
1
i
0
(1 .05)
0
(1.05)
2
2
X
3
(1 r)
2
500
(1 .05)
500
(1.05)
3
3
X
4
(1 r)
3
1000
(1.05)
X
5
(1 r)
1000
(1 .05)
4
4
4
5000
(1.05)
5000
(1 .05)
5
5
5
5000
1.2762815625
The present value of the project is $5,172.25 when using a social discount rate of 5%.
PV
PV
PV
PV
PV
PV
PV
n
X
i
(1 r)
i1
X
1
(1 r)
0
(1 .20)
0
(1.20)
1
1
0 0 500 1000 5000
1.20 1.44 1.728 2.0736 2.48832
0 0 289.35 482.25 2009.39
$2,780.99
X
2
(1 r)
1
i
0
(1 .20)
0
(1.20)
2
2
X
3
(1 r)
2
500
(1 .20)
500
(1.20)
3
3
X
4
(1 r)
3
1000
(1.20)
X
5
(1 r)
1000
(1 .20)
4
4
4
5000
(1.20)
5000
(1 .20)
5
5
5
The present value of the project is $2,780.99 when we use a social discount rate of 20%.
Math Camp 2011 Page 154

APPLICATIONS IN ENVIRONMENTAL SCIENCE
One important application in environmental science is to model the growth or decline of
kt
populations. We have used the exponential growth model A( t)
A0e
to model the growth of a
population. However, this growth function assumes uninhibited growth. In many real-life
situations, however, the environment has limited space and food, and that inhibits the growth of
populations. In these cases, we can use the logistic growth model to model these types of
situations.
The logistic growth model follows the formula:
c
P(
t)
1
bt
ae
where P is the size of the population, t is the time, b is the growth rate, and c is the carrying
capacity of the environment. The carrying capacity is the maximum number that the
environment can support.
Example: Fruit flies are placed in a half-pint milk bottle with a banana (for food) and yeast
plants (for food and to provide a stimulus to lay eggs). Suppose that the fruit fly population P
230
after t days is given by P( t)
.
0.
37t
1
56.5e
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 471)
a. What is the carrying capacity of the half-pint bottle?
b. How many fruit flies were initially placed in the half-pint bottle?
c. When will the population of fruit flies be 180?
Solutions:
a. The carrying capacity of the half-pint bottle is equal to the coefficient c, which in this
case is 230. Therefore, the half-pint bottle cannot support more than 230 fruit flies.
b. We need to know the fruit fly population when t 0 .
230
P(
t)
0.37t
1
56.5e
230
P(
t)
0.37(0)
1
56.5e
230
P(
t)
0
1
56.5e
230 230
P(
t)
4
1
56.5 57.5
c. To find the time when the fruit fly population reaches 180, we solve the following
equation for t.
Math Camp 2011 Page 155

230
P(
t)
1
56.5e
230
180
1
56.5e
180 1
56.5e
180 10170e
10170e
10170e
0.37t
0.37t
0.37t
0.37t
50
0.37t
0.37t
0.37t
50
e
10170
0.37t
50
ln e ln
10170
0.37t
ln e ln 50 ln10170
0.37t
ln 50 ln10170
ln 50 ln10170
t
14.37
0.37
230
230
230 180
The fruit fly population will reach 180 at approximately 14.37 days.
Math Camp 2011 Page 156

MATH CAMP
School of Public and Environmental Affairs
SOLVING WORD PROBLEMS
Day Five In-Class Exercises
1. A rectangular walk in a park surrounds a flower bed. The outer perimeter of the walk is
60- by 80-foot. If the walk is of uniform width and its area is equal to the area of the
flower bed, how wide is the walk?
(Johnson, How to Solve Word Problems in Algebra: A Solved Problem Approach, McGraw Hill, 2000)
APPLICATIONS IN STATISTICS
2. An educator wants to see how the number of absences a student in her class has affects
the student’s final grade. Draw a scatter plot of the data. Calculate r. Calculate a and b
and write the equation for the regression line.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 495)
No. of
absences,
x
Final
grade, y
10 12 2 0 8 5
70 65 96 94 75 82
APPLICATIONS IN ECONOMICS
3. General Motors (GM) produces light trucks in several Michigan factories, where its
annual fixed costs are $180 million, and its marginal cost per truck is approximately
$20,000. Regional demand for the trucks is given by P 30000 0. 1Q
, where P denotes
price in dollars and Q denotes annual sales of trucks.
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, pp. 115-116)
a. Find GM’s profit maximizing output level and price. Find the annual profit
generated by light trucks. {Hint: To find the maximizing output level and price,
we need to set marginal revenue equal to marginal cost and solve the resulting
equation. Marginal cost is given above. Marginal revenue is the first derivative
of the revenue equation, which is R PQ . Remember that profit is
R C PQ C ( P CPV ) Q FC where CPV is cost per vehicle and FC
is total fixed costs.}
b. GM is getting ready to export trucks to several markets in South America. Based
on several marketing surveys, GM has found the elasticity of demand in these
foreign markets to be EP
9
for a wide range of prices (between $20,000 and
$30,000). The additional cost of shipping (including paying some import fees) is
about $800 per truck. One manager argues that the foreign price should be set at
$800 above the domestic price (in part a) to cover the transportation cost. Do you
agree that this is the optimal price for foreign sales? Justify your answer. {Hint:
Math Camp 2011 Page 157

Use the optimal price markup formula to solve this problem. This formula is
EP
P MC
E
where MC is the marginal cost and EP
is the price elasticity.
1
P
Marginal cost was given in part a, and we need to add to it the price of shipping
the trucks.}
c. GM also produces an economy (“no frills”) version of its light truck at a marginal
cost of $12,000 per vehicle. However, at the price set by GM, $20,000 per truck,
customer demand has been very disappointing. GM has recently discontinued
production of this model but still finds itself with an inventory of 18,000 unsold
trucks. The best estimate of demand for the remaining trucks is: P 30000 Q .
One manager recommends keeping the price at $20,000; another favors cutting
the price to sell the entire inventory. What price (one of these or some other
price) should GM set and how many trucks should it sell? Justify your answer.
{Hint: We just need to maximize revenue in this problem. Remember that
R PQ .}
APPLICATIONS IN FINANCE
4. Suppose that your local government is interested in doing a capital works project that will
have the following benefit schedule:
Year 1 $0
Year 2 $8000
Year 3 $15,000
a. Calculate the present value of the project if the social discount rate is 5%
b. Calculate the present value of the project if the social discount rate is 10%
c. Calculate the present value of the project if the social discount rate is 15%
APPLICATIONS IN ENVIRONMENTAL SCIENCE
5. Fruit flies are placed in a half-pint milk bottle with a banana (for food) and yeast plants
(for food and to provide a stimulus to lay eggs). Suppose that the fruit fly population P
230
after t days is given by P( t)
.
0.
37t
1
56.5e
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 471)
a. How many fruit flies will there be after 3 days?
b. When will the population of fruit flies be 215?
Math Camp 2011 Page 158

MATH CAMP
School of Public and Environmental Affairs
SOLVING WORD PROBLEMS
Day Five In-Class Exercises Solutions
1. A rectangular walk in a park surrounds a flower bed. The outer perimeter of the walk is
60- by 80-foot. If the walk is of uniform width and its area is equal to the area of the
flower bed, how wide is the walk?
(Johnson, How to Solve Word Problems in Algebra: A Solved Problem Approach, McGraw Hill, 2000)
Let x = width of walk in feet
The area of the large rectangle is 80 × 60 = 4800, the area of the small rectangle is onehalf
the area of the large rectangle = 2400, and the formula for the area of the small rectangle
is (80 - 2x) (60 - 2x).
Math Camp 2011 Page 159

APPLICATIONS IN STATISTICS
2. An educator wants to see how the number of absences a student in her class has affects
the student’s final grade. Draw a scatter plot of the data. Calculate r. Calculate a and b
and write the equation for the regression line.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 495)
No. of
absences,
x
Final
grade, y
10 12 2 0 8 5
70 65 96 94 75 82
120
100
80
60
40
20
0
0 2 4 6 8 10 12 14
x 37
y 482
xy 2,682
x 2
337
y 2
39, 526
r
n
xy
x
y
2
2
2
x
x
n
y
2
n
y
6(2682) (37)(482)
2
2
6(337)
(37) 6(39526) (482)
.98
Math Camp 2011 Page 160

a
2
y
x
x
xy
2
2
n
x
x
(482)(337) (37)(2682)
96.78
2
6(337) (37)
n
b
xy
x
y
2
2
n
x
x
6(2682) (37)(482)
2.668
2
6(337) (37)
Y
2.668X
96.78
APPLICATIONS IN ECONOMICS
3. General Motors (GM) produces light trucks in several Michigan factories, where its
annual fixed costs are $180 million, and its marginal cost per truck is approximately
$20,000. Regional demand for the trucks is given by P 30000 0. 1Q
, where P denotes
price in dollars and Q denotes annual sales of trucks.
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, pp. 115-116)
a. Find GM’s profit maximizing output level and price. Find the annual profit
generated by light trucks. {Hint: To find the maximizing output level and price,
we need to set marginal revenue equal to marginal cost and solve the resulting
equation. Marginal cost is given above. Marginal revenue is the first derivative
of the revenue equation, which is R PQ . Remember that profit is R C .}
R PQ
R (30000 0.1Q ) Q
R 30000Q
0.1Q
MR R
30000 (2)(0.1) Q
MR 30000 0.2Q
C 20000Q
180,000,000
MC C
20000
2
Math Camp 2011 Page 161

MR MC
30000 0.2Q
20000
0.2Q
20000 30000
0.2Q
10000
10000
Q 50,000
0.2
Thus, the optimal output is 50,000 trucks. We insert this into the Price equation
to get the optimal price.
P 30,000 0.1Q
P 30,000 0.1(50,000)
P 30,000 5,000 $25,000
At this level of output and price, profit would be given by
( P CPV ) Q FC
(25,000 20,000)(50,000) 180,000,000
$70,000,000
b. GM is getting ready to export trucks to several markets in South America. Based
on several marketing surveys, GM has found the elasticity of demand in these
foreign markets to be EP
9
for a wide range of prices (between $20,000 and
$30,000). The additional cost of shipping (including paying some import fees) is
about $800 per truck. One manager argues that the foreign price should be set at
$800 above the domestic price (in part a) to cover the transportation cost. Do you
agree that this is the optimal price for foreign sales? Justify your answer. {Hint:
Use the optimal price markup formula to solve this problem. This formula is
EP
P MC
E
where MC is the marginal cost and EP
is the price elasticity.
1
P
Marginal cost was given in part a, and we need to add to it the price of shipping
the trucks.}
EP
P
1
MC
EP
9
P (20,000
800)
1
9
P (1.125)(20,800) $23,400
According to the markup formula, GM should charge $23,400 per truck in the
foreign market due to the elasticity of demand to maximize profit.
Math Camp 2011 Page 162

c. GM also produces an economy (“no frills”) version of its light truck at a marginal
cost of $12,000 per vehicle. However, at the price set by GM, $20,000 per truck,
customer demand has been very disappointing. GM has recently discontinued
production of this model but still finds itself with an inventory of 18,000 unsold
trucks. The best estimate of demand for the remaining trucks is: P 30000 Q .
One manager recommends keeping the price at $20,000; another favors cutting
the price to sell the entire inventory. What price (one of these or some other
price) should GM set and how many trucks should it sell to maximize profit?
Justify your answer. {Hint: We just need to maximize revenue in this problem.
Remember that R PQ .}
R PQ
R (30,000 Q)
Q
R 30,000Q
Q
R
30,000 2Q
30,000 2Q
0
2Q
30,000
30,000
Q 15,000
2
At this quantity, the price will be
P 30,000 Q
P 30,000 15,000
P $15,000
2
These numbers say that GM should discount the price to $15,000 per vehicle, but
that in order to maximize revenue, GM should only sell 15,000 of the 18,000
trucks on hand.
APPLICATIONS IN FINANCE
4. Suppose that your local government is interested in doing a capital works project that will
have the following benefit schedule:
Year 1 $0
Year 2 $8000
Year 3 $15,000
a. Calculate the present value of the project if the social discount rate is 5%
Math Camp 2011 Page 163

PV
PV
PV
PV
PV
PV
PV
n
X
i
(1 r)
i1
X
1
(1 r)
0
(1 .05)
0
(1.05)
1
1
0 8000 15000
1.05 1.1025 1.157625
0 7256.24 12957.56
$20,213.80
X
2
(1 r)
1
i
8000
(1 .05)
8000
(1.05)
2
2
X
3
(1 r)
2
15000
(1 .05)
15000
3
(1.05)
3
3
b. Calculate the present value of the project if the social discount rate is 10%
PV
PV
PV
PV
PV
PV
PV
n
X
i
(1 r)
i1
X
1
(1 r)
0
(1 .10)
0
(1.10)
1
1
0 8000 15000
1.10 1.21 1.331
0 6611.57 11269.72
$17,881.29
X
2
(1 r)
1
i
8000
(1 .10)
8000
(1.10)
2
2
X
3
(1 r)
2
3
15000
(1 .10)
15000
3
(1.10)
3
c. Calculate the present value of the project if the social discount rate is 15%
Math Camp 2011 Page 164

PV
PV
PV
PV
PV
PV
PV
n
X
i
(1 r)
i1
X
1
(1 r)
0
(1 .15)
0
(1.15)
1
1
0 8000 15000
1.15 1.3225 1.520875
0 6049.15 9862.74
$15,911.89
X
2
(1 r)
1
i
8000
(1 .15)
8000
(1.15)
2
2
X
3
(1 r)
2
15000
(1 .15)
15000
3
(1.15)
3
3
APPLICATIONS IN ENVIRONMENTAL SCIENCE
5. Fruit flies are placed in a half-pint milk bottle with a banana (for food) and yeast plants
(for food and to provide a stimulus to lay eggs). Suppose that the fruit fly population P
230
after t days is given by P( t)
.
0.
37t
1
56.5e
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 471)
a. How many fruit flies will there be after 3 days?
230
P(
t)
0.37t
1
56.5e
230
P(
t)
0.37(3)
1
56.5e
230
P(
t)
1.11
1
56.5e
230
P(
t)
1
56.5(.32956)
230
P(
t)
118.62
230
19.62
11.7
b. When will the population of fruit flies be 215?
Math Camp 2011 Page 165

230
P(
t)
1
56.5e
230
215
1
56.5e
215 1
56.5e
215 12147.5e
12147.5e
12147.5e
0.37t
0.37t
0.37t
0.37t
0.37t
0.37t
15
e
12147.5
0.37t
15
ln e ln
12147.5
0.37t
ln e ln15 ln12147.5
0.37t
ln15 ln12147.5
ln15 ln12147.5
t
18.1
0.37
0.37t
230 215
15
230
230
The fruit fly population will be 215 after 18.1 days.
Math Camp 2011 Page 166

MATH CAMP
School of Public and Environmental Affairs
SOLVING WORD PROBLEMS
Day Five Homework
1. A carpenter wants to cut a 30-foot board into two pieces. He wants the longer piece to be
3 feet longer than twice the length of the shortest piece. Where should he cut the board?
2. You manage an ice cream factory that makes two flavors: Creamy Vanilla and
Continental Mocha. Into each quart of Creamy Vanilla go two eggs and 3 cups of cream.
Into each quart of Continental Mocha go one egg and 3 cups of cream. You have in stock
500 eggs and 900 cups of cream. How many quarts of each flavor should you make to
use all the eggs and cream?
(Waner & Costenoble, Finite Mathematics, 2 nd Edition, Brooks/Cole 2001)
3. Ezekiel has some coins in his pocket consisting of dimes, nickels, and pennies. He has
two more nickels than dimes and three times as many pennies as nickels. How many of
each kind of coin does he have if the total value is 52 cents?
(Johnson, How to Solve Word Problems in Algebra: A Solved Problem Approach, McGraw Hill, 2000)
APPLICATIONS IN STATISTICS
4. Refer to the following frequency distribution.
Class
Frequency
0 up to 5 2
5 up to 10 7
10 up to 15 12
15 up to 20 6
20 up to 25 3
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th edition, McGraw-Hill: Boston. p. 111)
a. What is the range of times taken to assemble a dresser?
fx
b. What is the grouped mean? x where f is the frequency of each group and
n
x is the midpoint of each group and n is the total number of observations (or sum
x
of f). The midpoint can be calculate as 1
x
m 2
.
2
c. What is variance?
s
2
fx
2
n 1
n
fx
2
Math Camp 2011 Page 167

5. The following data were obtained from a sample of counties in Southwestern
Pennsylvania and indicate the number (in thousands) of tons of bituminous coal produced
in each county and the number of employees working in coal production in each county.
Draw a scatter plot of the data. Calculate r. Calculate a and b and write the equation for
the regression line. Predict the number of employees needed to produce 500 thousand
tons of coal. The data are given here.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 495)
Tons, x 227 5410 5328 147 729 8095 635 6157
No. of
employees,
y
110 731 1031 20 118 1162 103 752
APPLICATIONS IN ECONOMICS
6. Suppose a firm assess its profit function as
2 3
10 48Q
15Q
Q
a. Compute the firm’s profit for the following levels of output: Q = 2, 8 and 14
b. Derive an expression for marginal profit (take the first derivative of the profit
function with respect to Q). Compute marginal profit at Q = 2, 8, and 14.
Confirm that profit is maximized at Q = 8.
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006)
7. Firms A and B make up a cartel that monopolizes the market for a scarce natural resource.
The firms’ marginal costs are MC
A
6 2QA
and MCB
18 QB
, respectively. The
firms seek to maximize the cartel’s total profit.
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, pp. 463)
a. The firms have decided to limit their total output to Q 18 . What outputs should
the firms produce to achieve this level of output at total minimum cost? What is
each firm’s marginal cost? {Hint: To produce output at minimum total cost, the
first should set their outputs so that MC
A
MCB
. We also know that we want
Q A
Q B
18 . We can solve these two equations simultaneously for Q
A
Q
B
b. The market demand curve is P 86 Q , where Q is the total output of the cartel.
Show that the cartel can increase its profit by expanding its total output. (Hint:
Compare MR to MC at Q 18 ).
c. Find the cartel’s optimal outputs and optimal price. (Hint: At the optimum,
MR MC A
MC B
.)
Math Camp 2011 Page 168

APPLICATIONS IN FINANCE
8. Capital One Bank of Glen Allen, Virginia recently offered a certificate of deposit that
paid 3.62% compounded daily. If a $5,000 CD earns this rate for 5 years, how much will
it be worth?
(Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences, tenth edition, Upper
Saddle River, NJ: Prentice Hall, pp. 124)
9. A cost-benefit analysis of a new irrigation project indicates that the net benefits ( B C)
of the project in each of the first four years will be -$2 million. Thereafter, the project
will yield positive net benefits of $750,000 for the next 20 years. Calculate the present
value of benefits minus costs when the social rate of discount is 10 percent. Does the
program merit approval? How would the present value of the net benefits change if the
social rate of discount were 15 percent? (Use a spreadsheet to make things easier.)
(Hyman, Public Finance: A Contemporary Application of Theory to Policy, seventh edition, Fort Worth, TX: Harcount,
2002, p. 238)
APPLICATIONS IN ENVIRONMENTAL SCIENCE
10. Often environmentalists will capture an endangered species and transport the species to a
controlled environment where the species can produce offspring and regenerate its
population. Suppose that six American bald eagles are captured, transported to Montana,
and set free. Based on experience, the environmentalists expect the population to grow
500
according to the model t)
where P (t)
is the population after t years.
P( 0.
162t
1
83.33e
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 474)
a. What is the predicted population of this species of American bald eagle in 20
years?
b. When will the population be 300?
11. In a particular region, there are two lakes rich in fish. The quantity of fish caught in each
lake depends on the number of persons who fish in each, according to
Q
1
N1
.
1
10 N
2
and Q2 16N2
.
4N2
, where N
1
and N
2
denote the number of fishers at each lake. In
all, there are 40 fishers.
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, pp. 248-249)
a. Suppose N
1
16 and N
2
24. At which lake is the average catch per fisher
greater? In light of this fact, how would you expect the fishers to redeploy
themselves?
b. How many fishers will settle at each lake? (Hint: Find N
1
and N
2
such that the
average catch is equal between the two lakes.)
c. The commissioner of fisheries seeks a division of fishers that will maximize the
total catch at the two lakes. Explain how the commissioner should use
information on the marginal catch at each lake to accomplish this goal. What
division of the 40 fishers would you recommend? {Hint: Set the marginal catch at
each lake equal to each other. Use the fact that N
1
N2
40 to help you solve
for N
1
and N
2
}
2
1
Math Camp 2011 Page 169

MATH CAMP
School of Public and Environmental Affairs
SOLVING WORD PROBLEMS
Day Five Homework Solutions
1. A carpenter wants to cut a 30-foot board into two pieces. He wants the longer piece to be
3 feet longer than twice the length of the shortest piece. Where should he cut the board?
Let x = length of the shorter piece
Let y = length of the longer piece
So x + y = 30 feet
Because the longer piece is 3 feet longer than twice the shortest piece, we know that y =
2x + 3
x y 30
x 2x
3 30
3x
3 30
3x
27
x 9
y 2x 3 2(9) 3 21
So the length of the shorter piece is 9 feet, the length of the longer piece is 21 feet. The
carpenter should cut the board at 9 feet.
2. You manage an ice cream factory that makes two flavors: Creamy Vanilla and
Continental Mocha. Into each quart of Creamy Vanilla go two eggs and 3 cups of cream.
Into each quart of Continental Mocha go one egg and 3 cups of cream. You have in stock
500 eggs and 900 cups of cream. How many quarts of each flavor should you make to
use all the eggs and cream?
(Waner & Costenoble, Finite Mathematics, 2 nd Edition, Brooks/Cole 2001)
Let V = creamy vanilla
Let M = continental mocha
Each quart of vanilla uses 2 eggs and 3 cups of cream
Each quart of mocha uses 1 egg and 3 cups of cream
Total eggs = 500
Total cups of cream = 900
Math Camp 2011 Page 170

2V + M = 500 (egg equation)
3V +3M = 900 (cream equation
2V
M 500
M 500 2V
3V
3M
3V
3(500 2V
) 900
3V
1500
6V
3V
V 200
900
600
900
M
M
M
500 2V
500 2(200)
500 400 100
So you should make 100 quarts of Continental Mocha and 200 quarts of Creamy Vanilla.
3. Ezekiel has some coins in his pocket consisting of dimes, nickels, and pennies. He has
two more nickels than dimes and three times as many pennies as nickels. How many of
each kind of coin does he have if the total value is 52 cents?
(Johnson, How to Solve Word Problems in Algebra: A Solved Problem Approach, McGraw Hill, 2000)
Let x = number of dimes
y = number of nickels
z = number of pennies
Math Camp 2011 Page 171

APPLICATIONS IN STATISTICS
4. Refer to the following frequency distribution.
Class
Frequency
0 up to 5 2
5 up to 10 7
10 up to 15 12
15 up to 20 6
20 up to 25 3
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th edition, McGraw-Hill: Boston. p. 111)
a. What is the range of times taken to assemble a dresser? 25-0 = 25
fx
b. What is the grouped mean? x where f is the frequency of each group and
n
x is the midpoint of each group and n is the total number of observations (or sum
x
of f). The midpoint can be calculate as 1
x
m 2
.
2
fx 2.5 2 ... 22.5 3
380
x
12.6667
n
30 30
2
fx 380
2
fx 5,637.50
2
c. What is variance? s
n
30
28.42
n 1
30 1
5. The following data were obtained from a sample of counties in Southwestern
Pennsylvania and indicate the number (in thousands) of tons of bituminous coal produced
in each county and the number of employees working in coal production in each county.
Draw a scatter plot of the data. Calculate r. Calculate a and b and write the equation for
the regression line. Predict the number of employees needed to produce 500 thousand
tons of coal. The data are given here.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 495)
Tons, x 227 5410 5328 147 729 8095 635 6157
No. of
employees,
y
110 731 1031 20 118 1162 103 752
2
Math Camp 2011 Page 172

1400
1200
1000
800
600
400
200
0
0 2000 4000 6000 8000 10000
x 26,728
y 4,027
xy 23,663,669
x
2 162,101,162
y 2
3,550, 103
r
n
xy
x
y
2
2
2
x
x
n
y
2
n
y
8(23663669) (26728)(4027)
2
2
8(162101162)
(26728) 8(3550103) (4027)
.97
a
2
y
x
x
xy
2
2
n
x
x
(4027)(162101162) (26728)(23663669)
2
8(162101162) (26728)
n
b
xy
x
y
2
2
n
x
x
8(23663669) (26728)(4027)
.14
2
8(162101162) (26728)
34.85
Math Camp 2011 Page 173

Y
. 14X
34.85
To produce 500 thousand tons of coal, we would need
Y . 14X 34.85 .14(500) 34.85 105
employees
APPLICATIONS IN ECONOMICS
6. Suppose a firm assess its profit function as
2 3
10 48Q
15Q
Q
a. Compute the firm’s profit for the following levels of output: Q = 2, 8 and 14
Respective profits are: -10, -54, 54, and -486
b. Derive an expression for marginal profit (take the first derivative of the profit
function with respect to Q). Compute marginal profit at Q = 2, 8, and 14.
Confirm that profit is maximized at Q =8.
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006)
d
M
48
30Q
3Q
dQ
2
3(
Q 2)( Q 8)
Marginal profit is zero at Q = 2 and Q = 8. From part a, we can determine that
profit reaches a local minimum at Q = 2 and a maximum at Q = 8.
7. Firms A and B make up a cartel that monopolizes the market for a scarce natural resource.
The firms’ marginal costs are MC
A
6 2QA
and MCB
18 QB
, respectively. The
firms seek to maximize the cartel’s total profit.
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, pp. 463)
a. The firms have decided to limit their total output to Q 18 . What outputs should
the firms produce to achieve this level of output at total minimum cost? What is
each firm’s marginal cost? {Hint: To produce output at minimum total cost, the
first should set their outputs so that MC
A
MCB
. We also know that we want
Q A
Q B
18 . We can solve these two equations simultaneously for Q
A
and
MC
6 2Q
2Q
A
A
MC
A
Q
18 Q
B
B
12
B
Q
B
Math Camp 2011 Page 174

2QA
QB
12
Q
A
QB
18
2Q
Q Q Q
3Q
Q
Q
A
A
A
A
10 Q
A
30
B
30
10
3
Q 18
B
B
18
B
12 18
QB
18 10
8
To find each firm’s marginal cost, we only need to solve for one firm’s marginal
cost because they are both equal.
MC
A
6 2Q
6 2(10) 6 20 26
A
b. The market demand curve is P 86 Q , where Q is the total output of the cartel.
Show that the cartel can increase its profit by expanding its total output. (Hint:
Compare MR to MC at Q 18 ).
R PQ
R (86 Q)
Q
R 86Q
Q
MR R
86 2Q
2
Now we calculate marginal revenue at Q 18
MR 86 2Q
MR 86 2(18) 50
In part a, we found that MC 26 50 MR . Because the marginal cost is lower
than the marginal revenue, the cartel can profit by expanding output.
c. Find the cartel’s optimal outputs and optimal price. (Hint: At the optimum,
MR MC A
MC B
.)
From part a, we know that:
Math Camp 2011 Page 175

MC
6 2Q
2Q
Q
Q
B
A
A
B
MC
A
Q
12 2Q
2Q
18 Q
B
A
B
12
12
A
B
We also know that
MR MC
86 2( Q
86 2Q
A
A
A
Q ) 6 2Q
B
2Q
B
6 2Q
Now we can substitute in for
MR MC
86 2( Q
86 2Q
86 2Q
86 2Q
2Q
8Q
Q
A
A
A
A
A
A
A
A
2Q
2(2Q
4Q
4Q
Q
A
104
B
B
2Q
104
13
8
Now plug this value of
B
6 2Q
A
) 6 2Q
A
A
A
Q to find
12)
6 2Q
24 6 2Q
A
A
A
A
6 24 86
Q
A
.
A
Q
A
back into the first equation to get
Q
B
.
Q
Q
B
B
2Q
A
12
2(13) 12
26 12
14
The price at this equilibrium will be:
P 86 ( Q
A
Q
P 86 (13 14)
86 27 59
B
)
The common value for
MR MC A
MC will be:
B
MR 86 2( Q
A
Q
MR 86 2(13 14)
86 54 32
B
)
Math Camp 2011 Page 176

APPLICATIONS IN FINANCE
8. Capital One Bank of Glen Allen, Virginia recently offered a certificate of deposit that
paid 3.62% compounded daily. If a $5,000 CD earns this rate for 5 years, how much will
it be worth?
(Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences, tenth edition, Upper
Saddle River, NJ: Prentice Hall, pp. 124)
Compound Interest
A P
1
r
m
mt
where A = future value of the account
P = the principal or present value of the account
r = annual rate (as a decimal)
m = the number of times per year the interest is compounded (for instance, for interest
compounded quarterly m 4)
A P1
r
m
.0362
A 5,0001
365
365*5
1825
$5,992. 02
A 5,000 1.000099178
mt
9. A cost-benefit analysis of a new irrigation project indicates that the net benefits ( B C)
of the project in each of the first four years will be -$2 million. Thereafter, the project
will yield positive net benefits of $750,000 for the next 20 years. Calculate the present
value of benefits minus costs when the social rate of discount is 10 percent. Does the
program merit approval? How would the present value of the net benefits change if the
social rate of discount were 15 percent? (Use a spreadsheet to make things easier.)
(Hyman, Public Finance: A Contemporary Application of Theory to Policy, seventh edition, Fort Worth, TX: Harcount, 2002, p. 238)
Year Net Benefit 1 + r (1+r)^n x/(1+r)^n
1 -2,000,000 1.05 1.05 -1904761.905
2 -2,000,000 1.05 1.1025 -1814058.957
3 -2,000,000 1.05 1.157625 -1727675.197
4 -2,000,000 1.05 1.215506 -1645404.95
5 750,000 1.05 1.276282 587644.6249
6 750,000 1.05 1.340096 559661.5475
7 750,000 1.05 1.4071 533010.9976
8 750,000 1.05 1.477455 507629.5215
9 750,000 1.05 1.551328 483456.6872
10 750,000 1.05 1.628895 460434.9402
Math Camp 2011 Page 177

11 750,000 1.05 1.710339 438509.4668
12 750,000 1.05 1.795856 417628.0636
13 750,000 1.05 1.885649 397741.013
14 750,000 1.05 1.979932 378800.9647
15 750,000 1.05 2.078928 360762.8236
16 750,000 1.05 2.182875 343583.6415
17 750,000 1.05 2.292018 327222.5157
18 750,000 1.05 2.406619 311640.4912
19 750,000 1.05 2.52695 296800.4678
20 750,000 1.05 2.653298 282667.1122
21 750,000 1.05 2.785963 269206.7735
22 750,000 1.05 2.925261 256387.4033
23 750,000 1.05 3.071524 244178.4793
24 750,000 1.05 3.2251 232550.9327
PV $ 597,617.46
Year Net Benefit 1 + r (1+r)^n x/(1+r)^n
1 -2,000,000 1.15 1.15 -1739130.435
2 -2,000,000 1.15 1.3225 -1512287.335
3 -2,000,000 1.15 1.520875 -1315032.465
4 -2,000,000 1.15 1.749006 -1143506.491
5 750,000 1.15 2.011357 372882.5515
6 750,000 1.15 2.313061 324245.6969
7 750,000 1.15 2.66002 281952.7799
8 750,000 1.15 3.059023 245176.3304
9 750,000 1.15 3.517876 213196.809
10 750,000 1.15 4.045558 185388.5296
11 750,000 1.15 4.652391 161207.417
12 750,000 1.15 5.35025 140180.3626
13 750,000 1.15 6.152788 121895.9675
14 750,000 1.15 7.075706 105996.4935
15 750,000 1.15 8.137062 92170.8639
16 750,000 1.15 9.357621 80148.57731
17 750,000 1.15 10.76126 69694.41505
18 750,000 1.15 12.37545 60603.83917
19 750,000 1.15 14.23177 52698.99059
20 750,000 1.15 16.36654 45825.20921
21 750,000 1.15 18.82152 39848.008
22 750,000 1.15 21.64475 34650.44174
23 750,000 1.15 24.89146 30130.81891
24 750,000 1.15 28.62518 26200.71209
PV -$3,025,861.91
Math Camp 2011 Page 178

When using a social discount rate of 5%, the present benefit of the project is $597,617. 46.
When using a social discount rate of 15%, the present value of the project is -$3,025,861.91. If
the social discount rate if 15%, then the project shouldn’t be done.
APPLICATIONS IN ENVIRONMENTAL SCIENCE
10. Often environmentalists will capture an endangered species and transport the species to a
controlled environment where the species can produce offspring and regenerate its
population. Suppose that six American bald eagles are captured, transported to Montana,
and set free. Based on experience, the environmentalists expect the population to grow
500
according to the model P( t)
where P (t)
is the population after t years.
0.
162t
1
83.33e
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 474
a. What is the predicted population of this species of American bald eagle in 20
years?
500
P(
t)
0.162t
1
83.33e
500
P(
t)
1
83.33e
500
P(
t)
1
83.33(.039)
500
P(
t)
1
3.26
0.162(20)
500
4.26
117.3
After 20 years, there will be approximately 117 bald eagles.
Math Camp 2011 Page 179

. When will the population be 300?
500
P(
t)
0.162t
1
83.33e
500
300
0.162t
1
83.33e
0.162t
300(1 83.33e
) 500
300 24999e
24999e
24999e
0.162t
0.162t
0.162t
500 300
200
500
0.162t
200
e
24999
0.162t
200
ln e ln
24999
0.162t
ln e ln 200 ln 24999
0.162t
ln 200 ln 24999
ln 200 ln 24999 5.298 10.1266
t
29.80
0.162 0.162
It will take almost 30 years for the population to reach 300.
11. In a particular region, there are two lakes rich in fish. The quantity of fish caught in each
2
lake depends on the number of persons who fish in each, according to Q 10N
.
N
1 1
1
2
and Q2 16N2
.
4N2
, where N
1
and N
2
denote the number of fishers at each lake. In
all, there are 40 fishers.
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, pp. 248-249)
a. Suppose N
1
16 and N
2
24. At which lake is the average catch per fisher
greater? In light of this fact, how would you expect the fishers to redeploy
themselves?
The average catch at lake 1 is given by the formula:
1
AQ
AQ
AQ
AQ
1
1
1
1
Q
N
1
1
10N1
.1N
N
2
10(16) .1(16)
16
160 .1(256) 160 25.6
16 16
8.4
1
2
1
Math Camp 2011 Page 180

The average catch at the second lake will be:
2
Q2
16N
2
.4N
2
AQ2
N N
1
2
2
16(24) .4(24)
AQ1
24
384 .4(576) 384 230.4
AQ1
24
24
AQ 6.4
2
Because the average catch at the first lake is greater, more fishermen will
gravitate toward that lake.
b. How many fishers will settle at each lake? (Hint: Find N
1
and N
2
such that the
average catch is equal between the two lakes.)
AQ
1
10N1
.1N
N
1
10 .1N
10N1
.1N
N
1
2
1
1
2
1
16N
2
.4N
N
16 .4N
16N
2
.4N
N
2
2
2
2
2
2
2
AQ
2
We also know that N
1
N2
40 . Using these two equations, we can solve for
N
1
and N
2
.
N
N
1
1
.1N
.1N
.5N
N
2
40
40 N
10 .1N
10 4 .1N
6 .1N
.4N
10
2
10 .1(40 N
N
2
2
2
2
2
1
16 .4N
16 .4N
16 .4N
16 6 .4N
2
2
2
) 16 .4N
10
10
20
.5
2
2
2
2
2
N
N
1
1
40 N
40 20 20
2
Math Camp 2011 Page 181

The movement between the lakes will cease when the average catch at the two
lakes are the same. This happens when we have 20 fishers at each lake.
c. The commissioner of fisheries seeks a division of fishers that will maximize the
total catch at the two lakes. Explain how the commissioner should use
information on the marginal catch at each lake to accomplish this goal. What
division of the 40 fishers would you recommend? {Hint: Set the marginal catch at
each lake equal to each other. Then use the information that N
1
N2
40 to
help you solve the equation for N
1
and N
2
}
2
Q1
10N1
.1N1
MQ Q 10 .2N
1
1
1
2
Q2
16N
2
.4N
2
MQ Q 16 .8N
2
2
2
MQ
1
MQ
10 .2N
16 .8N
1
2
2
We know that
N
N
1
1
N
2
40
40 N
2
, so we can insert this value in for N1
to solve for N
2
MQ
10 .2N
10 .2(40 N
10 8 .2N
.2N
N
2
1
2
MQ
.8N
14
1
2
16 .8N
2
2
2
2
) 16 .8N
16 .8N
2
16 10
8
2
N
N
1
1
40 N
40 14
26
2
Thus, we want 26 fishers at lake one and 14 fishers at lake two.
Math Camp 2011 Page 182

500
P(
t)
1
83.33e
500
300
1
83.33e
300(1 83.33e
300 24999e
24999e
24999e
0.162t
0.162t
0.162t
0.162t
0.162t
0.162t
500 300
200
) 500
500
0.162t
200
e
24999
0.162t
200
ln e ln
24999
0.162t
ln e ln 200 ln 24999
0.162t
ln 200 ln 24999
ln 200 ln 24999 5.298 10.1266
t
29.80
0.162 0.162
It will take almost 30 years for the population to reach 300.
Math Camp 2011 Page 183

MATH CAMP
School of Public and Environmental Affairs
Supplemental Problems
ORDERS OF OPERATIONS
1. Evaluate the following expressions using the order of operations:
a.
2
49 75
4 316
b. 5 2 30 6
c.
2
( 5) 30 6
d. 2
16 2 3 7
4
e. 49 7 2 15
3 42
48
3
2
2 2
f. 2 5
4
3
2 36
23
1200
3 15
82
ROOTS AND EXPONENTS
2. Compute:
5
a. 3
1
b.
3
4
c. 3 -8
d.
0
9
1
1
e.
5
f.
40
1
g.
2
15
h. 5 -2
i. 16 -1
j. ( 2)
k. 5 -3
3. Evaluate the following:
3
2
a. 9
2
b. 8 3
3
1
1
3
c.
27
d. 3 2
e. 4
16
Math Camp 2011 Page 184

1
3
f. 125
g.
7 2
4. Simplify the following:
2 8
a. ( 3) ( 3)
b.
c. x x
d.
4
7 7
e.
0
2 2
f.
9
5
2
5
g.
3
3
6
3
3 6
5 5
1 9
2
h. 4
5
7
3
5
8
i.
2 2
3 4
j.
5
5
5
10
(Hint: the answer should be in the form of 2 n )
4
3 7
3
k.
6
5
l.
4
5
5
x
m.
5
x
3x
n.
5y
5. Put in radical form:
1
2
a. ( 4x
)
b.
2
0.8
x (Hint: Change 0.8 to a fraction.)
1 1
3
6
c. x x
3
2
d. x
6. Simplify:
a.
b.
3a
a
1
2
b
1
2
1
3
b
3
25x
y
3
9x
y
1
3
2
3
1
2
Math Camp 2011 Page 185

7. Express the following in scientific notation:
a. 6,780,000
b. 456
c. -9,312
d. 938,000
e. .00000276
f. .03
g. .00000000000000000000723
8. Write in common decimal form:
3
a. 4.2
10
b. (5 x 10 5 )(4 x 10 -3 )
c. 4.007 x 10 -6
9. Multiply the following:
a. (5)(10
5 12
) (2.5)(10 )
FRACTIONS
4
10
b. ( 3.4)(10
) (8.1)(10
)
10. Compute:
a.
3 3
4 4
b.
8 4
9 7
c.
6 11
8
d.
2 13
7 14
e.
4 3
6 6
f.
5 13
3 14
g.
4 3
9
4
h.
7
2
3
3
j. 6
7
3
4
i.
5
Math Camp 2011 Page 186

WORKING WITH ALGEBRAIC EXPRESSIONS
11. What degree are the following polynomials?
6 4
a. 5x
7x
2x
3
b. 3x – 2
c. x 3 x
12. Simplify the following expressions by collecting like terms:
3
2
3
a. x 5x
15
2x
2x
3x
8x
9 x
2 3
2
3
b. 4x
6x
x x 4( x 3) 4x
2
c. 17x
5x
3x
3
x 12
x
13. Multiply the following polynomials:
5x
3 7x
1
a.
b.
6x
2 5 x 4
14. Compute the following:
a.
7x 2 4 x 5
2x
1
x
b.
8x
3 9x
2
x 1
x 2
c.
3
5x
6 x 1
2
x 1
5x
d.
3x
2 5x
2
x 9 x 1
e.
3
2
6x
5x
1
7x
2x
6 x 1
15. Evaluate the following expressions:
a. 5x 2 6x
3 , where x = 5
b.
2
2
7a 5ab
2b
11a
3, where a = 2 and b = -3
c. 9y 17 y 6 , where y = -1
2
d. y 5x
3y
7( x 2)
where y = 3 and x = 5
2
e. ( y ) 5x
3y
7( x 2)
where y = 3 and x = 5
16. A culture of bacteria triples every hour. If it weighs one ounce at the beginning, what will it
weigh:
a. One hour later
b. Two hours later
c. Three hours later
d. Four hours later
17. A retail store faces a demand equation for Roller Blades given by:
Q = 180 – 1.5P,
where Q is the number of pairs sold per month and P is the price per pair in dollars.
a. The store currently charges P = $80 per pair. At this price, determine the number
of pairs sold.
Math Camp 2011 Page 187

. If management were to raise the price to $100, how many pairs would the store
sell?
18. The original revenue function for the microchip producer is R = 170Q -20Q 2 , where R is the
total revenue and Q is the number of microchips sold.
a. If the microchip producer sells 10 microchips, how much money will the producer
make?
b. If the microchip producer sells 5 microchips, how much money will the producer
make?
PERCENT CHANGE
19. Suppose that your car was worth $11,354 three years ago and it is now worth $3,221. What
is the percent change?
WORKING WITH SUMMATION SIGNS
20. Find
x
x for the following:
n
a. 5, 6, 17, 3, -5, -25, 6, -12, 9, 31
b. 3.5, 2.1, 2.8, 3.9, 4.0, 1.9, 3.0
c. $5.50, $10.71, $12.01, $1.35, $6.50, $8.98, $9.12, $8.80, $15.00, $2.36, $3.30,
$6.66
x
21. For each set of values, find x , x 2 , and 2
a. 80, 76, 42, 53, 77
b. -9, -12, 18, 0, -2, -15
c. 12, 52, 36, 81, 63, 74
22. Calculate 2
x i
x
a. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
b. -13, 15, -34, 23, 31, -40
for the following. To do this you will first need to compute
2
x
x .
n
2
x
2
23. Show that x
x
x
n
24. A sample of personnel files of eight male employees employed by Acme Carpet revealed
that, during a six-month period, they lost the following number of days due to illness: 2, 0, 6,
x
3, 10, 4, 1, and 2. Calculate the mean x and the mean deviation
x x
for these
n
n
data.
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th edition, McGraw-Hill: Boston.)
25. Each person who applies for an assembly job at Carolina Furniture, Inc. is given a
mechanical aptitude test. One part of the test involves assembling a dresser based on
numbered instructions. A sample of the lengths of time it took 42 persons to assemble the
dresser was organized into the following frequency distribution.
Math Camp 2011 Page 188

Length of Time
Frequency
(minutes)
2 - 4 4
4 – 6 8
6 – 8 14
8 – 10 9
10 – 12 5
12 – 14 2
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th edition, McGraw-Hill: Boston. p. 111)
a. What is the range of times taken to assemble a dresser?
fx
b. What is the grouped mean? x where f is the frequency of each group and x is
n
the midpoint of each group and n is the total number of observations (or sum of f).
x
The midpoint can be calculate as 1
x
m 2
.
2
fx
2
fx
2
c. What is variance? s
n
n 1
26. Merrill Lynch Securities and Health Care Retirement, Inc. are two large employers in
downtown Toledo, Ohio. They are considering jointly offering child care for their
employees. As a part of the feasibility study, they wish to estimate the mean weekly childcare
cost of their employees. A sample of 10 employees who use child care reveals the
following amounts spent last week--$107, $92, $97, $95, $105, $101, $91, $99, $95, $104.
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th edition, McGraw-Hill: Boston. p. 313)
a. What is the mean?
b. What is the variance?
c. If the standard deviation is the square root of the variance, what is the standard
deviation?
d. Given the following formula for a confidence interval, compute the confidence
s
interval for these data. x t . In this case, x is the mean you just calculated, s is
n
the standard deviation you just calculated, n is the sample size, and t =1.833.
27. There is a statistic called the chi-squared statistic that you will see again in your statistics
class. The formula for computing this statistic is:
2 O E
E
where O is the
observed value and E is the expected value. Given the following table of data, compute the
chi-squared statistic.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 515)
Frequency Cherry Strawberry Orange Lime Grape
Observed 32 28 16 14 10
Expected 20 20 20 20 20
2
2
Math Camp 2011 Page 189

28. Given the following data, calculate r. This is called the correlation coefficient, and you will
see more of this in your statistics class.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 481)
r
FACTORING
n
xy
x
y
2
2
2
x
x
n
y
n
y
Subject Age, x Pressure, y
A 43 128
B 48 120
C 56 135
D 61 143
E 67 141
F 70 152
29. Factor the given polynomial:
5 3 2 4
a. 15x y 24x
y
b.
3
3
15 5x
3y
yx
c.
3 2
x 1
x x
d.
2
4x
20xy
25y
e.
f.
g.
2
2
4x
101xy
25y
2 2
36x 16y
x
2n
y
4n
SOLVING EQUATIONS
2
30. Find the solutions to the following equations using the factoring formulas:
2
a. y 9y 20 0
b.
2
x 8x 12
c. 45x
2 57x 18
0
d. 9x
2 4 0
e.
2
( x 2) x(2
3x)
f. y 5 y 1
g. x 2 6
2
Math Camp 2011 Page 190

31. Find the solutions to the following equations by using the quadratic formula:
2
a. x 2x 3 0
2
b. 2x
5 x
c. 8w
2 12w
9
32. Find a number whose square exceeds fourteen times the number by 51.
33. An object is thrown upward from the top of a building, and if s feet is the distance of the
object from the ground t seconds after it is thrown, s 16t
2 48t
100
. How long will it
take the object to strike the ground?
SOLVING SYSTEMS OF EQUATIONS
34. Find the solution set to the following systems of equations using the method of substitution.
3x
y 5
a.
5x
y 3
x
3y
11
b.
x
y 1
35. Find the solution set to the following system of equations using the addition/subtraction
method. You might have to rearrange the equations to be in the correct format first. If the
system cannot be solved, then indicate whether it is dependent or inconsistent.
4x
3y
1
a.
6x
y 7
5x
3y
1
b.
x
3y
4
4x
3y
12
0
c. 4 3
y
x
3 2
x y
1
2 3
d.
x y
5
4
3
SOLVING INEQUALITIES
36. Solve the following inequalities for x:
a. 9x
36 x 64
b. x 2 49
c. 3x 10
2x
5
d. x 4 5x
12
32 x
e. 6x 2 6x
13
f. ( x 2)( x 3) ( x 1)(
x 1)
g. 6(2y 3) 2(6y
12)
Math Camp 2011 Page 191

h.
i.
j.
5x 3 1 3
2x
2 9x 5 0
x 2 x
37. Solve the following equations and inequalities for x:
a. 5y 8
b.
c.
d.
e.
38. In your Economics 101 class, you have scores of 68, 82, 87, and 89 on the first four of fives
tests. To get a grade of B, the average of the first five test scores must be greater than or
equal to 80 and less than 90. Solve an inequality to find the range of the score that you need
on the last test to get a B. (Michael Sullivan, College Algebra, Upper Saddle Rive, NJ: Prentice Hall, 2005, p. 135)
FUNCTIONS
3 2 2
x
2
x
2
3x 2 2
1
3
1
8x
5( x 4) 5 x 3 6
9x
6 x 12
x 2 x
39. Determine the domain and range of the following functions:
e.
5
f ( x)
2x
8x
7
f.
x 3
f ( x)
x 5
g. f ( x)
12 x
BASICS OF GRAPHING FUNCTIONS
40. Graph the following lines:
a. 4x
2y
20
b. 4x
2y
20
c. x 8 0
d. y 1
e. 2x
3y
0
41. A statistics instructor wishes to determine if a relationship exists between the final exam
scores in Statistics 101 and the final exam scores of the same students who took Statistics 102.
Draw a scatter plot and comment on the nature of the relationship.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 79)
Math Camp 2011 Page 192

Stat 101, x 87 92 68 72 95 78 83 98
Stat 102, y 83 88 70 74 90 74 83 99
42. The data shown indicate the number of tournaments and the earnings in thousands of dollars
of 10 randomly selected LPGA golfers. Draw a scatter plot for the data and determine the
nature of the relationship.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 79)
No. of 27 30 24 21 18 32 25 32 25 27
tournaments,x
Earnings, y $956 757 583 517 104 173 252 303 355 405
43. An educator wants to see if there is relationship between the number of absences a student
has and his or her final grade in a course. Draw a scatter plot and comment on the nature of
the relationship.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 79)
No. of
10 12 2 0 8 5
absences, x
Final grade, y 70 65 96 94 75 82
LINEAR FUNCTIONS AND THEIR GRAPHS
44. Draw a graph of the line connecting these two points and find the length of the line.
a. (2,1), (-1,5)
b. (7,-4), (-5,1)
45. Draw a graph of the line connecting these two points and find the slope of the line.
a. (-7,-1), (-3,4)
b. (0,-5), (-5,0)
c. (3,5), (5, 9)
46. Find the midpoint between the two points:
a. (8,9), (3,9)
b. (-5,0), (7,7)
c. (1,1), (3,3)
47. Find the lengths of the sides of the triangle having vertices at the three given points. Graph
the triangle. A(2,3), B(3,-3), C(-1,1)
48. Show that the line through the points A(3,9) and B(5,5) is parallel to the line through the
points C(1,7) and D(3,3). Draw a sketch of the two lines.
49. Find an equation of the line through the two points. Graph the line.
a. ( 3, 6),(
2,
3)
b. ( 1, 2),(7,
2)
c. ( 3,1),( 5,4)
50. Find an equation of the line through the given point with the given slope. Graph the line.
a. ( 3,
6),
m 3
Math Camp 2011 Page 193

2
b. ( 4,0),
m
5
51. Find an equation of the line through the point (6,3) and perpendicular to the line whose
equation is x 3y 15
0 . Graph the lines (on the same axis).
52. Find an equation of the line that is parallel to the given line and passes through the given
point. 5x 2y 1
0, ( 3,3)
53. A ramp connects the ground to the loading platform of a warehouse. If the platform is 5 feet
high and the ramp has a slope of 0.28, how far from the wall of the warehouse is the base of
the ramp?
QUADRATIC FUNCTIONS AND THEIR GRAPHS
54. Graph the following equations:
a. f ( x)
x
2 10
b. f ( x)
25x
2 13
c. f ( x)
10x
5
d. f ( x)
10x
5
e. f ( x)
10x
5
55. The percentage s of seats in the House of Representatives won by Democrats and the
percentage v of votes cast for Democrats (when expresses as decimal fractions) are related by
the equation:
5v 2s
1.4
0 s 1
0.28
v 0. 68
a. Express v as a function of x, and find the percentage of votes required for the
Democrats to win 51% of the seats.
b. Express s as a function of v, and find the percentage of seats won if the Democrats
receive 51% of the votes.
56. A projectile is fired from a cliff 500 feet above the water at an inclination of 45 degrees to
the horizontal, with a muzzle velocity of 400 feet per second. In physics, it is established that
2
32x
the height h of the projectile above the water is given by h ( x)
x 500 , where x is
2
(400)
the horizontal distance of the projectile from the base of the cliff. (Sullivan, College Algebra, seventh
edition, Upper Saddle River, NJ: Prentice Hall, pp. 302-303)
a. Find the maximum height of the projectile.
b. How far from the base of the cliff will the projectile strike the water?
Math Camp 2011 Page 194

57. The price p and the quantity x sold of a certain project obey the demand equation
x 5p
100 , 0 p 20
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, pp. 308)
a. Express the revenue R as a function of x.
b. What is the revenue if 15 units are sold?
c. What quantity x maximizes revenue? What is the maximum revenue?
d. What price should the company charge to maximize revenue?
POLYNOMIAL AND RATIONAL FUNCTIONS
58. The following data represent the number of motor vehicle thefts (in thousands) in the United
States for the years 1987-1997, where 1 represents 1987, 2 represents 1988, and so on.
Year, x Motor Vehicle
Thefts, T
1987,1 1289
1988,2 1433
1989,3 1565
1990,4 1636
1991,5 1662
1992,6 1611
1993,7 1563
1994,8 1539
1995,9 1472
1996,10 1394
1997,11 1354
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, pp. 329)
a. Draw a scatter diagram of the data. Comment on the type of relation that may
exist between the two variables.
b. The cubic function of best fit to these data is:
3
2
T ( x)
1.52x
39.81x
282.29x
1035.5
Use this function to predict the number of motor vehicle thefts in 1994.
59. A company manufacturing snow-boards has fixed costs of $200 per day and total costs of
$3,800 per day at a daily output of 20 boards. (Barnett, Ziegler and Byleen, Finite Mathematics for Business,
Economics, Life Sciences, and Social Sciences, tenth edition, Upper Saddle River, NJ: Prentice Hall, pp. 93)
a. Assuming that the total cost per day, C (x)
, is linearly related to the total output
per day, x, write an equation for the cost function.
b. The average cost per board for an output of x boards is given by C ( x)
C(
x)
x .
Find the average cost function.
c. Sketch a graph of the average cost function, including any asymptotes, for
1 x 30.
d. What does the average cost per board tend to as production increases?
60. Graph the following equations:
3
a. f ( x)
x 10x
6
b. f ( x)
x 3
c. f ( x)
x 2 10
Math Camp 2011 Page 195

EXPONENTIAL FUNCTIONS
61. Graph the following functions:
x
a. f ( x)
5
b.
f ( x)
3
x
x
1
c. f ( x)
10
d. f ( x)
(log
2
x)
3
e. f ( x)
log
10
x
f. f ( x)
ln( x 5)
62. Suppose that $2,500 is invested at 7% compounded quarterly. How much money will be in
the account in:
(Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences, tenth edition, Upper Saddle
River, NJ: Prentice Hall, pp. 107)
a. ¾ of a year?
b. 15 years?
63. If you invest $7,500 in an account paying 8.35% compounded continuously, how much
money will be in the account at the end of:
(Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences, tenth edition, Upper Saddle
River, NJ: Prentice Hall, pp. 107)
a. 5.5 years?
b. 12 years?
64. The Joint United Nations Program on HIV/AIDS reported that HIV had infected 60 million
people worldwide prior to 2002. Assume that number increases at an annual rate of 8%
compounded continuously.
(Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences, tenth edition, Upper Saddle
River, NJ: Prentice Hall, p. 108)
a. Write an equation that models the worldwide spread of HIV, letting 2002 be year
0.
b. Based on the model, how many people (to the nearest million) had been infected
prior to 1999? How many would be infected prior to 2010?
LOGARITHMIC FUNCTIONS
65. Express the following relationships in terms of logarithms.
a. 3
b. 10 5
. 00001
66. Express the following relationships using exponential notation.
a. log 7
49 2
9 2 1
b. log 16 4
1
2
Math Camp 2011 Page 196

67. Find the value of the following logarithms.
a.
1
log 4
4
b. log 3
3
c. log 10
. 000001
d.
3
log
5
25
e. log
e
e
68. Solve the following for x or b, respectively.
1 2
a. log
b
16 3
b. log x 4
1
3
c. log 81 2
b
d. log
6(
x 4) log
6(
x 9) 2
e.
2
log
4(
x 6x)
2
f. log
2(11
x ) log
2(
x 1)
3
69. Simplify the following logarithms (express as one logarithm):
a.
1
4log
10
x log
10
y
2
b.
3
4
log
b
x 6log
b
y log
b
z
4
5
70. Evaluate the following logarithms given that log 10
2 0. 3010 , log 10
3 0. 4771, and
log 10
7
a.
0.8451
log
log
10
10
2
3
b.
5
49
log
10
2
36
c. log 10
35
d.
14
log
10
3
84
e. log 10
30
f. log 10
49000
g. log 10
. 0000006
71. Shannon’s diversity index is a measure of the diversity of a population. The diversity
index is given by the formula
H p1 log p1
p2
log p2
...
p n
log p n
Math Camp 2011 Page 197

where p
1
is the proportion of the population that is species 1, p
2
is the proportion of the
population that is species 2, and so on.
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 439)
a. According to the U.S. Census Bureau, the distribution of race in the United States
in 2000 was as follows:
Race
Proportion
American Indian or 0.014
Native Alaskan
Asian 0.041
Black or African 0.128
American
Hispanic 0.124
Native Hawaiian or 0.003
Pacific Islander
White 0.690
Compute the diversity index of the United States in 2000.
b. The largest value of the diversity index is given by H log( S)
, where S is the
number of categories of race. Compute H
max
.
H
c. The evenness ratio is given by E H
, where 0 E 1. If E 1, there
is a complete evenness. Compute the evenness ratio for the United States.
72. How many years (to two decimal places) will it take $1,000 to grow to $1,800 if it is invested
at 6% compounded quarterly? Compounded continuously?
(Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences, tenth edition, Upper Saddle
River, NJ: Prentice Hall, pp. 120)
SOLVING WORD PROBLEMS
73. A ladder is leaning against a building. The base of the ladder is 8 feet away from the side of
the building and the ladder reaches 9 feet high on the building. How long is the ladder?
74. A farmer mixes milk containing 3% butterfat with cream containing 30% butterfat to obtain
900 gallons of milk which is 8% butterfat. How much of each must the farmer use?
(Johnson, How to Solve Word Problems in Algebra: A Solved Problem Approach, McGraw Hill, 2000)
75. The data shown indicate the number of wins and the number of points scored for teams in the
National Hockey League. Draw a scatter plot for the data and describe the nature of the
relationship.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 79)
Wins,x 10 9 6 5 4 12 11 8 7 5 9 8 6 6 4
Points,y 23 22 15 15 10 26 26 26 21 16 12 19 16 16 11
H max
max
H
H
76. Evaluate the following expression given that p =.8, n = 2000, and z = 1.96.
p( 1
p)
a. Evaluate. p z
n
Math Camp 2011 Page 198

p(1
p)
b. If p = .8 and z = 19.6 and p z . 75 , find n.
n
APPLICATIONS IN STATISTICS
77. A sample of eight companies in the aerospace industry was surveyed as to their return on
investment last year. The results are: 10.6, 12.06, 14.8, 18.2, 12.0, 14.8, 12.2, and 15.6.
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th edition, McGraw-Hill: Boston. p. 108)
x
a. Calculate the mean. x
n
2
x
x
2
b. Calculate the variance using the deviation formula. s
n 1
x
2
x
2
c. Calculate the variance using the direct formula. s
n
n 1
78. The formula for computing the chi-squared statistic is:
2
2 O E
where O is the
E
observed value and E is the expected value. Given the following table of data, compute the
chi-squared statistic.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 515)
Day Mon Tues Wed Thur Fri Sat Sun
Observed 28 32 15 14 38 43 19
Expected 20 34 17 15 30 45 20
2
79. Prove that the following two formulas are equal.
r
n
xy
x
y
2
2
2
x
x
n
y
n
y
2
x
x y y
r
( n 1)(
s x
)( s
y
)
and
x
2
x x
2
Remember that: x s
n
x
n
n 1
(Hint: It might be easier to start with the second formula and work backwards.)
2
Math Camp 2011 Page 199

80. A statistics instructor is interested in finding the strength of a relationship between the final
exam grades of students enrolled in Statistics I and Statistics II. Draw a scatter plot of the
data. Calculate r. Calculate a and b and write the equation for the regression line. The data
are given here in percentages.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 495)
Stat I, x 87 92 68 72 95 78 83 98
Stat II,
y
83 88 70 74 90 74 83 99
APPLICATIONS IN ECONOMICS
81. Management of McPablo’s Food Shops has completed a study of weekly demand for its
“old-fashioned” tacos in 53 regional markets. The study revealed that
Q 400 1,200P
.8A
55Pop
800P
0
where Q is the number of tacos sold per store per week, A is the level of local advertising
0
expenditure (in dollars), Pop denotes the local population (in thousands), and P is the
average taco price of local competitors. For the typical McPablo’s outlet, P = $1.50, A =
0
$1,000, Pop = 40 and P =$1.
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006)
a. Estimate the weekly sales for the typical McPablo’s outlet.
b. What is the current price elasticity for tacos? What is the advertising elasticity?
2
82. A firm’s long-run total cost function is: C 360 40Q
10Q
.
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, p. 293)
a. What is the shape of the long-run average cost curve? (Hint: To find the average
cost function, divide the cost function by Q.)
b. Find the output that minimizes average cost. (Hint: The minimum average costs
occurs where AC MC . Remember that MC is the first derivative of the cost
function.)
c. The firm faces the fixed market price of $140 per unit. At this price, can the firm
survive in the long run? Explain.
83. A manufacturing firm produces output using a single plant. The relevant cost function is
2
C 500 5Q
. The firm’s demand curve is P 600 5Q
.
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, pp. 296-297)
a. Find the level of output at which average cost is minimized. (Hint: Set AC equal
to MC.) What is the minimum level of average cost?
b. Find the firm’s profit-maximizing output and price. Find its profit. {Hint:
MR MC }
84. In a perfectly competitive market, industry demand is given by Q 1,000
20P
. The typical
firm’s average cost is AC 300
Q .
Q 3
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, pp. 429)
Confirm that Q
min
30. (Hint: Set AC equal to MC.) What is AC
min
?
Math Camp 2011 Page 200

APPLICATIONS IN FINANCE
ln m
85. The formula t
can be used to find the number of years t required to multiply
r
nln
1
n
and investment m times when r is the per annum interest rate compounded n times a year.
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 464)
a. How many years will it take to double the value of an IRA that compounds
annually at the rate of 12%?
b. How many years will it take to triple the value of a savings account that
compounds quarterly at an annual rate of 6%?
c. Giver a derivation of this formula.
86. You have $12,000 to invest. If part is invested at 10% and the rest at 15%, how much should
be invested at each rate to yield 12% on the total amount?
(Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences, tenth edition, Upper Saddle
River, NJ: Prentice Hall, pp. 690)
87. Suppose that your local government is interested in doing a capital works project that will
have the following benefit schedule:
Year 1 $0
Year 2 $0
Year 3 $50
Year 4 $500
Year 5 $5,000
Year 6 $15,000
a. Calculate the present value of the project if the social discount rate is 5%
b. Calculate the present value of the project if the social discount rate is 10%
c. Calculate the present value of the project if the social discount rate is 15%
APPLICATIONS IN ENVIRONMENTAL SCIENCE
88. The size P of a certain insect population at a time t (in days) obeys the function
0.02t
P( t)
500e
.
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 472)
a. Determine the number of insects at t 0 days.
b. What is the growth rate of the insect population?
c. What is the population after 10 days?
d. When will the insect population reach 800?
e. When will the insect population double?
89. A piece of charcoal is found to contain 30% of the carbon 14 that it originally had. When did
the tree from which the charcoal came die? Use 5600 years as the half-life of carbon 14. Use
kt
the formula ( t)
A e where A
0
is the original amount of the substance and k is the rate of
A
0
decay.
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 473)
Math Camp 2011 Page 201

MATH CAMP
School of Public and Environmental Affairs
Solutions to Supplemental Problems
ORDERS OF OPERATIONS
1. Evaluate the following expressions using the order of operations:
2
49 7 5
4 3
16
49 7 5
16
3
16
7 5
16
3
16
a.
35 16
3
16
35 48 16
83 16
67
b. 5 2 30 6 25
30 6 25
5 20
2
c. ( 5) 30 6 25 30 6 25 5 30
2
2
2
d. 16
2
3
7 8 3
7 4 (10) 4 100
108
e.
f.
49 7 2
49 7 2
49 7 2
2 5
4
2 5
2 5
2 5
2 5
2 5
2 5
2 5
2 5
2 5
4
15
3 42
4 8 49 7 2 15
8142
4 8
15
8142
32 49 7 2 15
8110 49 7 2 15
810
795 49 7 2 795 7 2 795 14 795 809
3 2
36 2 3
2
3
2 2
4 3 2 36
2 91200
3 15
82
3
2 2
4 3 2 18
91200
3 15
82
3
2 2
4 3 2 1621200
3 15
82
2 2
4
3
81621200
3 15
82
2 2
4
3
1296
1200
3 15
82
2 2
4
1299
1200
3 15
82
2 2
2
4 99
3 15
82 2 5
4 9801
3
4
9801
9 15
82 2 5
9797 9 15
9788 15
82 2 5
9803
82 2 (
15
49013
82 49,095
3
2
1200
3
2
15
82
82
82
49015) 82
ROOTS AND EXPONENTS
2. Compute:
5
a. 3 = 243
Math Camp 2011 Page 202

1 1
b.
4
3
64
c. 3 -8 1
= 6561
d.
0
9 = 1
1
1
e.
5
= 5
f.
40
1 = 1
g.
2
15 = 225
h. 5 -2 1
= 25
i. 16 -1 1
= 16
3
j. ( 2) = -8
k. 5 -3 1
= 125
3. Evaluate the following:
3
2
2
a. 9 9 3 27
2
3 3
b. 8 2
3
2
2
8
2
3
1
2
1
1
1
3
1
1
27
d.
1
3 2
9
e. 4 16 2
1
4
3 3
c. 27 27 3 3
1
3
3
f. 125 125 5
g.
7 2 7 7
4. Simplify the following:
8 2
( 3)
( 3)
3
8
a. 10
b.
c.
d.
e.
f.
2 3
3 6 8 368
5 5
5
5
1 9 19
10
x x x x
4 7 47
7 7
7 7
0 3 03
2 2 2 2
9
5 9 2 7
5
5
2
5
11
3
5
17
Math Camp 2011 Page 203

OR
g.
3 3
36
3
3 3
6
3
5
4 4 4
2 2 5 10
h.
i. 2 2
2
j.
3 2 4 3
4 12
5 5 5
5
5 5 1 5
2
5
5 5 5 5
10
2 5
2
5
4
7 47
3
k.
3 3 3 3
6
5 6(
4)
10
l. 5 5
4
5
5
x 55
0
m. x x 1
5
x
2
2 2 2
3x
( 3)
x 9x
n.
2 2
2
5y
5 y 25y
5. Put in radical form:
1
2
1
2
1
2
a. ( 4x)
4 x 4 x 2 x
8
10
0.8
10
b. x x x 8
1
3
1
6
1 1
2 1
3 6 6 6
c. x x x x x x x
3
2
x
6. Simplify:
x
d. 3
a.
a
1
2
3a
b
1
2
1
3
b
3a
a
1
2
1
3
b
b
1
2
3a
b
1
2
1
3
2
3
6
1
2
a
b
1
2
3a
b
1 1
2 2
1
1
3
1
3a
b
1 3
3 3
3a
b
4
3
a
1
2
3a
b
1
2
1
3
b
3a
1 1
2 2
b
1
1
3
1
3a
b
1 3
3 3
3ab
4
3
3a
b
4
3
Math Camp 2011 Page 204

.
3
25x
y
3
9x
y
3x
y
3
2
1
3
1
3
2
3
x
3
2
5y
1
2
1
6
3
9x
y
3
25x
y
3x
5y
3 3
2 2
1 1
3 6
2
3
1
3
5y
1
2
3x
6
2
1 2 1
3 2 6
9
1
2
25
1
2
x
5y
1
3
2
x
1
3
2
3x
3
y
2 1
6 6
2
1
3 2
y
1 1
3 2
3x
5y
3x
3
3
6
5x
3
2
y
3
2
5y
1
3
y
3x
1
6
3
1
2
3x
y
3
2
1
3
5y
x
1
6
3
2
OR
3
25x
y
3
9x
y
3x
1
3
2
3
3 3
2 2
y
5
1
2
3
9x
y
3
25x
y
1 1
3 6
3x
6
2
2
3
y
5
1
3
1
2
2 1
6 6
9
1
2
25
1
2
x
3
3x
y
5
1
3
2
x
1
3
2
3
6
y
2
1
3 2
y
1 1
3 2
3
3x
y
5
3x
5x
1
2
3
2
y
3
2
1
3
y
1
6
3x
5y
3
1
2
7. Express the following in scientific notation:
a. 6,780,000 = (6.78)(10 6 )
b. 456 = (4.56)(10 2 )
c. -9,312 = (-9.312)(10 3 )
d. 938,000 = (9.38)(10 5 )
e. .00000276 = (2.76)(10 -6 )
f. .03 = 3(10 -2 )
g. .00000000000000000000723 = (7.23)(10 -21 )
8. Write in common decimal form:
3
a. 4.2
10 = 4,200
b. (5 x 10 5 )(4 x 10 -3 53
2
) = 5
410
(20)(10 ) 2, 000
c. 4.007 x 10 -6 = .000004007
9. Multiply the following and express the results in scientific notation:
5
12
512
17
18
a. (5)(10 ) (2.5)(10
) (5
2.5)(10 ) (12.5)(10 ) (1.25)(10 )
4
10
410
6
7
b. ( 3.4)(10
) (8.1)(10
) (
3.4)(8.1)
10
( 27.54)(10
) ( 2.754)(10
)
Math Camp 2011 Page 205

FRACTIONS
10. Compute:
a.
3 3 3 3 6 3
4 4 4 4 2
b.
8 4 8 7 4 9 56 36 56 36 92
9 7 9 7 7 9 63 63 63 63
c.
6 6 11 6 11 8 6 88 6 88
11
8 8 1 8 1 8 8 8 8
d.
2 13 2 2 13 4 13 4 13
9
7 14 7 2 14 14 14 14 14
e.
4 3 43
12 1
6 6 6
6 36 3
f.
5 13 513
65
3 14 314
42
g.
4 4 3 43
12 4
3
9 9 1 91
9 3
h.
4 3 4 4 16
7 4 7 3 21
i.
2 2 5 2 1 21
2
5
3 3 1 3 5 3
( 5)
15
j.
3 3 6 3 1 31
3 1
6
7 7 1 7 6 7 6 42 14
94
8
2
15
47
4
WORKING WITH ALGEBRAIC EXPRESSIONS
11. What degree are the following polynomials?
6 4
a. 5x 7x
2x
3 = degree of 6
b. 3x – 2 = degree of 1
c. x 3 x = degree of 3
12. Simplify the following expressions by collecting like terms:
3
2
3
x 5x
15
2x
2x
3x
8x
9 x
a.
3 2
7x
2x
5x
24
b.
c.
4x
2
4x
10x
6x
2
6x
3
3
3
5x
x x
2
2
x x
17x
5x
3x
3 x
x
2
20x
9
4( x 3) 4x
2
4x
12
4x
x 4x
12
10x
2
12
x
3
3
3
5x
2
3x
12
Math Camp 2011 Page 206

Math Camp 2011 Page 207
13. Multiply the following polynomials:
a. 3
16
35
3
21
5
35
1
7
3
5
2
2
x
x
x
x
x
x
x
b. 20
5
24
6
4
5
6
2
3
2
x
x
x
x
x
14. Compute the following:
a.
1)
(2
5
5
2
7
1)
(2
5
5
2
7
1)
(2
5
10
2
4
7
1)
(2
5
10
2
1)
(2
4
7
1
2
1
2
5
1
2
4
7
5
1
2
4
7
2
3
2
3
2
3
2
3
2
2
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
b.
2
3
8
2
2
3
2)
6
(
)
2
9
3
16
(
)
9
(8
2
3
2
2
9
9
6
3
16
8
2
2
2)
2
9
(9
6
3
16
8
2
2
2
2
9
9
2
2
6
3
16
8
1
1
2
2
9
2
2
1
3
8
2
2
9
1
3
8
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
c. 2
3
3
4
2
3
3
4
2
3
5
5
6
5
6
5
5
5
6
6
5
5
5
1
1
6
5
x
x
x
x
x
x
x
x
x
x
x
x
x
x
d.
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
45
5
2
5
3
45
5
2
2
3
3
5
1
9
2
3
1
5
9
2
3
3
2
3
2
2
2
e.
2
3
2
3
4
2
3
3
2
4
2
3
2
3
42
14
1
4
5
6
6
42
14
1
5
6
5
6
7
1
6
2
1
5
6
1
7
6
2
1
5
6
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
15. Evaluate the following expressions:
a. 3
6
5 2
x
x , where x = 5
98
3
30
125
3
30
25
5
3
6(5)
5(5)
2
b. 3
11
2
5
7
2
2
a
b
ab
a , where a = 2 and b = -3

7(2)
2
5(2)( 3)
2( 3)
11(2)
28 30 6 44 3 51
c. 9y 17 y 6 , where y = -1
9( 1)
17
( 1)
6 9
1
6 2
2
d. y 5x
3y
7( x 2)
where y = 3 and x = 5
3
2
3
55
33
7(5 2)
2
55
33
7(7) 9
55
33
7 7
16 9 49 7 49 56
2
3 7 4 10(
3)
6 11
4 3
9
25 33
7 7 9
25 9 7 7 9
25 9 49
2
e. ( y ) 5x
3y
7( x 2)
where y = 3 and x = 5
( 3)
2
( 3)
55
33
7(5 2)
2
55
33
7(7) 9 55
33
7 7
9 25 33
7 7 9 25 9 7 7 9 25 9 49
34 9 49 25 49 74
16. A culture of bacteria triples every hour. If it weighs one ounce at the beginning, what
will it weigh:
a. One hour later = 1 3 3 ounces
b. Two hours later = 3 3 9 ounces
c. Three hours later 93
27 ounces
d. Four hours later 27 3
81ounces
17. A retail store faces a demand equation for Roller Blades given by:
Q = 180 – 1.5P, where Q is the number of pairs sold per month and P is the price per pair
in dollars.
a. The store currently charges P = $80 per pair. At this price, determine the number
of pairs sold.
Q = 180 – 1.5(80) = 180 – 120 = 60
b. If management were to raise the price to $100, how many pairs would the store
sell?
Q = 180 – 1.5(100) = 180 – 150 = 30
18. The original revenue function for the microchip producer is R = 170Q -20Q 2 , where R is
the total revenue and Q is the number of microchips sold.
a. If the microchip producer sells 10 microchips, how much money will the producer
make?
R = 170 (10) – 20 (10) 2 = 1700 – 20 (100) = 1700 – 2000 = -$300
b. If the microchip producer sells 5 microchips, how much money will the producer
make?
R = 170 (5) – 20 (5) 2 = 850 – 500 = $350
Math Camp 2011 Page 208

PERCENT CHANGE
19. Suppose that your car was worth $11,354 three years ago and it is now worth $3,221.
What is the percent change?
3221 .28 => your car is worth 28% of its original worth, so its worth
11354
decreased by 72%
WORKING WITH SUMMATION SIGNS
x
20. Find x for the following:
n
a. 5, 6, 17, 3, -5, -25, 6, -12, 9, 31 = 3.5
b. 3.5, 2.1, 2.8, 3.9, 4.0, 1.9, 3.0 = 3.03
c. $5.50, $10.71, $12.01, $1.35, $6.50, $8.98, $9.12, $8.80, $15.00, $2.36, $3.30,
$6.66 = $7.52
21. For each set of values, find x , x 2 , and x 2
a. 80, 76, 42, 53, 77
x 80 76 42 53 77 328
2
2 2 2 2 2
x 80 76 42 53 77 22,678
2
2 2
(80 76 42 53 77) 328 107, 584
x
b. -9, -12, 18, 0, -2, -15
x 9 12
18
0 2 15
20
2 2
2 2 2
x ( 9)
( 12)
18
0 ( 2)
2
( 15)
2
778
2
2
2
( 9
12
18
0 2 15)
( 20)
400
x
c. 12, 52, 36, 81, 63, 74
x 12 52 36 81
63 74 318
x
2
12
2
52
2
36
2
81
2
63
2
74
2
20,150
2
2 2
(12 52 36 81
63 74) 318 101, 124
x
Math Camp 2011 Page 209

22. Calculate 2
x i
x
x
x .
n
a. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
for the following. To do this you will first need to compute
x
x
n
5.5
x i
82.5
x
2
(1 5.5)
2
(2 5.5)
2
(3 5.5)
2
.... (10 5.5)
2
b. -13, 15, -34, 23, 31, -40
x
x
n
3
x i
4586
x
2
( 13
3)
2
(15 3)
2
( 34
3)
2
.... ( 40
3)
2
23. Show that x
x
2
x
2
n
x
2
Note:
S
2
2
(
X X ) = ( X X )( X X )
2
X XX
= XX
X
X
2 X
= X
= X
= X
X
X
n
(
2
2
( )
X X X
n 1
n 1
X
2
X
X n
2
n n n
2
2
2(
X ) (
X
n n
2
( X
n
2
)
2
)
X )
n
2
X
Math Camp 2011 Page 210

24. A sample of personnel files of eight male employees employed by Acme Carpet revealed
that, during a six-month period, they lost the following number of days due to illness: 2,
x
0, 6, 3, 10, 4, 1, and 2. Calculate the mean x and the mean deviation
x x
for
n
n
these data.
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th edition, McGraw-Hill: Boston.)
x
x
n
2 0 6 3 10
4 1
2
3.5
8
x x
n
2 3.5 0 3.5 6 3.5 ....
8
2 3.5
19
2.375
8
25. Each person who applies for an assembly job at Carolina Furniture, Inc. is given a
mechanical aptitude test. One part of the test involves assembling a dresser based on
numbered instructions. A sample of the lengths of time it took 42 persons to assemble
the dresser was organized into the following frequency distribution.
Length of Time
Frequency
(minutes)
2 - 4 4
4 – 6 8
6 – 8 14
8 – 10 9
10 – 12 5
12 – 14 2
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th edition, McGraw-Hill: Boston. p. 111)
a. What is the range of times taken to assemble a dresser? 14-2 = 12 minutes
fx
b. What is the grouped mean? x where f is the frequency of each group and
n
x is the midpoint of each group and n is the total number of observations (or sum
x
of f). The midpoint can be calculate as 1
x
m 2
.
2
The midpoints are: 3, 5, 7, 9, 11, and 13
Math Camp 2011 Page 211

fx 3
4 58
... 13
2 312
x
7.43
n
42 42
c. What is variance?
fx
2
43
2
85
2
...2
13
2
2,594
2
fx 312
2
fx 2,594
2
s
n
42
6.7387
n 1
42 1
26. Merrill Lynch Securities and Health Care Retirement, Inc. are two large employers in
downtown Toledo, Ohio. They are considering jointly offering child care for their
employees. As a part of the feasibility study, they wish to estimate the mean weekly
child-care cost of their employees. A sample of 10 employees who use child care reveals
the following amounts spent last week--$107, $92, $97, $95, $105, $101, $91, $99, $95,
$104.
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th edition, McGraw-Hill: Boston. p. 313)
a. What is the mean?
2
x 107 92 ... 104
x
98.6
n 10
b. What is the variance?
2
2
x x
2
107 98.6 ... 104 98.6
s
n 1
10 1
2
276.4
30.71
9
c. If the standard deviation is the square root of the variance, what is the standard
deviation? s = 5.54
d. Given the following formula for a confidence interval, compute the confidence
s
interval for these data. x t . In this case, x is the mean you just calculated,
n
s is the standard deviation you just calculated, n is the sample size, and t =1.833.
s
5.54
x t 98.6
(1.833) (95.39,101.81)
n
10
27. There is a statistic called the chi-squared statistic that you will see again in your statistics
class. The formula for computing this statistic is:
2 O E
where O is the
E
observed value and E is the expected value. Given the following table of data, compute
the chi-squared statistic.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 515)
Frequency Cherry Strawberry Orange Lime Grape
Observed 32 28 16 14 10
Expected 20 20 20 20 20
2
Math Camp 2011 Page 212

2
2
2
2
2
O
32
20 28
20 16
20 14
20 10
20
2 E
18.0
E 20 20 20 20 20
28. Given the following data, calculate r. This is called the correlation coefficient, and you
will see more of this in your statistics class.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 481)
r
n
xy
x
y
2
2
2
x
x
n
y
n
y
Subject Age, x Pressure, y
A 43 128
B 48 120
C 56 135
D 61 143
E 67 141
F 70 152
We first compute the following:
x 345
y 819
xy 47,634
x 2
20, 399
y
2 112,443
r
FACTORING
n
xy
x
y
2
2
2
x
x
n
y
2
2
n
y
(6)(47,634) (345)(819)
2
2
(6)(20,399)
(345) (6)(112,443) (819)
29. Factor the given polynomial:
5 3 2 4 2 3 3
a. 15x
y 24x
y 3x
y (5x
8y)
0.897
3
3
3
3
b. 15 5x
3y
yx 3(
y 5) x ( y 5) ( y 5)( x 3)
3 2
3 2
2
2
c. x 1
x x x x x 1
x ( x 1)
( x 1)
( x 1)(
x 1)
2
2
d. 4x
20xy
25y
(2x
5y)(2x
5y)
2
2
e. 4x
101xy
25y
(4x
y)(
x 25y)
2
Math Camp 2011 Page 213

2 2
f. 36x
16y
(6x
4y)(6x
4y)
2n
4n
n 2n
n 2n
g. x y ( x y )( x y )
SOLVING EQUATIONS
30. Find the solutions to the following equations using the factoring formulas:
2
y 9y
20 ( y 5)( y 4) 0
a.
y 4,5
b.
x
2
x
8x
12
2
8x
12
0
( x 6)( x 2) 0
x 2,6
c.
45x
2 57x
18
0
(5x
3)(9x
6)
5x
3 0
5x
3
3
x
5
9x
6 0
9x
6
6 2
x
9 3
3 2
x ,
5 3
0
d.
9x
2 4 0
(3x
2)(3x
2) 0
3x
2 0
3x
2
2
x
3
3x
2 0
3x
2
2
x
3
Math Camp 2011 Page 214

e.
f.
g.
( x 2)
( x 2)( x 2) 2x
3x
x
x
2
2
2x
2x
2x
4 2x
3x
3x
2
2
2
x(2
3x)
4x
2x
4 0
2x
4 0
( 2x
4)( x 1)
0
2x
4 0
2x
4
x 2
x 1
0
x 1
y 5 y 1
y 5 1
2
2
y 5 1
y
y 5 y 5 1
y 1
y
y 5 1
2 y y
y y 5 1
2 y
4 2
y
2
4 2
y
16 4y
y 4
x 2 6
x 6 2
2
x 4
x 16
2
y
2
2
2
Math Camp 2011 Page 215

31. Find the solutions to the following equations by using the quadratic formula:
2
b b 4ac
x
2a
a.
2
x 2x 3 0
( 2)
2 4(1)( 3)
x
2(1)
2 4 12
2 16 2 4
x
1,3
2 2 2
2
b.
2x
5 x
x
2
2
2x
5 0
x
b
2
b 4ac
2a
x
( 2)
( 2)
2(1)
2
4(1)( 5)
x
2
4 20
2
2 24
2
2 2
2
6
1
6
c.
8w
8w
2
2
12w
9
12w
9 0
2
b b 4ac
x
2a
(12)
x
(12)
2(8)
4(8)( 9)
12
144 288
x
16
12
432 12
3(144)
x
16
16
12
12
3 3 3 3
16 4
2
32. Find a number whose square exceeds fourteen times the number by 51.
Let x be the number we are looking for.
2
Then x 14x 51
Now we solve for x
Math Camp 2011 Page 216

x
2
14x
51
b
x
( 14)
2(1)
4(1)( 51)
2
x 14x
51 0
14 196 204 14 400
x
2
2
14 20
3,17
2
33. An object is thrown upward from the top of a building, and if s feet is the distance of the
object from the ground t seconds after it is thrown, s 16t
2 48t
100
. How long will
it take the object to strike the ground?
We need to solve this equation for t when s = 0, which is the value of s when the
object hits the ground.
s 16t
16t
2
2
48t
100
48t
100
0
b
x
48
x
48
x
48 16
x
32
2
b
2a
( 14)
x
2
b 4ac
2a
( 48)
2
2( 16)
2304 6400
32
4ac
4( 16)(100)
34 3 34
2
2
3 34
2
Math Camp 2011 Page 217

SOLVING SYSTEMS OF EQUATIONS
34. Find the solution set to the following systems of equations using the method of
substitution.
3x
y 5
y 5 3x
a.
3x
y 5
5x
y 3
5x
y 3
5x
(5 3x)
3
5x
5 3x
3
8x
8
x 1
y 5 3x
y 5 3(1)
y 2
x y 1
x y 1
b.
x
3y
11
x
y 1
( y 1)
3y
11
4y
12
y
12
4
3
x y 1
x 3 1
2
35. Find the solution set to the following system of equations using the addition/subtraction
method. You might have to rearrange the equations to be in the correct format first. If
the system cannot be solved, then indicate whether it is dependent or inconsistent.
4x
3y
1
6x
y 7
4x
3y
1
18x
3y
21
4x
3y
1
a.
4x
18x
3y
3y
1
21
6x
y 7
22x
22
x 1
Math Camp 2011 Page 218

Math Camp 2011 Page 219
1
3
3
3
3
4
1
3
1
3
4
1
3
1
4
y
y
y
y
y
b.
4
3
1
3
5
y
x
y
x
6
7
21
18
20
1
15
3
5
5
20
15
5
1
3
5
4
3
1
3
5
y
y
y
y
x
x
y
x
y
x
y
x
y
x
2
1
3
6
24
21
6
4
6
7
3
4
3
x
x
x
x
y
x
c.
2
3
3
4
0
12
3
4
x
y
y
x
15
0
9
24
6
6
8
8
9
6
8
24
6
8
9
6
8
12
3
4
9
8
6
12
3
4
2
3
3
4
0
12
3
4
y
y
x
x
y
x
y
x
y
x
y
x
x
y
y
x
x
y
y
x
This system of equations is inconsistent and cannot be solved.

Math Camp 2011 Page 220
d.
5
3
4
1
3
2
y
x
y
x
8
24
3
24
2
6
4
2
5
1
3
3
4
2
5
3
4
1
3
2
x
x
x
x
x
x
y
y
x
x
y
x
y
x
9
9
3
12
1
3
4
1
3
2
8
1
3
2
y
y
y
y
y
y
x
SOLVING INEQUALITIES
36. Solve the following inequalities for x:
a.
10
10
100
10
10
100
10
36
64
9
64
36
9
x
x
x
x
x
x
x
b.
7
7
7
49
49
49
2
2
x
x
x
x
x
c.
3
15
5
10
5
2
3
5
2
10
3
x
x
x
x
x
x
d.
5
4
20
4
16
12
32
4
12
4
32
12
5
4
32
12
5
4
x
x
x
x
x
x
x
x

6x
2 6x
13
e. 6x
6x
2 13
2 13
This statement is never true, so there is no solution to this inequality.
f.
( x 2)( x 3) ( x 1)(
x 1)
x
x
x
2
2
2
2x
3x
6 x
5x
6 x
5x
6 x
5x
1
6
5x
7
7
x
5
2
2
1
2
1
x x 1
6(2y
3) 2(6y
12)
12y
18
12y
24
g.
12y
18
12y
24
18
24
This statement is always true, so all numbers are solutions to this inequality.
h.
5x
3
1
1
5x
3
3
1
5x
3
3
1 9
5x
3 3
10
5x
3
10 2
x
15 3
3
Math Camp 2011 Page 221

i.
2x
2 9x
5 0
(2x
1)(
x 5) 0
2x
1
0
2x
1
x
1
2
x 5 0
x 5
(-)(-) (-)(+) (+)(+)
x = -5
The solutions to this inequality are x 5
and
1
x .
2
j.
3x
3x
2
2
2 x
x 2 0
(3x
2)( x 1)
0
3x
2 0
3x
2
x
2
3
x 1
0
x 1
(-)(-) (-)(+) (+)(+)
x = -1
Therefore, the solution to this inequality is
2
1 x .
3
Math Camp 2011 Page 222

37. Solve the following equations and inequalities for x:
a. 5y
8 2
5y
8 2
5y
2 8
5y
10
y 2
5y
8 2
5y
2
8
5y
6
y
6
5
Now we have to check the solutions to make sure that they work.
5y
8 2
5(2) 8 2
10 8 2
2 2
2 2
5y
8 2
6
5
8 2
5
6 8 2
2 2
2 2
Both
y 2
and
6
y are solutions to this equation.
5
2
b. x 3x 2 2
x
x
2
2
3x
2 2
3x
4 0
( x 4)( x 1)
0
x 4,1
x
x
2
2
3x
2 2
3x
0
x(
x 3) 0
x 0, 3
Now we need to check all these solutions.
x
2
( 4)
3x 2 2
2
3( 4)
2
16 12
2 2
2 2
2 2
2
x
2
(0)
3x 2 2
2
2 2
2 2
3(0) 2 2
Math Camp 2011 Page 223

2
x 3x 2 2
2
(1) 3(1) 2 2
1
3 2 2
2 2
x
2
( 3)
3x 2 2
2
3( 3)
2
9 9 2 2
2 2
2 2
2
The solutions to this equation are x 4
, x 1, x 0, and x 3
x 1
c. 1
2 3
x 1
1
2 3
3x
2 6
3x
6 2
3x
8
x
8
3
x 1
1
2 3
3x
2 6
3x
6
2
3x
4
x
4
3
Now we need to check both of these solutions.
x
2
8 3
2
8
6
4
3
3
3
1 1
1
3
1
3
1
3
1
1
1
3
1
1
1
x
2
4 3
2
4
6
2
3
3
3
1
1
1 1
1
3
1
3
1
3
1
1
1
3
1
1
1
Both solutions check out, so
8
x and
3
4
x are solutions to the equation.
3
Math Camp 2011 Page 224

d.
8x
5( x 4) 5 x 3 6
8x
5x
20 5 x 6 3
12x
15
9
12x
15
9
12x
9 15
12x
24
x 2
12x
15
9
12x
9
15
12x
6
x
1
2
e.
9x
6 x 12
x 2 x
10x
6 2 12
x x
10x
6 14
14
10x
6 14
14
6 10x
14 6
8 10x
20
8 20
x
10 10
4
x 2
5
38. In your Economics 101 class, you have scores of 68, 82, 87, and 89 on the first four of
fives tests. To get a grade of B, the average of the first five test scores must be greater
than or equal to 80 and less than 90. Solve an inequality to find the range of the score
that you need on the last test to get a B. (Michael Sullivan, College Algebra, Upper Saddle Rive, NJ: Prentice
Hall, 2005, p. 135)
68 82 87 89 x
80
90
5
400 68 82 87 89 x 450
400 326 x 450
400 326 x 450 326
74 x 124
Math Camp 2011 Page 225

FUNCTIONS
39. Determine the domain and range of the following functions:
a.
5
f ( x)
2x
8x
7 => Domain = all real numbers
=> Range = all real numbers
b.
x 3
f ( x)
x 5
=> Domain = all real numbers except x 5
=> Range = all real numbers except y 1
c. f ( x)
12 x => Domain = all real numbers where x 12
=> Range = y 0
BASICS OF GRAPHING FUNCTIONS
40. Graph the following lines:
a. 4x
2y
20
b. 4x
2y
20
15
10
5
0
-15 -10 -5 0
-5
5 10 15
-10
-15
-20
-25
-30
-35
35
30
25
20
15
10
5
0
-15 -10 -5 0
-5
5 10 15
-10
-15
Math Camp 2011 Page 226

c. x 8 0
d. y 1
e. 2x
3y
0
8
6
4
2
0
-15 -10 -5 0 5 10 15
-2
-4
-6
-8
Math Camp 2011 Page 227

41. A statistics instructor wishes to determine if a relationship exists between the final exam
scores in Statistics 101 and the final exam scores of the same students who took Statistics
102. Draw a scatter plot and comment on the nature of the relationship.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 79)
Stat 101, x 87 92 68 72 95 78 83 98
Stat 102, y 83 88 70 74 90 74 83 99
120
100
80
60
40
20
0
0 20 40 60 80 100 120
The graph shows a positive linear relationship.
42. The data shown indicate the number of tournaments and the earnings in thousands of
dollars of 10 randomly selected LPGA golfers. Draw a scatter plot for the data and
determine the nature of the relationship.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 79)
No. of 27 30 24 21 18 32 25 32 25 27
tournaments,x
Earnings, y $956 757 583 517 104 173 252 303 355 405
Math Camp 2011 Page 228

$1,200
$1,000
$800
$600
$400
$200
$0
0 5 10 15 20 25 30 35
The graph shows no real relationship between number of tournaments and earnings.
43. An educator wants to see if there is relationship between the number of absences a
student has and his or her final grade in a course. Draw a scatter plot and comment on
the nature of the relationship.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 79)
No. of
10 12 2 0 8 5
absences, x
Final grade, y 70 65 96 94 75 82
120
100
80
60
40
20
0
0 2 4 6 8 10 12 14
The graph shows a negative linear relationship between number of absences and the final grade.
Math Camp 2011 Page 229

LINEAR FUNCTIONS AND THEIR GRAPHS
44. Draw a graph of the line connecting these two points and find the length of the line.
a. (2,1), (-1,5)
6
5
4
3
2
1
0
-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
2
2
2 2
1
2 5
1 ( 3)
(4) 9 16
25 5
b. (7,-4), (-5,1)
2
1
0
-6 -4 -2 0 2 4 6 8
-1
-2
-3
-4
-5
2
2
2 2
5 7 1
( 4)
( 12)
(5) 144 25 169 13
Math Camp 2011 Page 230

45. Draw a graph of the line connecting these two points and find the slope of the line.
a. (-7,-1), (-3,4)
0
-8 -7 -6 -5 -4 -3 -2 -1 0
-1
5
4
3
2
1
-2
rise
m
run
y
x
2
2
y
x
1
1
4 ( 1)
3 ( 7)
5
4
b. (0,-5), (-5,0)
0
-6 -5 -4 -3 -2 -1 0
-1
-2
-3
-4
-5
-6
rise
m
run
y
x
2
2
y
x
1
1
0 ( 5)
5 0
5
5
1
Math Camp 2011 Page 231

c. (3,5), (5, 9)
10
9
8
7
6
5
4
3
2
1
0
0 1 2 3 4 5 6
m
rise
run
y
x
2
2
y
x
1
1
9 5
5 3
4
2
2
46. Find the midpoint between the two points:
a. (8,9), (3,9)
x x
2
y
,
2
1 2 1
y2
b. (-5,0), (7,7)
x x
2
y
,
2
1 2 1
y2
c. (1,1), (3,3)
x1x2
2
y
,
1
y
2
2
8 3 9 9 11
, ,9
2 2 2
5 7 0 7 7
, 1,
2 2 2
1
3 1
3
,
2 2
47. Find the lengths of the sides of the triangle having vertices at the three given points.
Graph the triangle. A(2,3), B(3,-3), C(-1,1)
2,2
2
2 2
Length side AB = 3
2
3 3 (1) ( 6)
1
36 37
Length side BC =
2
1
3 1
( 3)
32 4 2
2
2
2
( 4)
2
(4)
2
16 16
2 2
Length side CA = 2
( 1)
3
1 (3) (2) 9 4 13
2
Math Camp 2011 Page 232

4
3
2
1
0
-2 -1
-1
0 1 2 3 4
-2
-3
-4
48. Show that the line through the points A(3,9) and B(5,5) is parallel to the line through the
points C(1,7) and D(3,3). Draw a sketch of the two lines.
m
m
rise
y
y
5 9 4
5 3 2
2 1
1
run x2
x1
rise
y
y
3 7 4
3 1
2
2 1
2
run x2
x1
Because the two lines have the same slope, they are parallel.
2
2
10
9
8
7
6
5
4
3
2
1
0
0 1 2 3 4 5 6
49. Find an equation of the line through the two points. Graph the line.
a. ( 3, 6),(
2,
3)
Math Camp 2011 Page 233

m
rise
run
y
x
2
2
y
x
1
1
3 ( 6)
2 ( 3)
3
3
1
y y
1
m( x x ) 1
y ( 6)
3( x ( 3))
y 3x
9 6
y 3x
3
40
30
20
10
0
-15 -10 -5 0 5 10 15
-10
-20
-30
b. ( 1, 2),(7,
2)
rise
m
run
y
x
2
2
y
x
1
1
2 ( 2)
7 ( 1)
0
0
8
y y
1
m( x x1)
y ( 2)
0( x ( 1))
y 2
2.5
2
1.5
1
0.5
0
-15 -10 -5 -0.5 0 5 10 15
-1
-1.5
-2
-2.5
Math Camp 2011 Page 234

c. ( 3,1),( 5,4)
m
rise
run
y
x
2
2
y
x
1
1
4 1
5 3
3
8
y y
1
m( x x ) 1
3
y 1
( x 3)
8
3 9
y x 1
8 8
3 9 8
y x
8 8 8
3 17
y x
8 8
7
6
5
4
3
2
1
0
-15 -10 -5 0
-1
5 10 15
-2
50. Find an equation of the line through the given point with the given slope. Graph the line.
a. ( 3,
6),
m 3
y y
1
m( x x ) 1
y ( 6)
3(
x ( 3))
y 3x
9 6
y 3x
15
Math Camp 2011 Page 235

20
10
0
-15 -10 -5 0 5 10 15
-10
-20
-30
-40
-50
b.
( 4,0),
m
2
5
y y
1
m( x x1)
2
y (0) ( x ( 4))
5
2 8
y x
5 5
6
5
4
3
2
1
0
-15 -10 -5
-1
0 5 10 15
-2
-3
51. Find an equation of the line through the point (6,3) and perpendicular to the line whose
equation is x 3y 15
0 . Graph the lines (on the same axis).
We first need to find the slope of the second line. To do this, rewrite the equation in slopeintercept
form.
x 3y
15
0
3y
x
15
1
y x 5
3
The slope of this line is
1
3
Math Camp 2011 Page 236

In order for the lines to be perpendicular, the product of their slopes must equal -1. So
1
1
1
1
3 3
m1m2
1
m1
. Given that m2 , we have m1 3
m2
3
1
1 1
1
3
y y
1
m( x x ) 1
y (3) 3( x 6)
y 3x
18
3
y 3x
15
20
10
0
-15 -10 -5 0 5 10 15
-10
-20
-30
-40
-50
52. Find an equation of the line that is parallel to the given line and passes through the given
point. 5x 2y 1
0,
(3,3)
5x
2y
1
0
2y
5x
1
5 1
y x
2 2
5 1
y x
2 2
5
Thus the slope of the line parallel to this line must also have a slope of .
2
y y m( x x ) 1
1
5
y 3 ( x 3)
2
5 15
y x 3
2 2
5 15 6
y x
2 2 2
5 9
y x
2 2
Math Camp 2011 Page 237

53. A ramp connects the ground to the loading platform of a warehouse. If the platform is 5
feet high and the ramp has a slope of 0.28, how far from the wall of the warehouse is the
base of the ramp?
Pretend that the ramp and platform are placed on an axis. Then the endpoints of the ramp would
be (0,5) and (-x,0). We want to find what x is.
We know that
m 0.28
y
x
2
2
5
y1
x
1
0 5 5
x 0 x
5
x
x 17.9
0.28
Therefore the base of the ramp is 17.9 feet away from the platform.
QUADRATIC FUNCTIONS AND THEIR GRAPHS
54. Graph the following equations:
a. f ( x)
x
2 10
120
100
80
60
40
20
0
-15 -10 -5 0 5 10 15
b.
f ( x)
25x
2
13
0
-15 -10 -5 -100 0 5 10 15
-200
-300
-400
-500
-600
-700
-800
-900
-1000
Math Camp 2011 Page 238

c.
d.
e.
f ( x)
10x
5
f ( x)
10x
5
f ( x)
10x
5
50
40
30
20
10
0
-5 -3 -1 -10 1 3 5
-20
-30
-40
-50
50
40
30
20
10
0
-5 -3 -1 -10 1 3 5
-20
-30
-40
-50
50
40
30
20
10
0
-5 -3 -1 -10 1 3 5
-20
-30
-40
-50
55. The percentage s of seats in the House of Representatives won by Democrats and the
percentage v of votes cast for Democrats (when expresses as decimal fractions) are
related by the equation:
5v 2s
1.4
0 s 1
0.28
v 0. 68
a. Express v as a function of x, and find the percentage of votes required for the
Democrats to win 51% of the seats.
5v
2s
1.4
5v
2s
1.4
v .4s
.28
v .4(.51) .28
v .484
In order to win 51% of the seats, the Democrats would have to get 48.4% of the votes.
Math Camp 2011 Page 239

. Express s as a function of v, and find the percentage of seats won if Democrats
receive 51% of the votes.
5v
2s
1.4
2s
5v
1.4
s 2.5v
.7
s 2.5(.51) .7
s .575
If the Democrats receive 51% of the votes, then they will get 57.5% of the seats in the House.
56. A projectile is fired from a cliff 500 feet above the water at an inclination of 45 degrees to
the horizontal, with a muzzle velocity of 400 feet per second. In physics, it is established that
2
32x
the height h of the projectile above the water is given by h( x)
x 500 , where x is
2
(400)
the horizontal distance of the projectile from the base of the cliff. (Sullivan, College Algebra, seventh
edition, Upper Saddle River, NJ: Prentice Hall, pp. 302-303)
57.
a. Find the maximum height of the projectile.
The height of the projectile is given by the quadratic function.
2
32x
1
2
h(
x)
x 500 x x 500
2
(400)
5000
We are looking for the maximum value of h. Since the maximum value is obtained at
the vertex, we compute
b 1 5000
x 2500
2a
1
2
2
5000
The maximum height of the projectile is
1
2
h(2500)
2500
2500 500
5000
1250
2500 500 1750
=> 1750 feet
b. How far from the base of the cliff will the projectile strike the water?
The projectile will strike the water when the height is zero. To find the distance
traveled, we need to solve the equation
1
h(
x)
x
5000
2
x 500 0
Math Camp 2011 Page 240

We use the quadratic formula
1
1
1
4
(500)
5000
x
1
2
5000
458
x
5458
1
1.4
1
2
5000
We discard the negative solution and find that the projectile will strike the water
at a distance of about 5458 feet from the base of the cliff.
58. The price p and the quantity x sold of a certain project obey the demand equation
x 5p
100 , 0 p 20
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, pp. 308)
a. Express the revenue R as a function of x. Remember that R xp
x 5
p 100
5 p x
100
1
p x 20
5
R(
x)
xp
R(
x)
x
x 20
1
R(
x)
x
2 20x
5
b. What is the revenue if 15 units are sold?
R(15)
255
1
5
1 2
R(15)
(15) 20(15)
5
1
R(15)
(225) 300
5
R(15)
45
300
The revenue if 15 units are sold is $255.
c. What quantity x maximizes revenue? What is the maximum revenue?
Math Camp 2011 Page 241

Remember that the maximum will occur at the vertex of the revenue function, so
we must begin by finding the vertex.
b 20 100
x 50
2a
1 2
2
5
Thus, 50 is the quantity that will maximize revenue.
b
1 2
R
R (50) (50) 20(50)
2a
5
2500
1000
500
1000
500
5
The revenue when 50 units are sold is $500.
d. What price should the company charge to maximize revenue?
1
p x 20
5
1
p (50) 20 10
20 10
5
The price charged should be $10.
POLYNOMIAL AND RATIONAL FUNCTIONS
59. The following data represent the number of motor vehicle thefts (in thousands) in the
United States for the years 1987-1997, where 1 represents 1987, 2 represents 1988, and
so on.
Year, x
Motor Vehicle
Thefts, T
1987,1 1289
1988,2 1433
1989,3 1565
1990,4 1636
1991,5 1662
1992,6 1611
1993,7 1563
1994,8 1539
1995,9 1472
1996,10 1394
1997,11 1354
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, pp. 329)
a. Draw a scatter diagram of the data. Comment on the type of relation that may
exist between the two variables.
Math Camp 2011 Page 242

1700
1650
1600
1550
1500
1450
1400
1350
1300
1250
1200
1986 1988 1990 1992 1994 1996 1998
The relationship between these data is not linear, nor is it quite a quadratic relationship (a
parabola), so a cubic relationship would be a better model for these data.
b. The cubic function of best fit to these data is:
3
2
T(
x)
1.52x
39.81x
282.29x
1035.5
Use this function to predict the number of motor vehicle thefts in 1994.
The year 1994 corresponds to the number 8 in our dataset.
T(1994)
1.52(8)
1.52(512) 39.81(64) 2258.32 1035.5
778.24 2547.84 2258.32 1035.5
1524.22
3
39.81(8)
282.29(8) 1035.5
The predicted number of motor vehicle deaths in 1994 is approximately 1,524,220
60. A company manufacturing snow-boards has fixed costs of $200 per day and total costs of
$3,800 per day at a daily output of 20 boards. (Barnett, Ziegler and Byleen, Finite Mathematics for Business,
Economics, Life Sciences, and Social Sciences, tenth edition, Upper Saddle River, NJ: Prentice Hall, pp. 93)
a. Assuming that the total cost per day, C(x) , is linearly related to the total output
per day, x, write an equation for the cost function.
3800
C(
x)
x 200
20
C(
x)
190x
200
2
b. The average cost per board for an output of x boards is given by C ( x)
C(
x)
x .
Find the average cost function.
C(
x)
C ( x)
x
190x
200
x
Math Camp 2011 Page 243

c. Sketch a graph of the average cost function, including any asymptotes, for
1 x 30 .
425
375
325
275
225
175
0 5 10 15 20 25 30
d. What does the average cost per board tend to as production increases?
To find the average cost per board as production increases, we need to find the horizontal
asymptote for this function. To do this, we divide every term in the numerator and the
denominator by the highest value of x.
200
190
190x
200
C ( x)
x
x 1
200
As x , the value of 0 , so the value of C ( x)
190
x
The horizontal asymptote is y 190 . As production increases, the average cost will tend toward
$190.
61. Graph the following equations:
3
a. f ( x)
x 10x
6
800
600
400
200
0
-25 -15 -5
-200
5 15 25
-400
-600
-800
Math Camp 2011 Page 244

.
f ( x)
x 3
5
4
3
2
1
0
-1 -1 1 3 5 7 9
-2
-3
-4
-5
c.
f ( x)
x 2 10
4
2
0
-15 -10 -5
-2
0 5 10
-4
-6
-8
-10
-12
EXPONENTIAL FUNCTIONS
61. Graph the following functions:
x
a. f ( x)
5
Math Camp 2011 Page 245

.
f ( x)
3
x
c.
f ( x)
1
10
x
d.
f ( x)
(log
2
x)
3
Math Camp 2011 Page 246

e.
f ( x)
log
10
x
f.
f ( x)
ln( x 5)
62. Suppose that $2,500 is invested at 7% compounded quarterly. How much money will be in
the account in:
(Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences, tenth edition, Upper Saddle River,
NJ: Prentice Hall, pp. 107)
a. ¾ of a year?
Math Camp 2011 Page 247

. 15 years?
63. If you invest $7,500 in an account paying 8.35% compounded continuously, how much
money will be in the account at the end of:
(Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences, tenth edition, Upper Saddle River,
NJ: Prentice Hall, pp. 107)
a. 5.5 years?
b. 12 years?
64. The Joint United Nations Program on HIV/AIDS reported that HIV had infected 60 million
people worldwide prior to 2002. Assume that number increases at an annual rate of 8%
compounded continuously.
(Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences, tenth edition, Upper Saddle River,
NJ: Prentice Hall, p. 108)
Math Camp 2011 Page 248

c. Write an equation that models the worldwide spread of HIV, letting 2002 be year
0.
d. Based on the model, how many people (to the nearest million) had been infected
prior to 1999? How many would be infected prior to 2010?
LOGARITHMIC FUNCTIONS
65. Express the following relationships in terms of logarithms.
a.
9 2 1
3
b.
10 5 .00001
66. Express the following relationships using exponential notation.
a. log 7 2 7
49 2 = 49
b. log 16 4
1
2
67. Find the value of the following logarithms.
1
a. log 4
= -1
4
1
b. log 3
3 =
2
c. log 10
.000001 = -6
Math Camp 2011 Page 249

2
d. log 3
5
25
3
1
e. log
e
e
2
68. Solve the following for x or b, respectively.
1 2
a. log
b
16 3
b.
log
1
x
3
4
c.
log
b
81 2
d.
log
6(
x 4) log
6(
x 9) 2
Math Camp 2011 Page 250

e.
2
log
4(
x 6x)
2
f.
log
2(11
x)
log
2(
x 1)
3
11
x
log 2
x 1
3
69. Simplify the following logarithms (express as one logarithm):
1
4log
10
x log
10
y
a.
2
b.
3
log
4
b
x 6log
b
4
y log
5
b
z
70. Evaluate the following logarithms given that log 10
2 0.3010 , log 10
3 0. 4771, and
7 log 10
0.8451
a.
log
log
10
10
2 0.3010
0.6309
3 0.4771
Math Camp 2011 Page 251

.
log
10
5
49
2
36
c.
d.
log 10
35
10
log 10 7 · 5 = log 10 7· = log 10 7 + log 10 10 – log 10 2
2
= 0.8451 + 1 – 0.3010
=1.544
14
log
10
3
84
e. log 10
30 = log 10 3 · 10 = log 10 3 + log 10 10 = .4771 + 1 = 1.4771
f. log = log 10 7 2 · 10 3 10
49000
= 2 log 10 7 + 3 log 10 10 = 2(.8451) + 3 = 4.69
g. log = log 10 2 · 3 · 10 -7 = log 10 2 + log 10 3 + log 10 10 -7
10
. 0000006
= (.3010) + (.4771) – 7 = -6.2219
71. Shannon’s diversity index is a measure of the diversity of a population. The diversity index
is given by the formula
H p1 log p1
p2
log p2
...
p n
log p n
where p1
is the proportion of the population that is species 1, p2
is the proportion of the
population that is species 2, and so on.
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 439)
Math Camp 2011 Page 252

a. According to the U.S. Census Bureau, the distribution of race in the United States
in 2000 was as follows:
Race
Proportion
American Indian or 0.014
Native Alaskan
Asian 0.041
Black or African 0.128
American
Hispanic 0.124
Native Hawaiian or 0.003
Pacific Islander
White 0.690
Compute the diversity index of the United States in 2000.
b. The largest value of the diversity index is given by H log( S)
, where S is the
H max
number of categories of race. Compute .
H
max
log( S)
H
max
log(6) 0.778
H
c. The evenness ratio is given by E H
, where 0 E 1 . If 1, there
H max
is a complete evenness. Compute the evenness ratio for the United States.
max
H
E H
E H
H
H
max
.428
.550
.778
72. How many years (to two decimal places) will it take $1,000 to grow to $1,800 if it is invested
at 6% compounded quarterly? Compounded continuously?
(Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences, tenth edition, Upper Saddle
River, NJ: Prentice Hall, pp. 120)
Math Camp 2011 Page 253

lne .06t = ln1.8
Math Camp 2011 Page 254

Math Camp 2011 Page 255
DERIVATIVES
73. Find derivatives of the following:
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
m.
n.
3
10
36
)
'(
5
3
5
12
)
(
2
2
3
x
x
x
f
x
x
x
x
f
6
6
5
5
5
)
'(
1
)
(
x
x
x
f
x
x
f
7
4
1
2
8
2
1)
(2
4)
2(
)
'(
4)
1)(
(2
)
(
x
x
x
x
x
x
f
x
x
x
f
x
x
x
x
f
x
x
x
x
f 8
15
144
)
'(
990
4
3
16
)
(
4
8
2
5
9
x
x
f
)
(
x
x
f
x
x
f 18
)
'(
9
)
(
2
1
7
3
)
( 2
x x
x
f
3
2
3
2
1
3
1
3
)
3
(5
1
)
3
(5
3)
(
)
3
(5
3
1
)
'(
3
5
)
(
x
x
x
x
f
x
x
f
2
2
2
1)
(
2
1)
(
2
2
2
1)
(
)
(1)(2
1)
(2)(
)
'(
1
2
)
(
x
x
x
x
x
x
x
x
f
x
x
x
f
4
2
4
2
5
2
1)
(
10
)
(2
1)
5(
)
'(
1)
(
)
(
x
x
x
x
x
f
x
x
f
8
16
24
)
'(
8
8
8
8
)
(
2
2
3
x
x
x
f
x
x
x
x
f
9
10
7)
10(
7)
(
)
(
x
x
x
f
2
2 5
5
18
)
(
x
x
x
x
x
f
)
(
1
)
(
2
3 x
x
x
x
f

4
5
o. f ( x)
8x
f '( x)
32x
74. Find the tangent lines to the given curve at the given point:
3 2
a. f ( x)
5x
2x
9x
15
, at x = -4
4 2
b. f ( x)
8x
x 10
at x = 0
3 2
c. f ( x)
x 7x
6x
6 at x = 10
Math Camp 2011 Page 256

OPTIMIZATION
75. Find all the maxima and minima of the following curves. (Hint: Find the first derivative
and set it equal to zero. Then solve for x. Then you can use the second derivative to
determine whether each point is a maximum or minimum.)
3 2
a. f ( x)
5x
4x
6x
19
This function has no real roots
b.
f ( x)
3
x
3
4x
2
12x
15
Math Camp 2011 Page 257

At x = 9.3, the second derivative is positive, so x =2 is a local minimum.
At x = -1.3, the second derivative is negative, so x =-1.3 is a local maximum.
2
76. Suppose a ball is thrown into the air and its height after t seconds is 4 48t 16t
feet.
At what time does the ball reach its highest point?
f '( t)
48 32t
48 32t
0
48 3
t
32 2
Now we need to check the second derivative to make sure this is a maximum.
f "(
x)
32
Because the second derivative is everywhere negative, this is a local maximum.
750
77. Find the value of x that minimizes the cost C(x) , where C( x)
95x
for x 0.
x
Math Camp 2011 Page 258

78. A firm’s total-revenue and total-cost functions are
TR 4Q
TC 0.04Q
3
0.9Q
2
10Q
5
(Salvatore, Managerial Economics in a Global Economy, sixth edition, New York: Oxford University Press, 2007, p. 78)
a. Determine the best level of output. {This means to find the point where profit is
maximized. Remember that TR TC , where represents the firm’s profit.}
Now find the first and second derivatives of the profit function.
Now we set the first derivative of the profit function equal to zero and solve for
the quantity.
Math Camp 2011 Page 259

We have two points that can potentially maximize the profit function, x = 5 and
x = 10. We can use the second derivative to determine whether these points are a
minimum or a maximum. If the second derivative is negative at that point, then it
is a maximum, if it is positive then that point is a minimum.
At x = 5 the second derivative is positive, so x =5 is a minimum point. At x = 10
the second derivative is negative, so x = 10 is a maximum point.
Thus, x =10 is the point that maximizes the firm’s profit.
b. Determine the total profit of the firm at its best level of output.
Basically this means that the firm is losing money. If we look at the graph of the
firm’s profit function, we see that at no point is the firm making money. The firm
loses the least money at x = 10.
Math Camp 2011 Page 260

79. Suppose a firm’s inverse demand curve is given by P 120 . 5Q and its cost equation is
2
C 420 60Q
Q .
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, p. 56)
a. Find the firm’s optimal quantity, price, and profit (1) by using the profit and
marginal profit equations and (2) by setting MR equal to MC. Also provide a
graph of MR and MC. (Hint: R PQ . Then profit is R C . Marginal profit
is the first derivative of the profit function. Marginal revenue is the first derivative
of the revenue function. Marginal cost is the first derivative of the cost function.)
We can set the marginal profit equation equal to zero to find its roots.
Now we can use the second derivative to confirm that this is a maximum point.
Math Camp 2011 Page 261

Because the second derivative is negative, the point Q = 20 is a maximum point.
To find the optimal price and profit levels, we just substitute Q = 20 in the price
and profit equations.
The second method of solving this problem involves setting the marginal revenue
equal to the marginal cost. We find the marginal equations by taking the
derivative of the original equations.
Math Camp 2011 Page 262

80. Suppose you were producing SPEA-logo binders with the following cost schedule.
Quantity Total Cost Average Cost Marginal
Cost
0 150
1 233
2 290
3 327
4 350
5 365
6 378
7 395
8 422
9 465
10 530
11 623
12 750
13 917
14 1130
15 1395
Average
Variable Cost
a. What is the fixed cost? $150
b. Complete the table by computing the average cost, marginal cost and average
variable cost for each unit produced.
Math Camp 2011 Page 263

Quantity Total Cost Average Cost Marginal
Cost
Average
Variable Cost
0 150
1 233 233.0 83 83.0
2 290 145.0 57 28.5
3 327 109.0 37 12.3
4 350 87.5 23 5.8
5 365 73.0 15 3.0
6 378 63.0 13 2.2
7 395 56.4 17 2.4
8 422 52.8 27 3.4
9 465 51.7 43 4.8
10 530 53.0 65 6.5
11 623 56.6 93 8.5
12 750 62.5 127 10.6
13 917 70.5 167 12.8
14 1130 80.7 213 15.2
15 1395 93.0 265 17.7
c. Graph the average cost, marginal cost and average variable cost curves on the
same axis.
d. Suppose the total cost function is TC Q
3 16Q
2 98Q
150 . Graph this
function.
Math Camp 2011 Page 264

e. What is the average cost function?
f. What is the marginal cost function?
TC Q
3
16Q
MC TC'
3Q
2
2
98Q
150
32Q
98
Math Camp 2011 Page 265

SOLVING WORD PROBLEMS
81. A ladder is leaning against a building. The base of the ladder is 8 feet away from the
side of the building and the ladder reaches 9 feet high on the building. How long is the
ladder?
2 2 2
We know that a b c . And in this case a = 8 and b = 9, so we just solve for c.
a
8
2
2
9
64 81 c
c
2
b
2
2
145
c
c 12.04
c
2
2
2
Therefore the ladder is 12 feet long.
82. A farmer mixes milk containing 3% butterfat with cream containing 30% butterfat to
obtain 900 gallons of milk which is 8% butterfat. How much of each must the farmer
use?
(Johnson, How to Solve Word Problems in Algebra: A Solved Problem Approach, McGraw Hill, 2000)
Let x = number of gallons of milk
Let y = number of gallons of cream
x y 900
0.03x 0.30y
0.08(900)
0.03x
0.30y
0.08(900)
0.03x
0.30y
72
3x
30y
7200
x 10y
2400
x 2400 10y
x y 900
2400 10y
y 900
2400 9y
900
9y
1500
1500
y
9
500
3
166
2
3
gallons
Math Camp 2011 Page 266

x y 900
500
x 900
3
x 900
500
3
2700
3
500
3
2200
3
1
733
3
83. The data shown indicate the number of wins and the number of points scored for teams
in the National Hockey League. Draw a scatter plot for the data and describe the nature
of the relationship.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 79)
Wins,x 10 9 6 5 4 12 11 8 7 5 9 8 6 6 4
Points,y 23 22 15 15 10 26 26 26 21 16 12 19 16 16 11
30
25
20
15
10
5
0
0 2 4 6 8 10 12 14
The graph shows a positive linear relationship between number of wins and number of points
scored.
Math Camp 2011 Page 267

84. Evaluate the following expression given that p =.8, n = 2000, and z = 1.96.
a. Evaluate.
p z
p( 1
p)
b. If p = .8 and z = 19.6 and p z = .75, find n.
n
p z
.8 (1.96)
p(1
p)
n
.8(1 .8) .75 .8
n 1.96
.8(1 .8)
n
.16
.00065
n
n
.16
.00065
246
APPLICATIONS IN STATISTICS
p(1
p)
.8 (1.96)
n
.8(1 .8)
.75
n
2
(.0255)
2
.8(1 .8)
(.782,.818)
2000
85. A sample of eight companies in the aerospace industry was surveyed as to their
return on investment last year. The results are: 10.6, 12.6, 14.8, 18.2, 12.0, 14.8, 12.2,
and 15.6.
(Lind, Marchal and Mason, Statistical Techniques in Business & Economics, 11 th
edition, McGraw-Hill: Boston. p. 108)
x 10.6 12.6
... 15.6
110.8
a. Calculate the mean. x
13.85
n
8
8
b. Calculate the variance using the deviation formula.
2
2
2
x
x 2
10.6
13.85 ... 15.6
13.85
s
6.0086
n 1
8 1
c. Calculate the variance using the direct formula.
s
2
x
2
x
n
n 1
2
(110.8)
1576.64
8
8 1
6.0086
86. The formula for computing the chi-squared statistic is:
2 O E
where O is the
E
observed value and E is the expected value. Given the following table of data, compute
the chi-squared statistic.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 515)
2
2
Math Camp 2011 Page 268

Day Mon Tues Wed Thur Fri Sat Sun
Observed 28 32 15 14 38 43 19
Expected 20 34 17 15 30 45 20
2
2
2
O
28
30 19
20
E
E
20
...
20
2
5.89
87. Prove that the following two formulas are equal.
r
n
xy
x
y
2
2
2
x
x
n
y
n
y
2
x
x y y
r
( n 1)(
s x
)( s
y
)
2
2 x
x x
2
Remember that: x and s
n
x
n
n 1
(Hint: It might be easier to start with the second formula and work backwards.)
Math Camp 2011 Page 269

n
( n 1)
x
x y y xy
xy xy xy
( n 1)(
s )( s
xy xy xy
1
2
2
x
y
2
y
xy x
y y
x nxy
1
2
2
2
x
y
2
2
y
x
xy
x
xy
x
2
( n 1)
n
x
n
y
)
( n 1)
n
1
2
y
y
n
( n 1)
x n
1
2
2
2 2
n x x n
y
y
xy x
y x
y x
y
1
2
2 2
2
2
n
x
x
n
y
y
n
xy x
y
2
2 2
2
n
x
x
n
y
y
n
1
2
2
x
y
2
y
2
x
n
n 1
n
1
2
2
x
n
1
2
2
1
2
y
n
n
n 1
2
88. A statistics instructor is interested in finding the strength of a relationship between the
final exam grades of students enrolled in Statistics I and Statistics II. Draw a scatter
plot of the data. Calculate r. Calculate a and b and write the equation for the regression
line. The data are given here in percentages.
(Bluman, Elementary Statistics, 2 nd Edition, McGraw-Hill: Boston. p. 495)
Stat I, x 87 92 68 72 95 78 83 98
Stat II, y 83 88 70 74 90 74 83 99
Math Camp 2011 Page 270

120
100
80
60
40
20
0
0 20 40 60 80 100 120
x 673
y 661
xy 56, 318
x 2
57, 443
y 2
55, 275
r
n
xy
x
y
2
2
2
x
x
n
y
2
n
y
8(56318) (673)(661)
2
2
8(57443)
(673) 8(55275) (661)
.96
a
2
y
x
x
xy
2
2
n
x
x
(661)(57443) (673)(56318)
10.25
2
8(57443) (673)
n
b
xy
x
y
2
2
n
x
x
8(56318) (673)(661)
.86
2
8(57443) (673)
Math Camp 2011 Page 271

Y . 86X
10.25
APPLICATIONS IN ECONOMICS
89. Management of McPablo’s Food Shops has completed a study of weekly demand for its
“old-fashioned” tacos in 53 regional markets. The study revealed that
Q 400 1,200P
.8A
55Pop
800P
0
where Q is the number of tacos sold per store per week, A is the level of local advertising
0
expenditure (in dollars), Pop denotes the local population (in thousands), and P is the
average taco price of local competitors. For the typical McPablo’s outlet, P = $1.50, A =
0
$1,000, Pop = 40 and P =$1.
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006)
a. Estimate the weekly sales for the typical McPablo’s outlet.
Q 400 (1200)(1.5) (.8)(1000) (55)(40) (800)(1) 2,400
b. What is the current price elasticity for tacos? What is the advertising elasticity?
E dQ
P
( 1200)(1.50) / 2,400 .75
P
dP
Q
E dQ
A
(. 8)(1000) /(2,400) . 333
A
dA
Q
90. A firm’s long-run total cost function is: C 360 40Q
10Q
2 .
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, p. 293)
a. What is the shape of the long-run average cost curve? (Hint: To find the average
cost function, divide the cost function by Q.)
C 360 40Q
10Q
360 40Q
10Q
AC
Q
2
2
360
40 10Q
Q
This curve will be U-shaped.
b. Find the output that minimizes average cost. (Hint: The minimum average costs
occurs where AC MC . Remember that MC is the first derivative of the cost
function.)
C 360 40Q
10Q
MC C
40 2(10) Q 40 20Q
2
Math Camp 2011 Page 272

AC
360
40 10Q
Q
360 40Q
10Q
360
360 10Q
Q
2
20Q
MC
36
2
2
40 20Q
2
10Q
40Q
20Q
2
2
Q 6
Thus, the quantity that minimizes average cost is 6. At this quantity, the average
cost will be:
AC
AC
360
40 10Q
Q
360
40 10(6)
60 40 60 $160
6
c. The firm faces the fixed market price of $140 per unit. At this price, can the firm
survive in the long run? Explain.
At this price, the firm cannot survive in the long run. If the average cost is $160
and the firm can only sell at $140, the firm will lose money.
91. A manufacturing firm produces output using a single plant. The relevant cost function is
2
C 500 5Q
. The firm’s demand curve is P 600 5Q
.
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, pp. 296-297)
a. Find the level of output at which average cost is minimized. (Hint: Set AC equal
to MC.) What is the minimum level of average cost?
C 500 5Q
500
AC 5Q
Q
C 500 5Q
MC C
10Q
2
2
Math Camp 2011 Page 273

AC MC
500
5Q
10Q
Q
500 5Q
500 10Q
500 5Q
10Q
5Q
2 500
Q 100
5
Q 10
2
2
2
500
AC(
Q 10) 5(10)
10
AC 50 50 100
2
2
b. Find the firm’s profit-maximizing output and price. Find its profit. {Hint:
MR MC }
P 600 5Q
R PQ
R (600 5Q)
Q
R 600Q
5Q
MR R
600 10Q
MR MC
600 10Q
10Q
600 10Q
10Q
20Q
600
600
Q 30
20
2
The maximizing output is
Q 30 . Thus the maximizing price is:
P 600 5Q
P 600 5(30)
P $450
The firm’s profit at this output and price level will be:
Math Camp 2011 Page 274

R C (600Q
5Q
600Q
5Q
600Q
10Q
500
600(30) 10(30)
500 5Q
) (500 5Q
500
18,000 9,000 500 $8,500
92. In a perfectly competitive market, industry demand is given by Q 1,000
20P . The
typical firm’s average cost is 300 Q
2
AC . The firm’s marginal cost is MC Q
.
Q 3
3
(Samuelson and Marks, Managerial Economics, (5 th Edition), Wiley, 2006, pp. 429)
2
2
2
2
c. Confirm that Qmin 30 . (Hint: Set AC equal to MC.) What is AC
min
?
AC MC
300 Q 2
Q
Q 3 3
3(300) Q
900 2Q
Q
2
900
2
2
2Q
Q
2
Q 30
The average cost at this quantity is:
2
2
2
)
AC
AC
300 Q
Q 3
300
30
30
10 10
$20
3
APPLICATIONS IN FINANCE
ln m
93. The formula t
can be used to find the number of years t required to
r
nln
1
n
multiply and investment m times when r is the per annum interest rate compounded n
times a year.
a. How many years will it take to double the value of an IRA that compounds
annually at the rate of 12%?
Math Camp 2011 Page 275

ln m
t
r
nln1
n
ln 2 ln 2 .6931
t
6.12
.12 ln1.12 .1133
(1)ln1
1
b. How many years will it take to triple the value of a savings account that
compounds quarterly at an annual rate of 6%?
ln m
t
r
nln1
n
ln 3 ln 3 1.0986
t
18.45
.06 4ln1.015 .0596
(4)ln1
4
c. Giver a derivation of this formula.
Our formula for compound interest is:
mP
r
P
n
nt
1
where mP = future value of the account (the number of times P is multiplied)
P = the principal or present value of the account
r = annual rate (as a decimal)
n = the number of times per year the interest is compounded (for instance, for
interest compounded quarterly m 4)
mP P1
r
m 1
n
r
n
nt
r
ln m ln 1
n
ln m
nt ln 1
ln m
t
r
n ln 1
n
nt
nt
r
n
Math Camp 2011 Page 276

94. You have $12,000 to invest. If part is invested at 10% and the rest at 15%, how much
should be invested at each rate to yield 12% on the total amount?
(Barnett, Ziegler and Byleen, Finite Mathematics for Business, Economics, Life Sciences, and Social Sciences, tenth edition, Upper
Saddle River, NJ: Prentice Hall, pp. 690)
Let x be the amount to invest at 10%.
(.10) x (.15)(12,000 x)
(.12)12,000
(.10) x (.15)(12,000) (.15) x 1,440
( .05)
x 1800
1440
( .05)
x 360
360
x $7,200
.05
You should invest $7,200 at 10% and 12,000 – 7,200 = $4,800 at 15%.
95. Suppose that your local government is interested in doing a capital works project that will
have the following benefit schedule:
Year 1 $0
Year 2 $0
Year 3 $50
Year 4 $500
Year 5 $5,000
Year 6 $15,000
a. Calculate the present value of the project if the social discount rate is 5%
PV
PV
PV
PV
PV
PV
PV
n
X
i
(1 r)
i1
X
1
(1 r)
0
(1 .05)
0
(1.05)
1
1
0 0 50 500 5000
1.05 1.1025 1.157625 1.21550625 1.2762815625
0 0 43.19 411.35 3917.63 11193.23
$15,565.40
X
2
(1 r)
1
i
0
(1 .05)
0
(1.05)
2
2
X
3
(1 r)
2
50
(1 .05)
50
(1.05)
3
3
X
4
(1 r)
3
500
(1.05)
X
5
(1 r)
500
(1 .05)
4
4
4
5000
(1.05)
X
6
(1 r)
5000
(1 .05)
5
5
5
15000
6
(1.05)
6
15000
(1 .05)
6
15000
1.3400956
Math Camp 2011 Page 277

. Calculate the present value of the project if the social discount rate is 10%
n
X
i
PV i
(1 r)
PV
PV
i1
X
1
PV
1
(1 r)
0
PV
1
(1 .10)
0
PV
1
(1.10)
X
2
(1 r)
0
(1 .10)
0
(1.10)
X
3
(1 r)
50
(1 .10)
50
(1.10)
X
4
(1 r)
500
(1.10)
0 0 50 500 5000 15000
1.10 1.21 1.331 1.4641 1.61051 1.771561
0 0 37.57 341.51
3104.61
8467.11
PV $11,950.79
c. Calculate the present value of the project if the social discount rate is 15%
n
X
i
PV i
(1 r)
i1
X
1
PV
1
(1 r)
0
PV
1
(1 .15)
0
PV
1
(1.15)
PV
PV
0
(1.15)
APPLICATIONS IN ENVIRONMENTAL SCIENCE
500
(1 .10)
96. The size P of a certain insect population at a time t (in days) obeys the function
0.02t
P( t)
500e
.
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 472)
a. Determine the number of insects at t 0 days.
0.02t
P(
t)
500e
2
X
2
(1 r)
X
5
(1 r)
5000
(1.10)
b. What is the growth rate of the insect population?
The growth rate is the coefficient on t, or 2%.
2
0
(1 .15)
2
2
2
50
(1.15)
3
X
3
(1 r)
2
3
50
(1 .15)
3
500
(1.15)
4
X
4
(1 r)
4
500
(1 .15)
4
5000
(1.15)
X
6
(1 r)
5000
(1 .10)
5
X
5
(1 r)
5
5
15000
6
(1.10)
X
6
(1 r)
5000
(1 .15)
15000
6
(1.15)
0 0 50 500 5000
1.15 1.3225 1.520875 1.74900625 2.011357
0 0 32.88 285.88 2485.88 6484.91
PV $9,289.55
P(
t)
500e
0.02(0)
500e
0
3
3
500(1) 500
3
4
4
4
5
5
5
6
15000
(1 .10)
6
15000
(1 .15)
15000
2.313060
6
6
Math Camp 2011 Page 278

c. What is the population after 10 days?
0.02t
P(
t)
500e
P(
t)
500e
P(
t)
500e
0.02(10)
.2
P(
t)
500(1.22) 610.7 611
d. When will the insect population reach 800?
0.02t
P(
t)
500e
800 500e
800
500
1.6 e
0.02t
ln1.6 ln e
ln1.6 0.02t
ln e
0.02t
ln1.6
t
e
ln1.6
0.02
0.02t
0.02t
0.02t
.4700
23.5
0.02
The insect population will reach 800 after about 23.5 days.
e. When will the insect population double?
0.02t
P(
t)
500e
2(500) 500e
1000
e
500
0.02t
2 e
ln 2 ln e
ln 2 0.02t
ln e
0.02t
ln 2
t
ln 2
0.02
0.02t
0.02t
0.02t
..69314
34.7
0.02
The population will double in approximately 34.7 days.
Math Camp 2011 Page 279

97. A piece of charcoal is found to contain 30% of the carbon 14 that it originally had. When
did the tree from which the charcoal came die? Use 5600 years as the half-life of carbon
kt
14. Use the formula A( t)
A0e
where A0
is the original amount of the substance and k
is the rate of decay.
(Sullivan, College Algebra, seventh edition, Upper Saddle River, NJ: Prentice Hall, p. 473)
If the half-life of carbon 14 is 5600, years, then half of the original amount should be left
after 5600 years.
A(
t)
A e
1
5600k
A0
A0e
2
1 5600k
e
2
ln .5 5600k
ln e
0
5600k
ln .5
k
ln .5
5600
kt
.693
.000124
5600
A(
t)
A e
(.30) A
0
.30 e
0
( .000124)
t
ln .3 ( .000124)
t ln e
( .000124)
t ln .3
k
kt
A e
0
ln .3
.000124
( .000124)
t
1.20397
.000124
9,727
The tree is approximately 9,727 years old.
Math Camp 2011 Page 280