Plaintext
Applied
Calculus
Edition 1
Shana Calaway
Dale Hoffman
David Lippman
This book is also available to read free online at http://www.opentextbookstore.com/appcalc/
Introduction Business Calculus 2
Copyright © 2014 Shana Calaway, Dale Hoffman, David Lippman
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Chapter 1 was remixed from Precalculus: An Investigation of Functions by David Lippman
and Melonie Rasmussen. It was adapted for this text by David Lippman, and is used under the
Creative Commons Attribution license by permission of the authors.
Chapters 2-4 were created by Shana Calaway, remixed from Contemporary Calculus by Dale
Hoffman, and edited and extended by David Lippman.
Shana Calaway teaches mathematics at Shoreline Community College.
Dale Hoffman teaches mathematics at Bellevue College. He is the author of the open textbook
Contemporary Calculus.
David Lippman teaches mathematics at Pierce College Ft Steilacoom. He is the coauthor of the
open textbooks Precalculus: An Investigation of Functions and Math in Society.
Introduction Business Calculus 3
Introduction
A Preview of Calculus
Calculus was first developed more than three hundred years ago by Sir Isaac Newton and
Gottfried Leibniz to help them describe and understand the rules governing the motion of
planets and moons. Since then, thousands of other men and women have refined the basic
ideas of calculus, developed new techniques to make the calculations easier, and found ways to
apply calculus to problems besides planetary motion. Perhaps most importantly, they have
used calculus to help understand a wide variety of physical, biological, economic and social
phenomena and to describe and solve problems in those areas.
Part of the beauty of calculus is that it is based on a few very simple ideas. Part of the power of
calculus is that these simple ideas can help us understand, describe, and solve problems in a
variety of fields.
About this book
Chapter 1 Review contains review material that you should recall before we begin calculus.
Chapter 2 The Derivative builds on the precalculus idea of the slope of a line to let us find and
use rates of change in many situations.
Chapter 3 The Integral builds on the precalculus idea of the area of a rectangle to let us find
accumulated change in more complicated and interesting settings.
Chapter 4 Functions of Two Variables extends the calculus ideas of chapter 2 to functions of
more than one variable.
Supplements
An online course framework is available on MyOpenMath.com for this book. The course
framework features:
• Links to individual sections of the e-text.
• Overview videos.
• Algorithmic, auto-grading online homework for each section of the text. Most
problems have video help tied to the question.
• A collection of printable resources created by Shana Calaway for the Open Course
Library project.
Introduction Business Calculus 4
How is Applied Calculus Different?
Students who plan to go into science, engineering, or mathematics take a year-long sequence of
classes that cover many of the same topics as we do in our one-quarter or one-semester course.
Here are some of the differences:
No trigonometry
We will not be using trigonometry at all in this course. The scientists and engineers need
trigonometry frequently, and so a great deal of the engineering calculus course is devoted to
trigonometric functions and the situations they can model.
The applications are different
The scientists and engineers learn how to apply calculus to physics problems, such as work.
They do a lot of geometric applications, like finding minimum distances, volumes of
revolution, or arclengths. In this class, we will do only a few of these (distance/velocity
problems, areas between curves). On the other hand, we will learn to apply calculus in some
economic and business settings, like maximizing profit or minimizing average cost, finding
elasticity of demand, or finding the present value of a continuous income stream. Additionally
we will apply calculus in life and social science settings, like determining the rate at which
drug concentration in the body is changing, or exploring the rate at which a subject learns.
These are applications that are seldom seen in a course for engineers.
Fewer theorems, no proofs
The focus of this course is applications rather than theory. In this course, we will use the
results of some theorems, but we won’t prove any of them. When you finish this course, you
should be able to solve many kinds of problems using calculus, but you won’t be prepared to
go on to higher mathematics.
Less algebra
In this class, you will not need clever algebra. If you need to solve an equation, it will either be
relatively simple, or you can use technology to solve it. In most cases, you won’t need “exact
answers;” calculator numbers will be good enough.
Introduction Business Calculus 5
Simplification and Calculator Numbers
When you were in tenth grade, your math teacher may have impressed you with the need to
simplify your answers. I’m here to tell you – she was wrong. The form your answer should be
in depends entirely on what you will do with it next. In addition, the process of “simplifying,”
often messy algebra can ruin perfectly correct answers. From the teacher’s point of view,
“simplifying” obscures how a student arrived at his answer, and makes problems harder to
grade. Moral: don’t spend a lot of extra time simplifying your answer. Leave it as close to
how you arrived at it as possible.
When should you simplify?
1. Simplify when it actually makes your life easier. For example, in Chapter 2 it’s easier to
find a second derivative if you simplify the first derivative.
2. Simplify your answer when you need to match it to an answer in the book. You may need to
do some algebra to be sure your answer and the book answer are the same.
When you use your calculator
A calculator is required for this course, and it can be a wonderful tool. However, you should be
careful not to rely too strongly on your calculator. Follow these rules of thumb:
1. Estimate your answers. If you expect an answer of about 4, and your calculator says
2500, you’ve made an error somewhere.
2. Don’t round until the very end. Every time you make a calculation with a rounded
number, your answer gets a little bit worse.
3. When you answer an applied problem, find a calculator number. It doesn’t mean much
12100 (2.4 )
to suggest that the company should produce items; it’s much more
2.5
meaningful to report that they should produce about 106 items.
4. When you present your final answer, round it to something that makes sense. If you’ve
found an amount of US money, round it to the nearest cent. If you’ve computed the
number of people, round to the nearest person. If there’s no obvious context, show your
teacher at least two digits after the decimal place.
5. Occasionally in this course, you will need to find the “exact answer.” That means – not
a calculator approximation. (You can still use your calculator to check your answer.)
Introduction Business Calculus 6
Table of Contents
Chapter 1: Review .................................................................................................................... 7
Section 1: Functions............................................................................................................... 7
Section 2: Operations on Functions ..................................................................................... 19
Section 3: Linear Functions ................................................................................................. 31
Section 4: Exponents ........................................................................................................... 43
Section 5: Quadratics ........................................................................................................... 46
Section 6: Polynomials and Rational Functions .................................................................. 51
Section 7: Exponential Functions ........................................................................................ 60
Section 8: Logarithmic Functions ........................................................................................ 67
Chapter 2: The Derivative ....................................................................................................... 73
Precalculus Idea: Slope and Rate of Change ....................................................................... 73
Section 1: Limits and Continuity ......................................................................................... 74
Section 2: The Derivative .................................................................................................... 80
Section 3: Power and Sum Rules for Derivatives ................................................................ 97
Section 4: Product and Quotient Rules .............................................................................. 107
Section 5: Chain Rule ........................................................................................................ 114
Section 6: Second Derivative and Concavity..................................................................... 122
Section 7: Optimization ..................................................................................................... 129
Section 8: Curve Sketching................................................................................................ 141
Section 9: Applied Optimization ....................................................................................... 148
Section 10: Other Applications .......................................................................................... 158
Section 11: Implicit Differentiation and Related Rates .................................................... 162
Chapter 3: The Integral ......................................................................................................... 163
PreCalculus Idea – The Area of a Rectangle ..................................................................... 163
Section 1: The Definite Integral ......................................................................................... 164
Section 2: The Fundamental Theorem and Antidifferentiation ......................................... 183
Section 3: Antiderivatives of Formulas ............................................................................. 190
Section 4: Substitution ....................................................................................................... 197
Section 5: Additional Integration Techniques ................................................................... 203
Section 6: Area, Volume, and Average Value ................................................................... 207
Section 7: Applications to Business................................................................................... 215
Section 8: Differential Equations ....................................................................................... 222
Chapter 4: Functions of Two Variables................................................................................. 233
PreCalculus Idea -- Topographical Maps........................................................................... 233
Section 1: Functions of Two Variables.............................................................................. 234
Section 2: Calculus of Functions of Two Variables .......................................................... 255
Section 3: Optimization ..................................................................................................... 265
Chapter 1 Review Applied Calculus 7
Chapter 1: Review
Section 1: Functions
What is a Function?
The natural world is full of relationships between quantities that change. When we see these
relationships, it is natural for us to ask “If I know one quantity, can I then determine the other?”
This establishes the idea of an input quantity, or independent variable, and a corresponding
output quantity, or dependent variable. From this we get the notion of a functional relationship
in which the output can be determined from the input.
For some quantities, like height and age, there are certainly relationships between these
quantities. Given a specific person and any age, it is easy enough to determine their height, but
if we tried to reverse that relationship and determine height from a given age, that would be
problematic, since most people maintain the same height for many years.
Function
Function: A rule for a relationship between an input, or independent, quantity and an
output, or dependent, quantity in which each input value uniquely determines one output
value. We say “the output is a function of the input.”
Example 1
In the height and age example above, is height a function of age? Is age a function of height?
In the height and age example above, it would be correct to say that height is a function of
age, since each age uniquely determines a height. For example, on my 18th birthday, I had
exactly one height of 69 inches.
However, age is not a function of height, since one height input might correspond with more
than one output age. For example, for an input height of 70 inches, there is more than one
output of age since I was 70 inches at the age of 20 and 21.
Function Notation
To simplify writing out expressions and equations involving functions, a simplified notation is
often used. We also use descriptive variables to help us remember the meaning of the quantities
in the problem.
Rather than write “height is a function of age”, we could use the descriptive variable h to
represent height and we could use the descriptive variable a to represent age.
“height is a function of age” if we name the function f we write
“h is f of a” or more simply
h = f(a) we could instead name the function h and write
h(a) which is read “h of a”
This chapter was remixed from Precalculus: An Investigation of Functions, (c) 2013 David Lippman and Melonie
Rasmussen. It is licensed under the Creative Commons Attribution license.
Chapter 1 Review Applied Calculus 8
Remember we can use any variable to name the function; the notation h(a) shows us that h
depends on a. The value “a” must be put into the function “h” to get a result. Be careful - the
parentheses indicate that age is input into the function (Note: do not confuse these parentheses
with multiplication!).
Function Notation
The notation output = f(input) defines a function named f.
This would be read “output is f of input”
Example 2
A function N = f(y) gives the number of police officers, N, in a town in year y. What does
f(2005) = 300 tell us?
When we read f(2005) = 300, we see the input quantity is 2005, which is a value for the input
quantity of the function, the year (y). The output value is 300, the number of police officers
(N), a value for the output quantity. Remember N=f(y). So this tells us that in the year 2005
there were 300 police officers in the town.
Tables as Functions
Functions can be represented in many ways: Words (as we did in the last few examples), tables
of values, graphs, or formulas. Represented as a table, we are presented with a list of input and
output values.
This table represents the age of children in years and their corresponding heights. While some
tables show all the information we know about a function, this particular table represents just
some of the data available for height and ages of children.
(input) a, age in years 5 5 6 7 8 9 10
(output) h, height inches 40 42 44 47 50 52 54
Example 3
Which of these tables define a function (if any)?
Input Output Input Output Input Output
2 1 -3 5 1 0
5 3 0 1 5 2
8 6 4 5 5 4
The first and second tables define functions. In both, each input corresponds to exactly one
output. The third table does not define a function since the input value of 5 corresponds with
two different output values.
Chapter 1 Review Applied Calculus 9
Solving and Evaluating Functions:
When we work with functions, there are two typical things we do: evaluate and solve.
Evaluating a function is what we do when we know an input, and use the function to determine
the corresponding output. Evaluating will always produce one result, since each input of a
function corresponds to exactly one output.
Solving equations involving a function is what we do when we know an output, and use the
function to determine the inputs that would produce that output. Solving a function could
produce more than one solution, since different inputs can produce the same output.
Example 4
Using the table shown, where Q=g(n)
a) Evaluate g(3) n 1 2 3 4 5
Q 8 6 7 6 8
Evaluating g(3) (read: “g of 3”)
means that we need to determine the output value, Q, of the function g given the input value
of n=3. Looking at the table, we see the output corresponding to n=3 is Q=7, allowing us to
conclude g(3) = 7.
b) Solve g(n) = 6
Solving g(n) = 6 means we need to determine what input values, n, produce an output value of
6. Looking at the table we see there are two solutions: n = 2 and n = 4.
When we input 2 into the function g, our output is Q = 6
When we input 4 into the function g, our output is also Q = 6
Graphs as Functions
Oftentimes a graph of a relationship can be used to define a function. By convention, graphs are
typically created with the input quantity along the horizontal axis and the output quantity along
the vertical.
Example 5
Which of these graphs defines a function y=f(x)?
Chapter 1 Review Applied Calculus 10
Looking at the three graphs above, the first two define a function y=f(x), since for each input
value along the horizontal axis there is exactly one output value corresponding, determined by
the y-value of the graph. The 3rd graph does not define a function y=f(x) since some input
values, such as x=2, correspond with more than one output value.
Vertical Line Test
The vertical line test is a handy way to think about whether a graph defines the vertical
output as a function of the horizontal input. Imagine drawing vertical lines through the
graph. If any vertical line would cross the graph more than once, then the graph does not
define only one vertical output for each horizontal input.
Evaluating a function using a graph requires taking the given input and using the graph to look
up the corresponding output. Solving a function equation using a graph requires taking the given
output and looking on the graph to determine the corresponding input.
Example 6
Given the graph below,
a) Evaluate f(2)
b) Solve f(x) = 4
a) To evaluate f(2), we find the input of x=2 on the horizontal axis. Moving up to the graph
gives the point (2, 1), giving an output of y=1. So f(2) = 1
b) To solve f(x) = 4, we find the value 4 on the vertical axis because if f(x) = 4 then 4 is the
output. Moving horizontally across the graph gives two points with the output of 4: (-1,4) and
(3,4). These give the two solutions to f(x) = 4: x = -1 or x = 3
This means f(-1)=4 and f(3)=4, or when the input is -1 or 3, the output is 4.
Notice that while the graph in the previous example is a function, getting two input values for the
output value of 4 shows us that this function is not one-to-one.
Formulas as Functions
When possible, it is very convenient to define relationships using formulas. If it is possible to
express the output as a formula involving the input quantity, then we can define a function.
Chapter 1 Review Applied Calculus 11
Example 7
Express the relationship 2n + 6p = 12 as a function p = f(n) if possible.
To express the relationship in this form, we need to be able to write the relationship where p is
a function of n, which means writing it as p = [something involving n].
2n + 6p = 12 subtract 2n from both sides
6p = 12 - 2n divide both sides by 6 and simplify
12 − 2n 12 2n 1
p= = − =2 − n
6 6 6 3
Having rewritten the formula as p=, we can now express p as a function:
1
p= f (n)= 2 − n
3
Not every relationship can be expressed as a function with a formula.
As with tables and graphs, it is common to evaluate and solve functions involving formulas.
Evaluating will require replacing the input variable in the formula with the value provided and
calculating. Solving will require replacing the output variable in the formula with the value
provided, and solving for the input(s) that would produce that output.
Example 8
Given the function k (t )= t 3 + 2
a) Evaluate k(2)
b) Solve k(t) = 1
a) To evaluate k(2), we plug in the input value 2 into the formula wherever we see the input
variable t, then simplify
k (2)= 23 + 2
k (2)= 8 + 2
So k(2) = 10
b) To solve k(t) = 1, we set the formula for k(t) equal to 1, and solve for the input value that
will produce that output
k(t) = 1 substitute the original formula k (t )= t 3 + 2
t3 + 2 =
1 subtract 2 from each side
3
t = −1 take the cube root of each side
t = −1
Chapter 1 Review Applied Calculus 12
When solving an equation using formulas, you can check your answer by using your solution
in the original equation to see if your calculated answer is correct.
We want to know if k (t ) = 1 is true when t = −1 .
k (−1) =(−1)3 + 2
= −1 + 2
= 1 which was the desired result.
Basic Toolkit Functions
There are some basic functions that it is helpful to know the name and shape of. We call these
the basic "toolkit of functions." For these definitions we will use x as the input variable and f(x)
as the output variable.
Toolkit Functions
Linear
Constant: f ( x) = c , where c is a constant (number)
Identity: f ( x) = x
Absolute Value: f ( x) = x
Power
Quadratic: f ( x) = x 2
Cubic: f ( x) = x 3
1
Reciprocal: f ( x) =
x
1
Reciprocal squared: f ( x) = 2
x
Square root: f (=
x) x
=
2
x
Cube root: f ( x) = 3
x
Chapter 1 Review Applied Calculus 13
Graphs of the Toolkit Functions
Constant Function: f ( x) = 2 Identity: f ( x) = x Absolute Value: f ( x) = x
Quadratic: f ( x) = x 2 Cubic: f ( x) = x 3 Square root: f ( x) = x
1 1
Cube root: f ( x) = 3 x Reciprocal: f ( x) = Reciprocal squared: f ( x) =
x x2
One of our main goals in mathematics is to model the real world with mathematical functions. In
doing so, it is important to keep in mind the limitations of those models we create.
Chapter 1 Review Applied Calculus 14
This table shows a relationship between circumference and height of a tree as it grows.
Circumference, c 1.7 2.5 5.5 8.2 13.7
Height, h 24.5 31 45.2 54.6 92.1
While there is a strong relationship between the two, it would certainly be ridiculous to talk
about a tree with a circumference of -3 feet, or a height of 3000 feet. When we identify
limitations on the inputs and outputs of a function, we are determining the domain and range of
the function.
Domain and Range
Domain: The set of possible input values to a function
Range: The set of possible output values of a function
Example 9
Using the tree table above, determine a reasonable domain and range.
We could combine the data provided with our own experiences and reason to approximate the
domain and range of the function h = f(c). For the domain, possible values for the input
circumference c, it doesn’t make sense to have negative values, so c > 0. We could make an
educated guess at a maximum reasonable value, or look up that the maximum circumference
measured is about 119 feet. With this information we would say a reasonable domain is
0 < c ≤ 119 feet.
Similarly for the range, it doesn’t make sense to have negative heights, and the maximum
height of a tree could be looked up to be 379 feet, so a reasonable range is 0 < h ≤ 379 feet.
A more compact alternative to inequality notation is interval notation, in which intervals of
values are referred to by the starting and ending values. Curved parentheses are used for “strictly
less than,” and square brackets are used for “less than or equal to.” Since infinity is not a
number, we can’t include it in the interval, so we always use curved parentheses with ∞ and -∞.
The table below will help you see how inequalities correspond to interval notation:
Inequality Interval notation
5 < h ≤ 10 (5, 10]
5 ≤ h < 10 [5, 10)
5 < h < 10 (5, 10)
h < 10 (−∞,10)
h ≥ 10 [10, ∞)
all real numbers (−∞, ∞)
Chapter 1 Review Applied Calculus 15
Example 10
Describe the intervals of values shown on the line graph below using set builder and interval
notations.
To describe the values, x, that lie in the intervals shown above we would say, “x is a real
number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5.”
As an inequality it is: 1 ≤ x ≤ 3 or x > 5
In interval notation: [1,3] ∪ (5, ∞)
Example 11
3
Find the domain of each function: a) f ( x) = 2 x + 4 b) g ( x) =
6 − 3x
a) Since we cannot take the square root of a negative number, we need the inside of the square
root to be non-negative.
x + 4 ≥ 0 when x ≥ −4 .
The domain of f(x) is [−4, ∞) .
b) We cannot divide by zero, so we need the denominator to be non-zero.
6 − 3 x = 0 when x = 2, so we must exclude 2 from the domain.
The domain of g(x) is (−∞,2) ∪ (2, ∞) .
1.1 Exercises
1. The amount of garbage, G, produced by a city with population p is given by G
= f ( p ) . G is
measured in tons per week, and p is measured in thousands of people.
a. The town of Tola has a population of 40,000 and produces 13 tons of garbage each
week. Express this information in terms of the function f.
b. Explain the meaning of the statement f ( 5 ) = 2 .
2. The number of cubic yards of dirt, D, needed to cover a garden with area a square feet is
given by D = g (a ) .
a. A garden with area 5000 ft2 requires 50 cubic yards of dirt. Express this information
in terms of the function g.
b. Explain the meaning of the statement g (100 ) = 1 .
Chapter 1 Review Applied Calculus 16
3. Select all of the following graphs which represent y as a function of x.
a b c
d e f
4. Select all of the following graphs which represent y as a function of x.
a b c
d e f
5. Select all of the following tables which represent y as a function of x.
a. x 5 10 15 b. x 5 10 15 c. x 5 10 10
y 3 8 14 y 3 8 8 y 3 8 14
6. Select all of the following tables which represent y as a function of x.
a. x 2 6 13 b. x 2 6 6 c. x 2 6 13
y 3 10 10 y 3 10 14 y 3 10 14
Chapter 1 Review Applied Calculus 17
7. Given the function g ( x) graphed here, 8. Given the function f ( x) graphed here.
a. Evaluate g (2) a. Evaluate f ( 4 )
b. Solve g ( x ) = 2 b. Solve f ( x) 4
=
9. Based on the table below,
a. Evaluate f (3) b. Solve f ( x) =1
x 0 1 2 3 4 5 6 7 8 9
f ( x ) 74 28 1 53 56 3 36 45 14 47
10. Based on the table below,
a. Evaluate f (8) b. Solve f ( x) 7
=
x 0 1 2 3 4 5 6 7 8 9
f ( x ) 62 8 7 38 86 73 70 39 75 34
For each of the following functions, evaluate: f ( −2 ) , f (−1) , f (0) , f (1) , and f (2)
11. f ( x )= 4 − 2 x 12. f ( x )= 8 − 3 x
13. f ( x ) = 8 7 3
x2 − x + 14. f ( x ) = 6 7 4
x2 − x +
15. f ( x ) =3+ x +3 16. f ( x ) =4 − 3 x − 2
x −3 x−2
17. f ( x ) = 18. f ( x ) =
x +1 x+2
19. Let f ( t =
) 3t + 5
a. Evaluate f (0) b. Solve f ( t ) = 0
20. Let g ( p )= 6 − 2 p
a. Evaluate g (0) b. Solve g ( p ) = 0 L f(x)
t
r
21. Using the graph shown,
a b c x
a. Evaluate f (c)
b. Solve f ( x ) = p
c. What are the coordinates of points L and K? K p
Chapter 1 Review Applied Calculus 18
22. Match each graph with its equation.
i. ii. iii. iv.
a. y = x
b. y = x 3
c. y = 3 x
1
d. y =
x
e. y = x 2 v. vi. vii. viii.
f.
y= x
g. y = x
1
h. y =
x2
Write the domain and range of each graph as an inequality.
23. 24.
Find the domain of each function
25. f (=
x) 3 x − 2 26. f (=
x) 5 x + 3
9 6
27. f ( x ) = 28. f ( x ) =
x 6
− x 8
−
3x + 1 5x + 3
29. f ( x ) = 30. f ( x ) =
4x + 2 4x −1
Chapter 1 Review Applied Calculus 19
Section 2: Operations on Functions
Composition of Functions
Suppose we wanted to calculate how much it costs to heat a house on a particular day of the year.
The cost to heat a house will depend on the average daily temperature, and the average daily
temperature depends on the particular day of the year. Notice how we have just defined two
relationships: The temperature depends on the day, and the cost depends on the temperature.
Using descriptive variables, we can notate these two functions.
The first function, C(T), gives the cost C of heating a house when the average daily temperature
is T degrees Celsius, and the second, T(d), gives the average daily temperature of a particular city
on day d of the year. If we wanted to determine the cost of heating the house on the 5th day of
the year, we could do this by linking our two functions together, an idea called composition of
functions. Using the function T(d), we could evaluate T(5) to determine the average daily
temperature on the 5th day of the year. We could then use that temperature as the input to the
C(T) function to find the cost to heat the house on the 5th day of the year: C(T(5)).
Composition of Functions
When the output of one function is used as the input of another, we call the entire
operation a composition of functions. We write f(g(x)), and read this as “f of g of x” or “f
composed with g at x”.
An alternate notation for composition uses the composition operator:
( f g )( x) is read “f of g of x” or “f composed with g at x”, just like f(g(x)).
Example 1
Suppose c(s) gives the number of calories burned doing s sit-ups, and s(t) gives the number of
sit-ups a person can do in t minutes. Interpret c(s(3)).
When we are asked to interpret, we are being asked to explain the meaning of the expression
in words. The inside expression in the composition is s(3). Since the input to the s function is
time, the 3 is representing 3 minutes, and s(3) is the number of sit-ups that can be done in 3
minutes. Taking this output and using it as the input to the c(s) function will gives us the
calories that can be burned by the number of sit-ups that can be done in 3 minutes.
Composition of Functions using Tables and Graphs
When working with functions given as tables and graphs, we can look up values for the functions
using a provided table or graph. We start evaluation from the provided input, and first evaluate
the inside function. We can then use the output of the inside function as the input to the outside
function. To remember this, always work from the inside out.
This chapter was remixed from Precalculus: An Investigation of Functions, (c) 2013 David Lippman and Melonie
Rasmussen. It is licensed under the Creative Commons Attribution license.
Chapter 1 Review Applied Calculus 20
Example 2
Using the graphs below, evaluate f ( g (1)) .
g(x) f(x)
To evaluate f ( g (1)) , we again start with the inside evaluation. We evaluate g (1) using the
graph of the g(x) function, finding the input of 1 on the horizontal axis and finding the output
value of the graph at that input. Here, g (1) = 3 . Using this value as the input to the f
function, f ( g (1)) = f (3) . We can then evaluate this by looking to the graph of the f(x)
function, finding the input of 3 on the horizontal axis, and reading the output value of the
graph at this input. Here, f (3) = 6 , so f ( g (1)) = 6 .
Composition using Formulas
When evaluating a composition of functions where we have either created or been given
formulas, the concept of working from the inside out remains the same. First we evaluate the
inside function using the input value provided, then use the resulting output as the input to the
outside function.
Example 3
Given f (t ) = t 2 − t and h( x) = 3 x + 2 , evaluate f (h(1)) .
Since the inside evaluation is h(1) we start by evaluating the h(x) function at 1:
h(1) = 3(1) + 2 = 5
Then f (h(1)) = f (5) , so we evaluate the f(t) function at an input of 5:
f (h(1)) = f (5) = 5 2 − 5 = 20
We are not limited, however, to using a numerical value as the input to the function. We can put
anything into the function: a value, a different variable, or even an algebraic expression,
provided we use the input expression everywhere we see the input variable.
Chapter 1 Review Applied Calculus 21
Example 4
1
Let f ( x) = x 2 and g ( x) = − 2 x , find f(g(x)) and g(f(x)).
x
To find f(g(x)), we start by evaluating the inside, writing out the formula for g(x)
1
g ( x) = − 2 x
x
1
We then use the expression − 2x as input for the function f.
x
1
f ( g ( x)) = f − 2 x
x
We then evaluate the function f(x) using the formula for g(x) as the input.
2
2 1 1
Since f ( x) = x then f − 2 x = − 2 x
x x
2
1
This gives us the formula for the composition: f ( g ( x)) = − 2 x
x
Likewise, to find g(f(x)), we evaluate the inside, writing out the formula for f(x)
( )
g ( f ( x)) = g x 2
Now we evaluate the function g(x) using x2 as the input.
1
g ( f ( x)) = 2 − 2 x 2
x
Example 5
A city manager determines that the tax revenue, R, in millions of dollars collected on a
population of p thousand people is given by the formula R( p ) = 0.03 p + p , and that the
city’s population, in thousands, is predicted to follow the formula p (t ) = 60 + 2t + 0.3t 2 ,
where t is measured in years after 2010. Find a formula for the tax revenue as a function of
the year.
Since we want tax revenue as a function of the year, we want year to be our initial input, and
revenue to be our final output. To find revenue, we will first have to predict the city
population, and then use that result as the input to the tax function. So we need to find
R(p(t)). Evaluating this,
( ) ( )
R( p (t )) = R 60 + 2t + 0.3t 2 = 0.03 60 + 2t + 0.3t 2 + 60 + 2t + 0.3t 2
This composition gives us a single formula which can be used to predict the tax revenue
during a given year, without needing to find the intermediary population value.
Chapter 1 Review Applied Calculus 22
For example, to predict the tax revenue in 2017, when t = 7 (because t is measured in years
after 2010)
( )
R( p(7)) = 0.03 60 + 2(7) + 0.3(7) 2 + 60 + 2(7) + 0.3(7) 2 ≈ 12.079 million dollars
Later in this course, it will be desirable to decompose a function – to write it as a composition of
two simpler functions.
Example 6
Write f ( x) = 3 + 5 − x 2 as the composition of two functions.
We are looking for two functions, g and h, so f ( x) = g (h( x)) . To do this, we look for a
function inside a function in the formula for f(x). As one possibility, we might notice that
5 − x 2 is the inside of the square root. We could then decompose the function as:
h( x ) = 5 − x 2
g ( x) = 3 + x
We can check our answer by recomposing the functions:
( )
g (h( x)) = g 5 − x 2 = 3 + 5 − x 2
Note that this is not the only solution to the problem. Another non-trivial decomposition
would be h( x) = x 2 and g ( x) = 3 + 5 − x
Transformations of Functions
Transformations allow us to construct new equations from our basic toolkit functions. The most
basic transformations are shifting the graph vertically or horizontally.
Vertical Shift
Given a function f(x), if we define a new function g(x) as
g=( x) f ( x) + k , where k is a constant
then g(x) is a vertical shift of the function f(x), where all the output values have been
increased by k.
If k is positive, then the graph will shift up
If k is negative, then the graph will shift down
Horizontal Shift
Given a function f(x), if we define a new function g(x) as
g (=
x) f ( x + k ) , where k is a constant
then g(x) is a horizontal shift of the function f(x)
If k is positive, then the graph will shift left
Chapter 1 Review Applied Calculus 23
If k is negative, then the graph will shift right
Example 7
Given f ( x) = x , sketch a graph of h( x) = f ( x + 1) − 3 = x + 1 − 3 .
The function f is our toolkit absolute value function. We know that this graph has a V shape,
with the point at the origin. The graph of h has transformed f in two ways: f ( x + 1) is a
change on the inside of the function, giving a horizontal shift left by 1, then the subtraction by
3 in f ( x + 1) − 3 is a change to the outside of the function, giving a vertical shift down by 3.
Transforming the graph gives
Example 8
Write a formula for the graph shown, a transformation of
the toolkit square root function.
The graph of the toolkit function starts at the origin, so this
graph has been shifted 1 to the right, and up 2. In function
notation, we could write that as h( x)= f ( x − 1) + 2 . Using
the formula for the square root function we can write
h( x)= x −1 + 2
Note that this transformation has changed the domain and
range of the function. This new graph has domain [1, ∞)
and range [2, ∞) .
Another transformation that can be applied to a function is a reflection over the horizontal or
vertical axis.
Chapter 1 Review Applied Calculus 24
Reflections
Given a function f(x), if we define a new function g(x) as g ( x) = − f ( x) ,
then g(x) is a vertical reflection of the function f(x), sometimes called a reflection about
the x-axis
If we define a new function g(x) as g ( x=
) f (− x) ,
then g(x) is a horizontal reflection of the function f(x), sometimes called a reflection about
the y-axis
Example 9
A common model for learning has an equation similar to
−2− t + 1 , where k is the percentage of mastery that can be
k (t ) =
achieved after t practice sessions. This is a transformation of
the function f (t ) = 2t shown here. Sketch a graph of k(t).
This equation combines three transformations into one equation.
A horizontal reflection: f (−t ) =2− t combined with
A vertical reflection: − f (−t ) =−2− t combined with
−t
A vertical shift up 1: − f (−t ) + 1 =−2 + 1
We can sketch a graph by applying these transformations one at a time to the original
function:
The original graph Horizontally reflected Then vertically reflected
Then, after shifting up 1, we get the final graph:
k (t ) =− f (−t ) + 1 =−2− t + 1 .
Note: As a model for learning, this function
would be limited to a domain of t ≥ 0 , with
corresponding range [0,1) .
Chapter 1 Review Applied Calculus 25
With shifts, we saw the effect of adding or subtracting to the inputs or outputs of a function. We
now explore the effects of multiplying the outputs.
Vertical Stretch/Compression
Given a function f(x), if we define a new function g(x) as
g ( x) = kf ( x) , where k is a constant
then g(x) is a vertical stretch or compression of the function f(x).
If k > 1, then the graph will be stretched
If 0< k < 1, then the graph will be compressed
If k < 0, then there will be combination of a vertical stretch or compression with a vertical
reflection
Example 10
The graph to the right is a transformation of the toolkit
function f ( x) = x 3 . Relate this new function g(x) to f(x),
then find a formula for g(x).
When trying to determine a vertical stretch or shift, it is
helpful to look for a point on the graph that is relatively
clear. In this graph, it appears that g (2) = 2 . With the
basic cubic function at the same input, f (2) = 2 3 = 8 .
Based on that, it appears that the outputs of g are ¼ the
1
outputs of the function f, since g (2) = f (2) . From this
4
we can fairly safely conclude that:
1
g ( x) = f ( x)
4
We can write a formula for g by using the definition of the function f
1 1
g ( x) = f ( x) = x 3
4 4
Combining Transformations
When combining vertical transformations, it is very important to consider the order of the
transformations. For example, vertically shifting by 3 and then vertically stretching by 2 does
not create the same graph as vertically stretching by 2 and then vertically shifting by 3. The order
follows nicely from order of operations.
Combining Vertical Transformations
When combining vertical transformations written in the form af ( x) + k ,
Chapter 1 Review Applied Calculus 26
first vertically stretch by a, then vertically shift by k.
Example 11
Write an equation for the transformed graph of the
quadratic function shown.
Since this is a quadratic function, first consider what the
basic quadratic tool kit function looks like and how this
has changed. Observing the graph, we notice several
transformations:
The original tool kit function has been flipped over the x
axis, some kind of stretch or compression has occurred,
and we can see a shift to the right 3 units and a shift up 1
unit.
In total there are four operations:
Vertical reflection, requiring a negative sign outside the function
Vertical Stretch
Horizontal Shift Right 3 units, which tells us to put x-3 on the inside of the function
Vertical Shift up 1 unit, telling us to add 1 on the outside of the function
By observation, the basic tool kit function has a vertex at (0, 0) and symmetrical points at
(1, 1) and (-1, 1). These points are one unit up and one unit over from the vertex. The new
points on the transformed graph are one unit away horizontally but 2 units away vertically.
They have been stretched vertically by two.
Not everyone can see this by simply looking at the graph. If you can, great, but if not, we can
solve for it. First, we will write the equation for this graph, with an unknown vertical stretch.
f ( x) = x 2 The original function
− f ( x) = − x 2 Vertically reflected
− af ( x) = −ax 2 Vertically stretched
2
− af ( x − 3) = −a ( x − 3) Shifted right 3
− af ( x − 3) + 1 = −a ( x − 3) 2 + 1 Shifted up 1
We now know our graph is going to have an equation of the form g ( x) = −a ( x − 3) 2 + 1 . To
find the vertical stretch, we can identify any point on the graph (other than the highest point),
such as the point (2,-1), which tells us g (2) = −1 . Using our general formula, and
substituting 2 for x, and -1 for g(x)
− 1 = − a (2 − 3) 2 + 1
− 1 = −a + 1
− 2 = −a
2=a
This tells us that to produce the graph we need a vertical stretch by two.
Chapter 1 Review Applied Calculus 27
The function that produces this graph is therefore g ( x) = −2( x − 3) 2 + 1 .
Example 12
−2
On what interval(s) is the function g ( x) = + 3 increasing and decreasing?
(x − 1)2
1
This is a transformation of the toolkit reciprocal squared function, f ( x) = :
x2
−2
− 2 f ( x) = A vertical flip and vertical stretch by 2
x2
−2
− 2 f ( x − 1) = A shift right by 1
(x − 1)2
−2
− 2 f ( x − 1) + 3 = +3 A shift up by 3
(x − 1)2
The basic reciprocal squared function is increasing on (−∞,0) and
decreasing on (0, ∞) . Because of the vertical flip, the g(x)
function will be decreasing on the left and increasing on the right.
The horizontal shift right by 1 will also shift these intervals to the
right one. From this, we can determine g(x) will be increasing on
(1, ∞) and decreasing on (−∞,1) . We also could graph the
transformation to help us determine these intervals.
1.2 Exercises
Given each pair of functions, calculate f ( g ( 0 ) ) and g ( f ( 0 ) ) .
1. f ( x=
) 4 x + 8 , g ( x )= 7 − x 2 2. f ( x=
) 5 x + 7 , g ( x )= 4 − 2 x 2
1
)
3. f ( x= x + 4 , g ( x=
) 12 − x3 4. f ( x ) = , g ( x=
) 4x + 3
x+2
Use the table of values to evaluate each expression x f ( x) g( x)
5. f ( g (8))
0 7 9
6. f ( g ( 5) ) 1 6 5
7. g ( f (5)) 2 5 6
3 8 2
8. g ( f ( 3) )
4 4 1
9. f ( f (4)) 5 0 8
10. f ( f (1) ) 6 2 7
7 1 3
11. g ( g (2))
8 9 4
12. g ( g ( 6 ) ) 9 3 0
Chapter 1 Review Applied Calculus 28
Use the graphs to evaluate the expressions below.
13. f ( g (3))
14. f ( g (1) )
15. g ( f (1))
16. g ( f ( 0 ) )
17. f ( f (5))
18. f ( f ( 4 ) )
19. g ( g (2))
20. g ( g ( 0 ) )
For each pair of functions, find f ( g ( x ) ) and g ( f ( x ) ) . Simplify your answers.
1 7 1 2
21. f ( x ) = , g ( x )= +6 22. f ( x ) = , g ( x )= +4
x−6 x x−4 x
) x 2 + 1 , g ( x=)
23. f ( x= x+2 24. f ( =
x) ) x2 + 3
x + 2 , g ( x=
x +1
25. f ( x ) = x , g ( x=
) 5x + 1 26. f ( x ) = 3 x , g ( x ) =
x3
27. If f ( x=
) x 4 + 6 , g ( x)
= x − 6 and h( x)
= x , find f ( g (h( x)))
1
) x2 + 1, g ( x ) =
28. If f ( x= and h ( x )= x + 3 , find f ( g (h( x)))
x
29. The function D( p ) gives the number of items that will be demanded when the price is p. The
production cost, C ( x) is the cost of producing x items. To determine the cost of production
when the price is $6, you would do which of the following:
a. Evaluate D(C (6)) b. Evaluate C ( D(6))
c. Solve D(C ( x)) 6
= d. Solve C ( D( p )) 6
=
30. The function A(d ) gives the pain level on a scale of 0-10 experienced by a patient with d
milligrams of a pain reduction drug in their system. The milligrams of drug in the patient’s
system after t minutes is modeled by m(t ) . To determine when the patient will be at a pain
level of 4, you would need to:
a. Evaluate A ( m ( 4 ) ) b. Evaluate m ( A ( 4 ) )
c. Solve A ( m ( t ) ) 4
= d. Solve m ( A ( d ) ) 4
=
Chapter 1 Review Applied Calculus 29
Find functions f ( x) and g ( x) so the given function can be expressed as h ( x ) = f ( g ( x ) ) .
31. h ( x=
) ( x + 2) 32. h ( x=
) ( x − 5)
2 3
3 4
33. h ( x ) = 34. h ( x ) =
( x + 2)
2
x −5
35. h ( x ) =3+ x − 2 36. h ( x )= 4 + 3 x
Sketch a graph of each function as a transformation of a toolkit function.
37. f ( t ) =(t + 1) 2 − 3 38. h ( x ) = x − 1 + 4
39. k ( x ) =( x − 2 ) − 1 40. m ( t ) =3 + t + 2
3
41. f ( x ) = 4 ( x + 1) − 5 42. g ( x) = 5 ( x + 3) − 2
2 2
43. h ( x ) =−2 x − 4 + 3 44. k ( x ) =
−3 x − 1
Write an equation for each function graphed below.
45. 46.
47. 48.
49. 50.
Chapter 1 Review Applied Calculus 30
For each function graphed, estimate the intervals on which the function is increasing and
decreasing.
51. 52.
Chapter 1 Review Applied Calculus 31
Section 3: Linear Functions
As you hop into a taxicab in Allentown, the meter will immediately read $3.30; this is the “drop”
charge made when the taximeter is activated. After that initial fee, the taximeter will add $2.40
for each mile the taxi drives. In this scenario, the total taxi fare depends upon the number of
miles ridden in the taxi, and we can ask whether it is possible to model this type of scenario with
a function. Using descriptive variables, we choose m for miles and C for Cost in dollars as a
function of miles: C(m).
We know for certain that C (0) = 3.30 , since the $3.30 drop charge is assessed regardless of how
many miles are driven. Since $2.40 is added for each mile driven, we could write that if m miles
are driven, C (m) = 3.30 + 2.40m because we start with a $3.30 drop fee and then for each mile
increase we add $2.40.
It is good to verify that the units make sense in this equation. The $3.30 drop charge is measured
in dollars; the $2.40 charge is measured in dollars per mile. So
dollars
C (m) = 3.30dollars + 2.40 (m miles )
mile
When dollars per mile are multiplied by a number of miles, the result is a number of dollars,
matching the units on the 3.30, and matching the desired units for the C function.
Notice this equation C (m) = 3.30 + 2.40m consisted of two quantities. The first is the fixed
$3.30 charge which does not change based on the value of the input. The second is the $2.40
dollars per mile value, which is a rate of change. In the equation this rate of change is
multiplied by the input value.
Looking at this same problem in table format we can also see the cost changes by $2.40 for every
1 mile increase.
m 0 1 2 3
C(m) 3.30 5.70 8.10 10.50
It is important here to note that in this equation, the rate of change is constant; over any
interval, the rate of change is the same.
Graphing this equation, C (m) = 3.30 + 2.40m we see
the shape is a line, which is how these functions get
their name: linear functions
When the number of miles is zero the cost is $3.30,
giving the point (0, 3.30) on the graph. This is the
vertical or C(m) intercept. The graph is increasing in a
straight line from left to right because for each mile the
cost goes up by $2.40; this rate remains consistent.
This chapter was remixed from Precalculus: An Investigation of Functions, (c) 2013 David Lippman and Melonie
Rasmussen. It is licensed under the Creative Commons Attribution license.
Chapter 1 Review Applied Calculus 32
Linear Function
A linear function is a function whose graph produces a line. Linear functions can always
be written in the form
f ( x) = b + mx or f ( x) = mx + b ; they’re equivalent
where
b is the initial or starting value of the function (when input, x = 0), and
m is the constant rate of change of the function
This form of a line is called slope-intercept form of a line.
Many people like to write linear functions in the form f ( x) = b + mx because it
corresponds to the way we tend to speak:
“The output starts at b and increases at a rate of m.”
For this reason alone we will use the f ( x) = b + mx form for many of the examples, but
remember they are equivalent and can be written correctly both ways.
Slope and Increasing/Decreasing
m is the constant rate of change of the function (also called slope). The slope determines if
the function is an increasing function or a decreasing function.
f ( x) = b + mx is an increasing function if m > 0
f ( x) = b + mx is a decreasing function if m < 0
If m = 0 , the rate of change zero, and the function f ( x) =b + 0 x =b is just a horizontal
line passing through the point (0, b), neither increasing nor decreasing.
Example 1
Marcus currently owns 200 songs in his iTunes collection. Every month, he adds 15 new
songs. Write a formula for the number of songs, N, in his iTunes collection as a function of
the number of months, m. How many songs will he own in a year?
The initial value for this function is 200, since he currently owns 200 songs, so N (0) = 200 .
The number of songs increases by 15 songs per month, so the rate of change is 15 songs per
month. With this information, we can write the formula:
N (m) = 200 + 15m .
N(m) is an increasing linear function.
With this formula we can predict how many songs he will have in 1 year (12 months):
N (12) = 200 + 15(12) = 200 + 180 = 380 . Marcus will have 380 songs in 12 months.
Chapter 1 Review Applied Calculus 33
Calculating Rate of Change
Given two values for the input, x1 and x 2 , and two corresponding values for the output,
y1 and y 2 , or a set of points, ( x1 , y1 ) and ( x 2 , y 2 ) , if we wish to find a linear function
that contains both points we can calculate the rate of change, m:
change in output ∆y y 2 − y1
m= = =
change in input ∆x x 2 − x1
Rate of change of a linear function is also called the slope of the line.
Note in function notation, y1 = f ( x1 ) and y 2 = f ( x 2 ) , so we could equivalently write
f ( x2 ) − f ( x1 )
m=
x2 − x1
Example 2
The population of a city increased from 23,400 to 27,800 between 2002 and 2006. Find the
rate of change of the population during this time span.
The rate of change will relate the change in population to the change in time. The population
increased by 27800 − 23400 = 4400 people over the 4 year time interval. To find the rate of
change, the number of people per year the population changed by:
4400 people people
= 1100 = 1100 people per year
4 years year
Notice that we knew the population was increasing, so we would expect our value for m to be
positive. This is a quick way to check to see if your value is reasonable.
Example 3
The pressure, P, in pounds per square inch (PSI) on a diver depends upon their depth below
the water surface, d, in feet, following the equation P(d ) = 14.696 + 0.434d . Interpret the
components of this function.
output pressure PSI
The rate of change, or slope, 0.434 would have units = = . This tells us
input depth ft
the pressure on the diver increases by 0.434 PSI for each foot their depth increases.
The initial value, 14.696, will have the same units as the output, so this tells us that at a depth
of 0 feet, the pressure on the diver will be 14.696 PSI.
Chapter 1 Review Applied Calculus 34
We can now find the rate of change given two input-output pairs, and could write an equation for
a linear function if we had the rate of change and initial value. If we have two input-output pairs
and they do not include the initial value of the function, then we will have to solve for it.
Example 4
Write an equation for the linear function graphed to
the right.
Looking at the graph, we might notice that it passes
through the points (0, 7) and (4, 4). From the first
value, we know the initial value of the function is b =
7, so in this case we will only need to calculate the
rate of change:
4−7 −3
m= =
4−0 4
This allows us to write the equation:
3
f ( x) = 7 − x
4
Example 5
If f (x) is a linear function, f (3) = −2 , and f (8) = 1 , find an equation for the function.
3
In example 3, we computed the rate of change to be m = . In this case, we do not know the
5
initial value f (0) , so we will have to solve for it. Using the rate of change, we know the
3
equation will have the form f ( x) = b + x . Since we know the value of the function when x
5
= 3, we can evaluate the function at 3.
3
f (3) = b + (3) Since we know that f (3) = −2 , we can substitute on the left side
5
3
− 2 = b + (3) This leaves us with an equation we can solve for the initial value
5
9 − 19
b = −2 − =
5 5
Combining this with the value for the rate of change, we can now write a formula for this
function:
− 19 3
f ( x) = + x
5 5
Chapter 1 Review Applied Calculus 35
As an alternative to the approach used above to find the initial value, b, we can use the point-
slope form of a line instead.
Point-Slope Equation of a Line
An equation for the line passing through the point (x1, y1) with slope m can be written as
y − y1 = m( x − x1 )
This is called the point-slope form of a line. It is a little easier to write if you know a
point and the slope, but requires a bit of work to rewrite into slope-intercept form, and
requires memorizing another formula.
Example 6
Working as an insurance salesperson, Ilya earns a base salary and a commission on each new
policy, so Ilya’s weekly income, I, depends on the number of new policies, n, he sells during
the week. Last week he sold 3 new policies, and earned $760 for the week. The week before,
he sold 5 new policies, and earned $920. Find an equation for I(n), and interpret the meaning
of the components of the equation.
The given information gives us two input-output pairs: (3,760) and (5,920). We start by
finding the rate of change.
920 − 760 160
m= = = 80
5−3 2
Keeping track of units can help us interpret this quantity. Income increased by $160 when the
number of policies increased by 2, so the rate of change is $80 per policy; Ilya earns a
commission of $80 for each policy sold during the week.
We can now write the equation using the point-slope form of the line, using the slope we just
found and the point (3,760):
I − 760 = 80(n − 3)
If we wanted this in function form (slope intercept form), we could rewrite the equation into
that form:
I − 760 = 80(n − 3)
I − 760 = 80n − 240
I (n) = 520 + 80n
This form allows us to see the starting value for the function: 520. This is Ilya’s income when
n = 0, which means no new policies are sold. We can interpret this as Ilya’s base salary for
the week, which does not depend upon the number of policies sold.
Our final interpretation is: Ilya’s base salary is $520 per week and he earns an additional $80
commission for each policy sold each week.
Chapter 1 Review Applied Calculus 36
Graphs of Linear Functions
Graphical Interpretation of a Linear Equation
Graphically, in the equation f ( x) = b + mx
b is the vertical intercept of the graph and tells us we can start our graph at (0, b)
m is the slope of the line and tells us how far to rise & run to get to the next point
Once we have at least 2 points, we can extend the graph of the line to the left and right.
Example 7
2
Graph f ( x) = 5 − x using the vertical intercept and slope.
3
The vertical intercept of the function is (0, 5), giving us a point on the graph of the line.
2
The slope is − . This tells us that for every 3 units the graph “runs” in the horizontal, the
3
vertical “rise” decreases by 2 units. In graphing, we can use this by first plotting our vertical
intercept on the graph, then using the slope to find a second point. From the initial value (0,
5) the slope tells us that if we move to the right 3, we will move down 2, moving us to the
point (3, 3). We can continue this again to find a third point at (6, 1). Finally, extend the line
to the left and right, containing these points.
Another option for graphing is to use transformations of the identity function f ( x) = x . In the
equation f ( x) = mx , the m is acting as the vertical stretch of the identity function. When m is
negative, there is also a vertical reflection of the graph. Looking at some examples will also help
show the effect of slope on the shape of the graph.
Chapter 1 Review Applied Calculus 37
f ( x) = 3x f ( x) = 2 x
f ( x) = x
1
f ( x) = x
2
1
f ( x) = x
3
1
f ( x) = − x
2
f ( x) = − x
f ( x) = −2 x
In f ( x) = mx + b , the b acts as the vertical shift, moving the graph up and down without affecting
the slope of the line. Some examples:
f ( x)= x + 4 f ( x)= x + 2
f ( x) = x
f ( x)= x − 2
f ( x)= x − 4
Chapter 1 Review Applied Calculus 38
Example 8
Match each equation with one of the lines in the graph below
f ( x) = 2 x + 3
g ( x) = 2 x − 3
h( x) = −2 x + 3
1
j ( x) = x + 3
2
Only one graph has a vertical intercept of -3, so we can immediately match that graph with
g(x). For the three graphs with a vertical intercept at 3, only one has a negative slope, so we
can match that line with h(x). Of the other two, the steeper line would have a larger slope, so
we can match that graph with equation f(x), and the flatter line with the equation j(x).
1
j (=
x) x+3 g ( x=
) 2x − 3
2
f ( x=
) 2x + 3
h( x ) =−2 x + 3
In addition to understanding the basic behavior of a linear function (increasing or decreasing,
recognizing the slope and vertical intercept), it is often helpful to know the horizontal intercept
of the function – where it crosses the horizontal axis.
Finding Horizontal Intercept
The horizontal intercept of the function is where the graph crosses the horizontal axis. If
a function has a horizontal intercept, you can always find it by solving f(x) = 0.
Chapter 1 Review Applied Calculus 39
Example 9
1
Find the horizontal intercept of f ( x) = −3 + x
2
Setting the function equal to zero to find what input will put us on the horizontal axis,
1
0 = −3 + x
2
1
3= x
2
x=6
The graph crosses the horizontal axis at (6,0)
Intersections of Lines
The graphs of two lines will intersect if they are not parallel. They will intersect at the point that
satisfies both equations. To find this point when the equations are given as functions, we can
solve for an input value so that f ( x) = g ( x) . In other words, we can set the formulas for the
lines equal, and solve for the input that satisfies the equation.
Economics tells us that in a free market, the price for an item is related to the quantity that
producers will supply and the quantity that consumers will demand. Increases in prices will
decrease demand, while supply tends to increase with prices. Sometimes supply and demand are
modeled with linear functions
Example 10
The supply, in thousands of items, for custom phone cases can be modeled by the equation
s ( p ) = 0.5 + 1.2 p , while the demand can be modeled by d ( p ) = 8.7 − 0.7 p , where p is in
dollars. Find the equilibrium price and quantity, the intersection of the supply and demand
curves.
Setting s ( p ) = d ( p ) ,
0.5 + 1.2 p = 8.7 − 0.7 p
1.9 p = 8.2
8 .2
p= ≈ $4.32
1.9
We can find the output value of the intersection
point by evaluating either function at this input:
s (4.32) = 0.5 + 1.2(4.32) ≈ 5.68
These lines intersect at the point (4.32, 5.68).
Looking at the graph, this result seems
reasonable.
Chapter 1 Review Applied Calculus 40
1.3 Exercises
1. A town's population has been growing linearly. In 2003, the population was 45,000, and the
population has been growing by 1700 people each year. Write an equation, P ( t ) , for the
population t years after 2003.
2. A town's population has been growing linearly. In 2005, the population was 69,000, and the
population has been growing by 2500 people each year. Write an equation, P ( t ) , for the
population t years after 2005.
3. Timmy goes to the fair with $40. Each ride costs $2. How much money will he have left
after riding n rides?
4. At noon, a barista notices she has $20 in her tip jar. If she makes an average of $0.50 from
each customer, how much will she have in her tip jar if she serves n more customers during
her shift?
5. A phone company charges for service according to the formula: C (n=
) 24 + 0.1n , where n is
the number of minutes talked, and C (n) is the monthly charge, in dollars.
Find and interpret the rate of change and initial value.
6. A phone company charges for service according to the formula: C (n=
) 26 + 0.04n , where n
is the number of minutes talked, and C (n) is the monthly charge, in dollars.
Find and interpret the rate of change and initial value.
Given each set of information, find a linear equation satisfying the conditions, if possible
7. f (−5) 4
=− , and f (5) 2
= =, and f (5) =1
8. f (−1) 4
9. Passes through (2, 4) and (4, 10) 10. Passes through (1, 5) and (4, 11)
11. Passes through (-1,4) and (5, 2) 12. Passes through (-2, 8) and (4, 6)
13. x intercept at (-2, 0) and y intercept at (0, -3)
14. x intercept at (-5, 0) and y intercept at (0, 4)
Chapter 1 Review Applied Calculus 41
Find an equation for the function graphed
15. 16.
17. A clothing business finds there is a linear relationship between the number of shirts, n, it can
sell and the price, p, it can charge per shirt. In particular, historical data shows that 1000
shirts can be sold at a price of $30 , while 3000 shirts can be sold at a price of $22 . Find a
linear equation in the form = p mn + b that gives the price p they can charge for n shirts.
18. A farmer finds there is a linear relationship between the number of bean stalks, n, she plants
and the yield, y, each plant produces. When she plants 30 stalks, each plant yields 30 oz of
beans. When she plants 34 stalks, each plant produces 28 oz of beans. Find a linear
relationships in the form =y mn + b that gives the yield when n stalks are planted.
Match each linear equation with its graph
A
19. f ( x ) =− x − 1 B
20. f ( x ) =−2 x − 1
C
1
21. f ( x ) = − x −1
2
22. f ( x ) = 2
23. f ( x )= 2 + x
D
24. f ( x=
) 3x + 2
E
F
Sketch the graph of each equation
25. f ( x ) =−2 x − 1 26. g ( x ) =−3 x + 2
1 2
27. h ( =
x) x+2 28. k ( =
x) x −3
3 3
29. Find the point at which the line f ( x) =−2 x − 1 intersects the line g ( x) = − x
30. Find the point at which the line f ( x=
) 2 x + 5 intersects the line g ( x) =−3 x − 5
Chapter 1 Review Applied Calculus 42
31. A car rental company offers two plans for renting a car.
Plan A: 30 dollars per day and 18 cents per mile
Plan B: 50 dollars per day with free unlimited mileage
How many miles would you need to drive for plan B to save you money?
32. A cell phone company offers two data options for its prepaid phones
Pay per use: $0.002 per Kilobyte (KB) used
Data Package: $5 for 5 Megabytes (5120 Kilobytes) + $0.002 per addition KB
Assuming you will use less than 5 Megabytes, under what circumstances will the data
package save you money?
33. The Federal Helium Reserve held about 16 billion cubic feet of helium in 2010, and is being
depleted by about 2.1 billion cubic feet each year.
a. Give a linear equation for the remaining federal helium reserves, R, in terms of t, the
number of years since 2010.
b. In 2015, what will the helium reserves be?
c. If the rate of depletion doesn’t change, when will the Federal Helium Reserve be
depleted?
34. Suppose the world's current oil reserves are 1820 billion barrels. If, on average, the total
reserves is decreasing by 25 billion barrels of oil each year:
d. Give a linear equation for the remaining oil reserves, R, in terms of t, the number of
years since now.
e. Seven years from now, what will the oil reserves be?
f. If the rate of depletion isn’t change, when will the world’s oil reserves be depleted?
Chapter 1 Review Applied Calculus 43
Section 4: Exponents
The Laws of Exponents let you rewrite algebraic expressions that involve exponents. The last
three listed here are really definitions rather than rules.
Laws of Exponents:
All variables here represent real numbers and all variables in denominators are nonzero.
xa
1) x a ⋅ x b = x a +b 2) b = x a −b
x
( )
b
3) x a = x ab
b
x xb
4) ( xy ) = x y
a a a
5) = b 6) x 0 = 1 , provided x ≠ 0
y y
1
7) x − n = , provided x ≠ 0 8) x 1 / n = n x , provided x ≠ 0
xn
Example 1
Simplify (2 x 2 ) (4 x ) .
3
We'll begin by simplifying the (2x 2 ) portion. Using property #4, we can write
3
( ) (4 x )
23 x 2
3
Evaluate 23, and use property #3
8 x 6 (4 x ) Multiply the constants, and use property #1, recalling x = x1
32x 7
Being able to work with negative and fractional exponents will be very important later in this
course.
Example 2
5
Rewrite using negative exponents.
x3
1 1
Since x −n = n
, then x −3 = 3 .
x x
5
= 5 x −3
x3
This chapter was remixed from Precalculus: An Investigation of Functions, (c) 2013 David Lippman and Melonie
Rasmussen. It is licensed under the Creative Commons Attribution license.
Chapter 1 Review Applied Calculus 44
Example 3
2
x −2
Simplify as much as possible and write your answer using only positive exponents: −3
y
x −2
−3 =
2
x −2 ( ) 2
=
x −4
=
y6
y y −3 ( ) 2
y −6 x 4
Example 4
3
Rewrite 4 x − using exponents.
x
A square root is a radical with index of two. In other words, x = 2 x . Using the exponent
rule give above, x = 2 x = x1 / 2 . Rewriting the square roots using the fractional exponent,
3 3
4 x− = 4 x1 / 2 − 1 / 2
x x
Now we can use the negative exponent rule to rewrite the second term in the expression.
3
4 x1 / 2 − 1 / 2 = 4 x1 / 2 − 3 x −1 / 2
x
Example 5
Rewrite using only positive exponents: (p)5
−1 / 3
(p) 5
−1 / 3
((
= p5 ) )
1 / 2 −1 / 3
= p −5 / 6 =
1
p5/ 6
Example 6
Rewrite x −4 / 3 as a radical.
1
First, we can use the negative exponent rule to rewrite this as .
x4/3
4 1 1
Since = 4 ⋅ , we can use laws of exponents to rewrite this as .
3 3 x /3
1
( ) 4
1
Now using the radical equivalence, we can rewrite this as .
3
x
4
( )
Chapter 1 Review Applied Calculus 45
1.4 Exercises
Simplify each expression
1. x 3 x 5 2. x 4 x 2
3. ( x 3 ) 4. ( x 7 )
4 2
5. ( 2x 2 ) x 4 6. ( 5x 4 ) x 5
3 2
( 3x )
2
5x ( 4 x )
2 2
7. 8.
6 x3 2x2
Simplify, and rewrite without negative exponents
9. 4x −3 10. 2x −5
11. x −4 x 2 12. x −2 x
5 x −3 2 x −4
13. 14.
2 x −6 6 x −2
Rewrite using negative or fractional exponents
15. 4−5 16. 4−3
x x
17. 3 x 18. 4
x
19. 4 20. 1
3
x 5 x
Rewrite as a radical
21. 4x −1/ 2 22. 5x −1/3
23. 2x1/3 24. 5x 3/ 2
Chapter 1 Review Applied Calculus 46
Section 5: Quadratics
Quadratics
Quadratics are transformations of the f ( x) = x 2 function. Quadratics commonly arise from
problems involving area and projectile motion, providing some interesting applications.
Example 1
A backyard farmer wants to enclose a rectangular space for a new garden. She has purchased
80 feet of wire fencing to enclose 3 sides, and will put the 4th side against the backyard fence.
Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the
existing fence have length L.
In a scenario like this involving geometry, it is often Garden L
helpful to draw a picture. It might also be helpful to
introduce a temporary variable, W, to represent the side of
W
fencing parallel to the 4th side or backyard fence.
Since we know we only have 80 feet of fence available, Backyard we
know that L + W + L = 80 , or more simply,
2 L + W = 80 . This allows us to represent the width, W, in terms of L: W = 80 − 2 L
Now we are ready to write an equation for the area the fence encloses. We know the area of a
rectangle is length multiplied by width, so
A = LW = L(80 − 2 L)
A( L) = 80 L − 2 L2
This formula represents the area of the fence in terms of the variable length L.
Forms of Quadratic Functions
The standard form of a quadratic function is f ( x) = ax 2 + bx + c
The transformation form of a quadratic function is f ( x) = a ( x − h) 2 + k
The vertex of the quadratic function is located at (h, k), where h and k are the numbers in
the transformation form of the function. Because the vertex appears in the transformation
form, it is often called the vertex form.
This chapter was remixed from Precalculus: An Investigation of Functions, (c) 2013 David Lippman and Melonie
Rasmussen. It is licensed under the Creative Commons Attribution license.
Chapter 1 Review Applied Calculus 47
Example 2
Write an equation for the quadratic graphed below as a transformation of f ( x) = x 2 .
We can see the graph is the basic quadratic shifted to the left 2 and down 3, putting the vertex
at (-2, 3), giving a formula in the form g ( x) = a ( x + 2) 2 − 3 . By plugging in a point that falls
on the grid, such as (0,-1), we can solve for the stretch factor:
− 1 = a ( 0 + 2) 2 − 3
2 = 4a
1
a=
2
1
The equation for this formula is g ( x) = ( x + 2) 2 − 3 .
2
Short run Behavior: Intercepts
As with any function, we can find the vertical intercepts of a quadratic by evaluating the function
at an input of zero, and we can find the horizontal intercepts by solving for when the output will
be zero. Notice that depending upon the location of the graph, we might have zero, one, or two
horizontal intercepts.
zero horizontal intercepts one horizontal intercept two horizontal intercepts
Notice that in the standard form of a quadratic, the constant term c reveals the vertical intercept
of the graph, since f (0) = a (0) 2 + b(0) + c = c .
Chapter 1 Review Applied Calculus 48
Example 3
Find the vertical and horizontal intercepts of the quadratic f ( x) = 3 x 2 + 5 x − 2
We can find the vertical intercept by evaluating the function at an input of zero:
f (0) = 3(0) 2 + 5(0) − 2 = −2 Vertical intercept at (0,-2)
For the horizontal intercepts, we solve for when the output will be zero
0 = 3x 2 + 5 x − 2
In this case, the quadratic can be factored easily, providing the simplest method for solution
0 = (3 x − 1)( x + 2)
0 = 3x − 1
0= x+2 1
1 or Horizontal intercepts at ,0 and (-2,0)
x= x = −2 3
3
When a quadratic is not factorable or is hard to factor, we can turn to the quadratic formula.
Quadratic Formula
For a quadratic function given in standard form f ( x) = ax 2 + bx + c , the quadratic
formula gives the horizontal intercepts of the graph of this function.
− b ± b 2 − 4ac
x=
2a
Example 4
A ball is thrown upwards from the top of a 40 foot high building at a speed of 80 feet per
second. The ball’s height above ground can be modeled by the equation
H (t ) =−16t 2 + 80t + 40 . When does the ball hit the ground?
To find when the ball hits the ground, we need to determine when the height is zero – when
H(t) = 0. While we could do this using the transformation form of the quadratic, we can also
use the quadratic formula:
− 80 ± 80 2 − 4(−16)(40) − 80 ± 8960
t= =
2(−16) − 32
Since the square root does not simplify nicely, we can use a calculator to approximate the
values of the solutions:
− 80 − 8960 − 80 + 8960
t= ≈ 5.458 or t= ≈ −0.458
− 32 − 32
The second answer is outside the reasonable domain of our model, so we conclude the ball
will hit the ground after about 5.458 seconds.
Chapter 1 Review Applied Calculus 49
1.5 Exercises
Write an equation for the quadratic function graphed.
1. 2.
3. 4.
5. 6.
For each of the follow quadratic functions, find the vertical and horizontal intercepts.
7. y ( x ) = 2 x 2 + 10 x + 12 8. z ( p ) = 3 x 2 + 6 x − 9
9. f ( x ) = 2 x 2 − 10 x + 4 10. g ( x ) =
−2 x 2 − 14 x + 12
11. h ( t ) =−4t 2 + 6t − 1 12. k ( t ) = 2 x 2 + 4 x − 15
Chapter 1 Review Applied Calculus 50
1
13. The height of a ball thrown in the air is given by h ( x ) =
− x 2 + 6 x + 3 , where x is the
12
horizontal distance in feet from the point at which the ball is thrown.
a. How high is the ball when it was thrown?
b. What is the maximum height of the ball?
c. How far from the thrower does the ball strike the ground?
1
14. A javelin is thrown in the air. Its height is given by h ( x ) =
− x 2 + 8 x + 6 , where x is the
20
horizontal distance in feet from the point at which the javelin is thrown.
a. How high is the javelin when it was thrown?
b. What is the maximum height of the javelin?
c. How far from the thrower does the javelin strike the ground?
Chapter 1 Review Applied Calculus 51
Section 6: Polynomials and Rational Functions
Polynomial Functions
Terminology of Polynomial Functions
A polynomial is function that can be written as f ( x) = a 0 + a1 x + a 2 x 2 + + a n x n
Each of the ai constants are called coefficients and can be positive, negative, or zero, and
be whole numbers, decimals, or fractions.
A term of the polynomial is any one piece of the sum, that is any ai x i . Each individual
term is a transformed power function.
The degree of the polynomial is the highest power of the variable that occurs in the
polynomial.
The leading term is the term containing the highest power of the variable: the term with
the highest degree.
The leading coefficient is the coefficient of the leading term.
Because of the definition of the “leading” term we often rearrange polynomials so that the
powers are descending.
f ( x) = a n x n + ..... + a 2 x 2 + a1 x + a 0
Example 1
Identify the degree, leading term, and leading coefficient of the polynomial
f ( x) = 3 + 2 x 2 − 4 x 3
For the function f(x), the degree is 3, the highest power on x. The leading term is the term
containing that power, − 4x 3 . The leading coefficient is the coefficient of that term, -4.
Short run Behavior: Intercepts
As with any function, the vertical intercept can be found by evaluating the function at an input of
zero. Since this is evaluation, it is relatively easy to do it for a polynomial of any degree. To
find horizontal intercepts, we need to solve for when the output will be zero. For general
polynomials, this can be a challenging prospect. Consequently, we will limit ourselves to three
cases:
1) The polynomial can be factored using known methods: greatest common factor and
trinomial factoring.
2) The polynomial is given in factored form.
3) Technology is used to determine the intercepts.
This chapter was remixed from Precalculus: An Investigation of Functions, (c) 2013 David Lippman and Melonie
Rasmussen. It is licensed under the Creative Commons Attribution license.
Chapter 1 Review Applied Calculus 52
Example 2
Find the horizontal intercepts of f ( x) = x 6 − 3 x 4 + 2 x 2 .
We can attempt to factor this polynomial to find solutions for f(x) = 0.
x 6 − 3x 4 + 2 x 2 = 0 Factoring out the greatest common factor
2 4
( 2
x x − 3x + 2 = 0 ) Factoring the inside as a quadratic in x2
x 2
(x 2
)( 2
−1 x − 2 = 0 ) Then break apart to find solutions
(x 2
)
−1 = 0 (x 2
)
−2 =0
x2 = 0 2
x =1 2
or or x =2
x=0
x = ±1 x=± 2
This gives us 5 horizontal intercepts.
Example 3
Find the horizontal intercepts of h(t ) = t 3 + 4t 2 + t − 6
Since this polynomial is not in factored form, has no
common factors, and does not appear to be factorable
using techniques we know, we can turn to technology to
find the intercepts.
Graphing this function, it appears there are horizontal
intercepts at t = -3, -2, and 1.
We could check these are correct by plugging in these
values for t and verifying that h(−3) = h(−2) = h(1) = 0 .
Solving Polynomial Inequalities
One application of our ability to find intercepts and sketch a graph of polynomials is the ability
to solve polynomial inequalities. It is a very common question to ask when a function will be
positive and negative, and one we will use later in this course.
Example 4
Solve ( x + 3)( x + 1) 2 ( x − 4) > 0
As with all inequalities, we start by solving the equality ( x + 3)( x + 1) 2 ( x − 4) = 0 , which has
solutions at x = -3, -1, and 4. We know the function can only change from positive to
negative at these values, so these divide the inputs into 4 intervals.
Chapter 1 Review Applied Calculus 53
We could choose a test value in each interval and evaluate the function
f ( x) = ( x + 3)( x + 1) 2 ( x − 4) at each test value to determine if the function is positive or
negative in that interval
Interval Test x in interval f( test value) >0 or <0?
x < -3 -4 72 >0
-3 < x < -1 -2 -6 <0
-1 < x < 4 0 -12 <0
x>4 5 288 >0
On a number line this would look like:
positive 0 negative 0 negative 0 positive
From our test values, we can determine this function is positive when x < -3 or x > 4, or in
interval notation, (−∞,−3) ∪ (4, ∞)
Rational Functions
Rational functions are the ratios, or fractions, of polynomials. They can arise from both simple
and complex situations.
Example 5
You plan to drive 100 miles. Find a formula for the time the trip will take as a function of the
speed you drive.
You may recall that multiplying speed by time will give you distance. If we let t represent the
drive time in hours, and v represent the velocity (speed or rate) at which we drive, then
vt = distance . Since our distance is fixed at 100 miles, vt = 100 . Solving this relationship
for the time gives us the function we desired:
100
t (v ) =
v
1
Notice that this is a transformation of the reciprocal toolkit function, f ( x) = . Several natural
x
phenomena, such as gravitational force and volume of sound, behave in a manner inversely
proportional to the square of another quantity. For example, the volume, V, of a sound heard
k
at a distance d from the source would be related by V = 2 for some constant value k. These
d
1
functions are transformations of the reciprocal squared toolkit function f ( x) = 2 .
x
Chapter 1 Review Applied Calculus 54
We have seen the graphs of the basic reciprocal function and the squared reciprocal function
from our review of toolkit functions. These graphs have several important features.
1 1
f ( x) = f ( x) =
x x2
1
Let’s begin by looking at the reciprocal function, f ( x) = . As you well know, dividing by zero
x
is not allowed and therefore zero is not in the domain, and so the function is undefined at an
input of zero.
Short run behavior:
As the input values approach zero from the left side (taking on very small, negative values), the
function values become very large in the negative direction (in other words, they approach
negative infinity).
We write: as x → 0 − , f (x) → −∞ .
As we approach zero from the right side (small, positive input values), the function values
become very large in the positive direction (approaching infinity).
We write: as x → 0 + , f (x) → ∞ .
This behavior creates a vertical asymptote. An asymptote is a line that the graph approaches. In
this case the graph is approaching the vertical line x = 0 as the input becomes close to zero.
Long run behavior:
As the values of x approach infinity, the function values approach 0.
As the values of x approach negative infinity, the function values approach 0.
Symbolically: as x → ±∞ , f ( x) → 0
Based on this long run behavior and the graph we can see that the function approaches 0 but
never actually reaches 0, it just “levels off” as the inputs become large. This behavior creates a
horizontal asymptote. In this case the graph is approaching the horizontal line f ( x) = 0 as the
input becomes very large in the negative and positive directions.
Chapter 1 Review Applied Calculus 55
Vertical and Horizontal Asymptotes
A vertical asymptote of a graph is a vertical line x = a where the graph tends towards
positive or negative infinity as the inputs approach a. As x → a , f (x) → ±∞ .
A horizontal asymptote of a graph is a horizontal line y = b where the graph approaches
the line as the inputs get large. As x → ±∞ , f ( x) → b .
Example 6
Sketch a graph of the reciprocal function shifted two units to the left and up three units.
Identify the horizontal and vertical asymptotes of the graph, if any.
Transforming the graph left 2 and up 3 would result in the function
1 3x + 7
f ( x) = + 3 , or equivalently, by giving the terms a common denominator, f ( x) =
x+2 x+2
Shifting the toolkit function would give us this
graph. Notice that this equation is undefined at
x = -2, and the graph also is showing a vertical
asymptote at x = -2.
As x → −2− , f ( x) → −∞ , and as x → −2+ ,
f ( x) → ∞
As the inputs grow large, the graph appears to
be leveling off at output values of 3, indicating
a horizontal asymptote at y = 3 .
As x → ±∞ , f ( x) → 3 .
Notice that horizontal and vertical asymptotes get shifted left 2 and up 3 along with the
function.
A general rational function is the ratio of any two polynomials.
Rational Function
A rational function is a function that can be written as the ratio of two polynomials, P(x)
and Q(x).
2 p
P( x) a0 + a1 x + a2 x + + a p x
f ( x) =
=
Q( x) b0 + b1 x + b2 x 2 + + bq x q
Rational functions can arise from real situations.
Chapter 1 Review Applied Calculus 56
Example 7
A large mixing tank currently contains 100 gallons of water, into which 5 pounds of sugar
have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the
same time sugar is poured into the tank at a rate of 1 pound per minute. Find the
concentration (pounds per gallon) of sugar in the tank after t minutes.
Notice that the amount of water in the tank is changing linearly, as is the amount of sugar in
the tank. We can write an equation independently for each:
water = 100 + 10t sugar = 5 + 1t
The concentration, C, will be the ratio of pounds of sugar to gallons of water
5+t
C (t ) =
100 + 10t
Vertical Asymptotes of Rational Functions
The vertical asymptotes of a rational function will occur where the denominator of the
function is equal to zero and the numerator is not zero.
Horizontal Asymptote of Rational Functions
The horizontal asymptote of a rational function can be determined by looking at the
degrees of the numerator and denominator.
Degree of denominator > degree of numerator: Horizontal asymptote at y = 0
Degree of denominator < degree of numerator: No horizontal asymptote
Degree of denominator = degree of numerator: Horizontal asymptote at ratio of leading
coefficients.
Example 8
5+t
In the sugar concentration problem from earlier, we created the equation C (t ) = .
100 + 10t
Find the horizontal asymptote and interpret it in context of the scenario.
Both the numerator and denominator are linear (degree 1), so since the degrees are equal,
there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator,
the leading term is t, with coefficient 1. In the denominator, the leading term is 10t, with
coefficient 10. The horizontal asymptote will be at the ratio of these values: As t → ∞ ,
1 1
C (t ) → . This function will have a horizontal asymptote at y = .
10 10
This tells us that as the input gets large, the output values will approach 1/10. In context, this
means that as more time goes by, the concentration of sugar in the tank will approach one
tenth of a pound of sugar per gallon of water or 1/10 pounds per gallon.
Example 9
Find the horizontal and vertical asymptotes of the function
Chapter 1 Review Applied Calculus 57
( x − 2)( x + 3)
f ( x) =
( x − 1)( x + 2)( x − 5)
First, note this function has no inputs that make both the numerator and denominator zero, so
there are no potential holes. The function will have vertical asymptotes when the
denominator is zero, causing the function to be undefined. The denominator will be zero at x
= 1, -2, and 5, indicating vertical asymptotes at these values.
The numerator has degree 2, while the denominator has degree 3. Since the degree of the
denominator is greater than the degree of the numerator, the denominator will grow faster
than the numerator, causing the outputs to tend towards zero as the inputs get large, and so as
x → ±∞ , f ( x) → 0 . This function will have a horizontal asymptote at y = 0 .
As with all functions, a rational function will have a vertical intercept when the input is zero, if
the function is defined at zero. It is possible for a rational function to not have a vertical
intercept if the function is undefined at zero.
Likewise, a rational function will have horizontal intercepts at the inputs that cause the output to
be zero (unless that input corresponds to a hole). It is possible there are no horizontal intercepts.
Since a fraction is only equal to zero when the numerator is zero, horizontal intercepts will occur
when the numerator of the rational function is equal to zero.
Example 10
( x − 2)( x + 3)
Find the intercepts of f ( x) =
( x − 1)( x + 2)( x − 5)
We can find the vertical intercept by evaluating the function at zero
(0 − 2)(0 + 3) −6 3
f (0) = = =−
(0 − 1)(0 + 2)(0 − 5) 10 5
The horizontal intercepts will occur when the function is equal to zero:
( x − 2)( x + 3)
0= This is zero when the numerator is zero
( x − 1)( x + 2)( x − 5)
0 = ( x − 2)( x + 3)
x = 2, − 3
Chapter 1 Review Applied Calculus 58
1.6 Exercises
Find the degree and leading coefficient of each polynomial
1. 4x 7 2. 5x 6
3. 5 − x 2 4. 6 + 3 x − 4 x3
5. −2 x 4 − 3 x 2 + x − 1 6. 6 x5 − 2 x 4 + x 2 + 3
Find the vertical and horizontal intercepts of each function.
7. f ( t ) = 2 ( t − 1)( t + 2 ) (t − 3) 9. f ( x ) =3 ( x + 1)( x − 4 ) ( x + 5)
9. g ( n ) =
−2 ( 3n − 1) (2n + 1) 10. k ( u ) =
−3 ( 4 − n ) (4n + 3)
11. C ( t ) = 2t 4 − 8t 3 + 6t 2 12. C ( t ) =4t 4 + 12t 3 − 40t 2
Use your calculator or other graphing technology to solve graphically for the zeros of the
function.
13. f ( x ) = x3 − 7 x 2 + 4 x + 30 14. g ( x ) = x 3 − 6 x 2 + x + 28
Solve each inequality.
15. ( x − 3)( x − 2 ) > 0 16. ( x − 5 )( x + 1) > 0
2 2
17. ( x − 1)( x + 2 ) ( x − 3) < 0 18. ( x − 4 )( x + 3) ( x + 6 ) < 0
For each function, find the horizontal intercepts, the vertical intercept, the vertical asymptotes,
and the horizontal asymptote.
2x − 3 x −5
19. p ( x ) = 20. q ( x ) =
x+4 3x − 1
4 5
21. s ( x ) = 22. r ( x ) =
( x − 2) ( x + 1)
2 2
3 x 2 − 14 x − 5 2 x 2 + 7 x − 15
23. f ( x ) = 24. g ( x ) =
3 x 2 + 8 x − 16 3 x 2 − 14 + 15
2 x2 + x −1 2 x 2 − 3 x − 20
25. h ( x ) = 26. k ( x ) =
x−4 x −5
Chapter 1 Review Applied Calculus 59
27. A scientist has a beaker containing 20 mL of a solution containing 20% acid. To dilute this,
she adds pure water.
a. Write an equation for the concentration in the beaker after adding n mL of water.
b. Find the concentration if 10 mL of water has been added.
c. How many mL of water must be added to obtain a 4% solution?
d. What is the behavior as n → ∞ , and what is the physical significance of this?
28. A scientist has a beaker containing 30 mL of a solution containing 3 grams of potassium
hydroxide. To this, she mixes a solution containing 8 milligrams per mL of potassium
hydroxide.
e. Write an equation for the concentration in the tank after adding n mL of the second
solution.
f. Find the concentration if 10 mL of the second solution has been added.
g. How many mL of water must be added to obtain a 50 mg/mL solution?
h. What is the behavior as n → ∞ , and what is the physical significance of this?
Chapter 1 Review Applied Calculus 60
Section 7: Exponential Functions
Consider these two companies:
Company A has 100 stores, and expands by opening 50 new stores a year
Company B has 100 stores, and expands by increasing the number of stores by 50% of their total
each year.
Company A is exhibiting linear growth. In linear growth, we have a constant rate of change – a
constant number that the output increased for each increase in input. For company A, the
number of new stores per year is the same each year.
Company B is different – we have a percent rate of change rather than a constant number of
stores/year as our rate of change. To see the significance of this difference compare a 50%
increase when there are 100 stores to a 50% increase when there are 1000 stores:
100 stores, a 50% increase is 50 stores in that year.
1000 stores, a 50% increase is 500 stores in that year.
Calculating the number of stores after several years, we can clearly see the difference in results.
B
years Company A Company B
2 200 225
4 300 506
6 400 1139
8 500 2563
10 600 5767
A
This percent growth can be modeled with an exponential function.
Exponential Function
An exponential growth or decay function is a function that grows or shrinks at a constant
percent growth rate. The equation can be written in the form
f ( x) = a (1 + r ) x or f ( x) = ab x where b = 1+r
Where
a is the initial or starting value of the function
r is the percent growth or decay rate, written as a decimal
b is the growth factor or growth multiplier. Since powers of negative numbers behave
strangely, we limit b to positive values.
Example 1
India’s population was 1.14 billion in the year 2008 and is growing by about 1.34% each year.
Write an exponential function for India’s population, and use it to predict the population in
2020.
This chapter was remixed from Precalculus: An Investigation of Functions, (c) 2013 David Lippman and Melonie
Rasmussen. It is licensed under the Creative Commons Attribution license.
Chapter 1 Review Applied Calculus 61
Using 2008 as our starting time (t = 0), our initial population will be 1.14 billion. Since the
percent growth rate was 1.34%, our value for r is 0.0134.
Using the basic formula for exponential growth f ( x) = a (1 + r ) x we can write the formula,
f (t ) = 1.14(1 + 0.0134) t
To estimate the population in 2020, we evaluate the function at t = 12, since 2020 is 12 years
after 2008.
f (12) = 1.14(1 + 0.0134)12 ≈ 1.337 billion people in 2020
Example 2
A certificate of deposit (CD) is a type of savings account offered by banks, typically offering
a higher interest rate in return for a fixed length of time you will leave your money invested.
If a bank offers a 24 month CD with an annual interest rate of 1.2% compounded monthly,
how much will a $1000 investment grow to over those 24 months?
First, we must notice that the interest rate is an annual rate, but is compounded monthly,
meaning interest is calculated and added to the account monthly. To find the monthly interest
rate, we divide the annual rate of 1.2% by 12 since there are 12 months in a year: 1.2%/12 =
0.1%. Each month we will earn 0.1% interest. From this, we can set up an exponential
function, with our initial amount of $1000 and a growth rate of r = 0.001, and our input m
measured in months.
m
.012
f (m) = 10001 +
12
f (m) = 1000(1 + 0.001) m
After 24 months, the account will have grown to f (24) =1000(1 + 0.001) 24 =$1024.28
Example 3
Bismuth-210 is an isotope that radioactively decays by about 13% each day, meaning 13% of
the remaining Bismuth-210 transforms into another atom (polonium-210 in this case) each
day. If you begin with 100 mg of Bismuth-210, how much remains after one week?
With radioactive decay, instead of the quantity increasing at a percent rate, the quantity is
decreasing at a percent rate. Our initial quantity is a = 100 mg, and our growth rate will be
negative 13%, since we are decreasing: r = -0.13. This gives the equation:
Q(d ) = 100(1 − 0.13) d = 100(0.87) d
This can also be explained by recognizing that if 13% decays, then 87 % remains.
After one week, 7 days, the quantity remaining would be
Q(7) = 100(0.87) 7 = 37.73 mg of Bismuth-210 remains.
Chapter 1 Review Applied Calculus 62
Example 4
T(q) represents the total number of Android smart phone contracts, in thousands, held by a
certain Verizon store region measured quarterly since January 1, 2010,
Interpret all of the parts of the equation T (2) = 86(1.64) 2 = 231.3056 .
Interpreting this from the basic exponential form, we know that 86 is our initial value. This
means that on Jan. 1, 2010 this region had 86,000 Android smart phone contracts. Since b = 1
+ r = 1.64, we know that every quarter the number of smart phone contracts grows by 64%.
T(2) = 231.3056 means that in the 2nd quarter (or at the end of the second quarter) there were
approximately 231,305 Android smart phone contracts.
When working with exponentials, there is a special constant we must talk about. It arises when
we talk about things growing continuously, such as continuous compounding, or natural
phenomena like radioactive decay that happen continuously.
Euler’s Number: e
e ≈ 2.718282
Because e is often used as the base of an exponential, most scientific and graphing calculators
have a button that can calculate powers of e, usually labeled ex. Some computer software instead
defines a function exp(x), where exp(x) = ex. Since calculus studies continuous change, we will
almost always use the e-based form of exponential equations in this course.
Continuous Growth Formula
Continuous Growth can be calculated using the formula
f ( x) = ae rx
where
a is the starting amount
r is the continuous growth rate
Example 5
Radon-222 decays at a continuous rate of 17.3% per day. How much will 100mg of Radon-
222 decay to in 3 days?
Since we are given a continuous decay rate, we use the continuous growth formula. Since the
substance is decaying, we know the growth rate will be negative: r = -0.173
f (3) = 100e −0.173(3) ≈ 59.512 mg of Radon-222 will remain.
Chapter 1 Review Applied Calculus 63
Graphs of Exponential Functions
Graphical Features of Exponential Functions
Graphically, in the function f ( x) = ab x
a is the vertical intercept of the graph
b determines the rate at which the graph grows
the function will increase if b > 1
the function will decrease if 0 < b < 1
The graph will have a horizontal asymptote at y = 0
The graph will be concave up if a > 0; concave down if a < 0.
The domain of the function is all real numbers
The range of the function is (0, ∞)
When sketching the graph of an exponential function, it can be helpful to remember that the
graph will pass through the points (0, a) and (1, ab).
The value b will determine the function’s long run behavior:
If b > 1, as x → ∞ , f (x) → ∞ and as x → −∞ , f ( x) → 0 .
If 0 < b < 1, as x → ∞ , f ( x) → 0 and as x → −∞ , f (x) → ∞ .
Example 6
x
1
Sketch a graph of f ( x) = 4
3
This graph will have a vertical intercept at (0,4), and pass
4
through the point 1, . Since b < 1, the graph will be
3
decreasing towards zero. Since a > 0, the graph will be
concave up.
We can also see from the graph the long run behavior: as
x → ∞ , f ( x) → 0 and as x → −∞ , f (x) → ∞ .
To get a better feeling for the effect of a and b on the graph, examine the sets of graphs below.
Chapter 1 Review Applied Calculus 64
The first set shows various graphs, where a remains the same and we only change the value for
b. Notice that the closer the value of b is to 1, the less steep the graph will be.
( 13 )
x
3x
2x
( 12)
x
1.5 x
0.9 x
In the next set of graphs, a is altered and our value for b remains the same.
4 (1.2 x ) 3 (1.2 x )
2 (1.2 x )
1.2 x
0.5 (1.2 x )
Notice that changing the value for a changes the vertical intercept. Since a is multiplying the bx
term, a acts as a vertical stretch factor, not as a shift. Notice also that the long run behavior for
all of these functions is the same because the growth factor did not change and none of these a
values introduced a vertical flip.
Chapter 1 Review Applied Calculus 65
Example 7
Match each equation with its graph.
f ( x) = 2(1.3) x
g ( x) = 2(1.8) x
h( x) = 4(1.3) x
k ( x) = 4(0.7) x
The graph of k(x) is the easiest to identify, since it is the only equation with a growth factor
less than one, which will produce a decreasing graph. The graph of h(x) can be identified as
the only growing exponential function with a vertical intercept at (0,4). The graphs of f(x)
and g(x) both have a vertical intercept at (0,2), but since g(x) has a larger growth factor, we
can identify it as the graph increasing faster.
g(x)
f(x)
h(x)
k(x)
Chapter 1 Review Applied Calculus 66
1.7 Exercises
1. A population numbers 11,000 organisms initially and grows by 8.5% each year. Write an
exponential model for the population.
2. A population is currently 6,000 and has been increasing by 1.2% each day. Write an
exponential model for the population.
3. A vehicle purchased for $32,500 depreciates at a constant rate of 5% each year. Determine the
approximate value of the vehicle 12 years after purchase.
4. A business purchases $125,000 of office furniture which depreciates at a constant rate of 12%
each year. Find the residual value of the furniture 6 years after purchase.
5. If $4,000 is invested in a bank account at an interest rate of 7 per cent per year, find the
amount in the bank after 9 years if interest is compounded annually, quarterly, monthly, and
continuously.
6. If $6,000 is invested in a bank account at an interest rate of 9 per cent per year, find the
amount in the bank after 5 years if interest is compounded annually, quarterly, monthly, and
continuously.
Match each function with one of the graphs below. C
B D
7. f ( x ) = 2 ( 0.69 )
x
E
8. f ( x ) = 2 (1.28 )
x
A
F
9. f ( x ) = 2 ( 0.81)
x
10. f ( x ) = 4 (1.28 )
x
11. f ( x ) = 2 (1.59 )
x
12. f ( x ) = 4 ( 0.69 )
x
If all the graphs to the right have equations with form D
B
f ( t ) = ae kt , C
A E
13. Which graph has the largest value for k?
14. Which graph has the smallest value for k? F
15. Which graph has the largest value for a?
16. Which graph has the smallest value for a?
Chapter 1 Review Applied Calculus 67
Section 8: Logarithmic Functions
Logarithms are the inverse of exponential functions -- they allow us to undo exponential
functions and solve for the exponent. They are also commonly used to express quantities that
vary widely in size.
Logarithm Equivalent to an Exponential
The logarithm (base b) function, written log b ( x ) , is the inverse of the exponential
function (base b), b x .
This means the statement b a = c is equivalent to the statement log b (c) = a .
Properties of Logs: Inverse Properties
( )
log b b x = x
log b x
b =x
Example 1
Write these exponential equations as logarithmic equations:
1
a) 2 3 = 8 b) 5 2 = 25 c) 10 − 4 =
10000
a) 2 3 = 8 is equivalent to log 2 (8) = 3
b) 5 2 = 25 is equivalent to log 5 (25) = 2
1 1
c) 10−4 = is equivalent to log10 = −4
10000 10000
Example 2
Solve 2 x = 10 for x.
By rewriting this expression as a logarithm, we get x = log 2 (10)
While this does define a solution, and an exact solution at that, you may find it somewhat
unsatisfying since it is difficult to compare this expression to the decimal estimate we made
earlier. Also, giving an exact expression for a solution is not always useful – often we really
need a decimal approximation to the solution. Luckily, this is a task calculators and computers
are quite adept at. Unluckily for us, most calculators and computers will only evaluate
logarithms of two bases. Happily, this ends up not being a problem, as we’ll see briefly.
This chapter was remixed from Precalculus: An Investigation of Functions, (c) 2013 David Lippman and Melonie
Rasmussen. It is licensed under the Creative Commons Attribution license.
Chapter 1 Review Applied Calculus 68
Common and Natural Logarithms
The common log is the logarithm with base 10, and is typically written log(x) .
The natural log is the logarithm with base e, and is typically written ln(x) .
Example 3
Evaluate log(1000) using the definition of the
Values of the common log
common log. number number as log(number)
exponential
To evaluate log(1000) , we can say x = log(1000) , 1000 103 3
then rewrite into exponential form using the 100 10 2
2
common log base of 10. 10 101 1
10 x = 1000 1 10 0
0
From this, we might recognize that 1000 is the cube 0.1 10-1 -1
of 10, so x = 3. 0.01 10 -2
-2
We also can use the inverse property of logs to write 0.001 10-3 -3
( )
log10 10 3 = 3
Example 4
Evaluate log(500) using your calculator or computer.
Using a computer or calculator, we can evaluate log(500) ≈ 2.69897
Another property provides the basis for solving exponential equations.
Properties of Logs: Exponent Property
( )
log b A r = r log b ( A)
Solving exponential equations:
Isolate the exponential expressions when possible
Take the logarithm of both sides
Utilize the exponent property for logarithms to pull the variable out of the exponent
Use algebra to solve for the variable.
Example 5
In the last section, we predicted the population (in billions) of India t years after 2008 by
using the function f (t ) = 1.14(1 + 0.0134) t . If the population continues following this trend,
when will the population reach 2 billion?
Chapter 1 Review Applied Calculus 69
We need to solve for the t so that f(t) = 2
2 = 1.14(1.0134) t Divide by 1.14 to isolate the exponential expression
2
= 1.0134 t Take the logarithm of both sides of the equation
1.14
2
= ln (1.0134 ) Apply the exponent property on the right side
t
ln
1.14
2
ln = t ln(1.0134 ) Divide both sides by ln(1.0134)
1.14
2
ln
1.14
t= ≈ 42.23 years
ln(1.0134 )
If this growth rate continues, the model predicts the population of India will reach 2 billion
about 42 years after 2008, or approximately in the year 2050.
Example 6
Solve 5e −0.3t = 2 for t.
First we divide by 5 to isolate the logarithm
2
e −0.3t =
5
Since this equation involves e, it makes sense to use the natural log:
2
ln (e −0.3t ) = ln Utilizing the inverse property for logs,
5
2
− 0.3t = ln Now dividing by -0.3,
5
2
ln
5
t = ≈ 3.054
− 0 .3
In addition to solving exponential equations, logarithmic expressions are common in many
physical situations.
Chapter 1 Review Applied Calculus 70
Example 7
In chemistry, pH is a measure of the acidity or basicity of a liquid. The pH is related to the
concentration of hydrogen ions, [H+], measured in moles per liter, by the equation
(
pH = − log H + . )
If a liquid has concentration of 0.0001 moles per liber, determine the pH.
Determine the hydrogen ion concentration of a liquid with pH of 7.
To answer the first question, we evaluate the expression − log(0.0001) . While we could use
our calculators for this, we do not really need them here, since we can use the inverse property
of logs:
( )
− log(0.0001) = − log 10 −4 = −(−4) = 4
( )
To answer the second question, we need to solve the equation 7 = − log H + . Begin by
isolating the logarithm on one side of the equation by multiplying both sides by -1:
−7 = (
log H + )
Rewriting into exponential form yields the answer
H=
+
10
= −7
0.0000001 moles per liter.
While we don't often need to sketch the graph of a logarithm, it is helpful to understand the basic
shape.
Graphical Features of the Logarithm
Graphically, in the function g ( x) = log b ( x)
The graph has a horizontal intercept at (1, 0)
The graph has a vertical asymptote at x = 0
The graph is increasing and concave down
The domain of the function is x > 0, or (0, ∞)
The range of the function is all real numbers, or (−∞, ∞)
When sketching a general logarithm with base b, it can be helpful to remember that the graph
will pass through the points (1, 0) and (b, 1).
To get a feeling for how the base affects the shape of the graph, examine the graphs below.
Chapter 1 Review Applied Calculus 71
log 2 ( x)
ln( x)
log( x)
Another important observation made was the domain of the logarithm: x > 0. Like the reciprocal
and square root functions, the logarithm has a restricted domain which must be considered when
finding the domain of a composition involving a log.
Example 8
Find the domain of the function f ( x) = log(5 − 2 x)
The logarithm is only defined with the input is positive, so this function will only be defined
when 5 − 2 x > 0 . Solving this inequality,
− 2 x > −5
5
x<
2
5 5
The domain of this function is x < , or in interval notation, − ∞,
2 2
Chapter 1 Review Applied Calculus 72
1.8 Exercises
Rewrite each equation in exponential form
1. log ( v ) = t 2. log ( r ) = s 3. ln ( w ) = n 4. ln ( x ) = y
Rewrite each equation in logarithmic form.
5. 10a = b 6. 10 p = v 7. e k = h 8. e y = x
Solve each equation for the variable.
1 1
9. 5 x = 14 10. 3x = 23 11. 7 x = 12. 3x =
15 4
13. e5 x = 17 14. e3 x = 12 15. 34 x−5 = 38 16. 42 x−3 = 44
17. 1000 (1.03) = 5000 18. 200 (1.06 ) = 550
t t
19. 3 (1.04 ) = 8 20. 2 (1.08 ) = 7
3t 4t
21. 50e −0.12t = 10 22. 10e −0.03t = 4
x x
1 1
23. 10 − 8 =
5 24. 100 − 100 =
70
2 4
25. The population of Kenya was 39.8 million in 2009 and has been growing by about 2.6% each
year. If this trend continues, when will the population exceed 45 million?
26. The population of Algeria was 34.9 million in 2009 and has been growing by about 1.5%
each year. If this trend continues, when will the population exceed 45 million?
27. If $1000 is invested in an account earning 3% compounded monthly, how long will it take
the account to grow in value to $1500?
28. If $1000 is invested in an account earning 2% compounded quarterly, how long will it take
the account to grow in value to $1300?
29. Sketch a graph
= of: f ( x ) log
= ( x ) , g ( x ) ln ( x )
Find the domain of each function.
f ( x ) log ( x − 5 )
30. = ( x ) ln ( 3 − x )
31. f =
32. f (=
x ) ln (1 − 3 x ) f ( x ) log ( 2 x + 5 )
33. =
Chapter 2 The Derivative Applied Calculus 73
Chapter 2: The Derivative
Precalculus Idea: Slope and Rate of Change
The slope of a line measures how fast a line rises or falls as
we move from left to right along the line. It measures the
rate of change of the y-coordinate with respect to changes in
the x-coordinate. If the line represents the distance traveled
over time, for example, then its slope represents the
velocity. In the figure, you can remind yourself of how we
calculate slope using two points on the line:
rise y 2 − y1 ∆y
m = slope from P to Q = = =
run x 2 − x1 ∆x
We would like to be able to get that same sort of information (how fast the curve rises or falls,
velocity from distance) even if the graph is not a straight line. But what happens if we try to find
the slope of a curve at a point, as in the figure below? We need two points in order to determine
the slope of a line. How can we find a slope of a curve, at just one point?
The answer, as suggested in the figure, is to find the slope of the
tangent line to the curve at that point. Most of us have an intuitive
idea of what a tangent line is. Unfortunately, “tangent line” is hard to
define precisely.
Definition: A secant line is a line between two points on a
curve, as in the figure to the right.
Can’t-quite-do-it-yet Definition: A tangent line is a line at one
point on a curve …. that does its best to be the curve at that
point?
As you may be able to see, the closer the point Q is to the point P,
the closer the secant slope gets to the tangent slope. This will be
key to finding the tangent slope, but first we need to more carefully
define the idea of "getting closer to."
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 2 The Derivative Applied Calculus 74
Section 1: Limits and Continuity
In the last section, we saw that as the interval over which we calculated got smaller, the secant
slopes approached the tangent slope. The limit gives us better language with which to discuss the
idea of “approaches.”
The limit of a function describes the behavior of the function when the variable is
near, but does not equal , a specified number (Fig. 1). If the values of f(x) get
closer and closer , as close as we want, to one number L as we take values of x
very close to (but not equal to) a number c, then
We say "the limit of f(x), as x approaches c, is L " and we write
lim f ( x) = L. (The symbol " → " means "approaches" or "gets very close to.")
x→c
f(c) is a single number that describes the behavior (value) of f(x) AT the point x = c.
lim f(x) is a single number that describes the behavior of f(x) NEAR, BUT NOT AT, the
x →c
point x = c.
If we have a graph of the function near x = c , then it is usually easy to determine lim f ( x) .
x→c
Example 1
Use the graph of y = f(x) in Fig. 2 to determine the following limits:
(a) lim f(x) (b) lim f(x)
x →1 x →2
(c) lim f(x) (d) lim f(x)
x →3 x →4
(a) lim f(x) = 2 .
x →1
When x is very close to 1, the values of f(x) are
very close to y = 2. In this example, it happens that
f(1) = 2, but that is irrelevant for the limit. The only thing that matters is what happens for x
close to 1 but x ≠ 1.
(b) f(2) is undefined, but we only care about the behavior of f(x) for x close to 2 and not
equal to 2. When x is close to 2, the values of f(x) are close to 3. If we restrict x close
enough to 2, the values of y will be as close to 3 as we want, so lim f(x) = 3.
x →2
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 2 The Derivative Applied Calculus 75
(c) When x is close to 3 (or as x approaches the value 3), the values of f(x) are close to 1
(or approach the value 1), so lim f(x) = 1. For this limit it is completely irrelevant that f(3) =
x →3
2, We only care about what happens to f(x) for x close to and not equal to 3.
(d) This one is harder and we need to be careful. When x is close to 4 and slightly less than
4 (x is just to the left of 4 on the x–axis), then the values of f(x) are close to 2. But if x is
close to 4 and slightly larger than 4 then the values of f(x) are close to 3. If we only know
that x is very close to 4, then we cannot say whether y = f(x) will be close to 2 or close to 3
–– it depends on whether x is on the right or the left side of 4. In this situation, the f(x)
values are not close to a single number so we say lim f(x) does not exist. It is irrelevant that
x →4
f(4) = 1. The limit, as x approaches 4, would still be undefined if f(4) was 3 or 2 or
anything else.
We can also explore limits using tables and using algebra.
Example 2
2x 2 − x −1
Find lim .
x →1 x −1
2x 2 − x − 1
Solution: You might try to evaluate f ( x) = at x = 1, but f(x) is not defined at x = 1.
x −1
It is tempting, but wrong, to conclude that this function does not have a limit as x approaches 1.
Using Tables: Trying some "test" values for x which get closer and closer to 1 from both the
left and the right, we get
x f(x) x f(x)
0.9 2.82 1.1 3.2
0.9998 2.9996 1.003 3.006
0.999994 2.999988 1.0001 3.0002
0.9999999 2.9999998 1.000007 3.000014
↓ ↓ ↓ ↓
1 3 1 3
The function f is not defined at x = 1, but when x is close to 1, the values of f(x) are getting
very close to 3. We can get f(x) as close to 3 as we want by taking x very close to 1 so
2x 2 − x −1
lim = 3.
x →1 x −1
Chapter 2 The Derivative Applied Calculus 76
Using algebra: We could have found the same result by noting that
2 x 2 − x − 1 (2 x + 1)( x − 1)
f ( x) = = = 2x+1 as long as x ≠ 1. (If x≠1, then x–1 ≠ 0 so it is valid
x −1 ( x − 1)
to divide the numerator and denominator by the factor x–1.) The "x→1" part of the limit
means that x is close to 1 but not equal to 1, so our division step is valid and
2x 2 − x −1
lim = lim 2x + 1 = 3 , the correct answer.
x →1 x −1 x →1
2x 2 − x − 1
Using a graph: We can graph y = f ( x) = for
x −1
x close to 1, and notice that whenever x is close to 1,
the values of y = f(x) are close to 3. f is not defined at x
= 1, so the graph has a hole above x = 1, but we only
care about what f(x) is doing for x close to but not
equal to 1.
One Sided Limits
Sometimes, what happens to us at a place depends on the direction we use to approach that place.
If we approach Niagara Falls from the upstream side, then we will be 182 feet higher and have
different worries than if we approach from the downstream side. Similarly, the values of a
function near a point may depend on the direction we use to approach that point.
Definition of Left and Right Limits:
The left limit as x approaches c of f(x) is L if the values of f(x) get as close to L as
we want when x is very close to and left of c, x < c: lim f(x) = L .
x →c −
The right limit, written with x → c+ , requires that x lie to the right of c, x > c:
lim f ( x) = L
x →c +
Chapter 2 The Derivative Applied Calculus 77
Example 3
Evaluate the one sided limits of the function f(x) graphed here at
x = 0 and x = 1.
As x approach 0 from the left, the value of the function is getting
closer to 1, so lim f ( x) = 1 .
x →0 −
As x approaches 0 from the right, the value of the function is getting closer to 2, so
lim f ( x) = 2
x →0 +
Notice that since the limit from the left and limit from the right are different, the general limit,
lim f ( x) , does not exit.
x→0
At x approaches 1 from either direction, the value of the function is approaching 1, so
lim f ( x) = lim f ( x) = lim f ( x) = 1 .
x →1− x →1+ x →1
Continuity
A function that is “friendly” and doesn’t have any breaks or jumps in it is called continuous.
More formally,
Definition of Continuity at a Point
A function f is continuous at x = a if and only if lim f(x) = f(a) .
x →a
The graph to the right illustrates some of the different ways a
function can behave at and near a point, and the table contains
some numerical information about the function and its behavior.
Based on the information in the table, we can conclude that f is
continuous at 1 since lim f(x) = 2 = f(1).
x →1
We can also conclude from the information in the table that
f is not continuous at 2 or 3 or 4, because
a f(a) lim f(x)
lim f(x) ≠ f(2) , lim f(x) ≠ f(3) , and lim f(x) ≠ f(4). x→a
x →2 x →3 x →4
1 2 2
The behaviors at x = 2 and x = 4 exhibit a hole in the graph,
2 1 2
sometimes called a removable discontinuity, since the graph
could be made continuous by changing the value of a single 3 2 does not exist
point. The behavior at x = 3 is called a jump discontinuity,
4 undefined 2
since the graph jumps between two values.
Chapter 2 The Derivative Applied Calculus 78
So which functions are continuous? It turns out pretty much every function you’ve studied is
continuous where it is defined: polynomial, radical, rational, exponential, and logarithmic
functions are all continuous where they are defined. Moreover, any combination of continuous
functions is also continuous.
This is helpful, because the definition of continuity says that for a continuous function,
lim f ( x) = f (a ) . That means for a continuous function, we can find the limit by direct substitution
x→a
(evaluating the function) if the function is continuous at a.
Example 4
Evaluate using continuity, if possible:
x−4 x−4
a) lim x 3 − 4 x b) lim b) lim
x→2 x→2 x+3 x→2 x−2
a) The given function is polynomial, and is defined for all values of x, so we can find the limit
by direct substitution:
lim x 3 − 4 x = 2 3 − 4(2) = 0
x→2
b) The given function is rational. It is not defined at x = -3, but we are taking the limit as x
approaches 2, and the function is defined at that point, so we can use direct substitution:
x−4 2−4 2
lim = =−
x→2 x+3 2+3 5
c) This function is not defined at x = 2, and so is not continuous at x = 2. We cannot use direct
substitution.
2.1 Exercises
1. Use the graph to determine the following limits.
(a) lim f (x) (b) lim f (x)
x →1 x →2
(c) lim f (x) (d) lim f (x)
x →3 x →4
2. Use the graph to determine the following limits.
(a) lim f (x) (b) lim f (x)
x →1 x →2
(c) lim f (x) (d) lim f (x)
x →3 x →4
Chapter 2 The Derivative Applied Calculus 79
x 2 + 3x + 3 x 2 + 3x + 3
5. Evaluate (a) lim (b) lim
x →1 x −2 x →2 x −2
x+7 x+7
6. Evaluate (a) lim 2
(b) lim 2
x →0 x + 9x + 14 x →3 x + 9x + 14
x+7 x+7
(c) lim 2
(d) lim 2
x →−4 x + 9x + 14 x →−7 x + 9x + 14
7. At which points is the function shown
discontinuous?
8. At which points is the function shown discontinuous?
9. Find at least one point at which each function is not continuous and state which of the 3
conditions in the definition of continuity is
violated at that point.
x+5 x2 + x – 6 x
(a) (b) (c)
x–3 x–2 x
π
(d) 2 (e) ln( x2 )
x –6x + 9
Chapter 2 The Derivative Applied Calculus 80
Section 2: The Derivative
Definition of the Derivative
Suppose we drop a tomato from the top of a 100 foot building and time its fall.
Time (sec) Height (ft)
0.0 100
0.5 96
1.0 84
1.5 64
2.0 36
2.5 0
Some questions are easy to answer directly from the table:
(a) How long did it take for the tomato n to drop 100 feet? (2.5 seconds)
(b) How far did the tomato fall during the first second? (100 – 84 = 16 feet)
(c) How far did the tomato fall during the last second? (64 – 0 = 64 feet)
(d) How far did the tomato fall between t =.5 and t = 1? (96 – 84 = 12 feet)
Some other questions require a little calculation:
(e) What was the average velocity of the tomato during its fall?
distance fallen ∆ position –100 ft
Average velocity = = = = –40 ft/s .
total time ∆ time 2.5 s
(f) What was the average velocity between t=1 and t=2 seconds?
∆ position 36 ft – 84 ft –48 ft
Average velocity = ∆ time = 2s–1s = 1 s = –48 ft/s .
Some questions are more difficult.
(g) How fast was the tomato falling 1 second after it was dropped?
This question is significantly different from the previous two questions about average velocity.
Here we want the instantaneous velocity, the velocity at an instant in time. Unfortunately the
tomato is not equipped with a speedometer so we will have to give an approximate answer.
One crude approximation of the instantaneous velocity after 1 second is simply the average
velocity during the entire fall, –40 ft/s . But the tomato fell slowly at the beginning and
rapidly near the end so the "–40 ft/s" estimate may or may not be a good answer.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 2 The Derivative Applied Calculus 81
We can get a better approximation of the instantaneous velocity at t=1 by calculating the
average velocities over a short time interval near t = 1 . The average velocity between t = 0.5
–12 feet
and t = 1 is 0.5 s = –24 ft/s, and the average velocity between t = 1 and t = 1.5 is
–20 feet
.5 s = –40 ft/s so we can be reasonably sure that the instantaneous velocity is between –
24 ft/s and –40 ft/s.
In general, the shorter the time interval over which we
calculate the average velocity, the better the average
velocity will approximate the instantaneous velocity.
∆ position
The average velocity over a time interval is ∆ time ,
which is the slope of the secant line through two points
on the graph of height versus time. The instantaneous
velocity at a particular time and height is the slope of the
tangent line to the graph at the point given by that time
and height.
∆ position
Average velocity = = slope of the secant line through 2 points.
∆ time
Instantaneous velocity = slope of the line tangent to the graph.
GROWING BACTERIA
Suppose we set up a machine to count the
number of bacteria growing on a Petri plate. At
first there are few bacteria so the population
grows slowly. Then there are more bacteria to
divide so the population grows more quickly.
Later, there are more bacteria and less room and
nutrients available for the expanding population,
so the population grows slowly again. Finally,
the bacteria have used up most of the nutrients,
and the population declines as bacteria die.
Chapter 2 The Derivative Applied Calculus 82
The population graph can be used to answer a number of questions.
(a) What is the bacteria population at time t = 3 days?
From the graph, at t = 3, the population is about 0.5 thousand, or 500 bacteria.
(b) What is the population increment from t = 3 to t =10 days?
At t = 10, the population is about 4.5 thousand, so the increment is about 4000 bacteria
(c) What is the rate of population growth from t = 3 to t = 10 days?
The rate of growth from t = 3 to t = 10 is the average change in population during that
time:
change in population ∆ population
average change in population = change in time = ∆ time
4000 bacteria
= ≈ 570 bacteria/day .
7 days
This is the slope of the secant line through the two points (3, 500) and (10, 4500).
(d) What is the rate of population growth on the third day, at t = 3 ?
This question is asking for the
instantaneous rate of population change,
the slope of the line which is tangent to
the population curve at (3, 500). If we
sketch a line approximately tangent to the
curve at (3, 500) and pick two points
near the ends of the tangent line segment ,
we can estimate that instantaneous rate of
population growth is approximately 320
bacteria/day .
Chapter 2 The Derivative Applied Calculus 83
Tangent Lines
Do this!
The graph below is the graph of y = f ( x ) . We want to find the slope of the tangent line at the
point (1, 2).
First, draw the secant line between (1, 2) and (2, −1) and compute its slope.
Now draw the secant line between (1, 2) and (1.5, 1) and compute its slope.
Compare the two lines you have drawn. Which would be a better approximation of the tangent
line to the curve at (1, 2)?
Now draw the secant line between (1, 2) and (1.3, 1.5) and compute its slope. Is this line an even
better approximation of the tangent line?
Now draw your best guess for the tangent line and measure its slope. Do you see a pattern in the
slopes?
You should have noticed that as the interval got smaller and smaller, the secant line got closer to
the tangent line and its slope got closer to the slope of the tangent line. That’s good news – we
know how to find the slope of a secant line.
In some applications, we need to know where the graph of a function f(x) has horizontal tangent lines
(slopes = 0).
Example 1
At right is the graph of y = g(x). At what values of x does the graph of y = g(x) below have
horizontal tangent lines?
The tangent lines to the graph of g(x) are horizontal (slope = 0) when x ≈ –1, 1, 2.5, and 5.
Chapter 2 The Derivative Applied Calculus 84
Let's explore further this idea of finding the tangent slope based on the secant slope.
Example 2
Find the slope of the line L in the graph below which is tangent to f(x) = x2 at the point (2,4).
We could estimate the slope of L from the graph, but we won't. Instead, we will use the idea
that secant lines over tiny intervals approximate the tangent line.
We can see that the line through (2,4) and (3,9) on the graph of f is an approximation of the
slope of the tangent line, and we can calculate that slope exactly: m = ∆y/∆x = (9–4)/(3–2) = 5.
But m = 5 is only an estimate of the slope of the tangent line and not a very good estimate. It's
too big. We can get a better estimate by picking a second point on the graph of f which is closer
to (2,4) –– the point (2,4) is fixed and it must be one of the points we use.
From the second figure, we can see that the slope of the line through the points (2,4) and
(2.5,6.25) is a better approximation of the slope of the tangent line at (2,4):
∆y 6.25 − 4 2.25
=m = = = 4.5
∆x 2.5 − 2 0.5
a better estimate, but still an approximation. We can continue picking points closer and closer
to (2,4) on the graph of f, and then calculating the slopes of the lines through each of these
points and the point (2,4):
Points to the left of (2,4) Points to the left of (2,4)
x y = x2 Slope x y = x2 Slope
1.5 2.25 3.5 3 9 5
1.9 3.61 3.9 2.5 6.25 4.5
1.99 3.9601 3.99 2.01 4.0401 4.01
The only thing special about the x–values we picked is that they are numbers which are close,
and very close, to x = 2. Someone else might have picked other nearby values for x. As the
points we pick get closer and closer to the point (2,4) on the graph of y = x2 , the slopes of the
lines through the points and (2,4) are better approximations of the slope of the tangent line,
and these slopes are getting closer and closer to 4.
Chapter 2 The Derivative Applied Calculus 85
We can bypass much of the calculating by not picking the points one at a time: let's look at a
general point near (2,4). Define x = 2 + h so h is the increment from 2 to x. If h is small,
then x = 2 + h is close to 2 and the point ( 2 + h, f (2 + h) ) =( 2 + h, (2 + h) 2 ) is close to (2,4).
The slope m of the line through the points (2,4) and ( 2 + h, (2 + h) ) is a good approximation
2
of the slope of the tangent line at the point (2,4):
m=
∆y (2 + h) 2 − 4
= =
( 4 + 4h + h ) − 4=
2
4h + h 2
= 4+h
∆x (2 + h) − 2 h h
The value m = 4 + h is the slope of the secant line through the two points (2,4) and
( 2 + h, (2 + h)2 ) . As h gets smaller and smaller, this slope approaches the slope of the tangent
line to the graph of f at (2,4).
∆y
More formally, we could write: Slope of the tangent line = lim = lim(4 + h)
h →0 ∆x h →0
We can easily evaluate this limit using direct substitution, finding that as the interval h shrinks
towards 0, the secant slope approaches the tangent slope, 4.
The tangent line problem and the instantaneous velocity problem are the same problem. In each
problem we wanted to know how rapidly something was changing at an instant in time, and the
answer turned out to be finding the slope of a tangent line, which we approximated with the slope
of a secant line. This idea is the key to defining the slope of a curve.
Chapter 2 The Derivative Applied Calculus 86
The Derivative:
The derivative of a function f at a point (x, f(x)) is the instantaneous rate of change.
The derivative is the slope of the tangent line to the graph of f at the point (x, f(x)).
The derivative is the slope of the curve f(x) at the point (x, f(x)).
A function is called differentiable at (x, f(x)) if its derivative exists at (x, f(x)).
Notation for the Derivative:
The derivative of y = f(x) with respect to x is written as
f ' (x ) (read aloud as “f prime of x”), or y ' (“y prime”)
dy df
or (read aloud as “dee why dee ex”), or
dx dx
The notation that resembles a fraction is called Leibniz notation. It displays not only the
name of the function (f or y), but also the name of the variable (in this case, x). It looks
∆y
like a fraction because the derivative is a slope. In fact, this is simply written in Roman
∆x
letters instead of Greek letters.
Verb forms:
We find the derivative of a function, or take the derivative of a function, or differentiate
a function.
dy
We use an adaptation of the notation to mean “find the derivative of f(x):”
dx
d
( f (x )) = df
dx dx
Formal Algebraic Definition:
f (x + h ) − f (x )
f ' ( x) = lim
h →0 h
Practical Definition:
The derivative can be approximated by looking at an average rate of change, or the slope of
a secant line, over a very tiny interval. The tinier the interval, the closer this is to the true
instantaneous rate of change, slope of the tangent line, or slope of the curve.
Looking Ahead:
We will have methods for computing exact values of derivatives from formulas soon. If
the function is given to you as a table or graph, you will still need to approximate this way.
This is the foundation for the rest of this chapter. It’s remarkable that such a simple idea (the slope
of a tangent line) and such a simple definition (for the derivative f ' ) will lead to so many
important ideas and applications.
Chapter 2 The Derivative Applied Calculus 87
Example 3
1
Find the slope of the tangent line to f ( x) = when x = 3.
x
The slope of the tangent line is the value of the derivative 𝑓 ′ (3).
1 1
f (3) = , and f (3 + h) =
3 3+ h
Using the formal limit definition of the derivative,
1 1
−
f (3 + h) − f (3)
= = lim 3 + h 3
f ′(3) lim
h →0 h h →0 h
We can simplify by giving the fractions a common denominator.
1 3 1 3+ h
⋅ − ⋅
lim 3 + h 3 3 3+ h
h →0 h
3 3+ h
−
= lim 9 + 3h 9 + 3h
h →0 h
−h
= lim 9 + 3h
h →0 h
−h 1
= lim ⋅
h →0 9 + 3h h
−1
= lim
h →0 9 + 3h
We can evaluate this limit by direct substitution:
−1 1
lim = −
h →0 9 + 3h 9
1 1
The slope of the tangent line to f ( x) = at x = 3 is −
x 9
The Derivative as a Function
We now know how to find (or at least approximate) the derivative of a function for any x-value;
this means we can think of the derivative as a function, too. The inputs are the same x’s; the output
is the value of the derivative at that x value.
Chapter 2 The Derivative Applied Calculus 88
Example 4
Below is the graph of a function y = f ( x ) . We can use the information in the graph to fill in a
table showing values of f ' ( x ) :
At various values of x, draw your best guess at the tangent line and measure its slope. You
might have to extend your lines so you can read some points. In general, your estimate of the
slope will be better if you choose points that are easy to read and far away from each other.
Here are my estimates for a few values of x (parts of the tangent lines I used are shown):
x y = f (x ) f ' ( x ) = the estimated SLOPE
of the tangent line to the curve
at the point ( x, y ) .
0 0 1
1 1 0
2 0 −1
3 −1 0
3.5 0 2
4 1 1
5 2 0.5
We can estimate the values of f’(x) at some non-integer values of x, too: f’(.5) ≈ 0.5 and
f’(1.3) ≈ –0.3.
We can even think about entire intervals. For
example, if 0 < x < 1, then f(x) is increasing, all the
slopes are positive, and so f’(x) is positive.
The values of f’(x) definitely depend on the values
of x , and f’(x) is a function of x. We can use the
results in the table to help sketch the graph of f’(x) .
Chapter 2 The Derivative Applied Calculus 89
Example 5
Shown is the graph of the height h(t) of a rocket at time t. Sketch the graph of the velocity of
the rocket at time t. (Velocity is the derivative of the height function, so it is the slope of the
tangent to the graph of position or height.)
We can estimate the slope of the function at several points. The lower graph below shows the
velocity of the rocket. This is v(t) = h’(t).
We can also find derivative functions algebraically using limits.
Example 6
d
Find
dx
( 2 x 2 − 4 x + 1)
Setting up the derivative using a limit,
f ( x + h) − f ( x )
f ′( x) = lim
h →0 h
We will start by simplifying f ( x + h) .
f ( x + h) = 2( x + h) 2 − 4( x + h) − 1 Expand
Chapter 2 The Derivative Applied Calculus 90
= 2( x 2 + 2 xh + h 2 ) − 4( x + h) − 1
= 2 x 2 + 4 xh + 2h 2 − 4 x − 4h − 1
Now finding the limit,
f ( x + h) − f ( x )
f ′( x) = lim Substitute in the formulas
h →0 h
= lim
( 2 x + 4 xh + 2h 2 − 4 x − 4h − 1) − ( 2 x 2 − 4 x − 1)
2
Now simplify
h →0 h
2 x 2 + 4 xh + 2h 2 − 4 x − 4h − 1 − 2 x 2 + 4 x + 1
= lim
h →0 h
2
4 xh + 2h − 4h
= lim
h →0 h
h ( 4 x + 2h − 4 )
= lim
h →0 h Factor out the h and simplify
= lim ( 4 x + 2h − 4 )
h →0
We can find the limit of this expression by direct substitution:
f ′( x) = lim ( 4 x + 2h − 4 ) = 4 x − 4
h →0
Notice that the derivative depends on x, and that this formula will tell us the slope of the tangent
line to f at any value x. For example, if we wanted to know the tangent slope of f at x = 3, we
would simply evaluate: f ′(3) = 4 ⋅ 3 − 4 = 8 .
A formula for the derivative function is very powerful, but as you can see, calculating the
derivative using the limit definition is very time consuming. In the next section, we will identify
some patterns that will allow us to start building a set of rules for finding derivatives without
needing the limit definition.
Interpreting the Derivative
So far we have emphasized the derivative as the slope of the line tangent to a graph. That
interpretation is very visual and useful when examining the graph of a function, and we will
continue to use it. Derivatives, however, are used in a wide variety of fields and applications, and
some of these fields use other interpretations. The following are a few interpretations of the
derivative that are commonly used.
Chapter 2 The Derivative Applied Calculus 91
General
Rate of Change: f '(x) is the rate of change of the function at x. If the units for x are
df people
years and the units for f(x) are people, then the units for dx are year , a rate of
change in population.
Graphical
Slope: f '(x) is the slope of the line tangent to the graph of f at the point ( x, f(x) ).
Physical
Velocity: If f(x) is the position of an object at time x, then f '(x) is the velocity of the
object at time x. If the units for x are hours and f(x) is distance measured in miles,
df miles
then the units for f '(x) = dx are hour , miles per hour, which is a measure of
velocity.
Acceleration: If f(x) is the velocity of an object at time x, then f '(x) is the acceleration of
miles
the object at time x. If the units are for x are hours and f(x) has the units hour ,
df miles/hour miles
then the units for the acceleration f '(x) = dx are = , miles per
hour hour2
hour per hour.
Business
Marginal Cost, Marginal Revenue, and Marginal Profit: We'll explore these terms in more
depth later in the section. Basically, the marginal cost is approximately the additional
cost of making one more object once we have already made x objects. If the units for
df
x are bicycles and the units for f(x) are dollars, then the units for f '(x) = dx
dollars
are , the cost per bicycle.
bicycle
In business contexts, the word "marginal" usually means the derivative or rate of change of
some quantity.
Example 7
1
Suppose the demand curve for widgets was given by D( p ) = , where D is the quantity of
p
items widgets, in thousands, at a price of p dollars. Interpret the derivative of D at p = $3.
1
Note that we calculated D′(3) earlier to be D′(3) = − ≈ −0.111 .
9
Chapter 2 The Derivative Applied Calculus 92
Since D has units "thousands of widgets" and the units for p is dollars of price, the units for D′
thousands of widgets
will be . In other words, it shows how the demand will change as the
dollar of price
price increases.
Specifically, D′(3) ≈ −0.111 tells us that when the price is $3, the demand will decrease by
about 0.111 thousand items for every dollar the price increases.
2.2 Exercises
1. What is the slope of the line through (3,9) and (x, y) for y = x2 and x = 2.97? x = 3.001?
x = 3+h? What happens to this last slope when h is very small (close to 0)? Sketch the
graph of y = x2 for x near 3.
2. What is the slope of the line through (–2,4) and (x, y) for y = x2 and x = –1.98? x = –
2.03? x = –2+h? What happens to this last slope when h is very small (close to 0)? Sketch
the graph of y = x2 for x near –2.
3. What is the slope of the line through (2,4) and (x, y) for y = x2 + x – 2 and x = 1.99?
x = 2.004? x = 2+h? What happens to this last slope when h is very small? Sketch the
graph of y = x2 + x – 2 for x near 2.
4. What is the slope of the line through (–1,–2) and (x, y) for y = x2 +x – 2 and x = –.98?
x = –1.03? x = –1+h? What happens to this last slope when h is very small? Sketch the
graph of y = x2 + x – 2 for x near –1.
5. The graph to the right shows the temperature
during a day in Ames.
(a) What was the average change in
temperature from 9 am to 1 pm?
(b) Estimate how fast the temperature was
rising at 10 am and at 7 pm?
Chapter 2 The Derivative Applied Calculus 93
6. The graph shows the distance of a car from a measuring position located on the edge of a
straight road.
(a) What was the average velocity of the car
from t = 0 to t = 30 seconds?
(b) What was the average velocity of the car
from t = 10 to t = 30 seconds?
(c) About how fast was the car traveling at
t = 10 seconds? at t = 20 s ? at t = 30 s
?
(d) What does the horizontal part of the
graph between t = 15 and t = 20
seconds mean?
(e) What does the negative velocity at t =
25 represent?
7. The graph shows the distance of a car from a
measuring
position located on the edge of a straight
road.
(a) What was the average velocity of the
car from t = 0 to t = 20 seconds?
(b) What was the average velocity from t
= 10 to t = 30 seconds?
(c) About how fast was the car traveling at t = 10
seconds? at t = 20 s ? at t = 30 s ?
8. The graph shows the composite developmental skill
level of chessmasters at different ages as determined
by their performance against other chessmasters.
(From "Rating Systems for Human Abilities", by
W.H. Batchelder and R.S. Simpson, 1988. UMAP
Module 698.)
(a) At what age is the "typical" chessmaster
playing the best chess?
(b) At approximately what age is the chessmaster's skill level increasing most rapidly?
(c) Describe the development of the "typical" chessmaster's skill in words.
(d) Sketch graphs which you think would reasonably describe the performance levels
versus age for an athlete, a classical pianist, a rock singer, a mathematician, and a
professional in your major field.
Chapter 2 The Derivative Applied Calculus 94
10. Use the function in the graph to fill in the table and then graph m(x).
x y = f(x) m(x) = the estimated slope of the tangent
line to y=f(x) at the point (x,y)
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
11. Use the function in the graph to fill in the table and then graph m(x).
x y = g(x) m(x) = the estimated slope of the tangent
line to y=g(x) at the point (x,y)
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
12. (a) At what values of x does the graph of f in the
graph have a horizontal tangent line?
(b) At what value(s) of x is the value of f the
largest? smallest?
(c) Sketch the graph of m(x) = the slope of the line
tangent to the graph of f at the point (x,y).
Chapter 2 The Derivative Applied Calculus 95
13. (a) At what values of x does the graph of g have a
horizontal tangent line?
(b) At what value(s) of x is the value of g the
largest? smallest?
(c) Sketch the graph of m(x) = the slope of the line
tangent to the graph of g at the point (x,y).
14. Match the situation descriptions with the corresponding time–velocity graph.
(a) A car quickly leaving from a stop sign.
(b) A car sedately leaving from a stop sign.
(c) A student bouncing on a trampoline.
(d) A ball thrown straight up.
(e) A student confidently striding across
campus to take a calculus test.
(f) An unprepared student walking across
campus to take a calculus test.
For each function f(x) in problems 15 – 20,
perform steps (a) – (d):
f ( x + h) − f ( x )
(a) calculate msec = and simplify (b) determine mtan = lim msec
h h →0
(c) evaluate mtan at x = 2 , (d) find the equation of the line tangent to the graph of f at (2, f(2) )
15. f(x) = 3x – 7 16. f(x) = 2 – 7x 17. f(x) = ax + b where a and b are constants
18. f(x) = x2 + 3x 19. f(x) = 8 – 3x2 20. f(x) = ax2 + bx + c where a, b and c are constants
21. Match the graphs of the three functions below with the graphs of their derivatives.
Chapter 2 The Derivative Applied Calculus 96
22. Below are six graphs, three of which are derivatives of the other three. Match the functions
with their derivatives.
23. The graph below shows the temperature during a summer day in Chicago. Sketch the graph of
the rate at which the temperature is changing. (This is just the graph of the slopes of the lines
which are tangent to the temperature graph.)
24. Fill in the table with the appropriate units for f '(x).
units for x units for f(x) units for f '(x)
hours miles
people automobiles
dollars pancakes
days trout
seconds miles per second
seconds gallons
study hours test points
25. If C(x) is the total cost, in millions, of producing x thousand items, interpret C ′(4) = 2 .
26. Suppose P(t) is the number of individuals infected by a disease t days after it was first detected.
Interpret P′(50) = −200 .
Chapter 2 The Derivative Applied Calculus 97
Section 3: Power and Sum Rules for Derivatives
In the next few sections, we’ll get the derivative rules that will let us find formulas for derivatives
when our function comes to us as a formula. These are very algebraic section, and you should get
lots of practice. As we learn new rules, we will look at some basic applications.
Building Blocks
These are the simplest rules – rules for the basic functions. We won’t prove these rules; we’ll just
use them. But first, let’s look at a few so that we can see they make sense.
Example 1
Find the derivative of y = f ( x ) = mx + b
This is a linear function, so its graph is its own tangent line! The slope of the tangent line, the
derivative, is the slope of the line: f ' (x ) = m
Rule: The derivative of a linear function is its slope
Example 2
Find the derivative of f ( x ) = 135.
Think about this one graphically, too. The graph of f(x) is a horizontal line. So its slope is zero.
f ' (x ) = 0
Rule: The derivative of a constant is zero
Example 3
Find the derivative of f ( x ) = x
2
This question is challenging using limits, as you saw in the previous section. We will show you
the long way to do it, then give you a shorthand rule to bypass all this.
f (x + h ) − f (x )
Recall the formal definition of the derivative: f ' ( x) = lim .
h →0 h
Using our function f ( x ) = x , f ( x + h ) = ( x + h ) = x 2 + 2 xh + h 2 . Then
2 2
f (x + h ) − f (x ) x 2 + 2 xh + h 2 − x 2
f ' ( x) = lim = lim
h →0 h h →0 h
2 xh + h 2
h(2 x + h )
= lim = lim = lim(2 x + h ) = 2 x
h →0 h h →0 h h →0
From all that, we find the f ′(x) = 2x
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 2 The Derivative Applied Calculus 98
Luckily, there is a handy rule we use to skip using the limit:
Power Rule: The derivative f ( x ) = x is f ′(x ) = nx
n n −1
Example 4
Find the derivative of g (x ) = 4x .
3
Using the power rule, we know that if f ( x ) = x , then f ′(x ) = 3x . Notice that g is 4 times the
3 2
function f. Think about what this change means to the graph of g – it’s now 4 times as tall as
∆g 4∆f ∆f
the graph of f. If we find the slope of a secant line, it will be = =4 ; each slope will
∆x ∆x ∆x
be 4 times the slope of the secant line on the f graph. This property will hold for the slopes of
tangent lines, too:
d d 3
dx
( )
4x 3 = 4
dx
( )
x = 4 ⋅ 3x 2 = 12 x 2
d
Rule: Constants come along for the ride; (kf ) = kf '
dx
Here are all the basic rules in one place.
Chapter 2 The Derivative Applied Calculus 99
Derivative Rules: Building Blocks
In what follows, f and g are differentiable functions of x.
d
(a) Constant Multiple Rule: (kf ) = kf '
dx
d
(b) Sum (or Difference) Rule: ( f + g ) = f '+ g ' (or d ( f − g ) = f '− g ' )
dx dx
d n
(c) Power Rule:
dx
( )
x = nx n −1
d
Special cases: (k ) = 0 (because k = kx 0 )
dx
d
(x ) = 1 (because x = x 1 )
dx
d x
(d) Exponential Functions:
dx
( )
e = ex
d x
dx
( )
a = ln a ⋅ a x
d
(e) Natural Logarithm: (ln x ) = 1
dx x
The sum, difference, and constant multiple rule combined with the power rule allow us to easily
find the derivative of any polynomial.
Example 5
Find the derivative of p( x ) = 17 x + 13x − 1.8 x + 1003
10 8
d
dx
(17 x 10 + 13x 8 − 1.8 x + 1003 )
d d d d
=
dx
(
17 x 10 + )
dx
( )
13x 8 − (1.8 x ) +
dx dx
(1003)
d 10 d 8 d d
= 17
dx
( )
x + 13( ) dx
x − 1.8 ( x ) +
dx dx
(1003)
( ) ( )
= 17 10 x 9 + 13 8 x 7 − 1.8(1) + 0
= 170 x 9 + 104 x 7 − 1.8
Chapter 2 The Derivative Applied Calculus 100
You don’t have to show every single step. Do be careful when you’re first working with the rules,
but pretty soon you’ll be able to just write down the derivative directly:
Example 6
d
Find
dx
(
17 x 2 − 33 x + 12 )
Writing out the rules, we'd write
d
dx
( )
17 x 2 − 33 x + 12 = 17(2 x) − 33(1) + 0 = 34 x − 33
Once you're familiar with the rules, you can, in your head, multiply the 2 times the 17 and the
33 times 1, and just write
d
dx
( )
17 x 2 − 33 x + 12 = 34 x − 33
The power rule works even if the power is negative or a fraction. In order to apply it, first translate
all roots and basic rational expressions into exponents:
Example 7
4
Find the derivative of y = 3 t − + 5e t
t4
First step – translate into exponents:
4
y = 3 t − 4 + 5e t = 3t 1 / 2 − 4t − 4 + 5e t
t
Now you can take the derivative:
d 4 d 1/ 2
t
3 t − 4 + 5e =
dt
(
3t − 4t − 4 + 5e t )
t dt
1 3
( ) ( )
= 3 t −1 / 2 − 4 − 4t −5 + 5 e t = t −1 / 2 + 16t −5 + 5e t .
2 2
If there is a reason to, you can rewrite the answer with radicals and positive exponents:
3 −1/ 2 3 16
t + 16t −5 + 5e t = + 5 + 5e t
2 2 t t
Be careful when finding the derivatives with negative exponents.
We can immediately apply these rules to solve the problem we started the chapter with - finding a
tangent line.
Chapter 2 The Derivative Applied Calculus 101
Example 8
Find the equation of the line tangent to g (t ) = 10 − t 2 when t = 2.
The slope of the tangent line is the value of the derivative. We can compute g ′(t ) = −2t . To
find the slope of the tangent line when t = 3, evaluate the derivative at that point.
g ′(2) = −2(2) = −4 . The slope of the tangent line is -4.
To find the equation of the tangent line, we also need a point on the tangent line. Since the
tangent line touches the original function at t = 2, we can find the point by evaluating the
original function: g (3) = 10 − 2 2 = 6 . The tangent line must pass through the point (2, 6).
Using the point-slope equation of a line, the tangent line will
have equation y − 6 = −4(t − 2) .
Simplifying to slope-intercept form, the equation is
y = −4t + 14 .
Graphing, we can verify this line is indeed tangent to the
curve.
We can also use these rules to help us find the derivatives we need to interpret the behavior of a
function.
Example 9
In a memory experiment, a researcher asks the subject to memorize as many words from a list
as possible in 10 seconds. Recall is tested, then the subject is given 10 more seconds to study,
and so on. Suppose the number of words remembered after t seconds of studying could be
modeled by W (t ) = 4t 2/5 . Find and interpret W ′(20) .
2 8 −3/5
W ′(t ) =
4 ⋅ t −3/5 = t , so
5 5
8
( 20 ) ≈ 0.2652
−3/5
W ′(20)
=
5
Since W is measured in words, and t is in seconds, W' has units words per second.
W ′(20) ≈ 0.2652 means that after 20 seconds of studying, the subject is learning about 0.27
more words for each additional second of studying.
Chapter 2 The Derivative Applied Calculus 102
Business and Economics
Next we will delve more deeply into some business applications. To do that, we first need to
review some terminology.
Suppose you are producing and selling some item. The profit you make is the amount of money
you take in minus what you have to pay to produce the items. Both of these quantities depend on
how many you make and sell. (So we have functions here.) Here is a list of definitions for some
of the terminology, together with their meaning in algebraic terms and in graphical terms.
Your cost is the money you have to spend to produce your items.
The Fixed Cost (FC) is the amount of money you have to spend regardless of how many items
you produce. FC can include things like rent, purchase costs of machinery, and salaries for office
staff. You have to pay the fixed costs even if you don’t produce anything.
The Total Variable Cost (TVC) for q items is the amount of money you spend to actually
produce them. TVC includes things like the materials you use, the electricity to run the machinery,
gasoline for your delivery vans, maybe the wages of your production workers. These costs will
vary according to how many items you produce.
The Total Cost (TC, or sometimes just C) for q items is the total cost of producing them. It’s the
sum of the fixed cost and the total variable cost for producing q items.
The Marginal Cost (MC) at q items is the cost of producing the next item. Really, it’s
MC(q) = TC(q + 1) – TC(q).
In many cases, though, it’s easier to approximate this difference using calculus (see Example
below). And some sources define the marginal cost directly as the derivative,
MC(q) = TC'(q).
In this course, we will use both of these definitions as if they were interchangeable.
The units on marginal cost is cost per item.
Why is it OK that are there two definitions for Marginal Cost (and Marginal Revenue, and
Marginal Profit)?
We have been using slopes of secant lines over tiny intervals to approximate derivatives. In this
example, we’ll turn that around – we’ll use the derivative to approximate the slope of the secant
line.
Notice that the “cost of the next item” definition is actually the slope of a secant line, over an
interval of 1 unit:
C (q + 1) − 1
MC (q ) = C (q + 1) − 1 =
1
So this is approximately the same as the derivative of the cost function at q:
MC (q ) = C ' (q )
In practice, these two numbers are so close that there’s no practical reason to make a distinction.
For our purposes, the marginal cost is the derivative is the cost of the next item.
Chapter 2 The Derivative Applied Calculus 103
Example 10
The table shows the total cost (TC) of producing q items. Items, q Total Cost, TC
a) What is the fixed cost? 0 $20,000
b) When 200 items are made, what is the total variable cost? 100 $35,000
The average variable cost? 200 $45,000
c) When 200 items are made, estimate the marginal cost. 300 $53,000
a) The fixed cost is $20,000, the cost even when no items are made.
b) When 200 items are made, the total cost is $45,000. Subtracting the fixed cost, the total
variable cost is $45,000 - $20,000 = $25,000.
The average variable cost is the total variable cost divided by the number of items, so we would
divide the $25,000 total variable cost by the 200 items made. $25,000 ÷ 200= $125. On
average, each item had a variable cost of $125.
c) We need to estimate the value of the derivative, or the slope of the tangent line at q = 200.
45,000 − 35,000
Finding the secant line from q=100 to q=200 gives a slope of = 100 . Finding
200 − 100
53,000 − 45,000
the secant line from q=200 to q=300 gives a slope of = 80 . We could estimate
300 − 200
the tangent slope by averaging these secant slopes, giving us an estimate of $90/item.
This tells us that after 200 items have been made, it will cost about $90 to make one more item.
Example 11
The cost to produce x items is x hundred dollars.
(a) What is the cost for producing 100 items? 101 items? What is cost of the 101st item?
(b) For C(x) = x , calculate C'(x) and evaluate C' at x = 100. How does C '(100)
compare with the last answer in part (a)?
(a) Put C(x) = x = x1/2 hundred dollars, the cost for x items. Then C (100) = $1000 and
C(101) = $1004.99, so it costs $4.99 for that 101st item. Using this definition, the marginal
cost is $4.99.
1 −1/2 1 1 1
C ′( x)
(b) = = x ′(100) =
so C= hundred dollars = $5.00.
2 2 x 2 100 20
Note how close these answers are! This shows (again) why it’s OK that we use both definitions
for marginal cost.
Chapter 2 The Derivative Applied Calculus 104
Demand is the functional relationship between the price p and the quantity q that can be sold (that
is demanded). Depending on your situation, you might think of p as a function of q, or of q as a
function of p.
Your revenue is the amount of money you actually take in from selling your products. Revenue is
price × quantity.
The Total Revenue (TR, or just R) for q items is the total amount of money you take in for selling
q items.
The Marginal Revenue (MR) at q items is the cost of producing the next item,
MR(q) = TR(q + 1) – TR(q).
Just as with marginal cost, we will use both this definition and the derivative definition
MR(q) = TR’(q).
Your profit is what’s left over from total revenue after costs have been subtracted.
The Profit (P) for q items is TR(q) – TC(q), the difference between total revenue and total costs
The average profit for q items is P/q. The marginal profit at q items is P(q + 1) – P(q), or P′(q )
Graphical Interpretations of the Basic Business Math Terms
Illustration/Example:
Here are the graphs of TR and TC for producing and selling a certain item. The horizontal axis is
the number of items, in thousands. The vertical axis is the number of dollars, also in thousands.
First, notice how to find the fixed cost and variable cost from the graph here. FC is the y-
intercept of the TC graph. (FC = TC(0).) The graph of TVC would have the same shape as the
graph of TC, shifted down. (TVC = TC – FC.)
Chapter 2 The Derivative Applied Calculus 105
MC(q) = TC(q + 1) – TC(q), but that’s impossible to read on this graph. How could you
distinguish between TC(4022) and TC(4023)? On this graph, that interval is too small to see, and
our best guess at the secant line is actually the tangent line to the TC curve at that point. (This is
the reason we want to have the derivative definition handy.)
MC(q) is the slope of the tangent line to the TC curve at (q, TC(q)).
MR(q) is the slope of the tangent line to the TR curve at (q, TR(q)).
Profit is the distance between the TR and TC curve. If you experiment with your clear plastic
ruler, you’ll see that the biggest profit occurs exactly when the tangent lines to the TR and TC
curves are parallel. This is the rule “profit is maximized when MR = MC.” which we'll explore
later in the chapter.
Example 12
p ) 200 − 0.2 p 2 . Find the
The demand, D, for a product at a price of p dollars is given by D(=
marginal revenue when the price is $10.
First we need to form a revenue equation. Since Revenue = Price×Quantity, and the demand
equation shows the quantity of product that can be sold, we have
R( p= ( )
) D( p ) ⋅ p= 200 − 0.2 p 2 p= 200 p − 0.2 p3
Now we can find marginal revenue by finding the derivative
200 1 − 0.2 3 p 2 =
R′( p ) = () ( )
200 − 0.6 p 2
200 0.6 (10 ) =
2
At a price of $10, R′(10) =− 140 .
dollars of Revenue
Notice the units for R' are , so R′(10) = 140 means that when the price is
dollar of price
$10, the revenue will increase by $140 for each dollar the price was increased.
2.3 Exercises
1. Fill in the values in the table for
d
(3 f (x )) , d (2 f (x ) + g (x )) , and d (3g (x ) − f (x )) .
dx dx dx
d d
x f(x) f '(x) g(x) g '(x) (3 f (x )) d (2 f (x ) + g (x )) (3g (x ) − f (x ))
dx dx dx
0 3 –2 –4 3
1 2 –1 1 0
2 4 2 3 1
Chapter 2 The Derivative Applied Calculus 106
d 7 1 d xe
2. Find (a) D( x12 ) (b) dx ( x ) (c) D( ) (d) dx
x3
d x2/3 1
3. Find (a) D( x ) 9
(b) dx (c) D( 4 ) (d) D( xπ )
x
In problems 4 – 8, (a) calculate f '(1) and (b) determine when f '(x) = 0.
4. f(x) = x2 – 5x + 13
5. f(x) = 5x2 – 40x + 73
6. f(x) = x3 + 9x2 + 6
7. f(x) = x3 + 3x2 + 3x – 1
8. f(x) = x3 + 2x2 + 2x – 1
9. Where do f(x) = x2 – 10x + 3 and g(x) = x3 – 12x have horizontal tangent lines ?
10. It takes T(x) = x2 hours to weave x small rugs. What is the marginal production time to
weave a rug? (Be sure to include the units with your answer.)
11. It costs C(x) = x dollars to produce x golf balls. What is the marginal production cost to
make a golf ball? What is the marginal production cost when x = 25? when x= 100? (Include
units.)
12. An arrow shot straight up from ground level with an initial velocity of 128 feet per second will
be at height h(x) = –16x2 + 128x feet at x seconds.
(a) Determine the velocity of the arrow when x = 0, 1 and 2
seconds.
(b) What is the velocity of the arrow, v(x), at any time x?
(c) At what time x will the velocity of the arrow be 0?
(d) What is the greatest height the arrow reaches?
(e) How long will the arrow be aloft?
(f) Use the answer for the velocity in part (b) to determine the
acceleration, a(x) = v '(x), at any time x.
13. If an arrow is shot straight up from ground level on the moon with an initial velocity of 128
feet per second, its height will be h(x) = –2.65x2 + 128x feet at x seconds. Do parts (a) – (e)
of problem 40 using this new equation for h.
14. f(x) = x3 + A x2 + B x + C with constants A, B and C. Can you find conditions on the
constants A, B and C which will guarantee that the graph of y = f(x) has two distinct
"vertices"? (Here a "vertex" means a place where the curve changes from increasing to
decreasing or from decreasing to increasing.)
Chapter 2 The Derivative Applied Calculus 107
Section 4: Product and Quotient Rules
The basic rules will let us tackle simple functions. But what happens if we need the derivative of a
combination of these functions?
Example 1
( )
Find the derivative of h ( x ) = 4 x − 11 ( x + 3)
3
This function is not a simple sum or difference of polynomials. It’s a product of polynomials.
We can simply multiply it out to find its derivative:
h ( x ) = ( 4 x3 − 11) ( x + 3) = 4 x 4 − 11x + 12 x3 − 33
h ' ( x )= 16 x 3 − 11 + 36 x 2
Now suppose we wanted to find the derivative of
( )(
f (x ) = 4 x 5 + x 3 − 1.5 x 2 − 11 x 7 − 7.25 x 5 + 120 x + 3 )
This function is not a simple sum or difference of polynomials. It’s a product of polynomials. We
could simply multiply it out to find its derivative as before – who wants to volunteer? Nobody?
We’ll need a rule for finding the derivative of a product so we don’t have to multiply everything
out.
It would be great if we can just take the derivatives of the factors and multiply them, but
unfortunately that won’t give the right answer. to see that, consider finding derivative of
( )
g (x ) = 4 x 3 − 11 (x + 3) . We already worked out the derivative. It’s g ' (x ) = 16 x 3 − 11 + 36 x 2 .
( )
What if we try differentiating the factors and multiplying them? We’d get 12 x (1) = 12 x , which
2 2
is totally different from the correct answer.
The rules for finding derivatives of products and quotients are a little complicated, but they save us
the much more complicated algebra we might face if we were to try to multiply things out. They
also let us deal with products where the factors are not polynomials. We can use these rules,
together with the basic rules, to find derivatives of many complicated looking functions.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 2 The Derivative Applied Calculus 108
Derivative Rules: Product and Quotient Rules
In what follows, f and g are differentiable functions of x.
(a) Product Rule:
d
( fg ) = f ' g + fg '
dx
The derivative of the first factor times the second left alone, plus the first left alone times
the derivative of the second.
The product rule can extend to a product of several functions; the pattern continues – take
the derivative of each factor in turn, multiplied by all the other factors left alone, and add
them up.
(b) Quotient Rule:
d f f ' g − fg '
=
dx g g2
The numerator of the result resembles the product rule, but there is a minus instead of a
plus; the minus sign goes with the g’. The denominator is simply the square of the original
denominator – no derivatives there.
Example 2
(
Find the derivative of h ( x ) = 4 x − 11 ( x + 3)
3
)
This is the same function we found the derivative of in Example 1, but let's use the product rule
and check to see if we get the same answer. For this first example, we will provide a lot more
detail and steps than one usually actually shows when working a problem like this.
x) 4 x 3 − 11 and g ( x)= x + 3 .
Notice we can think of h(x) as the product of two functions f (=
Finding the derivative of each of these,
f ′( x) = 12 x 2
g ′( x) = 1
Using the product rule,
′( x)
h= ( f ′)( g ) + ( f )(=
g ′) (12 x ) ( x + 3) + ( 4 x
2 3
)
− 11 (1)
To check if this is equivalent to the answer we found in Example 1 we could simplify:
( ) ( )
h′ ( x ) = 12 x 2 ( x + 3) + 4 x3 − 11 (1) =12 x3 + 36 x12 + 4 x3 − 11 =16 x3 + 36 x12 − 11
From this, we can see the answers are equivalent.
Chapter 2 The Derivative Applied Calculus 109
Example 3
Find the derivative of F (t ) = e ln t
t
This is a product, so we need to use the product rule. I like to put down empty parentheses to
remind myself of the pattern; that way I don’t forget anything.
F ' (t ) = ( )( ) + ( )( )
Then I fill in the parentheses – the first set gets the derivative of et , the second gets ln t left
alone, the third gets et left alone, and the fourth gets the derivative of ln t .
t 1 et
( ) ( )
F ' (t ) = e (ln t ) + e = e ln t +
t t
t t
Notice that this was one we couldn’t have done by “multiplying out” before taking the derivative.
Example 4
x4 + 4x
Find the derivative of y =
3 + 16 x 3
This is a quotient, so we need to use the quotient rule. Again, you find it helpful to put down
the empty parentheses as a template:
( )( ) − ( )( )
y' = 2
()
Then fill in all the pieces:
y' =
( 4 x 3 + 4 x ⋅ ln 4 )( 3 + 16 x 3 ) − ( x 4 + 4 x )( 48 x 2 )
( 3 + 16 x3 )
2
Now for goodness’ sakes don’t try to simplify that! Remember that “simple” depends on what you
will do next; in this case, we were asked to find the derivative, and we’ve done that. Please STOP,
unless there is a reason to simplify further.
Example 5
Suppose a large tank contains 8 kg of a chemical dissolved in 50 liters of water. If a tap is
opened and water is added to the tank at a rate of 5 liters per minute, at what rate is the
concentration of chemical in the tank changing after 4 minutes?
Chapter 2 The Derivative Applied Calculus 110
First we need to set up a model for the concentration of chemical. The concentration would be
kg
measured as kg of chemical per liter of water, . The number of kg of chemical stays
L
constant at 8 kg, but the quantity of water in the tank is increasing by 5 L/min. The total
volume of water in the tank after t minutes is 50 + 5t, so the concentration after t minutes is
8
c(t ) =
50 + 5t
To find the rate at which the concentration is changing, we need the derivative:
d d
(8) ⋅ ( 50 + 5t ) − (8) ( 50 + 5t )
c′(t ) = dt dt
( 50 + 5t )
2
=
(=
0 ) ⋅ ( 50 + 5t ) − ( 8 )( 5 ) −40
( 50 + 5t ) ( 50 + 5t )
2 2
At t = 4,
−40
=c′(4) ≈ −0.00816
( 50 + 5 ⋅ 4 )
2
kg / L
Note that the units here are kg per liter, per minute, or . In other words, this tells us that
min
after 4 minutes, the concentration of chemical is decreasing by 0.00816 kg/L each minute.
Returning to our discussion of business and economics topics, in addition to total cost and
marginal cost, we often also want to talk about average cost or average revenue.
TC (q )
The Average Cost (AC) for q items is the total cost divided by q, or AC (q ) = . You can
q
also talk about the average fixed cost, FC/q, or the average variable cost, TVC/q.
The Average Revenue (AR) for q items is the total revenue divided by q, or TR/q.
We already know that we can find average rates of change by finding slopes of secant lines. AC,
AR, MC, and MR are all rates of change, and we can find them with slopes, too.
AC(q) is the slope of a diagonal line, from (0, 0) to (q, TC(q)).
AR(q) is the slope of the line from (0, 0) to (q, TR(q)).
Chapter 2 The Derivative Applied Calculus 111
slope = AC
slope = AR
Just as we found marginal Total Cost, we can also find marginal Average Cost.
Example 6
The cost, in thousands of dollars, for producing x thousand cellphone cases is given by
C ( x) = 22 + x − 0.004 x 2 . Find
a) The Fixed costs
b) The Average Cost when 5 thousand, 10 thousand, or 20 thousand cases are produced
c) The Marginal Average Cost when 5 thousand cases are produced
a) The fixed costs are the costs when no items are produced:
C (0) = 22 thousand dollars
b) The average cost function is total cost divided by number of items, so
C ( x) 22 + x − 0.004 x 2
AC= ( x) =
x x
Note the units are thousands of dollars per thousands of items, which simplifies to just dollars
per item.
22 + 5 − 0.004(5) 2
=
At a production of 5 thousand items: AC (5) = 5.38 dollars per item
5
22 + 10 − 0.004(10) 2
=
At a production of 10 thousand items: AC (10) = 3.16 dollars per item
10
22 + 20 − 0.004(20) 2
=
At a production of 20 thousand items: AC (20) = 2.02 dollars per item
20
Notice that while the total cost increases with production, the average cost per item decreases,
because the initial fixed costs are being distributed across more items.
Chapter 2 The Derivative Applied Calculus 112
c) For the marginal average cost, we need to find the derivative of the average cost function.
We can either calculate this using the quotient rule, or we could use algebra to simplify the
equation first:
22 + x − 0.004 x 2 22 x 0.004 x 2 22
AC ( x)= = + − = + 1 − 0.004 x= 22 x −1 + 1 − 0.004 x .
x x x x x
Taking the derivative,
22
AC ′( x) =
−22 x −2 − 0.004 =
− 2 − 0.004
x
When 5 thousand items are produced,
22
AC ′(5) =
− 2 − 0.004 = −0.884
5
Since the units on AC are dollars per item, and the units on x are thousands of items, the units
on AC' dollars per item per thousands of items. This tells us that when 5 thousand items are
produced, the average cost per item is decreasing by $0.884 for each additional thousand items
produced.
2.4 Exercises
1. Use the values in the table to fill in the rest of the table.
d d f (x ) d g (x )
x f(x) f '(x) g(x) g '(x) ( f (x ) ⋅ g (x ))
dx dx g (x ) dx f (x )
0 3 –2 –4 3
1 2 –1 1 0
2 4 2 3 1
2. Use the information in the graph to plot the
values of the functions f + g, f.g and f/g and
their derivatives at x = 1, 2 and 3 .
3. Use the information in the graph to plot the
values of the functions 2f, f – g and g/f and
their derivatives at x = 1, 2 and 3 .
d
4. Calculate ((x − 5)(3x + 7 )) by (a) using the product rule and (b) expanding the product
dx
and then differentiating. Verify that both methods give the same result.
Chapter 2 The Derivative Applied Calculus 113
d
( f (x ))
5. If the product of f and g is a constant ( f(x) ∙ g(x) = k for all x), then how are dx
f (x )
d
(g (x ))
and dx related?
g (x )
f (x )
6. If the quotient of f and g is a constant ( = k for all x), then how are g . f ' and f . g '
g (x )
related?
In problems 7 – 8, (a) calculate f '(1) and (b) determine when f '(x) = 0.
7x 3x 2
7. f(x) = 2 8. f(x) =
x +4 2x − 3
d 2 d 3t – 2
9. Determine
dx
( )
x + 1 (7 x − 3) and
dt
( 5t + 1 ) .
d 3 x d x 3
10. Find (a)
dx
( )
x e and (b)
dx
e .( )
d t
11. Find (a)
dt
( )
te , (b) d e x
5
( )
12. A manufacturer has determined that an employee with d days of production experience will
be able to
produce approximately P(d) = 3 + 15( 1 – e–0.2d ) items per day. Graph P(d).
(a) Approximately how many items will a beginning employee be able to produce each day?
(b) How many items will an experienced employee be able to produce each day?
(c) What is the marginal production rate of an employee with 5 days of experience? (What
are the units of your answer, and what does this answer mean?)
Chapter 2 The Derivative Applied Calculus 114
Section 5: Chain Rule
There is one more type of complicated function that we will want to know how to differentiate:
composition. The Chain Rule will let us find the derivative of a composition. (This is the last
derivative rule we will learn!)
Example 1
(
Find the derivative of y = 4 x 3 + 15 x . ) 2
This is not a simple polynomial, so we can’t use the basic building block rules yet. It is a
( ) (
2
)( )
product, so we could write it as y = 4 x 3 + 15 x = 4 x 3 + 15 x 4 x 3 + 15 x and use the product
rule. Or we could multiply it out and simply differentiate the resulting polynomial. I’ll do it the
second way:
(
y = 4 x 3 + 15 x )
2
= 16 x 6 + 120 x 4 + 225 x 2
y ' = 64 x 5 + 480 x 3 + 450 x
( )
20
Now suppose we want to find the derivative of y = 4 x 3 + 15 x . We could write it as a product with
20 factors and use the product rule, or we could multiply it out. But I don’t want to do that, do you?
We need an easier way, a rule that will handle a composition like this. The Chain Rule is a little
complicated, but it saves us the much more complicated algebra of multiplying something like this out.
It will also handle compositions where it wouldn’t be possible to “multiply it out.”
The Chain Rule is the most common place for students to make mistakes. Part of the reason is that the
notation takes a little getting used to. And part of the reason is that students often forget to use it when
they should. When should you use the Chain Rule? Almost every time you take a derivative.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 2 The Derivative Applied Calculus 115
Derivative Rules: Chain Rule
In what follows, f and g are differentiable functions with y = f (u ) and u = g ( x )
(h) Chain Rule (Leibniz notation):
dy dy du
= ⋅
dx du dx
Notice that the du’s seem to cancel. This is one advantage of the Leibniz notation; it can
remind you of how the chain rule chains together.
(h) Chain Rule (using prime notation):
f ' ( x ) = f ' (u ) ⋅ g ' ( x ) = f ' ( g ( x )) ⋅ g ' ( x )
(h) Chain Rule (in words):
The derivative of a composition is the derivative of the outside, with the inside staying the
same, TIMES the derivative of what’s inside.
I recite the version in words each time I take a derivative, especially if the function is complicated.
Example 2
Find the derivative of y = 4 x 3 + 15 x . ( )2
This is the same one we did before by multiplying out. This time, let’s use the Chain Rule: The
inside function is what appears inside the parentheses: 4 x 3 + 15 x . The outside function is the
first thing we find as we come in from the outside – it’s the square function, (inside)2.
The derivative of this outside function is (2*inside). Now using the chain rule, the derivative of
our original function is:
(2*inside) TIMES the derivative of what’s inside (which is 12 x 2 + 15 ):
(
y = 4 x 3 + 15 x )2
( )(
y ' = 2 4 x 3 + 15 x ⋅ 12 x 2 + 15 )
If you multiply this out, you get the same answer we got before. Hurray! Algebra works!
Example 3
Find the derivative of y = 4 x 3 + 15 x ( ) 20
Now we have a way to handle this one. It’s the derivative of the outside TIMES the derivative
of what’s inside.
Chapter 2 The Derivative Applied Calculus 116
The outside function is (inside)20, which has the derivative 20(inside)19.
(
y = 4 x 3 + 15 x )20
(
y ' = 20 4 x 3 + 15 x ) ⋅ (12 x
19 2
+ 15 )
Example 4
2
Differentiate e x +5
.
This isn’t a simple exponential function; it’s a composition. Typical calculator or computer
syntax can help you see what the “inside” function is here. On a TI calculator, for example,
when you push the e x key, it opens up parentheses: e^ ( This tells you that the “inside” of the
exponential function is the exponent. Here, the inside is the exponent x 2 + 5 . Now we can use
the Chain Rule: We want the derivative of the outside TIMES the derivative of what’s inside.
The outside is the “e to the something” function, so its derivative is the same thing. The
derivative of what’s inside is 2x. So
dx
e ( ) ( )
d x 2 +5
= e x + 5 ⋅ (2 x )
2
Example 5
The table gives values for f , f ' , g and g ' at a number of points. Use these values to
determine ( f°g )(x) and ( f°g ) '(x) at x = –1 and 0.
x f(x) g(x) f'(x) g'(x) ( f°g )(x) ( g°f )(x)
-1 2 3 1 0
0 -1 1 3 2
1 1 0 -1 3
2 3 -1 0 1
3 0 2 2 -1
( f°g )(–1) = f( g(–1) ) = f( 3 ) = 0
( f°g )(0) = f( g(0) ) = f( 1 ) = 1.
( f°g ) '(–1) = f '( g(–1) ).g '( –1 ) = f '( 3 ).(0) = (2)(0) = 0 and
( f°g ) '( 0 ) = f '( g( 0 ) ).g '( 0 ) = f '( 1 ).( 2 ) = (–1)(2) = –2 .
Chapter 2 The Derivative Applied Calculus 117
Example 6
If 2400 people now have a disease, and the number of people with the disease appears to double
=
every 3 years, then the number of people expected to have the disease in t years is y 2400 ⋅ 2t /3
(a) How many people are expected to have the disease in 2 years?
(b) When are 50,000 people expected to have the disease?
(c) How fast is the number of people with the disease expected to grow now and 2 years from
now?
(a) In 2 years, y = 2400 . 22/3 ≈ 3,810 people.
(b) We know y = 50,000 , and we need to solve 50,000 = 2400 . 2t/3 for t. We could start by
isolating the exponential by dividing both sides by 2400,
50000 t /3
=2 Taking the logarithm of both sides,
2400
50000
ln
2400
= ln 2
t /3
( ) Using the exponent property for logs,
50000 t
ln = ln ( 2 ) Solving for t,
2400 3
50000
3ln
2400
=t ≈ 13.14 years
ln(2)
We expect 50,000 people to have the disease about 13.14 years from now.
(c) This is asking for dy/dt when t = 0 and 2 years. Using the chain rule,
dy d 1
=
dt dt
( )
2400 ⋅ 2t /3= 2400 ⋅ 2t /3 ln(2) ⋅ ≈ 554.5 ⋅ 2t /3
3
Now, at t = 0, the rate of growth of the disease is approximately 554.5.20 ≈ 554.5 people/year.
In 2 years the rate of growth will be approximately 554.5 . 22/3 ≈ 880 people/year.
Derivatives of Complicated Functions
You’re now ready to take the derivative of some mighty complicated functions. But how do you
tell what rule applies first? Come in from the outside – what do you encounter first? That’s the
first rule you need. Use the Product, Quotient, and Chain Rules to peel off the layers, one at a
time, until you’re all the way inside.
Chapter 2 The Derivative Applied Calculus 118
Example 7
d 3x
Find
dx
(
e ⋅ ln (5 x + 7 ) )
Coming in from the outside, I see that this is a product of two (complicated) functions. So I’ll
need the Product Rule first. I’ll fill in the pieces I know, and then I can figure the rest as
separate steps and substitute in at the end:
d 3x d 3x d
dx
(
e ⋅ ln(5 x + 7 ) = ) ( ) ( )
e (ln(5 x + 7 )) + e 3 x (ln(5 x + 7 ))
dx dx
Now as separate steps, I’ll find
d 3x
dx
( )
e = 3e 3 x (using the Chain Rule) and
d
(ln(5 x + 7 )) = 1 ⋅ 5 (also using the Chain Rule).
dx 5x + 7
Finally, to substitute these in their places:
d 3x 1
dx
( ) ( ) ( )
e ⋅ ln (5 x + 7 ) = 3e 3 x (ln (5 x + 7 )) + e 3 x
⋅ 5
5x + 7
(And please don’t try to simplify that!)
Example 8
4
3t 3
Differentiate z = t
e (t − 1)
Don’t panic! As you come in from the outside, what’s the first thing you encounter? It’s that
4th power. That tells you that this is a composition, a (complicated) function raised to the 4th
power.
Step One: Use the Chain Rule. The derivative of the outside TIMES the derivative of what’s
inside.
4 3
dz d 3t 3 3t 3 d 3t 3
= t = 4 t ⋅ t
dt dt e (t − 1) e (t − 1) dt e (t − 1)
Chapter 2 The Derivative Applied Calculus 119
d 3t 3
Now we’re one step inside, and we can concentrate on just the part. Now, as you
dt e t (t −1)
come in from the outside, the first thing you encounter is a quotient – this is the quotient of two
(complicated) functions.
Step Two: Use the Quotient Rule. The derivative of the numerator is straightforward, so we
can just calculate it. The derivative of the denominator is a bit trickier, so we'll leave it for now.
d
d 3t 3
( )( ) ( ) ( )
9t 2 e t (t − 1) − 3t 3 e t (t − 1)
t = dt
dt e (t − 1) ( )
e (t − 1)
t 2
d t
Now we’ve gone one more step inside, and we can concentrate on just the
dt
( )
e (t − 1) part.
Now we have a product.
Step Three: Use the Product Rule:
d t
dt
( ) ( ) ( )
e (t − 1) = e t (t − 1) + e t (1)
And now we’re all the way in – no more derivatives to take.
Step Four: Now it’s just a question of substituting back – be careful now!
d t
dt
( ) ( ) ( )
e (t − 1) = e t (t − 1) + e t (1) , so
( )( ) ( )(( ) ( ) )
d 3t 3 9t 2 e t (t − 1) − 3t 3 e t (t − 1) + e t (1)
= , so
dt e t (t − 1) (
e t (t − 1)
2
)
dz d 3t 3
=
4
= 4 t
3
⋅
( )( ) ( )(( ) ( ) )
3t 3 9t 2 e t (t − 1) − 3t 3 e t (t − 1) + e t (1)
.
dt dt e t (t − 1) e (t − 1) ( )
e t (t − 1)
2
Phew!
Chapter 2 The Derivative Applied Calculus 120
What if the Derivative Doesn’t Exist?
A function is called differentiable at a point if its derivative exists at that point.
We’ve been acting as if derivatives exist everywhere for every function. This is true for most of
the functions that you will run into in this class. But there are some common places where the
derivative doesn’t exist.
Remember that the derivative is the slope of the tangent line to the curve. That’s what to think
about.
Where can a slope not exist? If the tangent line is vertical, the derivative will not exist.
Example 9
Show that f ( x) = 3 x = x1 / 3 is not differentiable at x = 0.
1 −2 / 3 1
Finding the derivative, f ( x) = x = 2 / 3 . At x = 0, this function is undefined. From the
3 3x
graph, we can see that the tangent line to this curve at x = 0 is vertical with undefined slope,
which is why the derivative does not exist at x = 0.
Where can a tangent line not exist? If there is a sharp corner (cusp) in the graph, the derivative
will not exist at that point because there is no well-defined tangent line (a teetering tangent, if you
will). If there is a jump in the graph, the tangent line will be different on either side and the
derivative can’t exist.
Example 10
Show that f ( x) = x is not differentiable at x = 0.
On the left side of the graph, the slope of the line is -1. On the
right side of the graph, the slope is +1. There is no well-
defined tangent line at the sharp corner at x = 0, so the function
is not differentiable at that point.
Chapter 2 The Derivative Applied Calculus 121
2.5 Exercises
1. The graph of y = f(x) is shown.
(a) At which integers is f continuous?
(b) At which integers is f differentiable?
2. The graph of y = g(x) is shown.
(a) At which integers is g continuous?
(b) At which integers is g differentiable?
Problems 3 and 4 refer to the values given in this table:
x f(x) g(x) f '(x) g '(x) ( f°g )(x) ( f°g )' (x)
–2 2 –1 1 1
–1 1 2 0 2
0 –2 1 2 –1
1 0 –2 –1 2
2 1 0 1 –1
3. Use the table of values to determine ( f°g )(x) and ( f°g )' (x) at x = 1 and 2.
4. Use the table of values to determine ( f°g )(x) and ( f°g )' (x) at x = –2, –1 and 0.
5. Use the graphs to estimate the values of g(x),
g '(x), (f°g)(x), f '( g(x) ), and ( f°g ) '( x ) at
x = 1.
6. Use the graphs to estimate the values of g(x),
g '(x), (f°g)(x), f '( g(x) ), and ( f°g ) '( x ) for
x = 2.
In problems 7 – 12 , find the derivative of each function.
7. f(x) = (2x – 8)5 8. f(x) = (6x – x2)10 9. f(x) = x .(3x + 7)5
x–5
10. f(x) = (2x + 3)6.(x – 2)4 11. f(x) = x2 + 6x – 1 12. f(x) =
(x + 3)4
13. If f is a differentiable function,
(a) how are the graphs of y = f(x) and y = f(x) + k related?
(b) how are the derivatives of f(x) and f(x) + k related?
Chapter 2 The Derivative Applied Calculus 122
Section 6: Second Derivative and Concavity
Second Derivative and Concavity
Graphically, a function is concave up if its graph is curved with the opening upward (a in the
figure). Similarly, a function is concave down if its graph opens downward (b in the figure).
This figure shows the concavity of a function at several points. Notice that a function can be
concave up regardless of whether it is increasing or decreasing.
For example, An Epidemic: Suppose an epidemic has started, and you, as a member of congress,
must decide whether the current methods are effectively fighting the spread of the disease or
whether more drastic measures and more money are needed. In the figure below, f(x) is the
number of people who have the disease at time x, and two different situations are shown. In both
(a) and (b), the number of people with the disease, f(now), and the rate at which new people are
getting sick, f '(now), are the same. The difference in the two situations is the concavity of f, and
that difference in concavity might have a big effect on your decision.
In (a), f is concave down at "now", the slopes are decreasing, and it looks as if it’s tailing off. We
can say “f is increasing at a decreasing rate.” It appears that the current methods are starting to
bring the epidemic under control.
In (b), f is concave up, the slopes are increasing, and it looks as if it will keep increasing faster and
faster. It appears that the epidemic is still out of control.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 2 The Derivative Applied Calculus 123
The differences between the graphs come from whether the derivative is increasing or decreasing.
The derivative of a function f is a function that gives information about the slope of f. The
derivative tells us if the original function is increasing or decreasing.
Because f ' is a function, we can take its derivative. This second derivative also gives us
information about our original function f. The second derivative gives us a mathematical way to
tell how the graph of a function is curved. The second derivative tells us if the original function
is concave up or down.
Second Derivative Let y = f ( x )
The second derivative of f is the derivative of y ' = f ' (x ) .
Using prime notation, this is f ' ' ( x ) or y ' ' . You can read this aloud as “y double prime.”
d2y d2 f
Using Leibniz notation, the second derivative is written or . This is read aloud as
dx 2 dx 2
“the second derivative of f.
If f ' ' ( x ) is positive on an interval, the graph of y = f ( x ) is concave up on that interval. We
can say that f is increasing (or decreasing) at an increasing rate.
If f ' ' ( x ) is negative on an interval, the graph of y = f ( x ) is concave up on that interval.
We can say that f is increasing (or decreasing) at a decreasing rate.
Example 1
Find f ' ' ( x ) for f ( x ) = 3x
7
First, we need to find the first derivative:
f ' (x ) = 21x 6
Then we take the derivative of that function:
d d
f ' ' (x ) = ( f ' (x )) =
dx dx
( )
21x 6 = 126 x 5
If f(x) represents the position of a particle at time x, then v(x) = f '(x) will represent the
velocity (rate of change of the position) of the particle and a(x) = v '(x) = f ''(x) will represent the
acceleration (the rate of change of the velocity) of the particle.
Chapter 2 The Derivative Applied Calculus 124
You are probably familiar with acceleration from driving or riding in a car. The speedometer tells
you your velocity (speed). When you leave from a stop and press down on the accelerator, you are
accelerating - increasing your speed.
Example 2
The height (feet) of a particle at time t seconds is f(t) = t3 – 4t2 + 8t . Find the height,
velocity and acceleration of the particle when t = 0, 1, and 2 seconds.
f(t) = t3 – 4t2 + 8t so f(0) = 0 feet, f(1) = 5 feet, and f(2) = 8 feet.
The velocity is v(t) = f '(t) = 3t2 – 8t + 8 so v(0) = 8 ft/s , v(1) = 3 ft/s, and v(2) = 4 ft/s. At
each of these times the velocity is positive and the particle is moving upward, increasing in
height.
The acceleration is a(t) = f ''(t) = 6t – 8 so a(0) = –8 ft/s2 , a(1) = –2 ft/s2 and a(2) = 4 ft/s2 .
At time 0 and 1, the acceleration is negative, so the particle's velocity would be decreasing at
those points - the particle was slowing down. At time 2, the velocity is positive, so the particle
was increasing in speed.
Inflection Points
Definition: An inflection point is a point on the graph of a function where the concavity of the
function changes, from concave up to down or from concave down to up.
Example 3
Which of the labeled points in the graph below are inflection points?
The concavity changes at points b and g. At points a and h, the graph is concave up on both
sides, so the concavity does not change. At points c and f, the graph is concave down on both
sides. At point e, even though the graph looks strange there, the graph is concave down on both
sides – the concavity does not change.
Chapter 2 The Derivative Applied Calculus 125
Inflection points happen when the concavity changes. Because we know the connection between
the concavity of a function and the sign of its second derivative, we can use this to find inflection
points.
Working Definition: An inflection point is a point on the graph where the second derivative
changes sign.
In order for the second derivative to change signs, it must either be zero or be undefined. So to
find the inflection points of a function we only need to check the points where f ''(x) is 0 or
undefined.
Note that it is not enough for the second derivative to be zero or undefined. We still need to check
that the sign of f’’ changes sign. The functions in the next example illustrate what can happen.
Example 4
Let f(x) = x3 , g(x) = x4 and h(x) = x1/3. For which of these functions is the point (0,0) an
inflection point?
Graphically, it is clear that the concavity of f(x) = x3 and h(x) = x1/3 changes at (0,0), so
(0,0) is an inflection point for f and h. The function g(x) = x4 is concave up everywhere so
(0,0) is not an inflection point of g.
We can also compute the second derivatives and check the sign change.
If f(x) = x3 , then f '(x) = 3x2 and f ''(x) = 6x . The only point at which f ''(x) = 0 or is
undefined (f ' is not differentiable) is at x = 0. If x < 0, then f ''(x) < 0 so f is concave down.
If x > 0 , then f ''(x) > 0 so f is concave up. At x = 0 the concavity changes so the point
(0,f(0)) = (0,0) is an inflection point of x3 .
If g(x) = x4 , then g '(x) = 4x3 and g ''(x) = 12x2 . The only point at which g ''(x) = 0 or is
undefined is at x = 0. If x < 0, then g ''(x) > 0 so g is concave up. If x > 0 , then g ''(x) > 0
so g is also concave up. At x = 0 the concavity does not change so the point (0, g(0)) =
(0,0) is not an inflection point of x4 . Keep this example in mind!.
Chapter 2 The Derivative Applied Calculus 126
1 2
If h(x) = x1/3 , then h '(x) = 3 x–2/3 and h ''(x) = – 9 x–5/3 . h'' is not defined if x = 0, but
h ''(negative number) > 0 and h ''(positive number) < 0 so h changes concavity at (0,0) and
(0,0) is an inflection point of h.
Example 5
Sketch the graph of a function with f(2) = 3, f '(2) = 1, and an inflection point at (2,3).
Two possible solutions are shown here.
2.6 Exercises
In problems 1 and 2, each quotation is a statement about a quantity of something changing over
time. Let f(t) represent the quantity at time t. For each quotation, tell what f represents and
whether the first and second derivatives of f are positive or negative.
1. (a) "Unemployment rose again, but the rate of increase is smaller than last month."
(b) "Our profits declined again, but at a slower rate than last month."
(c) "The population is still rising and at a faster rate than last year."
2. (a) "The child's temperature is still rising, but slower than it was a few hours ago."
(b) "The number of whales is decreasing, but at a slower rate than last year."
(c) "The number of people with the flu is rising and at a faster rate than last month."
4. On which intervals is the function in the graph
(a) concave up? (b) concave down?
5. On which intervals is the function in graph
(a) concave up? (b) concave down?
Chapter 2 The Derivative Applied Calculus 127
6. Sketch the graphs of functions which are defined and concave up everywhere and which have
(a) no roots. (b) exactly 1 root. (c) exactly 2 roots. (d) exactly 3 roots.
In problems 7 – 10, a function and values of x so that f '(x) = 0 are given. Use the Second
Derivative Test to determine whether each point (x, f(x)) is a local maximum, a local minimum or
neither
7. f(x) = 2x3 – 15x2 + 6 , x = 0 , 5 .
8. g(x) = x3 – 3x2 – 9x + 7 , x = –1 , 3 .
9. h(x) = x4 – 8x2 – 2 , x = –2, 0, 2 .
10. f(x) = x.ln(x) , x = 1/e .
11. Which of the labeled points in the graph are inflection points?
12. Which of the labeled points in the graph are inflection points?
13. How many inflection points can a (a) quadratic polynomial have? (b) cubic polynomial have?
(c) polynomial of degree n have?
14. Fill in the table with "+", "–", or "0" for the function shown.
x f(x) f '(x) f ''(x)
0
1
2
3
Chapter 2 The Derivative Applied Calculus 128
15. Fill in the table with "+", "–", or "0" for the function shown.
x g(x) g '(x) g ''(x)
0
1
2
3
In problems 16 – 22 , find the derivative and second derivative of each function.
16. f(x) = 7x2 + 5x – 3
17. f(x) = (2x – 8)5
18. f(x) = (6x – x2)10
19. f(x) = x .(3x + 7)5
20. f(x) = (2x3 + 3)6
21. f(x) = x2 + 6x – 1
(
22. f(x) = ln x 2 + 4 )
Chapter 2 The Derivative Applied Calculus 129
Section 7: Optimization
In theory and applications, we often want to maximize or minimize some quantity. An engineer
may want to maximize the speed of a new computer or minimize the heat produced by an
appliance. A manufacturer may want to maximize profits and market share or minimize waste. A
student may want to maximize a grade in calculus or minimize the hours of study needed to earn a
particular grade.
Without calculus, we only know how to find the optimum points in a few specific examples (for
example, we know how to find the vertex of a parabola). But what if we need to optimize an
unfamiliar function?
The best way we have without calculus is to examine the graph of the function, perhaps using
technology. But our view depends on the viewing window we choose – we might miss something
important. In addition, we’ll probably only get an approximation this way. (In some cases, that will
be good enough.)
Calculus provides ways of drastically narrowing the number of points we need to examine to find
the exact locations of maximums and minimums, while at the same time ensuring that we haven’t
missed anything important.
Local Maxima and Minima
Before we examine how calculus can help us find maximums and minimums, we need to define
the concepts we will develop and use.
Definitions: f has a local maximum at a if f(a) ≥ f(x) for all x near a
f has a local minimum at a if f(a) ≤ f(x) for all x near a
f has a local extreme at a if f(a) is a local maximum or minimum.
The plurals of these are maxima and minima. We often simply say “max”
or “min;” it saves a lot of syllables.
Some books say “relative” instead of “local.”
The process of finding maxima or minima is called optimization.
A point is a local max (or min) if it is higher (lower) than all the nearby
points. These points come from the shape of the graph.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 2 The Derivative Applied Calculus 130
Definitions: f has a global maximum at a if f(a) ≥ f(x) for all x in the domain of f.
f has a global minimum at a if f(a) ≤ f(x) for all x in the domain of f.
f has a global extreme at a if f(a) is a global maximum or minimum.
Some books say “absolute” instead of “global”
A point is a global max (or min) if it is higher (lower) than every point on the
graph. These points come from the shape of the graph and the window through
which we view the graph.
The local and global extremes of the function in Fig. 22 are labeled. You should notice that every
global extreme is also a local extreme, but there are local extremes that are not global extremes.
If h(x) is the height of the earth above sea level at the location x, then the global maximum of h is
h(summit of Mt. Everest) = 29,028 feet. The local maximum of h for the United States is h(summit
of Mt. McKinley) = 20,320 feet. The local minimum of h for the United States is h(Death Valley) =
– 282 feet.
Example 1
The table shows the annual calculus enrollments at a large university. Which years had local
maximum or minimum calculus enrollments? What were the global maximum and minimum
enrollments in calculus?
year 2000 2001 2002 2003 2004 2005 2006 20070 2008 2009 2010
enrollment 1257 1324 1378 1336 1389 1450 1523 1582 1567 1545 1571
There were local maxima in 2002 and 2007; the global maximum was 1582 students in 2007.
There were local minima in 2003 and 2009; the global minimum was 1336 students in 2003. I
choose not to think of 2000 as a local minimum or 2010 as a local maximum. However, some
books would include the endpoints.
Chapter 2 The Derivative Applied Calculus 131
Finding Maxima and Minima of a Function
What must the tangent line look like at a local max or min? Look at these two graphs again –
you’ll see that at all the extreme points, the tangent line is horizontal (so f’ = 0). There is one cusp
in the blue graph – the tangent line if vertical there (so f’ is undefined).
That gives us the clue how to find extreme values.
Definition: A critical number for a function f is a value x = a in the domain of f where
either f ’(a) = 0 or f ‘(a) is undefined.
Definition: A critical point for a function f is a point (a, f(a)) where a is a critical number
of f.
Useful Fact: A local max or min of f can only occur at a critical point.
Example 2
Find the critical points of f(x) = x3 – 6x2 + 9x + 2.
A critical number of f can occur only where f '(x) = 0 or where f’ does not exist.
f '(x) = 3x2 – 12x + 9 = 3(x2 – 4x + 3) = 3(x – 1)(x – 3)
f '(x) = 0 only at x = 1 and x = 3. There are no places where f’ is undefined.
The critical numbers are x = 1 and x= 3. So the critical points are (1, 6) and (3, 2).
These are the only possible locations of local extremes of f. We haven’t discussed yet how to
tell whether either of these points is actually a local extreme of f, or which kind it might be.
But we can be certain that no other point is a local extreme.
Chapter 2 The Derivative Applied Calculus 132
The graph of f (Fig. 23) shows that (1, f(1) ) = (1, 6) is a local maximum and (3, f(3) ) = (3,
2) is a local minimum. This function does not have a global maximum or minimum.
Example 3
Find all local extremes of f(x) = x3 .
f(x) = x3 is differentiable for all x, and f '(x) = 3x2. The only place where f '(x) = 0 is at x = 0,
so the only candidate is the critical point (0,0).
If x > 0 then f(x) = x3 > 0 = f(0), so f(0) is not a local maximum.
Similarly, if x < 0 then f(x) = x3 < 0 = f(0) so f(0) is not a local minimum.
The critical point (0,0) is the only candidate to be a local extreme of f, and this candidate did
not turn out to be a local extreme of f. The function f(x) = x3 does not have any local extremes.
Remember this example! It is not enough to find the critical points -- we can only say that f might
have a local extreme at the critical points.
First and Second Derivative Tests
Is that critical point a Maximum or Minimum (or Neither)?
Once we have found the critical points of f, we still have the problem of determining whether these
points are maxima, minima or neither.
Chapter 2 The Derivative Applied Calculus 133
All of the graphs in Fig. 25 have a critical point at (2, 3). It is clear from the graphs that the point
(2,3) is a local maximum in (a) and (d), (2,3) is a local minimum in (b) and (e), and (2,3) is not a
local extreme in (c) and (f).
The critical numbers only give the possible locations of extremes, and some critical numbers are
not the locations of extremes. The critical numbers are the candidates for the locations of maxima
and minima.
f ' and Extreme Values of f
Four possible shapes of graphs are shown here – in each graph, the point marked by an arrow is a
critical point, where f ‘(x) = 0. What happens to the derivative near the critical point?
At a local max, such as in the graph on the left, the function increases on the left of the local max,
then decreases on the right. The derivative is first positive, then negative at a local max. At a local
min, the function decreases to the left and increases to the right, so the derivative is first negative,
then positive. When there isn’t a local extreme, the function continues to increase (or decrease)
right past the critical point – the derivative doesn’t change sign.
Chapter 2 The Derivative Applied Calculus 134
The First Derivative Test for Extremes:
Find the critical points of f.
For each critical number c, examine the sign of f’ to the left and to the right of c. What
happens to the sign as you move from left to right?
If f '(x) changes from positive to negative at x = c, then f has a local maximum at (c, f(c)).
If f '(x) changes from negative to positive at x = c, then f has a local minimum at (c, f(c)).
If f '(x) does not change sign at x = c, then (c, f(c)) is neither a local max nor a local min.
Example 4
Find the critical points of f(x) = x3 – 6x2 + 9x + 2 and classify them as local max, local min,
or neither..
We already found the critical points; they are (1, 6) and (3, 2).
Now we can use the first derivative test to classify each. Recall that f’(x) = f '(x) = 3x2 – 12x +
9 = 3(x2 – 4x + 3) = 3(x – 1)(x – 3). The factored form is easiest to work with here, so let’s
use that.
(1, 6) – You could choose a number slightly less than 1 to plug into the formula for f’ – perhaps
use x = 0, or x = 0.9. Then you could examine its sign. But I don’t care about the numerical
value, all I’m interested in is its sign. And for that, you don’t have to do any plugging in:
If x is a little less than 1, then x – 1 is negative, and x – 3 is negative.
So f’ = 3(x – 1)(x – 3) will be pos(neg)(neg) = positive.
For x a little more than 1, you can evaluate f’ at a number more than 1 (but less than 3, you
don’t want to go past the next critical point!) – perhaps x = 2. Or you can make a quick
sign argument like what I did above.
For x a little more than 1, f’ = 3(x – 1)(x – 3) will be pos(pos)(neg) = negative.
f’ changes from positive to negative, so there is a local max at (1, 6)
As another approach, we could draw a number line,
and mark the critical numbers. 3
1
We already know the derivative is zero or undefined
at the critical numbers. On each interval between f' + – +
these values, the derivative will stay the same sign.
To determine the sign, we could pick a test value in 0 1 2 3 4
each interval, and evaluate the derivative at those
points (or use the sign approach used above).
(3, 2) – f’ changes from negative to positive, so there is a local min at (3, 2).
This confirms what we saw before in the graph.
Chapter 2 The Derivative Applied Calculus 135
f '' and Extreme Values of f
The concavity of a function can also help us determine whether a critical point is a maximum or
minimum or neither. For example, if a point is at the bottom of a concave up function, then the
point is a minimum.
The Second Derivative Test for Extremes:
Find all critical points of f. For those critical points where f’(c) = 0, find f’’(c).
(a) If f ''(c) < 0 then f is concave down and has a local maximum at x = c.
(b) If f ''(c) > 0 then f is concave up and has a local minimum at x = c.
(c) If f ''(c) = 0 then f may have a local maximum, a minimum or neither at x = c.
Chapter 2 The Derivative Applied Calculus 136
The cartoon faces can help you remember the Second Derivative Test.
Example 5
f(x) = 2x3 – 15x2 + 24x – 7 has critical numbers x = 1 and 4. Use the Second Derivative Test
for Extremes to determine whether f(1) and f(4) are maximums or minimums or neither.
We need to find the second derivative:
f ( x ) = 2 x 3 − 15 x 2 + 24 x − 7
f ' ( x ) = 6 x 2 − 30 x + 24
f ' ' ( x ) = 12 x − 30
Then we just need to evaluate f '' at each critical number:
x = 1: f ' ' (1) = 12(1) − 30 < 0 ; there is a local maximum at x = 1.
x = 4: f ' ' (4 ) = 12(4 ) − 30 > 0 ; there is a local minimum at x = 4.
Many students like the Second Derivative Test. The Second Derivative Test is often easier to use
than the First Derivative Test. You only have to find the sign of one number for each critical
number rather than two. And if your function is a polynomial, its second derivative will probably
be a simpler function than the derivative.
However, if you needed a product rule, quotient rule, or chain rule to find the first derivative,
finding the second derivative can be a lot of work. Also, even if the second derivative is easy, the
Second Derivative Test doesn’t always give an answer. The First Derivative Test will always give
you an answer.
Use whichever test you want to. But remember – you have to do some test to be sure that your
critical point actually is a local max or min.
Global Maxima and Minima
In applications, we often want to find the global extreme; knowing that a critical point is a local
extreme is not enough.
For example, if we want to make the greatest profit, we want to make the absolutely greatest profit
of all. How do we find global max and min?
Chapter 2 The Derivative Applied Calculus 137
There are just a few additional things to think about.
Endpoint Extremes
The local extremes of a function occur at critical points – these are points in the function that we can
find by thinking about the shape (and using the derivative to help us). But if we’re looking at a
function on a closed interval, the endpoints could be extremes. These endpoint extremes are not
related to the shape of the function; they have to do with the interval, the window through which
we’re viewing the function.
In the graph above, it appears that there are three critical points – one local min, one local max, and
one that is neither one. But the global max, the highest point of all, is at the left endpoint. The
global min, the lowest point of all, is at the right endpoint.
How do we decide if endpoints are global max or min? It’s easier than you expected – simply plug
in the endpoints, along with all the critical numbers, and compare y-values.
Example 6
Find the global max and min of f(x) = x3 – 3x2 – 9x + 5 for –2 ≤ x ≤ 6 .
f '(x) = 3x2 – 6x – 9 = 3(x + 1)(x – 3). We need to find critical points, and we need to check the
endpoints.
f'(x) = 3(x + 1)(x – 3) = 0 when x = –1 and x = 3.
The endpoints of the interval are x = –2 and x = 6.
Now we simply compare the values of f at these 4 values of x:
x f(x)
-2 3
-1 10
3 -22
6 59
The global minimum of f on [ –2, 6] is –22, when x = 3,
and the global maximum of f on [ –2, 6] is 59, when x = 6.
Chapter 2 The Derivative Applied Calculus 138
If there’s only one critical point
If the function has only one critical point and it’s a local max (or min), then it must be the global
max (or min). To see this, think about the geometry. Look at the graph on the left – there is a
local max, and the graph goes down on either side of the critical point. Suppose there was some
other point that was higher – then the graph would have to turn around. But that turning point
would have shown up as another critical point. If there’s only one critical point, then the graph can
never turn back around.
When in doubt, graph it and look.
If you are trying to find a global max or min on an open interval (or the whole real line), and there
is more than one critical point, then you need to look at the graph to decide whether there is a
global max or min. Be sure that all your critical points show in your graph, and that you go a little
beyond – that will tell you what you want to know.
Example 7
Find the global max and min of f(x) = x3 – 6x2 + 9x + 2.
We have previously found that (1, 6) is a local max and (3, 2) is a local min. This is not a
closed interval, and there are two critical points, so we must turn to the graph of the function to
find global max and min.
The graph of f shows that points to the left of x = 4 have y-values greater than 6, so (1, 6) is
not a global max. Likewise, if x is negative, y is less than 2, so (3, 2) is not a global min. There
are no endpoints, so we’ve exhausted all the possibilities. This function does not have a global
maximum or minimum.
Chapter 2 The Derivative Applied Calculus 139
To find Global Extremes:
The only places where a function can have a global extreme are critical points or endpoints.
(a) If the function has only one critical point, and it’s a local extreme, then it is also the global
extreme.
(b) If there are endpoints, find the global extremes by comparing y-values at all the critical
points and at the endpoints.
(c) When in doubt, graph the function to be sure.
2.7 Exercises
1. Find all of the critical points of the function shown
and identify them as local max, local min, or
neither. Find the global max and min on the
interval.
2. Find all of the critical points of the function
shown and identify them as local max, local min,
or neither. Find the global max and min on the
interval.
In problems 3 – 8, find all of the critical points and local maximums and minimums of each
function.
3. f(x) = x2 + 8x + 7 4. f(x) = 2x2 – 12x + 7
5. f(x) = x3 – 6x2 + 5 6. f(x) = (x – 1)2 (x – 3)
7. f(x) = ln( x2 – 6x + 11 ) 8. f(x) = 2x3 – 96x + 42
In problems 9 – 16 , find all critical points and global extremes of each function on the given
intervals.
9. f(x) = x2 – 6x + 5 on the entire real number line.
10. f(x) = 2 – x3 on the entire real number line.
11. f(x) = x3 – 3x + 5 on the entire real number line.
12. f ( x ) = x − e x on the entire real number line.
13. f(x) = x2 – 6x + 5 on [ –2, 5] .
14. f(x) = 2 – x3 on [ –2, 1] .
15. f(x) = x3 – 3x + 5 on [ –2, 1] .
16. f ( x ) = x − e x on [ 1, 2] .
Chapter 2 The Derivative Applied Calculus 140
17. Suppose f(1) = 5 and f '(1) = 0. What can we conclude about the point (1,5) if
(a) f '(x) < 0 for x < 1, and f '(x) > 0 for x > 1?
(b) f '(x) < 0 for x < 1, and f '(x) < 0 for x > 1?
(c) f '(x) > 0 for x < 1, and f '(x) < 0 for x > 1?
(d) f '(x) > 0 for x < 1, and f '(x) > 0 for x > 1?
18. Define A(x) to be the area bounded between the x–
axis, the graph of f, and a vertical line at x.
(a) At what value of x is A(x) minimum?
(b) At what value of x is A(x) maximum?
19. Define S(x) to be the slope of the line through the
points (0,0) and ( x, f(x) ) based on the graph of f
shown.
(a) At what value of x is S(x) minimum?
(b) At what value of x is S(x) maximum?
20. The graph of the derivative of a continuous function f.
(a) List the critical numbers of f.
(b) For what values of x does f have a local maximum?
(c) For what values of x does f have a local minimum?
21. The graph of the derivative of a continuous function g.
(a) List the critical numbers of g.
(b) For what values of x does g have a local maximum?
(c) For what values of x does g have a local minimum?
In problems 22 – 24, a function and values of x so that f '(x) = 0
are given. Use the Second Derivative Test to determine whether each point (x, f(x)) is a local
maximum, a local minimum or neither
22. f(x) = 2x3 – 15x2 + 6 , x = 0 , 5 .
23. g(x) = x3 – 3x2 – 9x + 7 , x = –1 , 3 .
24. h(x) = x4 – 8x2 – 2 , x = –2, 0, 2 .
Chapter 2 The Derivative Applied Calculus 141
Section 8: Curve Sketching
This section examines some of the interplay between the shape of the graph of f and the behavior
of f '. If we have a graph of f , we will see what we can conclude about the values of f '. If we
know values of f ', we will see what we can conclude about the graph of f. We will also utilize
the information from f '' that we learning in the last section
First Derivative Information
Definitions: The function f is increasing on (a,b) if a < x1 < x2 < b implies f( x1 ) < f( x2 ).
The function f is decreasing on (a,b) if a < x1 < x2 < b implies f( x1 ) > f( x2 ).
Graphically, f is increasing (decreasing) if, as we move from left to right along the graph of f,
the height of the graph increases (decreases).
These same ideas make sense if we consider h(t) to be the height (in feet) of a rocket at time t
seconds. We naturally say that the rocket is rising or that its height is increasing if the height h(t)
increases over a period of time, as t increases.
Example 1
List the intervals on which the function shown
increasing or decreasing.
f is increasing on the intervals [0,1] , [ 2,3] and [4,6].
f is decreasing on [1,2] and [6,8].
On the interval [3,4] the function is not increasing or
decreasing, it is constant.
First Derivative Information about Shape
For a function f which is differentiable on an interval (a,b);
(i) if f is increasing on (a,b) , then f '(x) ≥ 0 for all x in (a,b)
(ii) if f is decreasing on (a,b) , then f '(x) ≤ 0 for all x in (a,b)
(iii) if f is constant on (a,b) , then f '(x) = 0 for all x in (a,b).
Example 2
The graph shows the height of a helicopter during a period
of time. Sketch the graph of the upward velocity of the
helicopter, dh/dt.
Notice that the h(t) has a local maximum when t = 2 and
t = 5, and so v(2) = 0 and v(5) = 0. Similarly, h(t) has a
local minimum when t = 3, so v(3) = 0.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 2 The Derivative Applied Calculus 142
When h is increasing, v is positive. When h is decreasing, v is negative.
Using this information, we can sketch a graph of v(t) = dh/dt.
The next theorem is almost the converse of the First Shape Theorem and explains the
relationship between the values of the derivative and the graph of a function from a different
perspective. It says that if we know something about the values of f ' , then we can draw some
conclusions about the shape of the graph of f .
First Derivative Information about Shape (part 2)
For a function f which is differentiable on an interval I;
(i) if f '(x) > 0 for all x in the interval I , then f is increasing on I,
(ii) if f '(x) < 0 for all x in the interval I , then f is decreasing on I,
(iii) if f '(x) = 0 for all x in the interval I , then f is constant on I.
Example 3
Use information about the values of f ' to help graph f(x) = x3 – 6x2 + 9x + 1 .
f '(x) = 3x2 – 12x + 9 = 3(x – 1)(x – 3) so f '(x) = 0 only when x = 1 or x = 3.
f ' is a polynomial so it is always defined.
The only critical numbers for f are x = 1 and x = 3, and they divide the real number line into
three intervals: (–∞ , 1) , (1,3) and (3, ∞). On each of these intervals, the function is either
always increasing or always decreasing.
If x < 1, then f '(x) = 3(negative number)(negative number) > 0 so f is increasing.
If 1 < x < 3, then f '(x) = 3(positive number)(negative number) < 0 so f is decreasing.
If 3 < x, then f '(x) = 3(positive number)(positive number) > 0 so f is increasing.
Even though we don't know the value of f anywhere yet,
we do know a lot about the shape of the graph of f : as we
move from left to right along the x–axis, the graph of f
increases until x = 1, then the graph decreases until x = 3,
and then the graph increases again. The graph of f makes
"turns" when x = 1 and x = 3; it has a local maximum at
x = 1, and a local minimum at x = 3.
Chapter 2 The Derivative Applied Calculus 143
To plot the graph of f, we still need to
evaluate f at a few values of x, but only at
a very few values. f(1) = 5 , and (1,5) is a
local maximum of f. f(3) = 1 , and (3,1) is
a local minimum of f. The resulting graph
of f is shown here.
Second Derivative Information
Until now, we've only used first derivative information, but we could also use information from the
second derivative to provide more information about the shape of the function.
Second Derivative Information about Shape
(i) if f is concave up on (a,b) , then f ''(x) ≥ 0 for all x in (a,b)
(ii) if f is concave down on (a,b) , then f ''(x) ≤ 0 for all x in (a,b)
The converse of both of these are also true:
(i) if f ''(x) ≥ 0 for all x in (a,b), then f is concave up on (a,b)
(ii) if f ''(x) ≤ 0 for all x in (a,b), then f is concave down on (a,b)
Example 4
Use information about the values of f '' to help determine the intervals on which the function
f(x) = x3 – 6x2 + 9x + 1 is concave up and concave down.
For concavity, we need the second derivative: f ′ ( x) = 3 x 2 − 12 x + 9 , so f ′′ ( x) = 6 x − 12 .
To find possible inflection points, set the second derivative equal to zero.
6 x − 12 = 0 , so x = 2. This divides the real number line into two intervals: (−∞,2) and (2, ∞)
For x < 2, the second derivative is negative (for example, f ′′ (0) = 6(0) − 12 = −12 ), so f is
concave down. For x > 2, the second derivative is positive , so f is concave up.
We could have incorporated this concavity information when sketching the graph for the
previous example, and indeed we can see the concavity reflected in the graph shown.
Example 5
Use information about the values of f ' and f '' to help graph f ( x) = x 2 / 3 .
2 −1 / 3
f ′ ( x) = x . This is undefined at x = 0.
3
2
f ′′ ( x) = − x −4 / 3 . This is also undefined at x = 0.
9
Chapter 2 The Derivative Applied Calculus 144
This creates two intervals: x < 0, and x > 0.
2 2
On the interval x < 0, we could test out a value like x = -1. f ′ (−1) = − and f ′′ (−1) = − .
3 9
f ′ (x) is negative and f ′′(x) is negative, so we can conclude that the function is decreasing and
concave down on this interval.
2 2
On the interval x > 0, we could test out a value like x = 1. f ′ (1) = and f ′′ (1) = − .
3 9
f ′ (x) is positive and f ′′(x) is negative, so we can conclude that the function is increasing and
concave down on this interval.
We can also calculate that f (0) = 0 , giving us a base point for the graph. Using this
information, we can conclude the graph must look like this:
Sketching without an Equation
Of course, graphing calculators and computers are great at graphing functions. Calculus provides a
way to illuminate what may be hidden or out of view when we graph using technology. More
importantly, calculus gives us a way to look at the derivatives of functions for which there is no
equation given. We already saw the idea of this back in Section 3 where we sketched the
derivative of two graphs by estimating slopes on the curves.
We can summarize all the derivative information about shape in a table.
Summary of Derivative Information about the Graph
f(x) increasing Decreasing Concave up Concave down
f '(x) + - Increasing decreasing
f ''(x) + -
When f'(x) = 0, the graph of f may have a local max or min.
When f''(x) = 0, the graph of f may have an inflection point.
Chapter 2 The Derivative Applied Calculus 145
Example 6
A company's bank balance, B, in millions of dollars, t weeks after releasing a new product is
shown in the graph below. Sketch a graph of the marginal balance - the rate at which the bank
balance was changing over time.
Notice that since the tangent line will be horizontal at about t = 0.6 and t = 3.2, the derivative
will be 0 at those points.
We can then identifying intervals on which the original function is increasing or decreasing.
This will tell us when the derivative function is positive or negative.
Interval B(t) B'(t)
0<t<0.6 Decreasing Negative
0.6<t<3.2 Increasing Positive
t>3.2 Decreasing Negative
There appear to be inflection points at about t =1.5 and t = 5.5. At these points, the derivative
will be changing from increasing to decreasing or vice versa, so the derivative will have a local
max or min at those points.
Looking at the intervals of concavity:
Interval B(t) B'(t)
0<t<1.5 Concave up Increasing
1.5<t<5.5 Concave down Decreasing
t>5.5 Concave up Increasing
If we wanted a more accurate sketch of the derivative
function, we could also estimate the derivative at a
few points:
t B'(t)
0 -10
1.5 3
t>5.5 -1
Now we can sketch the derivative. We know a few values for the derivative function, and on
each interval we know the shape we need. We can use this to create a rough idea of what the
graph should look like.
Chapter 2 The Derivative Applied Calculus 146
Neg, Pos, Pos, Neg, Neg,
Inc Inc Dec Dec Inc
Smoothing this out gives us a good estimate for the graph of the derivative.
2.8 Exercises
1. Sketch the graph of a continuous function f so that
(a) f(1) = 3, f '(1) = 0 , and the point (1,3) is a local maximum of f .
(b) f(2) = 1, f '(2) = 0 , and the point (2,1) is a local minimum of f .
(c) f(5) = 4, f '(5) = 0, and the point (5,4) is not a local minimum or maximum of f .
In problems 2–4, sketch the graph of the derivative of each function.
2. 3. 4.
Chapter 2 The Derivative Applied Calculus 147
In problems 5–7, the graph of the height of a helicopter is shown. Sketch the graph of the upward
velocity of the helicopter.
6. 6. 7.
8. In the graphs to the right, match the graphs of the
functions with those of their derivatives
9. In the graphs below, match the graphs showing
the heights of rockets with those
showing their velocities.
In problems 10 – 14 , use information from the derivatives of each function to help you graph the
function. Find all local maximums and minimums of each function.
10. f(x) = x3 – 3x2 – 9x – 5 11. g(x) = 2x3 – 15x2 + 6 12. h(x) = x4 – 8x2 + 3
2 x2 + 3
13. r(t) = 14. f(x) =
t2 + 1 x
Chapter 2 The Derivative Applied Calculus 148
Section 9: Applied Optimization
We have used derivatives to help find the maximums and minimums of some functions given by
equations, but it is very unlikely that someone will simply hand you a function and ask you to find
its extreme values. More typically, someone will describe a problem and ask your help in
maximizing or minimizing something: "What is the largest volume package which the post office
will take?"; "What is the quickest way to get from here to there?"; or "What is the least expensive
way to accomplish some task?" In this section, we’ll discuss how to find these extreme values
using calculus.
Max/Min Applications
Example: The manager of a garden store wants to build a 600 square foot rectangular
enclosure on the store’s parking lot in order to display some equipment. Three sides of the
enclosure will be built of redwood fencing, at a cost of $7 per running foot. The fourth side
will be built of cement blocks, at a cost of $14 per running foot. Find the dimensions of the
least costly such enclosure.
The process of finding maxima or minima is called optimization. The function we’re optimizing is
called the objective function. The objective function can be recognized by its proximity to “est”
words (greatest, least, highest, farthest, most, …) Look at the garden store example; the cost
function is the objective function.
In many cases, there are two (or more) variables in the problem. In the garden store example
again, the length and width of the enclosure are both unknown. If there is an equation that relates
the variables we can solve for one of them in terms of the others, and write the objective function
as a function of just one variable. Equations that relate the variables in this way are called
constraint equations. The constraint equations are always equations, so they will have equals
signs. For the garden store, the fixed area relates the length and width of the enclosure. This will
give us our constraint equation.
Max-Min Story Problem Technique:
(a) Translate the English statement of the problem line by line into a picture (if that applies) and
into math. This is often the hardest step!
(b) Identify the objective function. Look for words indicating a largest or smallest value.
(b1) If you seem to have two or more variables, find the constraint equation. Think about
the English meaning of the word “constraint,” and remember that the constraint
equation will have an equals sign.
(b2) Solve the constraint equation for one variable and substitute into the objective function.
Now you have an equation of one variable.
(c) Use calculus to find the optimum values. (Take derivative, find critical points, test. Don’t
forget to check the endpoints!)
(d) Look back at the question to make sure you answered what was asked. Translate your
number answer back into English.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 2 The Derivative Applied Calculus 149
Example 1
The manager of a garden store wants to build a 600 square foot rectangular enclosure on the
store’s parking lot in order to display some equipment. Three sides of the enclosure will be
built of redwood fencing, at a cost of $7 per running foot. The fourth side will be built of
cement blocks, at a cost of $14 per running foot. Find the dimensions of the least costly such
enclosure.
First, translate line by line into math and a picture:
Text Translation
The manager of a garden store wants to Let x and y be the dimensions of the enclosure,
build a 600 square foot rectangular with y being the length of the side made of
enclosure on the store’s parking lot in blocks. Then:
order to display some equipment. Area = A = xy = 600
Three sides of the enclosure will be built 2x + y costs $7 per foot
of redwood fencing, at a cost of $7 per y costs $14 per foot
running foot. The fourth side will be So
built of cement blocks, at a cost of $14 Cost = C = 7(2x + y) + 14y = 14x + 21y
per running foot.
Find the dimensions of the least costly Find x and y so that C is minimized.
such enclosure.
The objective function is the cost function, and we want to minimize it. As it stands, though, it
has two variables, so we need to use the constraint equation. The constraint equation is the
600
fixed area A = xy = 600. Solve A for x to get x = , and then substitute into C:
y
600 8400
C = 14 + 21y = + 21y.
y y
Now we have a function of just one variable, so we can find the minimum using calculus.
8400
C ' = − 2 + 21
y
Chapter 2 The Derivative Applied Calculus 150
C’ is undefined for y = 0, and C’ = 0 when y = 20 or y = −20.
Of these three critical numbers, only y = 20 makes sense (is in the domain of the actual
function) – remember that y is a length, so it can’t be negative, and y = 0 would mean there was
no enclosure at all, so it couldn’t have an area of 600 square feet.
Test y = 20: (I chose the second derivative test)
16800
C' ' = > 0 , so this is a local minimum.
y3
Since this is the only critical point in the domain, this must be the global minimum. Going back
to our constraint function, we can find that when y = 20, x = 30. The dimensions of the
enclosure that minimize the cost are 20 feet by 30 feet.
When trying to maximize their revenue, businesses also face the constraint of consumer demand.
While a business would love to see lots of products at a very high price, typically demand
decreases as the price of goods increases. In simple cases, we can construct that demand curve to
allow us to maximize revenue.
Example 2
A concert promoter has found that if she sells tickets for $50 each, she can sell 1200 tickets, but
for each $5 she raises the price, 50 less people attend. What price should she sell the tickets at
to maximize her revenue?
We are trying to maximize revenue, and we know that R = pq , where p is the price per ticket,
and q is the quantity of tickets sold.
The problem provides information about the demand relationship between price and quantity -
as price increases, demand decreases. We need to find a formula for this relationship. To
investigate, let's calculate what will happen to attendance if we raise the price:
Price, p 50 55 60 65
Quantity, q 1200 1150 1100 1050
You might recognize this as a linear relationship. We can find the slope for the relationship by
1150 − 1200 − 50
using two points: m = = = −10 . You may notice that the second step in that
55 − 50 5
calculation corresponds directly to the statement of the problem: the attendance drops 50
people for every $5 the price increases.
Using the point-slope form of the line, we can write the equation relating price and quantity:
q − 1200 = −10( p − 50)
Simplifying to slope-intercept form gives the demand equation q = 1700 − 10 p
Chapter 2 The Derivative Applied Calculus 151
Substituting this into our revenue equation, we get an equation for revenue involving only one
variable:
R = pq = p(1700 − 10 p) = 1700 p − 10 p 2
Now, we can find the maximum of this function by finding critical numbers.
R′ = 1700 − 20 p , so R′ = 0 when p = 85.
Using the second derivative test, R′′ = −20 < 0 , so the critical number is a local maximum.
Since it's the only critical number, we can also conclude it's the global maximum.
The promoter will be able to maximize revenue by charging $85 per ticket. At this price, she
will sell q = 1700 − 10(85) = 850 tickets, generating $72,250 in revenue.
“Marginal Revenue = Marginal Cost”
You may have heard before that “profit is maximized when marginal cost and marginal revenue
are equal.” Now you can see why people say that! (Even though it’s not completely true.)
General Example 3
Suppose we want to maximize profit.
Now we know what to do – find the profit function, find its critical points, test them, etc.
But remember that Profit = Revenue – Cost. So Profit’ = Revenue’ – Cost’.
That is, the derivative of the profit function is MR – MC.
Now let’s find the critical points – those will be where Profit’ = 0 or is undefined.
Profit’ = 0 when MR – MC = 0, or where MR = MC.
That’s where the saying comes from! Here’s a more accurate way to express this:
Profit has critical points when Marginal Revenue and Marginal Cost are equal.
In all the cases we’ll see in this class, Profit will be very well behaved, and we won’t have to worry
about looking for critical points where Profit’ is undefined. But remember that not all critical
points are local max! The places where MR = MC could represent local max, local min, or neither
one.
Chapter 2 The Derivative Applied Calculus 152
Example 4
A company sells q ribbon winders per year at $p per ribbon winder. The demand function for
ribbon winders is given by: p = 300 − 0.02q . The ribbon winders cost $30 apiece to
manufacture, plus there are fixed costs of $9000 per year. Find the quantity where profit is
maximized.
We want to maximize profit, but there isn’t a formula for profit given. So let’s make one. We
can find a function for Revenue = pq using the demand function for p.
R(q ) = (300 − 0.02q )q = 300q − 0.02q 2
We can also find a function for Cost, using the variable cost of $30 per ribbon winder, plus the
fixed cost:
C (q ) = 9000 + 30q
Putting them together, we get a function for Profit:
( )
P(q ) = R(q ) − C (q ) = 300q − 0.02q 2 − (9000 + 30q ) = −0.02q 2 + 270q − 9000 .
Now we have two choices. We can find the critical points of Profit by taking the derivative of
P(q) directly, or we can find MR and MC and set them equal. (Naturally, you’ll get the same
answer either way.)
I’ll use MR = MC this time.
MR = 300 − 0.04q
MC = 30
300 − 0.04q = 30
270 = 0.04q
q = 6750
The only critical point is at q = 6750. Now we need to be sure this is a local max and not a
local min. In this case, I’ll look to the graph of P(q) – it’s a downward opening parabola, so this
must be a local max. And since it’s the only critical point, it must also be the global max.
Profit is maximized when they sell 6750 ribbon winders.
“Average Cost = Marginal Cost”
“Average cost is minimized when average cost = marginal cost” is another saying that isn’t quite
true; in this case, the correct statement is:
Average Cost has critical points when Average Cost and Marginal Cost are equal.
Let’s look at a geometric argument here:
Chapter 2 The Derivative Applied Calculus 153
Remember that the average cost
is the slope of the diagonal line,
the line from the origin to the
point on the total cost curve. If
you move your clear plastic
ruler around, you’ll see (and
feel) that the slope of the
diagonal line is smallest when
the diagonal line just touches the
cost curve – when the diagonal
line is actually a tangent line –
when the average cost is equal
to the marginal cost.
Example 5
The cost in dollars to produce q gift baskets is given by C (q ) = 160 + 2q + .1q 2 . Find the
quantity where the average cost is minimum.
C (q ) 160
A(q ) = = + 2 + .1q . We could find the critical points by finding A' , or by setting
q q
average cost to marginal cost; I’ll do the latter this time.
MC (q ) = 2 + .2q . So I want to solve:
160
+ 2 + .1q = 2 + .2q
q
160
= .1q
q
1600 = q 2
q = 40
The critical point of average cost is when q = 40.
Notice that we still have to confirm that the critical point is a minimum. For this, we can use
the first or second derivative test on A(q ) .
− 160
A' (q ) = 2 + .1
q
320
A' ' (q ) = 3 > 0
q
The second derivative is positive for all positive q, so that means this is a local min. Average
cost is minimized when they produce 40 gift baskets; at that quantity, the average cost is $10
per basket.
Chapter 2 The Derivative Applied Calculus 154
2.9 Exercises
1. (a) You have 200 feet of fencing available to construct a rectangular pen with a fence divider
down the middle (see below). What dimensions of the pen enclose the largest total area?
(b) If you need 2 dividers, what dimensions of the pen enclose the largest area?
(c) What are the dimensions in parts (a) and (b) if one edge of the pen borders on a river and
does not require any fencing?
2. You have 120 feet of fencing to construct a pen with 4 equal sized stalls. If the pen is
rectangular and shaped like the one below, what are the dimensions of the pen of largest
area and what is that area?
3. Suppose you decide to fence the rectangular garden in the corner of your yard. Then two sides
of the garden are bounded by the yard fence which is already there, so you only need to use the
80 feet of fencing to enclose the other two sides. What are the dimensions of the new garden of
largest area? What are the dimensions of the rectangular garden of largest area in the corner of
the yard if you have F feet of new fencing available?
4. (a) You have a 10 inch by 15 inch piece of tin which you plan
to form into a box (without a top) by cutting a square from
each corner and folding up the sides. How much should
you cut from each corner so the resulting box has the
greatest volume?
(b) If the piece of tin is A inches by B inches, how much
should you cut from each corner so the resulting box has
the greatest volume?
5. You have a 10 inch by 10 inch piece
of cardboard which you plan to cut
and fold as shown to form a box
with a top. Find the dimensions of
the box which has the largest
volume.
Chapter 2 The Derivative Applied Calculus 155
6. (a) You have been asked to bid on the construction of a square-bottomed box with no top
which will hold 100 cubic inches of water. If the bottom and sides are made from the same
material, what are the dimensions of the box which uses the least material? (Assume that
no material is wasted.)
(b) Suppose the box in part (a) uses different materials for the bottom and the sides. If the
bottom material costs 5¢ per square inch and the side material costs 3¢ per square inch,
what are the dimensions of the least expensive box which will hold 100 cubic inches of
water?
7. (a) Determine the dimensions of the least expensive cylindrical can which will hold 100 cubic
inches if the materials cost 2¢, 5¢ and 3¢ respectively for the top, bottom and sides.
(b) How do the dimensions of the least expensive can change if the bottom material costs more
than 5¢ per square inch?
8. You have 100 feet of fencing to build a pen in the shape of a circular
sector, the "pie slice" shown. The area of such a sector is rs/2. What
value of r maximizes the enclosed area?
9. (a) You have been asked to determine the least expensive route for a telephone cable which
connects Andersonville with Beantown. If it costs $5000
per mile to lay the cable on land and $8000 per mile to lay
the cable across the river and the cost of the cable is
negligible, find the least expensive route.
(b) What is the least expensive route if the cable costs $7000
per mile plus the cost to lay it.
10. You have been asked to determine where a water works
should be built along a river between Chesterville and Denton
to minimize the total cost of the pipe to the towns.
(a) Assume that the same size (and cost) pipe is used to each
town. (This part can be done quickly without using
calculus.)
(b) Assume that the pipe to Chesterville costs $3000 per mile
and to Denton it costs $7000 per mile.
13. U.S. postal regulations state that the sum of the length and girth
(distance around) of a package must be no more than 108 inches.
(a) Find the dimensions of the acceptable box with a square end
which has the largest volume.
(b) Find the dimensions of the acceptable box which has the
largest volume if its end is a rectangle twice as long as it is width.
(c) Find the dimensions of the acceptable box with a circular end which has the largest volume.
Chapter 2 The Derivative Applied Calculus 156
14. D. Simonton claims that the "productivity levels" of people in different fields can be described
as a function of their "career age" t by p(t) = e–at – e –bt where a and b are constants
which depend on the field of work, and career age is approximately 20 less than the actual age
of the individual.
(a) Based on this model, at what ages do mathematicians (a=.03, b=.05), geologists (a=.02,
b=.04), and historians (a=.02, b=.03) reach their maximum productivity?
(b) Simonton says "With a little calculus we can show that the curve ( p(t) ) maximizes at
1 b
t = b – a ln( a )." Use calculus to show that Simonton is correct.
Note: Models of this type have uses for describing the behavior of groups, but it is dangerous
and usually invalid to apply group descriptions or comparisons to individuals in the group.
(Scientific Genius, by Dean Simonton, Cambridge University Press, 1988, pp. 69 – 73)
15. You own a small airplane which holds a maximum of 20 passengers. It costs you $100 per
flight from St. Thomas to St. Croix for gas and wages plus an additional $6 per passenger for
the extra gas required by the extra weight. The charge per passenger is $30 each if 10 people
charter your plane (10 is the minimum number you will fly), and this charge is reduced by $1
per passenger for each passenger over 10 who goes (that is, if 11 go they each pay $29, if 12 go
they each pay $28, etc.). What number of passengers on a flight will maximize your profits?
16. In the planning of a coffee shop, we estimate that if there is seating for between 40 and 80 people,
the daily profit will be $50 per seat. However, if the seating capacity is more than 80 places, the
daily profit per seat will be decreased by $1 for each additional seat over 80. What should the
seating capacity be in order to maximize the coffee shop’s total profit?
17. In the planning of a taco restaurant, we estimate that if there is seating for between 10 and 40
people, the daily profit will be $10 per seat. However, if the seating capacity is more than 40
places, the daily profit per seat will be decreased by $0.20 per seat. What should the seating
capacity be in order to maximize the taco restaurant’s total profit?
18. The total cost in dollars for Alicia to make q oven mitts is given by C (q ) = 64 + 1.5q + .01q 2 .
(a) What is the fixed cost?
(b) Find a function that gives the marginal cost.
(c) Find a function that gives the average cost.
(d) Find the quantity that minimizes the average cost.
(e) Confirm that the average cost and marginal cost are equal at your answer to part (d).
19. Shaki makes and sells backpack danglies. The total cost in dollars for Shaki to make q danglies is
given by C (q ) = 75 + 2q + .015q 2 . Find the quantity that minimizes Shaki’s average cost for
making danglies.
Chapter 2 The Derivative Applied Calculus 157
Section 10: Other Applications
Tangent Line Approximation
Back when we first thought about the derivative, we used the slope of secant lines over tiny
intervals to approximate the derivative:
∆y f (x ) − f (a )
f ' (a ) ≅
=
∆x x−a
Now that we have other ways to find derivatives, we can exploit this approximation to go the other
way. Solve the expression above for f(x), and you’ll get the tangent line approximation:
The Tangent Line Approximation (TLA)
To approximate the value of f(x) using TLA, find some a where
1. a and x are “close,” and
2. You know the exact values of both f (a) and f ‘(a).
Then f ( x ) ≅ f (a ) + f ' (a )( x − a )
Another way to look at the same formula:
∆y ≅ f ' (a )∆x
How close is close? It depends on the shape of the graph of f. In general, the closer the better.
Example 1
Suppose we know that g(20) = 5 and g’(20) = 1.4. Use this information to approximate g(23)
and g(18).
Using the tangent line approximation:
g(23) ≈ 5 + (1.4)(23 – 20) = 9.2
g(18) ≈ 5 + (1.4)(18 – 20) = 2.2
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 2 The Derivative Applied Calculus 158
Note that we don’t know if these approximations are close – but they’re the best we can do with
the limited information we have to start with. Note also that 18 and 23 are sort of close to 20,
so we can hope these approximations are pretty good. We’d feel more confident using this
information to approximate g(20.003). We’d feel very unsure using this information to
approximate g(55).
Elasticity
We know that demand functions are decreasing, so when the price increases, the quantity
demanded goes down. But what about revenue = price × quantity? When the price increases will
revenue go down because the demand dropped so much? Or will revenue increase because
demand didn’t drop very much?
Elasticity of demand is a measure of how demand reacts to price changes. It’s normalized – that
means the particular prices and quantities don’t matter, and everything is treated as a percent
change. The formula for elasticity of demand involves a derivative, which is why we’re discussing
it here.
Elasticity of Demand
Given a demand function that gives q in terms of p,
p dq
The elasticity of demand is E = ⋅
q dp
(Note that since demand is a decreasing function of p, the derivative is negative. That’s why we
have the absolute values – so E will always be positive.)
If E < 1, we say demand is inelastic. In this case, raising prices increases revenue.
If E > 1, we say demand is elastic. In this case, raising prices decreases revenue.
If E = 1, we say demand is unitary. E = 1 at critical points of the revenue function.
Interpretation of elasticity:
If the price increases by 1%, the demand will decrease by E%.
Example 2
A company sells q ribbon winders per year at $p per ribbon winder. The demand function for
ribbon winders is given by p = 300 − 0.02q . Find the elasticity of demand when the price is $70
apiece. Will an increase in price lead to an increase in revenue?
dq
First, we need to solve the demand equation so it gives q in terms of p, so that we can find :
dp
dq
p = 300 − 0.02q, so q = 15000 − 50 p . Then = −50 .
dp
We need to find q when p = 70: q = 11500.
Chapter 2 The Derivative Applied Calculus 159
p dq 70
Now compute E = ⋅ = ⋅ (− 50 ) ≅ 0.3 .
q dp 11500
E < 1, so demand is inelastic. Increasing the price by 1% would only cause a 0.3% drop in
demand. Increasing the price would lead to an increase in revenue, so it seems that the
company should increase its price.
The demand for products that people have to buy, such as onions, tends to be inelastic. Even if the
price goes up, people still have to buy about the same amount of onions, and revenue will not go
down. The demand for products that people can do without, or put off buying, such as cars, tends
to be elastic. If the price goes up, people will just not buy cars right now, and revenue will drop.
Example 3
A company finds the demand q, in thousands, for their kites to be q = 400 − p 2 at a price of p
dollars. Find the elasticity of demand when the price is $5 and when the price is $15. Then
find the price that will maximize revenue.
dq
Calculating the derivative, = 2 p . The elasticity equation as a function of p will be:
dp
p dq p 2 p2
E= ⋅ = ⋅ 2 p =
q dp 400 − p 2 400 − p 2
Evaluating this to find the elasticity at $5 and at $15:
2(5) 2
E (5) = ≈ 0.133 . The demand is inelastic when the price is $5.
400 − (5) 2
At a price of $5, a 1% increase in price would decrease demand by only 0.133%. Revenue
could be raised by increasing prices.
2(15) 2
E (15) = ≈ 2.571 . The demand is elastic when the price is $15.
400 − (15) 2
At a price of $15, a 1% increase in price would decrease demand by 2.571%. Revenue could be
raised by decreasing prices.
To maximize the revenue, we could solve for when E = 1.
2 p2
=1
400 − p 2
2 p 2 = 400 − p 2
3 p 2 = 400
400
p= ≈ 11.55 .
3
A price of $11.55 will maximize the revenue.
Chapter 2 The Derivative Applied Calculus 160
2.10 Exercises
1. If g (20) = 35 and g ' (20) = −2 , estimate the value of g (22) .
2. If g (1) = −17 and g ' (1) = 5 , estimate the value of g (1.2) .
3. Use the Tangent Line Approximation to estimate the cube root of 9.
4. Use the Tangent Line Approximation to estimate the fifth root of 30.
5. A rectangle has one side on the x–axis, one side on the y–axis, and
a corner on the graph of y = x2 + 1 .
(a) Use Linear Approximation of the area formula to estimate the
increase in the area of the rectangle if the base grows from 2 to
2.3 inches.
(b) Calculate exactly the increase in the area of the rectangle as
the base grows from 2 to 2.3 inches.
6. You can measure the diameter of a circle to within 0.3 cm.
(a) How large is the "error" in the calculated area of a circle with a measured diameter of 7.4
cm?
(b) How large is the "error" in the calculated area of a circle with a measured diameter of 13.6
cm?
(c) How large is the percentage error in the calculated area of a circle with a measured
diameter of d?
7. The demand function for Alicia’s oven mitts is given by q = −8 p + 80 (q is the number of oven
mitts, p is the price in dollars). Find the elasticity of demand when p = $7.50. Will revenue
increase if Alicia raises her price from $7.50?
8. The demand function for Shaki’s danglies is given by q = −35 p + 205 (q is the number of danglies, p
is the price in dollars per dangly). Find the elasticity of demand when p = $5. Should Shaki raise
or lower his price to increase revenue?
Chapter 2 The Derivative Applied Calculus 161
Section 11: Implicit Differentiation and Related Rates
In our work up until now, the functions we needed to differentiate were either given explicitly,
such as y = x 2 + e x , or it was possible to get an explicit formula for them, such as solving
3 2
y 3 − 3 x 2 = 5 to get y = 5 + 3 x . Sometimes, however, we will have an equation relating x and
y which is either difficult or impossible to solve explicitly for y , such as y + e y = x 2 . In any
case, we can still find y ' = f '(x) by using implicit differentiation.
The key idea behind implicit differentiation is to assume that y is a function of x even if we
cannot explicitly solve for y. This assumption does not require any work, but we need to be very
careful to treat y as a function when we differentiate and to use the Chain Rule.
Example 1
Assume that y is a function of x . Calculate
d 3 d 3 2 d
(a)
dx
y ( ) (b)
dx
x y ( )
and (c)
dx
ln( y )
(a) We need the chain rule since y is a function of x:
d 3 d
dx
( )
y = 3 y 2 ( y) = 3 y 2 y′
dx
(b) We need to use the product rule and the Chain Rule:
d 3 2 d d dy
dx
( )
x y = x 3 ( y 2 ) + y 2 ( x 3 ) = x 3 2 y + y 2 3 x 2 = 2 x 3 2 yy ′ + 3 y 2 x 2
dx dx dx
d 1
(c) We know ln( x) = , so we use that and the Chain Rule:
dx x
d 1
ln( y ) = ⋅ y ′
dx y
IMPLICIT DIFFERENTIATION:
To determine y ' , differentiate each side of the defining equation, treating y
as a function of x , and then algebraically solve for y '.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 2 The Derivative Applied Calculus 162
Example 2
Find the slope of the tangent line to the circle x2 + y2 = 25 at the point (3,4) using implicit
differentiation.
We differentiate each side of the equation x2 + y2 = 25 and then
solve for y'
d 2 d
dx
( )
x + y 2 = (25)
dx
2 x + 2 yy ′ = 0
2x x
Solving for y', we have y ′ = − = − , and, at the point (3,4),
2y y
y' = – 3/4.
In the previous example, it would have been easy to explicitly solve for y, and then we could
differentiate y to get y '. Because we could explicitly solve for y , we had a choice of methods
for calculating y '. Sometimes, however, we can not explicitly solve for y , and the only way of
determining y ' is implicit differentiation.
Related Rates
If several variables or quantities are related to each other and some of the variables are changing at
a known rate, then we can use derivatives to determine how rapidly the other variables must be
changing.
Example 3
Suppose the border of a town is roughly circular, and the radius of that circle has been
increasing at a rate of 0.1 miles each year. Find how fast the area of the town has been
increasing when the radius is 5 miles.
We could get an approximate answer by calculating the area of the
circle when the radius is 5 miles ( A = πr2 = π(5 miles)2 ≈ 78.6
miles2 ) and 1 year later when the radius is 0.1 feet larger than
before ( A = πr2 = π(5.1 miles)2 ≈ 81.7 miles2 ) and then finding
∆Area/∆time = (81.7 mi2 – 78.6 mi2)/(1 year) = 3.1 mi2/yr. This
approximate answer represents the average change in area during
the 1 year period when the radius increased from 5 miles to 5.1
miles, and would correspond to the secant slope on the area graph.
To find the exact answer, though, we need derivatives. In this case both radius and area are
functions of time:
r(t) = radius at time t A(t) = area at time t
Chapter 2 The Derivative Applied Calculus 163
We know how fast the radius is changing, which is a statement about the derivative:
dr mile
= 0.1 . We also know that r = 5 at our moment of interest.
dt year
dA
We are looking for how fast the area is increasing, which is .
dt
Now we need an equation relating our variables, which is the area equation: A = πr 2 .
Taking the derivative of both sides of that equation with respect to t, we can use implicit
differentiation:
d
( A) = d πr 2
( )
dt dt
dA dr
= π 2r
dt dt
Plugging in the values we know for r and dr/dt,
dA miles miles 2
= π 2(5 miles) .1 = 3.14
dt year year
The area of the town is increasing by 3.14 square miles per year when the radius is 5 miles.
Related Rates
When working with a related rates problem,
1. Identify the quantities that are changing, and assign them variables
2. Find an equation that relates those quantities
3. Differentiate both sides of that equation with respect to time
4. Plug in any known values for the variables or rates of change
5. Solve for the desired rate.
Example 4
A company has determined the demand curve for their product is q = 5000 − p 2 , where p is
the price in dollars, and q is the quantity in millions. If weather conditions are driving the price
up $2 a week, find the rate at which demand is changing when the price is $40.
The quantities changing are p and q, and we assume they are both functions of time, t, in weeks.
We already have an equation relating the quantities, so we can implicitly differentiate it.
d
dt
(q ) = d
dt
( 5000 − p )
2
Chapter 2 The Derivative Applied Calculus 164
d
(q ) = d 5000 − p 2 1/ 2
( )
dt dt
dq 1 −1 / 2 d
dt 2
(
= 5000 − p 2 )dt
(
5000 − p 2 )
dq 1 dp
= (5000 − p 2 ) − 2 p
−1 / 2
dt 2 dt
Using the given information, we know the price is increasing by $2/week when the price is $40,
dp
giving = 2 when p = 40. Plugging in these values,
dt
dq 1
(
= 5000 − 40 2
dt 2
−1 / 2
)
(− 2 ⋅ 40 ⋅ 2) ≈ −1.37
Demand is falling by 1.37 million items per week.
Chapter 2 The Derivative Applied Calculus 165
2.11 Exercises
In problems 1 – 10 find dy/dx by differentiating implicitly then find the value of dy/dx at the
given point.
1. x2 + y2 = 100 , point (6, 8) 2. x2 + 5y2 = 45 , point (5, 2)
3. x2 – 3xy + 7y = 5 , point (2,1) 4. x + y = 5 , point (4,9)
x2 y2 x2 y2
5. 9 + 16 = 1 , point (0,4) 6. 9 + 16 = 1 , point (3,0)
7. ln(y) + 3x – 7 = 0 , point (2,e) 8. x2 – y2 = 16 , point (5,3)
9. x2 – y2 = 16 , point (5, –3) 10. y2 + 7x3 – 3x = 8 , point (1,2)
11. Find the slopes of the lines tangent to the graph in
shown at the points (3,1), (3,3), and (4,2) .
12. Find the slopes of the lines tangent to the graph in
shown where the graph crosses the y–axis.
13. Find the slopes of the lines tangent to the graph in graph shown at the points ((5,0),
(5,6), and (–4,3).
14. Find the slopes of the lines tangent to the graph in
the graph shown where the graph crosses the y–axis.
In problems 15 – 16 , find dy/dx using implicit differentiation and then find the slope of the line
tangent to the graph of the equation at the given point.
15. y3 – 5y = 5x2 + 7 , point (1,3) 16. y2 – 5xy + x2 + 21 = 0 , point (2,5)
Chapter 2 The Derivative Applied Calculus 166
17. An expandable sphere is being filled with liquid at a constant rate from a tap (imagine a water
balloon connected to a faucet). When the radius of the sphere is 3 inches, the radius is
4
increasing at 2 inches per minute. How fast is the liquid coming out of the tap? ( V = π r3 )
3
18. The 12 inch base of a right triangle is growing at 3 inches per hour,
and the 16 inch height is shrinking at 3 inches per hour.
(a) Is the area increasing or decreasing?
(b) Is the perimeter increasing or decreasing?
(c) Is the hypotenuse increasing or decreasing?
19. One hour later the right triangle in Problem 2 is 15 inches long and
13 inches high, and the base and height are changing at the same
rate as in Problem 18.
(a) Is the area increasing or decreasing now?
(b) Is the hypotenuse increasing or decreasing now?
(c) Is the perimeter increasing or decreasing now?
20. A young woman and her boyfriend plan to elope, but she must
rescue him from his mother who has locked him in his room. The
young woman has placed a 20 foot long ladder against his house
and is knocking on his window when his mother begins pulling
the bottom of the ladder away from the house at a rate of 3 feet
per second. How fast is the top of the ladder (and the young
couple) falling when the bottom of the ladder is
(a) 12 feet from the bottom of the wall?
(b) 16 feet from the bottom of the wall?
(c) 19 feet from the bottom of the wall?
21. The length of a 12 foot by 8 foot rectangle is increasing at a
rate of 3 feet per second and the width is decreasing at 2 feet per
second .
(a) How fast is the perimeter changing?
(b) How fast is the area changing?
22. An oil tanker in Puget Sound has sprung a leak, and a circular oil slick is forming. The oil slick
is 4 inches thick everywhere, is 100 feet in diameter, and the diameter is increasing at 12 feet
per hour. Your job, as the Coast Guard commander or the tanker's captain, is to determine how
fast the oil is leaking from the tanker.
Chapter 2 The Derivative Applied Calculus 167
Chapter 3 The Integral Applied Calculus 169
Chapter 3: The Integral
The previous chapters dealt with Differential Calculus. We started with the "simple"
geometrical idea of the slope of a tangent line to a curve, developed it into a combination of
theory about derivatives and their properties, techniques for calculating derivatives, and
applications of derivatives. This chapter deals with Integral Calculus and starts with the
"simple" geometric idea of area. This idea will be developed into another combination of
theory, techniques, and applications.
PreCalculus Idea – The Area of a Rectangle
If you look on the inside cover of nearly any traditional math book, you’ll find a bunch of area
and volume formulas – the area of a square, the area of a trapezoid, the volume of a right circular
cone, and so on. Some of these formulas are pretty complicated. But you still won’t find a
formula for the area of a jigsaw puzzle piece or the volume of an egg. There are lots of things
for which there is no formula. Yet we might still want to find their areas.
One reason areas are so useful is that they can represent quantities other than simple geometric
shapes. If the units for each side of the rectangle are meters, then the area will have the units
meters× meters = square meters = m2. But if the units of the base of a rectangle are hours and
the units of the height are miles/hour, then the units of the area of the rectangle are hours ×
miles/hour = miles, a measure of distance. Similarly, if the base units are centimeters and the
height units are grams, then the area units are gram-centimeters, a measure of work.
The basic shape we will use is the rectangle; the area of a rectangle is base × height. You should
1
also know the area formulas for triangles, A = bh and for circles, A = πr 2 .
2
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 3 The Integral Applied Calculus 170
Section 1: The Definite Integral
Distance from Velocity
Example 1
Suppose a car travels on a straight road at a constant speed of 40 miles per hour for two hours.
See the graph of its velocity below. How far has it gone?
We all remember distance = rate × time, so this one is easy. The car has gone 40 miles per
hour × 2 hours = 80 miles.
Example 2
Now suppose that a car travels so that its speed increases steadily from 0 to 40 miles per hour,
for two hours. (Just be grateful you weren’t stuck behind this car on the highway.) See the
graph of its velocity in below. How far has this car gone?
The trouble with our old reliable distance = rate × time relationship is that it only works if the
rate is constant. If the rate is changing, there isn’t a good way to use this formula.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 3 The Integral Applied Calculus 171
But look at graph from the last example again. Notice that distance = rate × time also
describes the area between the velocity graph and the t-axis, between t = 0 and t = 2 hours. The
rate is the height of the rectangle, the time is the length of the rectangle, and the distance is the
area of the rectangle. This is the way we can extend our simple formula to handle more
complicated velocities: And this is the way we can answer the second example.
The distance the car travels is the area between its velocity graph, the t-axis, t = 0 and t = 2.
This region is a triangle, so its area is ½bh = ½(2 hours)(40 miles per hour) = 40 miles. So the
car travels 40 miles during its annoying trip.
In our distance/velocity examples, the function represented a rate of travel (miles per hour), and
the area represented the total distance traveled. This principle works more generally:
For functions representing other rates such as the production of a factory (bicycles per day), or
the flow of water in a river (gallons per minute) or traffic over a bridge (cars per minute), or the
spread of a disease (newly sick people per week), the area will still represent the total amount of
something.
Example 3
The graph below shows the flow rate (cubic feet per second) of water in the Skykomish river at
the town of Goldbar in Washington state.
The area of the shaded region represents the total volume (cubic feet) of water flowing past the
town during the month of October. We can approximate this area to approximate the total water
by thinking of the shaded region as a rectangle with a triangle on top.
Chapter 3 The Integral Applied Calculus 172
Total water = total area ≈ area of rectangle + area of the “triangle”
1
≈ (2000 cubic feet/sec)(30 days) + 2 (1500 cf/s)(30 days) = (2750 cubic feet/sec)(30 days)
Note that we need to convert the units to make sense of our result:
Total water ≈ (2750 cubic feet/sec)(30 days) = (2750 cubic feet/sec)(2,592,000 sec)
≈ 7.128 x 109 cubic feet.
About 7 billion cubic feet of water flowed past Goldbar in October.
Approximating with Rectangles
How do we approximate the area if the rate curve is, well, curvy? We could use rectangles and
triangles, like we did in the last example. But it turns out to be more useful (and easier) to
simply use rectangles. The more rectangles we use, the better our approximation is.
Suppose we want to calculate the area between the graph of a positive function f and the x–axis
on the interval [a, b] (graphed above). The Riemann Sum method is to build several rectangles
with bases on the interval [a, b] and sides that reach up to the graph of f (see below). Then the
areas of the rectangles can be calculated and added together to get a number called a Riemann
Sum of f on [a, b]. The area of the region formed by the rectangles is an approximation of the
area we want.
Chapter 3 The Integral Applied Calculus 173
Example 4
Approximate the area in the graph on the
left between the graph of f and the x–
axis on the interval [2, 5] by summing
the areas of the rectangles in the graph
on the right.
The total area of rectangles is (2)(3) +
(1)(5) = 11 square units.
Example 5
Let A be the region bounded by the graph of f(x) = 1/x, the x–axis, and vertical lines at x = 1
and x = 5. We can’t find the area exactly (with what we know now), but we can approximate it
using rectangles.
When we make our rectangles, we have a lot of choices. We could pick any (non-overlapping)
rectangles whose bottoms lie within the interval on the x-axis, and whose tops intersect with the
curve somewhere. But it’s easiest to choose rectangles that – (a) have all the same width, and
(b) take their heights from the function at one edge. Below you'll see two ways to use four
rectangles to approximate this area. In the first graph, we used left-endpoints; the height of
each rectangle comes from the function value at its left edge. In the second graph on the next
page, we used right-hand endpoints.
Left-hand endpoints: The area is approximately the sum of the areas of the rectangles. Each
1
rectangle gets its height from the function f (x ) = and each rectangle has width = 1.
x
You can find the area of each rectangle using area = height × width. So the total area of the
rectangles, the left-hand estimate of the area under the curve, is
1 1 1 25
f (1)(1) + f (2 )(1) + f (3)(1) + f (4 )(1) = 1 + + + = ≅ 2.08
2 3 4 12
Notice that because this function is decreasing, all the left endpoint rectangles stick out above
the region we want – using left-hand endpoints will overestimate the area.
Right-hand endpoints: The right-hand estimate of the area is
1 1 1 1 77
f (2 )(1) + f (3)(1) + f (4 )(1) + f (5)(1) = + + + = ≅ 1.28
2 3 4 5 60
Chapter 3 The Integral Applied Calculus 174
All the right-hand rectangles lie completely under the curve, so this estimate will be an
underestimate.
We can see that the true area is actually in between these two estimates. So we could take their
average:
25 / 12 + 77 / 60 101
Average: = ≅ 1.68
2 60
In general, the average of the left-hand and right-hand estimates will be closer to the real area
than either individual estimate.
My estimate of the area under the curve is about 1.68. (The actual area is about 1.61.)
If we wanted a better answer, we could use even more, even narrower rectangles. But there’s a
limit to how much work we want to do by hand. In practice, it’s probably best to choose a
manageable number of rectangles. We’ll have better methods to get more accurate answers
before long.
These sums of areas of rectangles are called Riemann sums. You may see a shorthand notation
used when people talk about sums. We won’t use it much in this book, but you should know
what it means.
Riemann sum: A Riemann sum for a function f(x) over an interval [a, b] is a sum of areas
of rectangles that approximates the area under the curve. Start by dividing the interval [a,
b] into n subintervals; each subinterval will be the base of one rectangle. We usually
make all the rectangles the same width Δx. The height of each rectangle comes from the
function evaluated at some point in its sub interval. Then the Riemann sum is:
f (x1 )∆x + f (x2 )∆x + f (x3 )∆x + ⋅ ⋅ ⋅ + f (xn )∆x
Sigma Notation: The upper-case Greek letter Sigma Σ is used to stand for Sum. Sigma
notation is a way to compactly represent a sum of many similar terms, such as a Riemann
sum.
n
Using the Sigma notation, the Riemann sum can be written ∑ f (x )∆x .
i =1
i
This is read aloud as “the sum as i = 1 to n of f of x sub i Delta x.” The “i” is a counter,
like you might have seen in a programming class.
Chapter 3 The Integral Applied Calculus 175
Definition of the Definite Integral
Because the area under the curve is so important, it has a special vocabulary and notation.
The Definite Integral:
The definite integral of a positive function f(x) over an interval [a, b] is the area
between f, the x-axis, x = a and x = b.
The definite integral of a positive function f(x) from a to b is the area under the curve
between a and b.
If f(t) represents a positive rate (in y-units per t-units), then the definite integral of f from a
to b is the total y-units that accumulate between t = a and t = b.
Notation for the Definite Integral:
The definite integral of f from a to b is written
b
∫ f (x ) dx
a
The ∫ symbol is called an integral sign; it’s an elongated letter S, standing for sum.
(The ∫ corresponds to the Σ from the Riemann sum)
The dx on the end must be included; The dx tells what the variable is – in this example, the
variable is x. (The dx corresponds to the ∆x from the Riemann sum)
The function f is called the integrand.
The a and b are called the limits of integration.
Verb forms:
We integrate, or find the definite integral of a function. This process is called
integration.
b n
Formal Algebraic Definition: ∫ f ( x ) dx = lim ∑ f ( xi )∆x .
a n →∞
i =1
Practical Definition:
The definite integral can be approximated with a Riemann sum (dividing the area into
rectangles where the height of each rectangle comes from the function, computing the
area of each rectangle, and adding them up). The more rectangles you use, the narrower
the rectangles are, the better your approximation will be.
Looking Ahead:
We will have methods for computing exact values of some definite integrals from
formulas soon. In many cases, including when the function is given to you as a table or
graph, you will still need to approximate the definite integral with rectangles.
Chapter 3 The Integral Applied Calculus 176
Example 6
The graph shows y = r(t), the number of telephone calls made per hours on a Tuesday.
Approximately how many calls were made between 9 pm and 11 pm? Express this as a definite
integral and approximate with a Riemann sum.
We know that the accumulated calls will be the area under this rate graph over that two-hour
period, the definite integral of this rate from t = 9 to t = 11.
11
The total number of calls will be ∫ r (t )dt .
9
The top here is a curve, so we can’t get an exact answer. But we can approximate the area
using rectangles. I’ll choose to use 4 rectangles, and I’ll choose left-endpoints:
11
∫ r (t ) dt ≅ 100(.5) + 150(.5) + 180(.5) + 195(.5) = 312.5.
9
The units are calls per hour × hours = calls. My estimate is that about 312 calls were made
between 9 pm and 11 pm. Is this an under-estimate or an over-estimate?
Example 7
Describe the area between the graph of f(x) = 1/x, the x–axis, and the vertical lines at x = 1
and x = 5 as a definite integral.
This is the same area we estimated to be about 1.68 before. Now we can use the notation of the
5 1 5 1
definite integral to describe it. Our estimate of ∫ dx was 1.68. The true value of ∫ dx is
1 x 1 x
about 1.61.
Chapter 3 The Integral Applied Calculus 177
Example 8
3
Using the idea of area, determine the value of ∫ (1 + x )dx .
1
3
∫ (1 + x )dx represents the area between the graph of
1
f(x) = 1+x, the x–
axis, and the vertical lines at 1 and 3.
Since this area can be broken into a rectangle and a triangle, we can
find the area exactly. The area equals
4 + ½ (2)(2) = 6 square units.
Example 9
The table shows rates of population growth for Berrytown for several years. Use this table to
estimate the total population growth from 1970 to 2000:
Year (t) 1970 1980 1990 2000
Rate of population growth R(t) 1.5 1.9 2.2 2.4
(thousands of people per year)
The definite integral of this rate will give the total change in population over the thirty-year
period. We only have a few pieces of information, so we can only estimate. Even though I
haven’t made a graph, we’re still approximating the area under the rate curve, using rectangles.
How wide are the rectangles? I have information every 10 years, so the rectangles have a width
of 10 years. How many rectangles? Be careful here – this is a thirty-year span, so there are
three rectangles.
Using left-hand endpoints: (1.5)(10) + (1.9)(10) + (2.2)(10) = 56
Using right-hand endpoints: (1.9)(10) + (2.2)(10) + (2.4)(10) = 65
56 + 65
Taking the average of these two: = 60.5
2
Our best estimate of the total population growth from 1970 to 2000 is 60.5 thousand people.
Chapter 3 The Integral Applied Calculus 178
Signed Area
You may have noticed that until this point, we’ve insisted that the integrand (the function we’re
integrating) be positive. That’s because we’ve been talking about area, which is always positive.
If the “height” (from the function) is a negative number, then multiplying it by the width doesn’t
give us actual area, it gives us the area with a negative sign.
But it turns out to be useful to think about the possibility of negative area. We’ll expand our idea
of a definite integral now to include integrands that might not always be positive. The “heights”
of the rectangles, the values from the function, now might not always be positive.
The Definite Integral and Signed Area:
The definite integral of a function f(x) over an interval [a, b] is the signed area
between f, the x-axis, x = a and x = b.
The definite integral of a function f(x) from a to b is the signed area under the curve
between a and b.
If the function is positive, the signed area is positive, as before (and we can call it area.)
If the function dips below the x-axis, the areas of the regions below the x-axis come in with
a negative sign. In this case, we cannot call it simply “area.” These negative areas take
away from the definite integral.
b
∫ f (x ) dx = (Area above x-axis) – (Area below x-axis).
a
If f(t) represents a positive rate (in y-units per t-units), then the definite integral of f from a
to b is the total y-units that accumulate between t = a and t = b.
If f(t) represents any rate (in y-units per t-units), then the definite integral of f from a to b
is the net y-units that accumulate between t = a and t = b.
Example 10
Find the definite integral of of f(x) = –2 on the interval [1,4].
Chapter 3 The Integral Applied Calculus 179
4
⌠
⌡ –2 dx is the signed area of the region shown to the right. The region lies below the x-axis,
1
4
so the area, 6, comes in with a negative sign. So the definite integral is ⌠
⌡ –2 dx = –6.
1
Negative rates indicate that the amount is decreasing. For example, if f(t) is the velocity of a car
in the positive direction along a straight line at time t (miles/hour) , then negative values of f
indicate that the car is traveling in the negative direction, backwards. The definite integral of f is
the change in position of the car during the time interval. If the velocity is positive, positive
distance accumulates. If the velocity is negative, distance in the negative direction accumulates.
This is true of any rate. For example, if f(t) is the rate of population change (people/year) for a
town, then negative values of f would indicate that the population of the town was getting
smaller, and the definite integral (now a negative number) would be the change in the
population, a decrease, during the time interval.
Example 11
In 1980 there were 12,000 ducks nesting around a lake, and the rate of population change (in
ducks per year) is shown in Fig. 15. Write a definite integral to represent the total change in the
duck population from 1980 to 1990, and estimate the population in 1990.
The change in population
1990
= ⌡ ⌠ f(t) dt = – (area between f and axis)
1980
≈ – (200 ducks/year).(10 years) = – 2000 ducks.
Then (1990 duck population) = (1980 population) + (change from 1980 to 1990)
= (12,000) + ( –2000) = 10,000 ducks.
Example 12
A bug starts at the location x = 12 on the x–axis at 1 pm walks along the axis with the velocity
v(x) shown in the graph. How far does the bug travel between 1 pm and 3 pm, and where is the
bug at 3 pm?
Chapter 3 The Integral Applied Calculus 180
Note that the velocity is positive from 1 until 2:30, then becomes negative. So the bug moves in
the positive direction from 1 until 2:30, then turns around and moves back toward where it
started. The area under the velocity curve from 1 to 2:30 shows the total distance traveled by
the bug in the positive direction; the bug moved 12.5 feet in the positive direction. The area
between the velocity curve and the x-axis, between 2:30 and 3, shows the total distance traveled
by the bug in the negative direction, back toward home; the bug traveled 2.5 feet in the negative
3
direction. The definite integral of the velocity curve, ∫ v(t )dt , shows the net change in
1
distance:
3
∫ v(t )dt = 12.5 − 2.5 = 10
1
The bug ended up 10 feet further in the positive direction than he started. At 3 pm, the bug is at
x = 22.
Example 13
2 4 5 5
Use the graph to calculate ⌠
⌡ f(x) dx , ⌠
⌡ f(x) dx , ⌠
⌡ f(x) dx , and ⌠
⌡ f(x) dx .
0 2 4 0
2 4 5
Using the given areas, ⌠
⌡ f(x) dx = 2, ⌠
⌡ f(x) dx = – 5 , ⌠
⌡ f(x) dx = 2, and
0 2 4
5
⌠
⌡ f(x) dx = (area above) – (area below) = (2+2) – (5) = –1.
0
Chapter 3 The Integral Applied Calculus 181
Approximating with Technology
If your function is given as a graph or table, you will still have to approximate definite integrals
using areas, usually of rectangles. But if your function is given as a formula, you can turn to
technology to get a better approximate answer. For example, most graphing calculators have
some kind of numerical integration tool built in. You can also find many online tools that can do
this; type numerical integration into any search engine to see a selection of these.
Most numerical integration tools use rectangles to estimate the signed area, just as you would do
by hand. But they use many more rectangles than you would have the patience for, so they get a
better answer. Some of them use computer algebra systems to find exact answers; we will learn
how to do this ourselves later in this chapter.
When you turn to technology to find the value of a definite integral, be careful. Not every tool
will be able to give you a correct answer for every integral. I have had good luck with my TI 84.
You should make an estimate of the answer yourself first so you can judge whether the answer
you get makes sense.
Example 14
5 1
Use technology to approximate the definite integral ∫
1 x
dx . (This is the same definite integral
we approximated with rectangles before.)
I used my TI-84; the answer it gave me was 1.609437912. This agrees with the exact answer
for all the decimal digits displayed. WebMath said the answer was 1.60944, which is accurate
for all the decimal digits displayed. Microsoft Math said the answer was ln(5); that’s exactly
correct. Wolfram|Alpha says the answer is log(5); that’s not how everyone writes the natural
log, so that might trick you into writing the wrong answer.
Example 15
2
∫
2
Use technology to approximate the definite integral ex +x
dx
1
I asked WebMath, and it said the answer was zero – I know this is not correct, because the
function here is positive, so there must be some area under the curve here. I asked Microsoft
Math, and it simply repeated the definite integral; that’s because there isn’t an algebraic way to
find the exact answer. I asked my TI-84, and it said the answer was 86.83404047; that makes
sense with what I expected. Wolfram|Alpha also says the answer is about 86.834. So I believe:
2
∫
2
ex +x
dx ≅ 86.834.
1
Accumulation in Real Life
We have already seen that the "area" under a graph can represent quantities whose units are not the
usual geometric units of square meters or square feet. For example, if t is a measure of time in
seconds and f(t) is a velocity with units feet/second, then the definite integral has units
(feet/second) ∙ (seconds) = feet.
Chapter 3 The Integral Applied Calculus 182
b
⌡ f(x) dx are (y-units ) . (x-units). A quick check
In general, the units for the definite integral ⌠
a
of the units can help avoid errors in setting up an applied problem.
In previous examples, we looked at a function represented a rate of travel (miles per hour); in
that case, the area represented the total distance traveled. For functions representing other rates
such as the production of a factory (bicycles per day), or the flow of water in a river (gallons per
minute) or traffic over a bridge (cars per minute), or the spread of a disease (newly sick people
per week), the area will still represent the total amount of something.
Example 16
Suppose MR(q) is the marginal revenue in dollars/item for selling q items. What does
150
∫ MR(q ) dq represent?
0
150
∫ MR(q ) dq has units (dollars/item) ∙ (items) = dollars, and represents the accumulated dollars
0
150
for selling from 0 to 150 items. That is, ∫ MR(q ) dq = TR (150 ) , the total revenue from selling
0
150 items.
Example 17
Suppose r(t), in centimeters per year, represents how the diameter of a tree changes with time.
T2
What does ∫ r (t )dt
T1
represent?
T2
∫ r (t )dt has units of (centimeters per year) ∙ (years) = centimeters, and represents the
T1
T2
accumulated growth of the tree’s diameter from T1 to T2. That is, ∫ r (t )dt is the change in the
T1
diameter of the tree over this period of time.
Chapter 3 The Integral Applied Calculus 183
3.1 Exercises
1. Let A(x) represent the area bounded by the graph and the
horizontal axis and vertical lines at t=0 and t=x for the graph
shown. Evaluate A(x) for x = 1, 2, 3, 4, and 5.
2. Let B(x) represent the area bounded by the graph and the
horizontal axis and vertical lines at t=0 and t=x for the
graph shown. Evaluate B(x) for x = 1, 2, 3, 4, and 5.
3. Let C(x) represent the area bounded by the graph and the
horizontal axis and vertical lines at t=0 and t=x for the graph
shown. Evaluate C(x) for x = 1, 2, and 3 and find a formula for
C(x).
4. Let A(x) represent the area bounded by the graph and the
horizontal axis and vertical lines at t=0 and t=x for the graph
shown. Evaluate A(x) for x = 1, 2, and 3 and find a formula for
A(x).
5. A car had the velocity shown in the graph to the right. How far
did the car travel from t= 0 to t = 30 seconds?
6. A car had the velocity shown below. How afar did the car travel
from t = 0 to t = 30 seconds?
Chapter 3 The Integral Applied Calculus 184
7. The velocities of two cars are shown in the graph.
(a) From the time the brakes were applied, how many
seconds did it take each car to stop?
(b) From the time the brakes were applied, which car
traveled farther until it came to a complete stop?
8. You and a friend start off at noon and walk in the same direction
along the same path at the rates shown.
(a) Who is walking faster at 2 pm? Who is ahead at 2 pm?
(b) Who is walking faster at 3 pm? Who is ahead at 3 pm?
(c) When will you and your friend be together? (Answer in words.)
9. Police chase: A speeder traveling 45 miles per hour (in a 25
mph zone) passes a stopped police car which immediately
takes off after the speeder. If the police car speeds up
steadily to 60 miles/hour in 20 seconds and then travels at a
steady 60 miles/hour, how long and how far before the
police car catches the speeder who continued traveling at 45
miles/hour?
10. Water is flowing into a tub. The table shows the rate at which the water flows, in gallons per
minute. The tub is initially empty.
t, in 0 1 2 3 4 5 6 7 8 9 10
minutes
Flow rate, 0.5 1.0 1.2 1.4 1.7 2.0 2.3 1.8 0.7 0.5 0.2
in gal/min
Use the table to estimate how much water is in the tub after
a. five minutes
b. ten minutes
Chapter 3 The Integral Applied Calculus 185
11. The table shows the speedometer readings for a short car trip.
t , in minutes 0 5 10 15 20
Speed, in mph 0 30 40 65 40
a. Use the table to estimate how far the car traveled over the twenty minutes shown.
b. How accurate would you expect your estimate to be?
40
12. The table shows values of f (t ) . Use the table to estimate ∫ f (t ) dt .
0
t 0 10 20 30 40
f (t ) 17 22 18 11 35
13. The table shows values of g ( x ) .
x 0 1 2 3 4 5 6
g (x ) 140 142 144 152 154 165 200
Use the table to estimate
3 6 6
a. ∫ g (x ) dx
0
b. ∫ g (x ) dx
3
c. ∫ g (x ) dx
0
14. What are the units for the "area" of a rectangle with the given
base and height units?
Base units Height units "Area" units
miles per second seconds
hours dollars per hour
square feet feet
kilowatts hours
houses people per house
meals meals
In problems 15 – 17 , represent the area of each bounded region as a definite integral, and use
geometry to
determine the value of the definite integral.
15. The region bounded by y = 2x , the x–axis, the line x = 1, and x = 3.
16. The region bounded by y = 4 – 2x , the x–axis, and the y–axis.
17. The shaded region in the graph to the right.
Chapter 3 The Integral Applied Calculus 186
18. Using the graph of f shown and the given areas of
several regions, evaluate:
3 5 7
(a) ⌠
⌡ f(x) dx (b) ⌠
⌡ f(x) dx (c) ⌠
⌡ f(x) dx
0 3 5
19. Using the graph of f shown and the given areas of several regions, evaluate:
3 4
(a) ⌠
⌡ g(x) dx (b) ⌠
⌡ g(x) dx
1 3
8 8
(c) ⌠
⌡ g(x) dx (d) ⌠
⌡ g(x) dx
4 1
20. Use the graph to evaluate:
1 6
(a) ⌠
⌡ h(x) dx (b) ⌠
⌡ h(x) dx
–2 4
6 4
(c) ⌠
⌡ h(x) dx (d) ⌠
⌡ h(x) dx
–2 –2
21. Your velocity along a straight road is shown to the right.
How far did you travel in 8 minutes?
22. Your velocity along a straight road is shown below. How
many feet did you walk in 8 minutes?
Chapter 3 The Integral Applied Calculus 187
b
In problems 23 - 26, the units are given for x and for f(x) . Give the units of ⌠
⌡ f(x) dx .
a
23. x is time in "seconds", and f(x) is velocity in "meters per second."
24. x is time in "hours", and f(x) is a flow rate in "gallons per hour."
25. x is a position in "feet", and f(x) is an area in "square feet."
26. x is a position in "inches", and f(x) is a density in "pounds per inch."
In problems 27 – 31, represent the area with a definite integral and use technology to find the
approximate answer.
27. The region bounded by y = x 3, the x–axis, the line x = 1, and x = 5.
28. The region bounded by y = x , the x–axis, and the line x = 9.
29. The shaded region shown to the right.
30. The shaded region below.
3
31. Consider the definite integral ∫ (3 + x ) dx .
0
(a) Using six rectangles, find the left-hand Riemann sum for this definite integral.
(b) Using six rectangles, find the right-hand Riemann sum for this definite integral.
(c) Using geometry, find the exact value of this definite integral.
2
32. Consider the definite integral ∫ 0
x 3 dx .
(a) Using four rectangles, find the left-hand Riemann sum for this definite integral.
(b) Using four rectangles, find the right-hand Riemann sum for this definite integral.
33. Write the total distance traveled by the car in the graph
between 1 pm and 4 pm as a definite integral and
estimate the value of the integral.
Chapter 3 The Integral Applied Calculus 188
Problems 34 – 41 refer to the graph of f shown. Use the
graph to determine the values of the definite integrals.
(The bold numbers represent the area of each region.)
3 5 2
⌠ f(x) dx
34. ⌡ ⌠ f(x) dx
35. ⌡ ⌠ f(x) dx
36. ⌡
0 3 2
7 5 7 6 7
37. ⌠
⌡ f(w) dw 38. ⌠
⌡ f(x) dx 39. ⌠
⌡ f(x) dx 40. ⌠
⌡ f(t) dt 41. ⌠
⌡ f(x) dx
6 0 0 3 5
Problems 42 – 47 refer to the graph of g shown. Use the
graph to evaluate the integrals.
2 3 5
42. ⌠
⌡ g(x) dx 43. ⌠
⌡ g(t) dt 44. ⌠
⌡ g(x) dx
0 1 0
8 3 8
45. ⌠
⌡ g(s) ds 46.⌠
⌡ 2g(t) dt 47. ⌠
⌡ 1+g(x) dx
0 0 5
Chapter 3 The Integral Applied Calculus 189
Section 2: The Fundamental Theorem and Antidifferentiation
The Fundamental Theorem of Calculus
This section contains the most important and most used theorem of calculus, the Fundamental
Theorem of Calculus. Discovered independently by Newton and Leibniz in the late 1600s, it
establishes the connection between derivatives and integrals, provides a way of easily calculating
many integrals, and was a key step in the development of modern mathematics to support the rise
of science and technology. Calculus is one of the most significant intellectual structures in the
history of human thought, and the Fundamental Theorem of Calculus is a most important brick in
that beautiful structure.
The Fundamental Theorem of Calculus:
b
∫ F ' (x )dx = F (b ) − F (a )
a
This is actually not new for us; we’ve been using this relationship for some time; we just haven’t
written it this way. This says what we’ve said before: the definite integral of a rate from a to b
is the net y-units, the change in y, that accumulate between t = a and t = b. Here we’ve just made
it plain that that the rate is a derivative.
Thinking about the relationship this way gives us the key to finding exact answers for some
definite integrals. If the integrand is the derivative of some F, then maybe we could simply find F
and subtract – that would be easier than approximating with rectangles. Going backwards
through the differentiation process will help us evaluate definite integrals.
Example 1
Find f(x) if f ‘(x) = 2x.
Oooh, I know this one. It’s f (x ) = x 2 + 3 . Oh, wait, you were thinking something else? Yes, I
guess you’re right -- f (x ) = x 2 works too. So does f (x ) = x 2 − π , and f ( x ) = x 2 + 104,589.2 .
In fact, there are lots of answers.
In fact, there are infinitely many functions that all have the same derivative. And that makes
sense – the derivative tells us about the shape of the function, but it doesn’t tell about the
location. We could shift the graph up or down and the shape wouldn’t be affected, so the
derivative would be the same.
This leads to one of the trickier definitions – pay careful attention to the articles, because they’re
important.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 3 The Integral Applied Calculus 190
Antiderivatives
An antiderivative of a function f(x) is any function F(x) where F ‘(x) = f(x).
The antiderivative of a function f(x) is a whole family of functions, written F(x) + C,
where F ‘(x) = f(x) and
C represents any constant.
The antiderivative is also called the indefinite integral.
Notation for the antiderivative:
The antiderivative of f is written
∫ f (x ) dx
This notation resembles the definite integral, because the Fundamental Theorem of
Calculus says antiderivatives and definite integrals are intimately related. But in this
notation, there are no limits of integration.
The ∫ symbol is still called an integral sign; the dx on the end still must be included; you
can still think of ∫ and dx as left and right parentheses. The function f is still called
the integrand.
Verb forms:
We antidifferentiate, or integrate, or find the indefinite integral of a function. This
process is called antidifferentiation or integration.
There are no small families in the world of antiderivatives: if f has one antiderivative F, then f
has an infinite number of antiderivatives and every one of them has the form F(x) + C.
Example 2
Find an antiderivative of 2x.
I can choose any function I like as long as its derivative is 2x, so I’ll pick F ( x ) = x 2 − 5.2 .
Example 3
Find the antiderivative of 2x.
Now I need to write the entire family of functions whose derivatives are 2x. I can use the
notation:
∫ 2 x dx = x + C
2
Chapter 3 The Integral Applied Calculus 191
Example 4
Find ∫ e x dx .
This is likely one you remember -- e x is its own derivative, so it is also its own antiderivative.
The integral sign tells me that I need to include the entire family of functions, so I need that + C
on the end:
∫ e dx = e + C
x x
Antiderivatives Graphically or Numerically
Another way to think about the Fundamental Theorem of Calculus is to solve the expression for
F(b):
The Fundamental Theorem of Calculus (restated)
b
∫ F ' (x )dx = F (b ) − F (a )
a
The definite integral of a derivative from a to b gives the net change in the original function.
b
F (b ) = F (a ) + ∫ F ' ( x ) dx
a
The amount we end up is the amount we start with plus the net change in the function.
This lets us get values for the antiderivative – as long as we have a starting point, and we know
something about the area.
Example 5
Suppose F(t) has the derivative f(t) shown below, and suppose that we know F(0) = 5. Find
values for F(1), F(2), F(3), and F(4).
Using the second way to think about the Fundamental Theorem of Calculus,
b
F (b ) = F (a ) + ∫ F ' ( x ) dx -- we can see that
a
1 1
F (1) = F (0 ) + ∫ f ( x ) dx . We know the value of F(0), and we can easily find ∫ f (x ) dx from the
0 0
graph – it’s just the area of a triangle.
1
So F (1) = F (0 ) + ∫ f ( x ) dx = 5 + .5 = 5.5
0
Chapter 3 The Integral Applied Calculus 192
2
F (2 ) = F (0 ) + ∫ f ( x ) dx = 5 + 1 = 6
0
Note that we can start from any place we know the value of – now that we know F(2), we can
use that:
3
F (3) = F (2 ) + ∫ f ( x ) dx = 6 − .5 = 5.5
2
4
F (4 ) = F (3) + ∫ f ( x ) dx = 5.5 − 1 = 4.5
3
Example 6
F ‘(t) = f(t) is shown below. Where does F(t) have maximum and minimum values on the
interval [0, 4]?
b
Since F (b ) = F (a ) + ∫ f (t ) dt , we know that F is increasing as long as the area accumulating
a
under F ’ = f is positive (until t = 3), and then decreases when the curve dips below the x-axis so
that negative area starts accumulating. The area between t = 3 and t = 4 is much smaller than
the positive area that accumulates between 0 and 3, so we know that F(4) must be larger than
F(0). The maximum value is when t = 3; the minimum value is when t = 0.
Note that this is a different way to look at a problem we already knew how to solve – in Chapter
2, we would have found critical points of F, where f = 0 – there’s only one, when t = 3. f = F’
goes from positive to negative there, so F has a local max at that point. It’s the only critical
point, so it must be a global max. Then we would look at the values of F at the endpoints to find
which was the global min.
We can also attempt to sketch a function based on the graph of the derivative.
Example 7
The graph to the right shows f'(x) - the rate of change of f(x).
Use it sketch a graph of f(x) that satisfies f(0) = 0
Recall from the last chapter the relationships between the
function graph and the derivative graph:
f(x) increasing Decreasing Concave up Concave down
f '(x) + - Increasing decreasing
f ''(x) + -
In the graph shown, we can see the derivative is positive on the interval (0, 1) and (3, ∞), so the
graph of f should be increasing on those intervals. Likewise, f should be decreasing on the
interval (1,3).
Chapter 3 The Integral Applied Calculus 193
In the graph, f' is decreasing on the interval (0, 2), so f should be concave down on that interval.
Likewise, f should be concave up on the interval (2, ∞).
The derivative itself is not enough information to know where the function f starts, since there
are a family of antiderivatives, but in this case we are given a specific point to start at.
To start the sketch, we might note first the shapes we need
increasing decreasing decreasing increasing increasing
conc down conc down conc up conc up conc up
0 1 2 3 4
then sketch the basic shapes.
0 1 2 3 4
Now we can attempt to sketch the graph, starting at the point (0, 0). Notice we are very roughly
sketching this, as we don't have much information to work with. We can tell, though, from the
graph that the area from x = 0 to x = 1 is about the same as the area from x = 1 to x = 3, so we
would expect the net area from x = 0 to x = 3 to be close to 0.
0 1 2 3 4
It turns out this graph isn't horribly bad. Smoothing it out would give a graph closer to the
actual antiderivative graph, shown below.
Chapter 3 The Integral Applied Calculus 194
Derivative of the Integral
There is another important connection between the integral and derivative.
The Fundamental Theorem of Calculus (part 2):
x x
d
If A( x) = ∫
dx ∫a
f (t )dt , then A′( x) = f (t )dt = f ( x)
a
The derivative of the accumulation function is the original function.
Example 8
x
Let F ( x) = ∫ f (t )dt , where f is graphed below. Estimate F ′(3) .
0
The function F measures the area from t = 0 to some t = x. To estimate F ′(3) , we want to
estimate how much the area is increasing when t = 3. Since the value of the function f is 0 at t
= 3, the area will not be increasing or decreasing, so we can estimate F ′(3) = 0
Directly using the fundamental theorem of calculus part 2,
x
d
dx ∫0
F ′( x) = f (t )dt = f ( x) , so
F ′(3) = f (3) = 0
Chapter 3 The Integral Applied Calculus 195
3.2 Exercises
In problems 1 – 5, verify that F(x) is an antiderivative of the integrand f(x) and use Part 2 of the
Fundamental Theorem to evaluate the definite integrals.
1 4 3
1
⌠ 2x dx , F(x) = x2 + 5
1. ⌡ ⌠ 2 3
2. ⌡ 3x dx , F(x) = x + 2 3. ⌡ x2 dx , F(x) = 3 x3
⌠
0 1 1
5
3
2 1 3 2 ⌠ 1
⌡ (x + 4x – 3 ) dx , F(x) = 3 x + 2x – 3x
4. ⌠ 5. dx , F(x) = ln( x )
⌡x
0 1
x
6. Given A(x) = ⌠
⌡ 2t dt, find A'(x)
0
x
2
⌡ ( 3 – t ) dt, find A'(x)
7. Given A(x) = ⌠
0
x
8. Let A(x) = ⌠
⌡ f(t) dt for the function graphed here. Evaluate
0
A'(1), A'(2), A'(3).
For problems 9-10, the graph provided shows g'(x). Use it sketch a graph of g(x) that satisfies
g(0) = 0.
9. 10.
Chapter 3 The Integral Applied Calculus 196
Section 3: Antiderivatives of Formulas
Now we can put the ideas of areas and antiderivatives together to get a way of evaluating definite
b
integrals that is exact and often easy. To evaluate a definite integral ⌠
⌡ f(t) dt , we can find any
a
antiderivative F of f and evaluate F(b) – F(a). The problem of finding the exact value of a
definite integral reduces to finding some (any) antiderivative F of the integrand and then
evaluating F(b) – F(a). Even finding one antiderivative can be difficult, and we will stick to
functions that have easy antiderivatives.
Building Blocks
Antidifferentiation is going backwards through the derivative process. So the easiest
antiderivative rules are simply backwards versions of the easiest derivative rules. Recall from
Chapter 2:
Derivative Rules: Building Blocks
In what follows, f and g are differentiable functions of x and k and n are constants.
d
(a) Constant Multiple Rule: (kf ) = kf '
dx
d
(b) Sum (or Difference) Rule: ( f + g ) = f '+ g ' (or d ( f − g ) = f '− g ' )
dx dx
d n
(c) Power Rule:
dx
( )
x = nx n −1
d
Special cases: (k ) = 0 (because k = kx 0 )
dx
d
(x ) = 1 (because x = x 1 )
dx
d x
(d) Exponential Functions:
dx
( )
e = ex
d x
dx
( )
a = ln a ⋅ a x
d
(e) Natural Logarithm: (ln x ) = 1
dx x
Thinking about these basic rules was how we came up with the antiderivatives of 2x and e x
before.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 3 The Integral Applied Calculus 197
The corresponding rules for antiderivatives are next – each of the antiderivative rules is simply
rewriting the derivative rule. All of these antiderivatives can be verified by differentiating.
There is one surprise – the antiderivative of 1/x is actually not simply ln(x), it’s ln|x|. This is a
good thing – the antiderivative has a domain that matches the domain of 1/x, which is bigger
than the domain of ln(x), so we don’t have to worry about whether our x’s are positive or
negative. But you must be careful to include those absolute values – otherwise, you could end
up with domain problems.
Antiderivative Rules: Building Blocks
In what follows, f and g are differentiable functions of x and k, n, and C are constants.
(a) Constant Multiple Rule: ∫ kf ( x) dx = k ∫ f (x )dx
(b) Sum (or Difference) Rule: ∫ f ( x)± g (x ) dx = ∫ f (x )dx ± ∫ f (x )dx
x n +1
∫ x dx =
n
(c) Power Rule: + C , provided that n = −1
n +1
∫ k dx = kx + C (because k = kx )
0
Special case:
(d) Exponential Functions: ∫e
x
dx = e x + C
ax
∫ a dx =
x
+C
ln a
1
∫x dx = ∫ dx = ln x + C
−1
(e) Natural Logarithm:
x
Chapter 3 The Integral Applied Calculus 198
Example 1
14
Find the antiderivative of 3 x 7 − 15 x +
x2
14 x8 x3/ 2 x −1
7
∫
3 x − 15 x + dx = ∫ 3 x 7
− 15(x 1/ 2
+ 14 x −2
dx = 3) − 15 + 14 +C
x2 8 3/ 2 −1
That’s a little hard to look at, so you might want to simplify a little:
7 14 3x 8
∫ 3 x − 15 x + 2
x
dx =
8
− 10 x 3 / 2 − 14 x −1 + C.
Example 2
16
Find ∫ e x + 12 − dx
x
16
∫ e
x x
+ 12 − dx = e + 12 x − 16 ln x + C
x
Example 3
Find F(x) so that F ' ( x ) = e x and F (0 ) = 10 .
This time we are looking for a particular antiderivative; we need to find exactly the right
constant. Let’s start by finding the antiderivative:
∫e
x
dx = e x + C
So we know that F ( x ) = e x + some constant; we just need to find which one. For that, we’ll use
the other piece of information (the initial condition):
F (x ) = e x + C
F (0 ) = e 0 + C = 1 + C = 10
C =9
The particular constant we need is 9; F (x ) = e x + 9.
Chapter 3 The Integral Applied Calculus 199
The reason we are looking at antiderivatives right now is so we can evaluate definite integrals
exactly. Recall the Fundamental Theorem of Calculus:
b
∫ F ' (x )dx = F (b ) − F (a )
a
If we can find an antiderivative for the integrand, we can use that to evaluate the definite integral.
The evaluation F(b) – F(a) is represented by the symbol F ( x )] or F (x )| .
b b
a a
Example 4
3
Evaluate ∫ x dx in two ways:
1
(i) By sketching the graph of y = x and geometrically finding the area.
(ii) By finding an antiderivative of F(x) of the integrand and evaluating F(3)–F(1).
(i) The graph of y = x is shown to the right, and the shaded region
corresponding to the integral has area 4.
1 2
(ii) One antiderivative of x is F ( x) = x , and
2
3
1 2 3 1 2 1 2 9 1
∫ x dx = x ] = (3) − (1) = − = 4.
1
2 1
2 2 2 2
Note that this answer agrees with the answer we got geometrically.
1 2
If we had used another antiderivative of x, say F ( x) = x +7,
2
3
1 3 1 2 1 2 9 1
then ∫ x dx = x 2 + 7 ] = (3) + 7 − (1) + 7 = + 7 − − 7 = 4.
1 2 1 2 2 2 2
Whatever constant you choose, it gets subtracted away during the evaluation; we might as well
always choose the easiest one, where the constant = 0.
Example 5
Find the area between the graph of y = 3x2 and the horizontal axis for x between 1 and 2.
This is ∫
1
2
1
( ) ( )
3 x 2 dx = x 3 ] = 2 3 − 13 = 7.
2
Chapter 3 The Integral Applied Calculus 200
Example 6
A robot has been programmed so that when it starts to move, its velocity after t seconds will be
3t 2 feet/second.
(a) How far will the robot travel during its first 4 seconds of movement?
(b) How far will the robot travel during its next 4 seconds of movement?
(a) The distance during the first 4 seconds will be the area under
the graph of velocity, from t = 0 to t = 4.
4
That area is the definite integral ∫0
3t 2 dt . An antiderivative of
3t 2 is t 3 , so
0
4
∫ 3t
2
dt = t 3 ]
0
4
= 43 − 03 = 64 feet.
(b)
4
8
∫ 3t
2
dt = t 3 ] 8
4 = 8 3 − 4 3 = 512 − 64 = 448 feet.
Example 7
Suppose that t minutes after putting 1000 bacteria on a Petri plate the rate of growth of the
population is 6t bacteria per minute.
(a) How many new bacteria are added to the population during the first 7 minutes?
(b) What is the total population after 7 minutes?
(a) The number of new bacteria is the area under the rate of
growth graph, and one antiderivative of 6t is 3t2 .
7 7
So new bacteria = ⌠
⌡ 6t dt = 3t
0
2
0
= 3(7)2 – 3(0)2 = 147 |
(b) The new population = (old population) + (new bacteria)
= 1000 + 147 = 1147 bacteria.
Example 8
A company determines their marginal cost for production, in dollars per item, is
4
MC ( x) = + 2 when producing x thousand items. Find the cost of increasing production
x
from 4 thousand items to 5 thousand items.
Remember that marginal cost is the rate of change of cost, and so the fundamental theorem tells
b b
us that ∫ MC ( x)dx = ∫ C ′( x)dx = C (b) − C (a) .
a a
In other words, the integral of marginal cost will
Chapter 3 The Integral Applied Calculus 201
give us a net change in cost. To find the cost of increasing production from 4 thousand items to
5
5 thousand items, we need to integrate ∫ MC ( x)dx .
4
We can write the marginal cost as MC ( x) = 4 x −1 / 2 + 2 . We can then use the basic rules to find
an antiderivative:
x1 / 2
C ( x) = 4 + 2 x = 8 x + 2 x . Using this,
1/ 2
5
( )] ( ) ( )
5
4
Net change in cost = ∫ + 2 dx = 8 x + 2 x = 8 5 + 2 ⋅ 5 − 8 4 + 2 ⋅ 4 ≈ 3.889
4 x 4
It will cost 3.889 thousand dollars to increase production from 4 thousand items to 5 thousand
items.
Chapter 3 The Integral Applied Calculus 202
3.3 Exercises
For problems 1-10, find the indicated antiderivative.
1. ∫ (x − 14 x + 5)dx 2. ∫ (2.5 x − x − 1.25)dx
3 5
3. ∫ 12.3dy 4. ∫ π dw
2
1
5. ∫ e P dP 6. ∫ x + e x − 3 dx
4x
1 1
7. ∫ x dx 8. ∫ 2 dx
x
t5 − t2
9. ∫ (x − 2)(x + 2)dx 10. ∫
t
dt
For problems 11-18, find an antiderivative of the integrand and use the Fundamental Theorem to
evaluate the definite integral.
5 2 e
3
11. ⌠ 2
12. ⌠ 2 2 ⌠ 1 dx
⌡ 3x dx ⌡ x dx 13. ⌠
⌡ (x + 4x – 3 ) dx 14. ⌡x
2 –1 1 1
100 5 10 1000
1 1
15. ∫
25
x dx 16. ∫
3
x dx 17. ∫
1 x2
dx 18. ∫
1 x2
dx
For problems 19 - 21 find the area shown in the figure.
19. 20. 21.
Chapter 3 The Integral Applied Calculus 203
Section 4: Substitution
We don’t have many integration rules. For quite a few of the problems we see, the rules won’t
directly apply; we’ll have to do some algebraic manipulation first. In practice, it is much harder
to write down the antiderivative of a function than it is to find a derivative. (In fact, it’s really
easy to write a function that doesn’t have any antiderivative you can find with algebra.)
The Substitution Method is one way of algebraically manipulating an integrand so that the rules
apply. This is a way to unwind the Chain Rule for derivatives. When you find the derivative of
a function using the Chain Rule, you end up with a product of something like the original
function TIMES a derivative. We can reverse this to write an integral:
d
f ( g ( x) ) = f ′(g ( x) )g ′( x)dx , so f (g ( x) ) = ∫ f ′(g ( x) )g ′( x)dx
dx
du
With substitution, we will substitute u = g (x) . This means = g ′(x) , so du = g ′( x)dx .
dx
Making this substitutions, ∫ f ′(g ( x))g ′( x)dx becomes ∫ f ′(u )du , which will probably be easier
to ingegrate.
Try Substitution when you see a product in your integral, especially if you recognize one factor
as the derivative of some part of the other factor.
The Substitution Method for Antiderivatives:
The goal is to turn ∫ f (g ( x)) dx into ∫ f (u ) du , where f(u) is much less messy than f(g(x)).
1. Let u be some part of the integrand. A good first choice is “one step inside the messiest
bit.”
du
2. Compute du = dx
dx
3. Translate all your x’s into u’s everywhere in the integral, including the dx. When you’re
done, you should have a new integral that is entirely in u. If you have any x’s left, then
that’s an indication that the substitution didn’t work or isn't complete; you may need to
go back to step 1 and try a different choice for u.
4. Integrate the new u-integral, if possible. If you still can’t integrate it, go back to step 1
and try a different choice for u.
5. Finally, substitute back x’s for u’s everywhere in your answer.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 3 The Integral Applied Calculus 204
Example 1
x
∫ 4 − x2
dx
Evaluate .
This integrand is more complicated than anything in our list of basic integral formulas, so we’ll
have to try something else. The only tool we have is substitution, so let’s try that!
1. Let u be some part of the integrand. A good first choice is “one step inside the messiest bit.”
In this case, the square root in the denominator is the messiest part, so let’s let u be one step
inside:
Let u = 4 − x 2
du
2. Compute du = dx
dx
du = −2 xdx
There is x dx in the integrand, so that’s a good sign; that will be −½du.
3. Translate all your x’s into u’s everywhere in the integral, including the dx.
x 1 1 1 1 1 1
∫ dx = ∫ (xdx ) = ∫ − du = − ∫ du = − ∫ u −1 / 2 du
4 − x2 4 − x2 u 2 2 u 2
Alternatively, we could have solved du = −2 xdx for dx and substituted that and simplified:
du
dx = , so
− 2x
x x du 1 1 1 1 1 −1 / 2
∫ 4 − x2 dx = ∫ 4 − x 2 − 2x ∫ u 2 2 ∫ u
= − du = − du = −
2∫
u du
4. Integrate the new u-integral, if possible.
1 1 u1/ 2
− ∫ u −1 / 2 du = − + C = −u 1 / 2 + C
2 2 1/ 2
5. Finally, undo our u = 4 − x 2 substitution, putting back x’s for u’s everywhere in your answer.
− u 1 / 2 + C = − 4 − x 2 + C . So we have found
x
∫ 4− x 2
dx = − 4 − x 2 + C .
How would we check this? By differentiating:
d
dx
(
− 4 − x2 + C =
d
dx
) 1/ 2
( 1
2
−1 / 2
)
− (4 − x 2 ) + C = − (4 − x 2 ) (− 2 x ) = x(4 − x 2 )
−1 / 2
=
x
4 − x2
.
Chapter 3 The Integral Applied Calculus 205
Example 2
e x dx
Evaluate ∫ (e x
+ 15 )
3
This integral is not in our list of building blocks. But notice that the derivative of e x + 15 (that
we see in the denominator) is just e x (which I see in the numerator), so substitution will be a
good choice for this.
du
Let u = e x + 15 . Then du = e x dx , and this integral becomes ∫u 3
= ∫ u −3 du .
u −2 1
∫ u du =
−3
Luckily, that is on our list of building block formulas: + C = − 2 + C.
−2 u
Finally, translating back:
e x dx 1
∫ e x + 15 3 = − e x + 15 2 + C
( ) ( )
Example 3
x2 x3 + 5
Evaluate a) ∫ x3 + 5
dx b) ∫ x 2 dx
a) This is not a basic integral, but the composition is less obvious. Here, we can treat the
denominator as the inside of the 1/x function.
du
Let u = x 3 + 5 . Then du = 3 x 2 dx . Solving for dx, dx = . Substituting,
3x 2
x2 x 2 du 1 du 1 1
∫ x 3 + 5 dx =∫ u 3x 2 =∫ u 3 = 3 ∫ u du .
Using our basic formulas,
1 1 1
∫
3 u
du = ln u
3
Undoing the substitution,
x2 1
∫ x 3 + 5 dx = 3 ln x + 5 + C
3
b) It is tempting to start this problem the same way we did the last, but if we try it will not
work, since the numerator of this fraction is not the derivative of the denominator. Instead, we
need to try a different approach. For this problem, we can use some basic algebra.
Chapter 3 The Integral Applied Calculus 206
x3 + 5 x3 5
∫ x2 dx = ( )
∫ x 2 + x 2 dx = ∫ x + 5 x dx
−2
We can integrate this using our basic rules, without needing substitution.
x2 x −1 1 2 5
∫ (
x + 5 x −2
)
dx =
2
+ 5
−1 2
= x − +C
x
Substitution and Definite Integrals
When you use substitution to help evaluate a definite integrals, you have a choice for how to
handle the limits of integration. You can do either of these, whichever seems better to you. The
important thing to remember is – the original limits of integration were values of the original
variable (say, x), not values of the new variable (say, u).
(a) You can solve the antiderivative as a side problem, translating back to x’s, and then use
the antiderivative with the original limits of integration. Or
(b) You can substitute for the limits of integration at the same time as you’re substituting for
everything inside the integral, and then skip the “translate back into x” step. If the
original integral had endpoints x =a and
x =b, and we make the substitution u = g(x ) and du = g'(x )dx, then the new integral
will have endpoints
u= g(a) and u=g(b) and
x=b u=g(b)
⌠ (original integrand) dx
⌡ becomes ⌠ (new integrand) du .
⌡
x=a u=g(a)
Method (a) seems more straightforward for most students. But it can involve some messy
algebra. Method (b) is often neater and usually involves fewer steps.
Example 4
1
4
Evaluate ⌠
⌡ (3x –1) dx
0
We’ll need substitution to find an antiderivative, so we’ll need to handle the limits of
integration carefully. Let's solve this example both ways.
(a) Doing the antiderivative as a side problem:
Chapter 3 The Integral Applied Calculus 207
Step One – find the antiderivative, using substitution:
∫ (3x − 1) dx
4
5
41 1u
Let u = 3 x − 1 . Then du = 3dx and ∫ (3x − 1) dx = ∫ u du =
4
+C
3 3 5
(3x − 1) 5
∫ (3x − 1) dx =
4
Translating back to x: +C
15
Step Two – evaluate the definite integral:
1
1
( ) (3x − 1)5 (3(1) − 1)5 − (3(0) − 1)5 32 − 1 33
∫0
4
3 x − 1 dx = = = − = .
15 0 15 15 15 15 15
(b) Substituting for the limits of integration:
1
∫ (3x − 1)
4
dx
0
Let u = 3 x − 1 . Then du = 3dx , and (substituting for the limits of integration) when x = 0,
u = -1, when x = 1, u = 2.
1 u u = 2 (2 ) (− 1)
5 5 5
x =1 u =2 32 − 1 33
∫ x =0 ( ) ∫u =−1 3 15 ]u =−1 = 15 − 15 = 15 − 15 = 15 .
4 4
3 x − 1 dx = u du =
Example 5
10
(ln x )6 dx
Evaluate ∫ 2
x
I can see the derivative of ln x in the integrand, so I can tell that substitution is a good choice.
1
Let u = ln x . Then du = dx . When x = 2 , u = ln 2 . When x = 10 , u = ln 10 . So the new
x
definite integral is
(ln x )6 dx = u =ln 10u 6 du = u 7
u = ln 10
( )
x =10
1
∫ ∫ = (ln 10 ) − (ln 2 ) ≅ 49.01.
7 7
x=2
x u = ln 2
7 u =ln 2 7
Chapter 3 The Integral Applied Calculus 208
3.4 Exercises
For problems 1-8, find the indicated antiderivative.
1
1. ∫ (4 x + 1) 3
dx 2. ∫ e100 x dx
e10 / x
∫ (1.0003) dt ∫ x 2 dx
12 t
3. 4.
5. ∫ w + 5 dw 6. ∫ 6 x 2 3 x 3 −1 dx
dx x−3
7. ∫ x ln x 8. ∫x 2
− 6x + 5
dx
For problems 9-12, find an antiderivative of the integrand and use the Fundamental Theorem to
evaluate the definite integral.
2 4 1
2x 1
⌠ x
9. ∫ 2
dx 10. ∫e
2x
dx 11. ⌡ (x – 2)3 dx
⌠ 12. ⌡ 1 – x2 dx
−2 1 + x 0 2 0
Chapter 3 The Integral Applied Calculus 209
Section 5: Additional Integration Techniques
Integration By Parts
Integration by parts is an integration method which enables us to find antiderivatives of some
new functions such as ln(x) as well as antiderivatives of products of functions such as x 2 ln( x)
and xe x .
If the function we're trying to integrate can be written as a product of two functions, u, and dv,
then integration by parts lets us trade out a complicated integral for hopefully simpler one.
INTEGRATION BY PARTS FORMULA
∫ udv = uv − ∫ vdu
For definite integrals,
b b
∫ udv = uv]a −∫ vdu
b
a a
Example 1
∫ xe dx
x
Integrate
To use the By Parts method, we break apart the product into two parts:
u = x and dv = e x dx
We now calculate du, the derivative of u, and v, the integral of dv.
d
du = x dx = 1dx and v = ∫ e x dx = e x .
dx
Using the By Parts formula,
∫ xe dx = uv − ∫ vdu = xe − ∫ e dx
x x x
Notice the remaining integral is simpler that the original, and one we can easily evaluate.
∫ xe dx = xe − ∫ e dx = xe − e + C
x x x x x
We could have chosen either x or e x as our u in the last example, but had we chosen e x , the
second integral would have become messier, rather than simpler.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 3 The Integral Applied Calculus 210
RULE OF THUMB
When selecting the u for By Parts, select a logarithmic expression if one
is present. If not, select an algebraic expression (like x or dx).
Example 2
4
∫ 6x
2
Integrate ln x dx
1
Since this contains a logarithmic expression, we'll use it for our u.
u = ln x and dv = 6 x 2 dx
We now calculate du and v.
1 x3
du = dx and v = ∫ 6 x 2 dx = 6 = 2 x 3 .
x 3
Using the By Parts formula,
4 4
∫ 6 x ln x dx = 2 x ln x 1 − ∫ 6 x
2 3 2
]
4 1
x
dx
1 1
We can simplify the expression in the integral on the right:
4 4
∫ 6x
2
]
4
ln x dx = 2 x ln x 1 − ∫ 6 xdx
3
1 1
The remaining integral is a basic one we can now evaluate.
4
∫ 6x
2
]
4
ln x dx = 2 x 3 ln x 1 − 3 x 2 ]
4
1
1
Finally, we can evaluate the expressions
4
∫ 6x
2
[( ) ( )] [ ]
ln x dx = 2 ⋅ 43 ln 4 − 2 ⋅ 13 ln 1 − (3 ⋅ 4 2 ) − (3 ⋅ 12 ) = 128 ln (4 ) − 45 ≈ 132.446
1
Integration Using Tables of Integrals
There are many techniques of integration we will not be studying. Many of them lead to general
formulas which can be compiled into a Table of Integrals - a type of cheat-sheet for integration.
Chapter 3 The Integral Applied Calculus 211
For example, here are two entries you might find in a table of integrals:
TABLE OF INTEGRAL EXAMPLES
1 1 x−a
∫x 2
−a 2
= ln
2a x + a
+C
1
∫ 2
x +a 2
= ln x + x 2 + a 2 + C
Example 3
5
Integrate ∫x 2
−9
dx
This integral looks very similar to the form of the first integral in the examples table. By
employing the rule that allows us to pull out constants, and by rewriting 9 as 32, we can better
see the match.
5 1
∫ x 2 − 9dx = 5∫ x 2 − 32 dx
Now we simply use the formula from the table, with a = 3.
5 1 1 x−3 5 x−3
∫x 2
−9
dx = 5∫ 2
x −3 2
dx = 5 ln = ln
2⋅3 x + 3 6 x + 3
+C
Sometimes we have to combine the table with other techniques we've learned, like substitution.
Example 4
x2
Integrate ∫ x 6 + 16
dx
This integral looks somewhat like the second integral in the example table, but the power of x is
incorrect, and there is an x2 in the numerator which does not match. Trying to utilize this rule,
( )
we can try to rewrite the denominator to look like (something)2. Luckily, x 6 = x 3
2
x2 x2
∫ dx = ∫ dx
x 6 + 16 (x )3 2
+ 16
Now we can use substitution, letting u = x 3 , so du = 3 x 2 dx .
Chapter 3 The Integral Applied Calculus 212
Making the subsitution,
x2 1 du 1 1
∫ 3 2 dx = ∫ u 2 + 16 3 = 3 ∫ u 2 + 16 du
( )
x + 16
Now we can use the table entry.
1 1 1
3 ∫ u 2 + 16
du = ln u + u 2 + 16 + C
3
Undoing the substitution,
x2 1
∫ x 6 + 16 dx = 3 ln x + x + 16 + C
3 6
3.5 Exericses
In problems 1–4, a function u or dv is given. Find the piece u or dv which is not given,
calculate du and v, and apply the Integration by Parts Formula.
⌠ 12x.ln(x) dx
1. ⌡ u = ln(x) ⌠ x.e–x dx
2. ⌡ u=x
4 dv = x4 dx 19
3. ⌠
⌡ x ln(x) dx ⌡ x.(5x + 1) dx
4. ⌠ u=x
In problems 5 - 10 evaluate the integrals
1 1 3
x
⌠
5. ⌡ dx ⌠ 10x.e3x dx
6. ⌡ ⌠ ln(2x + 5) dx
7. ⌡
3x
0 e 0 1
2
3 ln(x)
8. ⌠
⌡ x ln(5x) dx 9. ⌠
⌡ x ln(x + 1) dx 10. ⌠
⌡ 2 dx
1 x
For problems 11 - 14 integrate each function.
1 2
11. ⌠
⌡ 12. ⌠
⌡ 13. ⌠
⌡ 4 + x2 14. ⌠
⌡ 9 + x2
4 – x2 9 – x2
Chapter 3 The Integral Applied Calculus 213
Section 6: Area, Volume, and Average Value
Area
We have already used integrals to find the area between the graph of a function and the
horizontal axis. Integrals can also be used to find the area between two graphs.
If f(x) ≥ g(x) for all x in [a,b], then we can approximate the area between f and g by partitioning
the interval [a,b] and forming a Riemann sum, as shown in the picture. The height of each
rectangle is top – bottom, f(ci) – g(ci) so the area of the ith rectangle is
(height).(base) = (f(ci) – g(ci)).∆x . Adding up this rectangles gives an approximation of the total
n
area as ∑ ( f (c ) − g (c ))∆x , a Riemann sum.
i =1
i i
The limit of this Riemann sum, as the number of rectangles gets larger and their width gets
b
smaller, is the definite integral ∫ ( f (x ) − g (x )) dx .
a
The area between two curves f(x) and g(x), where f(x) ≥ g(x), between x = a and x = b
is
b
∫ ( f (x ) − g (x )) dx
a
The integrand is “top – bottom.” Make a graph to be sure which curve is which.
Example 1
Find the area bounded between the graphs of f(x) = x and g(x) = 3 for 1 ≤ x ≤ 4.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 3 The Integral Applied Calculus 214
Always start with a graph so you can see which graph is the top and which is the bottom. In
this example, the two curves cross, and they change positions; we’ll need to split the area into
two pieces. Geometrically, we can see that the area is 2 + ½ = 2.5.
Writing the area as a sum of definite integrals, we get:
3 4
Area = ∫ (3 − x ) dx + ∫ (x − 3) dx
1 3
These integrals are easy to evaluate using antiderivatives:
3
2
(3 − x ) dx = 3x − x = 9 − 9 − 15 − 25 = 2.
3
∫1
2 0 2 2
4
4 x2 16 9 1
∫3 ( x − 3) dx = − 3x = − 12 − − 9 = .
2 3 2 2 2
The two integrals also tell us that the total area between f and g is 2.5 square units, which
we already knew.
4
Note that the single integral ∫ (3 − x ) dx = 1.5
1
is not the area we want in the last example. The
value of the integral is 1.5, and the value of the area is 2.5. That’s because for the triangle on
the right, the graph of y = x is above the graph of y = 3, so the integrand 3 – x is negative; in the
definite integral, the area of that triangle comes in with a negative sign.
In this example, it was easy to see exactly where the two curves crossed so we could break the
region into the two pieces to figure separately. In other examples, you might need to solve an
equation to find where the curves cross.
Example 2
Two objects start from the same location and travel along the
same path with velocities v A (t ) = t + 3 and v B (t ) = t 2 − 4t + 3
meters per second. How far ahead is A after 3 seconds?
Since v A (t ) ≥ v B (t ) , the "area" between the graphs of v A (t )
and v B (t ) represents the distance between the objects.
After 3 seconds, the distance apart
3
= ∫ (v A (t ) − v B (t )) dt = ∫
0
3
0
((t + 3) − (t 2
)) 3
( )
− 4t + 3 dt = ∫ 5t − t 2 dt
0
3
5 t3 5 27
= t 2 − = ⋅ 9 − − (0 ) = 13.5 meters.
2 3 0 2 3
Chapter 3 The Integral Applied Calculus 215
Volume
Just as we can partition an interval and imagine approximating an area with rectangles to find a
formula for the area between curves, we can partition an interval and imagine approximating a
volume with simple shapes to find a formula for the volume of a solid. While this approach
works for a variety of shapes, our focus will be on shapes formed by revolving a curve around
the horizontal axis.
We start with an area, the region below a function on the interval a ≤ x ≤ b. We are going to take
that region, and rotate it around the x axis, creating the solid shape shown.
Rotate about
the axis
a b
To find the volume of this solid, we can start by partitioning the
interval [0,1] and approximating the area with rectangles. As before,
the width of each rectangle would be ∆x and the height f(ci).
a b
If we took just one of these rectangles and rotated it about the
horizontal axis, it would form a cylindrical shape. The radius of that
cylinder would be f(ci), so the volume would be V = πr 2 h = π ( f (ci ) ) ∆x
2
The volume of the whole solid could be approximated by rotating each
of the rectangles about the x axis. Adding up the volume of each of the
little cylindrical discs gives an approximation of the total volume as a
n
∑ π ( f (c )) ∆x , a Riemann sum.
2 b
i
i =1
The limit of this sum as the width of the rectanges becomes small is the
b
definite integral ∫ π ( f ( x ))2 dx .
a
The volume of the solid obtained by rotating about the x-axis the area bounded by the
curve f(x), the x-axis, x = a, and x = b is
b
∫ π ( f (x ))
2
dx
a
Chapter 3 The Integral Applied Calculus 216
Example 3
Find the volume of the solid formed by rotating the area under f (x ) = e − x on the interval [0,1]
about the x-axis.
This is the region pictured in the earlier example. We substitute 1
in the function and bounds into the formula we derived to set up
the definite integral.
∫ π (e )
1
Volume = −x 2
dx
0
Using exponent rules, the integrand can be simplified. The 0 1
constant π can be pulled out of the integral.
1
π ∫ e −2 x dx
0
Using the substitution u = -2x, we can integrate this function.
1
1 1 1 1
π∫ e −2 x
dx = − πe −2 x = − πe −2 (1) − − πe −2 ( 0 ) ≈ 1.358 cubic units
0 2 0 2 2
Average Value
We know the average of n numbers, a1, a2, . . . , an , is their sum divided by n. But what if we need
to find the average temperature over a day’s time -- there are too many possible temperatures to add
them up. This is a job for the definite integral.
The average value of a function f(x) on the interval [a, b] is given by
1 b
f (x ) dx
b − a ∫a
The average value of a positive f has a nice
geometric interpretation. Imagine that the area
under f (Fig. a) is a liquid that can "leak"
through the graph to form a rectangle with the
same area (Fig. b).
If the height of the rectangle is H, then the area
of the rectangle is H ⋅ (b − a ) . We know the area of the rectangle is the same as the area under
1 b
f ( x ) dx , the average value of f on [a,b].
b
f so H ⋅ (b − a ) = ∫ f ( x ) dx . Then H =
a b − a ∫a
The average value of a positive function f is the height H of the rectangle whose area is the
same as the area under f.
Chapter 3 The Integral Applied Calculus 217
Example 4
During a 9 hour work day, the production rate at time t hours after the start of the shift was
given by the function r (t ) = 5 + t cars per hour. Find the average hourly production rate.
The average hourly production is
1
∫
9−0 0
9
( )
5 + t dt = 7 cars per hour.
A note about the units – remember that the definite integral has units (cars per hour) ∙ (hours) =
cars. But the 1/(b-a) in front has units 1/hours – the units of the average value are cars per hour,
just what we expect an average rate to be.
In general, the average value of a function will have the same units as the integrand.
Function averages, involving means and more complicated averages, are used to "smooth" data
so that underlying patterns are more obvious and to remove high frequency "noise" from signals.
In these situations, the original function f is replaced by some "average of f." If f is rather
1 x +5
jagged time data, then the ten year average of f is the integral g ( x ) = ∫ f (t ) dt , an average
10 x −5
of f over 5 units on each side of x. For
example, the figure here 1 shows the graphs of a
Monthly Average (rather “noisy” data) of surface
temperature data, an Annual Average (still rather
“jagged), and a Five Year Average (a much
smoother function). Typically the average
function reveals the pattern much more clearly
than the original data. This use of a “moving
average” value of “noisy” data (weather
information, stock prices) is a very common.
Example 5
1
http://commons.wikimedia.org/wiki/File:Short_Instrumental_Temperature_Record.png, CC-BY
Chapter 3 The Integral Applied Calculus 218
The graph to the right shows the amount
of water in a reservoir over a 12 hour
period. Estimate the average amount of
water in the reservoir over this period.
If V (t ) is the volume of the water (in
millions of liters) after t hours, then the
12
1
V (t ) dt . In order
12 ∫0
average amount is
to find the definite integral, we’ll have to
estimate. I’ll use 6 rectangles, and I’ll take the heights from their right edges.
My estimate of the integral is
12
∫ V (t ) dt ≅(18)(2) + (9.7 )(2) + (8.2)(2) + (12)(2) + (19.9)(2) + (22)(2) = 179.6 .
0
The units of this integral are millions of liters × feet. So my estimate of the average volume is
179.6
≅ 15 millions of liters. Your estimate might be a little different.
12
In the figure below, you can see the same graph with the line y = 15 drawn in. The area under
the curve and the area under the
rectangle are (approximately) the
same.
In fact, that would be a different way
to estimate the average value. We
could have estimated the placement of
the horizontal line so that the area
under the curve and under the line
were equal.
Chapter 3 The Integral Applied Calculus 219
3.6 Exercises
In problems 1 – 4, use the values in the table to estimate the areas.
x f (x ) g (x ) h( x )
0 5 2 5
1 6 1 6
2 6 2 8
3 4 2 6
4 3 3 5
5 2 4 4
6 2 0 2
1. Estimate the area between f and g, between x = 0 and x = 4.
2. Estimate the area between g and h, between x = 0 and x = 6.
3. Estimate the area between f and h, between x = 0 and x = 4.
4. Estimate the area between f and g, between x = 0 and x = 6.
5. Estimate the area of the island shown
In problems 6 – 15, find the area between the graphs of f
and g for x in the given interval. Remember to draw
the graph!
6. f(x) = x2 + 3 , g(x) = 1 and –1 ≤ x ≤ 2.
7. f(x) = x2 + 3 , g(x) = 1 + x and 0 ≤ x ≤ 3.
8. f(x) = x2 , g(x) = x and 0 ≤ x ≤ 2.
9. f(x) = (x –1)2 , g(x) = x + 1 and 0 ≤ x ≤ 3.
1
10. f(x) = x , g(x) = x and 1 ≤ x ≤ e.
11. f(x) = x , g(x) = x and 0 ≤ x ≤ 4.
12. f(x) = 4 – x 2 , g(x) = x + 2 and 0 ≤ x ≤ 2.
13. f(x) = ex , g(x) = x and 0 ≤ x ≤ 2.
14. f(x) = 3 , g(x) = 1 – x2 and 0 ≤ x ≤ 1.
15. f(x) = 2 , g(x) = 4 – x2 and –2 ≤ x ≤ 2.
Chapter 3 The Integral Applied Calculus 220
For problems 16-18, find the volume of the solid obtained by rotating the specified region about
the x axis.
16. Region under f(x) = x2 + 3 for –1 ≤ x ≤ 2.
17. Region under f(x) = 4 – x 2 for 0 ≤ x ≤ 2.
1
18. Region under f(x) = x for 1 ≤ x ≤ 2.
In problems 19 and 20 use the values in the table to estimate the average values.
x f (x ) g (x )
0 5 2
1 6 1
2 6 2
3 4 2
4 3 3
5 2 4
6 2 0
19. Estimate the average value of f on the interval [0, 6].
20. Estimate the average value of g on the interval [0, 6].
In problems 21 – 26, find the average value of f on the given interval.
21. f(x) from the graph for 0 ≤ x ≤ 2.
22. f(x) from the graph for 0 ≤ x ≤ 4.
23. f(x) from the graph for 1 ≤ x ≤ 6.
24. f(x) from the graph for 4 ≤ x ≤ 6.
25. f(x) = 2x + 1 for 0 ≤ x ≤ 4.
26. f(x) = x 2 for 0 ≤ x ≤ 2.
27. The graph shows the velocity of a car during a 5 hour trip.
(a) Estimate how far the car traveled during the 5 hours.
(b) At what constant velocity should you drive in order to
travel the same distance in 5 hours?
28. The graph shows the number of telephone calls per minute
at a large company. Estimate the average number of calls
per minute
(a) From 8 am to 5 pm.
(b) From 9 am to 1 pm.
Chapter 3 The Integral Applied Calculus 221
Section 7: Applications to Business
Consumer and Producer Surplus
Here are a demand and a supply curve for a product. Which is which?
The demand curve is decreasing – lower prices are associated with higher quantities demanded,
higher prices are associated with lower quantities demanded. Demand curves are often shown as
if they were linear, but there’s no reason they have to be.
The supply curve is increasing – lower prices are associated with lower supply, and higher prices
are associated with higher quantities supplied.
The point where the demand and supply curve cross is called the equilibrium point (q*, p*).
supply
Consumer p=s(q)
Surplus
(q*, p*)
Producer demand
Surplus p=d(q)
Suppose that the price is set at the equilibrium price, so that the quantity demanded equals the
quantity supplied. Now think about the folks who are represented on the left of the equilibrium
point. The consumers on the left would have been willing to pay a higher price than they ended
up having to pay, so the equilibrium price saved them money. On the other hand, the producers
represented on the left would have been willing to supply these goods for a lower price – they
made more money than they expected to. Both of these groups ended up with extra cash in their
pockets!
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 3 The Integral Applied Calculus 222
Graphically, the amount of extra money that ended up in consumers’ pockets is the area between
the demand curve and the horizontal line at p*. This is the difference in price, summed up over
all the consumers who spent less than they expected to – a definite integral. Notice that since the
area under the horizontal line is a rectangle, we can simplify the area integral:
q* q* q* q*
∫ (d (q ) − p *) dq = ∫
0 0
f (q )dq − ∫
0
p * dq = ∫ d (q )dq − p * q *
0
The amount of extra money that ended up in producers’ pockets is the area between the supply
curve and the horizontal line at p*. This is the difference in price, summed up over all the
producers who received more than they expected to. Similar to consumer surplus, this integral
can be simplified:
q* q* q* q*
∫ ( p * − s(q )) dq = ∫
0 0
p * dq − ∫ s (q )dq = p * q * − ∫ s (q )dq
0 0
Consumer and Producer Surplus
Given a demand function p = d(q) and a supply function p = s(q), and the equilibrium point
(q*, p*)
q*
The consumer surplus = ∫ d (q ) dq − p * q *
0
q*
The producer surplus = p * q * − ∫ s (q ) dq
0
The sum of the consumer surplus and producer surplus is the total gains from trade.
What are the units of consumer and producer surplus?
The units are (price per item)(quantity of items) = money!
Example 1
Suppose the demand for a product is given by p = d (q) = −0.8q + 150 and the supply for the
same product is given by p = s (q) = 5.2q . For both functions, q is the quantity and p is the
price, in dollars.
a. Find the equilibrium point.
b. Find the consumer surplus at the equilibrium price.
c. Find the producer surplus at the equilibrium price.
a. The equilibrium point is where the supply and demand functions are equal. Solving
− 0.8q + 150 = 5.2q gives q = 25.
The price when q = 25 is p = 130; the equilibrium point is (25, 130).
25
b. The consumer surplus is: ∫ (− 0.8q + 150)dq − (130)(25) = $250.
0
25
c. The producer surplus is: (130 )(25) − ∫ 5.2qdq = $1625.
0
Chapter 3 The Integral Applied Calculus 223
Example 2
The tables below show information about the demand and supply functions for a product. For
both functions, q is the quantity and p is the price, in dollars.
q 0 100 200 300 400 500 600 700
p 70 61 53 46 40 35 31 28
q 0 100 200 300 400 500 600 700
p 14 21 28 33 40 47 54 61
a. Which is which? That is, which table represents demand and which represents supply?
b. What is the equilibrium price and quantity?
c. Find the consumer and producer surplus at the equilibrium price.
a. The first table shows decreasing price associated with increasing quantity; that is the demand
function.
b. For both functions, q = 400 is associated with p = 40; the equilibrium price is $40 and the
equilibrium quantity is 400 units. Notice that we were lucky here, because the equilibrium
point is actually one of the points shown. In many cases with a table, we would have to
estimate.
c. The consumer surplus uses the demand function, which comes from the first table. We’ll
have to approximate the value of the integral using rectangles. There are 4 rectangles, and I
choose to use left endpoints.
400
The consumer surplus = ∫ (demand) dq − (40)(400) ≅
0
(70)(100) + (61)(100) + (53)(100) + (46)(100) − (40)(400) = 7000 .
The consumer surplus is about $7,000.
The producer surplus uses the supply function, which comes from the second table. I choose to
use left endpoints for this integral also.
400
The producer surplus = (40)(400) − ∫ (supply) dq ≅
0
(40)(400) − [(14)(100) + (21)(100) + (28)(100) + (33)(100)] = 6400 .
The producer surplus is about $6400.
Chapter 3 The Integral Applied Calculus 224
Continuous Income Stream
In precalculus, you learned about compound interest in that really simple situation where you
made a single deposit into an interest-bearing account and let it sit undisturbed, earning interest,
for some period of time. Recall:
Compound Interest Formulas
Let P = the principal (initial investment), r = the annual interest rate expressed as a
decimal, and let A(t) be the amount in the account at the end of t years.
nt
r
Compounding n times per year: A(t ) = P1 +
n
Compounded continuously: A(t ) = Pe rt
If you’re using this formula to find what an account will be worth in the future, t > 0 and
A(t) is called the future value.
If you’re using the formula to find what you need to deposit today to have a certain value
P sometime in the future, t < 0 and A(t) is called the present value.
You may also have learned somewhat more complicated annuity formulas to deal with slightly
more complicated situations – where you make equal deposits equally spaced in time.
But real life is not usually so neat.
Calculus allows us to handle situations where “deposits” are flowing continuously into an
account that earns interest. As long as we can model the flow of income with a function, we can
use a definite integral to calculate the present and future value of a continuous income stream.
The idea is – each little bit of income in the future needs to be multiplied by the exponential
function to bring it back to the present, and then we’ll add them all up (a definite integral).
Continuous Income Stream
Suppose money can earn interest at an annual interest rate of r, compounded continuously.
Let F(t) be a continuous income function (in dollars per year) that applies between
year 0 and year T.
T
Then the present value of that income stream is given by PV = ∫ F (t )e − rt dt .
0
The future value can be computed by the ordinary compound interest formula FV = PVe rt
This is a useful way to compare two investments – find the present value of each to see which is
worth more today.
Chapter 3 The Integral Applied Calculus 225
Example 3
You have an opportunity to buy a business that will earn $75,000 per year continuously over the
next eight years. Money can earn 2.8% per year, compounded continuously. Is this business
worth its purchase price of $630,000?
First, please note that we still have to make some simplifying assumptions. We have to assume
that the interest rates are going to remain constant for that entire eight years. We also have to
assume that the $75,000 per year is coming in continuously, like a faucet dripping dollars into
the business. Neither of these assumptions might be accurate.
But moving on:
The present value of the $630,000 is, well, $630,000. This is one investment, where we put our
$630,000 in the bank and let it sit there.
To find the present value of the business, we think of it as an income stream. The function F(t)
in this case is a constant $75,000 dollars per year, so F(t) = 75,000 . The interest rate is 2.8%
and the term we're interested in is 8 years, so r = .028, and T = 8:
8
PV = ∫ 75000e −.028t dt ≅ 672,511.66
0
The present value of the business is about $672,500, which is more than the $630,000 asking
price – this is a good deal.
While this integral could have been done using substitution, for many of the integrals in this
section, we don't have the techniques to use antiderivatives, or in some cases no antiderivative
exists. Technology will work quickly, and it will give you an answer that’s good enough.
Example 4
A company is considering purchasing a new machine for its production floor. The machine
costs $65,000. The company estimates that the additional income from the machine will be a
constant $7000 for the first year, then will increase by $800 each year after that. In order to buy
the machine, the company needs to be convinced that it will pay for itself by the end of 8 years
with this additional income. Money can earn 1.7% per year, compounded continuously.
Should the company buy the machine?
We’ll assume that the income will come in continuously over the 8 years. We’ll also assume
that interest rates will remain constant over that 8-year time period.
We’re interested in the present value of the machine, which we will compare to its $65,000
price tag. Let t be the time, in years, since the purchase of the machine. The income from the
machine is different depending on the time.
Chapter 3 The Integral Applied Calculus 226
From t = 0 to t = 1 (the first year), the income is constant $7000 per year. From t = 1 to t = 8,
the income is increasing by $800 each year; the income flow function F(t) will be
F (t ) = 7000 + 800(t − 1) = 6200 + 800t . To find the present value, we’ll have to divide the
integral into the two pieces, one for each of the functions:
1 8
PV = ∫ 7000e − 0.017 t
dt + ∫ (6200 + 800t )e −0.017 t dt ≅ 70166 .
0 1
The present value is greater than the cost of the machine, so the company should buy the
machine.
3.7 Exercises
1. The demand and supply functions for a certain product are given by p = 150 − .5q and
p = .002q 2 + 1.5 , where p is in dollars and q is the number of items.
(a) Which is the demand function?
(b) Find the equilibrium price and quantity
(c) Find the total gains from trade at the equilibrium price.
2. Still thinking about the product from Exercise 1, with its demand and supply functions,
suppose the price is set artificially at $70 (which is above the equilibrium price).
(a) Find the quantity supplied and the quantity demanded at this price.
(b) Compute the consumer surplus at this price, using the quantity demanded.
(c) Compute the producer surplus at this price, using the quantity demanded (why?).
(d) Find the total gains from trade at this price.
(e) What do you observe?
3. When the price of a certain product is $40, 25 items can be sold. When the price of the same
product costs $20, 185 items can be sold. On the other hand, when the price of this product
is $40, 200 items will be produced. But when the price of this product is $20, only 100 items
will be produced. Use this information to find supply and demand functions (assume for
simplicity that the functions are linear), and compute the consumer and producer surplus at
the equilibrium price.
4. Find the present and future values of a continuous income stream of $5000 per year for 12
years if money can earn 1.3% annual interest compounded continuously.
5. Find the present value of a continuous income stream of $40,000 per year for 35 years if
money can earn
(a) 0.8% annual interest, compounded continuously,
(b) 2.5% annual interest, compounded continuously,
(c) 4.5% annual interest, compounded continuously.
Chapter 3 The Integral Applied Calculus 227
6. Find the present value of a continuous income stream F (t ) = 20 + t , where t is in years and F
is in tens of thousands of dollars per year, for 10 years, if money can earn 2% annual interest,
compounded continuously.
7. Find the present value of a continuous income stream F (t ) = 12 + 0.3t 2 , where t is in years and
F is in thousands of dollars per year, for 8 years, if money can earn 3.7% annual interest,
compounded continuously.
8. Find the future value of a continuous income stream F (t ) = 8500 + 640t + 100 , where t is in
years and F is in dollars per year, for 15 years, if money can earn 6% annual interest,
compounded continuously.
9. A business is expected to generate income at a continuous rate of $25,000 per year for the next
eight years. Money can earn 3.4% annual interest, compounded continuously. The business
is for sale for $153,000. Is this a good deal?
Chapter 3 The Integral Applied Calculus 228
Section 8: Differential Equations
A differential equation is an equation involving the derivative of a function. They allow us to
express with a simple equation the relationship between a quantity and it's rate of change.
Example 1
A bank pays 2% interest on its certificate of deposit accounts, but charges a $20 annual fee.
Write an equation for the rate of change of the balance, B ′(t ) .
If the balance B(t ) has units of dollars, then B ′(t ) has units of dollars per year. When we think
of what is changing the balance of the account, there are two factors:
1) The interest, which increases the balance, and
2) The fee, which decreases the balance.
Considering the interest, we know each year the balance will increase by 2%, but 2% of what?
Each year that will change, since we earn interest on whatever the current balance is. We can
represent the amount of increase as 2% of the balance: 0.02 B(t ) dollars/year.
The fee already has the units of dollars/year. Since everything now has the same units, we can
put the two together, and create the equation:
B′(t ) = 0.02 B(t ) − 20
The result is an example of a differential equation. Notice this particular equation involves both
the derivative and the original function, and so we can't simplify find B(t ) using basic
integration.
Algebraic equations contain constants and variables, and the solutions
of an algebraic equation are typically numbers. For example, x = 3
and x = –2 are solutions of the algebraic equation x2 = x + 6.
Differential equations contain derivatives or differentials of functions.
Solutions of differential equations are functions. The differential
equation y' = 3x2 has infinitely many solutions, and two of those
solutions are the functions y = x3 + 2 and y = x3 – 4.
You have already solved lots of differential equations: every time you
found an antiderivative of a function f(x), you solved the differential
equation y' = f(x) to get a solution y. The differential equation
y' = f(x) , however, is just the beginning. Other applications generate
different differential equations, like in the bank balance example
above.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 3 The Integral Applied Calculus 229
Checking Solutions of Differential Equations
Whether a differential equation is easy or difficult to solve, it is important to be able to check
that a possible solution really satisfies the differential equation.
A possible solution of an algebraic equation can be checked by putting the solution into the
equation to see if the result is true: x = 3 is a solution of 5x + 1 = 16 since 5(3) + 1 = 16 is
true. Similarly, a solution of a differential equation can be checked by substituting the function
and the appropriate derivatives into the equation to see if the result is true: y = x2 is a solution
of xy' = 2y since y' = 2x and x(2x) = 2(x2) is true.
Example 2
Check (a) that y = x2 + 5 is a solution of y'' + y = x2 + 7 and
y
(b) that y = x + 5/x is a solution of y' + x = 2.
(a) y = x2 + 5 so y' = 2x and y'' = 2. Substituting these functions for y and y'' into the
differential equation y'' + y = x2 + 7, we have y'' + y = (2) + (x2 + 5) = x2 + 7, so y = x2 + 5 is
a solution of the differential equation.
(b) y = x + 5/x so y' = 1 – 5/x2 . Substituting these functions for y and y ' in the
y
differential equation y' + = 2 , we have
x
y 1
y' + = (1 – 5/x2) + (x + 5/x) = 1 – 5/x2 + 1 + 5/x2 = 2, the result we wanted to verify.
x x
Separable Differential Equations
A differential equation is called separable if the variables can be separated algebraically so that
all the x's and dx are one one side of the equation, and all the y's and dy are on the other side of
the equation. In other words, so the equation has the form f ( x)dx = g ( y )dy .
Once separated, separable differential equations can be solved by integrating both sides of the
equation.
Example 3
6x + 1
Find the solution of y ′ =
2y
Rewriting y' is a helpful first step.
dy 6 x + 1
=
dx 2y
Chapter 3 The Integral Applied Calculus 230
Now we can multiply both sides by dx and by 2y to separate the variables.
2 ydy = (6 x + 1)dx
Integrating each side,
∫ 2 ydy = ∫ (6 x + 1)dx
y 2 + C1 = 3x 2 + x + C2
Notice that we can combine the two constants to create a new, consolidated constant C, so we
usually only bother to put a constant on the right side.
y 2 = 3x 2 + x + C
As expected, there is a whole family of solutions to this differential equation.
An initial value problem is a differential equation that provides additional information about the
initial, or starting, value of the function. This allows us to then solve for the constant and find a
single solution.
Example 4
6x + 1
Find the solution of y ′ = , which satisfies y (2) = 3
2y
In the previous example we found the general solution, y 2 = 3 x 2 + x + C .
Substituting in the initial condition, y = 3 when x = 2,
32 = 3(2) 2 + 2 + C , so 9 = 12 + 2 + C , giving C = −5 .
The solution is y 2 = 3 x 2 + x − 5 . Sometimes it is desirable to solve for y.
y = ± 3 x 2 + x − 5 , but since the initial condition had a positive y value, we isolate the solution
y = 3x 2 + x − 5
Example 5
A bank pays 2% interest on its certificate of deposit accounts, but charges a $20 annual fee. If
you initially invest $3,000, how much will you have after 10 years?
You may recognize this as the example from the beginning of the section, for which we set up
dB
the equation B′(t ) = 0.02 B (t ) − 20 , or more simply, = 0.02 B − 20
dt
We can separate this equation by multiply by dt and dividing by the entire expression on the
right.
Chapter 3 The Integral Applied Calculus 231
dB
= dt
0.02 B − 20
Integrating the left side of this equation requires substitution. Let u = 0.02 B − 20 , so
u = 0.02dB . Making the substitution,
dB du / 0.02 1 du 1 1 1 1
∫ 0.02 B − 20 = ∫ u = ∫ u 0.02 = 0.02 ∫ u = 0.02 ln u = 0.02 ln 0.02 B − 20
Integrating on the right side of the differential equation is comparably easier:
∫ dt = t + C
Together, this gives us the general solution to the differential equation:
1
ln 0.02 B − 20 = t + C
0.02
Now we would like to solve for B. Start by multiplying by 0.02.
ln 0.02 B − 20 = 0.02t + 0.02C We can rename D = 0.02C for simplicity
ln 0.02 B − 20 = 0.02t + D Exponentiate both sides: eleft = eright
ln 0.02 B − 20
e = e 0.02t + D Use the log rule: e ln A = A
0.02 B − 20 = e 0.02t + D Since the right side is always positive, we can drop
the absolute value sign.
0.02 B − 20 = e 0.02t + D Using the rule e A+ B = e A + e B
0.02 B − 20 = e 0.02t e D Rename k = eD
0.02 B − 20 = ke 0.02t Add 20 and divide by 0.02
ke 0.02t + 20 ke 0.02t
B= = + 1000 Rename A = k/0.02
0.02 0.02
B = Ae 0.02t + 1000
Finally, we can substitute our initial value of B = 3000 when t = 0 to solve for the constant A:
3000 = Ae 0.02 ( 0 ) + 1000
A = 2000
This gives us the equation for the account balance after t years: B = 2000e 0.02t + 1000
To find the balance after 10 years, we can evaluate this equation at t = 10.
B(10) = 2000e 0.02 (10 ) + 1000 ≈ $3442.81
It's worth noting that this answer isn't exactly right. Differential equations assume continuous
changes, and it is unlikely interest is compounded continuously or the fee is extracted
continuously. However, the answer is likely close to the actual answer, and differential
equations provide a relatively simple model of a complicated situation.
Chapter 3 The Integral Applied Calculus 232
Models of Growth
The bank account example demonstrated one basic model of growth: growth proportional to the
existing quantity. Bank accounts and populations both tend to grow this way if not constrained.
This type of growth is called unlimited growth.
Unlimited Growth
If a quantity or population y grows at a rate proportional that quantity's size, it can be
modeled with unlimited growth, which has the differential equation:
y ′ = ry , where r is a constant
Example 6
A population grows by 8% each year. If the current population is 5,000, find an equation for
the population after t years.
The population is growing by a percent of the current population, so this is unlimited growth.
dy
= 0.08 y Separate the variables
dt
1
dy = 0.08dt Integrate both sides
y
ln y = 0.08t + C Exponentiate both sides
ln y
e = e 0.08t +C Simplify both sides, using the tricks we used in the bank example
y = Ae 0.08t
Now substitute in the initial condition
5000 = Ae 0.08( 0 ) , so A = 5000.
The population will grow following the equation y = 5000e 0.08t .
Notice that the solution to the unlimited growth equation is an exponential equation.
When a product is advertised heavily, sales will tend to grow very quickly, but eventually the
market will reach saturation, and sales will slow. In this type of growth, called limited growth,
the population grows at a rate proportional to the distance from the maximum value.
Limited Growth
If a quantity grows at a rate proportional to the distance from the maximum value, M, it can
be modeled with limited growth, which has the differential equation:
y ′ = k ( M − y ) , where k is a constant, and M is the maximum size of y.
Chapter 3 The Integral Applied Calculus 233
Example 7
A new cell phone is introduced. The company estimates they will sell 200 thousand phones.
After 1 month they have sold 20 thousand. How many will they have sold after 9 months?
In this case there is a maximum amount of phones they expect to sell, so M = 200 thousand.
Modeling the sales, y, in thousands of phones, we can write the differential equation
y ′ = k (200 − y )
Since it was a new phone, y (0) = 0 . We also know the sales after one month, y (1) = 20 .
Solving the differential equation,
dy
= k (200 − y ) Separate the variables
dt
dy
= kdt Integrate both sides. On the left use the substitution u = 200 − y
200 − y
− ln 200 − y = kt + C Multiply both sides by -1, and exponentiate both sides
ln 200 − y
e = e − kt −C Simplify
200 − y = Be − kt Subtract 200, divide by -1, and simplify
y = Ae −kt + 200
Using the initial condition y (0) = 0 ,
0 = Ae −k ( 0 ) + 200 , so 0 = A + 200 , giving A = −200
Using the value y (1) = 20 ,
20 = −200e −k (1) + 200 Subtract 200 and divide by -200
− 180
= 0.9 = e −k Take the ln of both sides
− 200
ln 0.9 = ln e − k = −k Divide by -1
k = − ln 0.9 ≈ 0.105
As a quick sanity check, this value is positive as we would expect, indicating that the sales are
growing over time. We now have the equation for the sales of phones over time:
A = −200e −0.105t + 200
Finally, we can evaluate this at t = 9 to find the sales after 9 months.
A = −200e −0.105(9 ) + 200 ≈ 122.26 thousand phones.
Limited growth is also commonly used for learning models, since when learning a new skill,
people typically learn quickly at first, then their rate of improvement slows down as they
approach mastery.
Chapter 3 The Integral Applied Calculus 234
Example 7 ** FIX**
Jim is learning a new set of product codes for work. Each day he studies, and tests his recall.
Suppose that after 4 days, Jim has mastered 70% of the new codes. How long will it take for
him to master 95% of the codes, if the limited growth model applies.
In this case there is a maximum amount of mastery possible, so M = 100%. Modeling his
learning, L, as percent of mastery, we can write the differential equation
=L′ k (100 − L)
=
Solving as we did in the previous example, we obtain L Ae − kt + 100 .
Since he started not knowing any of the codes L(0) = 0 , so
=0 Ae − k ⋅0 + 100 , so A = −100 .
Using that Jim mastered 70 of the codes in 4 days, L(4) = 70 , so
70 = −100e − k ⋅4 + 100
−30 = −100e − k ⋅4
0.3 = e − k ⋅4
( )
ln ( 0.3) = ln e − k ⋅4 = −4k
ln ( 0.3)
=k ≈ 0.301
−4
Now we have the equation L = −100e −0.301t + 100 , we can solve for when L = 95.
95 = −100e −0.301t + 100
−5 =−100e −0.301t
0.05 = e −0.301t
( )
ln ( 0.05 ) = ln e −0.301t = −0.301t
ln ( 0.05 )
=t ≈ 9.95
−0.301
Jim will have reached 95% mastery after about 10 days.
Earlier we used unlimited growth to model a population, but often a population will be
constrained by food, space, and other resources. When a population grows both proportional to
its size, and relative to the distance from some maximum, that is called logistic growth. This
leads to the differential equation y ′ = ky ( M − y ) , which is accurate but not always convenient to
use. We will use a slight modification. Since solving this differential equation requires
integration techniques we haven't learned, the solution form is given.
Chapter 3 The Integral Applied Calculus 235
Logistic Growth
If a quantity grows at a rate proportional to its size and to the distance from the maximum
value, M, it can be modeled with logistic growth, which has the differential equation:
y
y ′ = ry1 −
M
r can be interpreted as "the growth rate absent constraints" - how the population would
grow if there wasn't a maximum value.
This differential equation has solutions of the form
M
y=
1 + Ae −rt
Example 8
A colony of 100 rabbits is introduced to a reclaimed forest. After 1 year, the population has
grown to 300. It is estimated the forest can sustain 5000 rabbits. The forest service plans to
reintroduce wolves to the forest when the rabbit population reaches 3000 rabbits. When will
that occur?
The maximum sustainable population was given as M = 5000. Using the solution form,
5000
y=
1 + Ae −rt
Using the y (0) = 100 , we can solve for A
5000
100 = Simplify
1 + Ae −r ( 0 )
5000
100 = Multiply both sides by 1+A
1+ A
100(1 + A) = 5000 Divide by 100
1 + A = 50
A = 49
Now, using y (1) = 300 , we can solve for r.
5000
300 =
1 + 49e −r (1)
( )
300 1 + 49e −r = 5000
5000
1 + 49e −r =
300
50
−1
e −r = 3 ≈ 0.3197
49
Chapter 3 The Integral Applied Calculus 236
− r = ln 0.3197
r = − ln 0.3197 ≈ 1.1404
We now have the equation for the population after t years.
5000
y=
1 + 49e −1.1404t
To answer the original equation, of when the rabbit population will reach 3000, we need to
solve for t when y = 3000.
5000
3000 =
1 + 49e −1.1404t
(
3000 1 + 49e −1.1404t = 5000)
5
−1
e −1.1404 t
= 3 ≈ 0.01361
49
− 1.1404t = ln 0.01361
ln 0.01361
t= ≈ 3.77 years.
− 1.1404
Logistic growth is also a good model for unadvertised sales. A new product that is not
advertised will have sales increase slowly at first, then grow as word of mouth spreads and
people become familiar with the product. Sales will level off as they approach market saturation.
3.8 Exercises
In problems 1 – 4, check that the function y is a solution of the given differential equation.
1. y' + 3y = 6. y = e–3x + 2. 2. y' – 2y = 8. y = e2x – 4.
3. y' = – x/y. y = 7 – x2 . 4. y' = x – y. y = x – 1 + 2e–x .
In problems 5 – 8 check that the function y is a solution of the given initial value problem.
5. y' = 6x2 – 3 and y(1) = 2 . y = 2x3 – 3x + 3.
6. y' = 6x + 4 and y(2) = 3. y = 3x2 + 4x – 17.
7. y' = 5y and y(0) = 7. y = 7e5x.
8. y' = –2y and y(0) = 3. y = 3e–2x.
Chapter 3 The Integral Applied Calculus 237
In problems 9 – 12, a family of solutions of a differential equation is given. Find the value of the
constant C so the solution satisfies the initial value condition.
9. y' = 2x and y(3) = 7. y = x2 + C. 10. y' = 3x2 – 5 and y(1) = 2. y = x3 – 5x + C.
11. y' = 3y and y(0) = 5. y = Ce3x. 11. y' = –2y and y(0) = 3. y = Ce–2x.
In problems 13 – 18, solve the differential equation. (Assume that x and y are restricted so
that division by zero does not occur.)
13. y' = 2xy 14. y' = x/y 15. xy' = y + 3
16. y' = x2y + 3y 17. y' = 4y 18. y' = 5(2 – y)
In problems 19 – 22, solve the initial value separable differential equations.
19. y' = 2xy for y(0) = 3, y(0) = 5, and y(1) = 2.
20. y' = x/y for y(0) = 3, y(0) = 5, and y(1) = 2.
21. y' = 3y for y(0) = 4, y(0) = 7, and y(1) = 3.
22. y' = –2y for y(0) = 4, y(0) = 7, and y(1) = 3.
23. The rate of growth of a population P(t) which starts with 3,000 people and increases by 4% per year
is P '(t) = 0.0392.P(t). Solve the differential equation and use the solution to estimate the population
in 20 years.
24. The rate of growth of a population P(t) which starts with 5,000 people and increases by 3% per
year is P '(t) = 0.0296.P(t). Solve the differential equation and use the solution to estimate the
population in 20 years.
25. A manufacturer estimates that she can sell a maximum of 130 thousand cell phones in a city.
By advertising heavily, her total sales grow at a rate proportional to the distance below this
upper limit. If she enters a new market, and after 6 months her total sales are 59 thousand
phones, find a formula for the total sales (in thousands) t months after entering the market,
and use this to estimate the total sales at the end of the first year.
26. The temperature of a turkey in the oven will grow like limited growth. The turkey starts out
at 40 degrees Fahrenheit, and is placed into a 350 degree oven. After 30 minutes, the
Chapter 3 The Integral Applied Calculus 238
turkey's temperature has risen to 55 degrees. How long will it take until the turkey's
temperature reaches 165 degrees?
27. A new cell phone is introduced into the market. It is predicted that sales will grow
logistically. The manufacturer estimates that they can sell a maximum of 100 thousand cell
phones. After 44 thousand cell phones have been sold, sales are increasing by 4 thousand
phones per month. Use this to estimate the total sales at the end of the first year.
28. Biologists stocked a lake with 400 fish and estimated the carrying capacity of the lake to
be 8000 fish. The number of fish tripled in the first year. How long will it take the
population to increase to 4000?
Chapter 4 Functions of Two Variables Applied Calculus 239
Chapter 4: Functions of Two Variables
PreCalculus Idea -- Topographical Maps
If you’ve ever hiked, you have probably seen a topographical map. Here is part of a topographic
map of Stowe, Vermont, USA (courtesy of United States Geological Survey and
http://en.wikipedia.org/wiki/File:Topographic_map_example.png).
Points with the same elevation are connected with curves, so you can read not only your east-
west and your north-south location, but also your elevation. You may have also seen weather
maps that use the same principle – points with the same temperature are connected with curves
(isotherms), or points with the same atmospheric pressure are connected with curves (isobars).
These maps let you read not only a place’s location but also its temperature or atmospheric
pressure.
In this chapter, we’ll use that same idea to make graphs of functions of two variables.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 4 Functions of Two Variables Applied Calculus 240
Section 1: Functions of Two Variables
Real life is rarely as simple as one input – one output. Many relationships depend on lots of
variables. Examples:
• If I put a deposit into an interest-bearing account and let it sit, the amount I have at the
end of 3 years depends on P (how much my initial deposit is), r (the annual interest rate),
and n (the number of compoundings per year).
• The air resistance on a wing in a wind tunnel depends on the shape of the wing, the speed
of the wind, the wing’s orientation (pitch, yaw, and roll), plus a myriad of other things
that I can’t begin to describe.
• The amount of your television cable bill depends on which basic rate structure you have
chosen and how many pay-per-view movies you ordered.
Since the real world is so complicated, we want to extend our calculus ideas to functions of
several variables.
Functions of Two Variables
If x1 , x 2 , x 3 , ..., x n are real numbers, then ( x1 , x 2 , x 3 , ..., x n ) is called an n-tuple. This is an
extension of ordered pairs and triples. A function of n variables is a function whose domain is
some set of n-tuples and whose range is some set of real numbers.
For much of what we do here, everything would work the same if we were working with 2, 3, or
47 variables. Because we’re trying to keep things a little bit simple, we’ll concentrate on
functions of two variables.
A Function of Two Variables
A function of two variables is a function – that is, to each input is associated exactly one
output.
The inputs are ordered pairs, (x, y). The outputs are real numbers. The domain of a
function is the set of all possible inputs (ordered pairs); the range is the set of all possible
outputs (real numbers).
The function can be written z = f(x,y).
Functions of two variables can be described numerically (a table), graphically,
algebraically (a formula), or in English.
We will often now call the familiar y = f(x) a function of one variable.
Example 1
The cost of renting a car depends on how many days you keep it and how far you drive.
Represent this using a function.
Let d = the number of days you rent the car, and m = the number of miles you drive. Then the
cost of the car rental C(d, m) is a function of two variables.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 4 Functions of Two Variables Applied Calculus 241
Example 2
The demand for hot dog buns depends on the price for the hot dog buns and also on the price for
hot dogs. Represent this as a function.
The demand q B = f ( p B , p D ) is a function of two variables. (The demand for hot dogs also
depends on the price of both dogs and buns).
Formulas and Tables
Just as in the case of functions of one variable, we can display a function of two variables in a
table. The two inputs are shown in the margin (top row, left column), and the outputs are shown
in the interior cells.
Example 3
Here is a table that shows the cost C(d, m) in dollars for renting a car for d days and driving it m
miles:
d m → 100 200 300 400
↓
1 55 70 85 100
2 95 110 125 140
3 135 150 165 180
4 175 190 205 220
a. What is the cost of renting a car for 3 days and driving it 200 miles?
b. What is C(100, 4)? What is C(4, 100)?
c. Suppose we rent the car for 3 days. Is C an increasing function of miles?
a. According to the table, renting the car for 3 days (row with d = 3) and driving it 200 miles
(column with m = 200) will cost $150 (highlighted in aqua).
b. Careful now – the input is an ordered pair, so in C(100, 4), the 100 has to be a value of d and
the 4 has to be a value of m. C(100, 4) would be the cost of renting a car for 100 days and
driving it 4 miles. That cost is not in the table. (And that would be a pretty silly way to rent a
car.) On the other hand, C(4, 100) is the cost of renting for 4 days and driving 100 miles – the
table says that would cost $175.
c. If we know that d is fixed at 3, we’re looking at C(3, m). This is now a function of 1
variable, just m. We can see the table that displays values of this function by focusing our
attention on just the row where d = 3:
d m → 100 200 300 400
↓
3 135 150 165 180
Now we can see that if we rent for 3 days, the cost appears to be an increasing function of the
number of miles we drive, which shouldn't be surprising.
Chapter 4 Functions of Two Variables Applied Calculus 242
The idea of fixing one variable and watching what happens to the function as the other varies
will come up again and again.
It’s hard to display a function of more than two variables in a table. But it’s convenient to work
with formulas for functions of two variables, or as many variables as you like.
Example 4
The cost C(d,m) in dollars for renting a car for d days and driving it m miles is given by the
formula C (d , m) = 40d + .15m
a. What is the cost of renting a car for 3 days and driving it 200 miles?
b. What is C(100, 4)? What is C(4, 100)?
c. Suppose we rent the car for 3 days. Is C an increasing function of miles?
a. C (3,200) = 40(3) + 15(200) = $150 . This is the same value we got from the table. The
formula will give us the same answers for any of the table values.
b. C(100, 4) makes perfect sense to the formula (even if it doesn’t make sense for actually
renting a car). So now we can get an answer. To rent the car for 100 days and drive it for 4
miles should cost $4000.60. C(4, 100) = $175, as before.
c. If we fix d = 3, then C(d, m) becomes C(3, m) = 40(3) + .15m = 120 + .15m. Yes, this is an
increasing function of m; I can tell because it’s linear and its slope is .15 > 0.
Reality check – the formula that gives the cost for the rental car makes sense for all values of d
and m. But that’s not how the real cost works – you can’t rent the car for a negative number of
days or drive a negative number of miles. (That is, there are domain restrictions.) In addition,
most car rental agreements don’t compute a charge for fractions of days; they round up to the
next whole number of days.
Example 5
1
Let f ( x, y, z , w) = 35 x 2 w − + yz 2 . Evaluate f (0,1,2,3) .
z
Remember that this is an ordered 4-tuple; make sure the numbers get substituted into the correct
places:
1
f (0,1,2,3) = 35 ⋅ 0 2 3 − + 1 ⋅ 2 2 = 3.5
2
Graphs
The graph of a function of two variables is a surface in three-dimensional space. Let's start by
looking at the 3–dimensional rectangular coordinate system, how to locate points in three
dimensions, and distance between points in three dimensions.
Chapter 4 Functions of Two Variables Applied Calculus 243
In the 2–dimensional rectangular coordinate system we have two
coordinate axes that meet at right angles at the origin, and it takes two
numbers, an ordered pair (x, y), to specify the rectangular coordinate
location of a point in the plane (2 dimensions). Each ordered pair (x, y)
specifies the location of exactly one point, and the location of each point
is given by exactly one ordered pair (x, y). The x and y values are the coordinates of the point
(x, y).
The situation in three dimensions is very similar. In the 3–
dimensional rectangular coordinate system we have three
coordinate axes that meet at right angles, and three numbers, an
ordered triple (x, y, z), are needed to specify the location of a
point. Each ordered triple (x, y, z) specifies the location of
exactly one point, and the location of each point is given by exactly
one ordered triple (x, y, z). The x, y and z values are the
coordinates of the point (x, y, z). The figure below shows the
location of the point (4, 2, 3).
Typically we use a right-hand orientation. To see what this means, imagine
your right hand in front hand in front of you with the palm toward your
face, your thumb pointing up, you index finger straight out, and your next
finger toward your face (and the two bottom fingers bent into the palm.
Then, in the right hand coordinate system, your thumb points along the
positive z-axis, your index finger along the positive x-axis, and the other
finger along the positive y-axis. Other orientations of the axes are possible and valid (with
appropriate labeling), but the right–hand system is the most common orientation and is the one
we will generally use.
Each ordered triple (x, y, z) specifies the location of a single
point, and this location point can be plotted by locating the point
(x, y, 0) on the xy–plane and then going up z units. (We could
also get to the same (x, y, z) point by finding the point (x, 0, z)
on the xz–plane and then going y units parallel to the y–axis, or by
finding (0, y, z) on the yz–plane and then going x units parallel
to the x–axis.)
Chapter 4 Functions of Two Variables Applied Calculus 244
Example 6
Plot the locations of the points :
P = (0, 3, 4),
Q = (2, 0, 4),
R = (1, 4, 0),
S = (3, 2, 1), and
T(–1, 2, 1) .
The points are shown to the right.
Once we can locate points, we can begin to consider the graphs of various collections of points.
By the graph of "z = 2" we mean the collection of all points (x, y, z) which have the form "(x,
y, 2)". Since no condition is imposed on the x and y variables, they take all possible values.
The graph of z = 2 is a plane parallel to the xy–plane and 2 units above the xy–plane. Similarly,
the graph of y = 3 is a plane parallel to the xz–plane , and x = 4 is a plane parallel to the yz–
plane. (Note: The planes have been drawn as rectangles, but they actually extend infinitely far.)
Distance Between Points
In two dimensions we can think of the distance between points as the
length of the hypothenuse of a right triangle, and that leads to the
Pythagorean formula: distance = ∆x2 + ∆y2 .
Chapter 4 Functions of Two Variables Applied Calculus 245
In three dimensions we can also think of the distance between points
as the length of the hypothenuse of a right triangle, but in this situation
the calculations appear more complicated. Fortunately, they are
straightforward:
distance2 = base2 + height2 = ( ∆x2 + ∆y2 )2 + ∆z2
= ∆x2 + ∆y2 + ∆z2
so distance = ∆x2 + ∆y2 + ∆z2 .
If P = (x1 , y1 , z1 ) and Q = (x2 , y2 , z2 ) are points in space,
then the distance between P and Q is
distance = ∆x2 + ∆y2 + ∆z2
= (x2–x1)2 + (y2–y1)2 + (z2–z1)2 .
The 3–dimensional pattern is very similar to the 2–dimensional pattern with the additional piece
∆z2 .
Example 7
Find the distances between points A = (1, 2, 3) and B = (7, 5, –3)
Dist(A, B) = 62 + 32 + (–6)2 = 36 + 9 + 36 = 81 = 9.
In two dimensions, the set of points at a fixed distance from a given
point is a circle, and we used the distance formula to determine
equations describing circles: the circle with center (2, 3) and
radius 5 is given by (x–2)2 + (y–3)2 = 52 or x2 + y2 – 4x – 6y =
12.
The same ideas work for spheres in three dimensions.
Chapter 4 Functions of Two Variables Applied Calculus 246
Spheres
The set of points (x, y, z) at a fixed distance r
from a point (a, b, c) is a sphere with center
(a, b, c) and radius r.
The sphere is given by the equation
(x–a)2 + (y–b)2 + (z–c)2 = r2 .
Example 8
Write the equations of a sphere with center (2, –3, 4) and radius 3
(x–2)2 + (y+3)2 + (z–4)2 = 32 .
Now suppose that we want to graph a surface. We can think of each input (x,y) as a location on
the plane, and plot the point f(x,y) units above that point. Graphing that can be challenging. We
have a few options:
1. Use a fancy computer program to draw beautiful perspective drawings.
2. Try to draw a perspective drawing by hand. This is very challenging, and usually not worth
the effort.
3. Use level curves to draw contour diagrams, which is the approach we'll focus on here. A
contour diagram is like a topographical map – points with the same elevation (outputs) are
connected with curves. Each particular output is called a level, and these curves are called level
curves or contours. The closer the curves are to each other, the steeper that section of the surface
is. Topographical maps give hikers information about elevation, steep and shallow grades, peaks
and valleys. Contour diagrams give us the same kind of information about a function.
Chapter 4 Functions of Two Variables Applied Calculus 247
This is a contour diagram of the same surface shown in above. The level curves are graphs in
the xy-plane of curves f(x, y) = c for various constants c.
Each of the squares corresponds to one of the bumps on the surface. If the contours are positive,
as highlighted below, the bump is above the xy-plane. If the contours are negative, the bump
extends below the xy-plane.
Everywhere on the crisscrossed pattern of diagonal lines, the height of the surface is 0, so the
surface is on the xy-plane. This is a feature that I couldn’t see when I looked at the perspective
drawing.
Chapter 4 Functions of Two Variables Applied Calculus 248
To better understand contour diagrams, suppose we had a table of elevation data. We could
graph this by plotting the height at each point and connecting the dots with smooth curves, which
would result in the something like the graph shown.
If we "slice" the surface above with the plane z = 8, the points where the plane cuts the surface
are those points where the elevation of the surface is 8 units above the xy–plane. The figure
below shows the surface being sliced by the planes z = 8 and z = 4. Slicing the surface at
different elevations and sketching the curves where the plane intersects the surface results in the
second graph below.
Chapter 4 Functions of Two Variables Applied Calculus 249
If we move all of those curves to the xy–plane
(or, equivalently, view them from directly
overhead), the result is a 2–dimensional graph of
the level curves of the original surface. This is the
contour diagram.
Example 9
Create a contour diagram for our car rental example with cost function C (d , m) = 40d + .15m .
Draw curves for when the cost is 0, 100, 200, 300, and 400.
Set will set C (d , m) = 40d + .15m = c for
c = 0, 100, 200, 300, and 400 and draw the curves in the dm-plane.
The first coordinate of the ordered pair is d, so the d-axis will be horizontal; the m-axis will be
vertical. Remember that the domain for this function is really just where d ≥ 0 and m ≥ 0, so we
will only draw the curves in the first quadrant.
When c = 0:
C (d , m) = 40d + .15m = 0
.15m = −40d
40
m=− d ≅ −267d
.15
This is the equation of a line, with slope about − 267 , passing through the origin. Because of
the domain restrictions, the “curve” I will draw for this level is simply the origin.
Putting this back into the car rental context, the only point where I pay $0 for renting the car is
when I rent the car for 0 days and drive it 0 miles – that is, if I don’t rent it at all.
When c = 100:
C (d , m) = 40d + .15m = 100
.15m = −40d + 100
40 100
m=− d+ ≅ −267 d + 667
.15 .15
This is the equation of a line, with slope about − 267 , and d-intercept of about 667. This section
of this line that lies in the first quadrant is shown with 100 labeling it.
Chapter 4 Functions of Two Variables Applied Calculus 250
Putting this into context, any point on that
line represents a (d, m) combination of
days and miles that will make the cost
exactly $100. So, for example – if I rent
the car for 0 days and drive it 667 miles, it
will cost me $100. If I rent the car for 2.5
days and don’t drive any miles, it will cost
me $100.
We continue for c = 200, 300, and 400 and
sketch the curves in the plane, resulting in
the contour diagram shown to the right.
Example 10
The contour diagram for the cost C(d,m) in dollars for renting a car for d days and driving it m
miles is shown in the previous example. Use the diagram to answer the following questions.
a. What is the cost of renting a car for 3 days and driving it 200 miles?
b. What is C(100, 4)? What is C(4, 100)?
c. Suppose we rent the car for 3 days. Is C an increasing function of miles?
a. The point (3, 200) is between contours on this graph, so I can’t get an exact answer for C(3,
200). (But it’s typical for a graph that we would have to estimate). It looks to me as if (3, 200)
is halfway between the 100 and the 200 contours, so I will estimate that C(3, 200) is about
$150.
Estimates from the graph are necessarily very rough. The graph only shows a little information
(in this way, a contour diagram is like a table), so I have to extrapolate in between. But for
most graphs, I don’t actually know what happens between the contours. All I know for sure is
that the output at (3, 200) is between the two levels I see. For this car rental example, I also
know a formula, and my table showed this particular input, so I have other ways to get a better
answer.
b. I can’t find (100, 4) on this diagram, so I can’t make an estimate of C(100, 4) from this graph.
(4, 100) lies between the contours for 100 and 200. It looks closer to 200, so I’ll estimate that
C(4, 100) is about $180.
Chapter 4 Functions of Two Variables Applied Calculus 251
c. If we fix d = 3, we get a vertical line. What happens as m increases on this vertical line? As
m increases, the function values shown on the contours increase – C appears to be an increasing
function of miles.
Example 11
Here is a contour diagram for a function g(x,y).
Use the diagram to answer the following
questions:
a. What is g(3, 5)?
b. What is the highest point shown on the
diagram? What is the lowest point shown?
c. If you start at (3, 5) and head in the positive
x direction, do you go uphill or downhill
first?
a. g(3, 5) is 0.6. I can tell because the point is
right on one of the contours.
Chapter 4 Functions of Two Variables Applied Calculus 252
b. The highest contour shown is 0.9, and there would be a contour for 1.0 if the surface had ever
got that high. However, the height seems to be increasing as we move in toward the center, so
I’m guessing that the g gets to nearly 1 in the center. The lowest contour is 0.1. But again, I
will guess that the height continues to decrease, so I think g is nearly 0 around the outside.
c. Starting at the point (3, 5, 0.6) on the surface and traveling to the right along the horizontal
line shown in Fig. 9, you would cross the contour for 0.7 next. So the function increases first
(we go uphill), and then decreases again.
Note one more time – we don’t really know what happens between the contours. All we can do
is estimate from the information in the graph.
Example 12
Here is a contour diagram for a function
F(x,y).
a) Describe the shape of the surface.
b) Suppose you travel along the surface
in the positive y-direction, starting on the
surface at the point above (or below) the
point (x, y) = (-1, 1). Describe your
journey.
a) The surface is bumpy, with regularly
spaced oval bumps. Notice that some of
the bumps go up (positive contours), but
others go down. Between the bumps,
there are horizontal lines that are
completely level, with an elevation of 0.
b) It looks as if F(-1,1) is about 3. As I
head in the positive y-direction along the
line shown below, I first go uphill, nearly
to 4, then I start going downhill. As I keep going north, I keep descending, going into the dip,
until nearly -4. I’m starting to go uphill again just as I leave the graph.
Chapter 4 Functions of Two Variables Applied Calculus 253
What happens if you have a function of more than two variables? Its graph will be a hyper-
surface. For example, the graph of a function of four variables will be a hyper-surface in 5-
dimensional space. This is hard (impossible for most of us) to visualize. Even the contours are
hard to visualize – instead of curves in the plane, they’re hyper-surfaces in 4-dimensional space.
So – if you have more than two variables, the graph isn’t usually very useful.
Functions of Two Real-Life Variables
Complementary goods and substitute goods
The demand for some pairs of goods have a relationship, where the quantity demanded for one
product depends somehow on the prices for both.
Two goods are complementary if an increase in the price of either decreases the demand for both.
Examples:
• The demand for cars depends on both the price for cars and the price of gasoline.
• The demand for hot dog buns depends on both the price for the buns and the price for the
hot dogs.
Two goods are substitutes if an increase in the price of one increases the demand for the other.
Example:
• The demand for Brand A depends on its price and also on the price of its main competitor
Brand B. If the Brand B raises its price, consumers will switch brands – substitute – and
demand for Brand A will increase.
Think brands of soft drinks, detergent, or paper towels. A traditional example is coffee and tea –
the idea is that consumers are simply looking for a hot drink and they’ll buy whatever is cheaper.
Chapter 4 Functions of Two Variables Applied Calculus 254
But this has always seemed fishy to me – I’ve never met any coffee- or tea-drinkers who would
happily switch.
These demand functions are functions of two variables.
Example 13
The demand functions for two products are given below. p1, p2, q1, and q2 are the prices (in
dollars) and quantities for products 1 and 2.
q1 = 200 − 3 p1 − p 2
q 2 = 150 − p1 − 2 p 2
Are these two products complementary goods or substitute goods? What is the quantity
demanded for each when the price for product 1 is $20 per item and the price for product 2 is
$30 per item?
These products are complementary – an increase in either price decreases both demands. You
can see that because the coefficients are both negative in each demand function.
When p1 = 20 and p2 = 30, we have
q1 = 200 − 3(20 ) − (30 ) = 110
q 2 = 150 − (20) − 2(30) = 70
110 units are demanded for product 1 and 70 units are demanded for product 2 when the price
for product 1 is $20 per item and the price for product 2 is $30 per item
Cobb-Douglas Production function
Production functions are used to model the total output of a firm for a variety of inputs (doesn’t
this sound like a function of several variables?). One example is a Cobb-Douglas Production
function:
P = ALα K β
In this function, P is the total production, A is a constant, α and β are constants between 0 and 1,
L is the labor force, and K is the capital expenditure. (And the units must be massaged well.)
You can read more about Cobb-Douglas Production functions at
http://en.wikipedia.org/wiki/Cobb-Douglas. You can read about other kinds of production
functions at http://en.wikipedia.org/wiki/Production_function.
Chapter 4 Functions of Two Variables Applied Calculus 255
4.1 Exercises
1. F ( x, y ) = x 2 − y 2 . Find
a) F (0,4 ) b) F (4,0 ) c) F ( x,4 ) d) F (4, y )
e) F (800,800) f) F ( x, x ) g) F ( x,− x )
2. g (s, t ) = st 2 . Find
a) g (1,9 ) b) g (9,1) c) g (1, t ) d) g (s,9 ) e) g (w, z + 1)
1
3. Let f (x, y, z , w) = x 2 − + xyz 2 . Evaluate f (1,2,3,4 ) .
zw
4. Let f (x, y, z , w) = xy − w 2 + 102 yz . Evaluate f (1,2,3,4 ) .
5. Here is a table showing the function A(t , r )
t r → .03 .04 .05 .06 .07
↓
1 30.45 40.81 51.27 61.84 72.51
2 61.84 83.29 105.17 127.50 150.27
3 94.17 127.50 161.83 197.22 233.68
a) Find A(2,.05)
b) Find A(.05,.2 )
c) Is A(t ,.06) an increasing or decreasing function of t ?
d) Is A(3, r ) an increasing or decreasing function of r?
6. Here is a table showing values for the function H (t , h ) .
t h → 100 150 200
↓
0 100 150 200
1 110.1 160.1 210.1
2 110.4 160.4 210.4
3 100.9 150.9 200.9
4 81.6 131.6 181.6
5 52.5 102.5 152.5
a) Is H (t ,150) an increasing or decreasing function of t ?
b) Is H (4, h ) an increasing or decreasing function of h?
c) Fill in the blanks: The maximum value shown on this table is H (___, ___ ) = ____ .
d) Fill in the blanks: The minimum value shown on this table is H (___, ___ ) = ____ .
Chapter 4 Functions of Two Variables Applied Calculus 256
In problems 7 – 10, plot the given points.
7. A = (0,3,4), B = (1,4,0), C = (1,3,4), D = (1, 4,2)
8. E = (4,3,0), F = (3,0,1), G = (0,4,1), H = (3,3,1)
9. P = (2,3,–4), Q = (1,–2,3), R = (4,–1,–2), S = (–2,1,3)
10. T = (–2,3,–4), U = (2,0,–3), V = (–2,0,0), W = (–3,–1,–2)
In problems 11 – 14, calculate the distances between the given points
11. A = (5,3,4), B = (3,4,4)
12. A = (6,2,1), B = (3,2,1)
13. A = (3,4,2), B = (–1,6,–2)
14. A = (–1,5,0), B = (1,3,2)
In problems 15 – 18, graph the given planes.
15. y = 1 and z = 2 16. x = 4 and y = 2
17. x = 1 and y = 0 18. x = 2 and z = 0
In problems 19 – 22, the center and radius of a sphere are given. Find an equation for the sphere.
19. Center = (4, 3, 5), radius = 3 20. Center = (0, 3, 6), radius = 2
21. Center = (5, 1, 0), radius = 5 22. Center = (1, 2, 3), radius = 4
In problems 23 – 24, the equation of a sphere is given. Find the center and radius of the sphere.
23. (x–3)2 + (y+4)2 + (z–1)2 = 16 24. (x+2)2 + y2 + (z–4)2 = 25
For problems 25 through 30. Match the contour diagram to the computer-generated, perspective
drawing (a through f) it matches. Briefly explain your answer.
Chapter 4 Functions of Two Variables Applied Calculus 257
25. 26. 27.
28. 29. 30.
a. b. c.
d. e. f.
For problems 31 through 36. Match the contour diagram to the equation (a through f) it matches.
Briefly explain your answer.
Chapter 4 Functions of Two Variables Applied Calculus 258
31. 32. 33.
34. 35. 36.
a. f ( x, y ) = y − x b. f ( x, y ) = xy 2
c. f ( x, y ) = 25 − x 2 − y 2 d. f ( x, y ) = 5 − x − y
e. f ( x, y ) = 0.01x 2 y 2 f. f ( x, y ) = x 2 − y 2
Chapter 4 Functions of Two Variables Applied Calculus 259
37. The contour diagram shown is for
a function M(x, y). Use the
diagram to answer the following:
a) Estimate M(1, 3)
b) Estimate M(3, 1)
c) Is M(x, 3) an increasing or
decreasing function of x?
d) Is M(3, y) an increasing or
decreasing function of y?
e) Find a value of c so that M(c, y)
is a constant function of y.
38. The contour diagram shown is for
a function G(x, y). Use the
diagram to answer the following:
a) Estimate G(2, 3)
b) Suppose you travel north (in the
direction of increasing y) along
the surface, starting above (2, 3).
Describe your journey.
c) Suppose you travel east (in the
direction of increasing x) along
the surface, starting above (2, 3).
Describe your journey.
Chapter 4 Functions of Two Variables Applied Calculus 260
39. The demand functions for two products are given below. p1, p2, q1,and q2 are the prices (in
dollars) and quantities for products 1 and 2. Are these two products complementary goods
or substitute goods
q1 = 200 − 3 p1 + p2
q2 = 150 + p1 − 2 p2
40. The demand functions for two products are given below. p1, p2, q1,and q2 are the prices (in
dollars) and quantities for products 1 and 2. Are these two products complementary goods
or substitute goods
q1 = 350 + p1 + 2 p2
q2 = 225 + p1 + p2
41. Consider the Cobb-Douglas Production function: P ( L, K ) = 11L0.3 K 0.7 . Find the total units
of production when 19 units of labor and 12 units of capital are invested.
42. Consider the Cobb-Douglas Production function: P ( L, K ) = 6 L0.6 K 0.4 . Find the total units of
production when 24 units of labor and 8 units of capital are invested.
Chapter 4 Functions of Two Variables Applied Calculus 261
Section 2: Calculus of Functions of Two Variables
Now that you have some familiarity with functions of two variables, it’s time to start applying
calculus to help us solve problems with them. In Chapter 2, we learned about the derivative for
functions of two variables. Derivatives told us about the shape of the function, and let us find
local max and min – we want to be able to do the same thing with a function of two variables.
First let’s think. Imagine a surface, the graph of a function of two variables. Imagine that the
surface is smooth and has some hills and some valleys. Concentrate on one point on your
surface. What do we want the derivative to tell us? It ought to tell us how quickly the height of
the surface changes as we move …. Wait, which direction do we want to move? This is the
reason that derivatives are more complicated for functions of several variables – there are so
many directions we could move from any point.
It turns out that our idea of fixing one variable and watching what happens to the function as the
other changes is the key to extending the idea of derivatives to more than one variable.
Partial Derivatives
Partial Derivatives:
Suppose that z = f(x, y) is a function of two variables.
The partial derivative of f with respect to x is the derivative of the function f(x,y)
where we think of x as the only variable and act as if y is a constant.
The partial derivative of f with respect to y is the derivative of the function f(x,y)
where we think of y as the only variable and act as if x is a constant.
The “with respect to x” or “with respect to y” part is really important – you have to know
and tell which variable you are thinking of as THE variable.
Geometrically – the partial derivative with respect to x gives the slope of the curve as you
travel along a cross-section, a curve on the surface parallel to the x-axis. The partial
derivative with respect to y gives the slope of the cross-section parallel to the y-axis.
Notation for the Partial Derivative:
The partial derivative of y = f(x) with respect to x is written as
f x ( x, y ), or z x simply f x
∂f ∂z
The Leibniz notation is , or
dx dx
∂z
We use an adaptation of the notation to mean “find the partial derivative of f(x,y) with
dx
respect to x:”
∂
( f (x, y )) = ∂f
∂x ∂x
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 4 Functions of Two Variables Applied Calculus 262
To estimate a partial derivative from a table or contour diagram:
The partial derivative with respect to x can be approximated by looking at an average rate
of change, or the slope of a secant line, over a very tiny interval in the x-direction
(holding y constant). The tinier the interval, the closer this is to the true partial
derivative.
To compute a partial derivative from a formula:
If f(x,y) is given as a formula, you can find the partial derivative with respect to x
algebraically by taking the ordinary derivative thinking of x as the only variable (holding
y fixed).
Of course, everything here works the same way if we’re trying to find the partial
derivative with respect to y – just think of y as your only variable and act as if x is
constant.
The idea of a partial derivative works perfectly well for a function of several variables –
you focus on one variable to be THE variable and act as if all the other variables are
constants.
Example 1
Here is a contour diagram for a function
g(x,y). Use the diagram to answer the
following questions:
a. Estimate g x (3,5) and g y (3,5)
b. Where on this diagram is g x greatest?
Where is g y greatest?
a. g x (3,5) means we're thinking of x as the
only variable, so we’ll hold y fixed at y = 5.
That means we’ll be looking along the
horizontal line y = 5. To estimate gx, we need
two function values. (3, 5) lies on the contour
line, so we know that g(3, 5) = 0.6. The next
point as we move to the right is g(4.2,5) = 0.7.
Now we can find the average rate of change:
∆g 0.7 − 0.6 1
Average rate of change = (change in output) / (change in input) = = = ≅ .083 .
∆x 4.2 − 3 12
We can do the same thing by going to the next point we can read to the left, which is g(2.4,5) =
∆g 0.5 − 0.6 1
0.5. Then the average rate of change is = = = ≅ .167 .
∆x 2.4 − 3 6
Either of these would be a fine estimate of g x (3,5) given the information we have, or you could
take their average. We can estimate that g x (3,5) ≅ .125 .
Chapter 4 Functions of Two Variables Applied Calculus 263
Estimate g y (3,5) the same way, but moving on the vertical line. Using the next point up, we
∆g 0.7 − 0.6
get the average rate of change = = = .125 . Using the next point down, we get
∆y 5.8 − 5
∆g 0.5 − 0.6
= = .2 . Taking their average, we estimate g y (3,5) ≅ .1625 .
∆y 4.5 − 5
b. g x means x is my only variable, and we're thinking of y as a constant. So we're thinking
about moving across the diagram on horizontal lines. g x will be greatest when the contour lines
∆g
are closest together, when the surface is steepest – then the denominator in will be small, so
∆x
∆g
will be big. Scanning the graph, we can see that the contour lines are closest together when
∆x
we head to the left or to the right from about (0.5, 8) and (9, 8). So g x is greatest at about (0.5,
8) and (9, 8). For g y , I want to look at vertical lines. g y is greatest at about (5, 3.8) and (5,
12).
Example 2
Cold temperatures feel colder when the wind is blowing. Windchill is the perceived
temperature, and it depends on both the actual temperature and the wind speed – a function of
two variables! You can read more about windchill at http://www.nws.noaa.gov/om/windchill/.
Below is a table (courtesy of the National Weather Service) that shows the perceived
temperature for various temperatures and windspeeds. Note that they also include the formula,
but for this example we'll use the information in the table.
a. What is the perceived temperature when the actual temperature is 25˚F and the wind is
blowing at 15 miles per hour?
b. Suppose the actual temperature is 25˚F. Use information from the table to describe how the
perceived temperature would change if the wind speed increased from 15 miles per hour?
Chapter 4 Functions of Two Variables Applied Calculus 264
a. Reading the table, we see that the perceived temperature is 13˚F.
b. This is a question about a partial derivative. We’re holding the temperature (T) fixed at 25˚F,
and asking what happens as wind speed (V) increases from 15 miles per hour. We’re thinking
of V as the only variable, so we want WindChillV = WV when T = 25 and V = 15. We’ll find
the average rate of change by looking in the column where T = 25 and letting V increase, and
use that to approximate the partial derivative.
∆W 11 − 13
WV ≅ = = −0.4
∆V 20 − 15
What are the units? W is measured in ˚F and V is measured in mph, so the units here are
˚F/mph. And that lets us describe what happens:
The perceived temperature would decrease by about .4˚F for each mph increase in wind speed.
Example 3
Find f x and f y at the points (0, 0) and (1, 1) if f (x, y ) = x 2 − 4 xy + 4 y 2
To find f x , take the ordinary derivative of f with respect to x, acting as if y is constant:
f x ( x, y ) = 2 x − 4 y
Note that the derivative of the 4 y 2 term with respect to x is zero – it’s a constant.
Similarly, f y ( x, y ) = −4 x + 8 y .
Now we can evaluate these at the points:
Chapter 4 Functions of Two Variables Applied Calculus 265
f x (0,0) = 0 and f y (0,0 ) = 0 ; this tells us that the cross sections parallel to the x- and y- axes are
both flat at (0,0).
f x (1,1) = −2 and f y (1,1) = 4 ; this tells us that above the point (1, 1), the surface decreases if you
move to more positive x values and increases if you move to more positive y values.
Example 4
∂f ∂f e x+ y
Find and if f ( x, y ) = 3 + y (ln y )
∂x ∂y y +y
∂f
means x is our only variable, we’re thinking of y as a constant. Then we’ll just find the
∂x
ordinary derivative. From x’s point of view, this is an exponential function, divided by a
constant, with a constant added. The constant pulls out in front, the derivative of the
exponential function is the same thing, and we need to use the chain rule, so we multiply by the
derivative of that exponent (which is just 1):
∂f 1
= 3 e x+ y
∂x y + y
∂f
means that we’re thinking of y as the variable, acting as if x is constant. From y’s point of
∂y
view, f is a quotient plus a product – we’ll need the quotient rule and the product rule:
∂f ( )( ) − ( )( )
= + ( )( ) + ( )( )
∂y ( )2
=
(e (1))(y
x+ y 3
) ( )(
+ y − e x+ y 3 y 2 + 1 ) + (1)(ln y ) + ( y ) 1
y
(y + y )
3 2
Example 5
1
Find f z if f ( x, y, z , w) = 35 x 2 w − + yz 2
z
f z means we act as if z is our only variable, so we’ll act as if all the other variables (x, y and w)
are constants and take the ordinary derivative.
1
f z ( x, y, z , w) = 2 + 2 yz
z
Chapter 4 Functions of Two Variables Applied Calculus 266
Using Partial Derivatives to Estimate Function Values
We can use the partial derivatives to estimate values of a function. The geometry is similar to
the tangent line approximation in one variable. Recall the one-variable case: if x is close enough
to a known point a, then f ( x ) ≅ f (a ) + f ' (a )( x − a ) . In two variables, we do the same thing in
both directions at once:
Approximating Function Values with Partial Derivatives
To approximate the value of f(x, y), find some point (a, b) where
1. (x, y) and (a, b) are close – that is, x and a are close and y and b are close.
2. You know the exact values of f (a, b) and both partial derivatives there.
Then f ( x, y ) ≅ f (a, b ) + f x (a, b )(x − a ) + f y (a, b )( y − b )
Notice that the total change in f is being approximated by adding the approximate changes
coming from the x and y directions. Another way to look at the same formula:
∆f ≅ f x ∆x + f y ∆y
How close is close? It depends on the shape of the graph of f. In general, the closer the
better.
Example 6
Use partial derivatives to estimate the value of f ( x, y ) = x 2 − 4 xy + 4 y 2 at (0.9, 1.1)
Note that the point (0.9, 1.1) is close to an “easy” point, (1, 1). In fact, we already worked out
the partial derivatives at (1, 1): f x ( x, y ) = 2 x − 4 y ; f x (1,1) = −2 . f y ( x, y ) = −4 x + 8 y ;
f y (1,1) = 4 . We also know that f (1,1) = 1 .
So f (0.9,1.1) ≅ 1 − 2(− 0.1) + 4(0.1) = 1.6.
Note that it would have been possible in this case to simply compute the exact answer;
f (0.9,1.1) = (0.9 ) − 4(0.9 )(1.1) + 4(1.1) = 1.69 . Our estimate is not perfect, but it’s pretty
2 2
close.
Chapter 4 Functions of Two Variables Applied Calculus 267
Example 7
Here is a contour diagram for a function g(x,y). Use partial derivatives to estimate the value of
g(3.2, 4.7).
This is the same diagram from before, so we already estimated the value of the function and the
partial derivatives at the nearby point (3,5). g(3, 5) is 0.6, our estimate of g x (3,5) ≅ .125 , and
our estimate of g y (3,5) ≅ .1625 . So g (3.2,4.7 ) ≅ 0.6 + (.125)(.2 ) + (.1625)(− .3) = .57625. Note
that in this case, we have no way to know how close our estimate is to the actual value.
Chapter 4 Functions of Two Variables Applied Calculus 268
4.2 Exercises
For problems 1 through 16, find f x and f y for the function given
1. f ( x, y ) = x 2 − 5 y 2
x2 − 5y 2
2. f ( x, y ) =
x+4
3. f ( x, y ) = e x + 6 y
(
4. f ( x, y ) = x 2 − 5 y 2 e x )
1
(
5. f ( x, y ) = x 2 − 5 y 2 )
+ 4
3y
6. f ( x, y ) = x
7. f ( x, y ) = 6
8. f ( x, y ) = ln ( xy + 2 x − 6 y )
x2 − 5y 2
9. f ( x, y ) =
y 4 − 5x 4
10. f ( x, y ) = e x−4 y
(x − 4 y )
11. f ( x, y ) = y 5 e x
1
12. f ( x, y ) =
16 xy
(
13. f ( x, y ) = x + e y )
7
14. f (x, y ) = x 4 + 4 x 3 y − 6 x 2 y 2 − 4 xy 3 + y 4
15. f ( x, y ) = x + y
16. f ( x=
, y ) x 2 y 3 − 4 x3
Chapter 4 Functions of Two Variables Applied Calculus 269
17. Here is a table showing the function A(t , r )
t r → .03 .04 .05 .06 .07
↓
1 30.45 40.81 51.27 61.84 72.51
2 61.84 83.29 105.17 127.50 150.27
3 94.17 127.50 161.83 197.22 233.68
a. Estimate At (2,.05) .
b. Estimate Ar (2,.05)
c. Use your answers to parts a and b to estimate the value of A(2.5,.054)
( )
d. The values in the table came from A(t , r ) = 1000 e rt − 1 , which shows the interest
earned if 1000 dollars is deposited in an account earning r annual interest, compounded
continuously, and left there for t years. How close are your estimates from parts a, b, and
c?
18. Here is a table showing values for the function H (t , h ) .
t h → 100 150 200
↓
0 100 150 200
1 110.1 160.1 210.1
2 110.4 160.4 210.4
3 100.9 150.9 200.9
4 81.6 131.6 181.6
5 52.5 102.5 152.5
∂H
a. Estimate the value of at (3, 150).
dt
∂H
b. Estimate the value of at (3, 150).
dh
c. Use your answers to parts a and b to estimate the value of H (2.6,156) .
d. The values in the table came from H (t , h ) = h + 15t − 4.9t 2 , which gives the height in
meters above the ground after t seconds of an object that is thrown upward from an initial
height of h meters with an initial velocity of 15 meters per second. How close are your
estimates from parts a, b, and c?
19. Given the function f ( x, y ) = x 2 y
a. Calculate f ( 2, 4 ) , f x ( 2, 4 ) , and f y ( 2, 4 )
b. Use your answers from part a to estimate f (1.9, 4.1)
20. Given the function f ( x, y= (
) ln 10 − x 2 − y )
a. Calculate f ( 2,5 ) , f x ( 2,5 ) , and f y ( 2,5 )
b. Use your answers from part a to estimate f (1.8, 4.8 )
Chapter 4 Functions of Two Variables Applied Calculus 270
In problems 21 - 26, use the contour plot shown to estimate the desired value.
21. f x (1, −5 )
22. f x ( −5, 2 )
23. f x ( 5,5 )
24. f x ( 0, 0 )
25. f y (1, −5 )
26. f y ( −5, 2 )
Chapter 4 Functions of Two Variables Applied Calculus 271
Section 3: Optimization
The partial derivatives tell us something about where a surface has local maxima and minima.
Remember that even in the one-variable cases, there were critical points which were neither
maxima nor minima – this is also true for functions of many variables. In fact, as you might
expect, the situation is even more complicated.
Second Derivatives
When you find a partial derivative of a function of two variables, you get another function of two
variables – you can take its partial derivatives, too. We’ve done this before, in the one-variable
setting. In the one-variable setting, the second derivative gave information about how the graph
was curved. In the two-variable setting, the second partial derivatives give some information
about how the surface is curved, as you travel on cross-sections – but that’s not very complete
information about the entire surface.
Imagine that you have a surface that’s ruffled around a point, like what happens near a button on
an overstuffed sofa, or a pinched piece of fabric, or the wrinkly skin near your thumb when you
make a fist. Right at that point, every direction you move, something different will happen – it
might increase, decrease, curve up, curve down … A simple phrase like “concave up” or
“concave down” can’t describe all the things that can happen on a surface.
Surprisingly enough, though, there is still a second derivative test that can help you decide if a
point is a local max or min or neither, so we still do want to find second derivatives.
Second Partial Derivatives
Suppose f ( x, y ) is a function of two variables. Then it has four second partial derivatives:
∂ ∂
f xx = ( f x ) = ( f x )x ; f xy = ( f x ) = ( f x ) y ;
∂x ∂y
∂ ∂
f yx = ( f y ) = ( f y )x ; f yy = ( f y ) = ( f y ) y
∂x ∂y
f xy and f yx are called the mixed (second) partial derivatives of f
Leibniz notation for the second partial derivatives is a bit confusing, and we won’t use it as
often:
∂ ∂f ∂ 2 f ∂ ∂f ∂ 2 f
f xx = = ; f xy = = ;
∂x ∂x ∂x 2 ∂y ∂x ∂y∂x
∂ ∂f ∂ 2 f ∂ ∂f ∂ 2 f
f yx = = ; f yy = = 2
∂x ∂y ∂x∂y ∂y ∂y ∂y
Notice that the order of the variables for the mixed partials goes from right to left in the
Leibniz notation instead of left to right.
This chapter is (c) 2013. It was remixed by David Lippman from Shana Calaway's remix of Contemporary Calculus
by Dale Hoffman. It is licensed under the Creative Commons Attribution license.
Chapter 4 Functions of Two Variables Applied Calculus 272
Example 1
Find all four partial derivatives of f ( x, y ) = x 2 − 4 xy + 4 y 2
We have to start by finding the (first) partial derivatives:
f x ( x, y ) = 2 x − 4 y
f y ( x, y ) = −4 x + 8 y
Now we’re ready to take the second partial derivatives:
∂
f xx (x, y ) = (2 x − 4 y ) = 2
∂x
∂
f xy (x, y ) = (2 x − 4 y ) = −4
∂y
∂
f yx (x, y ) = (− 4 x + 8 y ) = −4
∂x
∂
f yy (x, y ) = (− 4 x + 8 y ) = 8
∂y
You might have noticed that the two mixed partial derivatives were equal in this last example. It
turns out that it’s not a coincidence – it’s a theorem.
Mixed Partial Derivative Theorem
If f , f x , f y , f xy , and f yx are all continuous (no breaks in their graph)
Then f xy = f yx .
In fact, as long as f and all its appropriate partial derivatives are continuous, the mixed
partials are equal even if they are of higher order, and even if the function has more than
two variables.
This theorem means that the confusing Leibniz notation for second derivatives is not a big
problem – in almost every situation, the mixed partials are equal, so it doesn’t matter in which
order we compute them.
Example 2
∂2 f e x+ y
Find for f ( x, y ) = 3 + y (ln y )
∂x∂y y +y
We already found the first partial derivatives in an earlier example:
∂f 1
= 3 e x+ y
∂x y + y
Chapter 4 Functions of Two Variables Applied Calculus 273
∂f
=
( )( ) ( )(
e x + y (1) y 3 + y − e x + y 3 y 2 + 1 ) 1
+ (1)(ln y ) + ( y )
∂y ( )
y3 + y
2
y
Now we need to find the mixed partial derivative – the Theorem says it doesn’t matter whether
∂f 1
we find the partial derivative of = 3 e x+ y with respect to y or the partial derivative of
∂x y + y
∂f
=
( )( ) ( )(
e x + y (1) y 3 + y − e x + y 3 y 2 + 1) 1
+ (1)(ln y ) + ( y ) with respect to x. Which would you
∂y ( )
y3 + y
2
y
rather do?
It looks like it will be easier to compute the mixed partial by finding the partial derivative of
∂f 1
= 3 e x+ y with respect to y – it still looks messy, but it looks less messy:
∂x y + y
∂2 f
=
∂2 f ∂ 1
= x+ y
e =
( )( ) ( )(
e x+ y y 3 + y − e x+ y 3 y 2 + 1 )
∂x∂y ∂y∂x ∂y y 3 + y y3 + y
2
( )
If you’d decided to do this the other way, you’d end up in the same place. Eventually.
Local Maxima, Minima, and Saddle Points
Let’s briefly review max-min problems in one variable.
A local max is a point on a curve that is higher than all the nearby points. A local min is lower
than all the nearby points. We know that local max or min can only occur at critical points,
where the derivative is zero or undefined. But we also know that not all critical points are max
or min, so we also need to test them, with the First Derivative or Second Derivative Test.
The situation with a function of two variables is much the same. Just as in the one-variable case,
the first step is to find critical points, places where both the partial derivatives are either zero or
undefined.
Definition:
f has a local maximum at (a, b) if f(a, b) ≥ f(x, y) for all points (x, y) near (a, b)
f has a local minimum at (a, b) if f(a, b) ≤ f(x, y) for all points (x, y) near (a, b)
A critical point for a function f(x, y) is a point (x, y) (or (x, y, f(x, y)) where both the
following are true:
f x = 0 or is undefined and f y = 0 or is undefined
Just as in the one-variable case, a local max or min of f can only occur at a critical point.
Chapter 4 Functions of Two Variables Applied Calculus 274
And then, just as in the one-variable setting, not all critical points are local max or min. For a
function of two variables, the critical point could be a local max, local min, or a saddle point.
A point on a surface is a local maximum if it’s higher than all the points nearby; a point is a local
minimum if it’s lower than all the points nearby.
A saddle point is a point on a surface that is a minimum along some paths and a maximum along
some others. It’s called this because it’s shaped a bit like a saddle you might use to ride a horse.
You can see a saddle point by making a fist – between the knuckles of your index and middle
fingers, you can see a place that is a minimum as you go across your knuckles, but a maximum
as you go along your hand toward your fingers.
Here is a picture of a saddle point from a few different angles. This is the surface
f ( x, y ) = 5 x 2 − 3 y 2 + 10 , and there is a saddle point above the origin. The lines show what the
surface looks like above the x- and y-axes. Notice how the point above the origin, where the
lines cross, is a local minimum in one direction, but a local maximum in the other direction.
Second Derivative Test
Just as in the one-variable case, we’ll need a way to test critical points to see whether they are
local max or min. There is a second derivative test for functions of two variables that can help –
but, just as in the one-variable case, it won’t always give an answer.
Chapter 4 Functions of Two Variables Applied Calculus 275
The Second Derivative Test for Functions of Two Variables:
Find all critical points of f(x,y).
Compute D = ( f xx )( f yy ) − ( f xy )( f yx ) , and evaluate it at each critical point.
(a) If D > 0, then f has a local max or min at the critical point. To see which, look at the sign
of f xx :
If f xx > 0 , then f has a local minimum at the critical point.
If f xx < 0 , then f has a local maximum at the critical point.
(b) If D < 0 then f has a saddle point at the critical point.
(c) If D = 0, there could be a local max, local min, or neither.
Example 3
Find all local maxima, minima, and saddle points for the function
f (x, y ) = x 3 + y 3 + 3x 2 − 3 y 2 − 8.
First we need the partial derivatives:
2
f x = 3 x 2 + 6 x and f y = 3 y − 6 y
Critical points are the places where both of these are zero (neither is ever undefined):
f x = 3x 2 + 6 x = 3 x( x + 2 ) = 0 when x = 0 or when x = −2.
f y = 3 y 2 − 6 y = 3 y ( y − 2 ) = 0 when y = 0 or when y = 2.
Putting these together, we get four critical points: (0, 0), (−2, 0), (0, 2), and (−2, 2).
Now to classify them, we’ll use the Second Derivative Test. We’ll need all the second partial
derivatives:
f xx = 6 x + 6 , f yy = 6 y − 6 , f xy = 0 = f yx
Then D = (6 x + 6 )(6 y − 6 ) − (0 )(0 ) = (6 x + 6 )(6 y − 6 ) .
Now look at each critical point in turn:
At (0, 0): D = (6(0 ) + 6 )(6(0 ) − 6 ) = (6 )(− 6 ) = −36 < 0 ; there is a saddle point at (0, 0).
At (−2, 0): D = (6(− 2 ) + 6 )(6(0 ) − 6 ) = (− 6)(− 6 ) = 36 > 0 , and f xx = 6(− 2 ) + 6 = −6 < 0 ; there is
a local maximum at (−2, 0).
At (0, 2): D = (6 )(6 ) > 0 and f xx = 6 > 0 ; there is a local minimum at (0, 2).
At (−2, 2): D = (− 6 )(6 ) < 0 ; there is another saddle point at (−2, 2).
Chapter 4 Functions of Two Variables Applied Calculus 276
Example 4
Find all local maxima, minima, and saddle points for the function
y3
z = 9x 3 + − 4 xy.
3
Solution: We’ll need all the partial derivatives and second partial derivatives, so let’s compute
them all first:
z x = 27 x 2 − 4 y; z y = y 2 − 4 x;
z xx = 54 x; z yy = 2 y; z xy = −4 = z yx
Now to find the critical points: We need both z x and z y to be zero (neither is ever undefined),
so we need to solve this set of equations simultaneously:
z x = 27 x 2 − 4 y = 0
z y = y 2 − 4x = 0
Perhaps it’s been a while since you solved systems of equations. Just remember the substitution
method – solve one equation for one variable and substitute into the other equation:
27 x 2 − 4 y = 0 2 y2
solve y − 4 x = 0 for x = , then substitute into the other equation
y 2 − 4x = 0 4
2
y2
27 − 4 y = 0
4
27 4
y − 4y = 0
16
Now we have just one equation in one variable to solve. Factoring out a y gives
27 3 27 3 4
y y − 4 = 0 , so y = 0 or y − 4 = 0 , giving y =
16 16 3
y2
Plugging back in to the equation x = to find x gives us the two critical points:
4
4 4
(0, 0) and , .
9 3
Now to test them. Compute D = ( f xx )( f yy ) − ( f xy )( f yx ) = (54 x )(2 y ) − (− 4 )(− 4 ) = 108 x − 16 .
Evaluate it at the two critical points, and see:
At (0,0): D = −16 < 0, so there is a saddle point at (0, 0).
4 4 4 4
At , : D = 48 > 0, and f xx > 0 , so there is a local minimum at , .
9 3 9 3
Chapter 4 Functions of Two Variables Applied Calculus 277
Applied Optimization
Example 5
A company makes two products. The demand equations for the two products are given below.
p1, p2, q1,and q2 are the prices and quantities for products 1 and 2.
q1 = 200 − 3 p1 − p 2
q 2 = 150 − p1 − 2 p 2
Find the price the company should charge for each product in order to maximize total revenue.
What is that maximum revenue?
Revenue is still price × quantity. If we’re selling two products, the total revenue will be the
sum of the revenues from the two products:
R ( p1 , p2 ) = p1q1 + p2 q2 = p1 (200 − 3 p1 − p2 ) + p2 (150 − p1 − 2 p2 )
2 2
R( p1 , p2 ) = 200 p1 − 3 p1 − 2 p1 p2 + 150 p2 − 2 p2
This is a function of two variables, the two prices, and we need to optimize it – just as in the
previous examples. First we find critical points. The notation here gets a bit hard to look at, but
hang in there – this is the same stuff we’ve done before.
R p = 200 − 6 p1 − 2 p 2 and R p = 150 − 2 p1 − 4 p 2
1 2
Solving these simultaneously gives the one critical point ( p1 , p 2 ) = (25,25) .
To confirm that this gives maximum revenue, we need to use the Second Derivative Test. Find
all the second derivatives:
R p p = −6 , R p p = −4 , and R p p = −2 = R p p
1 1 2 2 1 2 2 12
So D = (− 6 )(− 4 ) − (− 2 )(− 2 ) > 0 and R p p < 0 , so this really is a local maximum.
1 1
To maximize revenue, the company should charge $25 per unit for both products. This will
yield a maximum revenue of $4375.
Chapter 4 Functions of Two Variables Applied Calculus 278
4.3 Exercises
For problems 1 through 6, find f xx , f yy , f xy and f yx for the function given. Confirm that
f xy = f yx .
1. f (x, y ) = x 2 − 5 y 2
2. f (x, y ) = x 4 + 4 x 3 y − 6 x 2 y 2 − 4 xy 3 + y 4
3. f ( x, y ) = 5 x 2 y 2
4. f ( x, y ) = e x + 6 y
5. f ( x, y ) = ln ( xy + 2 x − 6 y )
x2
6. f ( x, y ) =
y4 − 5
7. Find the critical points of f (x, y ) = y 3 − x 3 + 15 x 2 − 12 y + 12 and use the Second Derivative
Test to classify them. If the test fails, say “the test fails.”
8. Find the critical points of f ( x, y ) = 2 xy − x 2 − 2 y 2 + 6 x + 4 and use the Second Derivative
Test to classify them. If the test fails, say “the test fails.”
9. Find the critical points of f ( x, y ) = y 2 − 4 ln( x ) + 4 x and use the Second Derivative Test to
classify them. If the test fails, say “the test fails.”
10. Find the critical points of f ( x, y ) = xy − 6 x 2 + 3 x − y + 2 and use the Second Derivative Test
to classify them. If the test fails, say “the test fails.”
11. The origin is a critical point for the function f ( x, y ) = x 3 + y 3 , and D = 0 there. That is, the
Second Derivative Test fails. Use what you know about shapes of functions to decide if there is
a local minimum, local maximum, or saddle point for this function at (0, 0).
12. The origin is a critical point for the function f ( x, y ) = 15 − x 2 y 2 , and D = 0 there. That is,
the Second Derivative Test fails. Use what you know about shapes of functions to decide if there
is a local minimum, local maximum, or saddle point for this function at (0, 0).
For problems 13 through 18, find all local maxima, minima, and saddle points for the function.
Chapter 4 Functions of Two Variables Applied Calculus 279
13. f ( x, y ) = xy − 5 x 2 − 5 y 2 + 33 y
14. f ( x, y ) = 10 xy − x 2 − y 2 + 3 x
15. f ( x, y ) = x 3 + y 3 − 3 xy
16. f ( x, y ) = 5 x 2 − 4 xy + 2 y 2 + 4 x − 4 y + 10
17. f ( x, y ) = y 2 e x + x 2
( )
18. f ( x, y ) = xy + 2 x − ln x 2 y , for x > 0 and y > 0 .
19. The demand functions for two products are given below. p1, p2, q1,and q2 are the prices (in
dollars) and quantities for products 1 and 2.
q1 = 200 − 3 p1 + p2
q2 = 150 + p1 − 2 p2
a. Are these two products complementary goods or substitute goods?
b. What is the quantity demanded for each when the price for product 1 is $20 per item
and the price for product 2 is $30 per item?
c. Write a function R( p1 , p 2 ) that expresses the total revenue from these two products.
d. Find the price and quantity for each product that maximizes the total revenue.
20. The demand functions for two products are given below. p1, p2, q1,and q2 are the prices (in
dollars) and quantities for products 1 and 2.
q1 = 350 + p1 + 2 p2
q2 = 225 + p1 + p2
a. Are these two products complementary goods or substitute goods?
b. What is the quantity demanded for each when the price for product 1 is $20 per item
and the price for product 2 is $30 per item?
c. Write a function R( p1 , p 2 ) that expresses the total revenue from these two products.
d. Find the price and quantity for each product that maximizes the total revenue.
21. Suppose the demand functions for two products are q1 = f ( p1 , p 2 ) and q 2 = g ( p1 , p 2 ) ,
where p1, p2, q1, and q2 are the prices (in dollars) and quantities for products 1 and 2. Consider
∂q ∂q ∂q ∂q 2
the four partial derivatives 1 , 1 , 2 , and . Tell the sign of each of these partial
∂p1 ∂p 2 ∂p1 ∂p 2
derivatives if
a. the products are complementary goods.
b. the products are substitute goods.