1 Understanding Derivative

Chapter 1
Understanding the Derivative
1.1 How do we measure velocity?

Motivating Questions
In this section, we strive to understand the ideas generated by the following important
questions:
• How is the average velocity of a moving object connected to the values of its
position function?
• How do we interpret the average velocity of an object geometrically with regard to
the graph of its position function?
• How is the notion of instantaneous velocity connected to average velocity?
Introduction
Calculus can be viewed broadly as the study of change. A natural and important question
to ask about any changing quantity is “how fast is the quantity changing?” It turns out that
in order to make the answer to this question precise, substantial mathematics is required.
We begin with a familiar problem: a ball being tossed straight up in the air from an
initial height. From this elementary scenario, we will ask questions about how the ball
is moving. These questions will lead us to begin investigating ideas that will be central
throughout our study of differential calculus and that have wide-ranging consequences.
In a great deal of our thinking about calculus, we will be well-served by remembering
this first example and asking ourselves how the various (sometimes abstract) ideas we are
considering are related to the simple act of tossing a ball straight up in the air.
1
2 1.1. HOW DO WE MEASURE VELOCITY?
Preview Activity 1.1. Suppose that the height s of a ball (in feet) at time t (in seconds) is
given by the formula s(t) = 64 − 16(t − 1)2 .
(a) Construct an accurate graph of y = s(t) on the time interval 0 ≤ t ≤ 3. Label at

least six distinct points on the graph, including the three points that correspond
to when the ball was released, when the ball reaches its highest point, and when
the ball lands.
(b) In everyday language, describe the behavior of the ball on the time interval
0 < t < 1 and on time interval 1 < t < 3. What occurs at the instant t = 1?
(c) Consider the expression
s(1) − s(0.5)
AV[0.5,1] = .
1 − 0.5
Compute the value of AV[0.5,1] . What does this value measure geometrically? What
does this value measure physically? In particular, what are the units on AV[0.5,1] ?
./
Position and average velocity
Any moving object has a position that can be considered a function of time. When this
motion is along a straight line, the position is given by a single variable, and we usually
let this position be denoted by s(t), which reflects the fact that position is a function of
time. For example, we might view s(t) as telling the mile marker of a car traveling on a
straight highway at time t in hours; similarly, the function s described in Preview Activity
1.1 is a position function, where position is measured vertically relative to the ground.
Not only does such a moving object have a position associated with its motion, but
on any time interval, the object has an average velocity. Think, for example, about driving
from one location to another: the vehicle travels some number of miles over a certain time
interval (measured in hours), from which we can compute the vehicle’s average velocity. In
this situation, average velocity is the number of miles traveled divided by the time elapsed,
which of course is given in miles per hour. Similarly, the calculation of AV[0.5,1] in Preview
Activity 1.1 found the average velocity of the ball on the time interval [0.5, 1], measured in
feet per second.
In general, we make the following definition: for an object moving in a straight line
whose position at time t is given by the function s(t), the average velocity of the object on the
interval from t = a to t = b, denoted AV[a, b] , is given by the formula
s(b) − s(a)
AV[a, b] = .
b−a
1.1. HOW DO WE MEASURE VELOCITY? 3
Note well: the units on AV[a, b] are “units of s per unit of t,” such as “miles per hour” or
“feet per second.”
Activity 1.1.
The following questions concern the position function given by s(t) = 64 − 16(t − 1)2 ,
which is the same function considered in Preview Activity 1.1.
(a) Compute the average velocity of the ball on each of the following time intervals:
[0.4, 0.8], [0.7, 0.8], [0.79, 0.8], [0.799, 0.8], [0.8, 1.2], [0.8, 0.9], [0.8, 0.81],
[0.8, 0.801]. Include units for each value.
(b) On the provided graph in Figure 1.1, sketch the line that passes through the
points A = (0.4, s(0.4)) and B = (0.8, s(0.8)). What is the meaning of the slope
of this line? In light of this meaning, what is a geometric way to interpret each
of the values computed in the preceding question?
(c) Use a graphing utility to plot the graph of s(t) = 64 − 16(t − 1)2 on an
interval containing the value t = 0.8. Then, zoom in repeatedly on the point
(0.8, s(0.8)). What do you observe about how the graph appears as you view it
more and more closely?
(d) What do you conjecture is the velocity of the ball at the instant t = 0.8? Why?
feet
s
64
B
A
56
48
sec
0.4 0.8 1.2
Figure 1.1: A partial plot of s(t) = 64 − 16(t − 1)2 .
Instantaneous Velocity
Whether driving a car, riding a bike, or throwing a ball, we have an intuitive sense that any
moving object has a velocity at any given moment – a number that measures how fast the
object is moving right now. For instance, a car’s speedometer tells the driver what appears
to be the car’s velocity at any given instant. In fact, the posted velocity on a speedometer
is really an average velocity that is computed over a very small time interval (by computing
how many revolutions the tires have undergone to compute distance traveled), since velocity
fundamentally comes from considering a change in position divided by a change in time.
But if we let the time interval over which average velocity is computed become shorter
and shorter, then we can progress from average velocity to instantaneous velocity.
Informally, we define the instantaneous velocity of a moving object at time t = a to be
the value that the average velocity approaches as we take smaller and smaller intervals
of time containing t = a to compute the average velocity. We will develop a more formal
definition of this momentarily, one that will end up being the foundation of much of our
work in first semester calculus. For now, it is fine to think of instantaneous velocity this
way: take average velocities on smaller and smaller time intervals, and if those average
velocities approach a single number, then that number will be the instantaneous velocity
at that point.
Activity 1.2.
Each of the following questions concern s(t) = 64 − 16(t − 1)2 , the position function
from Preview Activity 1.1.
(a) Compute the average velocity of the ball on the time interval [1.5, 2]. What is
different between this value and the average velocity on the interval [0, 0.5]?
(b) Use appropriate computing technology to estimate the instantaneous velocity
of the ball at t = 1.5. Likewise, estimate the instantaneous velocity of the ball
at t = 2. Which value is greater?
(c) How is the sign of the instantaneous velocity of the ball related to its behavior
at a given point in time? That is, what does positive instantaneous velocity tell
you the ball is doing? Negative instantaneous velocity?
(d) Without doing any computations, what do you expect to be the instantaneous
velocity of the ball at t = 1? Why?
C
At this point we have started to see a close connection between average velocity and
instantaneous velocity, as well as how each is connected not only to the physical behavior
of the moving object but also to the geometric behavior of the graph of the position
function. In order to make the link between average and instantaneous velocity more
formal, we will introduce the notion of limit in Section 1.2. As a preview of that concept,
we look at a way to consider the limiting value of average velocity through the introduction
of a parameter. Note that if we desire to know the instantaneous velocity at t = a of a
moving object with position function s, we are interested in computing average velocities
on the interval [a, b] for smaller and smaller intervals. One way to visualize this is to think
of the value b as being b = a + h, where h is a small number that is allowed to vary. Thus,
we observe that the average velocity of the object on the interval [a, a + h] is
s(a + h) − s(a)
AV[a, a+h] = ,
h
with the denominator being simply h because (a + h) − a = h. Initially, it is fine to think
of h being a small positive real number; but it is important to note that we allow h to
be a small negative number, too, as this enables us to investigate the average velocity of
the moving object on intervals prior to t = a, as well as following t = a. When h < 0,
AV[a, a+h] measures the average velocity on the interval [a + h, a].
To attempt to find the instantaneous velocity at t = a, we investigate what happens as
the value of h approaches zero. We consider this further in the following example.
Example 1.1. For a falling ball whose position function is given by s(t) = 16 − 16t 2 (where
s is measured in feet and t in seconds), find an expression for the average velocity of the
ball on a time interval of the form [0.5, 0.5 + h] where −0.5 < h < 0.5 and h , 0. Use this
expression to compute the average velocity on [0.5, 0.75] and [0.4, 0.5], as well as to make
a conjecture about the instantaneous velocity at t = 0.5.
Solution. We make the assumptions that −0.5 < h < 0.5 and h , 0 because h cannot be
zero (otherwise there is no interval on which to compute average velocity) and because the
function only makes sense on the time interval 0 ≤ t ≤ 1, as this is the duration of time
during which the ball is falling. Observe that we want to compute and simplify
s(0.5 + h) − s(0.5)
AV[0.5,0.5+h] = .
(0.5 + h) − 0.5
The most unusual part of this computation is finding s(0.5 + h). To do so, we follow the
rule that defines the function s. In particular, since s(t) = 16 − 16t 2 , we see that
s(0.5 + h) = 16 − 16(0.5 + h)2

= 16 − 16(0.25 + h + h2 )
= 16 − 4 − 16h − 16h2
= 12 − 16h − 16h2 .
Now, returning to our computation of the average velocity, we find that
s(0.5 + h) − s(0.5)
AV[0.5,0.5+h] =
(0.5 + h) − 0.5
(12 − 16h − 16h2 ) − (16 − 16(0.5)2 )
=
0.5 + h − 0.5
12 − 16h − 16h2 − 12
=
h
−16h − 16h2
= .
h
At this point, we note two things: first, the expression for average velocity clearly depends
on h, which it must, since as h changes the average velocity will change. Further, we note
that since h can never equal zero, we may further simplify the most recent expression.
Removing the common factor of h from the numerator and denominator, it follows that
AV[0.5,0.5+h] = −16 − 16h.
Now, for any small positive or negative value of h, we can compute the average velocity.
For instance, to obtain the average velocity on [0.5, 0.75], we let h = 0.25, and the average
velocity is −16 − 16(0.25) = −20 ft/sec. To get the average velocity on [0.4, 0.5], we let
h = −0.1, which tells us the average velocity is −16 − 16(−0.1) = −14.4 ft/sec. Moreover, we
can even explore what happens to AV[0.5,0.5+h] as h gets closer and closer to zero. As h
approaches zero, −16h will also approach zero, and thus it appears that the instantaneous
velocity of the ball at t = 0.5 should be −16 ft/sec.
Activity 1.3.
For the function given by s(t) = 64 − 16(t − 1)2 from Preview Activity 1.1, find the
most simplified expression you can for the average velocity of the ball on the interval
[2, 2 + h]. Use your result to compute the average velocity on [1.5, 2] and to estimate
the instantaneous velocity at t = 2. Finally, compare your earlier work in Activity 1.1.
C
Summary
In this section, we encountered the following important ideas:
• The average velocity on [a, b] can be viewed geometrically as the slope of the line
between the points (a, s(a)) and (b, s(b)) on the graph of y = s(t), as shown in Figure 1.2.
• Given a moving object whose position at time t is given by a function s, the average
s(b)−s(a)
m= b−a
(b, s(b))
(a, s(a))
Figure 1.2: The graph of position function s together with the line through (a, s(a)) and
(b, s(b)) whose slope is m = s(b)−s(a)
b−a . The line’s slope is the average rate of change of s on
the interval [a, b].
velocity of the object on the time interval [a, b] is given by AV[a,b] = s(b)−s(a)
b−a . Viewing
the interval [a, b] as having the form [a, a + h], we equivalently compute average velocity
by the formula AV[a,a+h] = s(a+h)−s(a)
h .
• The instantaneous velocity of a moving object at a fixed time is estimated by considering

average velocities on shorter and shorter time intervals that contain the instant of
interest.
Exercises
1. A bungee jumper dives from a tower at time t = 0. Her height h (measured in feet) at
time t (in seconds) is given by the graph in Figure 1.3.
In this problem, you may base your answers on estimates from the graph or use the
fact that the jumper’s height function is given by s(t) = 100 cos(0.75t) · e−0.2t + 100.
(a) What is the change in vertical position of the bungee jumper between t = 0
and t = 15?
(b) Estimate the jumper’s average velocity on each of the following time intervals:
[0, 15], [0, 2], [1, 6], and [8, 10]. Include units on your answers.
(c) On what time interval(s) do you think the bungee jumper achieves her greatest
average velocity? Why?
(d) Estimate the jumper’s instantaneous velocity at t = 5. Show your work and
explain your reasoning, and include units on your answer.
200
s
150
100
50
5 10 15 20
Figure 1.3: A bungee jumper’s height function.
(e) Among the average and instantaneous velocities you computed in earlier
questions, which are positive and which are negative? What does negative
velocity indicate?
2. A diver leaps from a 3 meter springboard. His feet leave the board at time t = 0,
he reaches his maximum height of 4.5 m at t = 1.1 seconds, and enters the water at
t = 2.45. Once in the water, the diver coasts to the bottom of the pool (depth 3.5 m),
touches bottom at t = 7, rests for one second, and then pushes off the bottom. From
there he coasts to the surface, and takes his first breath at t = 13.
(a) Let s(t) denote the function that gives the height of the diver’s feet (in meters)
above the water at time t. (Note that the “height” of the bottom of the pool
is −3.5 meters.) Sketch a carefully labeled graph of s(t) on the provided axes
in Figure 1.4. Include scale and units on the vertical axis. Be as detailed as
possible.
(b) Based on your graph in (a), what is the average velocity of the diver between
t = 2.45 and t = 7? Is his average velocity the same on every time interval
within [2.45, 7]?
(c) Let the function v(t) represent the instantaneous vertical velocity of the diver
at time t (i.e. the speed at which the height function s(t) is changing; note
that velocity in the upward direction is positive, while the velocity of a falling
object is negative). Based on your understanding of the diver’s behavior, as
well as your graph of the position function, sketch a carefully labeled graph of
v(t) on the axes provided in Figure 1.4. Include scale and units on the vertical
axis. Write several sentences that explain how you constructed your graph,
discussing when you expect v(t) to be zero, positive, negative, relatively large,
and relatively small.
s v
t t
2 4 6 8 10 12 2 4 6 8 10 12
Figure 1.4: Axes for plotting s(t) in part (a) and v(t) in part (c) of the diver problem.
(d) Is there a connection between the two graphs that you can describe? What
can you say about the velocity graph when the height function is increasing?
decreasing? Make as many observations as you can.
3. According to the U.S. census, the population of the city of Grand Rapids, MI, was
181,843 in 1980; 189,126 in 1990; and 197,800 in 2000.
(a) Between 1980 and 2000, by how many people did the population of Grand
Rapids grow?
(b) In an average year between 1980 and 2000, by how many people did the
population of Grand Rapids grow?
(c) Just like we can find the average velocity of a moving body by computing
change in position over change in time, we can compute the average rate of
change of any function f . In particular, the average rate of change of a function
f over an interval [a, b] is the quotient
f (b) − f (a)
.
b−a
f (b)− f (a)
What does the quantity b−a measure on the graph of y = f (x) over the
interval [a, b]?
(d) Let P(t) represent the population of Grand Rapids at time t, where t is measured
in years from January 1, 1980. What is the average rate of change of P on the
interval t = 0 to t = 20? What are the units on this quantity?
(e) If we assume the population of Grand Rapids is growing at a rate of ap-
proximately 4% per decade, we can model the population function with the
formula
P(t) = 181843(1.04)t /10 .
Use this formula to compute the average rate of change of the population on
the intervals [5, 10], [5, 9], [5, 8], [5, 7], and [5, 6].
(f) How fast do you think the population of Grand Rapids was changing on
January 1, 1985? Said differently, at what rate do you think people were being
added to the population of Grand Rapids as of January 1, 1985? How many
additional people should the city have expected in the following year? Why?
1.2. THE NOTION OF LIMIT 11
1.2 The notion of limit

questions:
• What is the mathematical notion of limit and what role do limits play in the study
of functions?
• What is the meaning of the notation lim f (x) = L?
x→a
• How do we go about determining the value of the limit of a function at a point?

• How does the notion of limit allow us to move from average velocity to instanta-
neous velocity?
Introduction
Functions are at the heart of mathematics: a function is a process or rule that associates
each individual input to exactly one corresponding output. Students learn in courses prior
to calculus that there are many different ways to represent functions, including through
formulas, graphs, tables, and even words. For example, the squaring function can be
thought of in any of these ways. In words, the squaring function takes any real number x
and computes its square. The formulaic and graphical representations go hand in hand,
as y = f (x) = x 2 is one of the simplest curves to graph. Finally, we can also partially
represent this function through a table of values, essentially by listing some of the ordered
pairs that lie on the curve, such as (−2, 4), (−1, 1), (0, 0), (1, 1), and (2, 4).
Functions are especially important in calculus because they often model important
phenomena – the location of a moving object at a given time, the rate at which an
automobile is consuming gasoline at a certain velocity, the reaction of a patient to the
size of a dose of a drug – and calculus can be used to study how these output quantities
change in response to changes in the input variable. Moreover, thinking about concepts
like average and instantaneous velocity leads us naturally from an initial function to a
related, sometimes more complicated function. As one example of this, think about the
falling ball whose position function is given by s(t) = 64 − 16t 2 and the average velocity of
the ball on the interval [1, x]. Observe that
s(x) − s(1) (64 − 16x 2 ) − (64 − 16) 16 − 16x 2

AV[1, x] = = = .
x−1 x−1 x−1
Now, two things are essential to note: this average velocity depends on x (indeed, AV[1, x] is
a function of x), and our most focused interest in this function occurs near x = 1, which is
2
where the function is not defined. Said differently, the function g(x) = 16−16x
x−1 tells us the
12 1.2. THE NOTION OF LIMIT
average velocity of the ball on the interval from t = 1 to t = x, and if we are interested in
the instantaneous velocity of the ball when t = 1, we’d like to know what happens to g(x)
as x gets closer and closer to 1. At the same time, g(1) is not defined, because it leads to
the quotient 0/0.
This is where the idea of limits comes in. By using a limit, we’ll be able to allow x to
get arbitrarily close, but not equal, to 1 and fully understand the behavior of g(x) near this
value. We’ll develop key language, notation, and conceptual understanding in what follows,
but for now we consider a preliminary activity that uses the graphical interpretation of a
function to explore points on a graph where interesting behavior occurs.
Preview Activity 1.2. Suppose that g is the function given by the graph below. Use the
graph to answer each of the following questions.
(a) Determine the values g(−2), g(−1), g(0), g(1), and g(2), if defined. If the function
value is not defined, explain what feature of the graph tells you this.
(b) For each of the values a = −1, a = 0, and a = 2, complete the following sentence:
“As x gets closer and closer (but not equal) to a, g(x) gets as close as we want to
.”
(c) What happens as x gets closer and closer (but not equal) to a = 1? Does the
function g(x) get as close as we would like to a single value?
g
3
-2 -1 1 2 3
-1
Figure 1.5: Graph of y = g(x) for Preview Activity 1.2.
./
The Notion of Limit
Limits can be thought of as a way to study the tendency or trend of a function as the input
variable approaches a fixed value, or even as the input variable increases or decreases
without bound. We put off the study of the latter idea until further along in the course
when we will have some helpful calculus tools for understanding the end behavior of
functions. Here, we focus on what it means to say that “a function f has limit L as x
approaches a.” To begin, we think about a recent example.
In Preview Activity 1.2, you saw that for the given function g, as x gets closer and
closer (but not equal) to 0, g(x) gets as close as we want to the value 4. At first, this may
feel counterintuitive, because the value of g(0) is 1, not 4. By their very definition, limits
regard the behavior of a function arbitrarily close to a fixed input, but the value of the
function at the fixed input does not matter. More formally1 , we say the following.
Definition 1.1. Given a function f , a fixed input x = a, and a real number L, we say that
f has limit L as x approaches a, and write
lim f (x) = L
x→a
provided that we can make f (x) as close to L as we like by taking x sufficiently close (but
not equal) to a. If we cannot make f (x) as close to a single value as we would like as x
approaches a, then we say that f does not have a limit as x approaches a.
For the function g pictured in Figure 1.5, we can make the following observations:
lim g(x) = 3, lim g(x) = 4, and lim g(x) = 1,

x→−1 x→0 x→2
but g does not have a limit as x → 1. When working graphically, it suffices to ask if the
function approaches a single value from each side of the fixed input, while understanding
that the function value right at the fixed input is irrelevant. This reasoning explains the
values of the first three stated limits. In a situation such as the jump in the graph of g at
x = 1, the issue is that if we approach x = 1 from the left, the function values tend to get
as close to 3 as we’d like, but if we approach x = 1 from the right, the function values get
as close to 2 as we’d like, and there is no single number that all of these function values
approach. This is why the limit of g does not exist at x = 1.
For any function f , there are typically three ways to answer the question “does f have
a limit at x = a, and if so, what is the limit?” The first is to reason graphically as we
have just done with the example from Preview Activity 1.2. If we have a formula for f (x),
there are two additional possibilities: (1) evaluate the function at a sequence of inputs that
approach a on either side, typically using some sort of computing technology, and ask if
the sequence of outputs seems to approach a single value; (2) use the algebraic form of the
function to understand the trend in its output as the input values approach a. The first
approach only produces an approximation of the value of the limit, while the latter can
1 What follows here is not what mathematicians consider the formal definition of a limit. To be completely
precise, it is necessary to quantify both what it means to say “as close to L as we like” and “sufficiently close
to a.” That can be accomplished through what is traditionally called the epsilon-delta definition of limits.
The definition presented here is sufficient for the purposes of this text.
often be used to determine the limit exactly. The following example demonstrates both of
these approaches, while also using the graphs of the respective functions to help confirm
our conclusions.
Example 1.2. For each of the following functions, we’d like to know whether or not the
function has a limit at the stated a-values. Use both numerical and algebraic approaches
to investigate and, if possible, estimate or determine the value of the limit. Compare the
results with a careful graph of the function on an interval containing the points of interest.
4 − x2
(a) f (x) = ; a = −1, a = −2
x+2
π
(b) g(x) = sin ; a = 3, a = 0
x
Solution. We first construct a graph of f along with tables of values near a = −1 and
a = −2.
x f (x) x f (x)
-0.9 2.9 -1.9 3.9 f
-0.99 2.99 -1.99 3.99 5
-0.999 2.999 -1.999 3.999
-0.9999 2.9999 -1.9999 3.9999
3
-1.1 3.1 -2.1 4.1
-1.01 3.01 -2.01 4.01
-1.001 3.001 -2.001 4.001 1
-1.0001 3.0001 -2.0001 4.0001
-3 -1 1
4 − x2
Figure 1.6: Tables and graph for f (x) = .
x+2
From the left table, it appears that we can make f as close as we want to 3 by taking x
sufficiently close to −1, which suggests that lim f (x) = 3. This is also consistent with the
x→−1
graph of f . To see this a bit more rigorously and from an algebraic point of view, consider
2
the formula for f : f (x) = 4−xx+2 . The numerator and denominator are each polynomial
functions, which are among the most well-behaved functions that exist. Formally, such
functions are continuous 2 , which means that the limit of the function at any point is equal
2 See Section 1.7 for more on the notion of continuity.
to its function value. Here, it follows that as x → −1, (4 − x 2 ) → (4 − (−1)2 ) = 3, and

(x + 2) → (−1 + 2) = 1, so as x → −1, the numerator of f tends to 3 and the denominator
3
tends to 1, hence lim f (x) = = 3.
x→−1 1
The situation is more complicated when x → −2, due in part to the fact that f (−2) is
not defined. If we attempt to use a similar algebraic argument regarding the numerator
and denominator, we observe that as x → −2, (4 − x 2 ) → (4 − (−2)2 ) = 0, and (x + 2) →
(−2 + 2) = 0, so as x → −2, the numerator of f tends to 0 and the denominator tends to
0. We call 0/0 an indeterminate form and will revisit several important issues surrounding
such quantities later in the course. For now, we simply observe that this tells us there is
somehow more work to do. From the table and the graph, it appears that f should have a
limit of 4 at x = −2. To see algebraically why this is the case, let’s work directly with the
form of f (x). Observe that
4 − x2
lim f (x) = lim
x→−2 x→−2 x + 2
(2 − x)(2 + x)
= lim .
x→−2 x+2
At this point, it is important to observe that since we are taking the limit as x → −2, we
are considering x values that are close, but not equal, to −2. Since we never actually allow
x to equal −2, the quotient 2+x
x+2 has value 1 for every possible value of x. Thus, we can
simplify the most recent expression above, and now find that
lim f (x) = lim 2 − x.

x→−2 x→−2
Because 2 − x is simply a linear function, this limit is now easy to determine, and its value
clearly is 4. Thus, from several points of view we’ve seen that lim f (x) = 4.
x→−2
Next we turn to the function g, and construct two tables and a graph.
x g(x) x g(x)
2.9 0.84864 -0.1 0
2.99 0.86428 -0.01 0 2
2.999 0.86585 -0.001 0 g
2.9999 0.86601 -0.0001 0
3.1 0.88351 0.1 0 -3 -1 1 3
3.01 0.86777 0.01 0
3.001 0.86620 0.001 0 -2
3.0001 0.86604 0.0001 0
π
Figure 1.7: Tables and graph for g(x) = sin .
x
First, as x → 3, it appears from the data (and the graph) that the function is
approaching approximately 0.866025. To be precise, we have to use the fact that πx → π3 ,
√
3
and thus we find that g(x) = sin( πx ) → sin( π3 ) as x → 3. The exact value of sin( π3 ) is 2 ,
which is approximately 0.8660254038. Thus, we see that
√
3
lim g(x) = .
x→3 2
As x → 0, we observe that πx does not behave in an elementary way. When x is

positive and approaching zero, we are dividing by smaller and smaller positive values,
and πx increases without bound. When x is negative and approaching zero, πx decreases
without bound. In this sense, as we get close to x = 0, the inputs to the sine function are
growing rapidly, and this leads to wild oscillations in the graph of g. It is an instructive
π
exercise to plot the function g(x) = sin x with a graphing utility and then zoom in on

x = 0. Doing so shows that the function never settles down to a single value near the
origin and suggests that g does not have a limit at x = 0.
How do we reconcile this with the righthand table above, which seems to suggest
that the limit of g as x approaches 0 may in fact be 0? Here we need to recognize that
the data misleads us because of the special nature ofthe sequence {0.1, 0.01, 0.001, . . .}:
when we evaluate g(10−k ), we get g(10−k ) = sin 10π−k = sin(10k π) = 0 for each positive
integer value of k. But if we take a different sequence of values approaching zero, say
{0.3, 0.03, 0.003, . . .}, then we find that
π √
10k π 3
!
g(3 · 10 ) = sin
−k
= sin = ≈ 0.866025.
3 · 10 −k 3 2
√
That sequence of data would suggest that the value of the limit is 23 . Clearly the function
cannot have two different values for the limit, and this shows that g has no limit as x → 0.
An important lesson to take from Example 1.2 is that tables can be misleading when
determining the value of a limit. While a table of values is useful for investigating the
possible value of a limit, we should also use other tools to confirm the value, if we think
the table suggests the limit exists.
Activity 1.4.
Estimate the value of each of the following limits by constructing appropriate tables of
values. Then determine the exact value of the limit by using algebra to simplify the
function. Finally, plot each function on an appropriate interval to check your result
visually.
x2 − 1
(a) lim
x→1 x−1
(2 + x)3 − 8
(b) lim
x→0 x
√
x+1−1
(c) lim
x→0 x
C
This concludes a rather lengthy introduction to the notion of limits. It is important to
remember that our primary motivation for considering limits of functions comes from our
interest in studying the rate of change of a function. To that end, we close this section by
revisiting our previous work with average and instantaneous velocity and highlighting the
role that limits play.
Instantaneous Velocity
Suppose that we have a moving object whose position at time t is given by a function s. We
know that the average velocity of the object on the time interval [a, b] is AV[a,b] = s(b)−s(a)
b−a .
We define the instantaneous velocity at a to be the limit of average velocity as b approaches
a. Note particularly that as b → a, the length of the time interval gets shorter and
shorter (while always including a). In Section 1.3, we will introduce a helpful shorthand
notation to represent the instantaneous rate of change. For now, we will write IVt=a for
the instantaneous velocity at t = a, and thus
s(b) − s(a)
IVt=a = lim AV[a, b] = lim .
b→a b→a b−a
Equivalently, if we think of the changing value b as being of the form b = a + h, where h is
some small number, then we may instead write
s(a + h) − s(a)
IVt=a = lim AV[a,a+h] = lim .
h→0 h→0 h
Again, the most important idea here is that to compute instantaneous velocity, we take a
limit of average velocities as the time interval shrinks. Two different activities offer the
opportunity to investigate these ideas and the role of limits further.
Activity 1.5.
Consider a moving object whose position function is given by s(t) = t 2 , where s is
measured in meters and t is measured in minutes.
(a) Determine the most simplified expression for the average velocity of the object
on the interval [3, 3 + h], where h > 0.
(b) Determine the average velocity of the object on the interval [3, 3.2]. Include
units on your answer.
(c) Determine the instantaneous velocity of the object when t = 3. Include units
on your answer.
C
The closing activity of this section asks you to make some connections among average
velocity, instantaneous velocity, and slopes of certain lines.
Activity 1.6.
For the moving object whose position s at time t is given by the graph below, answer
each of the following questions. Assume that s is measured in feet and t is measured in
seconds.
5 s
1
t
1 3 5
Figure 1.8: Plot of the position function y = s(t) in Activity 1.6.
(a) Use the graph to estimate the average velocity of the object on each of the
following intervals: [0.5, 1], [1.5, 2.5], [0, 5]. Draw each line whose slope
represents the average velocity you seek.
(b) How could you use average velocities or slopes of lines to estimate the instanta-
neous velocity of the object at a fixed time?
(c) Use the graph to estimate the instantaneous velocity of the object when t = 2.
Should this instantaneous velocity at t = 2 be greater or less than the average
velocity on [1.5, 2.5] that you computed in (a)? Why?
C
Summary
• Limits enable us to examine trends in function behavior near a specific point. In

particular, taking a limit at a given point asks if the function values nearby tend to
approach a particular fixed value.

• When we write lim f (x) = L, we read this as saying “the limit of f as x approaches a
x→a
is L,” and this means that we can make the value of f (x) as close to L as we want by
taking x sufficiently close (but not equal) to a.
• If we desire to know lim f (x) for a given value of a and a known function f , we can
x→a
estimate this value from the graph of f or by generating a table of function values that
result from a sequence of x-values that are closer and closer to a. If we want the exact
value of the limit, we need to work with the function algebraically and see if we can use
familiar properties of known, basic functions to understand how different parts of the
formula for f change as x → a.
• The instantaneous velocity of a moving object at a fixed time is found by taking the
limit of average velocities of the object over shorter and shorter time intervals that all
contain the time of interest.
Exercises
16 − x 4
1. Consider the function whose formula is f (x) = .
x2 − 4
(a) What is the domain of f ?
(b) Use a sequence of values of x near a = 2 to estimate the value of lim f (x), if
x→2
you think the limit exists. If you think the limit doesn’t exist, explain why.
4
(c) Use algebra to simplify the expression 16−x x 2 −4
and hence work to evaluate
lim x→2 f (x) exactly, if it exists, or to explain how your work shows the limit
fails to exist. Discuss how your findings compare to your results in (b).
(d) True or false: f (2) = −8. Why?
4
(e) True or false: 16−x
x 2 −4
= −4 − x 2 . Why? How is this equality connected to your
work above with the function f ?
(f) Based on all of your work above, construct an accurate, labeled graph of
y = f (x) on the interval [1, 3], and write a sentence that explains what you now
16 − x 4
know about lim 2 .
x→2 x − 4
|x + 3|
2. Let g(x) = − .
x+3
(a) What is the domain of g?
(b) Use a sequence of values near a = −3 to estimate the value of lim x→−3 g(x), if
you think the limit exists. If you think the limit doesn’t exist, explain why.
(c) Use algebra to simplify the expression |x+3|x+3 and hence work to evaluate
lim x→−3 g(x) exactly, if it exists, or to explain how your work shows the
limit fails to exist. Discuss how your findings compare to your results in (b).
(Hint: |a| = a whenever a ≥ 0, but |a| = −a whenever a < 0.)
(d) True or false: g(−3) = −1. Why?
(e) True or false: − |x+3|
x+3 = −1. Why? How is this equality connected to your work
above with the function g?
(f) Based on all of your work above, construct an accurate, labeled graph of
y = g(x) on the interval [−4, −2], and write a sentence that explains what you
now know about lim g(x).
x→−3
3. For each of the following prompts, sketch a graph on the provided axes of a function
that has the stated properties.
3 3
-3 3 -3 3
-3 -3
Figure 1.9: Axes for plotting y = f (x) in (a) and y = g(x) in (b).
(a) y = f (x) such that

• f (−2) = 2 and lim f (x) = 1
x→−2
• f (−1) = 3 and lim f (x) = 3
x→−1
• f (1) is not defined and lim f (x) = 0
x→1
• f (2) = 1 and lim f (x) does not exist.
x→2
(b) y = g(x) such that

• g(−2) = 3, g(−1) = −1, g(1) = −2, and g(2) = 3
• At x = −2, −1, 1 and 2, g has a limit, and its limit equals the value of the
function at that point.
• g(0) is not defined and lim g(x) does not exist.
x→0
4. A bungee jumper dives from a tower at time t = 0. Her height s in feet at time t in
seconds is given by s(t) = 100 cos(0.75t) · e−0.2t + 100.
(a) Write an expression for the average velocity of the bungee jumper on the
interval
[1, 1 + h].
(b) Use computing technology to estimate the value of the limit as h → 0 of the
quantity you found in (a).
(c) What is the meaning of the value of the limit in (b)? What are its units?
22 1.3. THE DERIVATIVE OF A FUNCTION AT A POINT
1.3 The derivative of a function at a point

questions:
• How is the average rate of change of a function on a given interval defined, and
what does this quantity measure?
• How is the instantaneous rate of change of a function at a particular point defined?
How is the instantaneous rate of change linked to average rate of change?
• What is the derivative of a function at a given point? What does this derivative
value measure? How do we interpret the derivative value graphically?
• How are limits used formally in the computation of derivatives?
Introduction
An idea that sits at the foundations of calculus is the instantaneous rate of change of a
function. This rate of change is always considered with respect to change in the input
variable, often at a particular fixed input value. This is a generalization of the notion
of instantaneous velocity and essentially allows us to consider the question “how do we
measure how fast a particular function is changing at a given point?” When the original
function represents the position of a moving object, this instantaneous rate of change is
precisely velocity, and might be measured in units such as feet per second. But in other
contexts, instantaneous rate of change could measure the number of cells added to a
bacteria culture per day, the number of additional gallons of gasoline consumed by going
one mile per additional mile per hour in a car’s velocity, or the number of dollars added
to a mortgage payment for each percentage increase in interest rate. Regardless of the
presence of a physical or practical interpretation of a function, the instantaneous rate of
change may also be interpreted geometrically in connection to the function’s graph, and
this connection is also foundational to many of the main ideas in calculus.
In what follows, we will introduce terminology and notation that makes it easier
to talk about the instantaneous rate of change of a function at a point. In addition,
just as instantaneous velocity is defined in terms of average velocity, the more general
instantaneous rate of change will be connected to the more general average rate of change.
Recall that for a moving object with position function s, its average velocity on the time
interval t = a to t = a + h is given by the quotient
s(a + h) − s(a)
AV[a,a+h] = .
h
1.3. THE DERIVATIVE OF A FUNCTION AT A POINT 23
In a similar way, we make the following definition for an arbitrary function y = f (x).
Definition 1.2. For a function f , the average rate of change of f on the interval [a, a + h]
is given by the value
f (a + h) − f (a)
AV[a, a+h] = .
h
Equivalently, if we want to consider the average rate of change of f on [a, b], we compute
f (b) − f (a)
AV[a, b] = .
b−a
It is essential to understand how the average rate of change of f on an interval is connected
to its graph.
Preview Activity 1.3. Suppose that f is the function given by the graph below and that
a and a + h are the input values as labeled on the x-axis. Use the graph in Figure 1.10 to
answer the following questions.
a a+h
Figure 1.10: Plot of y = f (x) for Preview Activity 1.3.
(a) Locate and label the points (a, f (a)) and (a + h, f (a + h)) on the graph.
(b) Construct a right triangle whose hypotenuse is the line segment from (a, f (a)) to
(a + h, f (a + h)). What are the lengths of the respective legs of this triangle?
(c) What is the slope of the line that connects the points (a, f (a)) and (a + h, f (a + h))?
(d) Write a meaningful sentence that explains how the average rate of change of the
function on a given interval and the slope of a related line are connected.
./
The Derivative of a Function at a Point
Just as we defined instantaneous velocity in terms of average velocity, we now define

the instantaneous rate of change of a function at a point in terms of the average rate of
change of the function f over related intervals. In addition, we give a special name to “the
instantaneous rate of change of f at a,” calling this quantity “the derivative of f at a,”
with this value being represented by the shorthand notation f 0(a). Specifically, we make
the following definition.
Definition 1.3. Let f be a function and x = a a value in the function’s domain. We define
the derivative of f with respect to x evaluated at x = a, denoted f 0(a), by the formula
f (a + h) − f (a)
f 0(a) = lim ,
h→0 h
provided this limit exists.
Aloud, we read the symbol f 0(a) as either “ f -prime at a” or “the derivative of f

evaluated at x = a.” Much of the next several chapters will be devoted to understanding,
computing, applying, and interpreting derivatives. For now, we make the following
important notes.
• The derivative of f at the value x = a is defined as the limit of the average rate of
change of f on the interval [a, a + h] as h → 0. It is possible for this limit not to
exist, so not every function has a derivative at every point.
• We say that a function that has a derivative at x = a is differentiable at x = a.
• The derivative is a generalization of the instantaneous velocity of a position function:

when y = s(t) is a position function of a moving body, s 0(a) tells us the instantaneous
velocity of the body at time t = a.
• Because the units on f (a+h)−

h
f (a)
are “units of f per unit of x,” the derivative has
these very same units. For instance, if s measures position in feet and t measures
time in seconds, the units on s 0(a) are feet per second.
• Because the quantity f (a+h)−

h
f (a)
represents the slope of the line through (a, f (a))
and (a + h, f (a + h)), when we compute the derivative we are taking the limit of a
collection of slopes of lines, and thus the derivative itself represents the slope of a
particularly important line.
While all of the above ideas are important and we will add depth and perspective to
them through additional time and study, for now it is most essential to recognize how the
derivative of a function at a given value represents the slope of a certain line. Thus, we
expand upon the last bullet item above.
As we move from an average rate of change to an instantaneous one, we can think of

one point as “sliding towards” another. In particular, provided the function has a derivative
at (a, f (a)), the point (a + h, f (a + h)) will approach (a, f (a)) as h → 0. Because this
process of taking a limit is a dynamic one, it can be helpful to use computing technology
to visualize what the limit is accomplishing. While there are many different options3 , one
of the best is a java applet in which the user is able to control the point that is moving.
See the examples referenced in the footnote here, or consider building your own, perhaps
using the fantastic free program Geogebra4 .
In Figure 1.11, we provide a sequence of figures with several different lines through the
points (a, f (a)) and (a + h, f (a + h)) that are generated by different values of h. These
lines (shown in the first three figures in magenta), are often called secant lines to the curve
y = f (x). A secant line to a curve is simply a line that passes through two points that lie
on the curve. For each such line, the slope of the secant line is m = f (a+h)−
h
f (a)
, where the
value of h depends on the location of the point we choose. We can see in the diagram
how, as h → 0, the secant lines start to approach a single line that passes through the
point (a, f (a)). In the situation where the limit of the slopes of the secant lines exists, we
say that the resulting value is the slope of the tangent line to the curve. This tangent line
(shown in the right-most figure in green) to the graph of y = f (x) at the point (a, f (a)) is
the line through (a, f (a)) whose slope is m = f 0(a).
y y y y
f f f f
x x x x
a a a a
Figure 1.11: A sequence of secant lines approaching the tangent line to f at (a, f (a)).
As we will see in subsequent study, the existence of the tangent line at x = a is

connected to whether or not the function f looks like a straight line when viewed up close
at (a, f (a)), which can also be seen in Figure 1.12, where we combine the four graphs in
Figure 1.11 into the single one on the left, and then we zoom in on the box centered at
(a, f (a)), with that view expanded on the right (with two of the secant lines omitted). Note
how the tangent line sits relative to the curve y = f (x) at (a, f (a)) and how closely it
3 For a helpful collection of java applets, consider the work of David Austin of Grand Valley State University
at http://gvsu.edu/s/5r, and the particularly relevant example at http://gvsu.edu/s/5s. For applets
that have been built in Geogebra, a nice example is the work of Marc Renault of Shippensburg University at
http://gvsu.edu/s/5p, with the example at http://gvsu.edu/s/5q being especially fitting for our work
in this section. There are scores of other examples posted by other authors on the internet.
4 Available for free download from http://geogebra.org.
resembles the curve near x = a.
x
a
Figure 1.12: A sequence of secant lines approaching the tangent line to f at (a, f (a)). At
right, we zoom in on the point (a, f (a)). The slope of the tangent line (in green) to f at
(a, f (a)) is given by f 0(a).
At this time, it is most important to note that f 0(a), the instantaneous rate of change
of f with respect to x at x = a, also measures the slope of the tangent line to the curve
y = f (x) at (a, f (a)). The following example demonstrates several key ideas involving the
derivative of a function.
Example 1.3. For the function given by f (x) = x − x 2 , use the limit definition of the
derivative to compute f 0(2). In addition, discuss the meaning of this value and draw a
labeled graph that supports your explanation.
Solution. From the limit definition, we know that

f (2 + h) − f (2)
f 0(2) = lim .
h→0 h
Now we use the rule for f , and observe that f (2) = 2 − 22 = −2 and f (2 + h) =
(2 + h) − (2 + h)2 . Substituting these values into the limit definition, we have that
(2 + h) − (2 + h)2 − (−2)
f 0(2) = lim .
h→0 h
Observe that with h in the denominator and our desire to let h → 0, we have to wait
to take the limit (that is, we wait to actually let h approach 0). Thus, we do additional
m = f ′ (2)
1 2
-2
-4
y = x − x2
Figure 1.13: The tangent line to y = x − x 2 at the point (2, −2).
algebra. Expanding and distributing in the numerator,
2 + h − 4 − 4h − h2 + 2
f 0(2) = lim .
h→0 h
Combining like terms, we have
−3h − h2
f 0(2) = lim .
h→0 h
Next, we observe that there is a common factor of h in both the numerator and denomina-
tor, which allows us to simplify and find that
f 0(2) = lim (−3 − h).

h→0
Finally, we are able to take the limit as h → 0, and thus conclude that f 0(2) = −3.
Now, we know that f 0(2) represents the slope of the tangent line to the curve y = x − x 2
at the point (2, −2); f 0(2) is also the instantaneous rate of change of f at the point (2, −2).
Graphing both the function and the line through (2, −2) with slope m = f 0(2) = −3, we
indeed see that by calculating the derivative, we have found the slope of the tangent line
at this point, as shown in Figure 1.3.
The following activities will help you explore a variety of key ideas related to derivatives.
Activity 1.7.
Consider the function f whose formula is f (x) = 3 − 2x.
(a) What familiar type of function is f ? What can you say about the slope of f at
every value of x?
(b) Compute the average rate of change of f on the intervals [1, 4], [3, 7], and
[5, 5 + h]; simplify each result as much as possible. What do you notice about
these quantities?
(c) Use the limit definition of the derivative to compute the exact instantaneous
rate of change of f with respect to x at the value a = 1. That is, compute f 0(1)
using the limit definition. Show your work. Is your result surprising?
(d) Without doing
√ any additional computations, what are the values of f 0(2), f 0(π),
and f 0(− 2)? Why?
C
Activity 1.8.
A water balloon is tossed vertically in the air from a window. The balloon’s height in
feet at time t in seconds after being launched is given by s(t) = −16t 2 + 16t + 32. Use
this function to respond to each of the following questions.
(a) Sketch an accurate, labeled graph of s on the axes provided in Figure 1.14. You
should be able to do this without using computing technology.
y
32
16
1 2
Figure 1.14: Axes for plotting y = s(t) in Activity 1.8.
(b) Compute the average rate of change of s on the time interval [1, 2]. Include
units on your answer and write one sentence to explain the meaning of the
value you found.
(c) Use the limit definition to compute the instantaneous rate of change of s with
respect to time, t, at the instant a = 1. Show your work using proper notation,
include units on your answer, and write one sentence to explain the meaning
of the value you found.
(d) On your graph in (a), sketch two lines: one whose slope represents the average
rate of change of s on [1, 2], the other whose slope represents the instantaneous
rate of change of s at the instant a = 1. Label each line clearly.
(e) For what values of a do you expect s 0(a) to be positive? Why? Answer the
same questions when “positive” is replaced by “negative” and “zero.”
C
Activity 1.9.
A rapidly growing city in Arizona has its population P at time t, where t is the number
of decades after the year 2010, modeled by the formula P(t) = 25000e t /5 . Use this
function to respond to the following questions.
(a) Sketch an accurate graph of P for t = 0 to t = 5 on the axes provided in
Figure 1.15. Label the scale on the axes carefully.
Figure 1.15: Axes for plotting y = P(t) in Activity 1.9.
(b) Compute the average rate of change of P between 2030 and 2050. Include units
on your answer and write one sentence to explain the meaning (in everyday
language) of the value you found.
(c) Use the limit definition to write an expression for the instantaneous rate of
change of P with respect to time, t, at the instant a = 2. Explain why this limit
is difficult to evaluate exactly.
(d) Estimate the limit in (c) for the instantaneous rate of change of P at the instant
a = 2 by using several small h values. Once you have determined an accurate
estimate of P 0(2), include units on your answer, and write one sentence (using
everyday language) to explain the meaning of the value you found.
(e) On your graph above, sketch two lines: one whose slope represents the average
rate of change of P on [2, 4], the other whose slope represents the instantaneous
rate of change of P at the instant a = 2.
(f) In a carefully-worded sentence, describe the behavior of P 0(a) as a increases in
value. What does this reflect about the behavior of the given function P?
C
Summary
f (b) − f (a)
• The average rate of change of a function f on the interval [a, b] is . The
b−a
units on the average rate of change are units of f per unit of x, and the numerical
value of the average rate of change represents the slope of the secant line between the
points (a, f (a)) and (b, f (b)) on the graph of y = f (x). If we view the interval as being
[a, a + h] instead of [a, b], the meaning is still the same, but the average rate of change
f (a + h) − f (a)
is now computed by .
h
• The instantaneous rate of change with respect to x of a function f at a value x = a
is denoted f 0(a) (read “the derivative of f evaluated at a” or “ f -prime at a”) and is
defined by the formula
f (a + h) − f (a)
f 0(a) = lim ,
h→0 h
provided the limit exists. Note particularly that the instantaneous rate of change at
x = a is the limit of the average rate of change on [a, a + h] as h → 0.
• Provided the derivative f 0(a) exists, its value tells us the instantaneous rate of change
of f with respect to x at x = a, which geometrically is the slope of the tangent line to
the curve y = f (x) at the point (a, f (a)). We even say that f 0(a) is the slope of the curve
y = f (x) at the point (a, f (a)).
• Limits are the link between average rate of change and instantaneous rate of change:
they allow us to move from the rate of change over an interval to the rate of change at
a single point.
Exercises
1. Consider the graph of y = f (x) provided in Figure 1.16.
(a) On the graph of y = f (x), sketch and label the following quantities:
• the secant line to y = f (x) on the interval [−3, −1] and the secant line to
y = f (x) on the interval [0, 2].
• the tangent line to y = f (x) at x = −3 and the tangent line to y = f (x) at
x = 0.
y
4
f
x
-4 4
-4
Figure 1.16: Plot of y = f (x).
(b) What is the approximate value of the average rate of change of f on [−3, −1]?
On [0, 2]? How are these values related to your work in (a)?
(c) What is the approximate value of the instantaneous rate of change of f at
x = −3? At x = 0? How are these values related to your work in (a)?
2. For each of the following prompts, sketch a graph on the provided axes in Figure 1.17 of
a function that has the stated properties.
(a) y = f (x) such that
• the average rate of change of f on [−3, 0] is −2 and the average rate of
change of f on [1, 3] is 0.5, and
• the instantaneous rate of change of f at x = −1 is −1 and the instantaneous
rate of change of f at x = 2 is 1.
(b) y = g(x) such that
• g(3)−g(−2)
5 = 0 and g(1)−g(−1)
2 = −1, and
• g (2) = 1 and g (−1) = 0
0 0
3. Suppose that the population, P, of China (in billions) can be approximated by the
function P(t) = 1.15(1.014)t where t is the number of years since the start of 1993.
(a) According to the model, what was the total change in the population of China
between January 1, 1993 and January 1, 2000? What will be the average rate of
change of the population over this time period? Is this average rate of change
3 3
-3 3 -3 3
-3 -3
Figure 1.17: Axes for plotting y = f (x) in (a) and y = g(x) in (b).
greater or less than the instantaneous rate of change of the population on

January 1, 2000? Explain and justify, being sure to include proper units on all
your answers.
(b) According to the model, what is the average rate of change of the population
of China in the ten-year period starting on January 1, 2012?
(c) Write an expression involving limits that, if evaluated, would give the exact
instantaneous rate of change of the population on today’s date. Then estimate
the value of this limit (discuss how you chose to do so) and explain the meaning
(including units) of the value you have found.
(d) Find an equation for the tangent line to the function y = P(t) at the point
where the t-value is given by today’s date.
4. The goal of this problem is to compute the value of the derivative at a point for several
different functions, where for each one we do so in three different ways, and then to
compare the results to see that each produces the same value.
For each of the following functions, use the limit definition of the derivative to compute
the value of f 0(a) using three different approaches: strive to use the algebraic approach
first (to compute the limit exactly), then test your result using numerical evidence (with
small values of h), and finally plot the graph of y = f (x) near (a, f (a)) along with the
appropriate tangent line to estimate the value of f 0(a) visually. Compare your findings
among all three approaches; if you are unable to complete the algebraic approach, still
work numerically and graphically.
(a) f (x) = x 2 − 3x, a = 2

(b) f (x) = x1 , a = 1
√
(c) f (x) = x, a = 1
(d) f (x) = 2 − |x − 1|, a = 1

π
(e) f (x) = sin(x), a = 2
34 1.4. THE DERIVATIVE FUNCTION
1.4 The derivative function

questions:
• How does the limit definition of the derivative of a function f lead to an entirely
new (but related) function f 0?
• What is the difference between writing f 0(a) and f 0(x)?
• How is the graph of the derivative function f 0(x) connected to the graph of f (x)?
• What are some examples of functions f for which f 0 is not defined at one or more
points?
Introduction
Given a function y = f (x), we now know that if we are interested in the instantaneous
rate of change of the function at x = a, or equivalently the slope of the tangent line to
y = f (x) at x = a, we can compute the value f 0(a). In all of our examples to date, we
have arbitrarily identified a particular value of a as our point of interest: a = 1, a = 3, etc.
But it is not hard to imagine that we will often be interested in the derivative value for
more than just one a-value, and possibly for many of them. In this section, we explore
how we can move from computing simply f 0(1) or f 0(3) to working more generally with
f 0(a), and indeed f 0(x). Said differently, we will work toward understanding how the
so-called process of “taking the derivative” generates a new function that is derived from
the original function y = f (x). The following preview activity starts us down this path.
Preview Activity 1.4. Consider the function f (x) = 4x − x 2 .
(a) Use the limit definition to compute the following derivative values: f 0(0), f 0(1),
f 0(2), and f 0(3).
(b) Observe that the work to find f 0(a) is the same, regardless of the value of a. Based
on your work in (a), what do you conjecture is the value of f 0(4)? How about
f 0(5)? (Note: you should not use the limit definition of the derivative to find either
value.)
(c) Conjecture a formula for f 0(a) that depends only on the value a. That is, in the
same way that we have a formula for f (x) (recall f (x) = 4x − x 2 ), see if you can
use your work above to guess a formula for f 0(a) in terms of a.
./
1.4. THE DERIVATIVE FUNCTION 35
How the derivative is itself a function
In your work in Preview Activity 1.4 with f (x) = 4x − x 2 , you may have found several
patterns. One comes from observing that f 0(0) = 4, f 0(1) = 2, f 0(2) = 0, and f 0(3) = −2.
That sequence of values leads us naturally to conjecture that f 0(4) = −4 and f 0(5) = −6.
Even more than these individual numbers, if we consider the role of 0, 1, 2, and 3 in the
process of computing the value of the derivative through the limit definition, we observe
that the particular number has very little effect on our work. To see this more clearly,
we compute f 0(a), where a represents a number to be named later. Following the now
standard process of using the limit definition of the derivative,
f (a + h) − f (a)
f 0(a) = lim
h→0 h
4(a + h) − (a + h)2 − (4a − a2 )
= lim
h→0 h
4a + 4h − a − 2ha − h2 − 4a + a2
2
= lim
h→0 h
4h − 2ha − h2
= lim
h→0 h
h(4 − 2a − h)
= lim
h→0 h
= lim (4 − 2a − h).
h→0
Here we observe that neither 4 nor 2a depend on the value of h, so as h → 0, (4−2a−h) →

(4 − 2a). Thus, f 0(a) = 4 − 2a.
This observation is consistent with the specific values we found above: e.g., f 0(3) =
4 − 2(3) = −2. And indeed, our work with a confirms that while the particular value of a
at which we evaluate the derivative affects the value of the derivative, that value has almost
no bearing on the process of computing the derivative. We note further that the letter
being used is immaterial: whether we call it a, x, or anything else, the derivative at a given
value is simply given by “4 minus 2 times the value.” We choose to use x for consistency
with the original function given by y = f (x), as well as for the purpose of graphing the
derivative function, and thus we have found that for the function f (x) = 4x − x 2 , it follows
that f 0(x) = 4 − 2x.
Because the value of the derivative function is so closely linked to the graphical
behavior of the original function, it makes sense to look at both of these functions plotted
on the same domain. In Figure 1.18, on the left we show a plot of f (x) = 4x − x 2 together
with a selection of tangent lines at the points we’ve considered above. On the right, we
show a plot of f 0(x) = 4 − 2x with emphasis on the heights of the derivative graph at the
same selection of points. Notice the connection between colors in the left and right graph:
the green tangent line on the original graph is tied to the green point on the right graph
in the following way: the slope of the tangent line at a point on the lefthand graph is the
m=0 (0, 4)
4 4
m=2 m = −2
3 3
(1, 2)
2 2
1 1
m=4 m = −4 (2, 0)
1 2 3 4 1 2 3 4
-1 -1
(3, −2)
-2 -2
y = f (x) y = f ′ (x)
-3 -3
-4 -4
(4, −4)
Figure 1.18: The graphs of f (x) = 4x − x 2 (at left) and f 0(x) = 4 − 2x (at right). Slopes on
the graph of f correspond to heights on the graph of f 0.
same as the height at the corresponding point on the righthand graph. That is, at each
respective value of x, the slope of the tangent line to the original function at that x-value
is the same as the height of the derivative function at that x-value. Do note, however, that
the units on the vertical axes are different: in the left graph, the vertical units are simply
the output units of f . On the righthand graph of y = f 0(x), the units on the vertical axis
are units of f per unit of x.
Of course, this relationship between the graph of a function y = f (x) and its derivative
is a dynamic one. An excellent way to explore how the graph of f (x) generates the graph
of f 0(x) is through a java applet. See, for instance, the applets at http://gvsu.edu/s/5C
or http://gvsu.edu/s/5D, via the sites of Austin and Renault5 .
In Section 1.3 when we first defined the derivative, we wrote the definition in terms of
a value a to find f 0(a). As we have seen above, the letter a is merely a placeholder, and it
often makes more sense to use x instead. For the record, here we restate the definition of
the derivative.
Definition 1.4. Let f be a function and x a value in the function’s domain. We define
the derivative of f with respect to x at the value x, denoted f 0(x), by the formula f 0(x) =
f (x + h) − f (x)
lim , provided this limit exists.
h→0 h
5 David Austin, http://gvsu.edu/s/5r; Marc Renault, http://gvsu.edu/s/5p.
We now may take two different perspectives on thinking about the derivative function:
given a graph of y = f (x), how does this graph lead to the graph of the derivative
function y = f 0(x)? and given a formula for y = f (x), how does the limit definition of
the derivative generate a formula for y = f 0(x)? Both of these issues are explored in the
following activities.
Activity 1.10.
For each given graph of y = f (x), sketch an approximate graph of its derivative
function, y = f 0(x), on the axes immediately below. The scale of the grid for the graph
of f is 1 × 1; assume the horizontal scale of the grid for the graph of f 0 is identical to
that for f . If necessary, adjust and label the vertical scale on the axes for f 0.
f g
x
x
f′ g′
x x
p q
x
x
p′ q′
x x
r s
x x
r′ s′
x x
w z
x x
w′ z′
x x
When you are finished with all 8 graphs, write several sentences that describe your
overall process for sketching the graph of the derivative function, given the graph
the original function. What are the values of the derivative function that you tend
to identify first? What do you do thereafter? How do key traits of the graph of the
derivative function exemplify properties of the graph of the original function?
C
For a dynamic investigation that allows you to experiment with graphing f0 when
given the graph of f , see http://gvsu.edu/s/8y.6
6 Marc Renault, Calculus Applets Using Geogebra.

Now, recall the opening example of this section: we began with the function y =
f (x) = 4x − x 2 and used the limit definition of the derivative to show that f 0(a) = 4 − 2a,
or equivalently that f 0(x) = 4 − 2x. We subsequently graphed the functions f and f 0
as shown in Figure 1.18. Following Activity 1.10, we now understand that we could have
constructed a fairly accurate graph of f 0(x) without knowing a formula for either f or f 0.
At the same time, it is ideal to know a formula for the derivative function whenever it is
possible to find one.
In the next activity, we further explore the more algebraic approach to finding f 0(x):
given a formula for y = f (x), the limit definition of the derivative will be used to develop
a formula for f 0(x).
Activity 1.11.
For each of the listed functions, determine a formula for the derivative function. For
the first two, determine the formula for the derivative by thinking about the nature of
the given function and its slope at various points; do not use the limit definition. For
the latter four, use the limit definition. Pay careful attention to the function names and
independent variables. It is important to be comfortable with using letters other than
f and x. For example, given a function p(z), we call its derivative p0(z).
(a) f (x) = 1
(b) g(t) = t
(c) p(z) = z 2
(d) q(s) = s3
1
(e) F(t) = t
√
(f) G(y) = y
C
Summary
• The limit definition of the derivative, f 0(x) = limh→0 f (x+h)− h

f (x)
, produces a value
for each x at which the derivative is defined, and this leads to a new function whose
formula is y = f 0(x). Hence we talk both about a given function f and its derivative
f 0. It is especially important to note that taking the derivative is a process that starts
with a given function ( f ) and produces a new, related function ( f 0).
• There is essentially no difference between writing f 0(a) (as we did regularly in Sec-
tion 1.3) and writing f 0(x). In either case, the variable is just a placeholder that is used
to define the rule for the derivative function.
• Given the graph of a function y = f (x), we can sketch an approximate graph of its
derivative y = f 0(x) by observing that heights on the derivative’s graph correspond to
slopes on the original function’s graph.
• In Activity 1.10, we encountered some functions that had sharp corners on their graphs,
such as the shifted absolute value function. At such points, the derivative fails to exist,
and we say that f is not differentiable there. For now, it suffices to understand this as a
consequence of the jump that must occur in the derivative function at a sharp corner
on the graph of the original function.
Exercises
1. Let f be a function with the following properties: f is differentiable at every value of x
(that is, f has a derivative at every point), f (−2) = 1, and f 0(−2) = −2, f 0(−1) = −1,
f 0(0) = 0, f 0(1) = 1, and f 0(2) = 2.
(a) On the axes provided at left in Figure 1.19, sketch a possible graph of y = f (x).
Explain why your graph meets the stated criteria.
(b) On the axes at right in Figure 1.19, sketch a possible graph of y = f 0(x). What
type of curve does the provided data suggest for the graph of y = f 0(x)?
(c) Conjecture a formula for the function y = f (x). Use the limit definition of the
derivative to determine the corresponding formula for y = f 0(x). Discuss both
graphical and algebraic evidence for whether or not your conjecture is correct.
3 3
-3 3 -3 3
-3 -3
Figure 1.19: Axes for plotting y = f (x) in (a) and y = f 0(x) in (b).
2. Consider the function g(x) = x 2 − x + 3.
(a) Use the limit definition of the derivative to determine a formula for g 0(x).
(b) Use a graphing utility to plot both y = g(x) and your result for y = g 0(x); does
your formula for g 0(x) generate the graph you expected?
(c) Use the limit definition of the derivative to find a formula for h 0(x) where
h(x) = 5x 2 − 4x + 12.
(d) Compare and contrast the formulas for g 0(x) and h 0(x) you have found. How
do the constants 5, 4, 12, and 3 affect the results?
3. Let g be a continuous function (that is, one with no jumps or holes in the graph) and
suppose that a graph of y = g 0(x) is given by the graph on the right in Figure 1.20.
2 2
-2 2 -2 2
-2 -2
Figure 1.20: Axes for plotting y = g(x) and, at right, the graph of y = g 0(x).
(a) Observe that for every value of x that satisfies 0 < x < 2, the value of g 0(x) is
constant. What does this tell you about the behavior of the graph of y = g(x)
on this interval?
(b) On what intervals other than 0 < x < 2 do you expect y = g(x) to be a linear
function? Why?
(c) At which values of x is g 0(x) not defined? What behavior does this lead you to
expect to see in the graph of y = g(x)?
(d) Suppose that g(0) = 1. On the axes provided at left in Figure 1.20, sketch an
accurate graph of y = g(x).
4. For each graph that provides an original function y = f (x) in Figure 1.21 (on the
following page), your task is to sketch an approximate graph of its derivative function,
y = f 0(x), on the axes immediately below. View the scale of the grid for the graph
of f as being 1 × 1, and assume the horizontal scale of the grid for the graph of f 0 is
identical to that for f . If you need to adjust the vertical scale on the axes for the graph
of f 0, you should label that accordingly.
f
f
x x
f′ f′
x x
f f
x x
f′ f′
x x
Figure 1.21: Graphs of y = f (x) and grids for plotting the corresponding graph of y = f 0(x).
1.5. INTERPRETING, ESTIMATING, AND USING THE DERIVATIVE 43
1.5 Interpreting, estimating, and using the derivative

questions:
• In contexts other than the position of a moving object, what does the derivative of
a function measure?
• What are the units on the derivative function f 0, and how are they related to the
units of the original function f ?
• What is a central difference, and how can one be used to estimate the value of the
derivative at a point from given function data?
• Given the value of the derivative of a function at a point, what can we infer about
how the value of the function changes nearby?
Introduction
An interesting and powerful feature of mathematics is that it can often be thought of both
in abstract terms and in applied ones. For instance, calculus can be developed almost
entirely as an abstract collection of ideas that focus on properties of arbitrary functions. At
the same time, calculus can also be very directly connected to our experience of physical
reality by considering functions that represent meaningful processes. We have already
seen that for a position function y = s(t), say for a ball being tossed straight up in the air,
the ball’s velocity at time t is given by v(t) = s 0(t), the derivative of the position function.
Further, recall that if s(t) is measured in feet at time t, the units on v(t) = s 0(t) are feet per
second.
In what follows in this section, we investigate several different functions, each with
specific physical meaning, and think about how the units on the independent variable,
dependent variable, and the derivative function add to our understanding. To start, we
consider the familiar problem of a position function of a moving object.
Preview Activity 1.5. One of the longest stretches of straight (and flat) road in North
America can be found on the Great Plains in the state of North Dakota on state highway
46, which lies just south of the interstate highway I-94 and runs through the town of Gackle.
A car leaves town (at time t = 0) and heads east on highway 46; its position in miles from
Gackle at time t in minutes is given by the graph of the function in Figure 1.22. Three
important points are labeled on the graph; where the curve looks linear, assume that it is
indeed a straight line.
(a) In everyday language, describe the behavior of the car over the provided time
44 1.5. INTERPRETING, ESTIMATING, AND USING THE DERIVATIVE
(104, 106.8)
s
100
80
(57, 63.8)
60
(68, 63.8)
40
20
t
20 40 60 80 100
Figure 1.22: The graph of y = s(t), the position of the car along highway 46, which tells its
distance in miles from Gackle, ND, at time t in minutes.
interval. In particular, discuss what is happening on the time intervals [57, 68]
and [68, 104].
(b) Find the slope of the line between the points (57, 63.8) and (104, 106.8). What are
the units on this slope? What does the slope represent?
(c) Find the average rate of change of the car’s position on the interval [68, 104].
Include units on your answer.
(d) Estimate the instantaneous rate of change of the car’s position at the moment
t = 80. Write a sentence to explain your reasoning and the meaning of this value.
./
Units of the derivative function
As we now know, the derivative of the function f at a fixed value x is given by
f (x + h) − f (x)
f 0(x) = lim ,
h→0 h
and this value has several different interpretations. If we set x = a, one meaning of f 0(a)
is the slope of the tangent line at the point (a, f (a)).
df dy
In alternate notation, we also sometimes equivalently write dx or dx instead of f 0(x),
and these notations helps us to further see the units (and thus the meaning) of the derivative
as it is viewed as the instantaneous rate of change of f with respect to x. Note that the units
on the slope of the secant line, f (x+h)−
h
f (x)
, are “units of f per unit of x.” Thus, when we
take the limit to get f 0(x), we get these same units on the derivative f 0(x): units of f per
unit of x. Regardless of the function f under consideration (and regardless of the variables
being used), it is helpful to remember that the units on the derivative function are “units
of output per unit of input,” in terms of the input and output of the original function.
For example, say that we have a function y = P(t), where P measures the population
of a city (in thousands) at the start of year t (where t = 0 corresponds to 2010 AD), and we
are told that P 0(2) = 21.37. What is the meaning of this value? Well, since P is measured
in thousands and t is measured in years, we can say that the instantaneous rate of change
of the city’s population with respect to time at the start of 2012 is 21.37 thousand people
per year. We therefore expect that in the coming year, about 21,370 people will be added
to the city’s population.
Toward more accurate derivative estimates
It is also helpful to recall, as we first experienced in Section 1.3, that when we want to
estimate the value of f 0(x) at a given x, we can use the difference quotient f (x+h)−
h
f (x)
with
a relatively small value of h. In doing so, we should use both positive and negative values
of h in order to make sure we account for the behavior of the function on both sides of
the point of interest. To that end, we consider the following brief example to demonstrate
the notion of a central difference and its role in estimating derivatives.
Example 1.4. Suppose that y = f (x) is a function for which three values are known:
f (1) = 2.5, f (2) = 3.25, and f (3) = 3.625. Estimate f 0(2).
Solution. We know that f 0(2) = limh→0 f (2+h)−h

f (2)
. But since we don’t have a graph for
y = f (x) nor a formula for the function, we can neither sketch a tangent line nor evaluate
the limit exactly. We can’t even use smaller and smaller values of h to estimate the limit.
Instead, we have just two choices: using h = −1 or h = 1, depending on which point we
pair with (2, 3.25).
So, one estimate is
f (1) − f (2) 2.5 − 3.25
f 0(2) ≈ = = 0.75.
1−2 −1
The other is
f (3) − f (2) 3.625 − 3.25
f 0(2) ≈ = = 0.375.
3−2 1
Since the first approximation looks only backward from the point (2, 3.25) and the second
approximation looks only forward from (2, 3.25), it makes sense to average these two
values in order to account for behavior on both sides of the point of interest. Doing so, we
find that
0.75 + 0.375
f 0(2) ≈ = 0.5625.
2
The intuitive approach to average the two estimates found in Example 1.4 is in fact the
best possible estimate to f 0(2) when we have just two function values for f on opposite
sides of the point of interest. To see why, we think about the diagram in Figure 1.23, which
3 3
2 2
1 1
1 2 3 1 2 3
Figure 1.23: At left, the graph of y = f (x) along with the secant line through (1, 2.5) and
(2, 3.25), the secant line through (2, 3.25) and (3, 3.625), as well as the tangent line. At
right, the same graph along with the secant line through (1, 2.5) and (3, 3.625), plus the
tangent line.
shows a possible function y = f (x) that satisfies the data given in Example 1.4. On the left,
we see the two secant lines with slopes that come from computing the backward difference
f (1)− f (2)
1−2 = 0.75 and from the forward difference f (3)−
3−2
f (2)
= 0.375. Note how the first such
line’s slope over-estimates the slope of the tangent line at (2, f (2)), while the second line’s
slope underestimates f 0(2). On the right, however, we see the secant line whose slope is
given by the central difference
f (3) − f (1) 3.625 − 2.5 1.125

= = = 0.5625.
3−1 2 2
Note that this central difference has the exact same value as the average of the forward
difference and backward difference (and it is straightforward to explain why this always
holds), and moreover that the central difference yields a very good approximation to the
derivative’s value, in part because the secant line that uses both a point before and after
the point of tangency yields a line that is closer to being parallel to the tangent line.
In general, the central difference approximation to the value of the first derivative is
given by
f (a + h) − f (a − h)
f 0(a) ≈ ,
2h
and this quantity measures the slope of the secant line to y = f (x) through the points
(a − h, f (a − h)) and (a + h, f (a + h)). Anytime we have symmetric data surrounding a
point at which we desire to estimate the derivative, the central difference is an ideal choice
for so doing.
The following activities will further explore the meaning of the derivative in several
different contexts while also viewing the derivative from graphical, numerical, and algebraic
perspectives.
Activity 1.12.
A potato is placed in an oven, and the potato’s temperature F (in degrees Fahrenheit) at
various points in time is taken and recorded in the following table. Time t is measured
in minutes.
t F(t)
0 70
15 180.5
30 251
45 296
60 324.5
75 342.8
90 354.5
(a) Use a central difference to estimate the instantaneous rate of change of the
temperature of the potato at t = 30. Include units on your answer.
(b) Use a central difference to estimate the instantaneous rate of change of the
temperature of the potato at t = 60. Include units on your answer.
(c) Without doing any calculation, which do you expect to be greater: F 0(75) or
F 0(90)? Why?
(d) Suppose it is given that F(64) = 330.28 and F 0(64) = 1.341. What are the units
on these two quantities? What do you expect the temperature of the potato to
be when t = 65? when t = 66? Why?
(e) Write a couple of careful sentences that describe the behavior of the temperature
of the potato on the time interval [0, 90], as well as the behavior of the
instantaneous rate of change of the temperature of the potato on the same time
interval.
C
Activity 1.13.
A company manufactures rope, and the total cost of producing r feet of rope is C(r)
dollars.
(a) What does it mean to say that C(2000) = 800?
(b) What are the units of C 0(r)?
(c) Suppose that C(2000) = 800 and C 0(2000) = 0.35. Estimate C(2100), and
justify your estimate by writing at least one sentence that explains your thinking.
(d) Which of the following statements do you think is true, and why?
• C 0(2000) < C 0(3000)
• C 0(2000) = C 0(3000)
• C 0(2000) > C 0(3000)
(e) Suppose someone claims that C 0(5000) = −0.1. What would the practical
meaning of this derivative value tell you about the approximate cost of the next
foot of rope? Is this possible? Why or why not?
C
Activity 1.14.
Researchers at a major car company have found a function that relates gasoline
consumption to speed for a particular model of car. In particular, they have determined
that the consumption C, in liters per kilometer, at a given speed s, is given by a
function C = f (s), where s is the car’s speed in kilometers per hour.
(a) Data provided by the car company tells us that f (80) = 0.015, f (90) = 0.02,
and f (100) = 0.027. Use this information to estimate the instantaneous rate of
change of fuel consumption with respect to speed at s = 90. Be as accurate as
possible, use proper notation, and include units on your answer.
(b) By writing a complete sentence, interpret the meaning (in the context of fuel
consumption) of “ f (80) = 0.015.”
(c) Write at least one complete sentence that interprets the meaning of the value
of f 0(90) that you estimated in (a).
C
In Section 1.4, we learned how use to the graph of a given function f to plot the
graph of its derivative, f 0. It is important to remember that when we do so, not only
does the scale on the vertical axis often have to change to accurately represent f 0, but
the units on that axis also differ. For example, suppose that P(t) = 400 − 330e−0.03t tells
us the temperature in degrees Fahrenheit of a potato in an oven at time t in minutes. In
Figure 1.24, we sketch the graph of P on the left and the graph of P 0 on the right.
◦F ◦ F/min
400 16
300 y = P(t) 12
200 8 y = P′ (t)
100 4
min min
20 40 60 80 20 40 60 80
Figure 1.24: Plot of P(t) = 400 − 330e−0.03t at left, and its derivative P 0(t) at right.
Note how not only are the vertical scales different in size, but different in units, as the
units of P are ◦ F, while those of P 0 are ◦ F/min. In all cases where we work with functions
that have an applied context, it is helpful and instructive to think carefully about units
involved and how they further inform the meaning of our computations.
Summary
• Regardless of the context of a given function y = f (x), the derivative always measures
the instantaneous rate of change of the output variable with respect to the input
variable.
• The units on the derivative function y = f 0(x) are units of f per unit of x. Again, this
measures how fast the output of the function f changes when the input of the function
changes.
• The central difference approximation to the value of the first derivative is given by
f (a + h) − f (a − h)
f 0(a) ≈ ,
2h
and this quantity measures the slope of the secant line to y = f (x) through the
points (a − h, f (a − h)) and (a + h, f (a + h)). The central difference generates a good
approximation of the derivative’s value any time we have symmetric data surrounding a
point of interest.
• Knowing the derivative and function values at a single point enables us to estimate
other function values nearby. If, for example, we know that f 0(7) = 2, then we know
that at x = 7, the function f is increasing at an instantaneous rate of 2 units of output
for every one unit of input. Thus, we expect f (8) to be approximately 2 units greater
than f (7). The value is approximate because we don’t know that the rate of change
stays the same as x changes.
Exercises
1. A cup of coffee has its temperature F (in degrees Fahrenheit) at time t given by the
function F(t) = 75 + 110e−0.05t , where time is measured in minutes.
(a) Use a central difference with h = 0.01 to estimate the value of F 0(10).
(b) What are the units on the value of F 0(10) that you computed in (a)? What is
the practical meaning of the value of F 0(10)?
(c) Which do you expect to be greater: F 0(10) or F 0(20)? Why?
(d) Write a sentence that describes the behavior of the function y = F 0(t) on the
time interval 0 ≤ t ≤ 30. How do you think its graph will look? Why?
2. The temperature change T (in Fahrenheit degrees), in a patient, that is generated by a

dose q (in milliliters), of a drug, is given by the function T = f (q).
(a) What does it mean to say f (50) = 0.75? Write a complete sentence to explain,
using correct units.
(b) A person’s sensitivity, s, to the drug is defined by the function s(q) = f 0(q).
What are the units of sensitivity?
(c) Suppose that f 0(50) = −0.02. Write a complete sentence to explain the
meaning of this value. Include in your response the information given in (a).
3. The velocity of a ball that has been tossed vertically in the air is given by v(t) = 16 − 32t,
where v is measured in feet per second, and t is measured in seconds. The ball is in
the air from t = 0 until t = 2.
(a) When is the ball’s velocity greatest?
(b) Determine the value of v 0(1). Justify your thinking.
(c) What are the units on the value of v 0(1)? What does this value and the
corresponding units tell you about the behavior of the ball at time t = 1?
(d) What is the physical meaning of the function v 0(t)?
4. The value, V , of a particular automobile (in dollars) depends on the number of miles,
m, the car has been driven, according to the function V = h(m).
(a) Suppose that h(40000) = 15500 and h(55000) = 13200. What is the average
rate of change of h on the interval [40000, 55000], and what are the units on
this value?
(b) In addition to the information given in (a), say that h(70000) = 11100. Deter-
mine the best possible estimate of h 0(55000) and write one sentence to explain
the meaning of your result, including units on your answer.
(c) Which value do you expect to be greater: h 0(30000) or h 0(80000)? Why?
(d) Write a sentence to describe the long-term behavior of the function V = h(m),
plus another sentence to describe the long-term behavior of h 0(m). Provide
your discussion in practical terms regarding the value of the car and the rate
at which that value is changing.
52 1.6. THE SECOND DERIVATIVE
1.6 The second derivative

questions:
• How does the derivative of a function tell us whether the function is increasing or
decreasing at a point or on an interval?
• What can we learn by taking the derivative of the derivative (to achieve the second
derivative) of a function f ?
• What does it mean to say that a function is concave up or concave down? How
are these characteristics connected to certain properties of the derivative of the
function?
• What are the units on the second derivative? How do they help us understand the
rate of change of the rate of change?
Introduction
Given a differentiable function y = f (x), we know that its derivative, y = f 0(x), is a related
function whose output at a value x = a tells us the slope of the tangent line to y = f (x)
at the point (a, f (a)). That is, heights on the derivative graph tell us the values of slopes
on the original function’s graph. Therefore, the derivative tells us important information
about the function f .
A
B
Figure 1.25: Two tangent lines on a graph demonstrate how the slope of the tangent line
tells us whether the function is rising or falling, as well as whether it is doing so rapidly or
slowly.
1.6. THE SECOND DERIVATIVE 53
At any point where f 0(x) is positive, it means that the slope of the tangent line to f is
positive, and therefore the function f is increasing (or rising) at that point. Similarly, if
f 0(a) is negative, we know that the graph of f is decreasing (or falling) at that point.
In the next part of our study, we work to understand not only whether the function
f is increasing or decreasing at a point or on an interval, but also how the function
f is increasing or decreasing. Comparing the two tangent lines shown in Figure 1.25,
we see that at point A, the value of f 0(x) is positive and relatively close to zero, which
coincides with the graph rising slowly. By contrast, at point B, the derivative is negative
and relatively large in absolute value, which is tied to the fact that f is decreasing rapidly
at B. It also makes sense to not only ask whether the value of the derivative function is
positive or negative and whether the derivative is large or small, but also to ask “how is
the derivative changing?”
We also now know that the derivative, y = f 0(x), is itself a function. This means that
we can consider taking its derivative – the derivative of the derivative – and therefore ask
questions like “what does the derivative of the derivative tell us about how the original
function behaves?” As we have done regularly in our work to date, we start with an
investigation of a familiar problem in the context of a moving object.
Preview Activity 1.6. The position of a car driving along a straight road at time t in
minutes is given by the function y = s(t) that is pictured in Figure 1.26. The car’s position
function has units measured in thousands of feet. For instance, the point (2, 4) on the
graph indicates that after 2 minutes, the car has traveled 4000 feet.
y s
14
10
2
t
2 6 10
Figure 1.26: The graph of y = s(t), the position of the car (measured in thousands of feet
from its starting location) at time t in minutes.
(a) In everyday language, describe the behavior of the car over the provided time
interval. In particular, you should carefully discuss what is happening on each of
the time intervals [0, 1], [1, 2], [2, 3], [3, 4], and [4, 5], plus provide commentary
overall on what the car is doing on the interval [0, 12].
(b) On the lefthand axes provided in Figure 1.27, sketch a careful, accurate graph of
y = s 0(t).
(c) What is the meaning of the function y = s 0(t) in the context of the given problem?
What can we say about the car’s behavior when s 0(t) is positive? when s 0(t) is
zero? when s 0(t) is negative?
(d) Rename the function you graphed in (b) to be called y = v(t). Describe the
behavior of v in words, using phrases like “v is increasing on the interval . . .” and
“v is constant on the interval . . ..”
(e) Sketch a graph of the function y = v 0(t) on the righthand axes provide in Fig-
ure 1.27. Write at least one sentence to explain how the behavior of v 0(t) is
connected to the graph of y = v(t).
y y
t t
2 6 10 2 6 10
Figure 1.27: Axes for plotting y = v(t) = s 0(t) and y = v 0(t).
./
Increasing, decreasing, or neither
When we look at the graph of a function, there are features that strike us naturally, and
common language can be used to name these features. In many different settings so far,
we have intuitively used the words increasing and decreasing to describe a function’s graph.
Here we connect these terms more formally to a function’s behavior on an interval of input
values.
Definition 1.5. Given a function f (x) defined on the interval (a, b), we say that f is
increasing on (a, b) provided that for all x, y in the interval (a, b), if x < y, then f (x) < f (y).
Similarly, we say that f is decreasing on (a, b) provided that for all x, y in the interval (a, b),
if x < y, then f (x) > f (y).
Simply put, an increasing function is one that is rising as we move from left to right
along the graph, and a decreasing function is one that falls as the value of the input
increases. For a function that has a derivative at a point, we will also talk about whether
or not the function is increasing or decreasing at that point. Moreover, the fact of whether
or not the function is increasing, decreasing, or neither at a given point depends precisely
on the value of the derivative at that point.
Let f be a function that is differentiable at x = a. Then f is increasing at x = a

if and only if f 0(a) > 0 and f is decreasing at x = a if and only if f 0(a) < 0. If
f 0(a) = 0, then we say f is neither increasing nor decreasing at x = a.
A
2
-2 2
y = f (x) -2 B
Figure 1.28: A function that is decreasing at A, increasing at B, and more generally,

decreasing on the intervals −3 < x < −2 and 0 < x < 2 and increasing on −2 < x < 0
and 2 < x < 3.
For example, the function pictured in Figure 1.28 is increasing at any point at which
f 0(x) is positive, and hence is increasing on the entire interval −2 < x < 0. Note that
at both x = ±2 and x = 0, we say that f is neither increasing nor decreasing, because
f 0(x) = 0 at these values.
The Second Derivative
For any function, we are now accustomed to investigating its behavior by thinking about
its derivative. Given a function f , its derivative is a new function, one that is given by the
rule
f (x + h) − f (x)
f 0(x) = lim .
h→0 h
Because f 0 is itself a function, it is perfectly feasible for us to consider the derivative of
the derivative, which is the new function y = [ f 0(x)]0. We call this resulting function the
second derivative of y = f (x), and denote the second derivative by y = f 00(x). Due to the
presence of multiple possible derivatives, we will sometimes call f 0 “the first derivative” of
f , rather than simply “the derivative” of f . Formally, the second derivative is defined by
the limit definition of the derivative of the first derivative:
f 0(x + h) − f 0(x)
f 00(x) = lim .
h→0 h
We note that all of the established meaning of the derivative function still holds, so
when we compute y = f 00(x), this new function measures slopes of tangent lines to the
curve y = f 0(x), as well as the instantaneous rate of change of y = f 0(x). In other words,
just as the first derivative measures the rate at which the original function changes, the
second derivative measures the rate at which the first derivative changes. This means that
the second derivative tracks the instantaneous rate of change of the instantaneous rate
of change of f . That is, the second derivative will help us to understand how the rate of
change of the original function is itself changing.
Concavity
In addition to asking whether a function is increasing or decreasing, it is also natural

to inquire how a function is increasing or decreasing. To begin, there are three basic
behaviors that an increasing function can demonstrate on an interval, as pictured in
Figure 1.29: the function can increase more and more rapidly, increase at the same rate, or
increase in a way that is slowing down. Fundamentally, we are beginning to think about
how a particular curve bends, with the natural comparison being made to lines, which
don’t bend at all. More than this, we want to understand how the bend in a function’s
graph is tied to behavior characterized by the first derivative of the function.
For the leftmost curve in Figure 1.29, picture a sequence of tangent lines to the curve.
As we move from left to right, the slopes of those tangent lines will increase. Therefore,
the rate of change of the pictured function is increasing, and this explains why we say this
function is increasing at an increasing rate. For the rightmost graph in Figure 1.29, observe
that as x increases, the function increases but the slope of the tangent line decreases,
hence this function is increasing at a decreasing rate.
Figure 1.29: Three functions that are all increasing, but doing so at an increasing rate, at a
constant rate, and at a decreasing rate, respectively.
Of course, similar options hold for how a function can decrease. Here we must be
extra careful with our language, since decreasing functions involve negative slopes, and
negative numbers present an interesting situation in the tension between common language
and mathematical language. For example, it can be tempting to say that “−100 is bigger
than −2.” But we must remember that when we say one number is greater than another,
this describes how the numbers lie on a number line: x < y provided that x lies to the
left of y. So of course, −100 is less than −2. Informally, it might be helpful to say that
“−100 is more negative than −2.” This leads us to note particularly that when a function’s
values are negative, and those values subsequently get more negative, the function must
be decreasing.
Now consider the three graphs shown in Figure 1.30. Clearly the middle graph
demonstrates the behavior of a function decreasing at a constant rate. If we think about a
sequence of tangent lines to the first curve that progress from left to right, we see that the
slopes of these lines get less and less negative as we move from left to right. That means
that the values of the first derivative, while all negative, are increasing, and thus we say
that the leftmost curve is decreasing at an increasing rate.
This leaves only the rightmost curve in Figure 1.30 to consider. For that function, the
slope of the tangent line is negative throughout the pictured interval, but as we move
from left to right, the slopes get more and more negative. Hence the slope of the curve is
decreasing, and we say that the function is decreasing at a decreasing rate.
This leads us to introduce the notion of concavity which provides simpler language to
describe some of these behaviors. Informally, when a curve opens up on a given interval,
like the upright parabola y = x 2 or the exponential growth function y = e x , we say that
the curve is concave up on that interval. Likewise, when a curve opens down, such as the
parabola y = −x 2 or the opposite of the exponential function y = −e x , we say that the
function is concave down. This behavior is linked to both the first and second derivatives
Figure 1.30: From left to right, three functions that are all decreasing, but doing so in
different ways.
of the function.
In Figure 1.31, we see two functions along with a sequence of tangent lines to each.
On the lefthand plot where the function is concave up, observe that the tangent lines to
the curve always lie below the curve itself and that, as we move from left to right, the
slope of the tangent line is increasing. Said differently, the function f is concave up on the
interval shown because its derivative, f 0, is increasing on that interval. Similarly, on the
righthand plot in Figure 1.31, where the function shown is concave down, there we see that
the tangent lines alway lie above the curve and that the value of the slope of the tangent
line is decreasing as we move from left to right. Hence, what makes f concave down on
the interval is the fact that its derivative, f 0, is decreasing.
Figure 1.31: At left, a function that is concave up; at right, one that is concave down.
We state these most recent observations formally as the definitions of the terms concave
up and concave down.
Definition 1.6. Let f be a differentiable function on an interval (a, b). Then f is concave
up on (a, b) if and only if f 0 is increasing on (a, b); f is concave down on (a, b) if and only
if f 0 is decreasing on (a, b).
The following activities lead us to further explore how the first and second derivatives
of a function determine the behavior and shape of its graph. We begin by revisiting
Preview Activity 1.6.
Activity 1.15.
The position of a car driving along a straight road at time t in minutes is given by
the function y = s(t) that is pictured in Figure 1.32. The car’s position function has
units measured in thousands of feet. Remember that you worked with this function and
sketched graphs of y = v(t) = s 0(t) and y = v 0(t) in Preview Activity 1.6.
y s
14
10
2
t
2 6 10
Figure 1.32: The graph of y = s(t), the position of the car (measured in thousands of feet
from its starting location) at time t in minutes.
(a) On what intervals is the position function y = s(t) increasing? decreasing?

Why?
(b) On which intervals is the velocity function y = v(t) = s 0(t) increasing? decreas-
ing? neither? Why?
(c) Acceleration is defined to be the instantaneous rate of change of velocity, as the
acceleration of an object measures the rate at which the velocity of the object
is changing. Say that the car’s acceleration function is named a(t). How is a(t)
computed from v(t)? How is a(t) computed from s(t)? Explain.
(d) What can you say about s 00 whenever s 0 is increasing? Why?
(e) Using only the words increasing, decreasing, constant, concave up, concave down,
and linear, complete the following sentences. For the position function s with
velocity v and acceleration a,
• on an interval where v is positive, s is .
• on an interval where v is negative, s is .
• on an interval where v is zero, s is .
• on an interval where a is positive, v is .
• on an interval where a is negative, v is .
• on an interval where a is zero, v is .
• on an interval where a is positive, s is .
• on an interval where a is negative, s is .
• on an interval where a is zero, s is .
C
The context of position, velocity, and acceleration is an excellent one in which to
understand how a function, its first derivative, and its second derivative are related to one
another. In Activity 1.15, we can replace s, v, and a with an arbitrary function f and its
derivatives f 0 and f 00, and essentially all the same observations hold. In particular, note
that f 0 is increasing if and only if both f is concave up, and similarly f 0 is increasing if
and only if f 00 is positive. Likewise, f 0 is decreasing if and only if both f is concave down,
and f 0 is decreasing if and only if f 00 is negative.
Activity 1.16.
A potato is placed in an oven, and the potato’s temperature F (in degrees Fahrenheit) at
in minutes. In Activity 1.12, we computed approximations to F 0(30) and F 0(60) using
central differences. Those values and more are provided in the second table below,
along with several others computed in the same way.
t F(t) t F 0(t)
0 70 0 NA
15 180.5 15 6.03
30 251 30 3.85
45 296 45 2.45
60 324.5 60 1.56
75 342.8 75 1.00
90 354.5 90 NA
(a) What are the units on the values of F 0(t)?

(b) Use a central difference to estimate the value of F 00(30).
(c) What is the meaning of the value of F 00(30) that you have computed in (b) in
terms of the potato’s temperature? Write several careful sentences that discuss,
with appropriate units, the values of F(30), F 0(30), and F 00(30), and explain
the overall behavior of the potato’s temperature at this point in time.
(d) Overall, is the potato’s temperature increasing at an increasing rate, increasing
at a constant rate, or increasing at a decreasing rate? Why?
C
Activity 1.17.
This activity builds on our experience and understanding of how to sketch the graph
of f 0 given the graph of f .
f f
x x
f′ f′
x x
f ′′ f ′′
x x
Figure 1.33: Two given functions f , with axes provided for plotting f 0 and f 00 below.
In Figure 1.33, given the respective graphs of two different functions f , sketch the
corresponding graph of f 0 on the first axes below, and then sketch f 00 on the second
set of axes. In addition, for each, write several careful sentences in the spirit of those in
Activity 1.15 that connect the behaviors of f , f 0, and f 00. For instance, write something
such as
f 0 is on the interval , which is connected

to the fact that f is on the same interval ,
and f 00 is on the interval as well
but of course with the blanks filled in. Throughout, view the scale of the grid for the
graph of f as being 1 × 1, and assume the horizontal scale of the grid for the graph of
f 0 is identical to that for f . If you need to adjust the vertical scale on the axes for the
graph of f 0 or f 00, you should label that accordingly.
C
Summary
• A differentiable function f is increasing at a point or on an interval whenever its first

derivative is positive, and decreasing whenever its first derivative is negative.
• By taking the derivative of the derivative of a function f , we arrive at the second
derivative, f 00. The second derivative measures the instantaneous rate of change of the
first derivative, and thus the sign of the second derivative tells us whether or not the
slope of the tangent line to f is increasing or decreasing.
• A differentiable function is concave up whenever its first derivative is increasing (or
equivalently whenever its second derivative is positive), and concave down whenever its
first derivative is decreasing (or equivalently whenever its second derivative is negative).
Examples of functions that are everywhere concave up are y = x 2 and y = e x ; examples
of functions that are everywhere concave down are y = −x 2 and y = −e x .
• The units on the second derivative are “units of output per unit of input per unit of
input.” They tell us how the value of the derivative function is changing in response to
changes in the input. In other words, the second derivative tells us the rate of change
of the rate of change of the original function.
Exercises
1. Suppose that y = f (x) is a differentiable function for which the following information
is known: f (2) = −3, f 0(2) = 1.5, f 00(2) = −0.25.
(a) Is f increasing or decreasing at x = 2? Is f concave up or concave down at
x = 2?
(b) Do you expect f (2.1) to be greater than −3, equal to −3, or less than −3? Why?
(c) Do you expect f 0(2.1) to be greater than 1.5, equal to 1.5, or less than 1.5? Why?
(d) Sketch a graph of y = f (x) near (2, f (2)) and include a graph of the tangent
line.
2. For a certain function y = g(x), its derivative is given by the function pictured in
Figure 1.34.
y = g′ (x)
4
-3 -1 1 3
Figure 1.34: The graph of y = g 0(x).
(a) What is the approximate slope of the tangent line to y = g(x) at the point
(2, g(2))?
(b) How many real number solutions can there be to the equation g(x) = 0? Justify
your conclusion fully and carefully by explaining what you know about how
the graph of g must behave based on the given graph of g 0.
(c) On the interval −3 < x < 3, how many times does the concavity of g change?
Why?
(d) Use the provided graph to estimate the value of g 00(2).
3. A bungee jumper’s height h (in feet ) at time t (in seconds) is given in part by the data
in the following table:
t 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
h(t) 200 184.2 159.9 131.9 104.7 81.8 65.5 56.8 55.5 60.4 69.8
t 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0
h(t) 81.6 93.7 104.4 112.6 117.7 119.4 118.2 114.8 110.0 104.7
(a) Use the given data to estimate h 0(4.5), h 0(5), and h 0(5.5). At which of these
times is the bungee jumper rising most rapidly?
(b) Use the given data and your work in (a) to estimate h 00(5).
(c) What physical property of the bungee jumper does the value of h 00(5) measure?
What are its units?
(d) Based on the data, on what approximate time intervals is the function y = h(t)
concave down? What is happening to the velocity of the bungee jumper on
these time intervals?
4. For each prompt that follows, sketch a possible graph of a function on the interval
−3 < x < 3 that satisfies the stated properties.
(a) y = f (x) such that f is increasing on −3 < x < 3, f is concave up on
−3 < x < 0, and f is concave down on 0 < x < 3.
(b) y = g(x) such that g is increasing on −3 < x < 3, g is concave down on
−3 < x < 0, and g is concave up on 0 < x < 3.
(c) y = h(x) such that h is decreasing on −3 < x < 3, h is concave up on
−3 < x < −1, neither concave up nor concave down on −1 < x < 1, and h is
concave down on 1 < x < 3.
(d) y = p(x) such that p is decreasing and concave down on −3 < x < 0 and p is
increasing and concave down on 0 < x < 3.
1.7. LIMITS, CONTINUITY, AND DIFFERENTIABILITY 65
1.7 Limits, Continuity, and Differentiability

questions:
• What does it mean graphically to say that f has limit L as x → a? How is this
connected to having a left-hand limit at x = a and having a right-hand limit at
x = a?
• What does it mean to say that a function f is continuous at x = a? What role do
limits play in determining whether or not a function is continuous at a point?
• What does it mean graphically to say that a function f is differentiable at x = a?
How is this connected to the function being locally linear?
• How are the characteristics of a function having a limit, being continuous, and
being differentiable at a given point related to one another?
Introduction
In Section 1.2, we learned about how the concept of limits can be used to study the trend
of a function near a fixed input value. As we study such trends, we are fundamentally
interested in knowing how well-behaved the function is at the given point, say x = a.
In this present section, we aim to expand our perspective and develop language and
understanding to quantify how the function acts and how its value changes near a
particular point. Beyond thinking about whether or not the function has a limit L at
x = a, we will also consider the value of the function f (a) and how this value is related to
lim x→a f (x), as well as whether or not the function has a derivative f 0(a) at the point of
interest. Throughout, we will build on and formalize ideas that we have encountered in
several settings.
We begin to consider these issues through the following preview activity that asks you
to consider the graph of a function with a variety of interesting behaviors.
Preview Activity 1.7. A function f defined on −4 < x < 4 is given by the graph in
Figure 1.35. Use the graph to answer each of the following questions. Note: to the right
of x = 2, the graph of f is exhibiting infinite oscillatory behavior similar to the function
sin( πx ) that we encountered in the key example early in Section 1.2.
(a) For each of the values a = −3, −2, −1, 0, 1, 2, 3, determine whether or not lim f (x)
x→a
exists. If the function has a limit L at a given point, state the value of the limit
using the notation lim f (x) = L. If the function does not have a limit at a given
x→a
point, write a sentence to explain why.
66 1.7. LIMITS, CONTINUITY, AND DIFFERENTIABILITY
f
3
2
1
-3 -2 -1 1 2 3
-1
-2
-3
Figure 1.35: The graph of y = f (x).
(b) For each of the values of a from part (a) where f has a limit, determine the value
of f (a) at each such point. In addition, for each such a value, does f (a) have the
same value as lim f (x)?
x→a
(c) For each of the values a = −3, −2, −1, 0, 1, 2, 3, determine whether or not f 0(a)
exists. In particular, based on the given graph, ask yourself if it is reasonable
to say that f has a tangent line at (a, f (a)) for each of the given a-values. If so,
visually estimate the slope of the tangent line to find the value of f 0(a).
./
Having a limit at a point
In Section 1.2, we first encountered limits and learned that we say that f has limit L as
x approaches a and write lim f (x) = L provided that we can make the value of f (x) as
x→a
close to L as we like by taking x sufficiently close (but not equal to) a. Here, we expand
further on this definition and focus in more depth on what it means for a function not to
have a limit at a given value.
Essentially there are two behaviors that a function can exhibit at a point where it fails
to have a limit. In Figure 1.36, at left we see a function f whose graph shows a jump at
a = 1. In particular, if we let x approach 1 from the left side, the value of f approaches
2, while if we let x go to 1 from the right, the value of f tends to 3. Because the value of
f does not approach a single number as x gets arbitrarily close to 1 from both sides, we
know that f does not have a limit at a = 1.
Since f does approach a single value on each side of a = 1, we can introduce the
notion of left and right (or one-sided) limits. We say that f has limit L 1 as x approaches a
from the left and write

lim f (x) = L 1
x→a −
provided that we can make the value of f (x) as close to L 1 as we like by taking x sufficiently
close to a while always having x < a. In this case, we call L 1 the left-hand limit of f as x
approaches a. Similarly, we say L 2 is the right-hand limit of f as x approaches a and write
lim f (x) = L 2
x→a +
provided that we can make the value of f (x) as close to L 2 as we like by taking x sufficiently
close to a while always having x > a. In the graph of the function f in Figure 1.36, we see
that
lim− f (x) = 2 and lim+ f (x) = 3
x→1 x→1
and precisely because the left and right limits are not equal, the overall limit of f as x → 1
fails to exist.
f g
3 3
2 2
1 1
Figure 1.36: Functions f and g that each fail to have a limit at a = 1.
For the function g pictured at right in Figure 1.36, the function fails to have a limit at
a = 1 for a different reason. While the function does not have a jump in its graph at a = 1,
it is still not the case that g approaches a single value as x approaches 1. In particular, due
to the infinitely oscillating behavior of g to the right of a = 1, we say that the right-hand
limit of g as x → 1+ does not exist, and thus lim g(x) does not exist.
x→1
To summarize, anytime either a left- or right-hand limit fails to exist or the left- and
right-hand limits are not equal to each other, the overall limit will not exist. Said differently,
A function f has limit L as x → a if and only if
lim f (x) = L = lim+ f (x).

x→a − x→a
That is, a function has a limit at x = a if and only if both the left- and right-hand
limits at x = a exist and share the same value.
In Preview Activity 1.7, the function f given in Figure 1.35 only fails to have a limit at
two values: at a = −2 (where the left- and right-hand limits are 2 and −1, respectively) and
at x = 2, where lim x→2+ f (x) does not exist). Note well that even at values like a = −1 and
a = 0 where there are holes in the graph, the limit still exists.
Activity 1.18.
Consider a function that is piecewise-defined according to the formula


 3(x + 2) + 2 for −3 < x < −2
2

3 (x + 2) + 1 for −2 ≤ x < −1





2

f (x) =  3 (x + 2) + 1 for −1 < x < 1


2 for x = 1






4 − x


for x > 1

Use the given formula to answer the following questions.
-2 -1 1 2
-1
Figure 1.37: Axes for plotting the function y = f (x) in Activity 1.18.
(a) For each of the values a = −2, −1, 0, 1, 2, compute f (a).

(b) For each of the values a = −2, −1, 0, 1, 2, determine lim− f (x) and lim+ f (x).
x→a x→a
(c) For each of the values a = −2, −1, 0, 1, 2, determine lim f (x). If the limit fails to
x→a
exist, explain why by discussing the left- and right-hand limits at the relevant
a-value.
(d) For which values of a is the following statement true?
lim f (x) , f (a)

x→a
(e) On the axes provided in Figure 1.37, sketch an accurate, labeled graph of
y = f (x). Be sure to carefully use open circles (◦) and filled circles (•) to
represent key points on the graph, as dictated by the piecewise formula.
Being continuous at a point
Intuitively, a function is continuous if we can draw it without ever lifting our pencil from
the page. Alternatively, we might say that the graph of a continuous function has no jumps
or holes in it. We first consider three specific situations in Figure 1.38 where all three
functions have a limit at a = 1, and then work to make the idea of continuity more precise.
f g h
3 3 3
2 2 2
1 1 1
Figure 1.38: Functions f , g, and h that demonstrate subtly different behaviors at a = 1.
Note that f (1) is not defined, which leads to the resulting hole in the graph of f at
a = 1. We will naturally say that f is not continuous at a = 1. For the next function g in
in Figure 1.38, we observe that while lim x→1 g(x) = 3, the value of g(1) = 2, and thus the
limit does not equal the function value. Here, too, we will say that g is not continuous, even
though the function is defined at a = 1. Finally, the function h appears to be the most
well-behaved of all three, since at a = 1 its limit and its function value agree. That is,
lim h(x) = 3 = h(1).

x→1
With no hole or jump in the graph of h at a = 1, we desire to say that h is continuous there.
More formally, we make the following definition.
Definition 1.7. A function f is continuous at x = a provided that
(a) f has a limit as x → a,
(b) f is defined at x = a, and
(c) lim f (x) = f (a).

x→a
Conditions (a) and (b) are technically contained implicitly in (c), but we state them
explicitly to emphasize their individual importance. In words, (c) essentially says that a
function is continuous at x = a provided that its limit as x → a exists and equals its
function value at x = a. If a function is continuous at every point in an interval [a, b], we
say the function is “continuous on [a, b].” If a function is continuous at every point in
its domain, we simply say the function is “continuous.” Thus, continuous functions are
particularly nice: to evaluate the limit of a continuous function at a point, all we need to
do is evaluate the function.
For example, consider p(x) = x 2 − 2x + 3. It can be proved that every polynomial is a
continuous function at every real number, and thus if we would like to know lim x→2 p(x),
we simply compute
lim (x 2 − 2x + 3) = 22 − 2 · 2 + 3 = 3.
x→2
This route of substituting an input value to evaluate a limit works anytime we know
function being considered is continuous. Besides polynomial functions, all exponential
functions and the sine and cosine functions are continuous at every point, as are many
other familiar functions and combinations thereof.
Activity 1.19.
This activity builds on your work in Preview Activity 1.7, using the same function f as
given by the graph that is repeated in Figure 1.39
(a) At which values of a does lim x→a f (x) not exist?
(b) At which values of a is f (a) not defined?
(c) At which values of a does f have a limit, but lim x→a f (x) , f (a)?
(d) State all values of a for which f is not continuous at x = a.
(e) Which condition is stronger, and hence implies the other: f has a limit at x = a
or f is continuous at x = a? Explain, and hence complete the following sen-
tence: “If f at x = a, then f
at x = a,” where you complete the blanks with has a limit and is continuous,
using each phrase once.
C
f
3
2
1
-3 -2 -1 1 2 3
-1
-2
-3
Figure 1.39: The graph of y = f (x) for Activity 1.19.
Being differentiable at a point
We recall that a function f is said to be differentiable at x = a whenever f 0(a) exists.

Moreover, for f 0(a) to exist, we know that the function y = f (x) must have a tangent line
at the point (a, f (a)), since f 0(a) is precisely the slope of this line. In order to even ask
if f has a tangent line at (a, f (a)), it is necessary that f be continuous at x = a: if f
fails to have a limit at x = a, if f (a) is not defined, or if f (a) does not equal the value of
lim x→a f (x), then it doesn’t even make sense to talk about a tangent line to the curve at
this point.
Indeed, it can be proved formally that if a function f is differentiable at x = a, then it
must be continuous at x = a. So, if f is not continuous at x = a, then it is automatically
the case that f is not differentiable there. For example, in Figure 1.38 from our early
discussion of continuity, both f and g fail to be differentiable at x = 1 because neither
function is continuous at x = 1. But can a function fail to be differentiable at a point where
the function is continuous?
In Figure 1.40, we revisit the situation where a function has a sharp corner at a point,
something we encountered several times in Section 1.4. For the pictured function f , we
observe that f is clearly continuous at a = 1, since lim x→1 f (x) = 1 = f (1).
But the function f in Figure 1.40 is not differentiable at a = 1 because f 0(1) fails to
exist. One way to see this is to observe that f 0(x) = −1 for every value of x that is less
than 1, while f 0(x) = +1 for every value of x that is greater than 1. That makes it seem that
either +1 or −1 would be equally good candidates for the value of the derivative at x = 1.
Alternately, we could use the limit definition of the derivative to attempt to compute f 0(1),
and discover that the derivative does not exist. A similar problem will be investigated in
Activity 1.20. Finally, we can also see visually that the function f in Figure 1.40 does not
have a tangent line. When we zoom in on (1, 1) on the graph of f , no matter how closely
we examine the function, it will always look like a “V”, and never like a single line, which
1 (1, 1)
Figure 1.40: A function f that is continuous at a = 1 but not differentiable at a = 1; at

right, we zoom in on the point (1, 1) in a magnified version of the box in the left-hand plot.
tells us there is no possibility for a tangent line there.

To make a more general observation, if a function does have a tangent line at a given
point, when we zoom in on the point of tangency, the function and the tangent line should
appear essentially indistinguishable7 . Conversely, if we have a function such that when we
zoom in on a point the function looks like a single straight line, then the function should
have a tangent line there, and thus be differentiable. Hence, a function that is differentiable
at x = a will, up close, look more and more like its tangent line at (a, f (a)), and thus we
say that a function is differentiable at x = a is locally linear.
To summarize the preceding discussion of differentiability and continuity, we make
several important observations.
• If f is differentiable at x = a, then f is continuous at x = a. Equivalently, if f fails

to be continuous at x = a, then f will not be differentiable at x = a.
• A function can be continuous at a point, but not be differentiable there. In particular,

a function f is not differentiable at x = a if the graph has a sharp corner (or cusp)
at the point (a, f (a)).
• If f is differentiable at x = a, then f is locally linear at x = a. That is, when a

function is differentiable, it looks linear when viewed up close because it resembles
its tangent line there.
7 See,
for instance, http://gvsu.edu/s/6J for an applet (due to David Austin, GVSU) where zooming in
shows the increasing similarity between the tangent line and the curve.
Activity 1.20.
In this activity, we explore two different functions and classify the points at which each
is not differentiable. Let g be the function given by the rule g(x) = |x|, and let f be the
function that we have previously explored in Preview Activity 1.7, whose graph is given
again in Figure 1.41.
(a) Reasoning visually, explain why g is differentiable at every point x such that
x , 0.
|h|
(b) Use the limit definition of the derivative to show that g 0(0) = limh→0 h .
(c) Explain why g 0(0) fails to exist by using small positive and negative values of
h.
f
3
2
1
-3 -2 -1 1 2 3
-1
-2
-3
Figure 1.41: The graph of y = f (x) for Activity 1.20.
(d) State all values of a for which f is not differentiable at x = a. For each, provide
a reason for your conclusion.
(e) True or false: if a function p is differentiable at x = b, then lim x→b p(x) must
exist. Why?
C
Summary
• A function f has limit L as x → a if and only if f has a left-hand limit at x = a, has a

right-hand limit at x = a, and the left- and right-hand limits are equal. Visually, this
means that there can be a hole in the graph at x = a, but the function must approach
the same single value from either side of x = a.
• A function f is continuous at x = a whenever f (a) is defined, f has a limit as x → a,

and the value of the limit and the value of the function agree. This guarantees that
there is not a hole or jump in the graph of f at x = a.
• A function f is differentiable at x = a whenever f 0(a) exists, which means that f has a
tangent line at (a, f (a)) and thus f is locally linear at the value x = a. Informally, this
means that the function looks like a line when viewed up close at (a, f (a)) and that
there is not a corner point or cusp at (a, f (a)).
• Of the three conditions discussed in this section (having a limit at x = a, being
continuous at x = a, and being differentiable at x = a), the strongest condition is
being differentiable, and the next strongest is being continuous. In particular, if f is
differentiable at x = a, then f is also continuous at x = a, and if f is continuous at
x = a, then f has a limit at x = a.
Exercises
1. Consider the graph of the function y = p(x) that is provided in Figure 1.42. Assume
that each portion of the graph of p is a straight line, as pictured.
p
3 3
-3 3 -3 3
-3 -3
Figure 1.42: At left, the piecewise linear function y = p(x). At right, axes for plotting
y = p0(x).
(a) State all values of a for which lim x→a p(x) does not exist.
(b) State all values of a for which p is not continuous at a.
(c) State all values of a for which p is not differentiable at x = a.
(d) On the axes provided in Figure 1.42, sketch an accurate graph of y = p0(x).
2. For each of the following prompts, give an example of a function that satisfies the
stated criteria. A formula or a graph, with reasoning, is sufficient for each. If no such
example is possible, explain why.
(a) A function f that is continuous at a = 2 but not differentiable at a = 2.

(b) A function g that is differentiable at a = 3 but does not have a limit at a = 3.
(c) A function h that has a limit at a = −2, is defined at a = −2, but is not
continuous at a = −2.
(d) A function p that satisfies all of the following:
• p(−1) = 3 and lim x→−1 p(x) = 2
• p(0) = 1 and p0(0) = 0
• lim x→1 p(x) = p(1) and p0(1) does not exist
3. Let h(x) be a function whose derivative y = h 0(x) is given by the graph on the right in
Figure 1.43.
(a) Based on the graph of y = h 0(x), what can you say about the behavior of the
function y = h(x)?
(b) At which values of x is y = h 0(x) not defined? What behavior does this lead
you to expect to see in the graph of y = h(x)?
(c) Is it possible for y = h(x) to have points where h is not continuous? Explain
your answer.
(d) On the axes provided at left, sketch at least two distinct graphs that are
possible functions y = h(x) that each have a derivative y = h 0(x) that matches
the provided graph at right. Explain why there are multiple possibilities for
y = h(x).
3 3
2
y = h′ (x)
1
-3 3 -3 -2 -1 1 2 3
-1
-2
-3 -3
Figure 1.43: Axes for plotting y = h(x) and, at right, the graph of y = h 0(x).
4. Consider the function g(x) =

p
|x|.
(a) Use a graph to explain visually why g is not differentiable at x = 0.
(b) Use the limit definition of the derivative to show that

p
|h|
g (0) = lim
0
.
h→0 h
(c) Investigate the value of g 0(0) by estimating the limit in (b) using
√ small positive
|−0.01|
and negative values of h. For instance, you might compute 0.01 . Be sure to
use several different values of h (both positive and negative), including ones
closer to 0 than 0.01. What do your results tell you about g 0(0)?
(d) Use your graph in (a) to sketch an approximate graph of y = g 0(x).
1.8. THE TANGENT LINE APPROXIMATION 77
1.8 The Tangent Line Approximation

questions:
• What is the formula for the general tangent line approximation to a differentiable
function y = f (x) at the point (a, f (a))?
• What is the principle of local linearity and what is the local linearization of a
differentiable function f at a point (a, f (a))?
• How does knowing just the tangent line approximation tell us information about
the behavior of the original function itself near the point of approximation? How
does knowing the second derivative’s value at this point provide us additional
knowledge of the original function’s behavior?
Introduction
Among all functions, linear functions are simplest. One of the powerful consequences
of a function y = f (x) being differentiable at a point (a, f (a)) is that, up close, the
function y = f (x) is locally linear and looks like its tangent line at that point. In certain
circumstances, this allows us to approximate the original function f with a simpler function
L that is linear: this can be advantageous when we have limited information about f
or when f is computationally or algebraically complicated. We will explore all of these
situations in what follows.
It is essential to recall that when f is differentiable at x = a, the value of f 0(a) provides
the slope of the tangent line to y = f (x) at the point (a, f (a)). By knowing both a point
on the line and the slope of the line we are thus able to find the equation of the tangent
line. Preview Activity 1.8 will refresh these concepts through a key example and set the
stage for further study.
Preview Activity 1.8. Consider the function y = g(x) = −x 2 + 3x + 2.
(a) Use the limit definition of the derivative to compute a formula for y = g 0(x).
(b) Determine the slope of the tangent line to y = g(x) at the value x = 2.
(c) Compute g(2).
(d) Find an equation for the tangent line to y = g(x) at the point (2, g(2)). Write your
result in point-slope form8 .
8 Recall that a line with slope m that passes through (x 0 , y0 ) has equation y − y0 = m(x − x 0 ), and this is
78 1.8. THE TANGENT LINE APPROXIMATION
Figure 1.44: Axes for plotting y = g(x) and its tangent line to the point (2, g(2)).
(e) On the axes provided in Figure 1.44, sketch an accurate, labeled graph of y = g(x)
along with its tangent line at the point (2, g(2)).
./
The tangent line
Given a function f that is differentiable at x = a, we know that we can determine the slope
of the tangent line to y = f (x) at (a, f (a)) by computing f 0(a). The resulting tangent line
through (a, f (a)) with slope m = f 0(a) has its equation in point-slope form given by
y − f (a) = f 0(a)(x − a),
which we can also express as y = f 0(a)(x − a) + f (a). Note well: there is a major difference
between f (a) and f (x) in this context. The former is a constant that results from using
the given fixed value of a, while the latter is the general expression for the rule that defines
the function. The same is true for f 0(a) and f 0(x): we must carefully distinguish between
these expressions. Each time we find the tangent line, we need to evaluate the function
and its derivative at a fixed a-value.
In Figure 1.45, we see a labeled plot of the graph of a function f and its tangent line
at the point (a, f (a)). Notice how when we zoom in we see the local linearity of f more
clearly highlighted as the function and its tangent line are nearly indistinguishable up
close. This can also be seen dynamically in the java applet at http://gvsu.edu/s/6J.
the point-slope form of the equation.

y
y = f (x)
y = f (x)
(a, f (a))
(a, f (a))
y = f ′ (a)(x − a) + f (a) y = L(x)
a x
Figure 1.45: A function y = f (x) and its tangent line at the point (a, f (a)): at left, from a
distance, and at right, up close. At right, we label the tangent line function by y = L(x)
and observe that for x near a, f (x) ≈ L(x).
The local linearization
A slight change in perspective and notation will enable us to be more precise in discussing
how the tangent line to y = f (x) at (a, f (a)) approximates f near x = a. Taking the
equation for the tangent line and solving for y, we observe that the tangent line is given by
y = f 0(a)(x − a) + f (a)
and moreover that this line is itself a function of x. Replacing the variable y with the
expression L(x), we call
L(x) = f 0(a)(x − a) + f (a)
the local linearization of f at the point (a, f (a)). In this notation, it is particularly important
to observe that L(x) is nothing more than a new name for the tangent line, and that for x
close to a, we have that f (x) ≈ L(x).
Say, for example, that we know that a function y = f (x) has its tangent line approxi-
mation given by L(x) = 3 − 2(x − 1) at the point (1, 3), but we do not know anything else
about the function f . If we are interested in estimating a value of f (x) for x near 1, such
as f (1.2), we can use the fact that f (1.2) ≈ L(1.2) and hence
f (1.2) ≈ L(1.2) = 3 − 2(1.2 − 1) = 3 − 2(0.2) = 2.6.
Again, much of the new perspective here is only in notation since y = L(x) is simply a
new name for the tangent line function. In light of this new notation and our observations
above, we note that since L(x) = f (a) + f 0(a)(x − a) and L(x) ≈ f (x) for x near a, it also
follows that we can write
f (x) ≈ f (a) + f 0(a)(x − a) for x near a.
The next activity explores some additional important properties of the local lineariza-
tion y = L(x) to a function f at given a-value.
Activity 1.21.
Suppose it is known that for a given differentiable function y = g(x), its local lineariza-
tion at the point where a = −1 is given by L(x) = −2 + 3(x + 1).
(a) Compute the values of L(−1) and L 0(−1).
(b) What must be the values of g(−1) and g 0(−1)? Why?
(c) Do you expect the value of g(−1.03) to be greater than or less than the value
of g(−1)? Why?
(d) Use the local linearization to estimate the value of g(−1.03).
(e) Suppose that you also know that g 00(−1) = 2. What does this tell you about the
graph of y = g(x) at a = −1?
(f) For x near −1, sketch the graph of the local linearization y = L(x) as well as a
possible graph of y = g(x) on the axes provided in Figure 1.46.
Figure 1.46: Axes for plotting y = L(x) and y = g(x).
C
As we saw in the example provided by Activity 1.21, the local linearization y = L(x)
is a linear function that shares two important values with the function y = f (x) that it is
derived from. In particular, observe that since L(x) = f (a) + f 0(a)(x − a), it follows that
L(a) = f (a). In addition, since L is a linear function, its derivative is its slope. Hence,
L 0(x) = f 0(a) for every value of x, and specifically L 0(a) = f 0(a). Therefore, we see that
L is a linear function that has both the same value and the same slope as the function f
at the point (a, f (a)).
In situations where we know the linear approximation y = L(x), we therefore know
the original function’s value and slope at the point of tangency. What remains unknown,
however, is the shape of the function f at the point of tangency. There are essentially four
possibilities, as enumerated in Figure 1.47.
Figure 1.47: Four possible graphs for a nonlinear differentiable function and how it can be
situated relative to its tangent line at a point.
These stem from the fact that there are three options for the value of the second
derivative: either f 00(a) < 0, f 00(a) = 0, or f 00(a) > 0. If f 00(a) > 0, then we know the
graph of f is concave up, and we see the first possibility on the left, where the tangent line
lies entirely below the curve. If f 00(a) < 0, then we find ourselves in the second situation
(from left) where f is concave down and the tangent line lies above the curve. In the
situation where f 00(a) = 0 and f 00 changes sign at x = a, the concavity of the graph will
change, and we will see either the third or fourth option9 . A fifth option (that is not very
interesting) can occur, which is where the function f is linear, and so f (x) = L(x) for all
values of x.
The plots in Figure 1.47 highlight yet another important thing that we can learn from
the concavity of the graph near the point of tangency: whether the tangent line lies above
or below the curve itself. This is key because it tells us whether or not the tangent line
approximation’s values will be too large or too small in comparison to the true value of f .
For instance, in the first situation in the leftmost plot in Figure 1.47 where f 00(a) > 0, since
the tangent line falls below the curve, we know that L(x) ≤ f (x) for all values of x near a.
We explore these ideas further in the following activity.
Activity 1.22.
This activity concerns a function f (x) about which the following information is known:
9 Itis possible to have f 00 (a) = 0 and have f 00 not change sign at x = a, in which case the graph will look
like one of the first two options.
• f is a differentiable function defined at every real number x

• f (2) = −1
• y = f 0(x) has its graph given in Figure 1.48
y = f ′ (x)
2 2 2
x x x
2 2 2
Figure 1.48: At center, a graph of y = f 0(x); at left, axes for plotting y = f (x); at right,
axes for plotting y = f 00(x).
Your task is to determine as much information as possible about f (especially near the
value a = 2) by responding to the questions below.
(a) Find a formula for the tangent line approximation, L(x), to f at the point
(2, −1).
(b) Use the tangent line approximation to estimate the value of f (2.07). Show your
work carefully and clearly.
(c) Sketch a graph of y = f 00(x) on the righthand grid in Figure 1.48; label it
appropriately.
(d) Is the slope of the tangent line to y = f (x) increasing, decreasing, or neither
when x = 2? Explain.
(e) Sketch a possible graph of y = f (x) near x = 2 on the lefthand grid in
Figure 1.48. Include a sketch of y = L(x) (found in part (a)). Explain how you
know the graph of y = f (x) looks like you have drawn it.
(f) Does your estimate in (b) over- or under-estimate the true value of f (2)? Why?
C
The idea that a differentiable function looks linear and can be well-approximated by a
linear function is an important one that finds wide application in calculus. For example, by
approximating a function with its local linearization, it is possible to develop an effective
algorithm to estimate the zeroes of a function. Local linearity also helps us to make further
sense of certain challenging limits. For instance, we have seen that a limit such as
sin(x)
lim
x→0 x
is indeterminate because both its numerator and denominator tend to 0. While there is no
algebra that we can do to simplify sin(x)
x , it is straightforward to show that the linearization
of f (x) = sin(x) at the point (0, 0) is given by L(x) = x. Hence, for values of x near 0,
sin(x) ≈ x. As such, for values of x near 0,
sin(x) x
≈ = 1,
x x
which makes plausible the fact that
sin(x)
lim = 1.
x→0 x
These ideas and other applications of local linearity will be explored later on in our work.
Summary
• The tangent line to a differentiable function y = f (x) at the point (a, f (a)) is given in
point-slope form by the equation
y − f (a) = f 0(a)(x − a).
• The principle of local linearity tells us that if we zoom in on a point where a function
y = f (x) is differentiable, the function should become indistinguishable from its tangent
line. That is, a differentiable function looks linear when viewed up close. We rename
the tangent line to be the function y = L(x) where L(x) = f (a) + f 0(a)(x − a) and note
that f (x) ≈ L(x) for all x near x = a.
• If we know the tangent line approximation L(x) = f (a) + f 0(a)(x − a), then because
L(a) = f (a) and L 0(a) = f 0(a), we also know both the value and the derivative of the
function y = f (x) at the point where x = a. In other words, the linear approximation
tells us the height and slope of the original function. If, in addition, we know the value
of f 00(a), we then know whether the tangent line lies above or below the graph of
y = f (x) depending on the concavity of f .
Exercises
1. A certain function y = p(x) has its local linearization at a = 3 given by L(x) = −2x + 5.
(a) What are the values of p(3) and p0(3)? Why?

(b) Estimate the value of p(2.79).
(c) Suppose that p00(3) = 0 and you know that p00(x) < 0 for x < 3. Is your
estimate in (b) too large or too small?
(d) Suppose that p00(x) > 0 for x > 3. Use this fact and the additional information
above to sketch an accurate graph of y = p(x) near x = 3. Include a sketch of
y = L(x) in your work.
2. A potato is placed in an oven, and the potato’s temperature F (in degrees Fahrenheit) at
in minutes.
t F(t)
0 70
15 180.5
30 251
45 296
60 324.5
75 342.8
90 354.5
(a) Use a central difference to estimate F 0(60). Use this estimate as needed in
subsequent questions.
(b) Find the local linearization y = L(t) to the function y = F(t) at the point where
a = 60.
(c) Determine an estimate for F(63) by employing the local linearization.
(d) Do you think your estimate in (c) is too large or too small? Why?
3. An object moving along a straight line path has a differentiable position function
y = s(t). It is known that at time t = 9 seconds, the object’s position is s = 4 feet
(measured from its starting point at t = 0). Furthermore, the object’s instantaneous
velocity at t = 9 is −1.2 feet per second, and its acceleration at the same instant is 0.08
feet per second per second.
(a) Use local linearity to estimate the position of the object at t = 9.34.
(b) Is your estimate likely too large or too small? Why?
(c) In everyday language, describe the behavior of the moving object at t = 9. Is it
moving toward its starting point or away from it? Is its velocity increasing or
decreasing?
2
4. For a certain function f , its derivative is known to be f 0(x) = (x − 1)e−x . Note that
you do not know a formula for y = f (x).
(a) At what x-value(s) is f 0(x) = 0? Justify your answer algebraically, but include a
graph of f 0 to support your conclusion.
(b) Reasoning graphically, for what intervals of x-values is f 00(x) > 0? What does
this tell you about the behavior of the original function f ? Explain.
(c) Assuming that f (2) = −3, estimate the value of f (1.88) by finding and using
the tangent line approximation to f at x = 2. Is your estimate larger or smaller
than the true value of f (1.88)? Justify your answer.

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Chapter 1

Understanding the Derivative

1.1 How do we measure velocity?

(a) Construct an accurate graph of y = s(t) on the time interval 0 ≤ t ≤ 3. Label at

(c) Consider the expression

Position and average velocity

Figure 1.1: A partial plot of s(t) = 64 − 16(t − 1)2 .

s(0.5 + h) = 16 − 16(0.5 + h)2

Now, returning to our computation of the average velocity, we ﬁnd that

AV[0.5,0.5+h] = −16 − 16h.

• The instantaneous velocity of a moving object at a ﬁxed time is estimated by considering

Figure 1.3: A bungee jumper’s height function.

1.2 The notion of limit

• How do we go about determining the value of the limit of a function at a point?

s(x) − s(1) (64 − 16x 2 ) − (64 − 16) 16 − 16x 2

Figure 1.5: Graph of y = g(x) for Preview Activity 1.2.

The Notion of Limit

lim g(x) = 3, lim g(x) = 4, and lim g(x) = 1,

to its function value. Here, it follows that as x → −1, (4 − x 2 ) → (4 − (−1)2 ) = 3, and

lim f (x) = lim 2 − x.

As x → 0, we observe that πx does not behave in an elementary way. When x is

Figure 1.8: Plot of the position function y = s(t) in Activity 1.6.

• Limits enable us to examine trends in function behavior near a speciﬁc point. In

approach a particular ﬁxed value.

(a) y = f (x) such that

(b) y = g(x) such that

1.3 The derivative of a function at a point

Figure 1.10: Plot of y = f (x) for Preview Activity 1.3.

The Derivative of a Function at a Point

Just as we deﬁned instantaneous velocity in terms of average velocity, we now deﬁne

Aloud, we read the symbol f 0(a) as either “ f -prime at a” or “the derivative of f

• We say that a function that has a derivative at x = a is diﬀerentiable at x = a.

• The derivative is a generalization of the instantaneous velocity of a position function:

• Because the units on f (a+h)−

• Because the quantity f (a+h)−

As we move from an average rate of change to an instantaneous one, we can think of

As we will see in subsequent study, the existence of the tangent line at x = a is

resembles the curve near x = a.

Solution. From the limit deﬁnition, we know that

Figure 1.13: The tangent line to y = x − x 2 at the point (2, −2).

algebra. Expanding and distributing in the numerator,

f 0(2) = lim (−3 − h).

Figure 1.14: Axes for plotting y = s(t) in Activity 1.8.

Figure 1.15: Axes for plotting y = P(t) in Activity 1.9.

Figure 1.16: Plot of y = f (x).

greater or less than the instantaneous rate of change of the population on

(a) f (x) = x 2 − 3x, a = 2

(d) f (x) = 2 − |x − 1|, a = 1

1.4 The derivative function

Preview Activity 1.4. Consider the function f (x) = 4x − x 2 .

How the derivative is itself a function

Here we observe that neither 4 nor 2a depend on the value of h, so as h → 0, (4−2a−h) →

6 Marc Renault, Calculus Applets Using Geogebra.

• The limit deﬁnition of the derivative, f 0(x) = limh→0 f (x+h)− h

2. Consider the function g(x) = x 2 − x + 3.

1.5 Interpreting, estimating, and using the derivative

Units of the derivative function

As we now know, the derivative of the function f at a ﬁxed value x is given by

Toward more accurate derivative estimates