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6.3
Contents
1
Understanding functions . . . . . . . . . . .
1.1
1.2
1.3
1.4
2
2.1
2.3
2.4
2.5
3.1
3.3
How crazy could it be? . . . . . . . . .
Polynomial functions . . . . . . . . . .
8
14
19
Exponential and logarithmic functions .
32
Stars and functions . . . . . . . . . . .
What is a limit? . . . . . . . . . . . . .
Continuity . . . . . . . . . . . . . . . .
Equal or not? . . . . . . . . . . . . . .
4.3
The Squeeze Theorem . . . . . . . . .
The limit laws . . . . . . . . . . . . . .
(In)determinate forms . . . . . . . . . . . . .
35
45
60
6.2
Vertical asymptotes . . . . . . . . . . .
63
64
65
Roxy and Yuri like food . . . . . . . . .
72
7.3
The Intermediate Value Theorem . . . .
76
Continuity of piecewise functions . . .
An application of limits . . . . . . . . . . . .
8.1
Limits and velocity . . . . . . . . . . .
Instantaneous velocity . . . . . . . . .
Definition of the derivative . . . . . . . . . .
9.1
Slope of a curve . . . . . . . . . . . . .
The definition of the derivative . . . . .
Derivatives as functions . . . . . . . . . . .
10.1
Wait for the right moment . . . . . . .
10.3
Differentiability implies continuity . .
The derivative as a function . . . . . .
73
79
80
81
85
86
87
92
93
94
97
Rules of differentiation . . . . . . . . . . . . 100
11.1
11.2
Patterns in derivatives . . . . . . . . . 101
Basic rules of differentiation . . . . . . 103
11.3 The derivative of the natural exponential function . . . . . . . . . . . . 107
54
Limits of the form nonzero over zero . .
Zoom out . . . . . . . . . . . . . . . .
11
51
56
7.1
10.2
47
5.3
6.1
10
46
55
71
9.2
37
Could it be anything? . . . . . . . . . .
Using limits to detect asymptotes . . . . . . .
9
36
42
Continuity and the Intermediate Value Theorem . . . . . . . . . . . . . . . . . . . .
8.2
25
5.1
Limits of the form zero over zero . . . .
8
20
21
67
7.2
12
23
Trigonometric functions . . . . . . . .
7
7
Rational functions . . . . . . . . . . . .
4.1
5.2
6
Inverses of functions . . . . . . . . . .
Limit laws . . . . . . . . . . . . . . . . . . . .
4.2
5
Compositions of functions . . . . . . .
What is a limit? . . . . . . . . . . . . . . . .
3.2
4
For each input, exactly one output . . .
Review of famous functions . . . . . . . . . .
2.2
3
Same or different? . . . . . . . . . . . .
6
Horizontal asymptotes . . . . . . . . .
12
11.4
The derivative of sine . . . . . . . . . 109
12.1
Derivatives of products are tricky . . . 112
Product rule and quotient rule . . . . . . . 111
12.2
13
The Product rule and quotient rule . . 113
Chain rule . . . . . . . . . . . . . . . . . . . 116
13.1
13.2
An unnoticed composition . . . . . . . 117
The chain rule . . . . . . . . . . . . . 118
14
13.3
Derivatives of trigonometric functions
14.1
Rates of rates . . . . . . . . . . . . . . 124
Higher order derivatives and graphs . . . . 123
14.2
14.3
14.4
15
Concavity . . . . . . . . . . . . . . . 128
Position, velocity, and acceleration . . 130
21.2
22
Concepts of graphing functions . . . . 182
Computations for graphing functions . . . . 184
22.1
22.2
23
What’s the graph look like? . . . . . . 181
Wanted: graphing procedure . . . . . 185
Computations for graphing functions . 186
Mean Value Theorem . . . . . . . . . . . . 192
15.3 Derivatives of inverse exponential functions . . . . . . . . . . . . . . . . 137
23.3
The Mean Value Theorem . . . . . . . 198
16.1
23.2
Implicit differentiation . . . . . . . . . 133
Logarithmic differentiation . . . . . . . . . 139
24
Multiplication to addition . . . . . . . 140
17.1
We can figure it out . . . . . . . . . . 145
17.3
The Inverse Function Theorem . . . . 149
A changing circle . . . . . . . . . . . 152
More than one rate . . . . . . . . . . . 153
25
Pizza and calculus, so cheesy . . . . . 159
Applied related rates . . . . . . . . . . 160
26
20.2
27
Linear approximation . . . . . . . . . 203
28
26.1
Basic optimization . . . . . . . . . . . 215
Volumes of aluminum cans . . . . . . 219
Applied optimization . . . . . . . . . 220
L’Hôpital’s rule . . . . . . . . . . . . . . . . 226
27.1
A limitless dialogue . . . . . . . . . . 227
L’Hôpital’s rule . . . . . . . . . . . . 228
Antiderivatives . . . . . . . . . . . . . . . . 233
28.1
Jeopardy! Of calculus . . . . . . . . . 234
28.3
Falling objects . . . . . . . . . . . . . 243
28.2
29
A mysterious formula . . . . . . . . . 214
Applied optimization . . . . . . . . . . . . . 218
27.2
More coffee . . . . . . . . . . . . . . 169
Maximums and minimums . . . . . . 170
25.1
26.2
Maximums and minimums . . . . . . . . . 168
20.1
Replacing curves with lines . . . . . . 202
Optimization . . . . . . . . . . . . . . . . . 213
25.2
Applied related rates . . . . . . . . . . . . . 158
19.1
24.1
24.3 Explanation of the product and chain rules210
More than one rate . . . . . . . . . . . . . . 151
18.1
The Extreme Value Theorem . . . . . 194
Linear approximation . . . . . . . . . . . . 201
24.2
Logarithmic differentiation . . . . . . 141
Derivatives of inverse functions . . . . . . . 144
19.2
20
21.1
Let’s run to class . . . . . . . . . . . . 193
18.2
19
Concepts of graphing functions . . . . . . . 180
23.1
17.2 Derivatives of inverse trigonometric
functions . . . . . . . . . . . . . . 146
18
21
Standard form . . . . . . . . . . . . . 132
15.1
16.2
17
Higher order derivatives and graphs . . 125
Implicit differentiation . . . . . . . . . . . . 131
15.2
16
120
Basic antiderivatives . . . . . . . . . . 235
Approximating the area under a curve . . . 245
29.1
What is area? . . . . . . . . . . . . . . 246
29.3
Approximating area with rectangles . . 251
29.2
30
Definite integrals . . . . . . . . . . . . . . . 259
30.1
30.2
31
Introduction to sigma notation . . . . . 247
Computing areas . . . . . . . . . . . . 260
The definite integral . . . . . . . . . . 261
Antiderivatives and area . . . . . . . . . . . 269
31.1
Meaning of multiplication . . . . . . . 270
31.2 Relating velocity, displacement, antiderivatives and areas . . . . . . . 271
32
First Fundamental Theorem of Calculus . . 276
32.1
What’s in a calculus problem? . . . . . 277
32.2 The First Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . 278
33
Second Fundamental Theorem of Calculus . 282
33.1
A secret of the definite integral . . . . 283
33.3
A tale of three integrals . . . . . . . . 287
34.1
What could it represent? . . . . . . . . 290
33.2 The Second Fundamental Theorem of
Calculus . . . . . . . . . . . . . . 284
34
Applications of integrals . . . . . . . . . . . 289
34.2
35
The idea of substitution . . . . . . . . . . . 296
35.1
35.2
36
Applications of integrals . . . . . . . . 291
Geometry and substitution . . . . . . . 297
The idea of substitution . . . . . . . . 298
Working with substitution . . . . . . . . . . 302
36.1
Integrals are puzzles! . . . . . . . . . 303
36.3
The Work-Energy Theorem . . . . . . 309
36.2
Working with substitution . . . . . . . 304
Index . . . . . . . . . . . . . . . . . . . . . . . . 313
Understanding functions
1
Understanding functions
After completing this section, students should be able to do the
following.
• State the definition of a function.
• Find the domain and range of a function.
• Distinguish between functions by considering their domains.
• Determine where a function is positive or negative.
• Plot basic functions.
• Perform basic operations and compositions on functions.
• Work with piecewise defined functions.
• Determine if a function is one-to-one.
• Recognize different representations of the same function.
• Define and work with inverse functions.
• Plot inverses of basic functions.
• Find inverse functions (algebraically and graphically).
• Find the largest interval containing a given point where
the function is invertible.
• Determine the intervals on which a function has an inverse.
6
Same or different?
Multiple Choice:
Break-Ground:
1.1
Same or di�erent?
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, I have a pressing question.
Riley: Tell me. Tell me everything.
Devyn: Think about the function
f (x) =
2
x * 3x + 2
.
x*2
Riley: OK.
Devyn: Is this function equal to g(x) = x * 1?
Riley: Well if I plot them with my calculator, they look the same.
Devyn: I know!
Riley: And I suppose if I write
x2 * 3x + 2
f (x) =
x*2
(x * 1)(x * 2)
=
x*2
=x*1
(a) These are the same function although they are represented
by different formulas.
(b) These are different functions because they have different
formulas.
Problem 5. Let f (x) = sin2 (x) and g(u) = sin2 (u). The domain of each of these functions is all real numbers. Which of
the following statements are true?
Multiple Choice:
(a) There is not enough information to determine if f = g.
(b) The functions are equal.
(c) If x ë u, then f ë g.
(d) We have f ë g since f uses the variable x and g uses the
variable u.
= g(x).
Devyn: Sure! But what about when x = 2? In this case
g(2) = 1
but
f (2) is undefined!
Riley: Right, f (2) is undefined because we cannot divide by zero.
Hmm. Now I see the problem. Yikes!
Problem 1. In the context above, are f and g the same function?
Problem 2. Suppose f and g are functions but the domain of
f is different from the domain of g. Could it be that f and g
are actually the same function?
Problem 3. Can the same function be represented by different
formulas?
˘
Problem 4. Are f (x) = x and g(x) = x2 the same function?
7
For each input, exactly one output
Dig-In:
1.2 For each input, exactly one
output
Life is complex. Part of this complexity stems from the fact
that there are many relationships between seemingly unrelated
events. Armed with mathematics, we seek to understand the
world. Perhaps the most relevant “real-world” relation is
the position of an object with respect to time.
Our observations seem to indicate that every instant in time is
associated to a unique positioning of the objects in the universe.
You may have heard the saying,
you cannot be two places at the same time,
and it is this fact that motivates our definition for functions.
Definition. A function is a relation between sets where for
each input, there is exactly one output.
Question 1. If our function is the “position with respect to
time” of some object, then the input is
Multiple Choice:
(a) position
(b) time
(c) none of the above
and the output is
Multiple Choice:
(a) position
(b) time
8
(c) none of the above
Something as simple as a dictionary could be thought of as
a relation, as it connects words to definitions. However, a
dictionary is not a function, as there are words with multiple
definitions. On the other hand, if each word only had a single
definition, then a dictionary would be a function.
Question 2. Which of the following are functions?
Select All Correct Answers:
(a) Mapping words to their definition in a dictionary.
(b) Given an object traveling through space, mapping time to
the position of the object in space.
(c) Mapping people to their birth date.
(d) Mapping mothers to their children.
What we are hoping to convince you is that the following are
true:
(a) The definition of a function is well-grounded in a real
context.
(b) The definition of a function is flexible enough that it can
be used to model a wide range of phenomena.
Whenever we talk about functions, we should explicitly state
what type of things the inputs are and what type of things the
outputs are. In calculus, functions often define a relation from
(a subset of) the real numbers (denoted by R) to (a subset of)
the real numbers.
Definition. We call the set of the inputs of a function the domain, and we call the set of the outputs of a function the range.
Example 1. Consider the function f that maps from the real
numbers to the real numbers by taking a number and mapping
For each input, exactly one output
it to its cube:
3
1≠1
2
*2 ≠ *8
1
1.5 ≠ 3.375
*2
and so on. This function can be described by the formula
f (x) = x3 or by the graph shown in the plot below:
5
*1.5
*1
*0.5
*1
*0.5
0.5
1
1.5
2
2.5
3
3.5
4
x
*1
*2
Observe that here we have multiple inputs that give the same
output. This is not a problem! To be a function, we merely need
to check that for each input, there is exactly one output, and
this condition is satisfied.
y
*2
*1.5
y
0.5
1
1.5
2
x
Question 3. Compute:
‚2.4„
*5
Question 4. Compute:
Warning. A function is a relation (such that for each input,
there is exactly one output) between sets. The formula and the
graph are merely descriptions of this relation.
• A formula describes the relation using symbols.
• A graph describes the relation using pictures.
The function is the relation itself, and is independent of how
it is described.
Our next example may be a function that is new to you. It is
the greatest integer function.
Example 2. Consider the greatest integer function. This function maps any real number x to the greatest integer less than or
equal to x. People sometimes write this as f (x) = ‚x„, where
those funny symbols mean exactly the words above describing
the function. For your viewing pleasure, here is a graph of the
greatest integer function:
‚*2.4„
Notice that both the functions described above pass the socalled vertical line test.
Theorem 1. The curve y = f (x) represents y as a function of
x at x = a if and only if the vertical line x = a intersects the
curve y = f (x) at exactly one point. This is called the vertical
line test.
Sometimes the domain and range are the entire set of real
numbers, denoted by R. In our next examples we show that
this is not always the case.
Example 3. Consider the function that maps non-negative
real numbers to their positive square root. This function can
be described by the formula
˘
f (x) = x.
The domain is 0 f x < ÿ, which we prefer to write as [0, ÿ)
in interval notation. The range is [0, ÿ). Here is a graph of
y = f (x):
9
For each input, exactly one output
Finally, we will consider a function whose domain is all real
numbers except for a single point.
y
4
2
*8
*7
*6
*5
*4
*3
*2
Example 5. Are
*1
1
2
3
4
5
6
7
8
x
f (x) =
*2
*4
and
To really tease out the difference between a function and its
description, let’s consider an example of a function with two
different descriptions.
Example 4. Explain why
˘
x2 = x.
g(x) = x * 1
the same function?
Let’s use a series of steps to think about this question.
First, what if we compare graphs? Here we see a graph
of f :
˘
Although x2 may appear to simplify to just x, let’s see
what happens when we plug in some values.
˘
˘
˘
˘
(*3)2 = 9
32 = 9
and
= 3.
= 3,
˘
Let f (x) = x2 . We see that for any negative
˘ x, x =
*x, and, therefore, f˘
(x) = f (*x) = (*x)2 =
˘
2
2
˘x = x. Hence x ë x. Rather
˘ we see that
2
x = x. The domain of f (x) = x2 is (*ÿ, ÿ)
and the range is [0, ÿ). For your viewing pleasure we’ve
included a graph of y = f (x):
3
10
*1
y
2
1
*2
*1
1
2
*1
*2
*3
On the other hand, here is a graph of g:
1
*2
3
y
2
*3
x2 * 3x + 2 (x * 2)(x * 1)
=
x*2
(x * 2)
1
2
3
x
3
4
x
For each input, exactly one output
3
y
2
1
*2
*1
1
2
3
4
x
*1
*2
*3
Second, what if we compare the domains? We cannot
evaluate f at x = 2. This is where f is undefined. On
the other hand, there is no value of x where we cannot
evaluate g. In other words, the domain of g is (*ÿ, ÿ).
Since these two functions do not have the same graph,
and they do not have the same domain, they must not be
the same function.
However, if we look at the two functions everywhere
except at x = 2, we can say that f (x) = g(x). In other
words,
f (x) = x * 1
when
x ë 2.
From this example we see that it is critical to consider
the domain and range of a function.
11
Compositions of functions
Dig-In:
1.3
Compositions of functions
Given two functions, we can compose them. Let’s give an
example in a “real context.”
Example 6. Let
g(m) = the amount of gas one can buy with m dollars,
and so
f (g(x)) = f (x + 7)
= (x + 7)2 + 5(x + 7) + 4.
Now let’s try an example with a more restricted domain.
Example 8. Suppose we have:
f (x) = x2
˘
g(x) = x
and let
f (g) = how far one can drive with g gallons of gas.
What does f (g(m)) represent in this setting?
With f (g(m)) we first relate how far one can drive with
g gallons of gas, and this in turn is determined by how
much money m one has. Hence f (g(m)) represents how
far one can drive with m dollars.
Composition of functions can be thought of as putting one
function inside another. We use the notation
(f ˝g)(x) = f (g(x)).
This means that
˘ the domain of f ˝g is 0 f x < ÿ. Next,
we substitute x for each instance of x found in
f (x) = x2
and so
˘
f (g(x)) = f ( x)
⇠˘ ⇡2
=
x .
{the range of g} is contained in or equal to {the domain of f }
f (x) = x2 + 5x + 4
g(x) = x + 7
Find f (g(x)) and state its domain.
for *ÿ < x < ÿ,
for *ÿ < x < ÿ.
The range of g is *ÿ < x < ÿ, which is equal to the
domain of f . This means the domain of f ˝g is *ÿ <
x < ÿ. Next, we substitute x + 7 for each instance of x
found in
f (x) = x2 + 5x + 4
12
for 0 f x < ÿ.
Find f (g(x)) and state its domain.
The domain of g is 0 f x < ÿ. From this we can see
that the range of g is 0 f x < ÿ. This is contained in
the domain of f .
Warning. The composition f ˝g only makes sense if
Example 7. Suppose we have
for *ÿ < x < ÿ,
We can summarize our results as a piecewise function,
which looks somewhat interesting:
h
nx
if 0 f x < ÿ
(f ˝g)(x) = l
nundefined otherwise.
j
Example 9. Suppose we have:
˘
f (x) = x
for 0 f x < ÿ,
g(x) = x2
for *ÿ < x < ÿ.
Compositions of functions
Find f (g(x)) and state its domain.
While the domain of g is *ÿ < x < ÿ, its range is only
0 f x < ÿ. This is exactly the domain of f . This means
that the domain of f ˝g is *ÿ < x < ÿ. Now we may
substitute x2 for each instance of x found in
˘
f (x) = x
and so
f (g(x)) = f (x2 )
˘
= x2 ,
= x.
Compare and contrast the previous two examples. We used the
same functions for each example, but composed them in different ways. The resulting compositions are not only different,
they have different domains!
13
Inverses of functions
So, we could rephrase these conditions as
Dig-In:
1.4
Inverses of functions
If a function maps every “input” to exactly one “output,” an
inverse of that function maps every “output” to exactly one “input.” We need a more formal definition to actually say anything
with rigor.
Definition. Let f be a function with domain A and range B:
f
A
These two simple equations are somewhat more subtle than
they initially appear.
Question 5. Let f be a function. If the point (1, 9) is on the
graph of f , what point must be the the graph of f *1 ?
Warning. This notation can be very confusing. Keep a watchful eye:
Multiple Choice:
g
A
We say that f and g are inverses of each other if f (g(b)) = b
for all b in B, and also g(f (a)) = a for all a in A. Sometimes
we write g = f *1 in this case.
f ˝f *1
B
(a) sin*1 (✓)
(b) sin(✓)*1
Question 7. Consider the graph of y = f (x) below
y
2
1.5
1
and
A
f *1 ˝f
0.5
A
0.5
14
f *1 (f (x)) = x.
Question 6. Which of the following is notation for the inverse
of the function sin(✓) on the interval [*⇡_2, ⇡_2]?
Let g be a function with domain B and range A:
B
and
f *1 (x) = the inverse function of f evaluated at x.
1
f (x)*1 =
.
f (x)
B
B
f (f *1 (x)) = x
1
1.5
2
x
Inverses of functions
Is f (x) invertible at x = 1?
that
So far, we’ve only dealt with abstract examples. Let’s see if we
can ground this in a real-life context.
since we solved for f *1 in our calculation. On the other
hand,
⇠⇠ ⇡
⇡
9
f *1 (f (t)) = f *1
t + 32
5
⇠⇠ ⇡
⇡
5
9
=
t + 32 * 32
9
5
=t
Example 10. The function
f (t) =
⇠ ⇡
9
t + 32
5
takes a temperature t in degrees Celsius, and converts it into
Fahrenheit. What does the inverse of this function tell you?
What is the inverse of this function?
If f converts Celsius measurements to Fahrenheit measurements of temperature, then f *1 converts Fahrenheit
measurements to Celsius measurements of temperature.
To find the inverse function, first note that
f (f *1 (t)) = t
by the definition of inverse functions.
Now write out the left-hand side of the equation
⇠ ⇡
9
f (f *1 (t)) =
f *1 (t) + 32
by the rule for f
5
and solve for f *1 (t).
⇠ ⇡
9
f *1 (t) + 32 = t
by the rule for f
5
⇠ ⇡
9
f *1 (t) = t * 32
5
⇠ ⇡
5
f *1 (t) =
(t * 32).
9
⇠ ⇡
5
So f *1 (t) =
(t * 32) is the inverse function of f ,
9
which converts a Fahrenheit measurement back into a
Celsius measurement.
Finally, we could check our work again using the definition of inverse functions. We have already guaranteed
f (f *1 (t)) = t,
which shows that f *1 (f (t)) = t.
We have examined several functions in order to determine their
inverse functions, but there is still more to this story. Not every
function has an inverse, so we must learn how to check for this
situation.
Question 8. Let f be a function, and imagine that the points
(2, 3) and (7, 3) are both on its graph. Could f have an inverse
function?
Look again at the last question. If two different inputs for a
function have the same output, there is no hope of that function
having an inverse function. Why? This is because the inverse
function must also be a function, and a function can only have
one output for each input. More specifically, we have the next
definition.
Definition. A function is called one-to-one if each output
value corresponds to exactly one input value.
Question 9. Which of the following are functions that are also
one-to-one?
Select All Correct Answers:
(a) Mapping words to their meaning in a dictionary.
(b) Given a runner racing forward on a straight path, mapping
time to the position of the runner.
15
Inverses of functions
(c) Mapping people to their birth date.
(d) Mapping mothers to their children.
Question 10. Which of the following functions are one to one?
Select all that apply.
Select All Correct Answers:
(a) f (x) = x
(b) f (x) = x2
(c) f (x) = x * 4x
3
(d) f (x) = x3 + 4
You may recall that a plot gives y as a function of x if every
vertical line crosses the plot at most once, and we called this the
vertical line test. Similarly, a function is one-to-one if every
horizontal line crosses the plot at most once, and we call this
the horizontal line test.
Theorem 2. A function is one-to-one on its domain if the horizontal line y = b intersects the curve y = f (x) in exactly
one point, for all values b in the range of f . This is called the
horizontal line test.
Below, we give a graph of f (x) = *5x2 + 30x + 60. While
this graph passes the vertical line test, and hence represents y
as a function of x, it does not pass the horizontal line test, so
the function is not one-to-one.
16
y
100
80
60
40
20
2
4
x
6
Although the line y = 20 intersects the curve y = f (x) exactly
once, the function f is not one-to-one on its domain, since the
line y = 80 intersects the curve twice.
As we have discussed, we can only find an inverse of a function
when it is one-to-one. If a function is not one-to-one, but we
still want an inverse, we must restrict the domain. Let’s see
what this means in our next examples.
Question 11. Consider the graph of the function f below:
f
y
f
A
B
C
D
E
On which of the following intervals is f one-to-one?
Select All Correct Answers:
x
Inverses of functions
(a) [A, B]
of the equation to find that
˘
x = f *1 (x).
(b) [A, C]
(c) [B, D]
(d) [C, E]
2
(e) [C, D]
This idea of restricting˘
the domain is critical for understanding
functions like f (x) = x.
˘
Warning. We define f (x) = x to be the positive square-root,
so that we can be sure that f is a function. Thinking of the
square-root as the inverse of the squaring function, we can
see the issue a little more clearly. There are two x-values that
square to 9.
x2 = 9
means x = ±3
Since we require that square-root is a function, we must have
only one output value when we plug in 9. We choose the positive
square-root, meaning that
˘
9 = 3.
y
f *1
1
f
*2
*1
1
2
x
*1
*2
Question 12. Consider the graph of y = f (x) below
y
Example 11. Consider the function
f (x) = x2 .
2
Does f have an inverse? If so, what is it? If not, attempt to
restrict the domain of f and find an inverse on the restricted
domain.
In this case f is not one-to-one. However, it is one-to-one
on the interval [0, ÿ). Hence we can find an inverse of
f (x) = x2 on this interval. We plug f *1 (x) into f and
write
f (f *1 (x)) = f *1 (x)
2
x = f *1 (x)
2
1.5
1
0.5
.
Since the domain of f is [0, ÿ), we know that x is positive. This means we can take the square-root of each side
0.5
1
1.5
2
x
Is the function f one-to-one on the interval [1, 2]?
17
Inverses of functions
Multiple Choice:
2
y
(a) yes
(b) no
1
*1
Question 13. Let f
to the domain [1, 2].
be the inverse function of f restricted
(a) Find the domain of f
*1
*2
.
*1
(b) Find the value of f f *1 (0.5) .
(c) Choose the correct answer regarding f
1
*1
*1
(0.5).
f
*2
Multiple Choice:
(i) f *1 (0.5) > 1.5
(ii) f *1 (0.5) f 1.5
(iii) f *1 (0.5) is not defined
Example 12. Consider the function
f (x) = x3 .
Does the function f have an inverse? If so, what is it? If not,
attempt to restrict the domain of f and find an inverse on the
restricted domain.
In this case f is one-to-one. We may write
f (f *1 (x)) = f *1 (x)
3
x = f *1 (x)
˘
3
x = f *1 (x).
3
For your viewing pleasure
we give a graph of y = f (x) =
˘
3
3
*1
x and y = f (x) = x. Note, the graph of f *1 is the
image of f after being flipped over the line y = x.
18
f *1
2
x
Review of famous functions
2
Review of famous functions
After completing this section, students should be able to do the
following.
• Know the graphs and properties of “famous” functions.
• Know and use the properties of exponential and logarithmic functions.
• Understand the relationship between exponential and logarithmic functions.
• Understand the definition of a rational function.
• Understand the properties of trigonometric functions.
• Evaluate expressions and solve equations involving
trigonometric functions and inverse trigonometric functions.
19
How crazy could it be?
Multiple Choice:
Break-Ground:
2.1
How crazy could it be?
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, do you remember when we first starting graphing
functions? Like with a “T-chart?”
Riley: I remember everything.
Devyn: I used to get so excited to plot stuff! I would wonder: “What
crazy curve would be drawn this time? What crazy picture will
I see?”
Riley: Then we learned about the slope-intercept form of a line.
Good-old
y = mx + b.
Devyn: Yeah, but lines are really boring. What about polynomials?
What could you tell me about
y = 5x6 * 5x5 * 5x4 + 5x3 + x2 * 1
just by looking at the equation?
Riley: Hmmmm. I’m not sure. . .
Problem 1. When x is a large number (furthest from zero),
which term of 5x6 * 5x5 * 5x4 + 5x3 + x2 * 1 is largest (furthest
from zero)?
Multiple Choice:
(a) *1
(b) x2
(c) 5x3
(d) *5x4
(e) *5x5
(f) 5x6
Problem 2. When x is a small number (near zero), which term
of 5x6 * 5x5 * 5x4 + 5x3 + x2 * 1 is largest (furthest from
zero)?
20
(a) *1
(b) x2
(c) 5x3
(d) *5x4
(e) *5x5
(f) 5x6
Problem 3. Very roughly speaking, what does the graph of
y = 5x6 * 5x5 * 5x4 + 5x3 + x2 * 1 look like?
Multiple Choice:
(a) The graph starts in the lower left and ends in the upper
right of the plane.
(b) The graph starts in the lower right and ends in the upper
left of the plane.
(c) The graph looks something like the letter “U.”
(d) The graph looks something like an upside down letter “U.”
Polynomial functions
Dig-In:
2.2
Polynomial functions
The functions you are most familiar with are probably polynomial functions.
What are polynomial functions?
Definition. A polynomial function in the variable x is a function which can be written in the form
f (x) = an xn + an*1 xn*1 + 5 + a1 x + a0
where the ai ’s are all constants (called the coefficients) and n is
a whole number (called the degree when n ë 0). The domain
of a polynomial function is (*ÿ, ÿ).
Question 14. Which of the following are polynomial functions?
Select All Correct Answers:
(a) f (x) = 0
What can the graphs look like?
Fun fact:
Theorem 3 (The Fundamental Theorem of Algebra). Every
polynomial of the form
an xn + an*1 xn*1 + 5 + a1 x + a0
where the ai ’s are real (or even complex!) numbers and an ë 0
has exactly n (possibly repeated) complex roots.
Remember, a root is where a polynomial is zero. The theorem
above is a deep fact of mathematics. The great mathematician
Gauss proved the theorem in 1799 for his doctoral thesis.
The upshot as far as we are concerned is that when we plot a
polynomial of degree n, its graph will cross the x-axis at most
n times.
Example 13. Here we see the the graphs of four polynomial
functions.
(b) f (x) = *9
2
(c) f (x) = 3x + 1
y
2
1
(d) f (x) = x1_2 * x + 8
*2
(e) f (x) = *4x*3 + 5x*1 + 7 * 18x2
is a polynomial in y, and
y2 * 4y + 1
2
sin (x) + sin(x) * 3
is a polynomial in sin(x).
2
x
*2
2
*1
1
2
x
*1
*2
y
2
1
*2
B
*1
*2
x2 * 3x + 2
(g) f (x) =
x*2
The phrase above “in the variable x” can actually change.
1
*1
(f) f (x) = (x + 1)(x * 1) + ex * ex
(h) f (x) = x7 * 32x6 * ⇡x3 + 45_84
1
A
*1
y
1
C
1
y
2
x
*2
*1
D
1
*1
*1
*2
*2
2
x
For each of the curves, determine if the polynomial has even
or odd degree, and if the leading coefficient (the one next to
the highest power of x) of the polynomial is positive or negative.
21
Polynomial functions
• Curve A is defined by an odd degree polynomial
with a positive leading term.
• Curve B is defined by an odd degree polynomial
with a negative leading term.
• Curve C is defined by an even degree polynomial
with a positive leading term.
• Curve D is defined by an even degree polynomial
with a negative leading term.
22
Rational functions
2.3
y
y
Dig-In:
Rational functions
What are rational functions?
Definition. A rational function in the variable x is a function
the form
p(x)
f (x) =
q(x)
where p and q are polynomial functions. The domain of a
rational function is all real numbers except for where the denominator is equal to zero.
Question 15. Which of the following are rational functions?
Select All Correct Answers:
(a) f (x) = 0
(b) f (x) =
3x + 1
x2 * 4x + 5
(c) f (x) = ex
(d) f (x) =
sin(x)
cos(x)
(e) f (x) = *4x*3 + 5x*1 + 7 * 18x2
(f) f (x) = x1_2 * x + 8
˘
x
(g) f (x) =
3
x *x
What can the graphs look like?
There is a somewhat wide variation in the graphs of rational
functions.
Example 14. Here we see the the graphs of four rational functions.
20
2
A
*20
B
20
x
*2
*20
2
4
x
*2
y
y
20
2
C
2
4
*20
x
*2
D
2
4
x
*2
Match the curves A, B, C, and D with the functions
x2 * 3x + 2
,
x*2
x*2
,
2
x * 3x + 2
x2 * 3x + 2
,
x+2
x+2
.
2
x * 3x + 2
x2 * 3x + 2
. This function is undefined only
x*2
at x = 2. Of the curves that we see above, D is undefined
exactly at x = 2.
Consider
x2 * 3x + 2
. This function is undefined
x+2
only at x = *2. The only function above that undefined
exactly at x = *2 is curve A.
Now consider
Now consider
at the roots of
x2
x*2
. This function is undefined
* 3x + 2
x2 * 3x + 2 = (x * 2)(x * 1).
Hence it is undefined at x = 2 and x = 1. It looks
like both curves B and C would work. Distinguishing
between these two curves is easy enough if we evaluate
23
Rational functions
at x = *2. Check it out.
4
5
x*2
*2 * 2
=
2
2
x * 3x + 2 x=*2 (*2) * 3(*2) + 2
*4
=
4+6+2
*4
=
.
12
Since this is negative, we see that
sponds to curve B.
x*2
correx2 * 3x + 2
Finally, it must be the case that curve C corresponds to
x+2
. We should note that if this function is evalx2 * 3x + 2
uated at x = *2, the output is zero, and this corroborates
our work above.
24
Trigonometric functions
Dig-In:
2.4
Trigonometric functions
and
What are trigonometric functions?
Definition. A trigonometric function is a function that relates a measure of an angle of a right triangle to a ratio of the
triangle’s sides.
The basic trigonometric functions are cosine and sine. They are
called “trigonometric” because they relate measures of angles
to measurements of triangles. Given a right triangle
sin(✓) =
k opposite
opposite
=
.
k hypotenuse hypotenuse
At this point we could simply assume that whenever we draw a
triangle for computing sine and cosine, that the hypotenuse will
be 1. We can do this because we are simply scaling the triangle,
and as we see above, this makes absolutely no difference when
computing sine and cosine. Hence, when the hypotenuse is 1,
we find that a convenient way to think about sine and cosine is
via the unit circle:
y
1
sin(✓)
✓
k adjacent
adjacent
=
k hypotenuse hypotenuse
opposite
use
n
ote
p
y
h
cos(✓) =
adjacent
we define
cos(✓) =
✓
adjacent
hypotenuse
and
sin(✓) =
opposite
.
hypotenuse
*1
Note, the values of sine and cosine do not depend on the scale
of the triangle. Being very explicit, if we scale a triangle by a
scale factor k,
✓
k adjacent
k opposite
se
u
ten
ypo
kh
cos(✓)
1
x
*1
If we consider all possible combinations of ratios of
adjacent,
opposite,
hypotenuse,
(allowing the adjacent and opposite to be negative, as on the
unit circle) we obtain all of the trigonometric functions.
25
Trigonometric functions
Definition. The trigonometric functions are:
adj
hyp
1
sec(✓) =
cos(✓)
sin(✓)
tan(✓) =
cos(✓)
cos(✓) =
opp
hyp
1
csc(✓) =
sin(✓)
cos(✓)
cot(✓) =
sin(✓)
sin(✓) =
where the domain of sine and cosine is all real numbers, and
the other trigonometric functions are defined precisely when
their denominators are nonzero.
Question 16. Which of the following expressions are equal to
sec(✓)?
Hence we want to be able to “undo” trigonometric functions.
However, since trigonometric functions are not one-to-one,
meaning there are are infinitely many angles with cos(✓) = .7,
it is impossible to find a true inverse function for cos(✓). Nevertheless, it is useful to have something like an inverse to these
functions, however imperfect. The usual approach is to pick
out some collection of angles that produce all possible values
exactly once. If we “discard” all other angles, the resulting
function has a proper inverse. In other words, we are restricting
the domain of the trigonometric function in order to find an
inverse. The function cos(✓) takes on all values between *1
and 1 exactly once on the interval [0, ⇡].
x
cos(✓)
Select All Correct Answers:
(a)
1
cos(✓)
(b)
1
sin(✓)
(c)
adj
hyp
(d)
hyp
adj
(e)
tan(✓)
sin(✓)
(f)
1
sin(✓) cot(✓)
*2⇡
*⇡
*⇡_2
⇡_2
⇡
3⇡_2
2⇡
✓
*1
If we restrict the domain of cos(✓) to this interval, then this
restricted function is one-to-one and hence has an inverse.
Question 17. What arc on the unit circle corresponds to the
restricted domain described above of cos(✓)?
Multiple Choice:
(a)
Connections to inverse functions
(b)
Trigonometric functions arise frequently in problems, and often we are interested in finding specific angle measures. For
instance, you may want to find some angle ✓ such that
(c)
cos(✓) = .7
26
*3⇡_2
1
Trigonometric functions
By examining both sine and cosine on restricted domains, we
can now produce functions arcsine and arccosine:
(d)
⇡
In a similar fashion, we need to restrict the domain of sine to be
able to take an inverse. The function sin(✓) takes on all values
between *1 and 1 exactly once on the interval [*⇡_2, ⇡_2].
⇡_2
x
1
*2⇡
*3⇡_2
*⇡
*⇡_2
✓
sin(✓)
⇡_2
⇡
3⇡_2
2⇡
✓
*1
*1
If we restrict the domain of sin(✓) to this interval, then this
restricted function is one-to-one and thus has an inverse.
1
x
Here we see a plot of arccos(x), the inverse
function of cos(✓) when the domain is restricted
to the interval [0, ⇡].
Question 18. What arc on the unit circle corresponds to the
restricted domain described above of sin(✓)?
✓
⇡_2
Multiple Choice:
(a)
(b)
(c)
*1
1
*⇡_2
Here we see a plot of arcsin(x), the inverse
function of sin(✓) when the domain is restricted
to the interval [*⇡_2, ⇡_2].
The functions
arccos(x)
(d)
x
and
arcsin(x)
are called “arc” because they give the angle that cosine or sine
used to produce their value. It is quite common to write
arccos(x) = cos*1 (x)
and
arcsin(x) = sin*1 (x).
27
Trigonometric functions
However, this notation is misleading as cos*1 (x) and sin*1 (x)
are not true inverse functions of cosine and sine. Recall that
a function and its inverse undo each other in either order, for
example,
˘
3
x3 = x
and
y
2⇡_3
⇡_3
2⇡_3
⇡_3
⇠˘ ⇡3
3
x = x.
Since arcsine is the inverse of sine restricted to the interval
[*⇡_2, ⇡_2], this does not work with sine and arcsine, for
example
arcsin(sin(⇡)) = 0.
⇡
0
⇡
x
though it is true that
sin(arcsin(x)) = x
and
cos(arccos(x)) = x.
4⇡_3
Question 19. Which of the following statements is true?
5⇡_3
Multiple Choice:
(a) sin*1 (x) is the inverse function of sin(x)
⇠
⇠ ⇡⇡
1
1
(b) sin sin*1
=
2
2
⇠ ⇠ ⇡⇡
5⇡
5⇡
(c) sin*1 sin
=
2
2
(d) sin*1 (x) =
1
sin(x)
Since the points 5⇡_3 and ⇡_3 have the same xcoordinate, cos(5⇡_3) = cos(⇡_3). Hence
arccos(cos(5⇡_3)) = ⇡_3.
Now that you have a feel for how arcsin(x) and arccos(x) behave, let’s examine tangent.
x
Example 15. Compute:
2
arccos(cos(5⇡_3))
tan(✓)
*2⇡
The issue here is that 5⇡_3 might not be in the range
of arccosine. To find our missing number, we’ll check
with a unit circle that we’ve decorated with the domain
of arccosine:
28
*3⇡_2
*⇡
*⇡_2
⇡_2
⇡
3⇡_2
2⇡
✓
*2
Question 20. What arc on the unit circle corresponds to the
restricted domain described above of tan(✓)?
Trigonometric functions
The issue here is that 5⇡_3 might not be in the range
of arctangent. To find our missing number, we’ll check
with a unit circle that we’ve decorated with the domain
of arctangent:
Multiple Choice:
(a)
y
⇡_3
2⇡_3
(b)
(c)
⇡_3
0
⇡
x
(d)
4⇡_3
Again, only working on a restricted domain of tangent, we can
produce an inverse function, arctangent. Here we see a plot
of arctan(x), the inverse function of tan(✓) when its domain is
restricted to the interval (*⇡_2, ⇡_2).
*⇡_3
Since the points 5⇡_3 and *⇡_3 have the same x and
y-coordinates,
✓
arctan(tan(5⇡_3)) = *⇡_3.
⇡_2
x
*⇡_2
Example 16. Compute:
arctan(tan(5⇡_3))
5⇡_3
Now we give some facts of other trigonometric and “inverse”
trigonometric functions.
Definition.
• arccos(x) = ✓ means that cos(✓) = x and 0 f ✓ f ⇡. The
domain of arccos(x) is *1 f x f 1.
⇡
⇡
• arcsin(x) = ✓ means that sin(✓) = x and * f ✓ f .
2
2
The domain of arcsin(x) is *1 f x f 1.
29
Trigonometric functions
⇡
⇡
• arctan(x) = ✓ means that tan(✓) = x and * < ✓ < .
2
2
The domain of arctan(x) is *ÿ < x < ÿ.
From the unit circle we can see
• arccot(x) = ✓ means that cot(✓) = x and 0 < ✓ < ⇡. The
domain of arccot(x) is *ÿ < x < ÿ.
sin(t)
• arcsec(x) = ✓ means that sec(✓) = x and 0 f ✓ f ⇡ with
✓ ë ⇡_2. The domain of arcsec(x) is all x with absolute
value greater than 1.
⇡
⇡
• arccsc(x) = ✓ means that csc(✓) = x and * f ✓ f
2
2
with ✓ ë 0. The domain of arccsc(x) is all x with absolute
value greater than 1.
t
cos(t)
The power of the Pythagorean Theorem
The Pythagorean Theorem is probably the most famous theorem
in all of mathematics.
Theorem 4 (Pythagorean Theorem). Given a right triangle:
via the Pythagorean Theorem that
c
a
✓
If we divide this expression by cos2 (t) we obtain
1 + tan2 (t) = sec2 (t)
b
We have that:
cos2 (t) + sin2 (t) = 1.
and if we divide cos2 (t)+sin2 (t) = 1 by sin2 (t) we obtain
a2 + b2 = c 2
cot 2 (t) + 1 = csc2 (t).
The Pythagorean Theorem gives several key trigonometric identities.
We can simplify expressions using the Pythagorean Theorem
Theorem 5 (Pythagorean Identities). The following hold:
cos2 ✓+sin2 ✓ = 1
30
1+tan2 ✓ = sec2 ✓
cot 2 ✓+1 = csc2 ✓
Example 17. Suppose that arctan(3_5) = ✓.
sin(✓).
Compute
Trigonometric functions
If arctan(3_5) = ✓, then
tan(✓). If cos(✓) = x, the triangle in question must be
similar to this triangle:
tan(arctan(3_5)) = tan(✓)
3_5 = tan(✓).
Now we will use the Pythagorean Theorem to deduce
sin(✓). If tan(✓) = 3_5, the triangle in question must be
similar to this triangle:
˘
1
˘
1 * x2
✓
x
32 + 52
3
✓
From this triangle and our work above, we see that
˘
1 * x2
tan(arccos(x)) = tan(✓) =
.
x
5
From this triangle and our work above, we see that
˘
sin(✓) = 3_ 32 + 52 .
We’ll also use the Pythagorean Theorem to help us simplify
abstract expressions into ones we can compute with ease.
Example 18. Simplify
tan(arccos(x))
Start by saying
✓ = arccos(x)
This means tan(arccos(x)) = tan(✓). Apply cosine to
both sides of the equation above,
cos(✓) = cos(arccos(x))
cos(✓) = x.
Now we will use the Pythagorean Theorem to deduce
31
Exponential and logarithmic functions
Dig-In:
What can the graphs look like?
2.5 Exponential and logarithmic
functions
Graphs of exponential functions
Exponential and logarithmic functions may seem somewhat
esoteric at first, but they model many phenomena in the realworld.
Example 19. Here we see the the graphs of four exponential
functions.
4
What are exponential and logarithmic functions?
y
3
C
B
Definition. An exponential function is a function of the form
2
A
f (x) = bx
where b ë 1 is a positive real number. The domain of an
exponential function is (*ÿ, ÿ).
D
1
Question 21. Is b*x an exponential function?
Definition. A logarithmic function is a function defined as
follows
logb (x) = y
means that
by = x
where b ë 1 is a positive real number. The domain of a logarithmic function is (0, ÿ).
In either definition above b is called the base.
Connections between exponential functions and logarithms Let b be a positive real number with b ë 1.
• blogb (x) = x for all positive x
• logb (bx ) = x for all real x
Question 22. What exponent makes the following expression
true?
0
1
x ?
3x = e
.
32
*2
*1
1
2
x
Match the curves A, B, C, and D with the functions
ex ,
⇠ ⇡x
1
,
2
⇠ ⇡x
1
,
3
2x .
One way to solve these problems is to compare these
functions along the vertical line x = 1,
Exponential and logarithmic functions
4
y
2
3
2
A
A
1
C
B
y
B
D
1
C
1
2
3
4
x
*1
*2
*1
Note
Hence we see:
1
2
D
x
*2
⇠ ⇡1 ⇠ ⇡1
1
1
<
< 21 < e1 .
3
2
⇠ ⇡x
1
corresponds to B.
3
⇠ ⇡x
1
•
corresponds to A.
2
•
• 2x corresponds to D.
• ex corresponds to C.
Match the curves A, B, C, and D with the functions
ln(x),
log1_2 (x),
log1_3 (x),
log2 (x).
First remember what logb (x) = y means:
logb (x) = y
means that
by = x.
Moreover, ln(x) = loge (x) where e = 2.71828 … . So
now examine each of these functions along the horizontal
line y = 1
Graphs of logarithmic functions
Example 20. Here we see the the graphs of four logarithmic
functions.
33
Exponential and logarithmic functions
2
Properties of exponents Let b be a positive real number
with b ë 1.
y
• bm bn = bm+n
A
1
• b*1 =
B
1
C
2
3
4
x
1
b
• (bm )n = bmn
Question 23. What exponent makes the following true?
24 23 = 2 ?
*1
D
*2
Note again (this is from the definition of a logarithm)
Hence we see:
⇠ ⇡1 ⇠ ⇡1
1
1
<
< 21 < e1 .
3
2
• log1_3 (x) corresponds to B.
• log1_2 (x) corresponds to A.
• log2 (x) corresponds to D.
• ln(x) corresponds to C.
Properties of logarithms Let b be a positive real number
with b ë 1.
• logb (m n) = logb (m) + logb (n)
• logb (mn ) = n logb (m)
⇠ ⇡
1
• logb
= logb (m*1 ) = * logb (m)
m
• loga (m) =
logb (m)
logb (a)
Question 24. What value makes the following expression true?
⇠ ⇡
8
log2
=3* ?
16
Question 25. What makes the following expression true?
Properties of exponential functions and logarithms
Working with exponential and logarithmic functions is often
simplified by applying properties of these functions. These
properties will make appearances throughout our work.
34
log3 (x) =
ln(x)
?
What is a limit?
3
What is a limit?
After completing this section, students should be able to do the
following.
• Consider values of a function at inputs approaching a
given point.
• Understand the concept of a limit.
• Limits as understanding local behavior of functions.
• Calculate limits from a graph (or state that the limit does
not exist).
• Understand possible issues when estimating limits using
nearby values.
• Define a one-sided limit.
• Explain the relationship between one-sided and two-sided
limits.
• Distinguish between limit values and function values.
• Identify when a limit does not exist.
• Define continuity in terms of limits.
• Famous functions are continuous on their domains.
35
Stars and functions
Problem 2. When you evaluate
Break-Ground:
3.1
Stars and functions
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, did you know I like looking at the stars at night?
Riley: Stars are freaking awesome balls of nuclear fire whose light
took thousands of years to reach us.
Devyn: I know! But did you know that the best way to see a very
dim star is to look near it but not exactly at it? It’s because
then you can use the “rods” in your eye, which work better in
low light than the “cones” in your eyes.
Riley: That’s amazing! Hey, that reminds me of when we were talking
about the two functions
f (x) =
x2 * 3x + 2
x*2
and
g(x) = x * 1,
which we now know are completely different functions.
Devyn: Whoa. How are you seeing a connection here?
Riley: If we want to understand what is happening with the function
f (x) =
x2 * 3x + 2
,
x*2
at x = 2, we can’t do it by setting x = 2. Instead we need to
look near x = 2 but not exactly at x = 2.
Devyn: Ah ha! Because if we are not exactly at x = 2, then
x2 * 3x + 2
= x * 1.
x*2
Problem 1. Let f (x) =
of the following is true?
x2 * 3x + 2
and g(x) = x * 1. Which
x*2
Multiple Choice:
(a) f (x) = g(x) for every value of x.
(b) There is no x-value where f (x) = g(x).
(c) f (x) = g(x) when x ë 2.
36
f (x) =
x2 * 3x + 2
,
x*2
at x-values approaching (but not equal to) 2, what happens to
the value of f (x)?
Problem 3. Just from checking some values, can you be absolutely certain that your answer to the previous problem is
correct?
What is a limit?
written
Dig-In:
3.2
What is a limit?
if the value of f (x) is as close as one wishes to L for all x
sufficiently close, but not equal to, a.
The basic idea
sin(x)
Question 27. Use the graph of f (x) =
above to finish
x
the following statement: “A good guess is that. . . ”
Consider the function
f (x) =
sin(x)
.
x
Multiple Choice:
While f (x) is undefined at x = 0, we can still plot f (x) at other
values near x = 0.
y
1
*2⇡
*3⇡_2
*⇡
*⇡_2
lim f (x) = L,
xôa
(a) lim
sin(x)
= 1.
x
(b) lim
sin(x)
= 0.
x
xô0
xô1
⇡_2
⇡
3⇡_2
2⇡
x
*1
Question 26. Use the graph of f (x) =
the following question: What is f (0)?
(c) lim f (x) =
sin(1)
.
1
(d) lim f (x) =
sin(0)
= ÿ.
0
xô1
sin(x)
above to answer
x
xô0
Question 28. Consider the following graph of y = f (x)
Multiple Choice:
3
(a) 0
y
2
(b) f (0)
(c) 1
1
(d) f (0) is undefined
(e) it is impossible to say
Nevertheless, we can see that as x approaches zero, f (x) approaches one. From this setting we come to our definition of a
limit.
Definition. Intuitively,
the limit of f (x) as x approaches a is L,
*4
*3
*2
*1
1
2
3
4
x
*1
*2
*3
Use the graph to evaluate the following. Write DNE if the value
does not exist.
37
What is a limit?
The function ‚x„ is the function that returns the greatest
integer less than or equal to x. Recall that
(a) f (*2)
(b) lim f (x)
xô*2
lim ‚x„ = L
(c) f (*1)
xô2
if ‚x„ can be made arbitrarily close to L by making x
sufficiently close, but not equal to, 2. So let’s examine x
near, but not equal to, 2. Now the question is: What is
L?
(d) lim f (x)
xô*1
(e) f (0)
(f) lim f (x)
If this limit exists, then we should be able to look sufficiently close, but not at, x = 2, and see that f is approaching some number. Let’s look at a graph:
xô0
(g) f (1)
(h) lim f (x)
xô1
y
Limits might not exist
3
Limits might not exist. Let’s see how this happens.
Example 21. Consider the graph of f (x) = ‚x„.
3
2
y
1
2
f (x)
1
*2
*1
1.5
1
*1
*2
Explain why the limit
does not exist.
38
lim f (x)
xô2
2
3
4
x
x 2
2.5
3
x
If we allow x values on the left of 2 to get closer and
closer to 2, we see that f (x) = 1. However, if we allow
the values of x on the right of 2 to get closer and closer
to 2 we see
What is a limit?
y
to a(= 0). But this does does not satisfy the definition of the
limit, at least, not yet.
3
But, wait! Fill in another table.
x
2
0.3
0.03
0.003
0.0003
f (x)
1
2 x
1.5
2.5
3
x
So just to the right of 2, f (x) = 2. We cannot find a
single number that f (x) approaches as x approaches 2,
and so the limit does not exists.
Tables can be used to help guess limits, but one must be careful.
⇠ ⇡
⇡
Question 29. Consider f (x) = sin
. Fill in the tables
x
below rounding to three decimal places:
x
f (x)
f (x)
0.1
0.01
0.001
0.0001
We may rush and say that, based on the table above,
⇠ ⇡
⇡
lim sin
= 0.
xô0
x
But, recall the definition of the limit: L is the limit if the value
of f (x) is as close as one wishes to L for all x sufficiently close,
but not equal to, a.
From this table we can see that f (x)(= 0) is as close as one
wishes to L(= 0) for some values x that are sufficiently close
What do these two tables tell us about
⇠ ⇡
⇡
lim sin
?
xô0
x
Multiple Choice:
⇠ ⇡
⇡
=0
xô0
x
⇠ ⇡
⇡
(b) lim sin
=1
xô0
x
⇠ ⇡
⇡
(c) lim sin
= *.866
xô0
x
⇠ ⇡
⇡
(d) lim sin
= *.433
xô0
x
(a) lim sin
(e) The limit does not exist.
One-sided limits
While we have seen that lim ‚x„ does not exist, its graph looks
xô2
much “nicer" near a = 2 than does the previous graph near
a = 0. More can be said about the function ‚x„ and its behavior
near a = 2.
Definition. Intuitively,
for the function f , L is the limit from the right as x approaches
a,
39
What is a limit?
written
lim f (x) = L,
xôa+
if the value of f (x) is as close as one wishes to L for all x > a
sufficiently close to a.
Similarly,
for the function f , L is the limit from the left as x approaches
a,
written
From the graph we can see that as x approaches 2 from
the left, ‚x„ remains at y = 1 up until the exact point that
x = 2. Hence
lim* f (x) = 1.
xô2
Also from the graph we can see that as x approaches 2
from the right, ‚x„ remains at y = 2 up to x = 2. Hence
lim f (x) = 2.
xô2+
lim f (x) = L,
xôa*
if the value of f (x) is as close as one wishes to L for all x < a
sufficiently close to a.
When you put this all together
Example 22. Compute:
One-sided limits help us talk about limits.
lim f (x)
xô2*
and
lim f (x)
Theorem 6. A limit
xô2+
by using the graph below
lim f (x)
xôa
3
y
exists if and only if
2
• lim* f (x) exists
xôa
y = f (x) = ‚x„
1
*2
*1
1
*1
*2
2
3
4
x
• lim+ f (x) exists
xôa
• lim* f (x) = lim+ f (x)
xôa
xôa
In this case, lim f (x) is equal to the common value of the two
xôa
one sided limits.
Question 30. Evaluate the expressions by referencing the
graph below. Write DNE if the limit does not exist.
40
What is a limit?
10
(i) lim f (x)
y
xô1
(j) f (0)
(k)
8
(l)
6
lim f (x)
xô*3*
lim f (x)
xô*2+
4
2
-4
-2
2
4
6
x
-2
(a) lim f (x)
xô4
(b) lim f (x)
xô*3
(c) lim f (x)
xô0
(d) lim* f (x)
xô0
(e) lim+ f (x)
xô0
(f) f (*2)
(g) lim* f (x)
xô2
(h)
lim f (x)
xô*2*
41
Continuity
It is very important to note that saying
Dig-In:
3.3
Continuity
Limits are simple to compute when they can be found by plugging the value into the function. That is, when
lim f (x) = f (a).
“a function f is continuous at a point a”
is really making three statements:
(a) f (a) is defined. That is, a is in the domain of f .
xôa
We call this property continuity.
(b) lim f (x) exists.
xôa
Definition. A function f is continuous at a point a if
(c) lim f (x) = f (a).
xôa
lim f (x) = f (a).
The first two of these statements are implied by the third statement.
xôa
Question 31. Consider the graph of the function f
Example 23. Find the discontinuities (the points x where a
function is not continuous) for the function described below:
y
2
5
1.5
y
4
1
3
0.5
2
0.5
1
1.5
2
x
1
Which of the following are true?
Multiple Choice:
(a) f is continuous at x = 0.5
(b) f is continuous at x = 1
(c) f is continuous at x = 1.5
42
2
4
6
8
x
10
To start, f is not even defined at x = 2, hence f cannot
be continuous at x = 2.
Next, from the plot above we see that lim f (x) does not
xô4
Continuity
exist because
Question 32. Compute: lim x3
lim f (x) = 1
xô4*
and
xô⇡
lim f (x) ˘ 3.3
xô4+
Since lim f (x) does not exist, f cannot be continuous at
xô4
x = 4.
Question 33. Compute: lim
xô⇡
Question 34. Compute: lim cos x
xô⇡
We also see that lim f (x) ˘ 2.9 while f (6) = 2. Hence
lim f (x) ë f (6), and so f is not continuous at x = 6.
xô6
xô6
Building from the definition of continuity at a point, we can
now define what it means for a function to be continuous on an
open interval.
Definition. A function f is continuous on an open interval
I if lim f (x) = f (a) for all a in I.
xôa
˘
2
Left and right continuity
At
˘ this point we have a small problem. For functions such as
x, the natural domain is 0 f x < ÿ. This is not an open
˘
interval. What does it mean to say that x is continuous at
˘
0 when x is not defined for x < 0? To get us out of this
quagmire, we need a new definition:
Definition. A function f is left continuous at a point a if
lim* f (x) = f (a).
Loosely speaking, a function is continuous on an interval I if
you can draw the function on that interval without any breaks
in the graph. This is often referred to as being able to draw the
graph “without picking up your pencil.”
A function f is right continuous at a point a if lim+ f (x) =
Theorem 7 (Continuity of Famous Functions). The following
functions are continuous on their natural domains, for k a real
number and b a positive real number:
This allows us to talk about continuity on closed and half-closed
intervals.
Constant function f (x) = k
Identity function f (x) = x
Power function f (x) = xb
Exponential function f (x) = bx
Logarithmic function f (x) = logb (x)
Sine and cosine Both sin(x) and cos(x)
In essence, we are saying that the functions listed above are
continuous wherever they are defined.
xôa
f (a).
xôa
Definition. A function f is
• continuous on a closed interval [a, b] if f is continuous
on (a, b), right continuous at a, and left continuous at b;
• continuous on a half-closed interval [a, b) if f is continuous on (a, b) and right continuous at a;
• continuous on a half-closed interval (a, b] if f is f is
continuous on (a, b) and left continuous at b.
Question 35. Here we give the graph of a function defined on
[0, 10].
43
Continuity
5
y
4
3
2
1
2
4
6
8
x
10
Select all intervals for which the following statement is true.
The function f is continuous on the interval I.
Select All Correct Answers:
(a) I = [0, 10]
(b) I = [0, 4]
(c) I = [4, 6]
(d) I = [4, 6)
(e) I = (4, 6]
(f) I = (4, 6)
(g) I = (6, 10]
(h) I = [6, 10)
44
Limit laws
4
Limit laws
After completing this section, students should be able to do the
following.
• Calculate limits using the limit laws.
• Calculate limits by replacing a function with a continuous
function.
• Understand the Squeeze Theorem and how it can be used
to find limit values.
• Calculate limits using the Squeeze Theorem.
45
Equal or not?
Problem 2. Using the context above,
Break-Ground:
4.1
Equal or not?
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, I’ve been thinking about limits.
Riley: So awesome!
Devyn: Think about
lim (f (x) + g(x)) .
large
=?
small
Multiple Choice:
(a) “large”
xôa
This is the number that f (x) + g(x) gets nearer and nearer to, as
x gets nearer and nearer to a.
Riley: You know it!
Devyn: So I think it is the same as
(b) “small”
(c) impossible to say
Problem 3. Using the context above,
small
=?
small
lim f (x) + lim g(x).
xôa
xôa
Riley: Yeah, that does make sense, since when you add two numbers,
say
(a number near 6) + (a number near 7)
you get
(a number near 13)
Riley: Right! And I think the same reasoning will work for multiplication! So we should be able to say
⇠
⇡ ⇠
⇡
lim (f (x) g(x)) = lim f (x)
lim g(x) .
xôa
xôa
xôa
Devyn: Yes, I think that’s right! But what about division? Can we
use similar reasoning to conclude
lim
xôa
lim f (x)
f (x)
xôa
=
.
g(x)
lim g(x)
xôa
Problem 1. Give an argument (similar to the one above) supporting the idea that
⇠
⇡ ⇠
⇡
lim (f (x) g(x)) = lim f (x)
lim g(x) .
xôa
xôa
xôa
For the next problems, suppose L is a number near 1 and that
M is a number near 0.
46
Multiple Choice:
(a) “large”
(b) “small”
(c) impossible to say
The limit laws
lim f (x)
f (x) xôa
L
=
=
, if M ë 0.
xôa g(x)
lim g(x)
M
Dig-In:
4.2
Quotient Law lim
The limit laws
In the previous section were able to compute the limits
3
3
lim x = ⇡ ,
xô⇡
˘
˘
lim 2 = 2,
lim cos x = *1,
xô⇡
xô⇡
˘
using continuity of the functions x3 , 2, and cos x, at x = ⇡.
Does this imply that we can compute the limits
In plain language, “limit of a sum equals the sum of the limits,”
“limit of a product equals the product of the limits,” etc.
Let’s examine how the Limit Laws can be used in computation
of limits.
Example 24. Compute the following limits using Limit Laws:
(a) lim (x3 * cos x)
lim (x3 * cos x),
˘
2
lim
,
xô⇡ cos x
˘
lim ( 2 x3 cos x), or
xô⇡
xô⇡
˘
2
(b) lim
xô⇡ cos x
˘
(c) lim ( 2 x3 cos x)
xô⇡
lim cos(x3 )?
xô⇡
xô⇡
Well, we cannot use continuity
˘ here, because we don’t know
2 ˘
if the functions x3 * cos x,
, 2 x3 cos x, and cos(x3 )
cos x
are continuous at x = ⇡, and
we have no other tools available, since the graphs and tables
are not reliable. Obviously, we need more tools to help us with
computation of limits.
In this section, we present a handful of rules, called the Limit
Laws, that allow us to find limits of various combinations of
functions.
Theorem 8 (Limit Laws). Suppose that lim f (x) = L,
xôa
lim g(x) = M.
xôa
Sum/Difference Law lim (f (x) ± g(x))
xôa
=
lim g(x) = L ± M.
lim f (x) ±
xôa
xôa
Product Law lim (f (x)g(x)) = lim f (x) lim g(x) = LM.
xôa
xôa
xôa
(d) lim cos(x3 )
xô⇡
For lim (x3 * cos x), write:
xô⇡
lim (x3 * cos x) = lim x3 * lim cos x by Difference Law
xô⇡
xô⇡
3
xô⇡
= ⇡ * (*1)
= ⇡3 + 1
For lim
xô⇡
˘
2
, write:
cos x
˘
˘
lim 2
2
xô⇡
lim
=
xô⇡ cos x
lim cos x
xô⇡
˘
2
=
*1
˘
=* 2
by Quotient Law
xôa
47
The limit laws
˘
For lim ( 2 x3 cos x), using the Product Law, we can
xô⇡
write:
˘
˘
lim ( 2 x3 cos x) = lim 2 lim x3 lim cos x
xô⇡
xô⇡
xô⇡
at x = 1?
We have to check whether
lim f (x) = f (1).
xô1
xô⇡
Since we already know that lim (5x2 + 3x * 2) = 6,
˘
= 2 ⇡ 3 (*1)
˘
= * 2⇡ 3
xô1
we only have to compute f (1).
f (1) = 5(1)2 + 3(1) * 2
For lim cos(x3 ), write:
=6
xô⇡
Therefore, the polynomial function f is continuous
at x = 1.
lim cos(x3 ) = 5
xô⇡
The function cos(x3 ) is a combination of the functions
cos(x) and x3 , but it is neither a sum/difference, nor a
product, nor a quotient of these two functions, so we
cannot apply any of the Limit Laws. This function is a
composition of the two functions, cos(x) and x3 .
Example 25. (a) Compute the following limit using limit
laws:
lim (5x2 + 3x * 2)
xô1
But what about continuity at any other value x? Is
the function f continuous on its entire domain?
And what about any other polynomial function?
Are polynomials continuous on their domains?
We can generalize the example above to get the following theorems.
Theorem 9 (Continuity of Polynomial Functions). All polynomial functions, meaning functions of the form
Well, get out your pencil and write with me:
lim (5x2 + 3x * 2) = lim 5x2 + lim 3x * lim 2
xô1
xô1
xô1
xô1
by the Sum/Difference Law. So now
= 5 lim x2 + 3 lim x * lim 2
xô1
xô1
xô1
by the Product Law. Finally by continuity of xb and
k,
= 5(1)2 + 3(1) * 2 = 6.
(b) Is the polynomial function f (x) = 5x2 +3x*2 continuous
48
f (x) = an xn + an*1 xn*1 + 5 + a1 x + a0
where n is a positive integer and each coefficient ai , i =
0, 1, ..., n, is a real number, are continuous for all real numbers.
In order to show that any polynomial, f , is continuous at
any real number, a, we have to show that
lim f (x) = f (a).
xôa
Write with me:
lim f (x) = lim (an xn + an*1 xn*1 + 5 + a1 x + a0 )
xôa
xôa
The limit laws
Now by the Sum Law,
x=a
= lim an x + lim an*1 x
n
xôa
n*1
xôa
+ 5 + lim a1 x + lim a0
xôa
xôa
Since we have shown that lim h(x) = h(a), we have
xôa
shown that h is continuous at x = a.
and by the Product Law,
= lim an lim xn + lim an*1 lim xn*1 + …
xôa
xôa
xôa
xôa
+ lim a1 lim x + lim a0
xôa
xôa
and by Continuity
= an an + an*1 an*1 + 5 + a1 a + a0
And this equals. . .
= f (a)
Since we have shown that lim f (x) = f (a), we have
xôa
shown that f is continuous at x = a.
Theorem 10 (Continuity of Rational Functions). A rational
function h, meaning a function of the form
h(x) =
f (x)
g(x)
where f and g are polynomials, is continuous for all real numbers except where g(x) = 0. That is, rational functions are
continuous wherever they are defined.
Let a be a real number such that g(a) ë 0. Then, since
g(x) is continuous at a, lim g(x) ë 0. Therefore, write
xôa
with me,
f (x)
lim h(x) = lim
xôa
xôa g(x)
and now by the Quotient Law,
lim f (x)
xôa
lim g(x)
xôa
and by the continuity of polynomials we may now set
f (a)
= h(a).
g(a)
xôa
Question 36. Where is f (x) =
x2 * 3x + 2
continuous?
x*2
Question 37. True or false: If f and g are continuous functions
on an interval I, then f ± g is continuous on I.
Question 38. True or false: If f and g are continuous functions
on an interval I, then f _g is continuous on I.
We still don’t know how to compute a limit of a composition of
two functions. Our next theorem provides basic rules for how
limits interact with composition of functions.
Theorem 11 (Composition Limit Law). If f (x) is continuous
at b = lim g(x), then
xôa
lim f (g(x)) = f (lim g(x)).
xôa
xôa
Because the limit of a continuous function is the same as the
function value, we can now pass limits inside continuous functions.
Corollary 1 (Continuity of Composite Functions). If g is continuous at x = a, and if f is continuous at g(a), then f (g(x))
is continuous at x = a.
Using the Composition Limit Law, we can compute the last
example from the beginning of this section.
Example 26. Compute the following limit using limit laws:
lim cos(x3 )
xô⇡
49
The limit laws
We will use the Composition Limit Law. Let
Question 42. Can this limit be directly computed by limit laws?
f (g(x)) = cos(x3 ),
where f (x) = cos x, and g(x) = x3 . Now, continuity
of x3 implies that lim g(x) = ⇡ 3 . Continuity of cos x
xô⇡
implies that f is continuous at ⇡ 3 = lim g(x). Now, the
xô⇡
Composition Limit Law implies that
⇠
⇡
lim cos(x3 ) = cos lim x3 = cos(⇡ 3 ) ˘ 0.91726.
xô⇡
xô⇡
Many of the Limit Laws and theorems about continuity in this
section might seem like they should be obvious. You may be
wondering why we spent an entire section on these theorems.
The answer is that these theorems will tell you exactly when
it is easy to find the value of a limit, and exactly what to do in
those cases.
The most important thing to learn from this section is whether
the limit laws can be applied for a certain problem, and when
we need to do something more interesting. We will begin
discussing those more interesting cases in the next section.
lim cot(x3 )
xô0
Question 43. Can this limit be directly computed by limit laws?
lim sec2 (ex * 1)
xô0
Question 44. Can this limit be directly computed by limit laws?
lim (x * 1) csc(ln(x))
xô1
Question 45. Can this limit be directly computed by limit laws?
lim x ln x
xô0
Question 46. Can this limit be directly computed by limit laws?
2x * 1
xô0 3x*1
lim
Question 47. Can this limit be directly computed by limit laws?
lim (1 + x)1_x
xô0
A list of questions
Let’s try this out.
Question 39. Can this limit be directly computed by limit laws?
x2 + 3x + 2
xô2
x+2
lim
Question 40. Can this limit be directly computed by limit laws?
x2 * 3x + 2
xô2
x*2
lim
Question 41. Can this limit be directly computed by limit laws?
lim x sin(1_x)
xô0
50
The Squeeze Theorem
We must show that lim f (x) = 0. First, let’s assume that
x g 0 and small. Since
Dig-In:
4.3
xô0
The Squeeze Theorem
In mathematics, sometimes we can study complex functions
by relating them for simpler functions. The Squeeze Theorem
tells us one situation where this is possible.
Theorem 12 (Squeeze Theorem). Suppose that
*1 f sin
by multiplying these inequalities by
˘
˘
* 5 x f f (x) f 5 x.
lim g(x) = L = lim h(x),
xôa
then lim f (x) = L.
xôa
Question 48. I’m thinking of a function f . I know that for all
x
0 f f (x) f x2 .
What is lim f (x)?
Now, let’s assume that x f 0 and small. Since
⇠ ⇡
1
*1 f sin
f1
x
˘
by multiplying these inequalities by 5 x f 0 , we obtain
⇠ ⇡
˘
˘
˘
1
5
x f 5 x sin
f*5 x
x
xô0
Example 27. Consider the function
which can be written as
˘
˘
5
x f f (x) f * 5 x.
if x ë 0,
if x = 0,
Therefore for all small values of x
Û˘ Û
Û˘ Û
*Û 5 xÛ f f (x) f Û 5 xÛ.
Û
Û
Û
Û
y
*0.2
*0.15
*0.1
*2
*5 10
5 10
x g 0 , we obtain
which can be written as
for all x close to a but not necessarily equal to a. If
⇠ ⇡
h˘
1
5
n x sin
x
f (x) = l
n0
j
˘
5
⇠ ⇡ ˘
˘
˘
1
* 5 x f 5 x sin
f 5x
x
g(x) f f (x) f h(x)
xôa
⇠ ⇡
1
f1
x
*2
0.1
0.15
x
0.2
Since
⇠ ˘ ⇡
Û
Û
Û˘ Û
lim *Û 5 xÛ = 0 = lim Û 5 xÛ
Û
Û
Û
xô0
xô0 Û
we apply the Squeeze Theorem and obtain that
Is this function continuous at x = 0?
51
The Squeeze Theorem
lim f (x) = 0. Hence f (x) is continuous.
and computing these areas we find
xô0
Here we see how the informal definition of continuity
being that you can “draw it” without “lifting your pencil”
differs from the formal definition.
y
y = x
1
5
cos(✓) sin(✓) ✓
tan(✓)
f f
.
2
2
2
Multiplying through by 2, and recalling that tan(✓) =
sin(✓)
we obtain
cos(✓)
cos(✓) sin(✓) f ✓ f
*0.2
*0.15
*0.1
*5 10*2
0.1
5 10*2
0.15
x
0.2
Example 28. Compute:
To compute this limit, use the Squeeze
⇠ Theorem.
⇡ First
*⇡ ⇡
note that we only need to examine ✓ À
,
and for
2 2
the present time, we’ll assume that ✓ is positive. Consider
the diagrams below:
y
y
tan(✓)
sin(✓)
✓
cos(✓)
Triangle A
x
✓
1
Sector
x
From our diagrams above we see that
✓
1
Triangle B
x
Area of Triangle A f Area of Sector f Area of Triangle B
52
sin(✓)
1
f
.
✓
cos(✓)
sin(*✓) sin(✓)
=
, so these
✓
⇠ *✓ ⇡
*⇡ ⇡
inequalities hold for all ✓ À
,
. Additionally, we
2 2
know
1
lim cos(✓) = 1 = lim
,
✓ô0
✓ô0 cos(✓)
Note, cos(*✓) = cos(✓) and
sin(✓)
lim
✓ô0
✓
y
Dividing through by sin(✓) and taking the reciprocals
(reversing the inequalities), we find
cos(✓) f
1
y = *x 5
sin(✓)
.
cos(✓)
and so we conclude by the Squeeze Theorem,
sin(✓)
lim
= 1.
✓ô0
✓
When solving a problem with the Squeeze Theorem, one
must write a sort of mathematical poem. You have to tell
your friendly reader exactly which functions you are using to
“squeeze-out” your limit.
Example 29. Compute:
⇠ ⇡1
0
cos 13
x
lim sin(x)e
xô0
Let’s graph this function to see what’s going on:
The Squeeze Theorem
and so
y
2
1
*4 *3.5 *3 *2.5 *2 *1.5 *1 *0.5
0.5
1
1.5
2
2.5
3
3.5
4
x
and 0 again ⇠ by⇡ 1 the
Squeeze
Theorem
cos 13
x
lim* sin(x)e
= 0. Hence we see that
0
*2
⇠ ⇡
cos 13
x
xô0
xô0
*1
The function sin(x)e
lim sin(x)e = 0 = lim* 0,
xô0*
lim sin(x)e
⇠ ⇡1
cos 13
xô0
has two factors:
x
= 0.
y
goes to zero as x ô 0
y = sin (x)e
2
1
⇠ ⇡
©Æ™
cos 13
x
sin(x) e
´Ø¨
y=0
y=0
*4 *3.5 *3 *2.5 *2 *1.5 *1 *0.5
0.5
1
1.5
2
2.5
3
3.5
4
x
*1
bounded between e*1 and e
y = sin (x)e
*2
Hence we have that when 0 < x < ⇡
⇠ ⇡
cos 13
0 f sin(x)e
x
f sin(x)e
and we see
lim 0 = 0 = lim+ sin(x)e
xô0+
xô0
and so by the Squeeze theorem,
⇠ ⇡1
0
cos 13
x
lim+ sin(x)e
= 0.
xô0
In a similar fashion, when *⇡ < x < 0,
⇠ ⇡
cos 13
sin(x)e f sin(x)e
x
f0
53
(In)determinate forms
5
(In)determinate forms
After completing this section, students should be able to do the
following.
• Understand what is meant by the form of a limit.
• Calculate limits of the form zero over zero.
• Calculate limits of the form nonzero number over zero.
• Identify determinate and indeterminate forms.
• Distinguish between determinate and indeterminate forms.
• Discuss why infinity is not a number.
54
Could it be anything?
Break-Ground:
5.1
Could it be anything?
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Hey Riley, remember
lim
✓ô0
sin(✓)
?
✓
Riley: It is equal to 1!
Devyn: But was that crazy proof with all the triangles really necessary? I mean, just plug in zero.
4
5
sin(✓)
sin(0)
0
=
= …
✓
0
0
✓=0
Riley: You were going to say “1,” right?
Devyn: Yeah, but now I’m not sure I was right.
Riley: Dividing by zero is usually a bad idea.
Devyn: You are right. I will never do it again! Also, don’t tell anyone
about this conversation.
Riley: What conversation?
Devyn: Exactly.
Problem 1. Consider the function
f (x) =
f (0) = ?
x
.
x
lim f (x) = ? .
xô0
Problem 2. Consider the function
f (x) =
f (0) = ?
4x
.
x
lim f (x) = ? .
xô0
Problem 3. Consider the function
f (x) =
f (0) = ?
x
.
*3x
lim f (x) = ? .
xô0
55
Limits of the form zero over zero
Dig-In:
5.2 Limits of the form zero over
zero
In the last section we computed limits using continuity and the
limit laws. What about limits that cannot be directly computed
using these methods? Consider the following limit,
lim
x2
xô2
Here
Warning. The symbol
0
0
is not the number 0 divided by 0. It
f (x)
is simply short-hand and means that a limit lim
has the
xôa g(x)
property that
lim f (x) = 0 and
lim g(x) = 0.
xôa
* 3x + 2
.
x*2
xôa
Let’s finish the example with the function above.
Example 30. Compute:
lim x2 * 3x + 2 = 0
xô2
and
Definition. A limit
is said to be of the form
0
0
x2 * 3x + 2 (x * 2)(x * 1)
=
= x * 1.
x*2
(x * 2)
3
Select All Correct Answers:
cos(x)
xô0
x
(b) lim
x2 * 3x + 2
xô0
x*2
(c) lim
x2 * 3x + 2
xô2
x*2
(d) lim
y
3
2
lim g(x) = 0.
*2
*1
*1
y
2
1
xôa
Question 49. Which of the following limits are of the form 00 ?
sin(x)
x
This limit is of the form 00 . However, note that if we
assume x ë 2, then we can write
if
lim f (x) = 0 and
xô0
lim
f (x)
lim
xôa g(x)
xôa
(a) lim
x2 * 3x + 2
xô2
x*2
lim (x * 2) = 0
xô2
in light of this, you may think that the limit is one or zero. Not
so fast. This limit is of an indeterminate form. What does this
mean? Read on, young mathematician.
56
x2 * 3x + 2
xô3
x*3
(e) lim
1
1
*2
2
3
4
x
*2
*1
*1
1
2
3
4
x
*2
*3
y=x*1
*3
x2 * 3x + 2
y=
x*2
So, “our function" is equal to the polynomial x * 1 everywhere, except at x = 2. Therefore, for all values of x
near 2, “our function" is equal to the polynomial x * 1!
This means that
x2 * 3x + 2
= lim (x * 1).
xô2
xô2
x*2
lim
Limits of the form zero over zero
But now, we have to take the limit of a polynomial,
lim (x * 1) = 1.
xô2
Hence
x2 * 3x + 2
= 1.
xô2
x*2
lim
Let’s consider a few more examples of the form 00 .
Example 31. Compute:
lim
xô1
x*1
.
x2 + 2x * 3
First note that
We find the form of this limit by looking at the limits of
the numerator and denominator separately. The limit of
the numerator is:
⇠
⇡
x + 5 * 3(x + 1)
1
3
lim
*
= lim
xô1 x + 1
xô1 (x + 1)(x + 5)
x+5
*2x + 2
= lim
xô1 (x + 1)(x + 5)
0
=
12
=0
The limit of the denominator is:
lim (x * 1) = 0
lim (x * 1) = 0 and
lim x2 + 2x * 3 = 0.
xô1
xô1
Hence, this limit is of the form 00 . Again, we cannot
apply the Quotient Law or any other Limit Law. We
cannot use continuity, either. Namely, “our function" is
not continuous at x = 1, since it is not defined at x = 1.
What can be done? We can hope to be able to cancel a
factor going to 0 out of the numerator and denominator.
Since (x * 1) is a factor going to 0 in the numerator, let’s
see if we can factor a (x * 1) out of the denominator as
well.
⇠
⇠
⇠
x*1
x*
1
lim
= lim
⇠
2
⇠
xô1 x + 2x * 3
xô1 ⇠
(x * 1) (x + 3)
1
= lim
xô1 x + 3
1
= .
4
Example 32. Compute:
lim
xô1
1
x+1
*
3
x+5
x*1
.
xô1
Our limit is therefore of the form 00 and we can probably
factor a term going to 0 out of both the numerator and
denominator.
1
3
* x+5
x+1
lim
xô1
x*1
When looking at the denominator, we hope that this term
is (x * 1). Unfortunately, it is not immediately obvious
how to factor an (x * 1) out of the numerator. So, we
should first simplify the complex fraction by multiplying
it by
(x + 1)(x + 5)
1=
(x + 1)(x + 5)
this will allow us to cancel immediately
lim
xô1
1
x+1
*
3
x+5
(x + 1)(x + 5)
(x + 1)(x + 5)
(x + 5) * 3(x + 1)
= lim
.
xô1 (x + 1)(x + 5)(x * 1)
x*1
Now we will multiply out the numerator. Note that we
do not want to multiply out the denominator because we
57
Limits of the form zero over zero
already have an (x * 1) factored out of the denominator
and that was the goal.
(x + 5) * 3(x + 1)
xô1 (x + 1)(x + 5)(x * 1)
lim
x + 5 * 3x * 3
(x + 1)(x + 5)(x * 1)
*2x + 2
= lim
xô1 (x + 1)(x + 5)(x * 1)
*2⇠
(x⇠
*⇠
1)
= lim
xô1 (x + 1)(x + 5)⇠
(x⇠
*⇠
1)
*2
= lim
.
xô1 (x + 1)(x + 5)
= lim
xô1
Now, we can see that the limit of “our function" is equal
*2
to the limit of a rational function
. This
(x + 1)(x + 5)
rational function is continuous on its domain, and, therefore, at x = 1. Hence
lim
xô1
1
x+1
*
3
x+5
x*1
*2
(x + 1)(x + 5)
*2
=
(1 + 1)(1 + 5)
*1
=
.
6
= lim
xô1
We will look at one more example.
Example 33. Compute:
˘
x+5*2
lim
.
xô*1
x+1
Note that
⇠˘
⇡
lim
x + 5 * 2 = 0 and
xô*1
58
lim (x + 1) = 0.
xô*1
Our limit is, again, of the form 00 and we can probably
factor a term going to 0 out of both the numerator and
denominator. We suspect from looking at the denominator that this term is (x + 1). Unfortunately, it is not
immediately obvious how to factor an (x + 1) out of the
numerator.
We will use an algebraic technique called multiplying
by the conjugate. This technique is useful when you are
trying to simplify an expression that looks like
˘
something ± something else.
It takes advantage of the difference of squares rule
a2 * b2 = (a * b)(a + b).
˘
In our case, we will use a = x + 5 and b = 2. Write
˘
x+5*2
lim
xô*1
x+1
Limits of the form zero over zero
⇠˘
⇡ ⇠˘
⇡
x+5*2
x+5+2
= lim
⇠˘
⇡
xô*1
(x + 1)
x+5+2
⇠˘
⇡2
x + 5 * 22
= lim
⇠˘
⇡
xô*1
(x + 1)
x+5+2
Finally, you may find it distressing that we introduced a form,
namely 00 , only to end up saying they give no information
on the value of the limit. But this is precisely what makes
indeterminate forms interesting. . . they’re a mystery!
x+5*4
⇠˘
⇡
(x + 1)
x+5+2
(x⇠
+⇠
1)
⇠
= lim
⇠˘
⇡
xô*1
(x⇠
+⇠
1)
x+5+2
⇠
= lim
xô*1
= lim ˘
xô*1
1
x+5+2
1
=˘
*1 + 5 + 2
1
= .
4
All of the examples in this section are limits of the form 00 .
When you come across a limit of the form 00 , you should try to
use algebraic techniques to come up with a continuous function
whose limit you can evaluate.
Notice that we solved multiple examples of limits of the form
0
and we got different answers each time. This tells us that
0
just knowing that the form of the limit is 00 is not enough to
compute the limit. The moral of the story is
Limits of the form
0
0
can take any value.
Definition. A form that gives us no information about whether
the limit exists or not, and if the limit exists, no information
about the value of the limit, is called an indeterminate form.
A form that gives information about whether the limit exists
or not, and if it exists gives information about the value of the
limit, is called a determinate form.
59
Limits of the form nonzero over zero
Fill in the table below:
Dig-In:
5.3 Limits of the form nonzero over
zero
Let’s cut to the chase:
Definition. A limit
lim
xôa
is said to be of the form
#
0
lim f (x) = k
xôa
if
f (x)
g(x)
and
lim g(x) = 0.
xôa
where k is some nonzero constant.
Question 50. Which of the following limits are of the form #0 ?
x
(x + 1)2
*1.1
*1.01
*1.001
*1.0001
*0.9
*0.99
*0.999
*0.9999
0.01
0.0001
0.000001
0.00000001
0.01
0.0001
0.000001
0.00000001
What does the table tell us about
lim
Select All Correct Answers:
xô*1
1
xô*1 (x + 1)2
(a) lim
1
?
(x + 1)2
It appears that the limit does not exist, since the expression
1
(x + 1)2
x2 * 3x + 2
xô2
x*2
(b) lim
becomes larger and larger as x approaches *1 . So,
sin(x)
xô0
x
(c) lim
lim
x2 * 3x * 2
(d) lim
xô2
x*2
xô*1
as
ex
(e) lim
xô1 ln(x)
1
(x + 1)2
lim 1 = 1
In our next example, let’s see what is going on with limits of
the form #0 .
Example 34. Consider the function
f (x) =
xô*1
xô*1
is of the form
and
#
0
lim (x + 1)2 = 0.
xô*1
Moreover, as x approaches *1:
• The numerator is positive.
• The denominator approaches zero and is positive.
1
.
(x + 1)2
Use a table of values to investigate lim
60
1
(x + 1)2
100
10000
?
?
?
?
?
?
f (x) =
1
.
(x + 1)2
Limits of the form nonzero over zero
Hence, the expression
Note: Saying “the limit is equal to infinity” describes more
precisely the behavior of the function f near a, than just saying
"the limit does not exist".
1
(x + 1)2
will become arbitrarily large as x approaches *1. We can
see this in the graph of f .
100
First let’s look at the form of this limit. We do this by
taking the limits of both the numerator and denominator:
lim ex =
xô*2
and lim (x + 2)4 = 0.
xô*2
• The numerator is a positive number.
20
*0.5
1
e2
So, this limit is of the form #0 . This form is determinate,
since it implies that the limit does not exist.
But, we can do better than that! As x approaches *2:
40
*1
ex
xô*2 (x + 2)4
y
60
*1.5
Example 35. Compute:
lim
80
*2
Let’s consider a few more examples.
0.5
1
x
• The denominator is positive and is approaching zero.
This means that
ex
= ÿ.
xô*2 (x + 2)4
We are now ready for our next definition.
Definition. If f (x) grows arbitrarily large as x approaches a,
we write
lim
Example 36. Compute:
lim f (x) = ÿ
lim+
xôa
and say that the limit of f (x) is infinity as x goes to a.
If f (x) grows arbitrarily large as x approaches a and f (x) is
negative, we write
lim f (x) = *ÿ
xôa
and say that the limit of f (x) is negative infinity as x goes to
a.
xô3
x2 * 9x + 14
x2 * 5x + 6
First let’s look at the form of this limit, which we do by
taking the limits of both the numerator and denominator.
lim x2 * 9x + 14 = *4
xô3+
and lim+ x2 * 5x + 6 = 0
xô3
This limit is of the form #0 . Next, we should factor the
61
Limits of the form nonzero over zero
numerator and denominator to see if we can simplify the
problem at all.
(x⇠
*⇠
2) (x * 7)
x2 * 9x + 14
⇠
lim+
= lim+
⇠
2
xô3 x * 5x + 6
xô3 ⇠
(x⇠
* 2) (x * 3)
x*7
= lim+
xô3 x * 3
Canceling a factor of x * 2 in the numerator and denominator means we can more easily check the behavior of
this limit. As x approaches 3 from the right:
• The numerator is a negative number.
• The denominator is positive and approaching zero.
This means that
lim+
xô3
x2 * 9x + 14
= *ÿ.
x2 * 5x + 6
Here is our final example.
Example 37. Compute:
x2 * 9x + 14
xô3 x2 * 5x + 6
lim
We’ve already considered part of this example, but now
we consider the two-sided limit. We already know that
x2 * 9x + 14
x*7
= lim
,
xô3 x2 * 5x + 6
xô3 x * 3
lim
and that this limit is of the form #0 . We also know that as
x approaches 3 from the right,
• The numerator is a negative number.
• The denominator is positive and approaching zero.
62
Hence our function is approaching *ÿ from the right.
As x approaches 3 from the left,
• The numerator is negative.
• The denominator is negative and approaching zero.
Hence our function is approaching ÿ from the left. This
means
x2 * 9x + 14
lim
= DNE.
xô3 x2 * 5x + 6
Some people worry that the mathematicians are passing into
mysticism when we talk about infinity and negative infinity.
However, when we write
lim f (x) = ÿ
xôa
and
lim g(x) = *ÿ
xôa
all we mean is that as x approaches a, f (x) becomes arbitrarily
large and g(x) becomes arbitrarily large, with g(x) taking
negative values.
Using limits to detect asymptotes
6 Using limits to detect
asymptotes
After completing this section, students should be able to do the
following.
• Recognize when a limit is indicating there is a vertical
asymptote.
• Evaluate the limit as x approaches a point where there is
a vertical asymptote.
• Match graphs of functions with their equations based on
vertical asymptotes.
• Discuss what it means for a limit to equal ÿ.
• Define a vertical asymptote.
• Find horizontal asymptotes using limits.
• Produce a function with given asymptotic behavior.
• Recognize that a curve can cross a horizontal asymptote.
• Understand the relationship between limits and vertical
asymptotes.
• Calculate the limit as x approaches ±ÿ of common functions algebraically.
• Find the limit as x approaches ±ÿ from a graph.
• Define a horizontal asymptote.
• Compute limits at infinity of famous functions.
• Find vertical asymptotes of famous functions.
• Identify horizontal asymptotes by looking at a graph.
• Identify vertical asymptotes by looking at a graph.
63
Zoom out
Break-Ground:
6.1
Zoom out
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, think about this function:
f (x) =
x2 * 2
.
x2 + 2
Riley: Hmmm. If you plot it, the graph looks like this:
y
0.5
*40
*20
20
x
40
*0.5
Devyn: Right! What I’ve noticed is that if x gets big, then our function looks like a horizontal line.
Riley: I wonder how we find the line?
Problem 1. Guess the y-intercept of this line. Come back and
check your answer after reading the Dig-In!
64
Vertical asymptotes
Question 51. Which of the following are correct?
Dig-In:
6.2
Vertical asymptotes
Select All Correct Answers:
Consider the function
f (x) =
1
.
(x + 1)2
100
(a) lim
1
=ÿ
(x + 1)2
(b) lim
1
ôÿ
(x + 1)2
xô*1
y
xô*1
80
60
(c) f (x) =
1
, so f (*1) = ÿ
(x + 1)2
(d) f (x) =
1
, so as x ô *1, f (x) ô ÿ
(x + 1)2
On the other hand, consider the function
40
f (x) =
20
*2
*1.5
*1
*0.5
1
.
(x * 1)
y
0.5
1
x
40
While the lim f (x) does not exist, something can still be said.
20
xô*1
Definition. If f (x) grows arbitrarily large as x approaches a,
we write
lim f (x) = ÿ
*1
xôa
and say that the limit of f (x) is equal to infinity as x goes to
a.
lim f (x) = *ÿ
and say that the limit of f (x) is equal to negative infinity as
x goes to a.
0.5
1
1.5
2
x
*20
If f (x) grows arbitrarily large as x approaches a and f (x) is
negative, we write
xôa
*0.5
*40
While the two sides of the limit as x approaches 1 do not agree,
we can still consider the one-sided limits. We see lim+ f (x) =
ÿ and lim* f (x) = *ÿ.
xô1
xô1
65
Vertical asymptotes
Definition. If at least one of the following hold:
negative, lim+ f (x) = *ÿ. Since lim* (x * 3) ap-
• lim f (x) = ±ÿ,
xôa
xô3
• lim+ f (x) = ±ÿ,
xôa
y
• lim* f (x) = ±ÿ,
xôa
40
then the line x = a is a vertical asymptote of f .
Example 38. Find the vertical asymptotes of
20
x2 * 9x + 14
f (x) =
.
x2 * 5x + 6
Since f is a rational function, it is continuous on its domain. So the only points where the function can possibly
have a vertical asymptote are zeros of the denominator.
Start by factoring both the numerator and the denominator:
x2 * 9x + 14 (x * 2)(x * 7)
=
(x * 2)(x * 3)
x2 * 5x + 6
Using limits, we must investigate what happens with f (x)
when x ô 2 and x ô 3, since 2 and 3 are the only zeros
of the denominator. Write
(x⇠
*⇠
2) (x * 7)
(x * 7)
⇠
lim
= lim
⇠
⇠
xô2 ⇠
(x * 2) (x * 3) xô2 (x * 3)
*5
=
*1
= 5.
Now write
(x⇠
*⇠
2) (x * 7)
(x * 7)
⇠
= lim
⇠
⇠
xô3 ⇠
xô3
(x * 2) (x * 3)
(x * 3)
*4
= lim
.
xô3 x * 3
lim
Consider the one-sided limits separately. Since lim+ (x *
xô3
3) approaches 0 from the right and the numerator is
66
xô3
xô3
proaches 0 from the left and the numerator is negative,
lim* f (x) = ÿ.
1.5
2
2.5
3
3.5
*20
*40
Hence we have a vertical asymptote at x = 3.
4
x
Horizontal asymptotes
Dig-In:
6.3
20
y
Horizontal asymptotes
Let’s start with an example:
10
Example 39. Consider the function:
1
g(x) =
x
0.5
with the table below
1
1.5
2
2.5
3
x
*10
g(x) =
x
10
100
1000
10000
100000
1000000
1
x
0.1
0.01
0.001
0.0001
0.00001
0.000001
What does the table tell us about g(x) as x grows bigger and
bigger?
As x grows bigger and bigger, it seems that g(x) approaches 0. It seems that we should write
lim
xôÿ
1
=0
x
Example 40. Now consider a different function:
f (x) =
6x * 9
x*1
Fill in the table below, rounding to 5 decimal places.
6x * 9
x*1
5.66667
5.96970
5.99700
5.99970
5.99997
f (x) =
x
10
100
1000
10000
100000
What does the table tell us about f (x) as x grows bigger and
bigger?
As x grows bigger and bigger, it seems like f (x) approaches 6. It seems that we should write
lim
xôÿ
6x * 9
=6
x*1
Definition. If f (x) becomes arbitrarily close to a specific value
L by making x sufficiently large, we write
lim f (x) = L
xôÿ
and we say, the limit at infinity of f (x) is L.
67
Horizontal asymptotes
If f (x) becomes arbitrarily close to a specific value L by making x sufficiently large and negative, we write
lim f (x) = L
xô*ÿ
and we say, the limit at negative infinity of f (x) is L.
In this case we multiply the numerator and denominator
by *1_x3 , which is a positive number as since x ô *ÿ,
x3 is a negative number.
x3 + 1
x3 + 1 *1_x3
lim ˘
= lim ˘
3
xô*ÿ
xô*ÿ
x6 + 5
x6 + 5 *1_x
*1 * 1_x3
= lim ˘
xô*ÿ
x6 _x6 + 5_x6
Example 41. Compute the limit.
6x * 9
.
x*1
We will confirm what we guessed earlier by computing the limit
at infinity.
We can assume that all the Limit Laws also apply to limits at
infinity.
Write
lim
xôÿ
lim
xôÿ
6x * 9
6x * 9 1_x
= lim
xôÿ
x*1
x * 1 1_x
=
Sometimes one must be careful, consider this example.
Example 42.
x3
lim ˘
xô*ÿ
xô*ÿ
*1 * lim (1_x)3
xô*ÿ
=u
⇠
⇡
lim 1 + 5_x6
6x
* x9
lim x
xôÿ x
* x1
x
⇠
⇡
9
lim 6 *
xôÿ
x
=
⇠
⇡
1
lim 1 *
xôÿ
x
1
6 * 9 lim
xôÿ x
=
1
1 * lim
xôÿ x
6 * 9(0)
=
1*0
6
=
1
= 6.
68
*1 * 1_x3
= lim ˘
xô*ÿ
1 + 5_x6
⇠
⇡
lim *1 * 1_x3
xô*ÿ
=
t
lim
1 + 5_x6
+1
x6 + 5
xô*ÿ
*1 * ( lim 1_x)3
xô*ÿ
=t
1 + 5( lim 1_x)6
xô*ÿ
*1 * (0)3
=˘
1 + 5(0)6
*1 * 0
=˘
1+0
= *1.
Note, since
and
lim f (x) = lim+ f
xôÿ
xô0
⇠ ⇡
1
x
⇠ ⇡
1
xô*ÿ
xô0
x
we can also apply the Squeeze Theorem when taking limits at
lim f (x) = lim* f
Horizontal asymptotes
infinity. Here is an example of a limit at infinity that uses the
Squeeze Theorem, and shows that functions can, in fact, cross
their horizontal asymptotes.
Example 43. Compute:
lim
xôÿ
5
sin(7x) + 4x
x
Since
*1 + 4x
1 + 4x
= 4 = lim
xôÿ
x
x
sin(7x)
we conclude by the Squeeze Theorem, lim
+4 =
xôÿ
x
4.
lim
xôÿ
Definition. If
y
or
lim f (x) = L
xôÿ
lim f (x) = L,
xô*ÿ
then the line y = L is a horizontal asymptote of f (x).
Example 44. Give the horizontal asymptotes of
4.5
f (x) =
4
6x * 9
x*1
From our previous work, we see that lim f (x) = 6, and
xôÿ
upon further inspection, we see that lim f (x) = 6.
3.5
xô*ÿ
Hence the horizontal asymptote of f (x) is the line y = 6.
5
10
15
We can bound our function
*1 + 4x sin(7x) + 4x 1 + 4x
f
f
.
x
x
x
Now write with me
lim
xôÿ
*1_x + 4
*1 + 4x 1_x
= lim
x
1_x xôÿ
1
= 4.
And we also have
lim
xôÿ
1_x + 4
1 + 4x 1_x
= lim
xôÿ
x
1_x
1
= 4.
x
20
It is a common misconception that a function cannot cross an
asymptote. As the next example shows, a function can cross a
horizontal asymptote, and in the example this occurs an infinite
number of times!
Example 45. Give a horizontal asymptote of
f (x) =
sin(7x) + 4x
.
x
Again from previous work, we see that lim f (x) = 4.
xôÿ
Hence y = 4 is a horizontal asymptote of f (x).
We conclude with an infinite limit at infinity.
Example 46. Compute
lim ln(x)
xôÿ
69
Horizontal asymptotes
4
y
3
2
1
5
10
15
x
20
*1
The function ln(x) grows very slowly, and seems like it
may have a horizontal asymptote, see the graph above.
However, if we consider the definition of the natural log
as the inverse of the exponential function
ln(x) = y means that ey = x and that x is
positive.
We see that we may raise e to higher and higher values
to obtain larger numbers. This means that ln(x) is unbounded, and hence lim ln(x) = ÿ.
xôÿ
70
Continuity and the Intermediate Value Theorem
7 Continuity and the
Intermediate Value Theorem
After completing this section, students should be able to do the
following.
• Identify where a function is, and is not, continuous.
• Understand the connection between continuity of a function and the value of a limit.
• Make a piecewise function continuous.
• State the Intermediate Value Theorem including hypotheses.
• Determine if the Intermediate Value Theorem applies.
• Sketch pictures indicating why the Intermediate Value
Theorem is true, and why all hypotheses are necessary.
• Explain why certain points exist using the Intermediate
Value Theorem.
• Bisection method.
71
Roxy and Yuri like food
Break-Ground:
7.1
Roxy and Yuri like food
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Yo Riley, I was watching my two cats Roxy and Yuri eat their
dry cat food last night.
Riley: Cats love food! It’s so weird that they swallow the pieces
whole!
Devyn: I know! I noticed something else kinda funny though: Both
Roxy and Yuri start and finish eating at the same times; and
while I gave Roxy a little more food than Yuri, less food was left
in Roxy’s bowl when they stopped eating.
I wonder, is there is a point in time when Roxy and Yuri have
the exact same amount of dry cat food in their bowls?
Riley: Hmmmmm. Do Roxy and Yuri both start and finish drinking
their water at the same times? And does Roxy start with a little
more water than Yuri, and finish with less water left than Yuri?
Devyn: Yes!
Riley: Interesting. I wonder, is there is a point in time when Roxy
and Yuri have the exact same amount of water in their bowls?
Problem 1. Is there a time when Roxy and Yuri have the same
amount of dry cat food in their bowls? Make the following
assumptions:
• They start and finish eating at the same times.
• Roxy starts with more food than Yuri, and leaves less food
uneaten than Yuri.
Problem 2. Is there a time when Roxy and Yuri have the same
amount of water in their bowls? Make the following assumptions:
• They start and finish drinking at the same times.
• Roxy starts with more water than Yuri, and leaves less
water left in her bowl than Yuri.
Problem 3. Within the context of the two problems above, what
is the difference between “dry cat food” and “water?”
72
Continuity of piecewise functions
Dig-In:
7.2 Continuity of piecewise
functions
Before we start talking about continuity of piecewise functions, let’s remind ourselves of all famous functions that are
continuous on their domains.
Theorem 13 (Continuity of Famous Functions). The following
functions are continuous on their natural domains:
domains.
Using the Limit Laws we can prove that given two functions,
both continuous on the same interval, then their sum, difference,
product, and quotient (where defined) are also continuous on
the same interval (where defined).
In this section we will work a couple of examples involving
limits, continuity and piecewise functions.
Rational functions
Example 47. Consider the following piecewise defined function
h x
if x < 0,
n
f (x) = l x * 1
ne*x + c if x g 0.
j
Power function
Find the constant c so that f is continuous at x = 0.
Constant function
Polynomials
Exponential function
Logarithmic function
Trigonometric functions
Inverse trigonometric functions
In essence, we are saying that the functions listed above are
continuous wherever they are defined.
We proved continuity of polynomials earlier using the Sum
Law, Product Law and continuity of power functions.
We proved continuity of rational functions earlier using the
Quotient Law and continuity of polynomials.
We can prove continuity of the remaining four trig functions
using the Quotient Law and continuity of sine and cosine
functions.
Since a continuous function and its inverse have “unbroken"
graphs, it follows that an inverse of a continuous function is
continuous on its domain.
This implies that inverse trig functions are continuous on their
To find c such that f is continuous at x = 0, we need to
find c such that
⇠ ⇡
lim f (x) = f 0 .
xô0
In this case, in order to compute the limit, we will have
to compute two one-sided limits, since the expression for
f (x) if x < 0 is different from the expression for f (x) if
x > 0. So,
lim f (x) = lim*
xô0*
xô0
0
*1
= 0.
x
x*1
=
and
lim f (x) = lim+ (e*x + c)
xô0+
xô0
= e0 + c
=1+c
73
Continuity of piecewise functions
Hence for our function to be continuous, the limit
lim f (x) has to exist, and, therefore,
xô0
so
1+c =0
c = *1.
So, if c = *1, the limit lim f (x) = 0. Now, we have find
lim f (x) = lim* (x + 4)
xô1*
= 5.
Looking at the limit from the right, we have
f (0) = 0.
⇠ ⇡
lim f (x) = f 0 Ç,
xô0
which proves that f is continuous at x = 0.
lim f (x) = lim+ ax2 + bx + 2
xô1+
xô1
= a + b + 2.
Hence, for the limit lim f (x) to exist, we must have that
xô1
5=a+b+2
Consider the next, more challenging example.
Example 48. Consider the following piecewise defined function
h
nx + 4
if x < 1,
n
2
f (x) = lax + bx + 2 if 1 f x < 3,
n
if x g 3.
n6x + a * b
j
Find the constants a and b so that f is continuous at both x = 1
and x = 3.
This problem is more challenging because we have more
unknowns. However, be brave intrepid mathematician.
To find a and b that make f continuous at x = 1, we need
to find a and b such that
lim f (x) = f (1).
xô1
Since f (1) = a + b, it follows that
lim f (x) = a + b + 2.
xô1
We have to compute two one-sided limits, since the ex74
xô1
xô0
the value
Therefore,
pression for f (x) if x < 1 is different from the expression
for f (x) if x > 1. Looking at the limit from the left, we
have
or
a + b = 3.
Also, Hmmmm. More work needs to be done.
To find a and b that make f is continuous at x = 3, we
need to find a and b such that
lim f (x) = f (3).
xô3
Since f (3) = 18 + a * b, it follows that
lim f (x) = 18 + a * b.
xô3
Looking at the limit from the left, we have
lim f (x) = lim* ax2 + bx + 2
xô3*
xô3
= a 9 + b 3 + 2.
Continuity of piecewise functions
Looking at the limit from the right, we have
lim f (x) = lim+ (6x + a * b)
xô3+
xô3
= 18 + a * b.
Hence
9a + 3b + 2 = 18 + a * b
8a + 4b * 16 = 0
2a + b * 4 = 0
So now we have two equations and two unknowns:
and
a+b=3
2a + b = 4.
Set b = 3 * a and write
2a + 3 * a = 4
a = 1,
hence
a=1
and so
b = 2.
Let’s check, so now plugging in values for both a and b
we find
h
nx + 4
n
f (x) = lx2 + 2x + 2
n
n6x * 1
j
Now
and
if x < 1,
if 1 f x < 3,
if x g 3.
lim f (x) = lim+ f (x) = f (1) = 5,
xô1*
xô1
lim f (x) = lim+ f (x) = f (3) = 17.
xô3*
xô3
So setting a = 1 and b = 2 makes f continuous at x = 1
and x = 3.
75
The Intermediate Value Theorem
y
Dig-In:
7.3
The Intermediate Value Theorem
The Intermediate Value Theorem should not be brushed off
lightly. Once it is understood, it may seem “obvious,” but
mathematicians should not underestimate its power.
Theorem 14 (Intermediate Value Theorem). If f is a continuous function for all x in the closed interval [a, b] and d is
between f (a) and f (b), then there is a number c in (a, b) such
that f (c) = d.
If you are more of a visual person, you should imagine a continuous function, where you know the value of the function at
two endpoints, x = a and x = b, but you don’t really know
what the function does between the points x = a and x = b:
f (b)
d
f (a)
a
b
x
and select all that are true:
y
Select All Correct Answers:
(a) f is continuous on (a, b).
f (b)
(b) f is continuous on [a, b).
(c) f is continuous on (a, b].
f (c) = d
(d) f is continuous on [a, b].
(e) There is a point c in [a, b] with f (c) = d.
f (a)
a
c
b
x
The Intermediate Value Theorem says that despite the fact that
you don’t really know what the function is doing between the
endpoints, a point x = c exists and gives an intermediate value
for f .
Now, let’s contrast this with a time when the conclusion of the
Intermediate Value Theorem does not hold.
Question 52. Consider the following situation,
76
Building on the question above, it is not difficult to see that
each of the hypothesis of the Intermediate Value Theorem is
necessary.
Let’s see the Intermediate Value Theorem in action.
Example 49. Explain why the function f (x) = x3 +3x2 +x*2
has a root between 0 and 1.
Since f is a polynomial, we see that f is continuous for
all real numbers. Since f (0) = *2 and f (1) = 3, and 0
The Intermediate Value Theorem
is between *2 and 3, by the Intermediate Value Theorem,
there is a point c in the interval [0, 1] such that f (c) = 0.
This example also points the way to a simple method for approximating roots.
Example 50. Approximate a root of f (x) = x3 + 3x2 + x * 2
between 0 and 1 to within one decimal place.
Again, since f is a polynomial, we see that f is continuous for all real numbers. Consider the table
x
f (x)
0.1
0.2
0.3
0.4
0.5
0.6
0.7
*1.869
*1.672
*1.403
*1.056
*0.625
*0.104
0.513
By the Intermediate Value Theorem, f has a root between
0.6 and 0.7. Repeating the process
x
f (x)
0.61
0.62
*0.046719
0.011528
and using the Intermediate Value Theorem, we can conclude that f has a root between 0.61 and 0.62, and the
root is 0.6 rounded to one decimal place.
To start, note that both f and g are continuous functions, and
hence h = f * g is also a continuous function. Now
h(1) = f (1) * g(1)
= (1)2 ln(1) * 2 1 cos(ln(1))
= *2.
and in a similar fashion
h(e) = f (e) * g(e)
= e2 1 * 2 e cos(1)
Since e > 2 and 0 < cos(1) < 1 we see that the expression
above is positive. Therefore, h(1) < 0 < h(e), and by the
Intermediate Value Theorem, there exist a number c in (1, e)
such that
h(c) = 0.
But this means that
f (c) * g(c) = 0,
and that f (c) = g(c). Therefore, the curves y = f (x) and
y = g(x) intersect at the point (c, f (c)).
Now we move on to a more subtle example:
Example 52. Suppose you have two cats, Roxy and Yuri. Is
there a time when Roxy and Yuri have the same amount of water
in their bowls assuming:
• They start and finish drinking at the same times.
• Roxy starts with more water than Yuri, and leaves less
water left in her bowl than Yuri.
To solve this problem, consider two functions:
The Intermediate Value Theorem can be used to show that
curves cross:
• WRoxy (t) = the amount of water in Roxy’s bowl at
time t.
Example 51. Explain why the graphs of the functions f (x) =
x2 ln(x) and g(x) = 2x cos(ln(x)) intersect on the interval
[1, e].
• WYuri (t) = the amount of water in Yuri’s bowl at
time t.
77
The Intermediate Value Theorem
Now if tstart is the time the cats start drinking and tf inish
is the time the cats finish drinking. Then we have
WRoxy (tstart ) * WYuri (tstart ) > 0
and
WRoxy (tf inish ) * WYuri (tf inish ) < 0.
Since the amount of water in a bowl at time t is a continuous function, as water is “lapped” up in continuous
amounts,
WRoxy * WYuri
is a continuous function, and hence the Intermediate
Value Theorem applies. Since WRoxy * WYuri is positive when at tstart and negative at tf inish , there is some
time tequal when the value is zero, meaning
However in this case, the amount of food in a bowl at
time t is not a continuous function! This is because dry
cat food consists of discrete kibbles, and is not eaten
in a continuous fashion. Hence the Intermediate Value
Theorem does not apply, and we can make no definitive
statements concerning the question above.
For some interesting extra reading check out:
• The intermediate value theorem is NOT obvious—and I
am going to prove it to you, S.M. Walk, College Math
Journal, September 20111 .
WRoxy (tequal ) * WYuri (tequal ) = 0
meaning there is the same amount of water in each of
their bowls.
And finally, an example when the Intermediate Value Theorem
does not apply.
Example 53. Suppose you have two cats, Roxy and Yuri. Is
there a time when Roxy and Yuri have the same amount of dry
cat food in their bowls assuming:
• They start and finish eating at the same times.
• Roxy starts with more food than Yuri, and leaves less food
uneaten than Yuri.
Here we could try the same approach as before, setting:
• FRoxy (t) = the amount of dry cat food in Roxy’s
bowl at time t.
• FYuri (t) = the amount of dry cat food in Yuri’s bowl
at time t.
78
1
See The intermediate value theorem is NOT obvious—and I am going to
prove it to you, S.M. Walk, College Math Journal, September 2011 at http:
//www.jstor.org/stable/10.4169/college.math.j.42.4.254
An application of limits
8
An application of limits
After completing this section, students should be able to do the
following.
• Compute average velocity.
• Approximate instantaneous velocity.
• Compare average and instantaneous velocity.
• Compute instantaneous velocity.
79
Limits and velocity
Break-Ground:
8.1
Limits and velocity
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Hey Riley, I’ve been thinking about limits.
Riley: That is awesome.
Devyn: I know! You know limits remind me of something. . . How a
GPS or a phone computes velocity!
Riley: Huh. A GPS can calculate our location. Then, to compute
velocity from position, it must look at
change in position
change in time
Devyn: And then we study this as the change in time gets closer and
closer to zero.
Riley: Just like with limits at zero, we can study something by looking
near a point, but not exactly at a point.
Devyn: O.M.G. Life’s a rich tapestry.
Riley: Poet, you know it.
Suppose you take a road trip from Columbus Ohio to UrbanaChampaign Illinois. Moreover, suppose your position is modeled by
s(t) = 36t2 * 4.8t3
(miles West of Columbus)
where t is measured in hours and runs from 0 to 5 hours.
Problem 1. What is the average velocity for the entire trip?
Problem 2. Use a calculator to estimate the instantaneous
velocity at t = 2.
Problem 3. Considering the work above, when we want to
compute instantaneous velocity, we need to compute
change in position
change in time
when (choose all that apply):
Select All Correct Answers:
80
(a) The “change in time” is zero.
(b) The “change in time” gets closer and closer to zero.
(c) The “change in time” approaches zero.
(d) The “change in time” is near zero.
(e) The “change in time” goes to zero.
Computing average velocities for smaller, and smaller, values
of t * 2 as we did above is tedious. Nevertheless, this is exactly
how a GPS determines velocity from position! To avoid these
tedious calculations, we would really like to have a formula.
Instantaneous velocity
Dig-In:
8.2
25
Instantaneous velocity
When we compute average velocity, we look at
change in position
.
change in time
s
20
15
10
5
To obtain the (instantaneous) velocity, we want the change in
time to “go to” zero. By this point we should know that “go to”
is a buzz-word for a limit. The change in time is often given as
the length of a time interval, and this length goes to zero.
The average velocity on the (time) interval [a, b] is given by
vav =
change in position s(b) * s(a)
=
.
change in time
b*a
Here s(t) denotes the position, at the time t, of an object moving
along a line.
Let’s put all of this together by working an example.
Example 54. A young mathematician throws a ball straight
into the air with a velocity of 40ft/sec. Its height (in feet) after
t seconds is given by
0
0.5
1
1.5
2
t
2.5
(a) When will the ball hit the ground?
To determine when the ball hits the ground we need
to solve the equation
s(t) = 0
for t. That is,
40t * 16t2 = 0
t(40 * 16t) = 0.
This has solutions t = 0 seconds and t = 2.5 seconds. Since the ball hits the ground after it’s thrown,
we know that the ball hits the ground at t = 2.5
seconds.
(b) What is the height of the ball after 2 seconds?
s(t) = 40t * 16t2
Here is the graph of s.
(feet above the ground).
To find the height of the ball after 2 seconds, we
simply need to plug 2 into the equation for s(t) to
find
s(2) = 40 2 * 16 22 = 16 f t.
81
Instantaneous velocity
(c) Consider the following points lying along the s axis.
on the time interval [a, b] is given by
s
25 D
20
15 C
10
s(b) * s (a)
.
b*a
Plugging in a = 1 and b = 2 we find that
vav =
s(2) * s(1) 16 * 24
=
= *8 f t_ s.
2*1
1
Check the figure below.
s
5
25
0 B
20
change in
position
*5 A
15
change in time
*10
Which points correspond to the height of the ball at times
0, 1 and 2 seconds?
The point that corresponds to s(0), the position
(height) of the ball at t = 0, is B.
The point that corresponds to s(1), the position
(height) of the ball at t = 1, is D.
The point that corresponds to s(2), the position
(height) of the ball at t = 2, is C.
Next let’s consider the average velocity of the ball.
(d) What is the average velocity of the ball on the time interval
[1, 2]?
In order to find the average velocity of the ball on
the interval [1, 2] we recall that the average velocity
82
vav =
(1, s(1))
(2, s(2))
10
5
0
0.5
1
1.5
2
t
2.5
(e) What is the average velocity of the ball on the time interval
[t, 2], for 0 < t < 2?
We use the same formula we used to find the average
velocity on the interval [1, 2] to find the average
Instantaneous velocity
velocity on the time interval [t, 2] for 0 < t < 2.
velocity formula one more time to find
s(2) * s (t)
2*t
16 * (40t * 16t2 )
=
2*t
8(2t2 * 5t + 2)
=
2*t
⇠⇠
8(2t * 1)⇠
(t *
2)
=
⇠
⇠2)
*⇠
(t *
= *8(2t * 1)
vav =
vav =
However, this is exactly the same expression we got
when calculating the average velocity on the interval
[t, 2] for 0 < t < 2. So the average velocity on the
interval [2, t], for 2 < t < 2.5, is given by
vav = 8 * 16t f t_ s.
(g) Using the results in parts (e) and (f), compute the average
velocity on the interval
= 8 * 16t f t_ s,
for 0 < t < 2. Check the figure below.
25
s(t) * s(2) s(2) * s(t)
=
.
t*2
2*t
(i) [2, 2.1]
vav = 8 * 16(2.1) = *25.6 f t_ s
s
(ii) [1.9, 2]
vav = 8 * 16(1.9) = *22.4 f t_ s
(t, s(t))
20
s
15
(2, s(2))
t
10
In our previous example, we computed average velocity on
several different intervals. If we let the size of the interval go
to zero, we get instantaneous velocity. Limits will allow us
to compute instantaneous velocity. Let’s use the same setting
as before.
Example 55. The height of a ball above the ground between 0
and 2.5 seconds is given by
5
s(t) = 40t * 16t2 .
0
0.5
1
1.5
t
2
t
2.5
(f) What is the average velocity of the ball on the interval
[2, t], for 2 < t < 2.5?
To calculate the average velocity on the interval
[2, t], for 2 < t < 2.5, we will use our average
Find the instantaneous velocity of the ball 2 seconds after it is
thrown.
From the previous example, we know that the average
velocity of the ball on the interval [t, 2], for 0 < t < 2, and
the average velocity on the interval [2, t], for 2 < t < 2.5,
are both given by
vav =
s(t) * s(2)
= 8 * 16t f t_ s.
t*2
83
Instantaneous velocity
All we need to do to find the instantaneous velocity at
t = 2 is take the limit as t goes to 2 of the expression
above. Doing so we find
v = lim vav = lim
tô2
84
tô2
s(t) * s(2)
= lim(8 * 16t)
tô2
t*2
= *24 f t_ s.
Definition of the derivative
9
De�nition of the derivative
After completing this section, students should be able to do the
following.
• Use limits to find the slope of the tangent line at a point.
• Understand the definition of the derivative at a point.
• Compute the derivative of a function at a point.
• Estimate the slope of the tangent line graphically.
• Write the equation of the tangent line to a graph of a
function at a given point.
• Recognize and distinguish between secant and tangent
lines.
• Recognize the the tangent line as a local approximation
for a differentiable function near a point.
85
Slope of a curve
Break-Ground:
9.1
Slope of a curve
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, do you remember “slope?’
Riley: Most definitely. “Rise over run.”
Devyn: You know it.
Riley: “Change in y over change in x.”
Devny: That’s right.
Riley: Brought to you by the letter “m.”
Devny: Enough! My important question is: could we define “slope”
for a curve that’s not a straight line?
Riley: Well, maybe if we “zoom in” on a curve, it would look like a
line, and then we could call it “the slope at that point.”
Devyn: Ah! And this “zoom in” idea sounds like a limit!
Riley: This is so awesome. We just made math!
The concept introduced above, of the “slope of a curve at a
point,” is in fact one of the central concepts of calculus. It will,
of course, be completely explained. Let’s explore Devyn and
Riley’s ideas a little more, first.
To find the “slope of a curve at a point,” Devyn and Riley spoke
of “zooming in” on a curve until it looks like a line. When you
zoom in on a smooth curve, it will eventually look like a line.
This line is called the tangent line.
y
f (x2 )
f (a)
f (x1 )
x1
a
x2
x
Problem 1. Which of the following approximate the slope of
the “zoomed line”? (You can select more than one.)
Select All Correct Answers:
(a)
86
f (x1 ) * f (a)
x2 * a
(b)
f (x1 ) * f (a)
a * x1
(c)
f (x1 ) * f (a)
x1 * a
(d)
f (x1 ) * a
x1 * a
(e)
f (a) * f (x2 )
a * x2
(f)
f (x2 ) * f (a)
x2 * a
(g)
x2 * a
f (x2 ) * f (a)
(h)
f (x1 ) * f (a)
f (x2 ) * f (a)
(i) a(f (x2 ) * f (x1 ))
(j)
f (a) * f (x1 )
a * x1
Problem 2. Let f (x) = 3x * 1. Zoom in on the curve around
a = *2 so that x1 = *1.9. Use one of the formulations in the
problem above to approximate the slope of the curve. The slope
of the curve at a = *2 is approximately. . .
Problem 3. Repeat the previous problem for f (x) = x2 * 1,
a = 0, and x2 = 0.2. Choose a formulation that will give you a
positive answer for the slope. The (positive) slope of the curve
at a = 0 is approximately. . .
Problem 4. Zoom in on the curve f (x) = x2 * 1 near x = 0
again. By looking at the graph, what is your best guess for the
actual slope of the curve at zero?
The definition of the derivative
Dig-In:
9.2
The de�nition of the derivative
Given a function, it is often useful to know the rate at which
the function changes. To give you a feeling why this is true,
consider the following:
• If s(t) represents the position of an object on a line (with
respect to time), the rate of change gives the velocity of
the object.
More precisely, the average rate of change of a function f is
given by
average rate of change of f =
as the input changes from a to x.
y
• If v(t) represents the velocity of an object with respect
to time, the rate of change gives the acceleration of the
object.
• If R(x) represents the revenue generated by selling x objects, the rate of change gives us the marginal revenue,
meaning the additional revenue generated by selling one
additional unit. Note, there is an implicit assumption that
x is quite large compared to 1.
• If C(x) represents the cost to produce x objects, the rate of
change gives us the marginal cost, meaning the additional
cost generated by selling one additional unit. Again, there
is an implicit assumption that x is quite large compared
to 1.
• The rate of change of a function can help us approximate
a complicated function with a simple function.
• The rate of change of a function can be used to help us
solve equations that we would not be able to solve via
other methods.
From slopes of secant lines to slopes of tangent lines
We’ve been computing average rates of change for a while now,
average rate of change=
change in the function (output)
.
change in the input
f (x) * f (a)
,
x*a
average rate
=
of change
y
f (x) * f (a)
=
x
x*a
f
f (x)
y
f (a)
x
0
a
x
x
What happens if we compute the average rate of change of f
for each value of x as x gets closer and closer to a? In other
words, what is the meaning of the limit
lim
xôa
f (x) * f (a)
,
x*a
provided that the limit exists?
Naturally, we call this limit the instantaneous rate of change
of the function f at a.
Example 56. (In this example we will use abreviation AvRCh
for the average rate of change, and InRCh for the instantaneous
rate of change.)
Let f (x) = 2x2 + 3.
87
The definition of the derivative
(a) Find the expression for the average rate of change of f
between the points a = 2 and x = *1.
AvRCh =
f (*1) * f (2)
*1 * (2)
We have already computed an expression for the average
rate of change for all x ë 2. Therefore,
Now evaluate the function,
AvRCh =
Simplify,
2(*1)2
+ 3 * (2(2)2
*1 * (2)
InRCh = lim 2(x + 2) = 8
+ 3)
xô2
.
5 * 11
AvRCh =
= 2.
*3
(b) Find the average rate of change of f between the points
a = 2 and x, x ë 2.
AvRCh =
So, the instantaneous rate of change is the limit, as x ô 2,
of average rates of change of f between points a = 2 and
x, for x ë 2.
f (x) * f (2)
x*2
We can arrive at the same limit,
lim
xôa
within a completely different context.
Have a look at the figure below. The figure depicts a graph of
the function f , two points on the graph, (a, f (a)) and (x, f (x)),
and a secant line that passes through these two points.
y
Now substitute in for the function,
AvRCh =
f (x) * f (a)
,
x*a
slope of the =
secant line
2x2 + 3 * 11
x*2
y
f (x) * f (a)
=
x
x*a
f
Simplify the top,
AvRCh =
2x2 * 8
.
x*2
AvRCh =
2(x2 * 4)
x*2
Factor,
Factor and cancel,
AvRCh =
2(x + 2)⇠
(x⇠
*⇠
2)
= 2(x + 2)
⇠
⇠2
⇠
x*
(c) Find the instantaneous rate of change of f at the point
x = 2.
f (x) * f (2)
InRCh = lim
xô2
x*2
88
f (x)
f (a)
(x, f (x))
(a, f (a))
a
x
x
f (x) * f (a)
!
x*a
This is exactly the expression for the average rate of change of
f as the input changes from a to x!
The slope of this secant line is equal to
Therefore,
The definition of the derivative
slope of secant line = average rate of change.
As before, we can ask ourselves: What happens as x gets closer
and closer to a? In other words, what is the meaning of the
limit of slopes of secant lines through the points (a, f (a)) and
(x, f (x)) as x gets closer and closer to a?
(a, f (a)) sufficiently so that it appears to be a straight line, then
that line is the tangent line to f (x) at the point (a, f (a)). This
is illustrated in the figure below.
y
How can we interpret the limit
lim
xôa
f (x) * f (a)
,
x*a
x
provided that the limit exists?
This scenario is illustrated in the figure below. What happens
as x ô a?
Question 53. Each of the following four figures depicts the
graph of the same function f and a line that passes through
the point (1, f (1)).
2
y
A
*2
y
2
y = f (x)
B
1
*1
1
2
x
*2
*1
C
*2
f (a)
a
x
x
Notice that all the secant lines in the figure pass through the
given point (a, f (a)). What distinguishes these lines are their
slopes; the slope of each line is determined by an additional
point on the line, (x, f (x)). The figure suggests that“the limit
of slopes of secant lines" is the slope of a special line, called
the tangent line to the curve y = f (x) at the point (a, f (a)).
Notice how the tangent line and the curve are indistinguishable
near the point (a, f (a)). If one can “zoom in” on the graph at
2
x
*2
y
2
y = f (x)
D
1
*1
1
*1
*2
2
y = f (x)
1
*1
f (x)
y
1
2
x
*2
y
y = f (x)
1
*1
1
*1
*1
*2
*2
2
x
Only one of the figures depicts the line tangent to the curve
y = f (x) at the point (1, f (1)). Which one?
Multiple Choice:
(a) Figure A
(b) Figure B
(c) Figure C
89
The definition of the derivative
(d) Figure D
Only the line in figure B looks indistinguishable from
the curve y = f (x) near the point (1, f (1)).
Obviously, the limit
lim
xôa
f (x) * f (a)
,
x*a
has many different interpretations, depending on the context,
such as the slope of the tangent line, the instantaneous rate of
change, and the instantaneous velocity.
Therefore, this limit deserves a special name that could be used
regardless of the context.
Definition. The derivative of f at a, denoted f ® (a), is given
by
f (x) * f (a)
f ® (a) = lim
,
xôa
x*a
provided that the limit exists. We say that f is differentiable
at a if this limit exists. Otherwise, we say that f is nondifferentiable at a.
The definition of the derivative allows us to define a tangent
line precisely.
Definition. Let f be a function differentiable at a. The line
tangent to the curve y = f (x) at the point (a, f (a)) is the line
that passes through the point (a, f (a)) whose slope is equal to
f ® (a).
Naturally, by the point-slope equation of the line, it follows that
the tangent line is given by the equation
y = f (a) + f ® (a)(x * a).
Question 54. Let f be a function given by f (x) = 4x * 3.
What is the instantaneous rate of change of f at a, where a is
any real number?
90
But, most functions are not linear, and their graphs are not
straight lines. Therefore, the computation of the derivative is
not as simple as in the previous example.
Example 57. Let f (x) = 2x2 + 3.
Find the slope of the tangent line to the curve y = f (x) at the
point (2, f (2)).
Finding the slope of the tangent line at the point (2, f (2))
means finding f ® (2)
f ® (2) = lim
xô2
f (x) * f (2)
x*2
But, this limit gives us the instantaneous rate of change
of f at a = 2, which was computed in our first example!
Therefore, the slope of the tangent line is
f ® (2) = lim
xô2
f (x) * f (2)
= 8.
x*2
1
. Find an equation for the line
3*x
tangent to the curve y = f (x) at the point (2, 1).
Example 58. Let f (x) =
To find an equation for a line, we need two pieces of
information. We need to know a point on the line, and we
need to know the slope of the line. In this question, we
are given that the point (2, 1) is on the line. So, we need
to find the slope of the tangent line. Finding the slope of
the tangent line at the point (2, 1) means finding f ® (2).
Start by writing out the definition of the derivative,
1
*1
f (x) * f (2)
3*x
f (2) = lim
= lim
.
xô2
xô2 x * 2
x*2
®
Multiply by
3*x
to clear the fraction in the numerator,
3*x
f ® (2) = lim
xô2
1 * (3 * x)
.
(x * 2)(3 * x)
The definition of the derivative
Combine like-terms in the numerator,
⇠
⇠
⇠
x*
2
f ® (2) = lim
,
⇠
xô2 ⇠
(x⇠
* 2) (3 * x)
Take the limit as x goes to 2,
1
f ® (2) = lim
= 1.
xô2 3 * x
We are looking for an equation of the line through the
point (2, 1) with slope m = f ® (2) = 1. The point-slope
formula tells us that the line has equation given by
y = 1 + x * 2,
or y = x * 1.
Example 59. An object
˘moving along a straight line has position given by s(t) = t + 3, where position is measured in
meters and time in seconds. Find the (instantaneous) velocity
of the object at time t = 6.
Now simplify the numerator,
s® (6) = lim
tô6
Combine like-terms,
s® (6) = lim
tô6
t+3*9
⇠˘
⇡.
(t * 6)
t+3+3
⇠
⇠
t *⇠
6
⇠˘
⇡.
⇠⇠
(t *
6)
t+3+3
⇠
Take the limit as t goes to 6,
1
1
s® (6) = lim ˘
= .
tô6
t+3+3 6
The object has velocity
1 m
at time t = 6.
6 s
Velocity is the instantaneous rate of change of position with respect to time. We are being asked to find
v(6) = s® (6). We have to adjust the definition of the the
derivative, namely, replace x with t,
˘
t+3*3
s(t)
*
s(6)
s® (6) = lim
= lim
.
tô6
tô6
t*6
t*6
˘
t+3+3
Multiply by ˘
, in order to rationalize the ext+3+3
pression,
H˘
I H˘
I
t
+
3
*
3
t
+
3
+
3
s® (6) = lim
.
˘
tô6
t*6
t+3+3
91
Derivatives as functions
10
Derivatives as functions
After completing this section, students should be able to do the
following.
• Understand the derivative as a function related to the original definition of a function.
• Find the derivative function using the limit definition.
• Relate the derivative function to the derivative at a point.
• Explain the relationship between differentiability and continuity.
• Relate the graph of the function to the graph of its derivative.
• Determine whether a piecewise function is differentiable.
92
Wait for the right moment
Select All Correct Answers:
Break-Ground:
10.1
Wait for the right moment
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, I might be a calculus genius.
Riley: Yeah? Explain this one to me.
Devyn: Let me first ask you a question. Say you have a function, like
f (x) = x2 , and you want to know f ® (3). Do you plug in the
number 3 before or after you find the derivative?
Riley: Hmmmm. Well, my next step is usually
f ® (3) = lim
hô0
(a) The domain of f ® is equal to the domain of f .
(b) The range of f ® is equal to the range of f .
(c) The domain of f ® is a subset of the domain of f .
(d) The range of f ® is a subset of the range of f .
Problem 2. Find g ® (2) for g(x) = x2 + 1 using both methods
described above.
f (3 + h) * f (3)
.
h
So I guess before.
Devyn: Aha! I think you’re wasting time. You see I write
f ® (x) = lim
hô0
f (x + h) * f (x)
.
h
and it means that I can look at the derivative of my function at
any point. So, I plug in the 3 after I’ve found the derivative.
Riley: That does seem like a pretty genius move. But doesn’t working
with x, instead of numbers, make all of this more difficult?
Devyn: Not at all. Let’s do the problems both ways, at the same time:
f (3 + h) * f (3)
hô0
h
(3 + h)2 * 9
= lim
hô0
h
9 + 6h + h2 * 9
= lim
hô0
h
6h + h2
= lim
hô0
h
= lim(6 + h)
f ® (3) = lim
hô0
= 6.
´≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠¨
plugging in
f (x + h) * f (x)
h
(x + h)2 * x2
= lim
hô0
h
x2 + 2xh + h2 * x2
= lim
hô0
h
2xh + h2
= lim
hô0
h
= lim(2x + h)
f ® (x) = lim
hô0
hô0
= 2x,
so f ® (3) = 6.
´≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠¨
working with x
Riley: Whoa. So now the derivative is a function. Wait, what’s its
domain? Its range?
Problem 1. Suppose you have a function f . Which of the
following are true?
93
The derivative as a function
writing the definition of the derivative at the point a, namely
Dig-in:
10.2
The derivative as a function
f ® (a) = lim
hô0
The derivative of a function, as a function
First, we have to find an alternate definition for f ® (a), the derivative of a function f at a.
Let’s start with the average rate of change of the function f
as the input changes from a to x. We will introduce a new
variable, h, to denote the difference between x and a. That is
x * a = h or x = a + h. Take a look at the figure below.
y
Example 60. Let f (x) = x2 * 2x. Using the alternate expression for the derivative, find the slope of the tangent line to the
curve y = f (x) at the point (2, f (2)).
The slope of the tangent line is given by the derivative,
f ® (2).
f (2 + h) * f (2)
f ® (2) = lim
.
hô0
h
Now substitute in for the function we know,
(2 + h)2 * 2(2 + h) * 0
.
hô0
h
f ® (2) = lim
y f (a + h) * f (a)
average rate
of change = h =
h
Now expand the numerator of the fraction,
f
4 + 4h + h2 * 4 * 2h
.
hô0
h
f ® (2) = lim
f (a + h)
Now combine like-terms,
y
2h + h2
.
hô0
h
f ® (2) = lim
f (a)
h
0
a
a+h
x
average rate of change =
Factor an h from every term in the numerator,
h (2 + h)
.
hô0
h
f ® (2) = lim
Now we can write
f (a + h) * f (a) f (a + h) * f (a)
=
(a + h) * a
h
What happens if h ô 0? In other words, what is the meaning
of the limit
f (a + h) * f (a)
lim
?
hô0
h
Obviously, this limit represents f ® (a), the instantaneous rate
of change of f at a! Therefore, we have an alternate way of
94
f (a + h) * f (a)
.
h
Compute the limit,
f ® (2) = lim (2 + h) = 2.
hô0
This alternate definition of the derivative of f at a, namely,
f (a + h) * f (a)
,
hô0
h
f ® (a) = lim
The derivative as a function
(provided that the limit exists), allows us to define f ® (x) for
any value of x,
Question 56. Can two different functions, say, f and g, have
the same derivative?
f (x + h) * f (x)
,
h
(provided that the limit exists).
Let’s compare the graphs of f and f ® for the derivatives we’ve
computed so far:
f ® (x) = lim
hô0
And this is how we define a new function, f ® , the derivative
of f . The domain of f ® consists of all points in the domain
of f where the function f is differentiable. f ® (x) gives us the
instantaneous rate of change of f at any point x in the domain
of f ® .
Given a function f from some set of real numbers to the real
numbers, the derivative f ® is also a function from some set of
real numbers to the real numbers. Understanding the relationship between the functions f and f ® helps us understand any
situation (real or imagined) involving changing values.
f (x) = x2 , f ® (x) = 2x
f®
f
Example 61. Given the function f (x) = 3x + 2, find f ® (x).
Start with the definition of f ® (x)
f ® (x) = lim
hô0
f (x + h) * f (x)
.
h
f (x) = 3x + 2, f ® (x) = 3
Replace f with its formula,
f
3(x + h) + 2 * (3x + 2)
f ® (x) = lim
.
hô0
h
f®
Simplify the top,
f ® (x) = lim
hô0
Evaluate the limit.
3h
.
h
f ® (x) = lim 3 = 3.
hô0
Question 55. Is it true that the domain of f ® is equal to the
domain of f ?
h
n1
for x > 0
f (x) = x, f ® (x) = l
n*1 for x < 0
j
95
The derivative as a function
f
f®
Question 57. For each of the three pairs of functions, describe
y = f (x) when f ® is positive, and when f ® is negative.
Question 58. Here we see the graph of f ® , the derivative of
some function f .
Which of the following graphs could be y = f (x)?
Multiple Choice:
(a)
(b)
(c)
96
Differentiability implies continuity
and use continuity of a constant to obtain that
Dig-In:
10.3 Di�erentiability implies
continuity
There are connections between continuity and differentiability.
Theorem 15 (Differentiability Implies Continuity). If f is a
differentiable function at x = a, then f is continuous at x =
a.
To explain why this is true, we are going to use the following definition of the derivative
f (x) * f (a)
f ® (a) = lim
.
xôa
x*a
Assuming that f ® (a) exists, we want to show that f (x) is
continuous at x = a, hence we must show that
lim f (x) = f (a).
xôa
Starting with
lim (f (x) * f (a))
Limit Law.
we apply the Difference Law to the left hand side
lim f (x) * lim f (a) = 0,
xôa
xôa
Now we see that lim f (x) = f (a), and so f is continuous
xôa
at x = a.
This theorem is often written as its contrapositive:
If f (x) is not continuous at x = a, then f (x) is not
differentiable at x = a.
Thus from the theorem above, we see that all differentiable
functions on R are continuous on R. Nevertheless there are
continuous functions on R that are not differentiable on R.
(a) x2
lim (f (x) * f (a)) = 0,
xôa
lim f (x) = f (a).
xôa
Multiple Choice:
= 0 f ® (a) = 0.
Since
Next, we add f (a) on both sides and get that
Question 59. Which of the following functions are continuous
but not differentiable on R?
xôa
we multiply and divide by (x * a) to get
0
1
f (x) * f (a)
= lim (x * a)
xôa
x*a
1
⇠
⇡0
f (x) * f (a)
= lim (x * a)
lim
xôa
xôa
x*a
lim f (x) * f (a) = 0.
xôa
(b) ‚x„
(c) x
sin(x)
x
Example 62. Consider
(d)
h
nx2
if x < 3,
f (x) = l
nmx + b if x g 3.
j
What values of m and b make f differentiable at x =
3?
97
Differentiability implies continuity
To start, we know that f must be continuous at x = 3,
since it has to be differentiable there. We will start by
making f continuous at x = 3. Write with me:
lim f (x) = 9
xô3*
lim f (x) = m3 + b
xô3+
f (3) = m3 + b
So for the function to be continuous, we must have
m 3 + b = 9.
We also must ensure that the function is differentiable
at x = 3. In other words, we have to ensure that the
following limit exists
f (x) * f (3)
.
x*3
In order to compute this limit, we have to compute the
two one-sided limits
and
lim
xô3+
f (x) * f (3)
mx + b * (3m + b)
= lim
xô3
x*3
x*3
mx * 3m
= lim+
xô3
x*3
m⇠
(x⇠
*⇠
3)
= lim+
⇠
⇠
⇠
xô3
x*
3
= lim+ m
xô3
= m.
Hence, we must have
m = 6.
Ah! So now the equation that must be satisfied
9 = m 3 + b,
lim
xô3
lim
xô3+
and
f (x) * f (3)
x*3
9 = 6 3 + b.
Therefore, b = *9. Thus setting m = 6 and b = *9
will give us a function that is differentiable (and hence
continuous) at x = 3.
f (x) * f (3)
,
xô3
x*3
since f (x) changes expression at x = 3. Write with me
Can we tell from its graph whether the function is differentiable
or not at a point a?
f (x) * f (3)
x2 * 9
lim*
= lim
xô3
xô3 x * 3
x*3
(x⇠
*⇠
3) (x + 3)
⇠
= lim*
⇠
⇠
⇠
xô3
x*
3
= lim* (x + 3)
Each of the figures A-D depicts a function that is not
differentiable at a = 1.
lim*
xô3
= 6,
98
becomes
Question 60. What does the graph of a function f possibly
look like when f is not differentiable at a?
Differentiability implies continuity
2
A
*2
y
2
B
1
*1
1
2
x
*2
1
*1
*1
C
*2
2
1
2
x
*2
y
2
D
1
*1
1
*1
*2
2
y
1
2
x
*2
y
1
*1
*1
*1
*2
*2
x
The function in figure A is not continuous at a, and, therefore, it is not differentiable there.
In figures B–D the functions are continuous at a, but in
each case the limit
f (x) * f (a)
xôa
x*a
lim
does not exist, for a different reason.
In figure B
lim
xôa+
f (x) * f (a)
f (x) * f (a)
ë lim*
.
xôa
x*a
x*a
In figure C
f (x) * f (a)
= ÿ.
x*a
In figure D the two one-sided limits don’t exist and neither
one of them is infinity.
lim
xôa
So, if at the point a a function either has a "jump" in the
graph, or a corner, or what looks like a “vertical tangent
line", or if it rapidly oscillates near a, then the function
is not differentiable at a.
99
Rules of differentiation
11
Rules of di�erentiation
After completing this section, students should be able to do the
following.
• Use the definition of the derivative to develop shortcut
rules to find the derivatives of constants and constant
multiples.
• Use the definition of the derivative to develop shortcut
rules to find the derivatives of powers of x.
• Use the definition of the derivative to develop shortcut
rules to find the derivatives of sums and differences of
functions.
• Compute the derivative of polynomials.
• Recognize different notation for the derivative.
• State the derivative of the natural exponential function.
• State the derivative of the sine function.
100
Patterns in derivatives
Devyn: Let’s check it. If f (x) =
Break-Ground:
˘
®
11.1
f (x) = lim
Patterns in derivatives
hô0
Check out this dialogue between two calculus students (based
on a true story):
f (x)
f ® (x)
x2
2 x1
x3
3 x2
x4
4 x3
So maybe if we have a function
f (x) = xn
then
Devyn: Hmmm does it work with square roots?
Riley: Oh that’s right, a square root is a power, just write
f (x) =
x = x1_2 .
So a square root is of the form xn .
H˘
hô0
˘
h
x+h*
x
˘
x
h
hô0
= lim
x,
x+h*x
˘
˘
h( x + h + x)
˘
˘ I
x
˘
˘
x+h+ x
x+h+
h
˘
˘
h( x + h + x)
1
= lim ˘
˘
hô0
x+h+ x
1
= ˘
˘
x+ x
1
= ˘
2 x
1 *1_2
=
x
.
2
= lim
hô0
Riley: Holy Cat Fur! It works! In this case f ® (x) = n xn*1 .
Devyn: I wonder if it always works? If so I want to know why it
works! I wonder what other patterns we can find?
The pattern
f ® (x) = n xn*1 .
˘
x+h*
= lim
Devyn: I hate the limit definition of derivative. I wish there were a
shorter way.
Riley: I think I might have found a pattern for taking derivatives.
Devyn: Really? I love patterns!
Riley: I know! Check this out, I’ve made a chart
˘
if
f (x) = xn
then
f ® (x) = n xn*1
holds whenever n is a constant. Explaining why it works in
generality will take some time. For now, let’s see if we can use
the problem to squash some derivatives with ease.
Problem 1. Using the pattern found above, compute:
d 101
x
dx
Problem 2. Using the pattern found above, compute:
d 1
dx x77
101
Patterns in derivatives
Problem 3. Using the pattern found above, compute:
d ˘
5
x
dx
Problem 4. Using the pattern found above, compute:
d e
x
dx
102
Basic rules of differentiation
Dig-In:
11.2
Basic rules of di�erentiation
It is tedious to compute a limit every time we need to know
the derivative of a function. Fortunately, we can develop a
small collection of examples and rules that allow us to quickly
compute the derivative of almost any function we are likely to
encounter. We will start simply and build up to more complicated examples.
First, we introduce a different notation for the derivative which
may be more convenient at times.
Definition. There are several different notations for the derivative. The two we’ll mainly be using are
f ® (x) =
• If v(t) represents the velocity of an object with respect to
time and v(t) is constant, then the object’s acceleration is
d
zero. Hence v(t) = 0.
dt
The examples above lead us to our next theorem. To gain
intuition, you should compute the derivative of f (x) = 6 using
the limit definition of the derivative.
Theorem 16 (The constant rule). Given a constant c,
d
c = 0.
dx
From the limit definition of the derivative, write
d
c*c
c = lim
hô0 h
dx
0
= lim
hô0 h
= lim 0 = 0.
d
f (x).
dx
The value of the derivative at x = a can also be written using
two different notations
4
5
d
®
f (a) =
f (x)
.
dx
x=a
The constant rule
The simplest examples of functions, hence the best place to start
our investigation, are constant functions. Recall that derivatives
measure the rate of change of a function at a given point. This
means the derivative of a constant function is zero. Here are
some ways to think about this situation.
• The constant function plots a horizontal line—so the slope
of the tangent line is 0.
• If s(t) represents the position of an object with respect to
time and s(t) is constant, then the object is not moving, so
d
its velocity is zero. Hence s(t) = 0.
dt
hô0
Question 61. What is
equal to?
d
e
dx
The power rule
Next, let’s examine derivatives of powers of a single variable.
To gain intuition, you should compute the derivative of f (x) =
x3 using the limit definition of the derivative. Before computing
this derivative, we should recall the Binomial Theorem.
Theorem 17 (Binomial Theorem). If n is a nonnegative integer,
then
0 1
0
1
n n*1 1
n
(a + b)n = an b0 +
a b +5+
a1 bn*1 + a0 bn
1
n*1
where
0 1
n
n!
=
,
k
k!(n * k)!
k = 0, 1, ..., n.
103
Basic rules of differentiation
The Binomial Theorem tells us the pattern of the coefficients
for the expanded form of
(a + b) ,
n
which will help us prove our next derivative rule.
Example 63. Expand (x + h)3 using the Binomial Theorem.
We apply the Binomial Theorem to the expression in
question, then simplify.
0 1
0 1
3 3*1 1
3 3*2 2
3
3 0
(x + h) = x h +
x h +
x h + x0 h3
1
2
= x3 + 3x2 h + 3xh2 + h3
Theorem 18 (The power rule). For any real number n,
At this point we will only prove this theorem for values
of n which are positive integers. Later we will give the
complete explanation. From the limit definition of the
derivative, write with me
(x + h)n * xn
d n
x = lim
.
hô0
dx
h
Expand the term (x + h)n :
hô0
xn +
n
1
xn*1 h +
n
2
xn*2 h2 + 5 +
n
n*1
xhn*1 + hn * xn
h
Note
0 1 that we are using the Binomial Theorem to write
n
for the coefficients. Canceling the terms xn and
k
0 1
n
n
*x , and noting
= n, write with me
1
104
hô0
= lim
hô0
= lim
hô0
nxn*1 h +
n
2
⇠
h nxn*1 +
0
nx
n*1
xn*2 h2 + 5 +
n
n*1
xhn*1 + hn
h
n
2
xn*2 h + 5 +
n
n*1
xhn*2 + hn*1
⇡
h 0
0 1
1
1
n n*2
n
+
x h+5+
xhn*2 + hn*1
2
n*1
= nxn*1 ,
since every term but the first has a factor of h.
Therefore,
d n
x = nxn*1 .
dx
Let’s consider several examples. We begin with something
basic.
Example 64. Compute:
d n
x = nxn*1 .
dx
= lim
= lim
d 13
x
dx
Applying the power rule, we write
d 13
x = 13x12 .
dx
Sometimes, it is not as obvious that one should apply the power
rule.
Example 65. Compute:
d 1
dx x4
Applying the power rule, we write
d 1
d *4
=
x = *4x*5 .
dx x4
dx
Basic rules of differentiation
The power rule also applies to radicals once we rewrite them
as exponents.
f (x + h) + g(x + h) * (f (x) + g(x))
d
f (x) + g(x) = lim
hô0
dx
h
f (x + h) + g(x + h) * f (x) * g(x)
= lim
hô0
h
f (x + h) * f (x) + g(x + h) * g(x)
= lim
hô0
h
0
1
f (x + h) * f (x) g(x + h) * g(x)
= lim
+
hô0
h
h
f (x + h) * f (x)
g(x + h) * g(x)
= lim
+ lim
hô0
hô0
h
h
®
®
= f (x) + g (x).
Example 66. Compute:
d ˘
5
x
dx
Applying the power rule, we write
d ˘
d 1_5 x*4_5
5
x=
x =
.
dx
dx
5
The sum rule
We want to be able to take derivatives of functions “one piece
at a time.” The sum rule allows us to do exactly this. The sum
rule says that we can add the rates of change of two functions
to obtain the rate of change of the sum of both functions. For
example, viewing the derivative as the velocity of an object, the
sum rule states that to find the velocity of a person walking on
a moving bus, we add the velocity of the bus and the velocity
of the walking person.
Theorem 19 (The sum rule). If f (x) and g(x) are differentiable
and c is a constant, then
(a)
d
f (x) + g(x) = f ® (x) + g ® (x),
dx
(b)
d
f (x) * g(x) = f ® (x) * g ® (x),
dx
(c)
d
c f (x) = c f ® (x).
dx
We will only prove part (a) above; the rest are similar.
Write with me
Question 62. Using different notation for the derivative can
be confusing. Which of the following are the same as part (a)
of the Sum Rule?
Select All Correct Answers:
(a) (f (x) + g(x))® = f ® (x) + g ® (x)
(b) (f (x) + g(x))® =
(c)
d
d
f (x) +
g(x)
dx
dx
d
d
f (x) +
g(x) = f ® (x) + g ® (x)
dx
dx
(d) (f ® (x) + g(x))® =
d
(f (x) + g(x))
dx
Example 67. Suppose we have two functions, f , and g, and we
know that f ® (2) = 6 and g ® (2) = *3. What is the slope of the
tangent line to the graph y = f (x) + g(x) at x = 2?
The slope of the tangent line to the graph of the
function
f (x) + g(x)
at x = 2 is the derivative
4
5
d
. By the Sum Rule, this deriva(f (x) + g(x))
dx
x=2
tive is given by f ® (2) + g ® (2). In this case, we have
f ® (2) + g ® (2) = 3.
We now have the tools to work some more complicated examples.
105
Basic rules of differentiation
Example 68. Compute:
⇠
⇡
d
1
x5 +
dx
x
Write with me
⇠
⇡
d
1
d 5
d *1
x5 +
=
x +
x
dx
x
dx
dx
= 5x4 * x*2 .
Example 69. Compute:
H
I
˘
d
3
1
˘ *2 x+ 7
dx 3 x
x
Write with me
H
I
˘
d
3
1
˘ *2 x+ 7
dx 3 x
x
d *1_3
d
d *7
x
* 2 x1_2 +
x
dx
dx
dx
= *x*4_3 * x*1_2 * 7x*8 .
=3
106
The derivative of the natural exponential function
Dig-In:
11.3 The derivative of the natural
exponential function
We don’t know anything about derivatives that allows us to
compute the derivatives of exponential functions without getting our hands dirty. Let’s do a little work with the definition
of the derivative:
d x
ax+h * ax
a = lim
hô0
dx
h
ax ah * ax
= lim
hô0
h
h*1
a
= lim ax
hô0
h
ah * 1
x
= a lim
hô0
h
= ax (constant)
´≠≠Ø≠≠¨
ah * 1
lim
hô0
h
There are two interesting things to note here: We are left with
a limit that involves h but not x, which means that whatever
lim (ah * 1)_h is, provided it exists, we know that it is a numhô0
ber, or in other words, a constant. This means that ax has a
remarkable property:
The derivative of an exponential function is a constant times itself.
Unfortunately it is beyond the scope of this text to compute the
limit
ah * 1
lim
.
hô0
h
h
(2h * 1)_h
h
(2h * 1)_h
h
(3h * 1)_h
h
(3h * 1)_h
*1
*0.1
*0.01
*0.001
*0.0001
*0.00001
.5
˘ 0.6700
˘ 0.6910
˘ 0.6929
˘ 0.6931
˘ 0.6932
1
0.1
0.01
0.001
0.0001
0.00001
1
˘ 0.7177
˘ 0.6956
˘ 0.6934
˘ 0.6932
˘ 0.6932
*1
*0.1
*0.01
*0.001
*0.0001
*0.00001
˘ 0.6667
˘ 1.0404
˘ 1.0926
˘ 1.0980
˘ 1.0986
˘ 1.0986
1
0.1
0.01
0.001
0.0001
0.00001
2
˘ 1.1612
˘ 1.1047
˘ 1.0992
˘ 1.0987
˘ 1.0986
While these tables don’t prove that we have a pattern, it turns
out that
2h * 1
˘ .7
hô0
h
and
lim
3h * 1
˘ 1.1.
hô0
h
lim
Moreover, if you do more examples, choosing other values for
the base a, you will find that the limit varies directly with the
value of a: bigger a, bigger limit; smaller a, smaller limit. As
we can already see, some of these limits will be less than 1 and
some larger than 1. Somewhere between a = 2 and a = 3 the
limit will be exactly 1. This happens when
a = e = 2.718281828459045 … .
We will define the number e by this property in the next definition:
Definition. The number denoted by e, called Euler’s number,
is defined to be the number satisfying the following relation
eh * 1
= 1.
hô0
h
lim
Using this definition, we see that the function ex has the following truly remarkable property.
Theorem 20 (The derivative of the natural exponential function). The derivative of the natural exponential function is the
natural exponential function itself. In other words,
d x
e = ex .
dx
However, we can look at some examples. Consider (2h * 1)_h
and (3h * 1)_h:
107
The derivative of the natural exponential function
From the limit definition of the derivative, write
d x
ex+h * ex
e = lim
hô0
dx
h
ex eh * ex
= lim
hô0
h
h
e *1
= lim ex
hô0
h
h*1
e
= ex lim
hô0
h
x
=e 1
= ex .
Hence ex is its own derivative. In other words, the slope of the
plot of ex is the same as its height, or the same as its second
coordinate. Said another way, the function f (x) = ex goes
through the point (a, ea ) and has slope ea at that point, no
matter what a is.
Question 63. What is the slope of the tangent line to the graph
of the function f (x) = ex at x = 5?
Example 70. Compute:
⇠ ˘
⇡
d
8 x + 7ex
dx
Write with me:
⇠ ˘
⇡
d
d
d
8 x + 7ex = 8 x1_2 + 7 ex
dx
dx
dx
= 4x*1_2 + 7ex .
108
The derivative of sine
After these delicious appetizers, we are now ready for the main
course.
Dig-In:
11.4
The derivative of sine
It is now time to visit our two friends who concern themselves
periodically with triangles and circles. In particular, we want
to show that
d
sin(✓) = cos(✓).
d✓
Before we tackle this monster, let’s remember a fact, and derive
a new fact. You may initially be uncomfortable because you
can’t quite see why we need these results, but this style of
exposition is a fact of technical writing; it is best to get used to
it.
First, recall the fact that
Next, we will use this fact to derive our new fact:
cos(✓) * 1
= lim
✓ô0
✓ô0
✓
0
cos(✓) * 1 cos(✓) + 1
✓
cos(✓) + 1
cos2 (✓) * 1
✓ô0 ✓(cos(✓) + 1)
= lim
* sin2 (✓)
✓ô0 ✓(cos(✓) + 1)
0
1
sin(✓)
sin(✓)
= * lim
✓ô0
✓
cos(✓) + 1
sin(✓)
sin(✓)
= * lim
lim
✓ô0
✓ô0 cos(✓) + 1
✓
0
= *1
= 0.
2
= lim
sin(✓ + h) * sin(✓)
d
sin(✓) = lim
hô0
d✓
h
sin(✓) cos(h) + sin(h) cos(✓) * sin(✓)
h
1
0
sin(✓) cos(h) * sin(✓) sin(h) cos ✓
= lim
+
hô0
h
h
0
1
cos(h) * 1
sin(h)
= lim sin(✓)
+ cos(✓)
hô0
h
h
cos(h) * 1
sin(h)
= lim sin(✓)
+ lim cos(✓)
hô0
hô0
h
h
cos(h) * 1
sin(h)
= sin(✓) lim
+ cos(✓) lim
hô0
hô0
h
h
= sin(✓) 0 + cos(✓) 1
hô0
cos(✓) * 1
= 0.
✓ô0
✓
lim
Using the definition of the derivative, write with me
= lim
lim
Write with me:
d
sin(✓) = cos(✓).
d✓
Now we get sneaky and apply the trigonometric addition
formula for sine, that says sin(↵ + ) = sin(↵) cos( ) +
sin( ) cos(↵), to write
sin(✓)
lim
= 1.
✓ô0
✓
Example 71.
Theorem 21 (The derivative of sine). For any angle ✓ measured in radians, the derivative of sin(✓) with respect to ✓ is
cos(✓). In other words,
1
= cos(✓).
Question 64. What is the slope of the line tangent to the graph
of sin(✓) at ✓ = ⇡_4?
The slope of the line tangent to the graph of sin(✓) at
109
The derivative of sine
✓ = ⇡_4 is given by
4
5
˘
d
sin(✓)
= cos(⇡_4) = 2_2.
dx
✓=⇡_4
y
®
f (x)
*2⇡
*⇡
*⇡_2
f (x)
⇡_2
⇡
3⇡_2
2⇡
x
*1
For your intellectual stimulation, consider the following geometric interpretation of the derivative of sin(✓).
y
Question 65. Using the graph above, what is the value of x in
the interval [0, 2⇡] where the tangent to the graph of f (x) =
sin(x) has slope *1?
Pro-tip: When working with trigonometric functions, you
should always keep their graphical representations in mind.
h sin(✓)
˘h
✓
h cos(✓)
sin(✓)
h
✓
cos(✓)
x
From this diagram, we see that increasing ✓ by a small amount
h increases sin(✓) by approximately h cos(✓). Hence,
y h cos(✓)
˘
= cos(✓).
✓
h
With all of this said, the derivative of a function measures the
slope of the plot of a function. If we examine the graphs of
the sine and cosine side by side, it should be clear that the
latter appears to accurately describe the slope of the former,
and indeed this is true.
110
*3⇡_2
1
Product rule and quotient rule
12 Product rule and quotient
rule
After completing this section, students should be able to do the
following.
• Identify products of functions.
• Use the product rule to calculate derivatives.
• Identify quotients of functions.
• Use the quotient rule to calculate derivatives.
• Combine derivative rules to take derivatives of more complicated functions.
• Explain the signs of the terms in the numerator of the
quotient rule.
• Use the product and quotient rule to calculate derivatives
from a table of values.
111
Derivatives of products are tricky
Break-Ground:
12.1 Derivatives of products are
tricky
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Hey Riley, remember the sum rule for derivatives?
Riley: You know I do.
Devyn: What do you think that the “product rule” will be?
Riley: Let’s give this a spin:
d
(f (x) g(x)) = f ® (x) g ® (x)?
dx
Devyn: Hmmm, let’s give this theory an acid test. Let’s try
f (x) = x2 + 1
and
g(x) = x3 * 3x
Now
f ® (x)g ® (x) = (2x)(3x2 * 3)
= 6x3 * 6x.
Riley: On the other hand,
f (x)g(x) = (x2 + 1)(x3 * 3x)
= x5 * 3x3 + x3 * 3x
= x5 * 2x3 * 3x.
Devyn: And so,
d
(f (x) g(x)) = 5x4 * 6x2 * 3.
dx
Riley: Wow. Hmmm. It looks like our guess was incorrect.
Devyn: I’ve got a feeling that the so-called “product rule” might be
a bit tricky.
Problem 1. Above, our intrepid young mathematicians guess
that the “product rule” might be:
d
(f (x) g(x)) = f ® (x) g ® (x)?
dx
Does this ever hold true?
112
The Product rule and quotient rule
Dig-In:
12.2 The Product rule and quotient
rule
The product rule
Consider the product of two simple functions, say
f (x) g(x)
where f (x) = x2 + 1 and g(x) = x3 * 3x. An obvious guess
for the derivative of f (x)g(x) is the product of the derivatives:
f ® (x)g ® (x) = (2x)(3x2 * 3)
= 6x3 * 6x.
Is this guess correct? We can check by rewriting f and g and
doing the calculation in a way that is known to work. Write
with me
f (x)g(x) = (x2 + 1)(x3 * 3x)
= x5 * 3x3 + x3 * 3x
Let’s return to the example with which we started.
Example 72. Let f (x) = (x2 + 1) and g(x) = (x3 * 3x).
Compute:
d
f (x)g(x)
dx
Write with me
d
f (x)g(x) = f (x)g ® (x) + f ® (x)g(x)
dx
= (x2 + 1)(3x2 * 3) + (2x)(x3 * 3x).
We could stop here, but we should show that expanding
this out recovers our previous result. Write with me
(x2 + 1)(3x2 * 3) + 2x(x3 * 3x)
= 3x4 * 3x2 + 3x2 * 3 + 2x4 * 6x2
= 5x4 * 6x2 * 3,
which is precisely what we obtained before.
Now that we are pros, let’s try one more example.
Example 73. Compute:
d
(xex * ex )
dx
= x5 * 2x3 * 3x.
Hence
d
d 5
f (x)g(x) =
(x * 2x3 * 3x) = 5x4 * 6x2 * 3,
dx
dx
so we see that
d
f (x)g(x) ë f ® (x)g ® (x).
dx
So the derivative of f (x)g(x) is not as simple as f ® (x)g ® (x).
Never fear, we have a rule for exactly this situation.
Theorem 22 (The product rule). If f and g are differentiable,
then
d
f (x)g(x) = f (x)g ® (x) + f ® (x)g(x).
dx
Using the sum rule and the product rule, write with me
d
d
d x
e
(xex * ex ) =
(xex ) *
dx
dx
dx
x
x
x
= (xe + e ) * e
= xex .
The quotient rule
We’d like to have a formula to compute
d f (x)
dx g(x)
113
The Product rule and quotient rule
This brings us to our next derivative rule.
First, we’ll compute the derivative using the quotient rule.
Write with me
⇠˘ ⇡
⇠
⇡
2 ) 1 x*1_2
x
*
(625
*
x
(*2x)
2
d 625 * x2
=
.
˘
dx
x
x
Example 74. Compute:
Second, we’ll compute the derivative using the product
rule:
Theorem 23 (The quotient rule). If f and g are differentiable,
then
d f (x) f ® (x)g(x) * f (x)g ® (x)
=
.
dx g(x)
g(x)2
d x2 + 1
dx x3 * 3x
Write with me
2x(x3 * 3x) * (x2 + 1)(3x2 * 3)
d x2 + 1
=
dx x3 * 3x
(x3 * 3x)2
=
*x4 * 6x2 + 3
.
(x3 * 3x)2
Example 75. Compute:
d sin x
dx x
Write with me
d sin x cos x x * sin x 1
=
dx x
x2
It is often possible to calculate derivatives in more than one
way, as we have already seen. Since every quotient can be
written as a product, it is always possible to use the product
rule to compute the derivative, though it is not always simpler.
Example 76. Compute:
d 625 * x2
˘
dx
x
in two ways. First using the quotient rule and then using the
product rule.
114
d 625 * x2
d
=
625 * x2 x*1_2
˘
dx
dx
x
0 *3_2 1
*x
= 625 * x2
+ (*2x) x*1_2 .
2
With a bit of algebra, both of these simplify to
*
3x2 + 625
.
2x3_2
Example 77. Suppose we have two functions, f , and g, and
we know that f (4) = 3, f ® (4) = 5, g(4) = *2, and g ® (4) = 2.
f (x)
What is the slope of the tangent line to the curve y =
at
g(x)
the point where x = 4?
f (x)
The slope of the tangent line to the curve y =
at
g(x)
4
5
d f (x)
x = 4 is given by
.
dx g(x) x=4
The Product rule and quotient rule
By the QuotientRule, this derivative is given by
4
5
4 ®
5
f (x)g(x) * f (x)g ® (x)
d f (x)
=
dx g(x) x=4
(g(x))2
x=4
f ® (4)g(4) * f (4)g ® (4)
=
(g(4))2
5(*2) * 3 2
=
(*2)2
*10 * 6
=
4
=4
Example 78. Suppose we have two functions, f , and g, and
we know that f (4) = 3, f ® (4) = 5, g(4) = *2, and g ® (4) = 2.
xf (x)
What is the slope of the tangent line to the curve y =
at
g(x)
the point where x = 4?
xf (x)
The slope of the tangent line to the curve y =
at
g(x)
4
5
d xf (x)
.
x = 4 is given by
dx g(x) x=4
By the QuotientRule and the Product Rule, this derivative
is given by
4
5
4
5
(xf ® (x) + f (x))g(x) * xf (x)g ® (x)
d xf (x)
=
dx g(x) x=4
(g(x))2
x=4
®
®
(4f (4) + f (4))g(4) * 4f (4)g (4)
=
(g(4))2
(4 5 + 3)(*2) * 4 3 2
=
(*2)2
(4 5 + 3)(*2) * 4 3 2
=
(*2)2
*35
=
2
115
Chain rule
13
Chain rule
After completing this section, students should be able to do the
following.
• Recognize a composition of functions.
• Take derivatives of compositions of functions using the
chain rule.
• Take derivatives that require the use of multiple rules of
differentiation.
• Use the chain rule to calculate derivatives from a table of
values.
• Understand rate of change when quantities are dependent
upon each other.
• Use order of operations in situations requiring multiple
rules of differentiation.
• Apply chain rule to relate quantities expressed with different units.
• Compute derivatives of trigonometric functions.
• Use multiple rules of differentiation to calculate derivatives from a table of values.
116
An unnoticed composition
(c) Riley was working in degrees, not radians.
Break-Ground:
13.1
An unnoticed composition
Check out this dialogue between two calculus students (based
on a true story):
(d) cos(0) ë 1
(e) Riley computed the slope incorrectly.
Devyn: Riley! Something is bothering me.
Riley: What is it?
Devyn: I have broken calculus.
Riley: How?
Devyn: Check this out, we know that:
d
sin(✓) = cos(✓)
d✓
Riley: Right.
Devyn: But I also know that the derivative is the slope of the tangent
line at a point.
Riley: Right!
Devyn: But check out this plot of sin(✓) that’s zoomed in around
zero:
1
*6
*5
*4
*3
*2
*1
1
2
3
4
5
6
*1
Riley: Ok. . .
Devyn: Well, cos(0) = 1, but the slope of this line is totally not one!
What’s going on here?
This problem that Riley and Devyn are having is somewhat
subtle.
Question 66. What mistake is being made?
Multiple Choice:
(a) Riley did not plot sin(✓).
(b) Riley did not take the derivative correctly.
117
The chain rule
Hence
Dig-In:
13.2
The chain rule
So far we have seen how to compute the derivative of a function
built up from other functions by addition, subtraction, multiplication and division. There is another very important way that
we combine functions: composition. The chain rule allows us
to deal with this case. Consider
h(x) = sin(1 + 2x).
While there are several different ways to differentiate this function, if we let f (x) = sin(x) and g(x) = 1 + 2x, then we can
express h(x) = f (g(x)). The question is, can we compute the
derivative of a composition of functions using the derivatives
of the constituents f (x) and g(x)? To do so, we need the chain
rule.
Theorem 24 (Chain Rule). If f and g are differentiable, then
d
f (g(x)) = f ® (g(x))g ® (x).
dx
It will take a bit of practice to make the use of the chain rule
come naturally, it is more complicated than the earlier differentiation rules we have seen. Let’s return to our motivating
example.
Example 79. Compute:
d
sin(1 + 2x)
dx
Set f (x) = sin(x) and g(x) = 1 + 2x, now
f ® (x) = cos(x)
and
g ® (x) = 2.
d
d
sin(1 + 2x) =
f (g(x))
dx
dx
= f ® (g(x))g ® (x)
= cos(1 + 2x) 2
= 2 cos(1 + 2x).
Let’s see a more complicated chain of compositions.
Example 80. Compute:
d
dx
Set f (x) =
t
1+
˘
˘
x and g(x) = 1 + x. Hence,
t
˘
1 + x = f (g(f (x)))
and by the chain rule we know
Since
d
f (g(f (x))) = f ® (g(f (x)))g ® (f (x))f ® (x).
dx
1
f ® (x) = ˘
2 x
and
g ® (x) = 1
We have that
t
˘
d
1
1
1+ x= t
1 ˘ .
˘
dx
2 x
2 1+ x
The chain rule allows to differentiate compositions of functions
that would otherwise be difficult to get our hands on.
Example 81. Compute:
d sin(x2 )
e
dx
118
x
The chain rule
Set f (x) = ex , g(x) = sin(x), and h(x) = x2 so that
2
f (g(h(x))) = esin(x ) . Now
d sin(x2 )
d
e
=
f (g(h(x)))
dx
dx
= f ® (g(h(x))) g ® (h(x)) h® (x)
2)
= esin(x
cos(x2 ) 2x.
Using the chain rule, the power rule, and the product rule it is
possible to avoid using the quotient rule entirely.
Example 82. Compute:
d x3
dx x2 + 1
Rewriting this as
d 3 2
x (x + 1)*1 ,
dx
set f (x) = x*1 and g(x) = x2 + 1 so that f (g(x)) =
(x2 + 1)*1 . Now
x3 (x2 + 1)*1 = x3 f (g(x)),
and by the product and chain rules
d 3
x f (g(x)) = 3x2 f (g(x)) + x3 f ® (g(x))g ® (x).
dx
Since f ® (x) =
*1
and g ® (x) = 2x, write
x2
d x3
3x2
2x4
=
*
.
2
2
2
dx x + 1 x + 1 (x + 1)2
119
Derivatives of trigonometric functions
Dig-In:
13.3 Derivatives of trigonometric
functions
Up until this point of the course we have been ignoring a large
class of functions: Trigonometric functions other than sin(x).
We know that
d
sin(x) = cos(x).
dx
Armed with this fact we will discover the derivatives of all of
the standard trigonometric functions.
Theorem 25 (The derivative of cosine).
d
cos(x) = * sin(x).
dx
4
d
cos
dx
0
x3
2
15
˘
x= 3 ⇡
40
0
0 3 111 5
3 2
x
x * sin
˘
2
2
x= 3 ⇡
⇠ ⇡
3 ˘
⇡
= * ( 3 ⇡)2 sin
2
2
3 23
=* ⇡ 1
2
=
*3⇡ 3
=
.
2
⇠
⇡
⇡
* x , and
2
⇠
⇡
⇡
• sin(x) = cos
*x .
2
• cos(x) = sin
Next we have:
Theorem 26 (The derivative of tangent).
⇡
d
d
⇡
cos(x) =
sin
*x
dx
dx ⇠ 2 ⇡
⇡
= * cos
*x
2
= * sin(x).
d
tan(x) = sec2 (x).
dx
⇠
Example 83. Compute:
4
0 315
d
x
cos
˘
dx
2
x= 3 ⇡
120
and so
2
Recall that
Now
Now that we know the derivative of cosine, we may combine this with the chain rule, so we have that
0 31
0
0 3 11
d
x
3x2
x
cos
=
* sin
dx
2
2
2
We’ll rewrite tan(x) as
Write with me:
sin(x)
and use the quotient rule.
cos(x)
d
d sin(x)
tan(x) =
dx
dx cos(x)
cos2 (x) + sin2 (x)
cos2 (x)
1
=
cos2 (x)
= sec2 (x).
=
Derivatives of trigonometric functions
Example 84. Compute:
d
dx
0
5x tan(x)
x2 * 3
Theorem 28 (The Derivatives of Trigonometric Functions).
1
Applying the quotient rule, and the product rule, and the
derivative of tangent:
0
1
d 5x tan(x)
dx
x2 * 3
=
=
(x2 * 3)
(x2
d
(5x tan(x)) * 5x tan(x)
dx
(x2 * 3)2
d
(x2
dx
* 3)(5 tan(x) + 5x sec2 (x)) * 5x tan(x)
* 3)
2x
(x2 * 3)2
2
5(x * 3)(tan(x) + x sec2 (x)) * 10x2 tan(x)
=
(x2 * 3)2
Finally, we have:
Theorem 27 (The derivative of secant).
d
sec(x) = sec(x) tan(x).
dx
We’ll rewrite sec(x) as (cos(x))*1 and use the power rule
and the chain rule. Write
d
d
sec(x) =
(cos(x))*1
dx
dx
= *1(cos(x))*2 (* sin(x))
sin(x)
=
cos2 (x)
sin(x)
1
=
cos(x) cos(x)
= sec(x) tan(x).
The derivatives of the cotangent and cosecant are similar and
left as exercises. Putting this all together, we have:
•
d
sin(x) = cos(x).
dx
•
d
cos(x) = * sin(x).
dx
•
d
tan(x) = sec2 (x).
dx
•
d
sec(x) = sec(x) tan(x).
dx
•
d
csc(x) = * csc(x) cot(x).
dx
•
d
cot(x) = * csc2 (x).
dx
Example 85. Compute:
4
5
d
(csc(x) cot(x))
dx
x= ⇡
3
Applying the product rule and the facts above, we know
that
d
(csc(x) cot(x)) = * csc3 (x) * cot 2 (x) csc(x)
dx
and so
4
5
d
(csc(x) cot(x))
dx
x= ⇡
3
4
=
3
2
* csc (x) * cot (x) csc(x)
˘
8
1
=* ˘ *
2_ 3
3 3 3
5
x= ⇡3
Warning. When working with derivatives of trigonometric
functions, we suggest you use radians for angle measure. For
121
Derivatives of trigonometric functions
example, while
2
sin (90˝ ) = sin
0⇠ ⇡ 1
⇡ 2
,
2
one must be careful with derivatives as
4
5
d
2
sin x
ë 2 90 cos(902 )
dx
´≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠¨
x=90˝
incorrect
Alternatively, one could think of x˝ as meaning
90˝ =
⇡
90 ⇡
= . In this case
180
2
2 90
122
˝
⇡
cos((90 ) ) = 2
cos
2
˝ 2
x ⇡
, as then
180
0⇠ ⇡ 1
⇡ 2
.
2
Higher order derivatives and graphs
14 Higher order derivatives
and graphs
After completing this section, students should be able to do the
following.
• Use the first derivative to determine whether a function is
increasing or decreasing.
• Define higher order derivatives.
• Compare differing notations for higher order derivatives.
• Identify the relationships between the function and its first
and second derivatives.
• Sketch a graph of the second derivative, given the original
function.
• Sketch a graph of the original function, given the graph
of its first and second derivatives.
• Sketch a graph of a function satisfying certain constraints
on its higher-order derivatives.
• State the relationship between concavity and the second
derivative.
• Interpret the second derivative of a position function as
acceleration.
• Calculate higher order derivatives.
123
Rates of rates
Multiple Choice:
Break-Ground:
14.1
Rates of rates
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, check out this plot:
A
B
C
Riley: Whoa, that looks cool!
Devyn: I know! Anyway, someone told me that this was a graph of
a function, its derivative, and the derivative of the derivative.
Which one is which?
Riley: Hmmmmm. I’m not sure.
Problem 1. Which of the following is true?
Multiple Choice:
(a) Curve A is increasing when curve B is positive.
(b) Curve A is increasing when curve C is positive.
(c) None of the above.
Problem 2. Which of the following is true?
Multiple Choice:
(a) Curve B is increasing when curve A is positive.
(b) Curve B is increasing when curve C is positive.
(c) None of the above.
Problem 3. Which of the following is true?
124
(a) Curve C is increasing when curve A is positive.
(b) Curve C is increasing when curve B is positive.
(c) None of the above.
Higher order derivatives and graphs
The function f is not increasing on the interval (*ÿ, b),
because if we pick a pair of numbers from (*ÿ, b), say,
x1 = a, and x2 = 0, then x1 < x2 , but f (x1 ) > f (x2 ).
Dig-In:
14.2 Higher order derivatives and
graphs
Definition. We say that a function f is increasing on an
interval I if f (x1 ) < f (x2 ), for all pairs of numbers x1 , x2 in
I such that x1 < x2 .
We say that a function f is decreasing on an interval I if
f (x1 ) > f (x2 ), for all pairs of numbers x1 , x2 in I such that
x1 < x2 .
Question 67. Consider the graph of the function f below:
f
b
On which of the following intervals is f increasing?
(a) (*ÿ, a)
(b) (*ÿ, b)
(c) (a, b)
(d) (a, +ÿ)
(e) (b, +ÿ)
(a) sin (x)
(b) sin (2x)
(c) cos (x)
(e) cot (x)
f
Select All Correct Answers:
Select All Correct Answers:
(d) tan (x)
y
a
Question 68. ⇠Which
⇡ of the following famous functions are
⇡
increasing on 0,
?
2
x
(f) f (x) = x2
(g) g(x) =
1
x
⇡
⇡
,
2
⇡
because if we take a pair of numbers in I, say, x1 = ,
6
⇡
and x2 = , then x1 < x2 , but
3
˘
⇠ ⇡
3
⇡
f (x1 ) > f (x2 ), since f (x1 ) = cos
=
, and
6
2
⇠ ⇡
⇡
1
f (x2 ) = cos
= .
3
2
The function cos (x) is not increasing on I =
⇠
0,
Since the derivative gives us a formula for the slope of a tangent
line to a curve, we can gain information about a function purely
from the sign of the derivative. In particular, we have the
following theorem
Theorem 29. A function f is increasing on any interval I
where f ® (x) > 0, for all x in I.
125
Higher order derivatives and graphs
A function f is decreasing on any interval I where f ® (x) < 0,
for all x in I.
A
B
C
Question 69. Below we have graphed y = f ® (x):
5
*2
*1.5
*1
*0.5
0.5
1
1.5
2
*5
Is the function f increasing or decreasing on the interval *1 <
x < 0?
From the graph of f ® we can see that f ® (x) > 0 for all
x in (*1, 0). Then, the Theorem above implies that the
function f is increasing on this interval.
We call the derivative of the derivative the second derivative,
the derivative of the second derivative (the derivative of the
derivative of the derivative) the third derivative, and so on.
We have special notation for higher derivatives, check it out:
First derivative:
B ® = C.
Hence f = A, f ® = B, and f ®® = C.
Example 87. Here we have unlabeled graphs of f , f ® , and
f ®® :
A
B
C
d2
f (x) = f ®® (x) = f (2) (x).
dx2
d3
f (x) = f ®®® (x) = f (3) (x).
dx3
We use the facts above in our next example.
Example 86. Here we have unlabeled graphs of f , f ® , and
f ®® :
126
Since B is increasing when C is positive and decreasing
when C is negative, we see
d
f (x) = f ® (x) = f (1) (x).
dx
Second derivative:
Third derivative:
Identify each curve above as a graph of f , f ® , or f ®® .
Here we see three curves, A, B, and C. Since A is increasing when B is positive and decreasing when B is
negative, we see
A® = B.
Identify each curve above as a graph of f , f ® , or f ®® .
Here we see three curves, A, B, and C. Since B is increasing when A is positive and decreasing when A is
negative, we see
B ® = A.
Higher order derivatives and graphs
Since A is increasing when C is positive and decreasing
when C is negative, we see
A® = C.
Hence f = B, f ® = A, and f ®® = C.
Example 88. Here we have unlabeled graphs of f , f ® , and
f ®® :
A
B
C
Identify each curve above as a graph of f , f ® , or f ®® .
Here we see three curves, A, B, and C. Since C is increasing when B is positive and decreasing when B is
negative, we see
C ® = B.
Since B is increasing when A is positive and decreasing
when A is negative, we see
B ® = A.
Hence f = C, f ® = B, and f ®® = A.
127
Concavity
f ® (x) < 0
Dig-In:
14.3
Concavity
The graphs of two functions, f and g, both increasing on the
given interval, are given below.
The function f is increasing,
while the rate itself is decreasing.
In this case the curve y = f (x)is
concave down.
The function g is increasing,
while the rate itself is increasing.
In this case the curve y = g(x)is
concave up.
Definition. Let f be a function differentiable on an open interval I.
• We say that the graph of f is concave up on I if f ® , the
derivative of f , is increasing on I.
• We say that the graph of f is concave down on I if f ® ,
the derivative of f , is decreasing on I.
We know that the sign of the derivative tells us whether a
function is increasing or decreasing at some point. Likewise,
the sign of the second derivative f ®® (x) tells us whether f ® (x) is
increasing or decreasing at x. If we are trying to understand the
shape of the graph of a function, knowing where it is concave
up and concave down helps us to get a more accurate picture.
This is summarized in a single theorem.
Theorem 30 (Test for Concavity). Let I be an open interval.
• If f ®® (x) > 0 for all x in I, then the graph of f is concave
up on I.
• If f ®® (x) < 0 for all x in I, then the graph of f is concave
down on I.
We summarize the consequences of this theorem in the table
below:
128
0 < f ® (x)
0 < f ®® (x)
f ®® (x) < 0
The function f is decreasing, while the rate itself is
increasing. In this case the
curve y = f (x) is concave
up.
The function f is increasing, while the rate itself is
increasing. In this case the
curve y = f (x) is concave
up.
The function f is decreasing, while the rate itself is
decreasing. In this case the
curve y = f (x) is concave
down.
The function f is increasing, while the rate itself is
decreasing. In this case the
curve y = f (x) is concave
down.
Example 89. Let f be a continuous function and suppose that:
• f ® (x) > 0 for *1 < x < 1.
• f ® (x) < 0 for *2 < x < *1 and 1 < x < 2.
• f ®® (x) > 0 for *2 < x < 0 and 1 < x < 2.
• f ®® (x) < 0 for 0 < x < 1.
Sketch a possible graph of f .
Start by marking points in the domain where the derivative changes sign and indicate intervals where f is increasing and intervals f is decreasing. The function f
has a negative derivative from *2 to x = *1. This means
that f is decreasing on this interval. The function f has
a positive derivative from x = *1 to x = 1. This means
that f is increasing on this interval. Finally, The function
f has a negative derivative from x = 1 to 2. This means
that f is decreasing on this interval.
Concavity
2
1
*2
*1.5
*1
*0.5
0.5
1
1.5
2
*1
*2
Now we should sketch the concavity: concave up when
the second derivative is positive, concave down when the
second derivative is negative.
2
1
*2
*1.5
*1
*0.5
0.5
1
1.5
2
0.5
1
1.5
2
*1
*2
Finally, we can sketch our curve:
2
1
*2
*1.5
*1
*0.5
*1
*2
129
Position, velocity, and acceleration
Dig-In:
14.4 Position, velocity, and
acceleration
Studying functions and their derivatives might seem somewhat
abstract. However, consider this passage from a physics book:
Assuming acceleration a is constant, we may write
velocity and position as
v(t) = v0 + at,
x(t) = x0 + v0 t + (1_2)at2 ,
where a is the (constant) acceleration, v0 is the velocity at time zero, and x0 is the position at time
zero.
These equations model the position and velocity of any object
with constant acceleration. In particular these equations can be
used to model the motion a falling object, since the acceleration
due to gravity is constant.
Calculus allows us to see the connection between these equations. First note that the derivative of the formula for position
with respect to time, is the formula for velocity with respect to
time.
x® (t) = v0 + at = v(t).
Moreover, the derivative of formula for velocity with respect
to time, is simply a, the acceleration.
Question 70. Suppose that s represents the position of a ball
tossed at time t = 0. Recalling that the acceleration for t > 0
is only due to gravity, and knowing that the acceleration due
to gravity is *9.8 m_s2 , what is s®® (t)?
Example 90. You recently took a road trip from Columbus
Ohio to Urbana-Champaign Illinois. The distance traveled
from Columbus Ohio is roughly modeled by:
s(t) = 36t2 * 4.8t3
(miles West of Columbus)
where t is measured in hours, and is between 0 and 6. Find a
formula for your acceleration.
130
Here we simply need to find the second derivative:
s® (t) = 72t * 14.4t2
and
s®® (t) = 72 * 28.8t
Hence our acceleration is 72 * 28t miles/hour2 .
Implicit differentiation
15
Implicit di�erentiation
After completing this section, students should be able to do the
following.
• Implicitly differentiate expressions.
• Find the equation of the tangent line for curves that are
not plots of functions.
• Understand how changing the variable changes how we
take the derivative.
• Understand the derivatives of expressions that are not
functions or not solved for y.
• Explain the formula for the derivative of the natural log
function.
• Calculate derivatives of logs in any base.
131
Standard form
The point
Break-Ground:
15.1
Standard form
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, I think we’ve been too explicit with each other. We
should try to be more implicit.
Riley: I. Um. Don’t really. . .
Devyn: I mean when plotting things!
Riley: Okay, but I still have no idea what you are talking about.
Devyn: Remember when we first learned the equation of a line, and
the “standard form” was
ax + by = c
or something, which is totally useless for graphing. Also a circle
is
x2 + y2 = r2
or something, and here y isn’t even a function of x.
Riley: Ah, I’m starting to remember. We can write the same thing in
two ways. For example, if you write
y = mx + b,
then y is explicity a function of x but if you write
ax + by = c,
then y is implicitly a function of x.
Devyn: What I’m trying to say is that we need to learn how to work
with these “implicit” functions.
Problem 1. Consider the unit circle
x2 + y2 = 1.
The point P = (0, 1) is on this circle. Reason geometrically to
determine the slope of the line tangent to x2 + y2 = 1 at P .
Problem 2. Consider the unit circle
x2 + y2 = 1.
132
H˘
P =
˘ I
2
2
,
2
2
is on this circle. Reason geometrically to determine the slope
of the line tangent to x2 + y2 = 1 at P .
Implicit differentiation
Dig-In:
15.2
Implicit di�erentiation
with such situations, we use implicit differentiation. We’ll start
with a basic example.
Example 91. Consider the curve defined by:
Review of the chain rule
x2 + y2 = 1
Implicit differentiation is really just an application of the chain
rule. So recall:
Theorem 31 (Chain Rule). If f (x) and g(x) are differentiable,
then
d
f (g(x)) = f ® (g(x))g ® (x).
dx
Of particular use in this section is the following. If y is a
differentiable function of x and if f is a differentiable function,
then
dy
d
d
(f (y)) = f ® (y)
(y) = f ® (y) .
dx
dx
dx
Implicit di�erentiation
The functions we’ve been dealing with so far have been explicit
functions, meaning that the dependent variable is written in
terms of the independent variable. For example:
2
y = 3x * 2x + 1,
3x
y=e ,
x*2
y=
.
2
x * 3x + 2
However, there is another type of function, called an implicit
function. In this case, the dependent variable is not stated
explicitly in terms of the independent variable. Some examples
are:
x2 +y2 = 4,
x3 +y3 = 9xy,
x4 +3x2 = x2_3 +y2_3 +1.
Your inclination might be simply to solve each of these for
y and go merrily on your way. However this can be difficult
and it may require two branches, ˘
for example to explicitly
˘ plot
x2 + y2 = 4, one needs both y = 4 * x2 and y = * 4 * x2 .
Moreover, it may not even be possible to solve for y. To deal
(a) Compute
dy
.
dx
H˘ ˘ I
2
2
(b) Find the slope of the tangent line at
,
.
2
2
Starting with
x2 + y2 = 1
we differentiate both sides of the equation with respect
to x to obtain
d
d
x2 + y2 =
1.
dx
dx
Applying the sum rule we see
d 2
d 2
x +
y = 0.
dx
dx
Let’s examine each of these terms in turn. To start
d 2
x = 2x.
dx
On the other hand,
d 2
y is somewhat different. Here
dx
you imagine that y = f (x), and hence by the chain rule
d 2
d
y =
(f (x))2
dx
dx
= 2 f (x) f ® (x)
dy
= 2y .
dx
133
Implicit differentiation
Putting this together we are left with the equation
2x + 2y
At this point, we solve for
(a) Compute
dy
=0
dx
(b) Find the slope of the tangent line at (4, 2).
Starting with
dy
. Write
dx
d
d
x3 + y3 =
9xy.
dx
dx
Applying the sum rule we see
For
˘ the second
˘part of the problem, we simply plug x =
2
2
and y =
into the formula above, hence the slope
2
2
of the tangent line at this point is *1.
Let’s see another illustrative example:
2
*4
*6
134
Considering the final term
2
*2
d 3
x = 3x2 .
dx
d 3
d
y =
(f (x))3
dx
dx
= 3(f (x))2 f ® (x)
dy
= 3y2 .
dx
y
4
*2
Let’s examine each of these terms in turn. To start
d 3
y is somewhat different. Here you
dx
imagine that y = f (x), and hence by the chain rule
x3 + y3 = 9xy
*4
d 3
d 3
d
x +
y =
9xy.
dx
dx
dx
On the other hand
Example 92. Consider the curve defined by:
*6
x3 + y3 = 9xy,
we differentiate both sides of the equation with respect
to x to obtain
dy
=0
dx
dy
2y
= *2x
dx
dy
*x
=
.
dx
y
2x + 2y
6
dy
.
dx
4
6
x
y = f (x). Hence
d
9xy, we again imagine that
dx
⇠
⇡
d
d
9xy = 9
x f (x)
dx
dx
= 9 x f ® (x) + f (x)
dy
= 9x
+ 9y.
dx
Implicit differentiation
Putting this all together we are left with the equation
3x2 + 3y2
dy
You might think that the step in which we solve for
could
dx
sometimes be difficult. In fact, this never happens. All occurdy
rences
arise from applying the chain rule, and whenever the
dx
dy
chain rule is used it deposits a single
multiplied by some
dx
dy
other expression. Hence our expression is linear in
, it will
dx
dy
always be possible to group the terms containing
together
dx
dy
and factor out the
, just as in the previous examples.
dx
One more last example:
dy
dy
= 9x
+ 9y.
dx
dx
At this point, we solve for
dy
. Write
dx
dy
dy
= 9x
+ 9y
dx
dx
dy
dy
3y2
* 9x
= 9y * 3x2
dx
dx
dy
3y2 * 9x = 9y * 3x2
dx
dy
9y * 3x2
3y * x2
=
=
.
2
dx 3y * 9x y2 * 3x
3x2 + 3y2
Example 93. Consider the curve defined by
For the second part of the problem, we simply plug x = 4
and y = 2 into the formula above, hence the slope of the
5
tangent line at (4, 2) is . We’ve included a plot for your
4
viewing pleasure:
6
y
4
2
*6
*4
*2
2
*2
*4
4
6
x
cos(xy) *
Compute
y
= 4x2 y3 .
x
dx
.
dy
dy
dx
, not
. So
dy
dx
we are considering x as a function of y. This means the
variables have changed places! Not to worry, everything
d
is exactly the same. We apply
to both sides of the
dy
equation to get
⇠
y⇡
d
d
cos(xy) *
=
(4x2 y3 )
dy
x
dy
First, notice that the problem asks for
which gives us
0
1 x * y dx
dy
dx
dx
* sin(xy) y
+x *
= 8xy3
+ 12x2 y2 .
2
dy
dy
x
*6
135
Implicit differentiation
Distributing and multiplying by x2 yields
*x2 y sin(xy)
dx
dx
* x3 sin(xy) * x + y
dy
dy
dx
= 8x3 y3
+ 12x4 y2 .
dy
Grouping terms, factoring, and dividing finally gives us
*x2 y sin(xy)
dx
dx
dx
+y
* 8x3 y3
dy
dy
dy
= x3 sin(xy) + x + 12x4 y2
so,
y * x2 y sin(xy) * 8x3 y3
dx
= x3 sin(xy)+x+12x4 y2
dy
and now we see
dx x3 sin(xy) + x + 12x4 y2
=
.
dy
y * x2 y sin(xy) * 8x3 y3
136
Derivatives of inverse exponential functions
Hence
Dig-In:
15.3 Derivatives of inverse
exponential functions
Geometrically, there is a close relationship between the plots
of ex and ln(x), they are reflections of each other over the line
y = x:
6
y
ey = x
d y
d
e =
x
dx
dx
dy
ey
=1
dx
dy
1
1
= y =
dx e
x
2
e
*4
*2
2
*2
Solve for
4
6
x
ln(x)
d
(* ln(cos(x)))
dx
From the derivative of the natural logarithm, we can deduce
another fact:
*4
Theorem 33 (The derivative of any logarithm). Let b be a
positive real number. Then
*6
d
1
logb (x) =
.
dx
x ln(b)
d x
e = ex , to dedx
duce the derivative of ln(x). We will use implicit differentiation
to exploit this relationship computationally.
Here we need to remember that
Theorem 32 (The Derivative of the Natural Logrithm).
So we may write
One may suspect that we can use the fact that
d
1
ln(x) = .
dx
x
exactly when
ey = x
logb (x) =
ln(x)
.
ln(b)
d
d ln(x)
logb (x) =
dx
dx ln(b)
1
=
.
x ln(b)
Recall
ln(x) = y
dy
.
dx
Question 71. Compute:
x
*6
Implicit differentiation.
d
1
ln(x) = .
dx
x
Since y = ln(x),
4
Differentiate both sides.
and
x > 0.
Question 72. Compute:
d
log7 (x)
dx
137
Derivatives of inverse exponential functions
We can also compute the derivative of an arbitrary exponential
function.
Theorem 34 (The derivative of an exponential function).
d x
a = ax ln(a).
dx
Here we need to be slightly sneaky. Note
ax = eln(a ) = ex ln(a) .
x
So we may write
d x
d x ln(a)
a =
e
dx
dx
= ex ln(a) ln(a)
= ax ln(a).
Question 73. Compute:
d x
7
dx
138
Logarithmic differentiation
16
Logarithmic di�erentiation
After completing this section, students should be able to do the
following.
• Identify situations where logs can be used to help find
derivatives.
• Use logarithmic differentiation to simplify taking derivatives.
• Take derivatives of functions raised to functions.
• Recognize the difference between a variable as the base
and a variable as the exponent.
139
Multiplication to addition
Problem 3. Compute
Break-Ground:
16.1
Multiplication to addition
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, why is the product rule so much harder than the sum
rule?
Riley: Ever since 2nd grade, I’ve known that multiplication is harder
than addition.
Devyn: I know! I was reading somewhere that a slide-rule somehow
turns “multiplication into addition.”
Riley: Wow! I wonder how that works?
Devyn: I think it has something to do with logs?
Riley: What? How does this work?
Devyn is right, logarithms are used (and were invented) to
convert difficult multiplication problems into simpler addition
problems.
˘
Problem 1. Let f (x) = x cos(x) ex . Compute
d
f (x)
dx
Now, let’s see what happens if we do the same problem but we
take the natural log of both sides first:
˘
x cos(x) ex
˘
ln(f (x)) = ln( x cos(x) ex )
1
ln(f (x)) = ln (x) + ln(cos(x)) + ln(ex )
2
f (x) =
Now we’ll take the derivative of both sides of the equation. By
the chain rule
f ® (x)
d
ln(f (x)) =
dx
f (x)
Problem 2. Compute
d
ln(x)
dx
140
d
ln(cos(x))
dx
Problem 4. Compute
d
ln(ex )
dx
So we have
f ® (x)
sin(x)
1
=
*
+1
f (x)
2x cos(x)
0
1
sin(x)
1
®
f (x) = f (x)
*
+1
2x cos(x)
0
1
˘
sin(x)
1
x
= x cos(x)e
*
+1
2x cos(x)
Logarithmic differentiation
From the table, we see that
Dig-In:
Logarithmic di�erentiation
Logarithms were originally developed as a computational tool.
The key fact that made this possible is that:
logb (xy) = logb (x) + logb (y).
log10 (1.38) ˘ 0.1399
and
log10 (2.34) ˘ 0.3692
Add these numbers together to get 0.5091. Essentially,
we know the following at this point:
˘
0.5091 =
2
1
2
3
4
5
6
7
x
˘
log10 (?) = log10 (1.38) + log10 (2.34)
y
˘
16.2
0.1399
+
0.3692
Using the table again, we see that log10 (3.23) ˘ 0.5091.
Since we were working in scientific notation, we need to
multiply this by 103 . Our final answer is
3230 ˘ 138 23.4
Since 138 23.4 = 3229.2, this is a good approximation.
*2
The moral is:
*4
Before the days of calculators and computers, this was critical
knowledge for anyone in a computational discipline.
Example 94. Compute 138 23.4 using logarithms.
Start by writing both numbers in scientific notation
1.38 102
2.34 101 .
Next we use a log-table, which gives log10 (N) for values
of N ranging between 0 and 9. We’ve reproduced part
of such a table below.
N
0
1
2
3
4
5
6
7
8
9
1.3 0.1139 0.1173 0.1206 0.1239 0.1271 0.1303 0.1335 0.1367 0.1399 0.1430
2.3 0.3617 0.3636 0.3655 0.3674 0.3692 0.3711 0.3729 0.3747 0.3766 0.3784
3.2 0.5052 0.5065 0.5079 0.5092 0.5105 0.5119 0.5132 0.5145 0.5159 0.5172
Logarithms allow us to use addition in place of
multiplication.
Logarithmic di�erentiation
When taking derivatives, both the product rule and the quotient
rule can be cumbersome to use. Logarithms will save the day.
A key point is the following
f ® (x)
d
1
ln(f (x)) =
f ® (x) =
dx
f (x)
f (x)
which follows from the chain rule. Let’s look at an illustrative
example to see how this is actually used.
Example 95. Compute:
d x9 e4x
˘
dx x * 4
141
Logarithmic differentiation
Example 96. For x > 0, compute
Recall the properties of logarithms:
d x
x
dx
• logb (xy) = logb (x) + logb (y)
The function xx is tricky to differentiate. We cannot
use the power rule, as the exponent is not a constant.
However, if we set f (x) = xx we can write
• logb (x_y) = logb (x) * logb (y)
• logb (xy ) = y logb (x)
While we could use the product and quotient rule to solve
this problem, it would be tedious. Start by taking the
logarithm of the function to be differentiated.
H
I
⇠˘
⇡
x9 e4x
ln ˘
= ln x9 e4x * ln
x*4
x*4
= ln x9 + ln e4x * ln (x * 4)1_2
1
= 9 ln(x) + 4x * ln(x * 4).
2
x9 e4x
Setting f (x) = ˘
, we can write
x*4
f (x) =
9
1
+4*
x
2(x * 4)
Now we can solve for f ® (x),
f ® (x) = xx + xx ln(x).
1H
x9 e4x
˘
x*4
d
d
ln (x) =
ln (x)
dx
dx
1
= .
x
I
.
The process above is called logarithmic differentiation. Logarithmic differentiation allows us to compute new derivatives
too.
142
f ® (x)
= 1 + ln(x).
f (x)
The function x is a piecewise defined function. If x > 0
f ® (x)
9
1
= +4*
.
f (x)
x
2(x * 4)
0
Differentiating both sides, we find
d
ln (x)
dx
Differentiating both sides, we find
®
= x ln(x).
Example 97. Compute the derivative.
1
ln(f (x)) = 9 ln(x) + 4x * ln(x * 4).
2
Finally we solve for f ® (x), write
ln(f (x)) = ln (xx )
Logarithmic differentiation
The function ln (x) is not defined at x = 0. If x < 0
d
d
ln (x) =
ln (*x)
dx
dx
*1
=
*x
1
= .
x
The next example will be useful when we want to use logarithmic differentiations for functions that assume negative values.
Example 98. Compute the derivative.
d
ln (f (x))
dx
By the previous example and the Chain Rule
d
1
ln (f (x)) =
f ® (x).
dx
f (x)
f ® (x)
=
f (x)
Proof of the power rule
Write
ln(f (x)) = ln (xn ) , x ë 0
= n ln(x).
Now differentiate both sides, and solve for f ® (x)
f ® (x)
n
=
f (x)
x
nf (x)
f ® (x) =
x
= nxn*1 .
Whenever x = 0 is in the domain of f , the definition of
the derivative at x = 0 implies that f ® (0) = 0, if n ë 1
and f ® (0) = 1, if n = 1 , so the power rule applies to
x = 0, too.
Thus we see that the power rule holds for all real-valued
(nonzero) exponents.
While logarithmic differentiation might seem strange and new
at first, with a little practice it will seem much more natural to
you.
Finally, recall that previously we only proved the power rule
for positive integer exponents. Now we’ll use logarithmic
differentiation to give a proof for all real-valued exponents.
We restate the power rule for convenience sake:
Theorem 35 (Power Rule). For any real number n,
d n
x = nxn*1 .
dx
We will use logarithmic differentiation. Set f (x) = xn .
143
Derivatives of inverse functions
17 Derivatives of inverse
functions
After completing this section, students should be able to do the
following.
• Find derivatives of inverse functions in general.
• Recall the meaning and properties of inverse trigonometric
functions.
• Derive the derivatives of inverse trigonometric functions.
• Understand how the derivative of an inverse function relates to the original derivative.
• Take derivatives which involve inverse trigonometric functions.
144
We can figure it out
Problem 2. What is the sign of f ® (x) = * csc2 (x) on the
interval (0, ⇡)?
Break-Ground:
17.1
We can �gure it out
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, I have a calculus question.
Riley: Hit me with it.
Devyn: What’s the derivative of arctan(x)?
Riley: Hmmm. . . we haven’t talked about that yet in our class.
Devyn: I know! But maybe we can figure it out.
Riley: Well
arctan(x) = tan*1 (x)
Problem 3. Multiple Choice:
(a) positive
(b) negative
Problem 4. Complete the sentence below:
Since the sign of f ® (x) = * csc2 (x) on the interval (0, ⇡) is
and now we can use the chain rule to take its derivative
Multiple Choice:
d
tan*1 (x) = * tan*2 (x) sec2 (x)
dx
cos2 x
1
=* 2
sin x cos2 x
*1
=
sin2 x
= * csc2 x
(a) positive
(b) negative
the function f must be
Multiple Choice:
Devyn: But is this right?
Let’s see if we can figure out if Devyn and Riley are correct.
Start by looking at a plot of ✓ = arctan(x):
(a) increasing on the interval (0, ⇡)
(b) decreasing on the interval (0, ⇡)
Problem 5. In light of the problems above, is it possible that
✓
⇡_2
d
arctan(x) = * csc2 (x)?
dx
x
Multiple Choice:
*⇡_2
Problem 1. Let f (x) = arctan(x). Use the plot to determine
the intervals(s) where the function f is increasing.
From the graph it seems that the function f is increasing on
the interval
(*ÿ, ÿ)
(a) yes
(b) no
Problem 6. When our friends wrote arctan(x) = tan*1 (x),
what do they think the “*1” represents? Are they correct?
On the other hand,
145
Derivatives of inverse trigonometric functions
Dig-In:
17.2 Derivatives of inverse
trigonometric functions
Now we will derive the derivative of arcsine, arctangent, and
arcsecant.
1
Theorem 36 (The derivative of arcsine).
✓
˘
1 * x2
d
1
arcsin(x) = ˘
.
dx
1 * x2
Recall
arcsin(x) = ✓
*⇡
⇡
means that sin(✓) = x and
f ✓ f . Implicitly
2
2
differentiating with respect x we see
sin(✓) = x
d
d
sin(✓) =
x
dx
dx
®
cos(✓) ✓ = 1
1
✓® =
cos(✓)
Differentiate both sides.
Implicit differentiation.
Solve for ✓ ® .
1
, we need our answer written in terms
cos(✓)
of x. Since we are assuming that
While ✓ ® =
sin(✓) = x,
consider the following triangle with the unit circle:
From the unit circle above, we see that
1
cos(✓)
hyp
=
adj
1
=˘
.
1 * x2
⇠
⇡
⇡ ⇡
Recall, cos x g 0 on the interval * ,
.
2 2
✓® =
To be completely explicit,
d
d
1
✓=
arcsin(x) = ˘
.
dx
dx
1 * x2
Question 74. Compute:
d
sin*1 (x)
dx
146
x
Derivatives of inverse trigonometric functions
We can do something similar with arctangent.
Theorem 37 (The derivative of arctangent).
d
1
arctan(x) =
.
dx
1 + x2
To start, note that the Inverse Function Theorem assures
us that this derivative actually exists. Recall
arctan(x) = ✓
*⇡
⇡
means that tan(✓) = x and
< ✓ < . Implicitly
2
2
differentiating with respect x we see
tan(✓) = x
d
d
tan(✓) =
x
dx
dx
sec2 (✓) ✓ ® = 1
1
✓® =
sec2 (✓)
Differentiate both sides.
Implicit differentiation.
Recall the trig identity: sec2 (✓) = 1 + tan2 (✓).
✓® =
1
1 + tan2 (✓)
Finally, we investigate the derivative of arcsecant.
Theorem 38 (The derivative of arcsecant).
d
1
arcsec(x) =
˘
dx
x x2 * 1
Recall
arcsec(x) = ✓
means that sec(✓) = x and 0 f ✓ f ⇡ with ✓ ë ⇡_2.
Implicitly differentiating with respect x we see
sec(✓) = x
d
d
sec(✓) =
x
dx
dx
sec(✓) tan(✓) ✓ ® = 1
1
✓ =
1 + x2
1
sec(✓) tan(✓)
cos2 (✓)
✓® =
.
sin(✓)
Solve for ✓ ® .
cos2 (✓)
, we need our answer written in terms
sin(✓)
of x. Since we are assuming that
sec(✓) =
1
= x,
cos(✓)
we may consider the following triangle:
˘
x2 * 1
x
d
d
1
✓=
arctan(x) =
.
dx
dx
1 + x2
˘
d
tan*1 ( x)
dx
Implicit differentiation.
While ✓ ® =
To be completely explicit,
Question 75. Compute:
Differentiate both sides.
✓® =
Recall from above: tan(✓) = x.
®
for x > 1.
✓
1
147
Derivatives of inverse trigonometric functions
1
We may now scale this triangle by a factor of to place
x
it on the unit circle:
To be completely explicit,
d
d
1
✓=
arcsec(x) =
˘
dx
dx
x x2 * 1
for x > 1.
Question 76. Compute:
1
✓
u
1*
1
x
From the unit circle above, we see that
✓® =
=
cos2 (✓)
sin(✓)
⇠ ⇡2
adj
hyp
opp
hyp
2
(adj)
=
opp
1_x2
=˘
1 * 1_x2
1
=
,
˘
x x2 * 1
148
Note, hyp = 1.
1
x2
d
sec*1 (3x)
dx
We leave it to you, the reader, to investigate the derivatives of
cosine, arccosecant, and arccotangent. However, as a gesture
of friendship, we now present you with a list of derivative
formulas for inverse trigonometric functions.
Theorem 39 (The Derivatives of Inverse Trigonometric Functions).
•
d
1
arcsin(x) = ˘
dx
1 * x2
•
d
*1
arccos(x) = ˘
dx
1 * x2
•
d
1
arctan(x) =
dx
1 + x2
•
d
1
arcsec(x) =
for x > 1
˘
dx
x x2 * 1
•
d
*1
arccsc(x) =
for x > 1
˘
dx
x x2 * 1
•
d
*1
arccot(x) =
dx
1 + x2
The Inverse Function Theorem
Dig-In:
17.3
The Inverse Function Theorem
There is one catch to all the explanations given above where we
computed derivatives of inverse functions. To write something
like
d y
(e ) = ey y®
dx
we need to know that the function y has a derivative. The
Inverse Function Theorem guarantees this.
Theorem 40 (Inverse Function Theorem). If f is a differentiable function that is one-to-one near a and f ® (a) ë 0, then
(a) f *1 (x) is defined for x near b = f (a),
y
5
4
f
f
3
2
f *1
1
where
f *1
x
b = f (a).
We will only explain the last result. We know
f (f *1 (x)) = x,
and now we use the chain rule to write
d
f (f *1 (x)) = f ® (f *1 (x))(f *1 )® (x)
dx
= 1.
Solving for (f *1 )® (x) we see
(f *1 )® (x) =
6
x
(b) f *1 (x) is differentiable near b = f (a),
(c) last, but not least:
4
5
d *1
1
f (x)
= ®
dx
f
(a)
x=b
increments in variable, x. Consider this plot of a function f
and its inverse:
1
.
f ® (f *1 (x))
This is what we have written above.
It is worth giving one more piece of evidence for the formula
above, this time based on increments in function, f , and
1
2
3
4
5
6
x
Since the graph of the inverse of a function is the reflection of
the graph of the function over the line y = x, we see that the
increments are “switched” when reflected. Hence we see that
f *1
=
x
x
1
=
.
f
f
x
Taking the limit as x goes to 0, we can obtain the expression
for the derivative of f *1 .
df *1
1
=
df
dx
dx
The inverse function theorem gives us a recipe for computing
the derivatives of inverses of functions at points.
149
The Inverse Function Theorem
Example 99. Let f be a differentiable function that has an
inverse. In the table below we give several values for both f
and f ® :
x f f®
Compute
2
0
2
3
1
5
4
3
0
d *1
f (x) at x = 1.
dx
From the table above we see that
1 = f (3).
From the table above we see that
0 = f (2).
Hence, by the inverse function theorem
®
f *1 (0) =
Finally, let’s see an example where the theorem does not apply.
Example 101. Let f be a differentiable function that has an
inverse. In the table below we give several values for both f
and f ® :
x f f®
Hence, by the inverse function theorem
f
*1 ®
1
1
(1) = ®
= .
f (3) 5
If one example is good, two are better:
Example 100. Let f be a differentiable function that has an
inverse. In the table below we give several values for both f
and f ® :
x f f®
Compute
Note,
150
2
0
2
3
1
5
4
3
0
®
f *1 (0)
®
f *1 (0) =
d *1
f (x) at x = 0.
dx
1
1
= .
f ® (2) 2
Compute
4
2
0
2
3
1
4
4
3
0
d *1
f (x)
dx
5
x=3
From the table above we see that
3 = f (4).
Ah! But here, f ® (4) = 0, so we have no guarantee that
the inverse exists near the point x = 3, but even if it did
the inverse would not be differentiable there.
More than one rate
18
More than one rate
After completing this section, students should be able to do the
following.
• Solve basic related rates word problems.
• Understand the process of solving related rates problems.
• Calculate derivatives of expressions with multiple variables implicitly.
151
A changing circle
Problem 5. Set r = 3 t. Now A(t) = ⇡ (3 t)2 . What is A® (t)
when r = 15?
Break-Ground:
18.1
A changing circle
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, I’ve been thinking about calculus.
Riley: YOLO.
Devyn: Consider a circle of some radius r.
Riley: Ha! What else would we ever call the “radius?”
Devyn: Exactly. Now the formula for the perimeter of a circle is?
Riley: P = 2 ⇡ r baby.
Devyn: And its area?
Riley: You know it’s A = ⇡ r2 .
Devyn: Right, but here’s what’s bugging me: If I know r® , what is
P ® ? What’s A® ?
Riley: Oooh. Ouch. Hmmm. I wanna say it’s
P ® = 2 ⇡ r®
and
A® = ⇡(r® )2
but I’m not sure that is right.
Devyn: Yeah. . . me too. But I’m not sure that’s right either. Are we
forgetting something?
Problem 1. Do you think our young mathematicians above
are correct?
Multiple Choice:
(a) Yes. P ® = 2 ⇡ r® and A® = ⇡(r® )2 .
(b) No. While P ® = 2 ⇡ r® , A® ë ⇡(r® )2 .
(c) No. While A® = ⇡(r® )2 , P ® ë 2 ⇡ r® .
(d) No. P ® ë 2 ⇡ r® and A® ë ⇡(r® )2 .
(e) There is no way to tell.
Problem 2. Set r(t) = 3 t. What is r® (t) when r = 15?
Problem 3. Set r = 3 t. Now P (t) = 2 ⇡ 3 t. What is P ® (t)
when r = 15?
Problem 4. Describe what P ® means in this context.
152
Problem 6. Describe what A® means in this context. Does it
make sense that A® is positive?
Problem 7. What, if anything, did our two young mathematicians forget about above?
More than one rate
Formulas
Dig-In:
18.2
More than one rate
Suppose we have two, variables x and y, which are both changing with respect to time. A related rates problem is a problem
dx
where we know one rate, say
, at a given instant, and wish
dt
to find the other (the unknown rate is "related" to the known
rate).
Here, the chain rule is key: If y is written in terms of x, and we
dy
dx
are given
, then it is easy to find
using the chain rule:
dt
dt
y = y(x)
dy
dy dx
=
.
dt
dx dt
In order to relate several variables we can use known formulas.
In our next example we consider an expanding circle and use
the formulas for perimeter and area of a circle.
Example 102. Imagine an expanding circle. If we know that
the perimeter is expanding at a rate of 4 m/s, at what rate is
the area changing when the radius is 3 meters?
In many cases, particularly the interesting ones, our functions
will be related in some other way. Nevertheless, in each case
we’ll use the chain rule to help us find the desired rate. In
this section, we will work several abstract examples in order
to emphasize the mathematical concepts involved. In each of
the examples below, we will follow essentially the same plan
of attack:
First, we introduce the variables P , r, and A, denoting
the perimeter, the radius, and the area of the circle, in
that order.
dP
We identify the given rate,
= 4 m/s, and the undt
dA
known rate,
, at the moment when r = 3.
dt
Next, we draw a picture.
Introduce variables, identify the given rate and the unknown rate.
Assign a variable to each quantity that changes in time.
Draw a picture. If possible, draw a schematic picture with all
the relevant information.
Find equations. Write equations that relate all relevant variables.
Differentiate with respect to t. Here we will often use implicit differentiation and obtain an equation that relates
the given rate and the unknown rate.
Evaluate and solve. Evaluate each quantity at the relevant instant and solve for the unknown rate.
dP
= 4 m/s
dt
r
dA
= ? , when r = 3
dt
Now we have to find equations that relate the variables
P , r, and A.
Naturally, we think of formulas for perimeter and area of
153
More than one rate
a circle
P =2 ⇡ r
and
d
A(t) = 2 ⇡ 3 2_⇡
dt
= 12 m2 _s.
A = ⇡ r2 .
We know that the perimeter of the circle is expanding.
This implies that both the radius and the area of the circle
are changing in time, too. So, we note that P , r, and A
are functions of time
P (t) = 2 ⇡ r(t)
and
A(t) = ⇡ r(t)2 .
We differentiate both sides of each equation with respect
to time, t. Using implicit differentiation, we get that
d
d
P (t) = 2 ⇡ r(t)
dt
dt
´Ø¨
and
d
A(t)
dt
´Ø¨
given rate
= 2 ⇡ r(t)
unknown rate
These two equations hold on some time interval. In particular, they are both true at an instant when r = 3.
d
We know that at that instant P (t) = 4 and r = 3.
dt
We now evaluate the two equations at the instant when
r = 3.
4=2 ⇡
d
r(t)
dt
and
d
d
A(t) = 2 ⇡ 3
r(t)
dt
dt
From the first equation we get that
d
r(t) = 2_⇡ m_s, when r = 3.
dt
Using this result and the equation on the right, we get, at
the instant when r = 3,
154
Hence, the area is expanding at a rate of 12 m2 _s at the
instant when r = 3 m.
Right triangles
In our next example, we consider an expanding right triangle
and use the Pythagorean Theorem to relate relevant variables.
d
r(t).
dt
3m
Example 103. Imagine an expanding right triangle. If one leg
has a fixed length of 3 m, one leg is increasing with a rate of
2 m/s, and the hypotenuse is expanding to accommodate the
expanding leg, at what rate is the hypotenuse expanding when
both legs are 3 m long?
First, we introduce the variables a, b, and c, denoting
the fixed length, the length of a leg that is increasing,
and the length of the hypotenuse, in that order. Then, we
db
identify the given rate
= 2 m/s and the unknown rate
dt
dc
, when b = 3. Now, we draw a picture.
dt
More than one rate
db
=2
dt
c
a = 3m
dc
= ? , when b = 3
dt
b
Next, we find equations that relate relevant variables.
Here we use the Pythagorean Theorem.
c 2 = a2 + b2
Note that a = 3 is a constant, and c and b are functions of
time, t. We, then, differentiate both sides of the equation
with respect to t, using implicit differentiation
d
d
c(t) = 2 b
b(t).
dt
dt
Now, we evaluate all the quantities at the instant when
db
b = 3, noting that
= 2 and that b = 3. Therefore,
dt
2 c(t)
2 c
dc
= 12, when b = 3.
dt
We need to compute c, the length of the hypotenuse at
the instant when b = 3. Here we use the Pythagorean
Theorem
c 2 = 32 + 32
= 18,
˘ dc
6 2
= 12
dt
dc ˘
= 2.
dt
Therefore, the hypotenuse is growing at a rate of
when both legs are 3m long.
˘
2 m/s
Angular rates
We can also investigate problems involving angular rates.
Example 104. Imagine an expanding right triangle. If one leg
has a fixed length of 3 m, one leg is increasing with a rate of
2 m/s, and the hypotenuse is expanding to accommodate the
expanding leg, at what rate is the angle opposite the fixed leg
changing when both legs are 3 m long?
First, we introduce the variables a, b, c, as in the previous example, and ✓, denoting the angle opposite the leg
with fixed length.
db
We identify the given rate
= 2 m/s and the unknown
dt
d✓
rate
, when b = 3.
dt
Next, we draw a picture.
db
=2
dt
a = 3m
c
˘
So, we see that c = 3 2, when b = 3.
Then, we solve for the rate at the moment when b = 3,
d✓
= ? , when b = 3
dt
✓
b
155
More than one rate
We now find equations that combine relevant variables.
Here we note that
Now we solve for the unknown rate, when b = 3,
2
3
tan(✓) = .
b
Since ✓ and b are functions of time, we differentiate both
sides of the equation using implicit differentiation
d✓
*3 db
=
.
dt
b2 dt
Now, we evaluate all the quantities at the instant when
db
b = 3, keeping in mind that
= 2 and b = 3,
dt
sec2 (✓)
sec2 (✓)
d✓
*3
*2
=
2=
.
dt
3
32
d✓
*2
=
dt
3
d✓
*1
=
dt
3
So, when b = 3, the angle is changing at *1_3 radians
per second.
Similar triangles
Finally, facts about similar triangles are often useful when
solving related rates problems.
Example 105. The figure shows two right triangles that share
an (acute) angle.
It is clear that we still need to compute sec2 (✓) at the
moment when b = 3. A picture will help.
sec2 ✓ =
a = 3m
⇠ ⇡2
c
c2
32 + 3 2
=
=
=2
b
b2
32
OR
c
sec2 ✓ = 1 + tan2 ✓ = 1 +
b = 3m
⇠ ⇡2
3
=2
3
✓
Since we know the lengths of both legs in the right triangle, it is easier to compute
tan(✓) than sec(✓). Recall the trig identity: sec2 (✓) =
1 + tan2 (✓).
sec2 (✓) = 1 + tan2 (✓)
⇠ ⇡2
3
=1+
3
=2
156
3
x
6
If x is growing from the vertex with a rate of 3 m/s, what
rate is the area of the smaller triangle changing when x =
5m?
First, we introduce the variables h and A, denoting the
height and the area of the smaller triangle, in that order.
dx
Next, we identify the given rate
= 3 m/s and the
dt
dA
unknown rate
, when x = 5. Despite the fact that
dt
a nice picture is given, we should do what we always
do and draw a picture. Note, we ad information to the
More than one rate
picture:
In particular, when x = 5,
dx
=3
dt
3
dA
= ? , when x = 5
dt
h
x
dA 1
=
2 5 3= ?.
dt
4
15 2
Hence, the area is changing at a rate of
m _s when
2
x = 5m.
6
Next, we find equations that combine relevant variables.
In this case there are two. The first is the formula for the
area of a triangle:
A=
1
x h
2
The second uses the fact that the larger triangle is similar
to the smaller triangle, meaning that the ratios between
the corresponding sides in both triangles are equal,
h 3
=
x 6
so
h=
1
x
2
At this point we could differentiate both equations, but
we choose a simpler path and express A in terms of x
A=
1
x
2
1
x .
2
´Ø¨
h
Therefore,
1 2
x .
4
Since A and x are both functions of time, we differentiate both sides of this equation with respect to t
A=
dA 1
dx
=
2 x
.
dt
4
dt
157
Applied related rates
19
Applied related rates
After completing this section, students should be able to do the
following.
• Identify word problems as related rates problems.
• Translate word problems into mathematical equations.
• Solve related rates word problems.
158
Pizza and calculus, so cheesy
Break-Ground:
19.1
Pizza and calculus, so cheesy
Check out this dialogue between two calculus students (based
on a true story):
(c) The surface area of the cylinder.
(d) The volume of the cylinder.
Devyn: Hey Riley, do you know what I love?
Riley: Calculus?
Devyn: And pizza! Last night I made my own pizza crust from
scratch!
Riley: Mmmmmmm. Calculus.
Devyn: I know! The best part of making the crust is tossing it. But
during my pizza tossing, I noticed something. The dough is
basically a cylinder with a radius that is expanding, and a
height that is getting smaller. Then I started wondering: how
are those two rates related?
Riley: Pizza. So delicious. So cheesy. So calculus.
The problem above is an example of a related rates word problem. To solve it, we will need the tools in the Dig-In. For now,
let’s see if we can reason through some questions related to this
problem.
Problem 1. In the context above, which of the following should
be assumed to be functions that vary?
Select All Correct Answers:
(a) The radius of the cylinder.
(b) The height of the cylinder.
(c) The surface area of the cylinder.
(d) The volume of the cylinder.
Problem 2. In the context above, which of the following should
be assumed to be constant?
Select All Correct Answers:
(a) The radius of the cylinder.
(b) The height of the cylinder.
159
Applied related rates
Dig-In:
19.2
Applied related rates
Now we are ready to solve related rates problems in context.
Just as before, we are going to follow essentially the same plan
of attack in each problem.
Introduce variables, identify the given rate and the unknown rate.
Assign a variable to each quantity that changes in time.
First, we introduce the variables V , r, and h, denoting
the volume, the radius, and the thickness of the pizza, in
dr
that order. We identify the given rate
= 15 cm/min
dt
dh
and the unknown rate
, when r = 20.
dt
Next, we draw a picture. Note that our pizza dough is
shaped as a cylinder.
Draw a picture. If possible, draw a schematic picture with all
the relevant information.
dr
= 15
dt
Find equations. Write equations that relate all relevant variables.
dh
= ?, when r = 20
dt
Differentiate with respect to t. Here we will often use implicit differentiation and obtain an equation that relates
the given rate and the unknown rate.
Evaluate and solve. Evaluate each quantity at the relevant instant and solve for the unknown rate.
Formulas
In our next problem, we will stretch and flatten a cylindershaped pizza dough.
Example 106. A hand-tossed pizza crust starts off as a ball
of dough with a volume of 400⇡ cm3 . First, the cook stretches
the dough to the shape of a cylinder of radius 12 cm. Next the
cook tosses the dough.
If during tossing, the dough maintains the shape of a cylinder
and the radius is increasing at a rate of 15 cm/min, how fast is
its thickness changing when the radius is 20 cm?
160
r
h
Next we need to find equations. Recall the formula for
the volume of the cylinder, V = ⇡ r2 h and note that
V = 400⇡ cm3 , and write
400⇡ = ⇡ r2 h,
which immediately simplifies to
400 = r2 h.
Since both r and h are functions of time, we now differentiate both sides of the equation using implicit differentiation,
dr
dh
h + r2
.
dt
dt
Now we’ll evaluate all the quantities at the instant when
r = 20. But, first, we have to compute h at that moment.
Since
400 = 202 h,
0=2 r
it follows that h = 1, when r = 20. Now, we are ready to
solve.
Applied related rates
0 = 2 20 15 1 + 202
*2 20 15 1 dh
=
dt
202
Therefore, when r = 20,
dh
dt
dh
= *1.5.
dt
Hence, the thickness of the dough is changing at a rate
of *1.5 cm/min when r = 20 cm.
In our next example we consider a melting snowball.
dV
ÌA
dt
r
dr
=?
dt
In order to find equations that relate all relevant variables, we recall formulas for the volume and the surface
area of the sphere,
(1) V =
Example 107. Consider a melting snowball. We will assume
that the rate at which the snowball is melting is proportional
to its surface area. Show that the radius of the snowball is
changing at a constant rate.
First, we introduce the variables V , r, and A, denoting
the volume, the radius, and the surface area of the snowdV
ball, in that order . Then, we identify the given rate
dt
dr
and the unknown rate , the rate to be determined. This
dt
problem is a bit unusual, because“the given rate" is not
explicitly given. We will deal with this issue below. Next,
we draw a picture.
4
⇡ r3
3
and
(2) A = 4⇡ r2 .
Now the key words are “the rate at which the snowball
is melting (the given rate) is proportional to its surface
area.” From this we have the following equation:
is proportional to
dV
=
dt
´Ø¨
©Æ™
k
rate the snowball is melting
A
´Ø¨
its surface area
We need to compute
dV
. Since r is a function of t, we
dt
differentiate both sides of equation (1). So
dV
dr
= 4 ⇡ r2
.
dt
dt
161
Applied related rates
when a = 3 km, b = 4 km
On the other hand,
dV
= k A.
dt
Therefore,
4 ⇡ r2
da
= 20 km/h
dt
A
dr
= k 4 ⇡ r2 .
´≠Ø≠¨
dt
c
a
dc
=?
dt
A
Now, we solve for
dr
and obtain that
dt
P
dr
= k.
dt
Hence, the radius is changing at a constant rate. Notice,
in this example we did not have to evaluate quantities at
particular time, because the unknown rate did not depend
on time.
Right triangles
Example 108. A road running north to south crosses a road
going east to west at the point P . Cyclist A is riding north along
the first road, and cyclist B is riding east along the second road.
At a particular time, cyclist A is 3 kilometers to the north of
P and traveling at 20 km/h, while cyclist B is 4 kilometers to
the east of P and traveling at 15 km/h. How fast is the distance
between the two cyclists changing at that time?
First, we introduce the variables a, b, and c, denoting
the distance of cyclist A from the point P , the distance of
cyclist B from the point P , and the distance between the
two cyclists, in that order. We identify the given rates
da
db
= 20 km/h,
= 15 km/h, when a = 3, b = 4 and
dt
dt
dc
the unknown rate
, when a = 3, b = 4. Now, we draw
dt
a picture.
162
db
= 15 km/h
dt
b
B
We find equations relating the variables a, b, and c. By
the Pythagorean Theorem,
(1)
c 2 = a2 + b2 .
Since a, b, and c are functions of time, we differentiate
both sides of the equation with respect to t.
(2)
2 c
dc
da
db
=2 a
+2 b
.
dt
dt
dt
This equation holds on some time interval, and , in particular, at the instant when a = 3 and b = 4. Therefore,
at that instant,
(3)
2 c
dc
=2 3
20 +2 4
15 .
´Ø¨
´Ø¨
dt
´Ø¨
given rate
given rate
unknown rate
In order to evaluate c, the distance between the cyclists,
at the instant when a = 3 and b = 4, we draw a picture
“taken" at that instant.
Applied related rates
Next, we draw a picture.
A
p
2
2
plane
dp
= 500 mph
dt
2
c = 3 + 4 = 25
3 km
3 miles
c
You
4 km
P
B
By the the Pythagorean Theorem and the picture above,
when a = 3, b = 4, it follows that
c = 5km.
Now we can evaluate all the quantities in equation (3)
and solve
2 5
dc
= 2 3 20 + 2 4 15.
dt
dc
We find that
= 24 km/hr at the moment when a = 3,
dt
b = 4.
Example 109. A plane is flying at an altitude of 3 miles directly
away from you at 500 mph . How fast is the plane’s distance
from you increasing at the moment when the plane is flying
over a point on the ground 4 miles from you?
ds
= ?, when p = 4 mi
dt
s
ground
Next we find equations relating the variables p and s.
By the Pythagorean Theorem we know that
p2 + 3 2 = s 2 .
Since p and s are functions of time, we now differentiate
both sides of the equation
2 p
dp
ds
=2 s
.
dt
dt
At the instant when p = 4 mi, we get that
unknown value
2 4
500 = 2
´Ø¨
given rate
©Æ™
s
ds
.
dt
´Ø¨
unknown rate
This implies that s has to be evaluated at time when p = 4.
In order to do that, let’s draw the picture at the given
instant, when p = 4 mi.
First, we introduce variables p, the distance the plane
has traveled after it flew right above you, and s, the
distance between you and the plane. The given rate is
dp
ds
= 500 mph and the unknown rate is
, when p = 4.
dt
dt
163
Applied related rates
p = 4 miles
3 miles
You
point on the ground
4 miles away from you
So, s = 5. Putting together all the information we get
2(4)(500) = 2(5)
Finally, we solve:
p = 4 miles.
ds
= 400 mph, at the moment when
dt
Example 110. A plane is flying at an altitude of 3 miles directly
away from you at 500 mph . Let ✓ be the angle of elevation
of the plane, i.e., the angle between the ground and your line
of sight to the plane. How fast is the angle ✓ changing at the
moment when the plane is flying over a point on the ground 4
miles from you?
First, we introduce a variable p, the distance the plane
has traveled after it flew right above you. So, the given
dp
rate is
= 500 mph and the unknown (related) rate is
dt
d✓
, at the instant when p = 4 miles. This rate should
dt
be measured in rad/s. Therefore, we have to convert the
dp
units of the given rate, mph, into mi/s:
= 500mph=
dt
500
mi/s.
60 60
Next, we draw a picture.
dp
500
=
mi/s
dt
60 60
You
d✓
= ?, when p = 4 mi
dt
✓
ground
Now we find an equation relating variables p and ✓.
From the picture we can see that
ds
.
dt
Angular rates
164
plane
3 miles
s2 = 42 + 32 = 25 mi2
s
4 miles
p
plane
tan ✓ =
3
.
p
Since p and ✓ are both functions of time, we now differentiate both sides of the equation. We write
sec2 ✓
d✓
3
=*
dt
p2
dp
.
dt
The above equation holds over some interval of time. In
particular, it is true when p = 4. Therefore, when p = 4,
unknown value
©Æ™
sec2 ✓
3
d✓
=*
dt
42
´Ø¨
unknown rate
500
.
3600
´Ø¨
given rate
This implies that we have to compute the unknown value,
sec2 ✓, when p = 4. To this end, we draw and label the
picture at the time when p = 4 mi.
Applied related rates
p = 4 miles
3 miles
You
✓
plane
dp
= 3ft/s
dt
sec2 ✓ = 1 + tan2 ✓ = 1 +
4 miles
⇠ ⇡2
3
25
=
4
16
15
ds
= ?, when p = 7ft
dt
6
point on the ground
4 miles away from you
p
So, at the moment when p = 4 miles,
⇠ ⇡2
3
25
sec2 ✓ = 1+
=
.
4
16
Now, by substituting all the known values into the equation above, we get
s
Now we find equations that relate the variables p and s.
We use the fact that we have similar triangles to write:
s+p
s
= ,
15
6
6 s + 6 p = 15 s,
6 p = 9 s,
25 d✓
3 500
=*
.
16 dt
16 3600
We solve for
d✓
, when p = 4, and get that
dt
d✓
1
= * rad_s.
dt
60
Similar triangles
Example 111. It is night. Someone who is 6 feet tall is walking
away from a street light at a rate of 3 feet per second. The street
light is 15 feet tall. The person casts a shadow on the ground
in front of them. How fast is the length of the shadow growing
when the person is 7 feet from the street light?
First, we introduce two variables, p, the distance from
the person to the lamp, and s, the length of the shadow.
dp
The given rate is
= 3 ft/s and the unknown (related)
dt
ds
rate is
at the moment when p = 7 feet. Next, we draw
dt
a picture.
2 p = 3 s.
Since p and s are both functions of time, we differentiate both sides of the equation above. We use implicit
differentiation and write
2
dp
ds
=3
dt
dt
At this point we evaluate all the quantities at time when
p = 7 ft
ds
2 3=3
.
dt
ds
ds
Now we solve for
and get that
= 2, when p = 7
dt
dt
ft.
Therefore, the shadow is growing at a rate of 2 ft/s when
the person is 7 ft from the lamp.
Example 112. Water is poured into a conical container at the
rate of 10 cm3 /s. The cone points directly down, and it has a
height of 30 cm and a base radius of 10 cm. How fast is the
water level rising when the water is 4 cm deep?
165
Applied related rates
First, we introduce several variables. Let V be the volume of the water in the container, let r be the radius of
the circular surface of the water, and let h be the depth
dV
of the water in the container. The given rate is
= 10
dt
dh
cm3 _s, and the unknown rate is
, at the moment when
dt
h = 4 cm. Now, we draw a picture.
dV
= 10 cm3 /sec;
dt
r cm
h cm
Note, no attempt was made to draw this picture to scale,
rather we want all of the relevant information to be available to the mathematician.
Now we find equations that relate all the variables. Notice that water in the container assumes the shape of the
container. In this example, the shape is a cone. Therefore,
we use the formula for the volume of a cone
h
Notice a big right triangle and a smaller right triangle
inside the big one. These two right triangles are similar,
because their corresponding angles are equal. Since the
ratios of corresponding sides in similar triangles are equal,
we get
r
10
h
=
so
r= .
h 30
3
At this point we could differentiate both sides of each
equation, but we can take a simpler approach by combining these two equations and expressing V in terms of
h.
QUESTION: Why is it more convenient to express V in
terms of h than in terms of r?
Therefore,
⇡
V = r2 h.
3
Let’s draw a cross section of the cone.
r
30 cm
dh
= ?, when h = 4 cm
dt
10 cm
30 cm
10 cm
and
r2
©Æ™
⇠ ⇡
⇡ h 2
V =
h,
3 3
⇡
h3 .
27
Now, we differentiate with respect to t.
V =
dV
⇡
dh
=
3 h2
.
dt
27
dt
This equation holds over some time interval. In particular,
the equation is true at time when h = 4 cm. Now we
166
Applied related rates
evaluate all the quantities at that time.
10 =
⇡
dh
3 42
.
27
dt
So, at the instant when the water in the container is 4 cm
45
deep, the water level is rising at the rate of
cm/s.
8⇡
167
Maximums and minimums
20
Maximums and minimums
After completing this section, students should be able to do the
following.
• Define a critical point.
• Find critical points.
• Define local maximum and local minimum.
• Classify critical points.
• State the First Derivative Test.
• Apply the First Derivative Test.
• Define inflection points.
• Find inflection points.
• State the Second Derivative Test.
• Apply the Second Derivative Test.
168
More coffee
Break-Ground:
20.1
More co�ee
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley!
Riley: Yes Devyn?
Devyn: Do you like coffee? I like coffee! Sometimes I feel really
“bad,” sluggish and tired. Then I drink coffee and I feel good!
Sometimes I drink a lot of coffee!
Riley: Um?
Devyn: But here’s the problem, see: If I drink too much, I become
over excited and can’t stop talking. I just drink coffee, then talk.
Then drink more coffee. Then I start to feel sick. Ugh. I have a
love-hate relationship with coffee.
Riley: If only there were a calculus solution to this problem!
Remember, calculus is about studying functions. If we can
“see” a function in the work above, maybe we can figure out
how to solve it.
Problem 1. If we were to try to solve Devyn’s coffee problem,
what would be the best function to know?
Multiple Choice:
(a) How many donuts Devyn eats.
(b) How “good” Devyn feels after x cups of coffee.
(c) How many cups of coffee Devyn drinks when Devyn feels
x “good.”
(d) Impossible to say.
Problem 2. If we let f (x) be “How ‘good’ Devyn feels after x
cups of coffee,” and we think about what Devyn says above, is
there an amount Devyn can drink and feel maximally “good?”
Multiple Choice:
(a) yes
(b) no
169
Maximums and minimums
f
y
Dig-In
20.2
I
Maximums and minimums
Whether we are interested in a function as a purely mathematical object or in connection with some application to the real
world, it is often useful to know what the graph of the function looks like. We can obtain a good picture of the graph
using certain crucial information provided by derivatives of the
function.
0.5
x
1
1.5
2
x
*2
f (x)
f (1)
*4
Extrema
Local extrema of a function are points on the graph where the
y-coordinate is larger (or smaller) than all other y-coordinates
on the graph at points “close to” (x, y).
Definition.
(a) A function f has a local maximum at x = a, if f (a) g
f (x) for every x in some open interval I containing a.
(b) A function f has a local minimum at x = a, if f (a) f
f (x) for every x in some open interval I containing a.
A local extremum is either a local maximum or a local minimum.
In our next example, we clarify the definition of a local minimum.
Example 113. Consider the graph of a function f :
170
Identify the local extrema of f and give an explanation.
In the figure above the function f has a local minimum
at,
a = 1,
because were able to find an open interval I (marked in
the figure) that contains the number
a = 1,
and for all numbers x in I the following statement is true:
f (1) f f (x) .
Local maximum and minimum points are quite distinctive on
the graph of a function, and are, therefore, useful in understanding the shape of the graph. In many applied problems we want
to find the largest or smallest value that a function achieves (for
example, we might want to find the minimum cost at which
some task can be performed) and so identifying maximum and
minimum points will be useful for applied problems as well.
Maximums and minimums
the derivative is zero or the derivative is undefined. As an
illustration of the first scenario, consider the plots of f (x) =
x3 * 4.5x2 + 6x and f ® (x) = 3x2 * 9x + 6.
Critical points
Consider the graph of the function f .
y
6
y
y = f (x)
2
f®
f
4
2
-6
-4
-2
0.5
2
4
6
1
1.5
2
x
2.5
x
-2
The function f has four local extremums: at x = *4, x =
*2, x = 0 and x = 4. Notice that the function f is not
differentiable at x = *4 and x = *2. Notice that f ® (0) = 0
and f ® (4) = 0.
After this example, the following theorem should not come as
a surprise.
Theorem 41 (Fermat’s Theorem). If f has a local extremum
at x = a and f is differentiable at a, then f ® (a) = 0.
Question 77. Does Fermat’s Theorem say that if f ® (a) = 0,
then f has a local extrema at x = a?
Multiple Choice:
(a) yes
(b) no
Fermat’s Theorem says that the only points at which a function
can have a local maximum or minimum are points at which
*2
Question 78. Make a correct choice that completes the
sentence below.
At the point (1, f (1)), the function f has
Multiple Choice:
(a) a local maximum
(b) a local minimum
(c) no local extremum
Question 79. Select the correct statement.
Multiple Choice:
(a) f ® (1) is undefined
(b) f ® (1) > 0
171
Maximums and minimums
(c) f ® (1) = 0
Question 83. Select the correct statement.
(d) f ® (1) < 0
Multiple Choice:
Question 80. Make a correct choice that completes the
sentence below.
At the point (1.5, f (1.5)), the function f has
Multiple Choice:
(a) a local maximum
(b) a local minimum
(a) f ® (2) is undefined
(b) f ® (2) > 0
(c) f ® (2) = 0
(d) f ® (2) < 0
As an illustration of the second scenario, consider the plots of
2
f (x) = x2_3 and f ® (x) =
:
3x1_3
(c) no local extremum
Question 81. Select the correct statement.
2
f
Multiple Choice:
y
1
f®
(a) f ® (1.5) is undefined
(b) f ® (1.5) > 0
(c) f ® (1.5) = 0
*3
*2
*1
(d) f ® (1.5) < 0
(b) a local minimum
(c) no local extremum
172
3
x
*2
Question 84. Make a correct choice that completes the
sentence below.
Multiple Choice:
(a) a local maximum
2
*1
Question 82. Make a correct choice that completes the
sentence below.
At the point (2, f (2)), the function f has
1
At the point (*2, f (*2)), the function f has
Multiple Choice:
(a) a local maximum
Maximums and minimums
(b) a local minimum
(c) no local extremum
Question 85. Select the correct statement.
Multiple Choice:
(a) f ® (*2) is undefined
(b) f (*2) > 0
®
(c) f ® (*2) = 0
(d) f (*2) < 0
®
Question 86. Make a correct choice that completes the
sentence below.
Definition. Assume that a function f is defined on an open
interval I that contains a point a. The function f has a critical
point at x = a if
or
f ® (a) = 0
f ® (a) does not exist.
Warning. When looking for local maximum and minimum
points, you are likely to make two sorts of mistakes:
• You may forget that a maximum or minimum can occur
where the derivative does not exist, and so forget to check
whether the derivative exists everywhere.
• You might assume that any place that the derivative is zero
is a local maximum or minimum point, but this is not true,
consider the plots of f (x) = x3 and f ® (x) = 3x2 .
y
At the point (0, 0), the function f has
2
Multiple Choice:
f®
f
(a) a local maximum
(b) a local minimum
*1
*0.5
0.5
1
x
(c) no local extremum
Question 87. Select the correct statement.
*2
Multiple Choice:
(a) f ® (0) is undefined
(b) f ® (0) > 0
(c) f ® (0) = 0
(d) f ® (0) < 0
This brings us to our next definition.
While f ® (0) = 0, there is neither a maximum nor minimum
at x = 0.
Since both local maximum and local minimum occur at a critical point, when we locate a critical point, we need to determine
which, if either, actually occurs.
Example 114. Find all local maximum and minimum points
for the function f (x) = x3 * x.
173
Maximums and minimums
Write
d
f (x) = 3x2 * 1.
dx
We can easily express f ® (x) as a product of its factors
H
˘ IH
˘ I
3
3
d
f (x) = 3 x +
x*
,
dx
3
3
which implies that˘the function f˘has only two criti3
3
cal points, x = *
and x =
. Notice that the
3
3
®
derivative f (x) is a polynomial, and polynomials do not
change signs except possibly at their zeros. This implies
®
that derivative
f˘
(x)I
does
on the I
inH
H not
H˘
˘ change
˘ Ithe sign
3
3
3
3
tervals *ÿ, *
, *
,
, and
,ÿ ,
3
3
3
3
because these intervals do not contain any zeros of f (x).
®
Question 88. Select the correct statement
H
I the
˘about
3
sign of f ® (x) on the intervals *ÿ, *
and
3
H ˘ ˘ I
3
3
*
,
.
3
3
Multiple Choice:
H
(a) f (x) > 0 on
H ˘ ˘ I
3
3
*
,
.
3
3
®
(b) f (x) > 0 on x in
®
174
˘ I
3
*ÿ, *
and f ® (x) > 0 on
3
H
˘ I
3
*ÿ, *
and f ® (x) < 0 on
3
H ˘ ˘ I
3
3
*
,
.
3
3
H
˘ I
3
*ÿ, *
and f ® (x) > 0 on
3
H
˘ I
3
*ÿ, *
and f ® (x) < 0 on
3
(c) f (x) < 0 on
H ˘ ˘ I
3
3
*
,
.
3
3
®
(d) f (x) < 0 on
H ˘ ˘ I
3
3
*
,
3
3
®
If we know the sign of the derivative on an interval,
we also know whether the function is increasing or
decreasing on that interval. This will help us determine
whether the function
˘has a local extremum at the critical
3
point where x = *
.
3
˘
3
Question 89. At the critical point where x = *
, the
3
function f has
Multiple Choice:
(a) no
extremum,
because f isH increasing
on
H local ˘
I
˘ ˘ I
3
3
3
*ÿ, *
and increasing on *
,
.
3
3
3
(b) a
because f isHincreasing
on
H local maximum,
˘ I
˘ ˘ I
3
3
3
*ÿ, *
and decreasing on *
,
.
3
3
3
Maximums and minimums
(c) a
because f isHdecreasing
on
H local minimum,
˘ I
˘ ˘ I
3
3
3
*ÿ, *
and increasing on *
,
.
3
3
3
(d) H
no local ˘
extremum,
because f isHdecreasing
I
˘ ˘ Ion
3
3
3
*ÿ, *
and decreasing on *
,
.
3
3
3
Question 90. Select theH
correct
H ˘the sign
I
˘ statement
˘ I about
3
3
3
of f ® (x) on the intervals *
,
and
,ÿ .
3
3
3
Again, the sign of the derivative on an interval determines whether the function is increasing or decreasing
on that interval. This will help us determine whether the
function
˘ has a local extremum at the critical point where
3
x=
.
3
˘
3
Question 91. At the critical point where x =
, the
3
function f has
Multiple Choice:
Multiple Choice:
H ˘ ˘ I
3
3
®
(a) f (x) > 0 on *
,
and f ® (x) > 0 on
3
3
H˘
I
3
,ÿ .
3
H ˘ ˘ I
3
3
(b) f (x) > 0 on *
,
and f ® (x) < 0 on
3
3
H˘
I
3
,ÿ .
3
®
H ˘ ˘ I
3
3
(c) f (x) < 0 on *
,
and f ® (x) > 0 on
3
3
H˘
I
3
,ÿ .
3
®
H ˘ ˘ I
3
3
(d) f ® (x) < 0 on *
,
and f ® (x) < 0 on
3
3
H˘
I
3
,ÿ .
3
(a) no
extremum,
because f isHincreasing
H local
I
I on
˘ ˘
˘
3
3
3
*
,
and increasing on
,ÿ .
3
3
3
(b) a
because f isHincreasing
H local
I on
˘ maximum,
˘ I
˘
3
3
3
*
,
and decreasing on
,ÿ .
3
3
3
(c) a
because f isHdecreasing
H local
I on
˘ minimum,
˘ I
˘
3
3
3
*
,
and increasing on
,ÿ .
3
3
3
(d) noHlocal
because f isH decreasing
I
˘ extremum,
˘ I
˘
3
3
3
on *
,
and decreasing on
,ÿ .
3
3
3
Do your answers agree with the graphs of f and f ® given
in the picture below?
175
Maximums and minimums
2
f
®
• If f ® (x) has the same sign to the left and right of a, then
f (a) is not a local extremum.
y
Example 115. Consider the function
1
f (x) =
f
*1.5
*1
*0.5
0.5
1
x
1.5
Find the intervals on which f is increasing and decreasing and
identify the local extrema of f .
Start by computing
*1
*2
The �rst derivative test
We will further explore and refine the method of the previous section for deciding whether there is a local maximum or
minimum at a critical point. Recall that
• If f ® (x) > 0 on an interval, then f is increasing on that
interval.
• If f ® (x) < 0 on an interval, then f is decreasing on that
interval.
So how exactly does the derivative tell us whether there is
a maximum, minimum, or neither at a point? Use the first
derivative test.
Theorem 42 (First Derivative Test). Suppose that f is continuous on an interval, and that f ® (a) = 0 for some value of a in
that interval.
• If f ® (x) > 0 to the left of a and f ® (x) < 0 to the right of
a, then f (a) is a local maximum.
• If f ® (x) < 0 to the left of a and f ® (x) > 0 to the right of
a, then f (a) is a local minimum.
176
x4 x3
+
* x2
4
3
d
f (x) = x3 + x2 * 2x.
dx
Now we need to find on what intervals is f ® positive and
what intervals it is negative. To do this, solve
f ® (x) = x3 + x2 * 2x = 0.
Factor f ® (x)
f ® (x) = x3 + x2 * 2x
= x(x2 + x * 2)
= x(x + 2)(x * 1).
So the critical points (when f ® (x) = 0) are when x =
*2, x = 0, and x = 1. Since the derivative, f ® (x), is
a polynomial, it does not change the sign on intervals
between its zeros, i.e., between the critical points. Now
we can check the sign of f ® (x) at some points between
the critical points to find where f ® (x) is positive and
where negative:
f ® (*3) = *12,
f ® (.5) = *0.625,
f ® (*1) = 2,
f ® (2) = 8.
From this we can make a sign table:
Maximums and minimums
In�ection points
*2
f ® (x) < 0
f ® (x) > 0
0
1
f ® (x) < 0
If we are trying to understand the shape of the graph of a
function, knowing where it is concave up and concave down
helps us to get a more accurate picture. It is worth summarizing
what we have seen already in to a single theorem.
f ® (x) > 0
Hence f is increasing on (*2, 0) and (1, ÿ) and f is
decreasing on (*ÿ, *2) and (0, 1). Moreover, from the
first derivative test, the local maximum is at x = 0 while
the local minimums are at x = *2 and x = 1, see the
graphs of f (x) = x4 _4 + x3 _3 * x2 and f ® (x) = x3 +
x2 * 2x.
y
*3
*2
2
*1
f
(a) If f ®® (x) > 0 on an interval, then f is concave up on that
interval.
(b) If f ®® (x) < 0 on an interval, then f is concave down on
that interval.
Of particular interest are points at which the concavity changes
from up to down or down to up.
Definition. If f is continuous at x = a and its concavity
changes either from up to down or down to up at x = a, then f
has an inflection point at x = a.
4
f®
Theorem 43 (Test for Concavity). Suppose that f ®® (x) exists
on an interval.
It is instructive to see some examples of inflection points:
1
2
x
*2
*4
Hence we have seen that if f ® is zero at a point and increasing on
an interval containing that point, then f has a local minimum at
the point. If f ® is zero at a point and decreasing on an interval
containing that point, then f has a local maximum at the point.
Thus, we see that we can gain information about f by studying
how f ® changes. This leads us to our next section.
This is an inflection point. The
concavity changes from concave
up to concave down.
This is an inflection point. The
concavity changes from concave
up to concave down.
It is also instructive to see some nonexamples of inflection
points:
This is not an inflection point.
The curve is concave down on
either side of the point.
This is not an inflection point.
The curve is concave down on
either side of the point.
177
Maximums and minimums
We identify inflection points by first finding x such that f ®® (x)
is zero or undefined and then checking to see whether f ®® (x)
does in fact go from positive to negative or negative to positive
at these points.
Warning. Even if f ®® (a) = 0, the point determined by x = a
might not be an inflection point.
Example 116. Describe the concavity of f (x) = x3 * x.
To start, compute the first and second derivative of f (x)
with respect to x,
®
and
2
f (x) = 3x * 1
®®
f (x) = 6x.
Since f ®® (0) = 0, there is potentially an inflection point
at x = 0. Using test points, we note the concavity does
change from down to up, hence there is an inflection point
at x = 0. The curve is concave down for all x < 0 and
concave up for all x > 0, see the graphs of f (x) = x3 * x
and f ®® (x) = 6x.
y
*1.5
*1
f
*0.5
Recall the first derivative test:
• If f ® (x) > 0 to the left of a and f ® (x) < 0 to the right of
a, then f (a) is a local maximum.
• If f ® (x) < 0 to the left of a and f ® (x) > 0 to the right of
a, then f (a) is a local minimum.
If f ® changes from positive to negative it is decreasing. In
this case, f ®® might be negative, and if in fact f ®® is negative
then f ® is definitely decreasing, so there is a local maximum at
the point in question. On the other hand, if f ® changes from
negative to positive it is increasing. Again, this means that
f ®® might be positive, and if in fact f ®® is positive then f ® is
definitely increasing, so there is a local minimum at the point
in question. We summarize this as the second derivative test.
Theorem 44 (Second Derivative Test). Suppose that f ®® (x) is
continuous on an open interval and that f ® (a) = 0 for some
value of a in that interval.
• If f ®® (a) < 0, then f has a local maximum at a.
2
f
The second derivative test
• If f ®® (a) > 0, then f has a local minimum at a.
®®
0.5
1
x
1.5
*2
• If f ®® (a) = 0, then the test is inconclusive. In this case, f
may or may not have a local extremum at x = a.
The second derivative test is often the easiest way to identify
local maximum and minimum points. Sometimes the test fails
and sometimes the second derivative is quite difficult to evaluate. In such cases we must fall back on one of the previous
tests.
Example 117. Once again, consider the function
Note that we need to compute and analyze the second derivative
to understand concavity, so we may as well try to use the second
derivative test for maxima and minima. If for some reason this
fails we can then try one of the other tests.
178
f (x) =
x4 x3
+
* x2
4
3
Use the second derivative test, to locate the local extrema of f .
Maximums and minimums
Start by computing
and
f ® (x) = x3 +x2 *2x
f ®® (x) = 3x2 +2x*2.
(c) It gives no information on whether the function has a local
extremum at x = a.
Using the same technique as we used before, we find that
f ® (*2) = 0,
f ® (0) = 0,
f ® (1) = 0.
Now we’ll attempt to use the second derivative test,
f ®® (*2) = 6,
f ®® (0) = *2,
f ®® (1) = 3.
Hence we see that f has a local minimum at x = *2, a
local maximum at x = 0, and a local minimum at x = 1,
see below for a plot of f (x) = x4 _4 + x3 _3 * x2 and
f ®® (x) = 3x2 + 2x * 2:
y
6
4
f ®®
*3
*2
2
*1
f
1
2
x
*2
*4
Question 92. If f ®® (a) = 0, what does the second derivative
test tell us?
Multiple Choice:
(a) The function has a local extremum at x = a.
(b) The function does not have a local extremum at x = a.
179
Concepts of graphing functions
21 Concepts of graphing
functions
After completing this section, students should be able to do the
following.
• Understand what information the derivative gives concerning when a function is increasing or decreasing.
• Understand what information the second derivative gives
concerning concavity of the graph of a function.
• Interpret limits as giving information about functions.
• Determine how the graph of a function looks based on an
analytic description of the function.
180
What’s the graph look like?
Break-Ground:
21.1
What’s the graph look like?
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, I’ve been thinking about the derivative.
Riley: It’s all about change. It’s some “change-detector” tool for
math.
Devyn: I know! What’s crazy is that you can use it as a tool for
sniffing out dirt on functions.
Riley: First f ® tells us increasing or decreasing.
Devyn: Then f ®® tells us concavity.
Riley: From just that we know all local maxes and mins.
Devyn: And if we use limits, we can find any asymptotes!
Riley: You know, I’d like to make up a procedure based on all these
facts, that would tell me what the graph of any function would
look like.
Devyn: Me too! Let’s get to work!
Problem 1. On some interval, we know that f ® (x) is positive
and f ®® (x) is positive. Which of the following is the best option
for the shape of the graph on that interval?
Multiple Choice:
(a)
(d)
Problem 2. On some interval, we know that f ® (x) is negative
and f ®® (x) is positive. Which of the following is the best option
for the shape of the graph on that interval?
Multiple Choice:
(a)
(b)
(c)
(d)
(b)
(c)
181
Concepts of graphing functions
• f ® (x) = *2 on (6, ÿ)
Dig-In:
21.2
Concepts of graphing functions
In this section, we review the graphical implications of limits,
and the sign of the first and second derivative. You already
know all this stuff: it is just important enough to hit it more
than once, and put it all together.
Example 118. Sketch the graph of a function f which has the
following properties:
• f (0) = 0
• f ®® (x) < 0 on (2.5, 5)
• f ®® (x) > 0 on (*ÿ, 2.5) ‰ (5, 6)
Try this on your own first, then either check with a friend
or check the online version.
Example 120. The graph of f ® (the derivative of f ) is shown
below.
Assume f is continuous for all real numbers.
• lim + f (x) = +ÿ
xô10
y
• lim * f (x) = *ÿ
xô10
4
• f ® (x) < 0 on (*ÿ, 0) ‰ (6, 10) ‰ (10, 14)
• f ® (x) > 0 on (0, 6) ‰ (14, ÿ)
2
®®
• f (x) < 0 on (4, 10)
• f ®® (x) > 0 on (*ÿ, 4) ‰ (10, ÿ)
*2
*1
1
Try this on your own first, then either check with a friend
or check the online version.
*2
Example 119. Sketch the graph of a function f which has the
following properties:
*4
• f (0) = 1
• f (6) = 2
• lim+ f (x) = 3
xô6
Select All Correct Answers:
(a) (*ÿ, 0)
• f ® (x) < 0 on (*ÿ, 1)
(b) (0, 1)
• f ® (x) > 0 on (1, 6)
(c) (1, 2)
182
3
4
x
Question 93. On which of the following intervals is f increasing?
• lim* f (x) = 1
xô6
2
Concepts of graphing functions
(d) (2, 3)
(a) (*ÿ, 0)
(e) (3, ÿ)
(b) (0, 1)
Question 94. Which of the following are critical points of f ?
(c) (1, 2)
Select All Correct Answers:
(d) (2, 3)
(a) x = 0
(e) (3, ÿ)
(b) x = 1
(c) x = 2
(d) x = 3
Question 95. Where do the local maxima occur?
Select All Correct Answers:
(a) x = 0
(b) x = 1
(c) x = 2
(d) x = 3
Question 96. Where does a point of inflection occur?
Select All Correct Answers:
(a) x = 0
(b) x = 1
(c) x = 2
(d) x = 3
Question 97. On which of the following intervals is f concave
down?
Select All Correct Answers:
183
Computations for graphing functions
22 Computations for graphing
functions
After completing this section, students should be able to do the
following.
• Find the intervals where a function is increasing or decreasing.
• Find the intervals where a function is concave up or down.
• Determine how the graph of a function looks without
using a calculator.
184
Wanted: graphing procedure
Break-Ground:
22.1
Wanted: graphing procedure
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, OK I know how to plot something if I’m given a
description.
Riley: Yes, it’s kinda fun right?
Devyn: I know! But now I’m not sure how to get the information I
need.
Riley: You know, I’d like to make up a procedure based on all these
facts, that would tell me what the graph of any function would
look like.
Devyn: Me too! Let’s get to work!
Use either the first or second derivative test to identify local
extrema and/or find the intervals where your function is increasing/decreasing. In what order should we take these steps?
For example, one must compute f ® before computing f ®® . Also,
one must compute f ® before finding the critical points. There
is more than one correct answer.
Problem 1. Below is a list of features of a graph of a function.
(a) Find the points x = a where f (x) goes to infinity as x
goes to a (from the right, left, or both). These are the
points where f has a vertical asymptote.
(b) Find the critical points (the points where f ® (x) = 0 or
f ® (x) is undefined).
(c) Identify inflection points and concavity.
(d) Determine an interval that shows all relevant behavior.
(e) Find the y-intercept, this is the point (0, f (0)). Place this
point on your graph.
(f) Find the candidates for inflection points, the points where
f ®® (x) = 0 or f ®® (x) is undefined.
(g) If possible, find the x-intercepts, the points where f (x) =
0. Place these points on your graph.
(h) Compute f ® and f ®® .
(i) Analyze end behavior: as x ô ±ÿ, what happens to the
graph of f ? Does it have horizontal asymptotes, increase
or decrease without bound, or have some other kind of
behavior?
185
Computations for graphing functions
Dig-In:
22.2 Computations for graphing
functions
Example 121. Sketch the graph of the function defined by
f (x) = 2x3 * 3x2 * 12x.
The y-intercept is (0, 0). Place this point on your plot.
Let’s get to the point. Here we use all of the tools we know to
sketch the graph of y = f (x):
y
20
• Find the y-intercept, this is the point (0, f (0)). Place this
point on your graph.
• Find any vertical asymptotes, these are points x = a where
f (x) goes to infinity as x goes to a (from the right, left,
or both).
10
*2
*1
• If possible, find the x-intercepts, the points where f (x) =
0. Place these points on your graph.
• Analyze end behavior: as x ô ±ÿ, what happens to the
graph of f ? Does it have horizontal asymptotes, increase
or decrease without bound, or have some other kind of
behavior?
• Compute f ® and f ®® .
• Find the critical points (the points where f ® (x) = 0 or
f ® (x) is undefined).
• Use either the first or second derivative test to identify
local extrema and/or find the intervals where your function
is increasing/decreasing.
• Find the candidates for inflection points, the points where
f ®® (x) = 0 or f ®® (x) is undefined.
• Identify inflection points and concavity.
• Determine an interval that shows all relevant behavior.
At this point you should be able to sketch the plot of your
function.
186
1
2
3
4
x
*10
*20
Which of the following are vertical asymptotes? Select all that
apply.
Select All Correct Answers:
(a) x = 0
(b) x = 1
(c) x = *1
˘
(d) x = 2
(e) There are no vertical asymptotes
In this case, f (x) = 2x3 * 3x2 * 12x, we can find the xintercepts. There are three x intercepts. Call them a, b, and c,
and order them such that a < b < c. Then
Computations for graphing functions
On the other hand
a=
3*
105
4
b = 0,
c=
˘
3+
˘
105
4
f ®® (x) = 12x * 6
⇠
⇡
1
= 12 x *
.
2
,
.
Which of the following best describes the end behavior of f as
x ô ÿ?
Multiple Choice:
(a) f increases without bound.
The critical points are where f ® (x) = 0, thus we need to solve
6(x + 1)(x * 2) = 0 for x. This equation has two solutions. If
we call them a and b, with a < b, then what are a and b?
a = *1
y
20
(c) f has a horizontal asymptote.
(d) f has some other behavior at ÿ.
Multiple Choice:
b = 2.
Mark the critical points x = 2 and x = *1 on your plot.
(b) f decreases without bound.
Which of the following best describes the end behavior of f as
x ô *ÿ?
and
10
*2
*1
1
2
3
4
x
*10
(a) f increases without bound.
(b) f decreases without bound.
(c) f has a horizontal asymptote.
(d) f has some other behavior at ÿ.
Compute f ® (x) and f ®® (x),
®
2
f (x) = 6x * 6x * 12
= 6 x2 * x * 2
= 6(x + 1) (x * 2) .
*20
Since f ® (x) = 6(x + 1)(x * 2), and both factors (x + 1) and
(x * 2) are negative for x < *1, it follows that f ® (x) > 0, and,
therefore, our function is increasing on (*ÿ, *1).
Similarly, since for *1 < x < 2, (x + 1) > 0 and (x * 2) < 0,
the derivative is negative there, and, therefore, our function is
decreasing on (*1, 2).
And, for x > 2, both factors (x + 1) and (x * 2) are positive,
and, therefore, our function is increasing on (2, ÿ).
187
Computations for graphing functions
Hence x = *1, corresponding to the point (*1, 7) is a local
maximum and x = 2, corresponding to the point (2, *20) is
local minimum of f (x). Identify this on your plot.
interval, see the figure below.
y
20
y
20
10
10
*2
*1
*2
1
2
3
4
x
The candidates for the inflection points are where f ®® (x) = 0,
1
thus we need to solve 12(x * ) = 0 for x.
2
The solution to this is x = ? .
This is only a possible inflection point, since the concavity
needs to change to make it a true inflection point.
1
We have that f ®® (x) < 0 for x < , therefore f is concave
2
1
downon (*ÿ, ).
2
1
Similarly, f ®® (x) > 0 for x > , therefore f is concave up on
2
1
( , ÿ).
2
So this point is a point of inflection.
Since all of this behavior as described above occurs on the
interval [*2, 4], we now have a complete sketch of f (x) on this
188
1
2
3
4
x
*10
*10
*20
*1
*20
Example 122. Sketch the plot of
h
nxex + 2
f (x) = l
nx4 * x2 + 3
j
if x < 0
if x g 0.
NOTE: lim xex = 0. You will need this limit when answering
xô*ÿ
some questions below.
Try this on your own first, then either check with a friend,
a graphing calculator (like Desmos2 ), or check the online
version.
Since this function is piecewise defined, we will analyze the
cases (*ÿ, 0) and [0, ÿ) separately.
Because f is piecewise defined, and potentially discontinuous
at 0, it is important to understand the behavior of f near x = 0.
lim f (x) = 2
xô0*
lim f (x) = 3
2
xô0+
See Desmos at http://www.desmos.com
Computations for graphing functions
Moreover,
Select All Correct Answers:
(a) x = 0
f (0) = 3
Record this information on our graph with filled and unfilled
circles.
y
(b) x = 1
(c) x = *1
˘
(d) x = 2
(e) There are no vertical asymptotes
4
Which of the following best describes the end behavior of f as
x ô ÿ?
2
Multiple Choice:
(a) f increases without bound.
*5
*4
*3
*2
*1
1
2
x
Which of the following are vertical asymptotes on (*ÿ, 0)?
Select all that apply.
Select All Correct Answers:
(a) x = 0
(b) x = 1
(c) x = *1
˘
(d) x = 2
(e) There are no vertical asymptotes
Which of the following are vertical asymptotes on (0, ÿ)? Select all that apply.
(b) f decreases without bound.
(c) f has a horizontal asymptote of y = 2.
(d) f has some other behavior at ÿ.
Which of the following best describes the end behavior of f as
x ô *ÿ?
Multiple Choice:
(a) f increases without bound.
(b) f decreases without bound.
(c) f has a horizontal asymptote of y = 2.
(d) f has some other behavior at ÿ.
We mark the location of the horizontal asymptote:
189
Computations for graphing functions
y
*5
*4
*3
*2
y
4
4
2
2
*1
1
2
x
*5
*4
*3
*2
*1
1
2
x
The derivative of f on (*ÿ, 0) is
f ® (x) = xex + ex = ex (x + 1)
The derivative of f on (0, ÿ) is
⇠
⇡
1
f ® (x) = 4x3 * 2x = 4x x2 *
2
The critical points are where f ® (x) = 0 or does not exist. 0
is a critical point, since we have already seen it is a point of
discontinuity for f , and thus f ® (0) does not exist there. On
(*ÿ, 0), f has a critical point at x = *1. On (0, ÿ), f has a
1
critical point at x = ˘ . Mark the critical points x = *1 and
2
1
x = ˘ on your plot. would be nice
2
190
Using the first derivative, we can see that on (*ÿ, *1), f ® (x) =
ex (x + 1) and has the same sign as the factor (x + 1), which
is negative. Therefore, f is decreasing on (*ÿ, *1). On
(*1, 0), f ® (x) has the same sign as the factor (x + 1), which is
1
positive. Therefore, increasing on (*1, 0). On (0, ˘ ), since
2I
H
⇠ ⇡
1
1 1
1
f®
=
* < 0, f is decreasing. On ˘ , ÿ , since
4
16 2
2
⇠
⇡
1
®
f (1) = 4 1 *
> 0, f is increasing.
2
Computations for graphing functions
y
*5
*4
*3
*2
y
4
4
2
2
*1
1
2
x
*4
*3
*2
*1
1
x
x
y
f (x) = xe + 2e = e (x + 2)
x
2
Since all of this behavior as described above occurs on the
interval [*5, 2], we now have a complete sketch of y = f (x)
on this interval, see the figure below.
The second derivative of f on (*ÿ, 0) is
®®
*5
x
4
The second derivative of f on (0, ÿ) is
⇠
f ®® (x) = 12x2 * 2 = 12 x2 *
1
6
2
⇡
The candidates for the inflection points are where f ®® (x) = 0.
*5
*4
*3
*2
*1
1
2
x
On (*ÿ, 0), f ®® has one zero, namely x = *2. The sign of
f ®® changes from negative to positive] through this point. On
1
(0, ÿ), f ®® has one zero, namely x = ˘ . The sign of f ®®
6
changes from negative to positive] through this point.
191
Mean Value Theorem
23
Mean Value Theorem
After completing this section, students should be able to do the
following.
• Understand the statement of the Extreme Value Theorem.
• Understand the statement of the Mean Value Theorem.
• Sketch pictures to illustrate why the Mean Value Theorem
is true.
• Determine whether Rolle’s Theorem or the Mean Value
Theorem can be applied.
• Find the values guaranteed by Rolle’s Theorem or the
Mean Value Theorem.
• Use the Mean Value Theorem to solve word problems.
• Compare and contrast the Intermediate Value Theorem,
Mean Value Theorem, and Rolle’s Theorem.
• Identify calculus ideas which are consequences of the
Mean Value Theorem.
192
Let’s run to class
Break-Ground:
(b) Riley has the larger average velocity.
23.1
(c) Their average velocities are equal.
Let’s run to class
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, I want to go to math class. Now. Look, this straight,
long hall is empty. Let’s race!
Riley: Yes. Let’s race! On, your mark. . . . Ready. Steady. Go!
Devyn: You may think you’re fast, but I’m catching up!
Riley: Noooooo!
Devyn: Now I’m winning! I’ve never won a foot race in my life!
Riley: Never. . . give. . . up!
Devyn: Whew! We both made it to math class at exactly the same
time!
Riley: Wow. We should run to every class. Hey I have a question,
was there a time during our race that we were running at exactly
the same speed?
(d) None of the above.
Problem 3. Record your guess to Riley’s question: is there a
moment during the race where Devyn and Riley were running
at exactly the same speed?
Problem 1. Which of the following describes the race above?
Multiple Choice:
(a) Devyn was leading until the end, when the race finished
in a tie.
(b) Riley was leading until the end, when the race finished in
a tie.
(c) Devyn was leading, then Riley was leading until the end,
when the race finished in a tie.
(d) Riley was leading, then Devyn was leading until the end,
when the race finished in a tie.
(e) None of the above.
Problem 2. What can you say about Devyn’s and Riley’s average velocities?
Multiple Choice:
(a) Devyn has the larger average velocity.
193
The Extreme Value Theorem
Dig-In:
23.2
The Extreme Value Theorem
Definition.
(a) A function f has a global maximum at x = a, if f (a) g
f (x) for every x in the domain of the function.
(b) A function f has a global minimum at x = a, if f (a) f
f (x) for every x in the domain of the function.
Question 100. Find the x-coordinate(s) of the point(s) where
the function f has a local minimum. Observe that f (1) f f (x)
for all x in the interval, say, (0, 2). But it is not true that f (1) f
f (x) for all x in the domain of f . For example, f (4.5) < f (1).
Question 101. Does every function attain a global extremum
on its domain? Select the correct answer.
Multiple Choice:
(a) All functions must attain both global minimum and global
maximum on their domain.
A global extremum is either a global maximum or a global
minimum.
(b) All functions attain a global minimum on their domain.
Let f be the function given by the graph below.
(d) Some functions have no global extremums on their domain.
(c) All functions attain a global maximum on their domain.
Check the following graph.
y
2
y
1
1
1
2
3
4
5
x
*1
Question 98. Find the x-coordinate of the point where the
function f has a global maximum.
Observe that f (2) g f (x) for all x in the domain of f . Notice
that the function f has also a local maximum at x = 2.
Question 99. Find the x-coordinate of the point where the
function f has a global minimum. Observe that f (5) f f (x)
for all x in the domain of f . Notice that the function f does
not have a local minimum at x = 5. Recall, a function cannot
not have a local extremum at a boundary point.
194
*1
1
2
3
4
5
x
*1
*2
Notice, the function is not continuous at x = 1, and, therefore,
f is not continuous on its domain, (*1, 5).
Does the function given by the graph above attain a global
extremum on its domain? Select the correct answer.
Multiple Choice:
(a) The function attains both global minimum and global
maximum on its domain.
The Extreme Value Theorem
(b) The function attains a global minimum, but has no global
maximum on its domain.
(c) The function attains a global maximum, but has no global
minimum on its domain.
(d) The function has no global extremum on its domain.
Check the graph below.
y
1
*1
1
2
3
4
5
x
*1
*2
y
1
*1
1
2
3
4
5
x
*1
*2
Notice, the function is continuous on its domain (*1, 5). Does
the function given by the graph above attain a global extremum
on its domain? Select the correct answer.
Multiple Choice:
(a) The function attains both global minimum and global
maximum on its domain.
(b) The function attains a global minimum, but has no global
maximum on its domain.
(c) The function attains a global maximum, but has no global
minimum on its domain.
(d) The function has no global extremum on its domain.
Check the following graph.
Notice, the function is continuous on a closed interval [*1, 5].
Does the function given by the graph above attain a global
extremum on its domain? Select the correct answer.
Multiple Choice:
(a) The function attains both global minimum and global
maximum on its domain.
(b) The function attains a global minimum, but has no global
maximum on its domain.
(c) The function attains a global maximum, but has no global
minimum on its domain.
(d) The function has no global extremum on its domain.
Question 102. Find the x-coordinate(s) of the point(s) where
the function f has a global minimum.
Question 103. Find the x-coordinate(s) of the point(s) where
the function f has a global maximum.
Sometimes it is important to know whether a function attains a
global extremum on its domain. The following theorem, which
comes as no surprise after the previous three examples, gives a
simple answer to that question.
Theorem 45 (Extreme Value Theorem). If f is continuous
on the closed interval [a, b], then there are points c and d in
195
The Extreme Value Theorem
[a, b], such that (c, f (c)) is a global maximum and (d, f (d)) is
a global minimum on [a, b].
(a) Does the function f satisfy the conditions of the Extreme
Value Theorem on its domain?
Below, we see a geometric interpretation of this theorem.
Multiple Choice:
y
(i) Yes, because f is continuous on (*2, 1).
f (c)
(ii) Yes, because f is continuous on [*2, 1].
(iii) No, because f is not continuous on [*2, 1].
(iv) No, because f is not differentiable on (*2, 1).
Now we know that the Extreme Value Theorem guarantees
that the function f attains both global extremums on its
domain!
f (d)
a
c
d
b
x
Question 104. Would this theorem hold if we were working on
an open interval?
Question 105. Would this theorem hold if we were working
on a closed interval [a, b], but a function is not continuous on
[a, b]?
Assume that a function f is continuous on a closed interval
[a, b]. By the Extreme Value Theorem, f attains both global
extremums on the interval [a, b]. How can we locate these
global extrema? We have seen that they can occur at the end
points or in the open interval (a, b). If a global extremum
occurs at a point x in the open interval (a, b), then f has a
local extremum at x. That means that f has a critical point at
x. So, the global extrema of a function f occur either at the
end points, a or b, or at critical points. If we want to locate
the global extrema, we have to evaluate the function at the end
points and at critical points, and compare the values.
Question 106. Let f (x) = x2 e*x , for *2 f x f 1.
196
(b) Locate the global extremums of f on the closed interval
[*2, 1].
The global extremums occur at the end points or at
critical points.
Let’s find the critical points of f . First, compute the
derivative of f . In order to find the critical points
of f , we have to solve the equation It follows that
the function f has only one critical point (c, f (c)).
Find c. In order to locate the global extremums of
f , we have to evaluate f at the end points and at the
critical point.
Order the three values, f (*2), f (c), and f (1), from
smallest to largest. You should replace c with its
value, when you write f (c) in your answer below.
Based on this comparison, the function f has the
global minimum at x = 0, and the global maximum
at x = *2.
For your convenience, the graph of f on the interval [*2, 1] is
given below.
The Extreme Value Theorem
y
30
20
y = x2 e*x
*2
*1.5
*1
10
*0.5
0.5
1
x
197
The Mean Value Theorem
Then
Dig-In:
23.3
The Mean Value Theorem
Here are some interesting questions involving derivatives:
(a) Suppose you toss a ball upward into the air and then catch
it. Must the ball’s velocity have been zero at some point?
(b) Suppose you drive a car on a straight stretch of highway
from toll booth to another toll booth 30 miles away in half
of an hour. Must you have been driving at 60 miles per
hour at some point?
(c) Suppose two different functions have the same derivative.
What can you say about the relationship between the two
functions?
While these problems sound very different, it turns out that the
problems are very closely related. We’ll start simply:
Theorem 46 (Rolle’s Theorem). Suppose that f is differentiable on the interval (a, b), continuous on the interval [a, b],
and that f (a) = f (b).
f ® (c) = 0
for some c in the open interval (a, b).
If the function f happens to be a constant, then f ® (c) = 0
for all points c in the open interval (a, b).
If f is not a constant, then it must attain both global
extrema in the interval [a, b], by the Extreme Value Theorem. One of these two different extrema is attained in
the open interval (a, b) at some point c, and the function
f has a critical point there. Since f is differentiable on
(a, b), it follows that f ® (c) = 0.
We can now answer our first question above.
Example 123. Suppose you toss a ball upward into the air and
then catch it at the same point where you tossed it from. Must
the ball’s velocity have been zero at some point?
Let p(t) be the position of the ball (on the vertical line)
at time t. Our time interval in question will be
y
[tstart , tf inish ].
We may assume that p is continuous on [tstart , tf inish ] and
differentiable on (tstart , tf inish ). Since p(tstart ) = p(tf inish ),
we may now apply Rolle’s Theorem and conclude that at
some time c in the given time interval, p® (c) = 0. Hence
the velocity must be zero at some point.
f
Rolle’s Theorem is a special case of a more general theorem.
a
198
c
b
x
Theorem 47 (Mean Value Theorem). Suppose that f has a
derivative on the interval (a, b) and is continuous on the interval
[a, b].
The Mean Value Theorem
some point.
y
Now we will address the unthinkable: could there be a continuous function f on [a, b] whose derivative is zero on (a, b) that
is not constant? As we will see, the answer is “no.”
Theorem 48. If f ® (x) = 0 for all x in an interval I, then f (x)
is constant on I.
Let a < b be two points in I. Since f is continuous
on [a, b] and differentiable on (a, b), by the Mean Value
Theorem we know
f (b) * f (a)
= f ® (c)
b*a
a
Then
c
f ® (c) =
for some c in (a, b).
b
x
f (b) * f (a)
b*a
We can now answer our second question above.
Example 124. Suppose you drive a car on a straight stretch
of highway from toll booth to another toll booth 30 miles away
in half of an hour. Must you have been driving at 60 miles per
hour at some point?
Let p(t) is the position of the car at time t, and 0 hours
is the starting time with 1_2 hours being the final time,
where p(0) = 0, and p(1_2) = 30. We can assume that p
is continuous on [0, 1_2] and differentiable on (0, 1_2).
Now the Mean Value Theorem states that there is a time
c such that
30 * 0
p® (c) =
= 60
1_2 * 0
where 0 < c < 1_2.
for some c in the interval (a, b). Since f ® (c) = 0 we
see that f (b) = f (a). Moreover, since a and b were
arbitrarily chosen, f (x) must be the constant function.
Now let’s answer our third question.
Example 125. Suppose two different functions have the same
derivative on some interval I. What can you say about the
relationship between the two functions?
Set h(x) = f (x) * g(x), so h® (x) = f ® (x) * g ® (x). Now
h® (x) = 0 on the interval I. This means that h(x) = k,
for all x in I, where k is some constant. Hence, for all x
in I
f (x) = g(x) + k.
Example 126. Describe all functions whose derivative is
sin(x).
One such function is * cos(x), so all such functions have
the form * cos(x) + k,
Since the derivative of position is velocity, this says that
the car must have been driving at 60 miles per hour at
199
The Mean Value Theorem
4
In fact, this shows us that the average rate of change of
y
f (t) = PDevyn (t) * PRiley (t)
[tstart , tf inish ]
is 0. Hence by the mean value theorem, there is a point c
2
tstart f c f tf inish
1
2
3
4
5
6
x
with f ® (c) = 0. However,
®
®
0 = f ® (c) = PDevyn
(c) * PRiley
(c).
*2
Hence at c,
*4
Finally, let us investigate two young mathematicians who run
to class.
Example 127. Two students Devyn and Riley raced to class
down a straight hall. They start at the same time and finish in
a tie. Was there a point during the race that Devyn and Riley
were running at exactly the same velocity?
Let PDevyn represent Devyn’s position with respect to
time, and let PRiley represent Riley’s position with respect
to time. Let tstart be the starting time of the race, and
tf inish be the end of the race. Set
f (t) = PDevyn (t) * PRiley (t).
Note, we may assume that PDevyn and PRiley (t) are continuous on [tstart , tf inish ] and that they are differentiable on
(tstart , tf inish ). Hence the same is true for f . Since both
runners start and finish at the same place,
f (tstart ) = PDevyn (tstart ) * PRiley (tstart ) = 0
f (tf inish ) = PDevyn (tf inish ) * PRiley (tf inish ) = 0.
200
on
and
®
®
PDevyn
(c) = PRiley
(c),
this means that there was a time when they were running
at exactly the same velocity.
In conclusion, the Mean Value Theorem relates the function f
and its derivative, f ® . Since the derivative has many interpretations, e.g. instantaneous rate of change, slope of the tangent
line, velocity, it is no surprise that we can use the MVT in
different contexts.
Therefore, if the the function f is continuous on [a, b] and
differentiable on (a, b), then the MVT guarantees that
(a) there is a point c in (a, b) where the instantaneous rate of
change of f is equal to the average rate of change of f
over the interval [a, b];
or
(b) there is a point c in (a, b) where the slope of the tangent
line to the curve y = f (x) is equal to the slope of the
secant line through the points (a, f (a)) and (b, f (b));
or
(c) there is a point c in the time interval (a, b) where the
instantaneous velocity is equal to the average velocity
over the time interval [a, b]. Here, we assume that f is the
position function for the object moving along a straight
line.
Linear approximation
24
Linear approximation
After completing this section, students should be able to do the
following.
• Define linear approximation as an application of the tangent to a curve.
• Find the linear approximation to a function at a point and
use it to approximate the function value.
• Identify when a linear approximation can be used.
• Label a graph with the appropriate quantities used in linear
approximation.
• Find the error of a linear approximation.
• Compute differentials.
• Use the second derivative to discuss whether the linear
approximation over or underestimates the actual function
value.
• Contrast the notation and meaning of dy versus y.
• Understand that the error shrinks faster than the displacement in the input.
• Justify the chain rule via the composition of linear approximations.
201
Replacing curves with lines
Break-Ground:
24.1
Replacing curves with lines
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Hmmmm. Riley, I just thought of something. . .
Riley: What is it?
Devyn: When we compute derivatives, we are looking at the slope
of tangent lines right?
Riley: You know it.
Devyn: Well, I wonder: Instead of studying curves, could we just
study “zoomed-in” lines?
Riley: I’m not sure. . .
You read someplace that
1
(x * 4) + 2
4
˘
is a good approximation for f (x) = x when x is close to 4.
™(x) =
Problem 1. Plot ™(x) and f (x). Explain how this shows that
™(x) is a good approximation when x is close to 4.
Problem 2. Explain (if you can) using concepts of calculus to
explain why ™(x) is a good approximation for f (x) when x is
close to 4.
202
Linear approximation
to the curve y = f (x) at the point (a, f (a)). Select the correct
choice.
Dig-In:
24.2
Linear approximation
Given a function, a linear approximation is a fancy phrase for
something you already know:
Multiple Choice:
The line tangent to the graph of a function at a
point is very close to the graph of the function
near that point.
(b) m = f ® (a)
This tangent line is the graph of a linear function, called the
linear approximation.
Example 128. Let f be a function that is differentiable on some
interval I that contains the point a. The graph of a function f
and the line tangent to the curve
y = f (x) at the point where x = a are given in the figure below.
Find the equation of the tangent line.
(a) m = f (a)
(c) m =
f (a) * f (0)
a*0
(d) m =
f (a + h) * f (a)
h
(e) We don’t have enough information to determine the slope.
Since we know that the point (a, f (a)) lies on the tangent line,
we can write an equation of the tangent line.
y = f ® (a)(x * a) + f (a).
y
y = L(x)
Now, we define a function, L, by L(x) = f ® (a)(x * a) + f (a).
This function is linear and its graph is the line tangent to the
curve y = f (x) at the point where x = a. This function
deserves a special name.
y = f (x)
Definition. If f is a function differentiable at x = a, then a
linear approximation to the function f at x = a is given by
(a, f (a))
L(x) = f ® (a)(x * a) + f (a).
Note that the graph of L is just the tangent line to the graph of
f at x = a.
x
First, find the expression for m, the slope of the tangent line
A linear approximation of f is a “good” approximation as
long as x is “not too far” from a. If one “zooms in” on the
graph of f sufficiently, then the graphs of f and L are nearly
indistinguishable.
As a first example, we will see how linear approximations allow
us to approximate “difficult” computations.
203
Linear approximation
Example 129. Let f be a function defined by
˘
f (x) = 3 x.
˘
3
Approximate 50, using L, a linear approximation to the function f at a = 64.
To start, write
d
d 1_3
1
f (x) =
x =
.
dx
dx
3x2_3
1
(x * 64)
3 642_3
1
= 4 + (x * 64)
48
x
8
=
+ .
48 3
while perhaps inexact, is computationally easier than
computing the cube root.
What would happen if we chose a = 27 instead?
Then we would use L27 , the linear approximation to the function
f at a = 27. In that case, L27 = f (27) + f ® (27)(x * 27) =
1
3 + (x * 27). The graph of L27 , together with the graphs of
27
f and L = L64 is given in the figure below.
5
y
L(x) = 4 +
5
y
2
L64
3
L27
2
1
4
3
4
L
f
27
f
40 50
64
x
80 90 100
From the picture we can see that
L27 (50) > L64 (50) > f (50).
So, our choice, a = 64, was better!
1
20
40 50
64
x
80 90 100
˘
3
Now we evaluate L(50) ˘ 3.71 and compare it to 50 ˘
3.68. From this we see that the linear approximation,
204
With modern calculators and computing software, it may not
appear necessary to use linear approximations. In fact they are
quite useful. In cases requiring an explicit numerical approximation, they allow us to get a quick rough estimate which can
be used as a “reality check” on a more complex calculation.
In some complex calculations involving functions, the linear
approximation makes an otherwise intractable calculation possible, without serious loss of accuracy.
Linear approximation
Example 130. Use a linear approximation of f (x) = sin(x)
at a = 0 to approximate sin(0.3).
To start, write
Di�erentials
The graph of a function f and the graph of L, the linear approximation of f at a, are shown in the figure below. Also,
two quantities, dx and df , and a point P are marked in the figure. Look carefully at the figure when answering the questions
below.
y
d
f (x) = cos(x),
dx
so our linear approximation is
L(x) = 0 + cos(0) (x * 0)
= x.
(x, f (x))
P
y
1
L
y = f (x)
f
y = L(x)
df
(a, f (a))
dx
x
0.3
*⇡_2
x
⇡_2
Question 107. Select all the correct expressions for the quantity dx.
Select All Correct Answers:
*1
(a) dx = f (x) * f (a)
(b) dx = f (x) * L(x)
Hence, a linear approximation for sin(x) at a = 0 is
L(x) = x, and so L(0.3) = 0.3. Comparing this to
sin(.3) ˘ 0.295, we see that the approximation is quite
good. For this reason, it is common to approximate sin(x)
with its linear approximation L(x) = x when x is near
zero.
(c) dx = x * a
(d) dx = L(x) * f (a)
(e) dx = L(x) * L(a)
Question 108. Select all the correct expressions for the quantity df .
205
Linear approximation
Select All Correct Answers:
y
(a) df = f (x) * f (a)
(b) df = f (x) * L(x)
(c) df = x * a
(x + dx, f (x + dx))
(d) df = L(x) * f (a)
(e) df = L(x) * L(a)
f
Question 109. Based on the figure and the expression for L(x),
select all the correct expressions for df .
df
(x, f (x))
dx
L
x
Select All Correct Answers:
(a) df = f ® (a)(x * a)
Note, it is now the case (by definition!) that
f ® (x) =
(b) df = f ® (a) dx
(c) df = f ® (x * a)
(d) df = f (x) dx
®
(e) df = f ® (x)(x * a)
So, we can write df = f ® (a) dx and call it a differential of f
at a. Notice that we can define a differential at any point x of
the domain of f , provided that f ® (x) exists. We will do that in
our next definition.
Definition. Let f be a differentiable function, let x be a point
in the domain of f , and let dx be some quantity, called a
differential of x. We define df , a differential of f , at a point
x by
df = f ® (x) dx
Geometrically, differentials can be interpreted via the diagram
below.
206
df
.
dx
We should not be surprised, since the slope of the tangent line
df
in the figure is f ® (x), and this slope is also given by
.
dx
Essentially, differentials allow us to solve the problems presented in the previous examples from a slightly different point
of view. Recall, when h is near but not equal zero,
f ® (x) ˘
Hence,
f (x + h) * f (x)
.
h
f ® (x)h ˘ f (x + h) * f (x).
We can replace a quantity h with a quantity dx to write
f ® (x) dx ˘ f (x + dx) * f (x)
df ˘ f (x + dx) * f (x).
Adding f (x) to both sides we see
f (x) + df ˘ f (x + dx)
Linear approximation
or, equivalently
4.5
y
f (x + dx) ˘ f (x) + df .
There are contexts where the language of differentials is common. Here is the basic strategy:
what you want
L
3.5
f
what you compute
We will repeat our previous examples using differentials.
˘
3
Example 131. Use differentials to approximate 50.
˘
˘
3
Set f (x) = 3 x. We want to know 50. Since 43 = 64,
we set x = 64. Setting dx = 50 * 64 = *14, we have
˘
3
50 = f (x + dx) ˘ f (x) + df
˘
3
˘ 64 + df .
Here we see a plot of y =
marked:
(64, 4)
df
what you know
©Æ™
f (x + dx) ˘ f (x) + df
´≠≠Ø≠≠¨
´Ø¨
dx
4
˘
3
45
50
55
60
x
70
65
Now we must compute df :
df = f ® (x) dx
1
=
dx
3x2_3
1
=
(*14)
3 642_3
1
=
(*14)
3 642_3
*7
=
24
x with the differentials above
Hence f (50) ˘ f (64) +
Example
sin(0.3).
132.
Use
*7
˘ 3.71.
24
differentials
to
approximate
Set y = sin(x). We want to know sin(0.3). Since sin(0) =
0, we will set x = 0 and dx = 0.3. Write with me
sin(0.3) = sin(x + dx) ˘ sin(x) + dy
˘ 0 + dy.
207
Linear approximation
Here we see a plot of y = sin(x) with the differentials
above marked:
1
y
f
(0, 0)
dx
Differentials also help us estimate error in real life settings.
Example 133. The cross-section of a 250 ml glass can be
x4
modeled by the function r(x) = :
3
L
dy
Error approximation
y
20
x
15
10
5
y=
*1
Now we must compute dy:
⇠
⇡
d
dy =
sin(x) dx
dx
= cos(0) dx
= 1 (0.3)
= 0.3
Hence sin(0.3) ˘ sin(0) + 0.3 ˘ 0.3.
The upshot is that linear approximations and differentials are
simply two slightly different ways of doing the exact same
thing.
208
*4
*2
2
x4
3
4
x
At 16.8 cm from the base of the glass, there is a mark indicating
when the glass is filled to 250 ml. If the glass is filled within ±2
millimeters of the mark, what are the bounds on the volume?
As a gesture of friendship, we will tell you that the volume in
milliliters, as a function of the height of water in centimeters,
y, is given by
2⇡y3_2
V (y) = ˘ .
3
Note: If you persist in your quest to learn calculus, you will be
able to derive the formula above like it’s no-big-deal.
We want to know what a small change in the height,
y, does to the volume V . These small changes can be
Linear approximation
modeled by the differentials dV and dy. Since
dV = V ® (y) dy
Note: L(x + dx) = f (x) + f ® (x) dx.
So, the change
˘
and V ® (y) = ⇡ 3y we use the fact that dy = ±0.2 with
y = 16.8 to see
˘
dV = ⇡ 3 16.8 0.2.
dy = L(x + dx) * L(x)
Hence the volume will vary by roughly ±4.46062
milliliters.
Question 110. Suppose f (x) = x2 . If we are at the point
x = 1 and x = dx = 0.1, what is y? What is dy?
New and old friends
You might be wondering, given a plot y = f (x),
What’s the difference between
about y and dy?
= f (x) + f ® (x) dx * L(x)
= f (x) + f ® (x) dx * f (x)
= f ® (x) dx
Differentials can be confusing at first. However, when you
master them, you will have a powerful tool at your disposal.
x and dx? What
Regardless, it is now a pressing question. Here’s the deal:
y
x
is the average rate of change of y = f (x) with respect to x.
On the other hand:
dy
dx
is the instantaneous rate of change of y = f (x) with respect
to x. Essentially, x and dx are the same type of thing, they
are (usually small) changes in x. However, y and dy are very
different things.
•
y = f (x + x) * f (x); it is the change in y = f (x)
associated to x.
• dy = L(x + dx) * L(x), it is the change in y = L(x)
associated to x = dx.
dy = f ® (x) dx
209
Explanation of the product and chain rules
Dig-In:
24.3 Explanation of the product and
chain rules
Now that we know about differentials, let’s use them to give
some intuition as to why the product and chain rules are true.
To understand the derivative of the product, we must
understand how the area, A, changes as x changes. If we
change the inputs of f and g by dx, then the size of the
rectangle changes:
Explanation of the product rule
Linear approximations can help us explain why the product
rule works.
f (x + dx)
A(x) = f (x) g(x)
Theorem 49 (The product rule). If f and g are differentiable,
then
d
f (x)g(x) = f (x)g ® (x) + f ® (x)g(x).
dx
To start, we need some way to understand the function
A(x) = f (x) g(x).
One interpretation of multiplication is it is the area of a
f (x) ù g(x) rectangle:
g(x + dx)
However, we know from our previous work that
f (x + dx) ˘ f (x) + df ,
g(x + dx) ˘ g(x) + dg,
so now our picture becomes:
f (x)
A(x) = f (x) g(x)
g(x)
210
Explanation of the product and chain rules
df
g(x) df
df dg
we see that
dA = f (x) dg + g(x) df + df dg
= f (x)g ® (x) dx + g(x)f ® (x) dx + f ® (x)g ® (x) (dx)2 .
Dividing both sides by dx we see
f (x)
A(x) = f (x) g(x)
f (x) dg
dA
= f (x)g ® (x) + g(x)f ® (x) + f ® (x)g ® (x) dx
dx
and letting dx go to zero we see
A® (x) = f (x)g ® (x) + g(x)f ® (x).
g(x)
dg
Note, if we think of A(x) = f (x) g(x), then we can also
label our picture as follows:
dA
A(x)
Explanation of the chain rule
Now we’ll use linear approximations to help explain why the
chain rule is true.
Theorem 50 (Chain Rule). If f and g are differentiable, then
d
f (g(x)) = f ® (g(x))g ® (x).
dx
We’ll try to understand this geometrically. In what follows, the functions f and g look like lines; however,
the young mathematician should realize that we are not
looking a true lines, instead we are looking at f and g
sufficiently “zoomed-in” so that they appear to be lines.
First consider a graph of g with respect to x:
Finally, from the pictures above and recalling that
df = f ® (x) dx
dg = g ® (x) dx,
211
Explanation of the product and chain rules
df
dg
x
dg
g(x)
g(x)
dx
dg
dx
Ah! From this we see that
df = f ® (g) dg
= f ® (g(x))g ® (x) dx,
x
Now consider a graph of f with respect to g:
df
f (g)
dg
g
If we combine these graphs, by laying the graph of g on
its side, we obtain:
212
so
df
= f ® (g(x))g ® (x).
dx
These “explanations” are not meant to be the end of the story
for the product rule and chain rule, rather they are hopefully the
beginning. As you learn more mathematics, these explanations
will be refined and made precise.
Optimization
25
Optimization
After completing this section, students should be able to do the
following.
• Describe the goals of optimization problems generally.
• Find all local maximums and minimums using the First
and Second Derivative tests.
• Identify when we can find an absolute maximum or minimum on an open interval.
• Contrast optimization on open and closed intervals.
• Describe the objective function and constraints in a given
optimization problem.
• Solve optimization problems by finding the appropriate
extreme values.
213
A mysterious formula
Break-Ground:
25.1
A mysterious formula
Check out this dialogue between two calculus students:
Devyn: Riley, what do you think is the maximum value of
f (x) =
10
?
x2 * 2.8x + 3
Riley: Where did that function come from?
Devyn: It’s just some, um, random function.
Riley: Wait, does this have to do with coffee?
Devyn: Um, uh, no?
Riley: Well what interval are we on?
Devyn: Let’s say [0, 10], I mean there’s no way I could possibly drink
ten cups of coff. . .
Riley: I knew this was about coffee.
Here Devyn has made a function, that is supposed to record
Devyn’s “well-being” with respect to the number of cups of
coffee consumed in one day.
Problem 1. Graph Devyn’s function. Where do you estimate
the maximum on the interval [0, 10] to be?
Problem 2. If you wanted to argue that this is the (global)
maximum value on [0, 10] without plotting, what arguments
could you use?
214
Basic optimization
Now we find the critical points, solving the equation
Dig-In:
25.2
Basic optimization
An optimization problem is a problem where you need to
maximize or minimize some quantity given some constraints.
This can be accomplished using the tools of differential calculus
that we have already developed.
Perhaps the most basic optimization problems is generated by
the following question:
Among all rectangles of a fixed perimeter, which has
the greatest area?
Let’s not do this problem in the abstract, let’s do it with numbers.
Example 134. Of all rectangles of perimeter 12, which side
lengths give the greatest area?
6 * 2x = 0,
we see that the only critical point of A is at x = 3
Since A® (x) = 6 * 2x is positive on (0, 3) and negative on
(3, 6), x = 3 is where the maximum value of A happens.
This is exactly when the rectangle is a square!
A key step to note, is when we explained why x = 3 is actually the maximum. Above we basically used facts about the
derivative. Below we use a similar argument.
Example 135. Of all rectangles of area 100, which has the
smallest perimeter?
First we draw a picture, Here is a rectangle with an area
of 100.
If a rectangle has perimeter 12 and one side is length x,
then the length of the other side is 6 * x. Hence the area
of a rectangle of perimeter 12 can be given by
A(x) = x(6 * x).
However, for the side lengths to be physically relevant,
we must assume that x is in the interval (0, 6).
A = 100
So to maximize the area of the rectangle, we need to find
the maximum value of A(x) on the appropriate interval.
At this point, you should graph the function
if you can.
We’ll continue on without the aid of a graph, and use the
derivative. Write
A® (x) = 6 * 2x
x
If x denotes one of the sides of the rectangle, then the
adjacent side must be 100_x.
215
Basic optimization
The perimeter of this rectangle is given by
P (x) = 2x + 2 100_x.
We wish to minimize P (x). Note, not all values of x make
sense in this problem: lengths of sides of rectangles must
be positive, so x > 0. If x > 0 then so is 100_x, so we
need no second condition on x.
Example 136. Find the rectangle with largest area that fits
inside the graph of the parabola y = x2 below the line y = a,
where a is an unspecified positive constant, with the top side
of the rectangle on the horizontal line y = a. See the figure
below:
y
At this point, you should graph the function
if you can.
(x, a)
We next find P ® (x) and set it equal to zero. Write
P ® (x) = 2 * 200_x2 = 0.
A(x) = area
Solving for x gives us x = ±10. We are interested only
in x > 0, so only the value x = 10 is of interest. Since
P ® (x) is defined everywhere on the interval (0, ÿ), there
are no more critical values, and there are no endpoints. Is
there a local maximum, minimum, or neither at x = 10?
The second derivative is
P ®® (x) = 400_x3 ,
and P (10) > 0, so there is a local minimum. Since there
is only one critical point, this is also the global minimum,
so the rectangle with smallest perimeter is the 10 ù 10
square.
®®
Hence, calculus gives a reason for why a square is the rectangle
with both
• the largest area for a given perimeter.
• the smallest perimeter for a given area.
We may be done with rectangles, but they aren’t done with
us. Here is a problem where there are more constraints on the
possible side lengths of the rectangle.
216
(x, x2 )
x
We want to maximize value of A(x). The lower right corner of the rectangle is at (x, x2 ), and once this is chosen
the rectangle is completely determined. Then the area is
A(x) = (2x)(a * x2 ).
˘
We want the maximum value of A(x) when x is in [0, a].
˘
You might object to allowing x = 0 or x = a, since
then the “rectangle” has either no width or no height, so
is not “really” a rectangle. But the problem is somewhat
easier if we simply allow such rectangles, which have zero
area as we may then apply the Extreme Value Theorem
and see that we indeed have a maximum and minimum
value.
At this point, you should graph the function
Basic optimization
if you can.
˘
Setting 0 = A® (x) = *6x2 + 2a we find x = a_3 as
the only critical point. Testing this and the two endpoints
(as the maximum could also be there),
˘ we have A(0) =
˘
˘
A( a) = 0 and A( a_3) = (4_9) 3a3_2 . Hence, the
maximum
area occurs when the rectangle has dimensions
˘
2 a_3 ù (2_3)a.
Again, note that above we used the Extreme Value Theorem to
guarantee that we found the maximum.
217
Applied optimization
26
Applied optimization
After completing this section, students should be able to do the
following.
• Recognize optimization problems.
• Translate a word problem into the problem of finding the
extreme values of a function.
• Set up an optimization problem by identifying the objective function and appropriate constraints.
• Identify the appropriate domain for functions which are
models of real-world phenomena.
• Solve optimization problems by finding the appropriate
absolute extremum.
• Solve basic word problems involving maxima or minima.
218
Volumes of aluminum cans
Break-Ground:
26.1
Volumes of aluminum cans
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, have you ever noticed aluminum cans?
Riley: So very recyclable!
Devyn: I know! But I’ve also noticed that there are some that are
short and fat, and others that are tall and skinny, and yet they
can still have the same volume!
Riley: So very observant!
Devyn: This got me wondering, if we want to make a can with volume
V , what shape of can uses the least aluminum?
Riley: Ah! This sounds like a job for calculus! The volume of a
cylindrical can is given by
(d) r
Problem 2. In the context above, what should be considered a
constant?
Select All Correct Answers:
(a) A
(b) V
(c) h
(d) r
V = ⇡ r2 h
As Devyn and Riley noticed, when we work out this type of
problem, we need to reduce the problem to a single variable.
where r is the radius of the cylinder and h is the height of the
cylinder. Also the surface area is given by
Problem 3. Consider r to be the variable, and express A as a
function of r.
A = ⇡ r2 + 2 ⇡ r h + ⇡ r2
´Ø¨ ´≠≠≠Ø≠≠≠¨ ´Ø¨
Problem 4. Now consider h to be the variable, and express A
as a function of h.
bottom
sides
top
= 2 ⇡ r2 + 2 ⇡ r h.
Somehow we want to minimize the surface area, because that’s
the amount of aluminum used, but we also want to keep the
volume constant.
Devyn: Whoa, we have way too many letters here.
Riley: Yeah, somehow we need only one variable. Yikes. Too many
letters.
Notice that we’ve reduced (one way or another) this function of
two variables to a function of one variable. This process will
be a key step in nearly every problem in this next section.
Problem 1. Suppose we wish to construct an aluminum can
with volume V that uses the least amount of aluminum. In the
context above, what do we want to minimize?
Select All Correct Answers:
(a) A
(b) V
(c) h
219
Applied optimization
Dig-In:
26.2
r
Applied optimization
In this section, we will present several worked examples of
optimization problems. Our method for solving these problems
is essentially the following:
Draw a picture. If possible, draw a schematic picture with all
the relevant information.
Determine your goal. We need identify what needs to be optimized.
h
Find constraints. What limitations are set on our optimization?
V = ⇡r2 h
Solve for a single variable. Now you should have a function
to optimize.
Use calculus to find the extreme values. Be sure to check
your answer!
Example 137. You are making cylindrical containers to contain a given volume. Suppose that the top and bottom are made
of a material that is N times as expensive (cost per unit area)
as the material used for the lateral side of the cylinder. Find
(in terms of N) the ratio of height to base radius of the cylinder
that minimizes the cost of making the containers.
First we draw a picture:
Letting c represent the cost of the lateral side, we can
write an expression for the cost of materials:
C = 2⇡crh + 2⇡r2 Nc.
Since we know that V = ⇡r2 h, we can use this relationship to eliminate h (we could eliminate r, but it’s a little
easier if we eliminate h, which appears in only one place
in the above formula for cost). We find
C(r) = 2c⇡r
=
V
+ 2Nc⇡r2
⇡r2
2cV
+ 2Nc⇡r2 .
r
Since we have expressed the cost as a function of a single
variable, the problem becomes a“calculus problem": Find
the minimum of the function C on the interval (0, ÿ).
Although the function C is continuous on its domain, its
domain is an open, unbounded interval and the Extreme
220
Applied optimization
Value Theorem does not apply; there is no guarantee that
the minimum even exists. But we don’t give up! Our
strategy will be to search for a critical point and check if
the minimum is attained there. By solving the equation
C ® (r) = *2cV _r2 + 4Nc⇡r = 0
˘
we find that the point where r = 3 V _(2N⇡) is the only
critical point of C. Since C ®® (r) = 4cV _r3 + 4Nc⇡ is
positive when r is positive, there is a local minimum at
the critical point, and hence a global minimum since there
is only one critical point.
Another way to reach this conclusion is to examine the
sign of the derivative C ® (r) on the interval (0, ÿ):
®
2
C (r) = *2cV _r + 4Nc⇡r =
4Nc⇡(r3 *
r2
V
)
2N⇡
.
From
this factorization it is clear that˘C ® (r) < 0 on
˘
3
(0, V _(2N⇡)), and C ® (r) > 0 on ( 3 V _(2N⇡), ÿ).
This
˘ means that the function C is˘decreasing on
(0, 3 V _(2N⇡)) and increasing on ( 3 V _(2N⇡), ÿ),
and, therefore, has a global minimum at the critical point.
will be able to sell another 1000 items. Suppose that your fixed
costs (“start-up costs”) total $2000, and the per item cost of
production (“marginal cost”) is $0.50. Find the price to set per
item and the number of items sold in order to maximize profit,
and also determine the maximum profit you can get.
The first step is to convert the problem into a function
maximization problem. The revenue for selling n items
at x dollars is given by
r(x) = n(x)x
and the cost of producing n items is given by
c(x) = 2000 + 0.5n(x).
However, from the problem we see that the number of
items sold is itself a function of x,
n(x) = 5000 +
So profit is give by:
P (x) = r(x) * c(x)
= nx * (2000 + 0.5n)
Finally, since h = V _(⇡r2 ),
h
V
=
r
⇡r3
V
=
⇡(V _(2N⇡))
= 2N,
so the minimum cost occurs when the height h is 2N
times the radius. If, for example, there is no difference in
the cost of materials, the height is twice the radius.
Example 138. You want to sell a certain number n of items in
order to maximize your profit. Market research tells you that if
you set the price at $1.50, you will be able to sell 5000 items,
and for every 10 cents you lower the price below $1.50 you
1000(1.5 * x)
0.10
= *10000x2 + 25000x * 12000.
We want to know the maximum value of the function
P on the interval [0, 1.5]. The Extreme Value Theorem
now guarantees that the maximum is attained, since the
function P is continuous on the closed interval, [0, 1.5].
We have to find the critical points:
P ® (x) = 0,
or
*20000x + 25000 = 0,
221
Applied optimization
which implies that the only critical point occurs at x =
1.25. Now we have to evaluate the function at the critical
point and both endpoints, and compare the values:
P (0) = *12000, P (1.25) = 3625, and P (1.5) = 3000.
Note that P (1.25) is the maximum of these. Thus the
maximum profit is $3625, attained when we set the price
at $1.25 and sell 7500 items.
of only one variable to maximize. In this problem, the
condition is apparent in the figure below.
R2 = r2 + (h * R)2
r
h*R
Alternately, we could apply the Second Derivative Test at
the critical point and conclude that there must be a local
maximum there, since P ®® (x) = *20000 < 0. Since this
is the only critical point, it must be a global maximum as
well.
Example 139. If you fit the largest possible right circular cone
inside a sphere, what fraction of the volume of the sphere is
occupied by the cone?
R
h
R
Apply the Pythagorean Theorem to the right triangle in
the figure above to obtain
r2 + (h * R)2 = R2 .
⇡r2 h
Vc =
3
Solve for r2 , since r2 is found in the formula for the
volume of the cone, and find that
r
r2 = R2 * (h * R)2 = 2hR * h2 .
h
R
Vs =
4⇡R3
3
Let R be the radius of the sphere, and let r and h be the
base radius and height of the cone inside the sphere. Our
goal is to maximize the volume of the cone: Vc = ⇡r2 h_3.
The largest r could be is R and the largest h could be is
2R.
Notice that the function we want to maximize, ⇡r2 h_3,
depends on two variables. Our next step is to find the
relationship between r and h and use it to solve for one of
the variables in terms of the other, so as to have a function
222
Substitute this into the formula for the volume of the cone
to find that
Vc (h) = ⇡(R2 * (h * R)2 )h_3.
By simplifying, we were able to express the volume of
the cone as a function of a single variable, h,
⇡
2
Vc (h) = * h3 + ⇡h2 R.
3
3
We want to maximize the function Vc on the interval
[0, 2R]. Since this is a continuous function on a closed
Applied optimization
interval, the maximum is attained either at a critical point
or an end point. By solving the equation
Vc® (h) = *⇡h2 + (4_3)⇡hR = 0,
we find that the function has at h = 4R_3 its only critical
point, since h = 0 is an end point. We evaluate the
function at all the critical and end points,
Vc (0) = Vc (2R) = 0,
3
Vc (4R_3) = (32_81)⇡R .
The maximum is obviously attained at h = 4R_3 . Since
the volume of the sphere is (4_3)⇡R3 , the fraction of the
sphere occupied by the cone is
(32_81)⇡R3
8
=
˘ 30%.
27
(4_3)⇡R3
Example 140. A power line needs to be run from a power
station located on the beach to an offshore facility.
facility
1000 ft
station
5000 ft
It costs $50/ft. to run a power line along the land, and $130/ft.
to run a power line under water. How much of the power line
should be run along the land to minimize the overall cost?
What is the minimal cost?
There are two immediate solutions that we could consider, each of which we will reject through “common
sense.” First, we could minimize the distance by directly
connecting the two locations with a straight line. However, this requires that all the wire be laid underwater,
the most costly option. Second, we could minimize the
underwater length by running a wire all 5000 ft. along
the beach, directly across from the offshore facility. This
has the undesired effect of having the longest distance of
all, probably ensuring a nonminimal cost.
The optimal solution likely has the line being run along
the ground for a while, then underwater, as the figure
implies. We need to label our unknown distances: the
distance run along the ground and the distance run underwater. Recognizing that the underwater distance can be
measured as the hypotenuse of a right triangle, we can
label our figure as follows
facility
˘
x2 + 10002
5000 * x
station
1000 ft
x
5000 ft
By choosing x as we did, we make the expression under
the square root simple. We now create the cost function:
Cost =
land cost
+
$50 ù land distance +
50(5000 * x)
+
water cost
$130 ù
˘water distance
130 x2 + 10002 .
223
Applied optimization
So we have
˘
c(x) = 50(5000 * x) + 130 x2 + 10002 .
This function only makes sense on the interval [0, 5000].
While we are fairly certain the endpoints will not give a
minimal cost, we still evaluate c(x) at each to verify.
c(0) = 380000
Recognize that this is never undefined. Setting c ® (x) = 0
and solving for x, we have:
˘
130x
x2 + 10002
130x
=0
= 50
x2 + 10002
1302 x2
= 502
2
x + 10002
1302 x2 = 502 (x2 + 10002 )
1302 x2 * 502 x2 = 502 10002
(1302 * 502 )x2 = 500002
500002
1302 * 502
50000
x= ˘
1302 * 502
50000
x=
,
120
x2 =
Evaluating c(x) at x = 416.67 gives a cost of about
$370000. The distance the power line is laid along land is
5000
˘ * 416.67 = 4583.33 f t and the underwater distance
is 416.672 + 10002 ˘ 1083 f t.
224
Example 141. Suppose you want to reach a point A that is
located across the sand from a nearby road.
A
c(5000) ˘ 662873.
We now find the critical points of c(x). We compute c ® (x)
as
130x
c ® (x) = *50 + ˘
.
x2 + 10002
*50 + ˘
We now work a similar problem without concrete numbers.
w
v
D
b
x
B
C
a
Suppose that the road is straight, and b is the distance from A
to the closest point C on the road. Let v be your speed on the
road, and let w, which is less than v, be your speed on the sand.
Right now you are at the point D, which is a distance a from C.
At what point B should you turn off the road and head across
the sand in order to minimize your travel time to A?
Let x be the distance short of C where you turn off, the
distance from B to C. We want to minimize the total
travel time. Recall that when traveling at constant velocity, time is distance divided by velocity.
You travel the distance from D to B at speed v, and then
the distance from B to A at speed w. The distance from
D to B is a*x. By the Pythagorean theorem, the distance
from B to A is
˘
x2 + b2 .
Hence the total time for the trip is
˘
x2 + b2
a*x
T (x) =
+
.
v
w
We want to find the minimum value of T when x is be-
Applied optimization
tween 0 and a. As usual we set T ® (x) = 0 and solve for
x. Write
T ® (x) = *1_v +
We find that
x
= 0.
˘
w x2 + b2
wb
x= ˘
v2 * w2
Notice that a does not appear in the last expression, but
a is not irrelevant, since we are interested
only in criti˘
cal values that are in [0, a], and wb_ v2 * w2 is either
in this interval or not. If it is, we can use the second
derivative to test it:
b2
T (x) =
.
(x2 + b2 )3_2 w
®®
So the upshot
˘ is this: If you start farther away from C
than wb_ v2 * w2 then you always want
˘ to cut across
the sand when you are a distance wb_ v2 * w2 from
point C. If you start closer than this to C, you should cut
directly across the sand.
With optimization problems you will see a variety of situations
that require you to combine problem solving skills with calculus.
Focus on the process. One must learn how to form equations
from situations that can be manipulated into what you need.
Forget memorizing how to do “this kind of problem” as opposed
to “that kind of problem.”
Learning a process will benefit one far more than
memorizing a specific technique.
Since this is always positive there is a local minimum at
the critical point, and so it is a global minimum as well.
If the critical value is not in [0, a] it is larger than a. In
this case the minimum must occur at one of the endpoints.
We can compute
a
b
+
v w
˘
a2 + b2
T (a) =
w
T (0) =
but it is difficult to determine which of these is smaller by
direct comparison. If, as is likely in practice, we know the
values of v, w, a, and b, then it is easy to determine this.
With a little cleverness, however, we can determine the
minimum in general. We have seen that T ®® (x) is always
positive, so the derivative
T ® (x) is always increasing. We
˘
know that at wb_ v2 * w2 the derivative is zero, so for
values of x less than that critical value, the derivative is
negative. This means that T (0) > T (a), so the minimum
occurs when x = a.
225
L’Hôpital’s rule
27
L’Hôpital’s rule
After completing this section, students should be able to do the
following.
• Recall how to find limits for forms that are determinate.
• Define an indeterminate form.
• Determine if a form is indeterminate.
• Convert some indeterminate forms to the form zero over
zero or infinity over infinity.
• Identify when can Define l’Hôpital’s Rule be used.
• Use l’Hôpital’s Rule to find limits.
226
A limitless dialogue
Break-Ground:
27.1
A limitless dialogue
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Yo Riley, guess what I did last night?
Riley: What?
Devyn: I was doing some calculus.
Riley: That. Is. Awesome.
Devyn: I know! Anyway, I noticed something kinda funny. I think
you can sometimes take limits by taking the derivative of the
numerator and the denominator.
Riley: That’s crazy.
Devyn: I know! But check it:
lim
xô0
sin(x)
= lim
xô0
x
= lim
xô0
= 1.
d
dx
sin(x)
d
x
dx
cos(x)
1
Riley: Woah. That. Is. Awes. . . weird. Hmmm, but it seems like
cheating. Wait, it doesn’t always work, check this out:
lim
xô0
but
d
dx
lim d
xô0
dx
x2 + 1
= 1,
x+1
x2 + 1
(x + 1)
= lim
xô0
2x
1
= 0.
Problem 1. Find five examples where this “trick” works, and
five examples where it doesn’t work.
Problem 2. What is the pattern for when the “trick” works
and when it does not work?
227
L’Hôpital’s rule
ÿ * ÿ This refers to a limit of the form lim (f (x) * g(x))
Dig-In:
27.2
xôa
where f (x) ô ÿ and g(x) ô ÿ as x ô a.
L’Hôpital’s rule
Derivatives allow us to take problems that were once difficult
to solve and convert them to problems that are easier to solve.
Let us consider L’Hôpital’s rule:
Theorem 51 (L’Hôpital’s Rule). Let f (x) and g(x) be functions that are differentiable near a. If
lim f (x) = lim g(x) = 0
xôa
or ± ÿ,
xôa
f ® (x)
exists, and g ® (x) ë 0 for all x near a, then
xôa g ® (x)
and lim
f (x)
f ® (x)
= lim ® .
xôa g(x)
xôa g (x)
lim
1ÿ This refers to a limit of the form lim f (x)g(x) where
f (x) ô 1 and g(x) ô ÿ as x ô a.
00 This refers to a limit of the form lim f (x)g(x) where f (x) ô
0 and g(x) ô 0 as x ô a.
Remark. L’Hôpital’s rule applies even when lim f (x) = ±ÿ
ÿ0 This refers to a limit of the form lim f (x)g(x) where
f (x) ô ÿ and g(x) ô 0 as x ô a.
Our first example is the computation of a limit that was somewhat difficult before.
Example 142. Compute
xôa
lim
xô0
L’Hôpital’s rule allows us to investigate limits of indeterminate
form.
Definition (List of Indeterminate Forms).
ÿ
ÿ
f (x)
This refers to a limit of the form lim
where f (x) ô 0
xôa g(x)
and g(x) ô 0 as x ô a.
This refers to a limit of the form lim
and g(x) ô ÿ as x ô a.
xôa
f (x)
where f (x) ô ÿ
g(x)
0 ÿ This refers to a limit of the form lim (f (x) g(x)) where
xôa
f (x) ô 0 and g(x) ô ÿ as x ô a.
228
xôa
In each of these cases, the value of the limit is not immediately
obvious. Hence, a careful analysis is required!
xôa
0
0
xôa
Basic indeterminant forms
This theorem is somewhat difficult to prove, in part because
it incorporates so many different possibilities, so we will not
prove it here.
and lim g(x) = -ÿ.
xôa
sin(x)
.
x
Set f (x) = sin(x) and g(x) = x. Since both f (x) and
g(x) are differentiable functions at 0, and
lim f (x) = lim g(x) = 0,
xô0
xô0
this situation is ripe for L’Hôpital’s Rule. Now
f ® (x) = cos(x)
and
g ® (x) = 1.
L’Hôpital’s rule
L’Hôpital’s rule tells us that
L’Hôpital’s rule tells us that
sin(x)
cos(x)
= lim
= 1.
xô0
xô0
x
1
lim
Remark. Note, the astute mathematician will notice that in
our example above, we are somewhat cheating. To apply
L’Hôpital’s rule, we need to know the derivative of sine; however, to know the derivative of sine we must be able to compute
the limit:
sin(x)
lim
xô0
x
Hence using L’Hôpital’s rule to compute this limit is a circular
argument! We encourage the gentle reader to view L’Hôpital’s
rule a “reminder” as to what is true, not as the formal derivation
of the result.
Our next set of examples will run through the remaining indeterminate forms one is likely to encounter.
Example 143. Compute
lim
xô⇡_2+
sec(x)
.
tan(x)
Set f (x) = sec(x) and g(x) = tan(x). Both f (x) and
g(x) are differentiable near ⇡_2. Additionally,
lim f (x) =
xô⇡_2+
lim g(x) = *ÿ.
xô⇡_2+
lim +
xô⇡_2
xô⇡_2
= 1.
Example 144. Compute
lim x ln x.
xô0+
This doesn’t appear to be suitable for L’Hôpital’s Rule.
As x approaches zero, ln x goes to *ÿ, so the product
looks like
(something very small) (something very large and negative).
This product could be anything. A careful analysis is
required. Write
ln x
x ln x =
.
x*1
Set f (x) = ln(x) and g(x) = x*1 . Since both functions
are differentiable near zero and
lim ln(x) = *ÿ
xô0+
and
and
lim x*1 = ÿ,
xô0+
we may apply L’Hôpital’s rule. Write with me
This situation is ripe for L’Hôpital’s Rule. Now
f ® (x) = sec(x) tan(x)
sec(x)
sec(x) tan(x)
= lim +
tan(x) xô⇡_2
sec2 (x)
= lim + sin(x)
f ® (x) = x*1
and
g ® (x) = *x*2 ,
g ® (x) = sec2 (x).
229
L’Hôpital’s rule
so
and
ln x
lim x ln x = lim+
xô0+
xô0 x*1
x*1
= lim+
xô0 *x*2
= lim+ *x
g ® (x) = cos(x),
so
cos(x) * 1
sin(x)
* sin(x)
= lim
xô0 cos(x)
= 0.
lim (cot(x) * csc(x)) = lim
xô0
xô0
xô0
= 0.
One way to interpret this is that since lim+ x ln x = 0,
xô0
the function x approaches zero much faster than ln x
approaches *ÿ.
Indeterminate forms involving subtraction
There are two basic cases here, we’ll do an example of each.
Example 145. Compute
lim (cot(x) * csc(x)) .
xô0
Here we simply need to write each term as a fraction,
0
1
cos(x)
1
lim (cot(x) * csc(x)) = lim
*
xô0
xô0 sin(x)
sin(x)
cos(x) * 1
= lim
xô0
sin(x)
Setting f (x) = cos(x) * 1 and g(x) = sin(x), both functions are differentiable near zero and
lim (cos(x) * 1) = lim sin(x) = 0.
xô0
Example 146. Compute
⇠˘
⇡
lim
x2 + x * x .
xôÿ
Again, this doesn’t appear to be suitable for L’Hôpital’s
Rule. A bit of algebraic manipulation will help. Write
with me
⇡
⇠ ⇠˘
⇠˘
⇡⇡
lim
x2 + x * x = lim x
1 + 1_x * 1
xôÿ
xôÿ
˘
1 + 1_x * 1
= lim
xôÿ
x*1
˘
Now set f (x) = 1 + 1_x * 1, g(x) = x*1 . Since both
functions are differentiable for large values of x and
˘
lim ( 1 + 1_x * 1) = lim x*1 = 0,
xôÿ
®
f (x) = * sin(x)
xôÿ
we may apply L’Hôpital’s rule. Write with me
f ® (x) = (1_2)(1 + 1_x)*1_2 (*x*2 )
xô0
We may now apply L’Hôpital’s rule. Write with me
230
Sometimes one must be slightly more clever.
and
g ® (x) = *x*2
L’Hôpital’s rule
so
lim
xôÿ
⇠˘
⇡
x2 + x * x = lim
˘
1 + 1_x * 1
x*1
(1_2)(1 + 1_x)*1_2 (*x*2 )
= lim
xôÿ
*x*2
1
= lim ˘
xôÿ 2 1 + 1_x
1
= .
2
xôÿ
There is a standard trick for dealing with the indeterminate
forms
1ÿ , 00 , ÿ0 .
Given u(x) and v(x) such that
lim u(x)v(x)
falls into one of the categories described above, rewrite as
and then we rewrite as
lim ev(x) ln (u(x)) .
xôa
Recall that the exponential function is continuous. Therefore
lim e
=e
5
lim v(x) ln(u(x))
xôa
(b)
ÿ
ÿ
(d) ÿ * ÿ
(e) 1ÿ
(f) ÿ0
Write
L
⇠
xôÿ
xôa
xôa
0
0
lim 1 +
u(x)v(x)
4
(a)
(g) 00
xôa
v(x) ln (u(x))
Multiple Choice:
(c) 0 ÿ
Exponential Indeterminate Forms
lim eln
Example 147. First determine the form of the limit, then compute the limit.
⇠
⇡
1 x
lim 1 +
.
xôÿ
x
Select the correct choice. The form of the limit is
.
Now we will focus on the limit
1
x
⇡x
⇠
1
lim x ln 1 +
xôÿ
x
=e
⇡M
.
So now look at the limit of the exponent
⇠
⇡
ln 1 + x1
⇠
⇡
1
lim x ln 1 +
= lim
.
xôÿ
xôÿ
x
x*1
⇠
⇡
1
Setting f (x) = ln 1 +
and g(x) = x*1 , both funcx
tions are differentiable for large values of x and
⇠
⇡
1
lim ln 1 +
= lim x*1 = 0.
xôÿ
xôÿ
x
lim v(x) ln(u(x))
xôa
using L’Hôpital’s rule. We will only give a single example.
231
L’Hôpital’s rule
We may now apply L’Hôpital’s rule. Write
f ® (x) =
and
*x*2
1+
1
x
g ® (x) = *x*2 ,
so
lim
⇠
⇡
ln 1 + x1
x*1
xôÿ
= lim
xôÿ
= lim
xôÿ
= 1.
Hence,
4
⇠
lim 1 +
xôÿ
1
x
⇡x
*x*2
1
1+
1
x
⇠
⇡5
1
lim x ln 1 +
x
= e xôÿ
⇠
⇡
4
ln 1 + x1 5
lim
x*1
= e xôÿ
= e1
= e.
232
*x*2
1+ x1
Antiderivatives
28
Antiderivatives
After completing this section, students should be able to do the
following.
• Define an antiderivative.
• Compute basic antiderivatives.
• Compare and contrast finding derivatives and finding antiderivatives.
• Define initial value problems.
• Solve basic initial value problems.
• Use antiderivatives to solve simple word problems.
• Discuss the meaning of antiderivatives of the velocity and
acceleration.
233
Jeopardy! Of calculus
Break-Ground:
28.1
Jeopardy! Of calculus
Check out this dialogue between two calculus students (based
on a true story):
Devyn: (Pretending to Alex Trebek) I’ve got a new costume.
Riley: Whoa! You look just like Trebek!
Devyn: In Jeopardy!, I, Trebek, give you an answer, and you must
tell me the question.
Riley: Uh Alex, ‘What are the rules of Jeopardy!?’
Devyn: Ha. Exactly! Let’s play a different version where I’ll tell you
a derivative, and you tell me the function. Are you ready?
Riley: I’ll take “Formulas for slope” for $200.
Devyn: 3 e3x
Riley: I’ve got an answer! Actually, I’ve got three different answers,
I mean questions!
(a) “What’s the derivative of e3x ?
(b) “What’s the derivative of e3x + 1?
(c) “What’s the derivative of e3x * 1?
Devyn: Hmmm. Now I’m not sure which one it was.
sin(x)
Riley: What about if you had given me
?
x
Problem 1. How many functions whose derivative is 3 e3x
are there?
Problem 2. How many functions whose derivative is 3 e3x
that equal 1 at x = 0 are there?
234
Basic antiderivatives
It follows that
Dig-In:
28.2
Basic antiderivatives
Computing derivatives is not too difficult. At this point, you
should be able to take the derivative of almost any function
you can write down. However, undoing derivatives is much
harder. This process of undoing a derivative is called taking an
antiderivative.
Definition. A function F is called an antiderivative of f on
an interval if
F ® (x) = f (x)
for all x in the interval.
Question 111. How many antiderivatives does f (x) = 2x
have? Recall: Any function whose derivative is equal to 0
on some interval is equal to a constant on that interval. This
means that the function G(x) * x2 = C, for some constant C.
This implies that G(x) = x2 + C, for all x.
Theorem 52 (The Family of Antiderivatives). If F is an antiderivative of f , then the function f has a whole family of
antiderivatives. Each antiderivative of f is the sum of F and
some constant C.
So, when we write F (x) + C, we denote the entire family of
antiderivatives of f . Alternative notation for this family of
antiderivatives was introduced by G.W. Leibniz3 (1646-1716):
Definition. Let f be a function. The family of of all antiderivatives of f is denoted by
f (x) dx.
This is called the indefinite integral of f .
3
See G.W. Leibniz at
Gottfried_Wilhelm_Leibniz
https://en.wikipedia.org/wiki/
f (x) dx = F (x) + C,
where F is any antiderivative of f and C is an arbitrary constant.
Now we are ready to "integrate" some famous functions.
Fill out these basic antiderivatives. Note each of these examples
comes directly from our knowledge of basic derivatives.
Theorem 53 (Basic Indefinite Integrals).
•
k dx = kx + C
•
1
dx = ln x + C
x
•
xn dx =
•
ex dx = ex + C
•
ax dx =
•
cos(x) dx = sin(x) + C
•
sin(x) dx = * cos(x) + C
•
sec2 (x) dx = tan(x) + C
•
csc2 (x) dx = * cot(x) + C
•
sec(x) tan(x) dx = sec(x) + C
xn+1
+C
n+1
(n ë *1)
ax
+C
ln(a)
235
Basic antiderivatives
•
•
•
csc(x) cot(x) dx = * csc(x) + C
x2
1
dx = arctan (x) + C
+1
1
dx = arcsin (x) + C
˘
1 * x2
It may seem that one could simply memorize these antiderivatives and antidifferentiating would be as easy as differentiating. This is not the case. The issue comes up when trying to
combine these functions. When taking derivatives we have
the product rule and the chain rule. The analogues of these
two rules are much more difficult to deal with when taking
antiderivatives. However, not all is lost.
Consider the following example.
Example 148. Find an antiderivative of the function h defined
by expression
1
h(x) = cos x + , for all x in some interval I.
x
Multiple Choice:
(a) H(x) = cos x +
1
x
(b) H(x) = sin x +
1
x2
(c) H(x) = sin x + ln x
It is easy to recognize an antiderivative: we just have to differentiate it, and check whether H ® (x) = h(x), for all x in
I.
Notice, that the function h is the sum of the two functions, f
1
and g, where f (x) = cos x and g(x) = , for x in I.
x
We know antiderivatives of both functions: F (x) = sin x and
G(x) = ln x, for x in I, are antiderivatives of f and g, respectively. So, in this example we see that the function F + G
236
is an antiderivative of f + g. In other words, "the sum of
antiderivatives is an antiderivative of a sum".
Is this true in general?
Let’s check whether this rule holds for any eligible pair of
functions f and g defined on some interval I.
Question 112. Let f , g, F and G be four functions defined on
some interval I such that F is an antiderivative of f and G is
an antiderivative of g, i.e.
F ® (x) = f (x) and G® (x) = g(x) for all x in some interval I.
Find an antiderivative of the function f + g.
Since
(F (x) + G(x))® = F ® (x) + G® (x) = f (x) + g(x), it follows
that F + G is an antiderivative of f + g.
To summarize: The sum of antiderivatives is an antiderivative
of a sum.
We have proved the following theorem.
Theorem 54 (The Sum Rule for Antiderivatives). If F is an
antiderivative of f and G is an antiderivative of g, then F + G
is an antiderivative of f + g.
We can write equivalently, using indefinite integrals,
(f (x) + g(x)) dx =
f (x) dx +
g(x) dx.
Next, we will try to prove an analogue of the constant multiple
rule for derivatives. Let’s consider the following example.
Example 149. Find an antiderivative of the function h, where
h(x) = 5 sec2 x, for all x in some interval I.
Multiple Choice:
(a) H(x) = 5 tan x
(b) H(x) = 5 sec2 x
(c) H(x) = sec2 5x
Basic antiderivatives
It is easy to recognize an antiderivative: we just have to differentiate it, and check whether H ® (x) = h(x), for all x in
I.
By the theorems above , we see that
3x7 dx = 3
Notice, in this example the function h is a constant multiple of
f,
where f (x) = sec2 x. On the other hand, we know that the
function F , defined by F (x) = tan x is an antiderivative of f .
If we differentiate the function 5F , we get that
(5F (x))® = (5 tan x)® = 5 (tan x)® = 5 sec2 x = 5f (x), for x in
I.
In other words, “a constant multiple of an antiderivative is an
antiderivative of a constant multiple of a function.” Is this
always true?
Let’s check whether this rule holds for any constant k and any
eligible function f defined on some interval I.
Question 113. Let k be a constant, let f be a function defined
on some interval I and let F be an antiderivative of f , i.e.
F ® (x) = f (x), for all x in some interval I.
®
Find an antiderivative of the function kf . Since (kF (x)) =
kF ® (x) = kf (x), it follows that kF is an antiderivative of kf .
To summarize: The constant multiple of an antiderivative is an
antiderivative of a constant multiple of a function.
We have proved the following theorem.
Theorem 55 (The Constant Multiple Rule for Antiderivatives).
If F is an antiderivative of f , and k is a constant, then kF is
an antiderivative of kf .
We can write equivalently, using indefinite integrals,
kf (x) dx = k
=3
x7 dx
x8
+ C.
8
The sum rule for antiderivatives allows us to integrate term-byterm. Let’s see an example of this.
Example 151. Compute:
x4 + 5x2 * cos(x) dx
Let’s start by simplifying the problem using the sum rule
and constant multiple rule for antiderivatives,
x4 + 5x2 * cos(x) dx
=
x4 dx + 5
x2 dx *
cos(x) dx.
Now we may integrate term-by-term to find
=
x5 5x3
+
* sin(x) + C.
5
3
Warning. While the sum rule for antiderivatives allows us to
integrate term-by-term, we cannot integrate factor-by-factor,
meaning that in general
f (x)g(x) dx ë
f (x) dx
g(x) dx.
f (x) dx.
Let’s put these rules and our knowledge of basic derivatives to
work.
Example 150. Find the antiderivative of 3x7 .
Example 152. A student claims that
2x cos(x) dx =
x2 sin(x) + C. Determine whether the student is correct or
incorrect.
237
Basic antiderivatives
If the student were correct, then the derivative of
x2 sin(x)+C with respect to x would have to be 2x cos(x).
However:
d 2
x sin(x) = 2x sin(x) + x2 cos(x)
dx
Hence, the student is incorrect!
Computing antiderivatives
Unfortunately, we cannot tell you how to compute every antiderivative; we view them as a sort of puzzle. Later we will
learn a hand-full of techniques for computing antiderivatives.
However, a robust and simple way to compute antiderivatives
is guess-and-check.
Tips for guessing antiderivatives
(a) If possible, express the function that you are integrating
in a form that is convenient for integration.
(b) Make a guess for the antiderivative.
(c) Take the derivative of your guess.
(d) Note how the above derivative is different from the function whose antiderivative you want to find.
(e) Change your original guess by multiplying by constants
or by adding in new functions.
Example 153. Compute:
˘
x+1+x
x
Before guessing the solution, let’s express the function
in the form convenient for integration. Due to Sum Rule,
it is more convenient to have a sum of functions, instead
of a single, complicated term.
H˘
I
˘
x+1+x
x 1 x
dx =
+ +
dx
x
x
x x
H
I
1
1
=
˘ + + 1 dx
x x
⇠ 1
⇡
1
=
x* 2 + + 1 dx
x
Now, we can apply the Sum Rule.
⇠
1
x* 2 +
⇡
1
+ 1 dx =
x
1
x* 2 dx+
1
dx+
x
1 dx
Now we can finish the problem.
˘
˘
x+1+x
dx = 2 x + ln x + x + C.
x
Example 154. Compute:
cos (4x) dx
Start with a guess of
cos (4x) dx ˘ sin (4x).
Take the derivative of your guess to see if it is correct:
dx
d
sin (4x) = 4 cos (4x).
dx
1
We’re off by a factor of , so multiply your guess by this
4
238
Basic antiderivatives
constant to get the solution,
cos (4x) dx =
constant to get the solution,
1
sin (4x) + C.
4
Example 155. Compute:
1
x2 sin (x3 * 6) dx = * cos (x3 * 6) + C.
3
Example 157. Compute:
x3
dx
˘
x4 * 6
3x2 sin (x3 * 6) dx
Start with a guess of
Start by rewriting the indefinite integral as
3x2 sin (x3 * 6) dx ˘ * cos (x3 * 6).
Take the derivative of your guess to see if it is correct:
d
* cos (x3 * 6) = 3x2 sin (x3 * 6).
dx
Therefore,
x3 x4 * 6
*1_2
dx.
Now start with a guess of
x3 x4 * 6
*1_2
dx ˘ x4 * 6
1_2
.
Take the derivative of your guess to see if it is correct:
3x2 sin (x3 * 6) dx = * cos (x3 * 6) + C.
Example 156. Compute:
x2 sin (x3 * 6) dx
Start with a guess of
x2 sin (x3 * 6) dx ˘ * cos (x3 * 6).
Take the derivative of your guess to see if it is correct:
d
* cos (x3 * 6) = 3x2 sin (x3 * 6).
dx
We’re off by a factor of 1_3, so multiply our guess by this
d
x4 * 6
dx
1_2
= (4_2)x3 x4 * 6
*1_2
.
We’re off by a factor of 2_4, so multiply our guess by this
constant to get the solution,
x3
1
dx = (x4 * 6)1_2 + C.
˘
2
4
x *6
Example 158. Compute:
2
xex dx
We try to guess the antiderivative. Start with a guess of
2
2
xex dx ˘ ex .
239
Basic antiderivatives
Take the derivative of your guess to see if it is correct:
2
d x2
e = 2xex .
dx
1
Ah! So we need only multiply our original guess by .
2
We now find
2
xex dx =
1 x2
e + C.
2
Example 159. Compute:
21x2
dx
7x3 + 3
We notice that the numerator is the derivative of the
denominator, i.e., the function that we are integrating has
f ® (x)
the form
.
f (x)
21x2
dx = ln(7x3 + 3).
7x3 + 3
Take the derivative of your guess to see if it is correct:
Therefore,
We’ll start with a guess of
2x2
dx ˘ ln(7x3 + 3).
7x3 + 3
Take the derivative of your guess to see if it is correct:
d
21x2
ln(7x3 + 3) =
.
dx
7x3 + 3
We are only off by a factor of 2_21, so we need to multiply
our original guess by this constant to get the solution,
2x2
dx = (2_21) ln(7x3 + 3) + C.
7x3 + 3
Final thoughts
Computing antiderivatives is a place where insight and rote
computation meet. We cannot teach you a method that will
always work. Moreover, merely understanding the examples
above will probably not be enough for you to become proficient in computing antiderivatives. You must practice, practice,
practice!
d
21x2
ln(7x3 + 3) =
.
dx
7x3 + 3
Di�erential equations
21x2
dx = ln (7x3 + 3) + C.
7x3 + 3
A differential equation is simply an equation with a derivative
in it. Here is an example:
Differential equations show you relationships between rates of
functions.
a f ® (x) + b f (x) = g(x).
Example 160. Compute:
2x2
dx
7x3 + 3
Question 114. Which one is a differential equation?
Multiple Choice:
(a) x2 + 3x * 1 = 0
240
Basic antiderivatives
The figure below shows several solutions of the differential equation.
(b) f ® (x) = x2 + 3x * 1
(c) f (x) = x2 + 3x * 1
When a mathematician solves a differential equation, they are
finding functions satisfying the equation.
6
4
Question 115. Which of the following functions solve the differential equation
f ® (x) = f (x)?
Select All Correct Answers:
2
*3
(a) f (x) = sin(x)
*2
(c) f (x) = ex
(e) f (x) = e*x
(f) f (x) = tan(x)
Remark. We can directly check that any function f (x) = Cex
is a solution to our differential equation f ® (x) = f (x). Could
there be any others? It turns out that these are the only solutions.
But showing that we didn’t miss any is a bit tricky.
A function Cex is called a general solution of the differential
equation. Since there are infinitely many solutions to a differential equation, we can impose an additional condition (say
f (0) = 1), called an initial condition. When we are asked to
find a function f that satisfies both the differential equation
(DE) and the initial condition (IC), this is called an initial value
problem (IVP). Let’s try one out.
Example 161. Solve the initial value problem (IVP):
f (0) = 1
1
*4
y = *2ex
*6
x
f ® (x) = f (x)
*1
f (x) = ex
2
3
x
*2
(b) f (x) = x2
(d) f (x) = 6e
y
y = 2ex
(DE)
(IC)
y = *ex
The figure suggests that the only solution to the initial
value problem is the function f (x) = ex . We can verify
this result.
Since all solutions of the differential equation have the
form f (x) = Cex and f (0) = 1, it follows that
Ce0 = 1.
Therefore,
C = 1.
So, the unique solution to the given initial value problem
is the function f (x) = ex .
Example 162. Solve the initial value problem (IVP):
f ® (x) = sin x
f (0) = *1
241
Basic antiderivatives
First, we have to solve the differential equation. The
solution is clearly an antiderivative of sin x.
f (x) = * cos x + C,
y = 2ex
where C is an arbitrary constant. We have found the
general solution of the DE and this family of solutions is
illustrated by the figure below.
y
y = *2cos x + 3
y = * cos x + 2
*3
*2
*1
y = * cos x + 1
1
2
3
x
f (x) = * cos x
y =*2
* cos x * 1
Now we must find the solution that also satisfies the initial
condition f (0) = *1.
Since
it follows that
Therefore,
The function
f (0) = *1 + C,
*1 = *1 + C.
C = 0.
f (x) = * cos x.
is the solution of the initial value problem and it is shown
in the figure above.
242
Falling objects
Dig-In:
28.3
Now when t = 1, v(1) = 5.2 m/s, and the ball is rising,
and at t = 2, v(2) = *4.6 m/s, and the ball is falling.
Falling objects
A differential equation is simply an equation with a derivative
in it like this:
f ® (x) = kf (x).
Now let’s do a similar problem, but instead of finding the velocity, we will find the position.
Consider a falling object. Recall that the acceleration due
to gravity is about *9.8 m/s2 . Since the first derivative of the
velocity function is the acceleration and the second derivative of
a position function is the acceleration, we have the differential
equations
Knowing that the acceleration due to gravity is *9.8 m/s2 ,
we write
v® (t) = *9.8.
When a mathematician solves a differential equation, they are
finding a function that satisfies the equation.
v® (t) = *9.8,
s®® (t) = *9.8.
From these simple equations, we can derive expressions for the
velocity and for the position of the object using antiderivatives.
Example 163. A ball is tossed into the air with an initial velocity of 15 m/s. What is the velocity of the ball after 1 second?
How about after 2 seconds?
Knowing that the acceleration due to gravity is *9.8 m/s2 ,
we write
v® (t) = *9.8.
To solve this differential equation, take the antiderivative
of both sides
v® (t) dt =
*9.8 dt
v(t) = *9.8t + C.
Since it is tossed up with an initial velocity of 15 m/s,
15 = v(0) = *9.8 0 + C,
Example 164. A ball is tossed into the air with an initial velocity of 15 m/s from a height of 2 meters. When does the ball
hit the ground?
Start by taking the antiderivative of both sides of the
equation
v® (t) dt =
*9.8 dt
v(t) = *9.8t + C.
Here C represents the initial velocity of the ball. Since
it is tossed up with an initial velocity of 15 m/s, C = 15
and
v(t) = *9.8t + 15.
Since the velocity is the derivative of the position, we
can write, alternatively
s® (t) = *9.8t + 15.
Now let’s take the antiderivative again.
s® (t) dt =
s(t) =
*9.8t + 15 dt
*9.8t2
+ 15t + D.
2
and we see that C = 15. Therefore, C represents the
initial velocity of the ball. Hence v(t) = *9.8t + 15.
243
Falling objects
Since we know the initial height was 2 meters, write
2 = s(0) =
*9.8 02
+ 15 0 + D.
2
*9.8t2
Hence s(t) =
+ 15t + 2. We need to know when
2
the ball hits the ground, this is when s(t) = 0. Solving
the equation
*9.8t2
+ 15t + 2 = 0
2
we find two solutions t ˘ *0.1 and t ˘ 3.2. Discarding
the negative solution, we see the ball will hit the ground
after approximately 3.2 seconds.
The power of calculus is that it frees us from rote memorization
of formulas and enables us to derive what we need.
244
Approximating the area under a curve
29 Approximating the area
under a curve
After completing this section, students should be able to do the
following.
• Express the sum of n terms using sigma notation.
• Apply the properties of sums when working with sums in
sigma notation.
• Understand the relationship between area under a curve
and sums of areas of rectangles.
• Approximate area of the region under a curve.
• Compute left, right, and midpoint Riemann sums with 10
or fewer rectangles.
• Understand how Riemann sums with n rectangles are computed and how the exact value of the area is obtained by
taking the limit as n ô ÿ .
245
What is area?
Break-Ground:
29.1
What is area?
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, I’ve have a troubling question.
Riley: Yes?
Devyn: What is area?
Riley: Oh, for a rectangle it is just
height ù width
Devyn: Right. But shouldn’t it mean more?
Area is a funny thing. What is it really? Well the idea is this:
Define something as having a “unit” area, and see
how many times it “covers” something.
The most obivious thing to use for our “unit” area is a unit
square. From this we can quickly move on to find the area of
any rectangle as
height ù width
Problem 1. Sometimes area is described as “how much paint
will cover a surface.” Is this accurate? What do you think?
246
Introduction to sigma notation
Example 166. Write out what is meant by the following:
Dig-In:
29.2
Introduction to sigma notation
8
…
(*1)i
i=1
The notation itself
Sigma notation is a way of writing a sum of many terms, in a
concise form. A sum in sigma notation looks something like
this:
5
…
(*1)i = (*1)1 + (*1)2 + (*1)3 + (*1)4
+ (*1)5 + (*1)6 + (*1)7 + (*1)8
3k
The ⌃ (sigma) indicates that a sum is being taken. The variable
k is called the index of the sum. The numbers at the top and
bottom of the ⌃ are called the upper and lower limits of the
summation. In this case, the upper limit is 5, and the lower
limit is 1. The notation means that we will take every integer
value of k between 1 and 5 (so 1, 2, 3, 4, and 5) and plug them
each into the summand formula (here that formula is 3k). Then
those are all added together.
3k = 3 1 + 3 2 + 3 3 + 3 4 + 3 5 = 45
k=1
This is equal to:
= *1 + 1 + *1 + 1 + *1 + 1 + *1 + 1
=0
Try one on your own.
Question 116. Write out what is meant by the following (no
need to simplify):
4 ˘
˘
˘
˘
˘
˘
˘
…
n+1= 0+ 1+ 2+ 3+ 4+ 5
n=*1
Example 165. Write out what is meant by the following:
3
…
1
k
+
1
k=0
Here, the index k takes the values 0, 1, 2, and 3. We’ll
1
plug those each into
and add them together.
k+1
3
…
8
…
i=1
k=1
5
…
The index variable here is written as i instead of k. That’s
ok. The most common variables to use for indexes include i, j, k, m, and n.
1
1
1
1
1
=
+
+
+
k
+
1
0
+
1
1
+
1
2
+
1
3
+
1
k=0
Let’s try going the other way around.
Example 167. Write the following sum in sigma notation.
2 + 4 + 6 + 8 + … + 22 + 24
Notice that we can factor a 2 out of each term to rewrite
this sum as
2 1 + 2 2 + 2 3 + 2 4 + … + 2 11 + 2 12
That means that we are adding together 2 times every
247
Introduction to sigma notation
number between 1 and 12. If we use k as our index, the
sigma notation could be
12
…
2k
k=1
There is no need to use k as our index variable. We could
have just as easily used m or j instead.
12
…
k=1
2k =
12
…
2m =
m=1
12
…
j=1
2m
k=1
Example 168. Write the following sum in sigma notation.
1*
1 1 1
1
1
+ * +…*
+
2 4 8
64 128
This one is a little more complicated. We’ll worry about
the signs later, first we’ll deal with the numbers themselves. Do you notice a pattern in the terms? Sure, we
get from one term to the next term by dividing by 2. That
is:
⇠ ⇡0
1
1
1=
=
2
20
⇠ ⇡1
1
1
1
=
=
1
2 2
2
⇠ ⇡2
1
1
1
=
=
4 22
2
4
⇠ ⇡7
1
1
1
= 7 =
128 2
2
248
7
…
2j
Notice, that these are NOT the same as
12
…
If we call our index variable k, then k should go from 0
⇠ ⇡k
1
to 7, and the numbers themselves are just
. Now
2
we need to deal with the signs. We say above that (*1)k
will alternate between +1 and *1. That means, if we
multiply the terms we just found by (*1)k , they will
alternate between + and *. We are starting with k = 0,
so (*1)0 = +1 will give us the alternation starting at the
sign we want.
(*1)k
k=0
7 ⇠
⇠ ⇡k …
⇡
*1 k
1
=
2
2
k=0
Try one on your own.
Question 117. Write the following sum in sigma notation.
1
2
3
4
51
52
˘ * ˘ + ˘ * ˘ +…+ ˘ * ˘
3
5
52
53
2
4
Calculating with sigma notation
We want to use sigma notation to simplify our calculations. To
do that, we will need to know some basic sums. First, let’s talk
about the sum of a constant. (Notice here, that our upper limit
of summation is n. n is not the index variable, here, but the
highest value that the index variable will take.)
n
…
k=1
C =C +C +5+C
´≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠¨
n terms
This is a sum of n terms, each of them having a value C. That
is, we are adding n copies of C. This sum is just nC. The other
basic sums that we need are much more complicated to derive.
Rather than explaining where they come from, we’ll just give
you a list of the final formulas, that you can use.
Formula 1.
n
…
k=1
C = nC.
Introduction to sigma notation
Formula 2.
n
…
k=1
n
…
First, we’ll use the properties above to split this into two
sums, then factor the 2 out of the first sum.
n(n + 1)
.
2
k=
10
…
n(n + 1)(2n + 1)
Formula 3.
k =
.
6
k=1
2
2k2 + 5 =
k=1
5
…
9.
This is just the sum of a constant, with C = 9 and n = 5.
The value is nC = 45.
Example 170. Find the value of the sum
k.
k=1
Because sigma notation is just a new way of writing addition,
the usual properties of addition still apply, but a couple of the
important ones look a little different.
Commutativity and Associativity:
(ak + bk ) =
k=1
Distribution:
n
…
k=1
n
…
k=1
c ak = c
ak +
n
…
k=1
n
…
k=1
ak
10
…
10
…
Similarly:
k2 =
k2 +
10
…
5
k=1
10(10 + 1)(2 10 + 1)
= 385
6
10
…
5 = 50
k=1
Putting all that together,
10
…
2k2 + 5 = 2 385 + 50 =
k=1
820.
Example 172. Find the value of the sum:
200
…
bk
5
The two sums we have left, can be found using formulas
1 and 3 above! We see that:
k=1
This is the sum 1+2+3+…+100. According to Formula
100(101)
2 above (with n = 100), this is
= 5050.
2
n
…
=2
10
…
k=1
k=1
k=1
100
…
2k2 +
k=1
Now that we have this list, let’s use them to compute.
Example 169. Find the value of the sum
10
…
*6k2 + 3
k=1
Let’s use the same approach as in the previous example.
First, we’ll use the properties to split this into individual
sums, then factor out the coefficients. After that, we’ll
Example 171. Find the value of the sum:
10
…
2k2 + 5
k=1
249
Introduction to sigma notation
use the formulas above to evaluate it.
200
…
*6k2 + 3 =
k=1
200
…
200
…
*6k2 +
k=1
= *6
200
…
2
k +
k=1
0
= *6
3
k=1
200
…
3
k=1
200(201)(401)
6
1
+ 200 3
= *16119600
The numbers in this example were horribly ugly, but we
were able to evaluate the sum without having to actually
calculate all 200 terms, then add them all up. In 4 small
lines, we were able to add 200 numbers.
Try one on your own.
Question 118. Find
50
…
4k2 * 18k + 2(*1)k
k=1
250
the
value
of
the
sum
Approximating area with rectangles
y
Dig-In:
29.3 Approximating area with
rectangles
1
2
a
Rectangles and areas
We want to compute the area between the curve y = f (x) and
the horizontal axis on the interval [a, b]:
b
x
With even more, we get a closer, and closer, approximation:
y
y
1
2
3
4
5
a
a
b
x
b
x
Definition. If we are approximating the area between a curve
and the x-axis on [a, b] with n rectangles of width x, then
One way to do this would be to approximate the area with
rectangles. With one rectangle we get a rough approximation:
x=
b*a
.
n
y
y
1
a
a
b
Two rectangles might make a better approximation:
x
x
x
x
x
x
b
x
Question 119. Suppose we wanted to approximate area between the curve y = x2 + 1 and the x-axis on the interval
[*1, 1], with 8 rectangles. What is x?
251
Approximating area with rectangles
As we add rectangles, we are more closely approximating the
area we are interested in:
y
Question 120. If we are approximating the area between a
curve and the horizontal axis with 11 rectangles, how many
grid points will we have?
But which set of rectangles?
a
b
x
We could find the area exactly if we could compute the limit
as the width of the rectangles goes to zero and the number of
rectangles goes to infinity.
Let’s setup some notation to help with these calculations:
Definition. When approximating an area with n rectangles, the
grid points
x0 , x1 , x2 , … , xn
are the x-coordinates that determine the edges of the rectangles.
In the graph below, we’ve numbered the rectangles to help you
see the relation between the indices of the grid points and the
kth rectangle.
y
When we use n rectangles to compute the area under a curve,
b*a
the width of each rectangle is x =
. It is clear that
n
x = xk * xk*1 , for k = 1, … , n.
But how do we determine the height of the rectangle?
We choose a sample point x<k and evaluate the function at that
point. The value f (x<k ) determines the height of a rectangle.
y
y = f (x)
x
x0 = a
x0 = a
3
2
x1
x2
5
4
x3
x4
x5 = b
x
Note, if we are approximating the area between a curve and the
horizontal axis on [a, b] with n rectangles, then it is always the
case that
x0 = a
and
xn = b.
252
xk*1 x<
k
xk
xn = b
x
Definition. When approximating an area with rectangles, a
sample point is the x-coordinate that determines the height of
kth rectangle. For k = 1, … , n, we denote a sample point as:
and the value
1
k
f (x<k )
x<k
f (x<k )
is the height of the kth rectangle.
Question 121. What is the area of the kth rectangle shown in
the figure above?
Select All Correct Answers:
(a) A = x
Approximating area with rectangles
(b) A = f (x) x
In the graph above, the kth rectangle’s right-endpoint of the
base determines the height.
(c) A = f (x<k )x<k
(d) A = f (x<k ) x
Rectangles de�ned by midpoints We can set the rectangles up so that the midpoint of the base determines the height.
(e) A = f (x)x
(f) A = k k
y
(g) A = f (x<k )(xk * xk*1 )
Here are three options for sample points that we consider:
Rectangles de�ned by left-endpoints We can set the rectangles up so that the sample point is the left-endpoint.
y
1
2
3
4
5
x<1
x<2
x<3
x<4
x<5
x
In the graph above, the midpoint of the base of the kth rectangle
determines the height.
1
x<1
3
2
x<2
x<3
x<4
Riemann sums and approximating area
5
4
x
x<5
In the graph above, the kth rectangle’s left-endpoint determines
the height of the rectangle.
Rectangles de�ned by right-endpoints We can set the
rectangles up so that the right-endpoint determines the height.
Once we know how to identify our rectangles, we can compute
approximations of some areas. If we are approximating area
with n rectangles, then
Area ˘
=
y
n
…
(height of kth rectangle) ù (width of kth rectangle)
k=1
n
…
k=1
f (x<k ) x
= f (x<1 ) x + f (x<2 ) x + f (x<3 ) x + 5 + f (x<n ) x.
Definition. A sum of the form:
n
…
1
3
2
x<1
x<2
5
4
x<3
x<4
k=1
x<5
x
f (x<k ) x = f (x<1 ) x + f (x<2 ) x + 5 + f (x<n ) x
is called a Riemann sum, pronounced “ree-mahn” sum.
253
Approximating area with rectangles
A Riemann sum computes an approximation of the area between a curve and the x-axis on the interval [a, b]. It can be
defined in several different ways. In our class, it will be defined
via left-endpoints, right-endpoints, or midpoints. Here we see
the explicit connection between a Riemann sum defined by
left-endpoints and the area between a curve and the x-axis on
the interval [a, b]:
y
x0 = *1
y
f (x<5 )
f (x<4 )
f (x<1 )
x2 = 0.6
x3 = 1.4
x4 = 2.2
x5 = 3
x
On the other hand, the sample points identify which endpoints we use:
f (x<3 )
f (x<1 )
x1 = *0.2
f (x<2 )
f (x<5 )
y
f (x<1 ) x f (x<2 ) x f (x<3 ) x f (x<4 ) x f (x<5 ) x
x
x
x
x
x
x
and here is the associated Riemann sum
5
…
k=1
f (x<k ) x = f (x<1 ) x+f (x<2 ) x+f (x<3 ) x+f (x<4 ) x+f (x<5 ) x.
3
2
x<2 = *0.2
x<3 = 0.6
5
4
x<4 = 1.4
x<5 = 2.2
x
It is helpful to collect all of this data into a table:
Left Riemann sums
Example 173. Consider f (x) = x3 _8 * x + 2. Approximate
the area between the curve y = f (x) and the x-axis on the
interval [*1, 3] using a left-endpoint Riemann sum with n = 5
rectangles.
First note that the width of each rectangle is
x=
1
x<1 = *1
3 * (*1)
= 4_5.
5
The grid points define the edges of the rectangle and are
seen below:
k
0
1
2
3
4
5
xk
*1
*0.2
0.6
1.4
2.2
3
x<k
NA
*1
*0.2
0.6
1.4
2.2
f (x<k )
NA
2.875
2.199
1.427
0.943
1.131
Now we may write a left Riemann sum
f (x<1 ) x + f (x<2 ) x + f (x<3 ) x + f (x<4 ) x + f (x<5 ) x,
254
Approximating area with rectangles
which evaluates to
y
= 2.875 (4_5) + 2.199 (4_5) + 1.427 (4_5)
+ 0.943 (4_5) + 1.131 (4_5)
and we find
= 6.86.
1
Right Riemann sums
Example 174. Consider f (x) = x3 _8 * x + 2. Approximate
the area between the graph of f and the x-axis on the interval [*1, 3] using a right-endpoint Riemann sum with n = 5
rectangles.
x<2 = 0.6
5
4
x<3 = 1.4
x<4 = 2.2
x<5 = 3
x
It is helpful to collect all of this data into a table:
k
0
1
2
3
4
5
First note that the width of each rectangle is
x=
3
2
x<1 = *0.2
3 * (*1)
= 4_5.
5
The grid points define the edges of the rectangle and are
seen below:
xk
*1
*0.2
0.6
1.4
2.2
3
x<k
NA
*0.2
0.6
1.4
2.2
3
f (x<k )
NA
2.199
1.427
0.943
1.131
2.375
Now we may write a right Riemann sum
y
f (x<1 ) x + f (x<2 ) x + f (x<3 ) x + f (x<4 ) x + f (x<5 ) x,
which evaluates to
= 2.199 (4_5) + 1.427 (4_5) + 0.943 (4_5)
x0 = *1
x1 = *0.2
x2 = 0.6
x3 = 1.4
x4 = 2.2
x5 = 3
x
On the other hand, the sample points identify which endpoints we use:
+ 1.131 (4_5) + 2.375 (4_5)
and we find
= 6.46.
Midpoint Riemann sums
Example 175. Consider f (x) = x3 _8 * x + 2. Approximate
the area between the graph of f and the x-axis on the interval
[*1, 3] using a midpoint Riemann sum with n = 5 rectangles.
255
Approximating area with rectangles
First note that the width of each rectangle is
x=
It is helpful to collect all of this data into a table:
3 * (*1)
= 4_5.
5
k
0
1
2
3
4
5
The grid points define the edges of the rectangle and are
seen below:
y
xk
*1
*0.2
0.6
1.4
2.2
3
x<k
NA
*0.6
0.2
1
1.8
2.6
f (x<k )
NA
2.573
1.801
1.125
0.929
1.597
Now we may write a midpoint Riemann sum
f (x<1 ) x + f (x<2 ) x + f (x<3 ) x + f (x<4 ) x + f (x<5 ) x,
x0 = *1
x1 = *0.2
x2 = 0.6
x3 = 1.4
x4 = 2.2
x5 = 3
x
On the other hand, the sample points identify which endpoints we use:
which evaluates to
= 2.573 (4_5) + 1.801 (4_5) + 1.125 (4_5)
+ 0.929 (4_5) + 1.597 (4_5)
and we find
y
= 6.42.
Example 176. Consider the function
1
2
3
4
5
x<1 = *0.6
x<2 = 0.2
x<3 = 1
x<4 = 1.8
x<5 = 2.6
x
f (x) = 16 * x2
on the interval [0, 4]. We will approximate the area between
the graph of f and the x-axis on the interval [0, 4]. See the
figure below.
256
Approximating area with rectangles
16
y
each rectangle.
x=
y = f (x)
12
4
n
Next, we will determine the grid-points.
8
x0 = 0
4
x1 = 0 + x =
0.5
1
1.5
2
2.5
3
3.5
4
4
n
4
x3 = x2 + x = 3
n
4
x2 = x1 + x = 2
x
The image depicts a Right Riemann sum with n = 8 subintervals.
xk*1 = xk*2 + x = (k * 1)
xk = xk*1 + x = k
x<1 = 0.5
f (x<1 ) = 15.75
x<6
f (x<6 )
x<n
f (x<n )
n
…
k=1
4
=4
=0
1
x=
2
f (x<k ) x = 38.5
This approximation is an underestimate.
Example 177. Consider the function
f (x) = 16 * x2
on the interval [0, 4]. We will approximate the area between the
graph of f and the x-axis on the interval [0, 4] using a right
Riemann sum with n rectangles. First, determine the width of
4
n
4
n
xn = 4.
=3
=7
4
n
For a right Riemann sum, for k = 1, … , n, we determine the
sample points as follows:
4
x<k = xk = k .
n
Now, we can approximate the area with a right Riemann sum.
A˘
n
…
k=1
f (x<k ) x =
n
…
k=1
n
…
f (xk )
4
n
⇠ ⇡
4 4
f k
n n
k=1
0
1
n
…
4
2 16
=
16 * k
2
n
n
k=1
=
We can now simplify the last sum by using the distribution,
257
Approximating area with rectangles
commutativity and associativity properties of a sum.
1
1
n 0
n 0
…
4 4…
2 16
2 16
A˘
16 * k
=
16 * k
n2 n n k=1
n2
k=1
H n
I
n
…
4 …
2 16
=
16 *
k
n k=1
n2
k=1
16
is a common factor, so we can
n2
again apply the distribution property and obtain the following
H n
I
n
4 …
16 … 2
A˘
16 *
k
n k=1
n2 k=1
In the last sum, the constant
In the first sum above, the constant 16 is added n times, and
n
…
we have a formula for the second sum. Recall:
k2 =
n(n + 1)(2n + 1)
. Therefore,
6
0
1
4
16 n(n + 1)(2n + 1)
A˘
n(16) *
n
6
n2
We can simplify this expression and obtain
k=1
32(n + 1)(2n + 1)
3n2
This approximation is an underestimate.
Remark. In our previous example we have computed a right
Riemann sum for n = 8. Check whether your result was correct
by plugging in n = 8 in the formula above.
Now, we can take the limit of Riemann sums as n ô ÿ to
find the exact value of the area of the region under the curve
y = f (x) on the interval [0, 4]. Namely,
nôÿ
and, therefore
A = lim
nôÿ
258
0
64 *
n
…
k=1
f (x<k ) x,
32(n + 1)(2n + 1)
3n2
• What interval are we on? In our discussion above we call
this [a, b].
• How many rectangles will be used? In our discussion
above we called this n.
• What is the width of each individual rectangle? In our
discussion above we called this x.
• What points will determine the height of the rectangle?
In our discussion above we called these sample points,
x<k , and they can be left-endpoints, right-endpoints, or
midpoints.
• What is the actual height of the rectangle? This will always
be f (x<k ).
• We approximate the area A with a Riemann sum
A˘
n
…
k=1
A ˘ 64 *
A = lim
Summary Riemann sums approximate the area between
curves and the x-axis via rectangles. When computing this
area via rectangles, there are several things to know:
1
=
128
.
3
f (x<k ) x.
• As n gets bigger and bigger, x gets smaller and smaller,
and approximation gets better and better. We compute the
exact value of A by taking the limit of Riemann sums
A = lim
nôÿ
n
…
k=1
f (x<k ) x.
Definite integrals
30
De�nite integrals
After completing this section, students should be able to do the
following.
• Use integral notation for both antiderivatives and definite
integrals.
• Compute definite integrals using geometry.
• Compute definite integrals using the properties of integrals.
• Justify the properties of definite integrals using algebra or
geometry.
• Understand how Riemann sums are used to find exact area.
• Define net area.
• Approximate net area.
• Split the area under a curve into several pieces to aid with
calculations.
• Use symmetry to calculate definite integrals.
• Explain geometrically why symmetry of a function simplifies calculation of some definite integrals.
259
Computing areas
Break-Ground:
30.1
Computing areas
Check out this dialogue between two calculus students (based
on a true story):
Devyn: I’m tired of Riemann sums.
Riley: Can’t we use high school geometry to compute integrals?
Problem 1. Which curves are good for using high school geometry to compute the area?
Problem 2. Which curves are good for using Riemann sums
to compute the area?
260
The definite integral
Dig-In:
10
y
5
30.2
f (2) 2
The de�nite integral
f (4) 2
x<2 = 2
The process of approximating areas under curves led to the
notion of a Riemann sum
f (6) 2
x<3 = 4
y = f (x)
x<4 = 6
f (8) 2
f (10) 2
x<5 = 8
x
*5
*10
A˘
n
…
k=1
*15
f (x<k ) x,
where f is a nonnegative, continuous function on the interval
[a, b], and
x<k is a sample point for the kth rectangle, k = 1, 2, ..., n.
The limit of Riemann sums, as n ô ÿ, gives the exact area
between the curve y = f (x) and the interval on the x*axis:
A = lim
nôÿ
n
…
k=1
f (x<k ) x.
It does not matter whether we consider only right Riemann
sums, or left Riemann sums, or midpoint Riemann sums, or
others: the limit of any kind of Riemann sum as n ô ÿ is
equal to the area, as long as f is nonnegative and continuous
on [a, b].
What happens when a continuous function f assumes negative
values on the interval [a, b]?
We can still form Riemann sums and take the limit.
The question is: What is the meaning of a Riemann sum in that
case?
Example 178. The graph of the function f on the interval
[0, 10] is given in the figure below.
The figure illustrates a right Riemann sum with n = 5 rectangles
for f on [0, 10].
5
…
k=1
f (x<k ) x = f (2) x+f (4) x+f (6) x+f (8) x+f (10) x
So, the Riemann sum is the sum of signed areas of rectangles:
rectangles that lie above the x-axis contribute positive values,
and rectangles that lie below the x-axis contribute negative
values to the Riemann sum.
5
…
k=1
f (x<k )
`
a
r
s
x = r f (2) x + f (4) x + f (6) x s
´Ø¨
´Ø¨ s
r ´Ø¨
pnonnegative nonnegative nonnegativeq
`
a
r
s
+ rf (8) x + f (10) xs
r´Ø¨ ´≠Ø≠¨s
p negative
negative q
When we take the limit of Riemann sums, it seems that we
should get that
lim
nôÿ
n
…
k=1
f (x<k ) x = A1 + (*A2 ),
where the areas A1 and A2 are depicted in the figure below.
In other words, the limit of Riemann sums seems to be equal to
the sum of signed areas of the regions that lie entirely above or
261
The definite integral
below the x-axis. The signed area of regions that lie above the
x-axis is positive, and the signed area of regions that lie below
the x-axis is negative.
10
y
y = f (x)
5
This sum of signed areas is called the net area of the region
between the graph of f and the interval on the x- axis .
A2
A4
3
A1
6
A3
9
x
*5
10
y
*10
y = f (x)
5
A1
2
4
6
8
10
x
9
A2
*5
*10
0
f (x) dx
Express the integral in terms of areas A1 , A2 , A3 and
A4 .
*15
The limit of Riemann sums will exist for any continuous functions on the interval [a, b], even if f assumes negative values
on [a, b]. The limit of Riemann sums gives the net area of the
region between the graph of f and an interval on the x- axis.
This leads to the following definition.
b
a
f (x) dx = lim
nôÿ
n
…
k=1
9
0
f (x) dx = *A1 + A2 * A3 + A4
Example 180. Consider the integral
Definition. Let f be a function which is continuous on the
interval [a, b]. We define the definite integral of f on [a, b]
by
f (x<k ) x.
The definite integral is a number that gives the net area of
the region between the curve y = f (x) and the x-axis on the
interval [a, b].
Example 179. The graph a function f on the interval [0, 9]
is given in the figure. The areas of four regions that lie either
above or below the x-axis are labeled in the figure.
262
Consider the integral
10
0
*x dx.
Compute the integral in two ways:
(a) by interpreting the integral as the net area of the region
between the curve y = *x and the interval [0, 10] on the
x-axis;
(b) using the definition of the definite integral, i.e. by computing the limit of Riemann sums.
(a) The area between the x-axis and the curve can be
easily computed, since it is the area of a triangle.
The definite integral
10
y
10
5
5
2
4
6
8
10
x<1 = 2
x<2 = 4
x<3 = 6
x<4 = 8
x
x<5 = 10
*5
*10
*10
*15
We can apply the constant multiple rule for sums
and limits:
1
A=
10 10 = 50.
2
10
Then, it follows that
10
0
0
*x dx = *A
*x dx = lim
nôÿ
n
…
k=1
*x dx = lim
nôÿ
n
…
k=1
nôÿ
f (x<k ) x = lim
nôÿ
n
…
k=1
*x<k x
*xk x
H
= lim
(b) We use the definition of the definite integral and
write
0
x
A
*5
10
y
= * lim
nôÿ
*
n
…
k=1
n
…
k=1
I
xk x
xk x
We can now finish our computation of the limit of
It does not matter what type of a Riemann sum we
use, so we choose a right Riemann sum.
10
0
*x dx = lim
nôÿ
n
…
k=1
*xk x
A right Riemann sum with n = 5 is illustrated in the
figure below.
263
The definite integral
right Riemann sums.
10
0
and
*x dx = * lim
nôÿ
n
…
k=1
n
…
Hence, we may rewrite this as
xk x
10 10
= * lim
k
nôÿ
n n
k=1
= * lim
⇠
nôÿ
⇠
10
n
n
⇡2 …
k
k=1
⇡
10 2 n(n + 1)
= * lim
nôÿ
n
2
n+1
= * lim 50
nôÿ
n
= *50
Example 181. (a) Express the limit as a definite integral.
v
n `
⇠
⇡2 a ⇠ ⇡
…
r 1 * *1 + 2k s 2
lim
nôÿ
r
n s n
k=1 p
q
(b) Compute this limit:
v
n `
⇠
⇡2 a ⇠ ⇡
…
r 1 * *1 + 2k s 2
lim
nôÿ
r
n s n
k=1 p
q
(i) This is a limit of Riemann sums! Specifically,
it is a limit of Riemann sums of n rectangles,
where
2
x= ,
n
2k
,
n
a = x<0 = *1,
b = x<n = 1.
lim
nôÿ
1
n 0t
…
1 * (x<k )2 =
k=1
1˘
*1
1 * x2 dx.
(ii) The limit computes the area between the xaxis ˘
and the curve
y=
1 * x2 . Let’s see it:
y
1
0.8
0.6
0.4
0.2
*1
*0.8
*0.6
*0.4
*0.2
0.2
0.4
0.6
0.8
By geometry, we know that this semicircle has
area ⇡_2. Hence
lim
nôÿ
1
n 0t
…
1 * (x<k )2
x = ⇡_2.
k=1
The definite integral computes the net area (sum of signed
areas) between y = f (x) and the x-axis on the interval [a, b].
x<k = *1 +
264
Question 122. Consider the graph of a function f on the interval [0, 5].
1
x
The definite integral
This looks like a property of the definite integral. Are there
other properties?
y
2
1
A1
*0.5
0.5
1
1.5
2
2.5
3
3.5
4
4.5
A2
5
x
5.5
*1
Compute the definite integral
5
0
0
f (x) dx.
The areas A1 and A2 are easily computed, since both are
the areas of triangles. Therefore,
0
f (x) dx = 1.25
There is more to this example, than just a value of the integral.
Notice that
3
0
5
3
and
5
It follows that
5
0
0
(c)
(d)
f (x) dx = A1 * A2
5
(a)
(b)
The net area in the figure above is given by
5
Theorem 56 (Properties of the definite integral). Let f and g
be defined on a closed interval [a, b] that contains the value c,
and let k be a constant. The following hold:
f (x) dx = A1 ,
(e)
a
a
c
a
b
a
b
a
b
a
f (x) dx = 0
f (x) dx +
b
c
f (x) dx = *
f (x) dx =
a
b
b
a
f (x) dx
f (x) dx
k f (x) dx = k
f (x) ± g(x) dx =
b
a
f (x) dx
b
a
f (x) dx ±
b
a
g(x) dx
We will address each property in turn:
(a) Here, there is no “area under the curve” when the
region has no width; hence this definite integral is
0.
(b) This states that total area is the sum of the areas
of subregions. Here a picture is worth a thousand
words:
f (x) dx = *A2 ,
f (x) dx = A1 * A2 .
f (x) dx =
3
0
f (x) dx +
5
3
f (x) dx.
265
The definite integral
y
4
y
2
*0.5
c
a
b
f (x) dx
a
c
c
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
x
6.5
4
4.5
5
5.5
6
x
6.5
*2
f (x) dx
b
x
It is important to note that this still holds true even
if a < b < c. We discuss this in the next point.
(c) For now, this property can be viewed a merely a convention to make other properties work well. However, later we will see how this property has a justification all its own.
(d) This states that when one scales a function by, for
instance, 7, the area of the enclosed region also is
scaled by a factor of 7.
(e) This states that the integral of the sum is the sum of
the integrals.
Due to the geometric nature of integration, geometric properties
of functions can help us compute integrals.
Example 182. Compute:
*4
Now we can graph y = x * 3:
4
y
2
*0.5
0.5
1
1.5
2
2.5
3
3.5
*2
*4
Now we see that we really have two triangles, each with
base 3 and height 3. Hence
6
0
x * 3 dx =
3
0
3 * x dx +
3 3 3 3
+
2
2
= 9.
6
3
x * 3 dx
=
6
0
0.5
x * 3 dx
This may seem difficult at first. Perhaps the first thing to
do is look at a graph of y = x * 3:
Definition. A function f is an odd function if
f (*x) = *f (x),
and a function g is an even function if
g(*x) = g(x).
The names odd and even come from the fact that these properties are shared by functions of the form xn where n is either
266
The definite integral
odd or even. For example, if f (x) = x3 , then
(b)
f (*7) = *f (7),
and if g(x) = x4 , then
(c)
g(*7) = g(7).
Geometrically, even functions have symmetry with respect to
the y*axis . Cosine is an even function:
x
*3⇡_2
*⇡
*⇡_2
⇡_2
⇡
3⇡_2
2⇡
✓
*1
On the other hand, odd functions have 180˝ rotational symmetry around the origin. Sine is an odd function:
x
1
*2⇡
*3⇡_2
*⇡
(e)
1
cos(✓)
*2⇡
(d)
*⇡_2
⇡
3⇡_2
2⇡
✓
*1
Question 123. Let f be an odd function defined for all real
numbers. Compute:
*2
(g)
*4
*4
*2
4
*2
4
*4
2
2
*4
(h) *
f (x) dx
(i) *
f (x) dx
f (x) dx
f (x) dx
*2
sin(✓)
⇡_2
2
(f)
*2
f (x) dx
f (x) dx
f (x) dx
2
*4
f (x) dx
*2
*4
f (x) dx
Net area versus geometric area
We know that the net area of the region between a curve y =
f (x) and the x-axis on [a, b] is given by
b
Question 124. Let f be an odd function defined for all real
numbers. Which of the following are equal to
4
2
Select All Correct Answers:
(a)
2
4
f (x) dx
f (x) dx?
a
f (x) dx.
On the other hand, if we want to know the geometric area,
meaning the “actual” area, we compute
b
a
f (x) dx.
Example 183. The graph of a function f is given in the figure
below.
267
The definite integral
10
can write
y
9
y = f (x)
5
A2
A1 1
2
3
0
A4
4
5
6
A3 7
8
9
f (x) dx =
1
0
f (x) dx +
x
+
*5
+
*10
1
=
0
(b) Express the geometric area of the region between the curve
y = f (x) and the x-axis on the interval [0, 9] in terms of
definite integrals of f .
(*f (x)) dx +
+
+
(c) Express the geometric area of the region between the curve
y = f (x) and the x-axis on the interval [0, 9] in terms of
areas A1 , A2 , A3 and A4 .
=*
f (x) dx
1
0
f (x) dx +
*
The figure below depicts the graph of f .
10
+
y
9
A1
1
2
A2
3
A3
4
5
6
7
*5
8
A4
9
x
0
5
1
8
f (x) dx
5
8
9
8
f (x) dx
(*f (x)) dx
f (x) dx
5
1
5
8
8
9
f (x) dx
f (x) dx
f (x) dx
(ii) Using the properties of definite integrals, we
f (x) dx = A1 + A2 + A3 + A4
Question 125. True or false:
b
*10
268
9
f (x) dx
(iii)
y = f (x)
5
5
8
f (x) dx
Pulling out negative signs:
(i) The geometric area is given by the integral
0
1
Rewriting the absolute value signs:
(a) Express the geometric area of the region between the curve
y = f (x) and the x-axis on the interval [0, 9] as a definite
integral.
9
5
a
Û
Û
f (x) dx = Û
Û
Û
b
a
Û
Û
f (x) dxÛ
Û
Û
Antiderivatives and area
31
Antiderivatives and area
After completing this section, students should be able to do the
following.
• Interpert the product of rate and time as area.
• Approximate position from velocity.
• Recognize Riemann sums.
269
Meaning of multiplication
Break-Ground:
31.1
Meaning of multiplication
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Hey Riley, I was reading about the history of mathematics.
Riley: Really? Tell me about it!
Devyn: Apparently, back in the day, mathematicians worried about
writing things like:
3ù4+5
Riley: Why? What’s the matter here?
Devyn: Well, they thought of 3 ù 4 as an area. But 5 was thought of
as a length. Apparently they worried whether it made sense to
add “areas” and “lengths.”
Riley: Hmmm. We don’t seem to worry about that now. I wonder
why?
Problem 1. What are some ways to interpret 3 ù 4?
270
Relating velocity, displacement, antiderivatives and areas
Dig-In:
31.2 Relating velocity, displacement,
antiderivatives and areas
We have a geometric interpretation of the derivative as the slope
of a tangent line at a point. We have not yet found a geometric
interpretation of antiderivatives.
More than one perspective
We’ll start with a question:
Question 126. Suppose you are in slow traffic moving at 4
mph from 2pm to 5pm. How far have you traveled?
Since the displacement of an object moving at constant velocity
over a time interval can be represented by the area of a rectangle, we can approximate displacement of an object moving
at nonconstant velocity using Riemann sums. We explain this
process in detail in our next example.
Example 184. Assume an object is moving along a straight
t2
line with the velocity v(t) =
+ 1, for 0 f t f 4, where t is
4
ft
measured in seconds, and v in .
s
Find the displacement of the object over the time interval
[0, 4].
Since we know how to compute the displacement when
the velocity is constant, we will start by asuming that the
velocity is constant and equal to v(0), the initial velocity,
for the entire four-second time interval, t = 4.
This scenario will give us a very rough approximation of
the displacement, but it is a good starting point.
displacement ˘ v(0)
t = 1 (4 * 0) = 4 f t_ s
5
y
4
y = v(t)
3
2
1
v(0)
1
2
t
3
4
t
Now it should be clear how to proceed. We will keep
approximating the velocity with a constant over smaller
and smaller intervals of time.
In our next step we will assume that the velocity is constant during two-second time intervals [0, 2] and [2, 4].
For each of these two intervals, we assume that the velocity is equal to the velocity at the beginning of the time
interval.
In this case, the object would move by v(0) 2 f t during
the time interval [0, 2], and during the next two seconds,
the object would move by v(2) 2 f t.
If we let t denote the length of the time interval, we can
approximate the displacement and write
displacement ˘ v(0)
t+v(2)
t = 1 2+2 2 = 6 f t_ s
This scenario is illustrated in the figure below.
271
Relating velocity, displacement, antiderivatives and areas
Using sigma notation, we write
displacement ˘
2
…
time intervals, and that velocity is equal to the velocity
at the beginning of each time interval.
v((k * 1)
t) t
k=1
Since we evaluate the velocity at the sample points t<k =
(k * 1) t, k = 1, 2, we can also write
displacement ˘
2
…
k=1
v(t<k )
t.
This is a left Riemann sum for the function v on the
interval [0, 4], when n = 2. This scenario is illustrated
in the figure below.
5
y
Under this assumption, during the first second, the object
would move by v(0) 1 f t; during the second time interval
of one second, the object would move by v(1) 1 f t, during
the third second, the object would move by v(2) 1 f t,
and during the last second, it would move by v(3) 1 f t.
During the entire time interval, the object would move
by
displacement ˘ v(0) t + v(1)
15
=
f t_ s
4
Using sigma notation and the fact that t =
we can write
displacement ˘
4
4
…
v((k * 1)
2
displacement ˘
4
…
k=1
1
v(2) t
2
3
4
t
Next, we approximate the displacement of the object
assuming that the velocity is constant during one-second
272
4*0
= 1,
4
t) t
Again, we denote the sample points by t<k = (k * 1)
k = 1, 2, 3, 4, and write
3
1
t + v(3)
k=1
y = v(t)
v(0) t
t + v(2)
t,
v(t<k ) t
This is a left Riemann sum for the function v on the
interval [0, 4], when n = 4. This scenario is illustrated
in the figure below.
t
Relating velocity, displacement, antiderivatives and areas
5
Using sigma notation and the fact that t =
we can write
y
4
displacement ˘
y = v(t)
8
…
v((k * 1)
4*0 1
= ,
8
2
t) t
k=1
3
Using the standard notation for sample points, we can
write
8
…
displacement ˘
v(t<k ) t
2
k=1
1
v(0) t
v(1) t
1
v(2) t
2
v(3) t
3
4
t
Now, we approximate the displacement of the object
assuming that the velocity is constant during half-second
time intervals, and that velocity is equal to the velocity
at the beginning of each time interval.
Under this assumption, during the first half-second, the
1
object would move by v(0)
f t; during the second
2
time
interval of half a second, the object would move by
⇠ ⇡
1
1
v
f t, during the third half- second interval, the
2
2
1
object would move by v(1)
f t,etc.
2
During the entire time interval, the object would move
by
⇠ ⇡
⇠ ⇡
1
7
displacement ˘ v(0) t + v
t+5+v
t
2
2
= 8.375 f t
and, This is a left Riemann sum for the function v on the
interval [0, 4],when n = 8. This scenario is illustrated in
the figure below.
5
y
4
y = v(t)
3
2
1
v(0) t
v(0.5) t
0.5
v(1) t
1
v(1.5) t
1.5
v(2) t
2
v(2.5) t
2.5
v(3) t
3
v(3.5) t
3.5
4
t
We pause here to illustrate the kth rectangle when computing a left Riemann sum of the function v on the interval
[0, 4].
273
Relating velocity, displacement, antiderivatives and areas
and
y
y = v(t)
k=1
v(t<k )
k
t<k
tn = 4
tk
t
We continue by approximating the displacement of the
object assuming that the velocity is constant during 4_nsecond time intervals, with velocity equal to the velocity
at the beginning of each time interval.
4
Under this assumption, during the first -second interval,
n
4
the object would move by v(0)
f t; during the second
n
⇠ ⇡
4 4
such time interval, the object would move by v
f t,
n n
during the third time interval, the object would move by
⇠ ⇡
4
4
v 2
f t, etc.
n
n
During the entire time interval, the object would move
by
⇠ ⇡
⇠ ⇡
4
4
displacement ˘ v(0) t + v
t+v 2
t+5
n
n
⇠
⇡
⇠ ⇡
4
4
+ v (n * 1)
t+v n
t
n
n
Using sigma notation, notation for the length of each time
4*0 4
interval, t =
= , and notation for sample pints,
n
n
we write
displacement ˘
n
…
k=1
274
n
…
v(t<k ) t.
This is a left Riemann sum for the function v on the
interval [0, 4].
t
tk*1 =
displacement ˘
v((k * 1)
t) t
What happens when we let n ô ÿ?
Approximations get better and better, because we approximate the velocity with a constant over smaller and smaller
time intervals. Therefore, it seems reasonable to write
displacement = lim
nôÿ
n
…
k=1
v(t<k ) t
By the definition of the definite integral, we can write
displacement =
4
0
v(t) dt.
This gives us another interpretation of the definite integral.
Namely,
“definite integral of the velocity over [a, b]= the displacement over [a, b].”
On the other hand, the displacement of an object during
the time interval [0, 4] is given by
displacement = change of position
The change of position can be written in terms of s, the
position function of the object
displacement = s(4) * s(0).
Relating velocity, displacement, antiderivatives and areas
But, we know that s is an antiderivative of v. Therefore,
0 2
1
t
t3
s(t) =
v(t) dt =
+ 1 dt =
+t+C
4
12
If t = 0, we get
s(0) = C
So, the position function is given by
s(t) =
t3
+ t + s(0).
12
Now we can compute the displacement
displacement = s(4) * s(0) =
28
.
3
Much more important than this computation is the fact that we
were able to express the displacement as a definite integral and
compute the definite integral of v using the antiderivative of v.
4
0
v(t) dt = s(4) * s(0)
Is this always so?
Is it true that for any continuous function f and its antiderivative F , we can compute the definite integral of f on the interval
[a, b] by this simple formula
b
a
f (t) dt = F (b) * F (a)?
275
First Fundamental Theorem of Calculus
32 First Fundamental
Theorem of Calculus
After completing this section, students should be able to do the
following.
• Define accumulation functions.
• Calculate and evaluate accumulation functions.
• State the First Fundamental Theorem of Calculus.
• Take derivatives of accumulation functions using the First
Fundamental Theorem of Calculus.
• Use accumulation functions to find information about the
original function.
• Understand the relationship between the function and the
derivative of its accumulation function.
276
What’s in a calculus problem?
Break-Ground:
32.1
What’s in a calculus problem?
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley! How do you know when you are doing a calculus
problem?
Riley: Hmmm. I guess if you take a derivative you are doing a
calculus problem.
Devyn: What else?
Problem 1. Let’s hear your answer to Devyn’s question above.
277
The First Fundamental Theorem of Calculus
Question 128. What is F (*5)?
Dig-In:
32.2 The First Fundamental
Theorem of Calculus
Example 185. Consider the following accumulation function
for f (x) = x3 .
F (x) =
Accumulation functions
While the definite integral computes a signed area, which is a
fixed number, there is a way to turn it into a function.
Definition. A given a function f , an accumulation function
is
x
F (x) =
a
x
*1
t3 dt.
Considering the interval [*1, 1], where is F increasing? Where
is F decreasing? When does F have local extrema?
We can see a graph of f along with the signed area measured by the accumulation function below
1
f (t) dt.
y
0.5
One thing that you might notice is that an accumulation function
seems to have two variables: x and t. Let’s see if we can explain
this. Consider the following graph:
+
*1
x
*
t
*0.5
y
*1
f (x)
The accumulation function starts off at zero, and then
as x grows, F is decreasing as the function accumulates
negatively signed area.
f
F (x) =
f (a)
x
a
f (t) dt
a
x
t
An accumulation function F measures the signed area in the
region [a, x] between f and the t-axis. Hence t is playing the
role of a “place-holder” that allows us to evaluate f . On the
other hand, x is the specific number that we are using to bound
the region that will determine the area between f and the t-axis,
and hence the value of F .
Question 127. Given
F (x) =
what is F (5)?
278
x
*3
4 dt,
However when x > 0, F starts to accumulate positively
signed area, and hence is increasing. Thus F is increasing
on (0, 1), decreasing on (*1, 0) and hence has a local
minimum at (0, 0).
Working with the accumulation function leads us to a question,
what is
x
a
f (t) dt
when x < a? The general convention is that
b
a
f (t) dt = *
a
b
f (t) dt.
With this in mind, let’s consider one more example.
The First Fundamental Theorem of Calculus
Example 186. Consider the following accumulation function
for f (x) = x3 .
x
F (x) =
*1
3
F (x) =
From our previous example, we know that F is increasing
on (0, 1). Since f continues to be positive at t = 1 and
beyond, F is increasing on (0, ÿ). On the other hand, we
know from our previous example that F is decreasing on
(*1, 0).
y
+
Let f be a continuous function on the real numbers and consider
t dt.
Where is F increasing? Where is F decreasing? When does F
have local extrema?
x
The First Fundamental Theorem of Calculus
t
*1
*2
For values to the left of t = *1, F is still decreasing,
as less and less positively signed area is accumulated.
Hence F is increasing on (0, ÿ), decreasing on (*ÿ, 0)
and hence has an absolute minimum at (0, 0).
The key point to take from these examples is that an accumulation function
F (x) =
x
a
f (t) dt
is increasing precisely when f is positive and is decreasing precisely when f is negative. In short, it seems that f is behaving
in a similar fashion to F ® .
f (t) dt.
Theorem 57 (First Fundamental Theorem of Calculus). Suppose that f is continuous on the real numbers and let
F (x) =
*6
*10
a
From our previous work we know that F is increasing when f
is positive and F is decreasing when f is negative. Moreover,
with careful observation, we can even see that F is concave
up when f ® is positive and that F is concave down when f ®
is negative. Thinking about what we have learned about the
relationship of a function to its first and second derivatives, it is
not too hard to guess that there must be a connection between
F ® and the function f . This is a good guess, check out our next
theorem:
*4
*8
x
x
a
f (t) dt.
Then F ® (x) = f (x).
The First Fundamental Theorem of Calculus says that an accumulation function of f is an antiderivative of f . Another way
of saying this is:
d
dx
x
a
f (t) dt = f (x)
This could be read as:
The rate that accumulated area under a
curve grows is described identically by
that curve.
279
The First Fundamental Theorem of Calculus
Now that we are working with accumulation functions, let’s see
what happens when we compose them with other functions.
G® (x) = *x3 . The chain rule gives us
F ® (x) = G® (h(x))h® (x)
Example 187. Find the derivative of
F (x) =
x2
2
Consider
x
G(x) =
2
and set h(x) = x2 . Now
= *h(x)3 h® (x)
ln t dt.
= * cos3 (x)(* sin(x)).
Example 189. Given the graph of a function f on the interval
[*1, 5], sketch the graph of the accumulation function
ln t dt
F (x) =
®
®
F (x) = G (h(x))h (x)
= ln(h(x))h® (x)
= ln(x2 )2x.
Let’s practice this once more.
Example 188. Find the derivative of
5
cos x
Consider
G(x) = *
and set h(x) = cos(x). Now
t3 dt.
x
5
y = f (t)
2
1
*1
*
+
1
t3 dt
The First Fundamental Theorem of Calculus states that
2
x
3
4
*1
*2
First, we evaluate F at some significant points.
F (*1) =
F (0) =
F (1) =
F (x) = G(h(x)).
280
*1 f x f 5.
f (t) dt,
y
The First Fundamental Theorem of Calculus states that
G® (x) = ln x. The chain rule gives us
F (x) =
*1
3
F (x) = G(h(x)).
®
x
F (3) =
F (4) =
*1
*1
0
*1
f (t) dt = 0.
f (t) dt = *1.5.
1
*1
f (t) dt = *2.
3
*1
4
*1
f (t) dt = 0.
f (t) dt = 2.
5
t
The First Fundamental Theorem of Calculus
F (5) =
5
*1
straight line, i.e. f (t) = v(t), a f t f b?
f (t) dt = 4.
Since F ® (x) = f (x), it follows that the function F is
increasing on the interval
(1, 5)
What is the meaning of an accumulation function in that case?
We use different variables in that case, since we want F to
be a function of time, t. With this adjustment, we define the
accumulation function F as follows.
and decreasing on the interval
Since the integral
(*1, 1) .
Since the function f , the derivative of F , is increasing
on (1, 3), F is concave up on the interval
(1, 3) .
Since f is constant on the interval [3, 5], F is linear on
[3, 5].
The function F has a local and global minimum at
x = 1,
Now we are ready to sketch the graph of F , on the same
set of axes as the graph of f .
v(z) dz
gives the displacement of the object on the time interval [a, t],
it follows that
t
s(t) * s(a) =
a
v(z) dz,
where s(t) gives the position of the object at the time t. If we
differentiate this equation with respect to t, we get that
d
dt
t
a
v(z) dz.
v(t) =
d
dt
t
a
v(z) dz.
This is the the First Fundamental Theorem of Calculus!
y
f
2
1
*2
a
v(z) dz.
Since s® (t) = v(t), we have that
x = 5.
*1
a
t
s® (t) =
and the global maximum at
4
t
F (t) =
2
F
3
4
5
t
What if f is the velocity function for an object moving along a
281
Second Fundamental Theorem of Calculus
33 Second Fundamental
Theorem of Calculus
After completing this section, students should be able to do the
following.
• State the Second Fundamental Theorem of Calculus.
• Evaluate definite integrals using the Second Fundamental
Theorem of Calculus.
• Understand how the area under a curve is related to the
antiderivative.
• Understand the relationship between indefinite and definite integrals.
282
A secret of the definite integral
Break-Ground:
33.1
A secret of the de�nite integral
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Ah. So now we have a connection between derivatives and
integrals.
Riley: Right, the derivative of the accumulation function is the “inside” function.
Devyn: So how do we use this to compute area?
Sometimes it helps to think about the most basic examples.
Consider
5
2
4 dt
We know (by geometry) that this computes the area of a 3 ù 4
rectangle which equals 12. On the other hand, if we consider
the accumulation function
F (x) =
x
2
we see that
F (5) =
5
2
4 dt,
4 dt.
Problem 1. What is F (2)?
Problem 2. On the other hand, the First Fundamental Theorem
of Calculus says that if
F (x) =
x
2
4 dt,
then F ® (x) = 4. Armed with this knowledge, and the fact that
F (2) = 0, what must F (x) be?
283
The Second Fundamental Theorem of Calculus
Dig-In:
33.2 The Second Fundamental
Theorem of Calculus
There is a another common form of the Fundamental Theorem
of Calculus:
Theorem 58 (Second Fundamental Theorem of Calculus). Let
f be continuous on [a, b]. If F is any antiderivative of f , then
b
a
f (x) dx = F (b) * F (a).
a
f (x) dx =
=
c
a
c
b
f (x) dx +
f (x) dx *
b
c
c
a
f (x) dx
f (x) dx.
By the First Fundamental Theorem of Calculus, we have
F (b) =
b
c
f (x) dx
and
F (a) =
for some antiderivative F of f . So
b
a
a
c
f (x) dx
f (x) dx = F (b) * F (a)
for this antiderivative. However, any antiderivative could
have be chosen, as antiderivatives of a given function
differ only by a constant, and this constant always cancels
out of the expression when evaluating F (b) * F (a).
From this you should see that the two versions of the Fundamental Theorem are very closely related. In reality, the two
forms are equivalent, just differently stated. Hence people
often simply call them both “The Fundamental Theorem of
284
b
a
f ® (x) dx = f (b) * f (a)
This could be read as:
The accumulation of a rate is given by the
change in the amount.
Let a f c f b and write
b
Calculus.” One way of thinking about the Second Fundamental
Theorem of Calculus is:
When we compute a definite integral, we first find an antiderivative and then evaluate at the limits of integration. It is convenient to first display the antiderivative and then evaluate. A
special notation is often used in the process of evaluating definite integrals using the Fundamental Theorem of Calculus.
Instead of explicitly writing F (b) * F (a), we often write
4
5b
F (x)
a
meaning that one should evaluate F (x) at b and then subtract
F (x) evaluated at a
4
F (x)
5b
a
= F (b) * F (a).
Let’s see some examples of the fundamental theorem in action.
Example 190. Compute:
2
*2
x3 dx
The Second Fundamental Theorem of Calculus
We start by finding an antiderivative of x3 . A correct
x4
choice is , one could verify this by taking the deriva4
tive. Hence
4 4 52
2
x
x3 dx =
4 *2
*2
24 (*2)4
*
4
4
= 0.
=
Hence
5
0
Example 193. Compute:
2⇠
1
0
We start by finding an antiderivative of sin ✓. A correct
choice is * cos ✓, one could verify this by taking the
derivative. Hence
4
5⇡
⇡
sin ✓ d✓ = * cos ✓
=
0
= * cos(⇡) * (* cos(0))
= 2.
This is interesting: It says that the area under one “hump”
of a sine curve is 2.
Example 192. Compute:
5
0
⇡
1
dx
x
210
1
+ ln(2) * .
10
10
Understanding motion with the Fundamental
Theorem of Calculus
We know that
• The derivative of a position function is a velocity function.
et dt
We start by finding an antiderivative of et . A correct
choice is et , one could verify this by taking the derivative.
x9 +
1
We start by finding an antiderivative of x9 + . A correct
x
x10
choice is
+ ln(x), one could verify this by taking the
10
derivative. Hence
4 10
52
⇡
2⇠
1
x
9
x +
dx =
+ ln(x)
x
10
1
1
sin ✓ d✓
0
0
= e5 * e0
= e5 * 1.
Example 191. Compute:
⇡
4 55
et dt = et
• The derivative of a velocity function is an acceleration
function.
Now consider definite integrals of velocity and acceleration
functions. Specifically, if v(t) is a velocity function, what does
b
a
v(t) dt mean?
285
The Second Fundamental Theorem of Calculus
The Second Fundamental Theorem of Calculus states that
b
a
v(t) dt = V (b) * V (a),
where V (t) is any antiderivative of v(t). Since v(t) is a velocity
function, V (t) must be a position function, and V (b) * V (a)
measures a change in position, or displacement.
Example 194. A ball is thrown straight up with velocity given
by v(t) = *32t + 20ft/s, where t is measured in seconds. Find,
and interpret,
1
0
v(t) dt.
Using the Second Fundamental Theorem of Calculus, we
have
1
0
1
v(t) dt =
4
=
0
(*32t + 20) dt
* 16t2 + 20t
= 4.
51
0
Thus if a ball is thrown straight up into the air with velocity
v(t) = *32t + 20,
the height of the ball, 1 second later, will be 4 feet above
the initial height. Note that the ball has traveled much
farther. It has gone up to its peak and is falling down, but
the difference between its height at t = 0 and t = 1 is 4ft.
Now we know that to solve certain kinds of problems, those
that involve accumulation of some form, we “merely” find an
antiderivative and substitute two values and subtract. Unfortunately, finding antiderivatives can be quite difficult. While
there are a small number of rules that allow us to compute the
derivative of any common function, there are no such rules
for antiderivatives. There are some techniques that frequently
prove useful, but we will never be able to reduce the problem
to a completely mechanical process.
286
A tale of three integrals
Which student is correct?
Dig-In:
33.3
A tale of three integrals
At this point we have three different “integrals.” Let’s see if we
can sort out the differences.
Multiple Choice:
(a) Devyn is correct
Inde�nite integrals
(b) Riley is correct
An indefinite integral, also called an antiderivative computes
classes of functions:
(d) Neither student is correct
f (x) dx = “a class of functions whose derivative is f ”
Here there are no limits of integration, and your answer will
have a “+C” at the end. Pay attention to the notation:
f (x) dx = F (x) + C
Where F ® (x) = f (x).
Indefinite integrals do not have limits of integration, and
they compute a class of antiderivatives.
Question 129. Two students, say Devyn and Riley, are working
with the following indefinite integral:
2
dx
x ln(x2 )
Devyn computes the integral as
2
dx = ln  ln x2  + C
x ln(x2 )
and Riley computes the integral as
2
dx = ln  ln x + C.
x ln(x2 )
(c) Both students are correct
Accumulation functions
An accumulation function, also called an area function computes accumulated area:
x
a
f (t) dt = “a function F whose derivative is f ”
This is a function of x whose derivative is f , with the additional
property that F (a) = 0. Pay attention to the notation:
x
F (x) =
a
f (t) dt
Where F ® (x) = f (x).
Accumulation functions have limits of integration, and
they compute an antiderivative.
Question 130. True or false: There exists a function f such
that
x
0
f (t) dt = ex
287
A tale of three integrals
De�nite integrals
A definite integral computes signed area:
b
a
f (x) dx = “the signed area between the x-axis and f ”
Here we always have limits of integration, both of which are
numbers. Moreover, definite integrals have definite values, the
signed area between f and the x-axis. Pay attention to the
notation:
b
a
f (x) dx = F (b) * F (a)
Where F ® (x) = f (x).
Definite integrals have limits of integration, and they
compute signed area.
288
Applications of integrals
34
Applications of integrals
After completing this section, students should be able to do the
following.
• Given a velocity function, calculate displacement and
distance traveled.
• Given a velocity function, find the position function.
• Given an acceleration function, find the velocity function.
• Understand the difference between displacement and distance traveled.
• Understand the relationship between position, velocity
and acceleration.
• Calculate the change in the amount.
• Compute the average value of the function on an interval.
• Understand that the average value of the function on an
interval is attained by the function on that interval.
289
What could it represent?
Break-Ground:
34.1
What could it represent?
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley, I like integrals.
Riley: I feel fancy when I make an integral sign.
Devyn: I know! An integral computes the signed area between a
curve y = f (x) and the x-axis. But why signed area? Maybe
we should just compute plain old area.
Riley: Makes sense to me!
Deyvn: Unless. . . maybe there are other applications where “signed”
area makes more sense.
One really great way to think about integrals is that they “accumulate rates.”
Problem 1. Write down as many examples of “rates” and
“accumulated rates” as you can. For example:
5 miles per hour is a rate, and 5 miles is then an
accumulated rate.
290
Applications of integrals
Dig-In:
34.2
What is the displacement of the particle from t = 0 to t = ⇡?
That is, compute:
Applications of integrals
⇡
Velocity and displacement, speed and distance
Some values include “direction” that is relative to some fixed
point.
• v(t) is the velocity of an object at time t. This represents
the “change in position” at time t.
What is the displacement of the particle from t = 0 to t = 2⇡?
That is, compute:
2⇡
0
t
a
v(z) dz.
• s(b) * s(a) is the displacement, the distance between the
starting and finishing locations.
On the other hand speed and distance are values without “direction.”
• v(t) is the speed.
b
•
a
v(t) dt is the distance traveled.
Question 131. Consider a particle whose velocity at time t is
given by v(t) = sin(t).
y
sin(t) dt
What is the distance traveled by the particle from t = 0 to
t = ⇡? That is, compute:
• s(t) is the position of an object at time t. This gives location with respect to the origin.
s(t) = s(a) +
sin(t) dt
0
⇡
0
sin(t) dt
What is the distance traveled by the particle from t = 0 to
t = 2⇡? That is, compute:
2⇡
0
sin(t) dt
Change in the amount
We can apply the Fundamental Theorems of Calculus to a
variety of problems where both accumulation and rate of change
play important roles. For example, we can consider a tank that
is being filled with fuel at some rate. Given the rate, we can
ask what is the amount of fuel in the tank at a certain time.
Or, a tank that is being emptied at a given rate, or a culture of
bacteria growing in a Petri dish, or a population of a city, etc.
This brings us to our next theorem.
Theorem 59.
⇡
2⇡
x
Let A(t) denote the amount of some substance/population at
the time t.
Assume that the function A is continuous, differentiable and its
derivative, A® , continuous on some time interval [a, b].
291
Applications of integrals
Then, the change in the amount A over the time interval [a, b]
is given by
b
A(b) * A(a) =
a
A® (t)dt.
The equation follows directly from the Second Theorem of Calculus. The left hand side obviously gives
the change in the amount A over the time interval [a, b],
which is the difference between the amount at the end of
the time interval and the amount at the beginning of the
time interval.
The right hand side of the equation gives the “accumulation of a rate". You can think of this definite integral as
the limit of Riemann sums, where each Riemann sum is
the“sum of the changes of the amount over small intervals
of time".
Example 195. An experiment is conducted in which a culture
of bacteria is grown in a Petri dish. The initial population was
estimated at 1200 cells. The growth rate of the population is
estimated at P ® (t) = 100e*0.2t cells/day.
(a) By how much has the population grown during the first
three days of the experiment?
The growth of population during the first three days
of the experiment is given by:
3
P (3) * P (0) =
0
=
4
=
0
3
This Riemann sum approximates the change in population during the first three days of the experiment.
We expect it to be an underestimate, because the
function P ® is decreasing and this is a right Riemann sum.
Since n = 3, t = 1.
P (3) * P (0) ˘
3
…
k=1
= P ® (1) + P ® (2) + P ® (3)
= 100e*0.2 + 100e*0.4 + 100e*0.6
˘ 204
(c) Find the population at any time t g 0.
100e*0.2t dt
53
0
˘ 226
(b) Compute the right Riemann sum of the function P ® and the
t
P (t) * P (0) =
0
=
4
P ® (t) dt
100 *0.2t
e
*0.2
P ® (k)
=
During the first three days, the population grew approximately by 226 cells.
292
interval [0, 3] with n = 3. What does this Riemann sum
approximate? Is this approximation and underestimate or
an overestimate and why?
Therefore, for t g 0,
0
t
P ® (z) dz
100e*0.2z dz
100 *0.2z
e
*0.2
5t
0
= 500 1 * e*0.2t
P (t) = P (0) + 500 1 * e*0.2t = 1700 * 500e*0.2t
Average value
Conceptualizing definite integrals as “signed area” works great
as long as one can actually visualize the “area.” In some cases,
Applications of integrals
a better metaphor for integrals comes from the idea of average
value. Looking back to your days as an even younger mathematician, you may recall the idea of an average:
Definition. Let f be continuous on [a, b]. The average value
of f on [a, b] is given by
f=
n
f1 + f2 + 5 + fn
1…
=
f
n
n k=1 k
If we want to know f , the average value of a function on the
interval [a, b], a naive approach might be to introduce n + 1
equally spaced grid points on the interval [a, b]
a = x0 < x1 < 5 < xn = b
and choose a sample point x<k in each interval [xk , xk+1 ], k =
1, … , n.
f (x<1 ) + f (x<2 ) + 5 + f (x<n )
n
Multiply this last expression by 1 =
n
=
n
1…
f (x<k ).
n k=1
(b * a)
:
(b * a)
n
(b * a) …
1…
1 (b * a)
f (x<k )
=
f (x<k )
n k=1
(b * a) k=1
n (b * a)
=
=
n
…
1
b*a
f (x<k )
b * a k=1
n
n
…
1
f (x<k ) x
b * a k=1
where x = (b * a)_n. Ah! On the right we have a Riemann
Sum!
What will happen as n ô ÿ?
We take the limit as n ô ÿ:
n
1 …
1
f = lim
f (x<k ) x =
nôÿ b * a
b
*
a
k=1
This leads us to our next definition:
b
a
b
a
f (x) dx.
Multiplying this equation by (b * a), we obtain that
f (b * a) =
b
a
f (x) dx.
If f is positive, the average value of a function gives the height
of a single rectangle whose area is equal to
We will approximate the average value of f on the interval
[a, b] with the average of f (x<1 ), f (x<2 ), . . . , and f (x<n ):
f˘
1
b*a
b
a
f (x) dx.
y
f (b * a) =
f
a
b
a
f (x) dx
b*a
b
x
An application of this definition is given in the next example.
Example 196. An object moves back and forth along a straight
line with a velocity given by v(t) = (t * 1)2 on [0, 3], where t
is measured in seconds and v(t) in ft/s.
What is the average velocity of the object?
f (x) dx.
293
Applications of integrals
By our definition, the average velocity is:
v=
1
3*0
=
1
3
=
1
3
=
3
0
4
3
0
the average value of a function, however, we need to be more
careful.
3
0
For instance, there are at least two different ways to make sense
of a vague phrase like “The average height of a point on the
unit semi circle”
v(t) dt
(t * 1)2 dt
(t2 * 2t + 1) dt
y
=
53
1 t3
* t2 + t
3 3
0
= 1 ft/s.
Question 132. Choose all the correct expressions for v, the
average velocity of an object moving along a straight line over
the time interval [a, b].
(Reminder: s is the position function, and a the acceleration).
(b) v =
1
b*a
1
b*a
1
(c) v =
b*a
b
a
b
a
b
a
x
s(t) dt
v(t) dt
on the interval [*1, 1].
a(t) dt
(d) v =
a(b) * a(a)
b*a
(e) v =
v(b) * v(a)
b*a
s(b) * s(a)
(f) v =
b*a
When we take the average of a finite set of values, it does
not matter how we order those values. When we are taking
294
1
One way we can make sense of “The average height of a point
on the unit semi circle” is to compute the average value of the
function
˘
Select All Correct Answers:
(a) v =
*1
1 * x2
f (x) =
Example 197. Compute the average value of the function
f (x) =
˘
1 * x2
on the interval [*1, 1].
By definition, we wish to compute
1
2
1˘
*1
1 * x2 dx.
Computing this integral geometrically, we find the aver⇡
age value to be .
4
Applications of integrals
Another way we can make sense of “The average height of a
point on the unit semi circle” is the average value of the function
g(✓) = sin(✓)
on [0, ⇡], since sin(✓) is the height of the point on the unit circle
at the angle ✓.
Example 198. Compute the average value of the function
We demonstrate the principles involved in this version of the
Mean Value Theorem in the following example.
1
⇡
⇡
0
sin(✓) d✓.
Computing this integral geometrically, we find the aver2
age value to be .
⇡
See if you can understand intuitively why the average using f
should be larger than the average using g.
Mean value theorem for integrals
Just as we have a Mean Value Theorem for Derivatives, we also
have a Mean Value Theorem for Integrals.
Theorem 60 (The Mean Value Theorem for integrals). Let f
be continuous on [a, b]. There exists a value c in [a, b] such
that
b
a
0
sin x dx. Find
a value c guaranteed by the Mean Value Theorem.
We first need to evaluate
g(✓) = sin(✓)
on the interval [0, ⇡].
By definition, we wish to compute
⇡
Example 199. Let f (x) = sin x. Consider
⇡
0
⇡
0
4
sin x dx =
sin x dx.
5⇡
* cos x = 2.
0
Thus we seek a value c in [0, ⇡] such that ⇡ sin c = 2.
⇡ sin c = 2 Ÿ sin c = 2_⇡ Ÿ c1 = arcsin(2_⇡) ˘ 0.69.
There is another such point: c2 = ⇡ * 0.69. Indeed,
f (⇡ * 0.69) = sin (⇡ * 0.69) = sin (0.69) =
2
= f.
⇡
A graph of sin x, along with a rectangle with height
sin(0.69), is pictured below. The area of the rectangle
is the same as the area under sin x on [0, ⇡]. The two
points c1 and c2 are also shown in the figure. Recall,
f (c1 ) = f (c2 ) = f .
y
y = f (x)
f
c1
c2
⇡
x
f (x) dx = f (c)(b * a).
This is an existential statement. The Mean Value Theorem for
Integrals tells us:
The average value of a continuous function is in
the range of the function.
295
The idea of substitution
35
The idea of substitution
After completing this section, students should be able to do the
following.
• Undo the chain rule.
• Calculate indefinite integrals (antiderivatives) using basic
substitution.
• Calculate definite integrals using basic substitution.
296
Geometry and substitution
Then, using this transformation idea, we can evaluate
Break-Ground:
35.1
Geometry and substitution
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Riley! We should be able to figure some integrals geometrically using transformations of functions.
Riley: That sounds
˘ like a cool idea. Maybe, since we know the graph
of f (x) = 1 * x2 is a semicircle, we get an ellipse defined
on [*2, 2] just by stretching the graph of f by a factor of 2
horizontally. The equation of this ellipse would be
u
⇠ ⇡2
x
g(x) = 1 *
2
b
a
f (3x + 1) dx
if a = ? and b = ? . The value of the integral on this interval
is ? .
Devyn: Exactly! So since we know that
1
*1
˘
⇡
1 * x2 dx =
2
geometrically. . .
Riley: And we know that the area under g from [*2, 2] is twice the
under f . . .
Devyn and Riley: We must have
2
˘
*2
1 * (x_2)2 dx = ⇡!
Devyn and Riley: Jinx!
Devyn: It is kind of like we just stretched out our whole coordinate
system, and that helped us solve an integral.
Riley: In this case, everything got stretched out by a constant factor
of 2 in the horizontal direction. I wonder if we could ever say
anything useful about cases where we stretch the x-axis by a
different amount at each point?
Devyn: Whao, that is a wild thought. That seems really hard. Since
derivatives measure how much a function stretches a little piece
of the domain, maybe the derivative will come into play here?
Riley: Hmmmm, but I do not see exactly how. Maybe we should ask
our TA about this?
Problem 1. Say we know that
4
1
f (x) dx = 5.
297
The idea of substitution
Dig-In:
35.2
The idea of substitution
Computing antiderivatives is not as easy as computing derivatives. One issue is that the chain rule can be difficult to “undo.”
We have a general method called “integration by substitution”
that will somewhat help with this difficulty.
If the functions are differentiable, we can apply the chain rule
and obtain
d
f (g(x)) = f ® (g(x))g ® (x)
dx
If the derivatives are continuous, we can use this equality to
evaluate a definite integral. Namely,
4
5b
b
®
®
f (g(x))g (x) dx = f (g(x))
a
a
= f (g(b)) * f (g(a)).
a
g(b)
g(a)
f ® (u) du.
This simple observation leads to the following theorem.
Theorem 61 (Integral Substitution Formula). Let g ® be the
derivative of a function g and let f ® be the derivative of a
function f . If g ® is continuous on the interval [a, b] and f ® is
continuous on the interval [g(a), g(b)], then
b
a
298
f ® (g(x))g ® (x) dx =
g(b)
g(a)
2
0
2
xex dx
Why do we think that the ISF can be useful in this example?
Does this integral have the structure
f ® (g(x))g ® (x) dx,
so that we can apply the ISF?
g(a)
Since the right hand sides of these equalities are equal, the left
hand sides must be equal, too. So, it follows that
f ® (g(x))g ® (x) dx =
Example 200. Compute:
a
= f (g(b)) * f (g(a)).
b
We will apply the Integral Substitution Formula (ISF) to the
following example.
b
On the other hand, it is also true that
4
5g(b)
g(b)
f ® (u) du = f (u)
g(a)
The integral on the right appears to be much simpler than the
original integral. So, when evaluating a difficult integral, we
try to apply this theorem and replace the original integral with
a simpler one. We can do this as long as the conditions of the
theorem are satisfied.
f ® (u) du.
We set g(x) = x2 , so g ® (x)dx = 2xdx, and note that
2
0
2
1
xe dx =
2
x2
e
0
x2
´Ø¨
1
2x dx =
´Ø¨ 2
2
0
eg(x) g ® (x) dx
So we wrote the integral in a form suitable for application
of the ISF. Therefore, we can substitute u for g(x) and du
for g ® (x) dx and write
2
0
2
xex dx =
1
2
2
0
eg(x) g ® (x) dx =
1
2
g(2)
g(0)
eu du
The idea of substitution
Now, compute the endpoints:
Example 201. Compute:
2
g(0) = 02 = 0
2
g(2) = 2 = 4.
0
We can easily evaluate the last integral
4 54
g(2)
4
1
1
1 u
1 4
e4 * 1
u
u
e du =
e du =
e
=
e * e0 =
2 g(0)
2 0
2
2
2
0
In summary,
2
0
1
xe dx =
2
4
x2
0
Here we will set
u = g(x) = x2 .
Then
du = g ® (x) dx = 2x dx.
Here, we are thinking in terms of differentials. We can
solve for dx to get
u
e du
The figure below illustrates both integrals.
y
dx =
A=
2
0
100
2
B=
xex dx
4
0
1 u
e du
2
2
2
y = xex
y=
A
2
4
x
The two shaded areas in the figure, A =
4
du
.
2x
Using this equality, we can write
y
100
2
xex dx
1 u
e
2
2
B
4
2
0
0
2
1
B=
eu du, appear to be equal, and this confirms
2 0
what was established by computation before.
Therefore, whenever we are faced with a problem of
evaluating a difficult integral, we try to replace it with a
simpler one, as long as both integrals represent the same
net area.
We can solve a problem like this in a slightly different way.
Let’s do the same example again, this time we will think in
terms of differentials.
xeu
g(0)
4 u
e
=
u
xex dx and
g(2)
2
xex dx =
0
2
du
2x
du.
At this point, we can continue as we did before and write
4
0
eu
e4 * e0
du =
.
2
2
Finally, sometimes we simply want to deal with the antiderivative on its own. We will repeat the example one more time in
oder to demonstrate this approach.
Example 202. Compute:
2
0
2
xex dx
299
The idea of substitution
We can apply the Second Fundamental Theorem of Calculus and, in order to do that, we first need to compute
an indefinite integral.
2
xex dx.
As before, we set u = g(x) = x2 , and compute du =
2x dx, thinking in terms of differentials. Now we see
that
2
du
eu
xex dx =
xeu
=
du.
2x
2
Hence
eu
xe dx =
+ C.
2
x2
This does not look right, since we need to find an an2
tiderivative of the function xex , and our result is a function of u. We can easily fix this, by simply substituting u
with g(x) = x2 . Therefore,
2
eu
ex
xe dx =
+C =
+ C.
2
2
x2
Finally, we can apply the SFTOC
2
0
4
2
xex dx =
=
2
ex
2
e4
52
0
* e0
2
.
More examples
With some experience, it is (usually) not too hard to see what to
substitute as u. We will work through the following examples
in the same way that we did for Example 201.
Example 203. Compute:
x4 (x5 + 1)9 dx
300
Here we set u = x5 + 1, so du = 5x4 dx. Then it follows
du
that dx =
. Therefore
5x4
x4 (x5 + 1)9 dx =
x4 (u)9
du
5x4
1
u9 du
5
u10
=
.
50
=
Notice that this example is an indefinite integral and not
a definite integral, meaning that there are no limits of
integration. So we do not need to worry about changing
the endpoints of the integral. However, we do need to
back-substitute into our answer, so that our final answer
is a function of x. Recalling that u = x5 + 1, we have our
final answer
x4 (x5 + 1)9 dx =
(x5 + 1)10
+ C.
50
Reminder: you can always verify your result by differentiating.
If substitution works to solve an integral (and that is not always
the case!), a common trick to find what to substitute for is to
locate the “ugly” part of the function being integrated. We
then substitute for the “inside” of this ugly part. While this
technique is certainly not rigorous, it can prove to be very
helpful. This is especially true for students new to the technique
of substitution. The next two problems are really good examples
of this philosophy.
Example 204. Compute:
0
*1
12x3 cos(x4 ) dx
The idea of substitution
The “ugly” part of the function being integrated is cos(x4 ).
The “inside” of this term is then x4 . So a good possibility
is to try
u = g(x) = x4 .
Then
Then we substitute back into the original integral and
solve:
⇡
e4
1
3
du = (4x ) dx
1
dx =
du
4x3
Ÿ
and so
⇡
4
cos(ln x)
dx =
x
0
=
4
0
⇡
4
cos(u)
x du
x
cos(u) du
= sin(u)
0
g(0)
1
12x cos(x ) dx =
12x cos(u)
du
4x3
*1
g(*1)
3
4
=
0
1
3
3 cos(u) du
4
50
= 3 sin(u)
1
= 3(0 * sin(1)).
Example 205. Compute:
⇡
e4
1
5⇡
4
0
˘
˘
2
2
=
*0=
.
2
2
To summarize, if we suspect that a given function is the derivative of another via the chain rule, we introduce a new variable
u = g(x), where g is a likely candidate for the inner function.
We rewrite the integral entirely in terms of u, with no x remaining in the expression. If we can integrate this new function of
u, then the antiderivative of the original function is obtained
by replacing u by g(x).
cos(ln x)
dx
x
Here the “ugly” part here is cos(ln x). So we substitute
for the inside:
u = g(x) = ln x.
Then
du =
Notice that
1
dx
x
Ÿ
dx = x du.
g(1) = ln(1) = 0
⇠ ⇡⇡
⇡⇡
⇡
g e 4 = ln e 4 = .
4
⇠
301
Working with substitution
36
Working with substitution
After completing this section, students should be able to do the
following.
• Determine when a function is a composition of two or
more functions.
• Calculate indefinite and definite integrals requiring complicated substitutions.
• Recognize common patterns in substitutions.
• Evaluate indefinite and definite integrals through a change
of variables.
302
Integrals are puzzles!
Break-Ground:
36.1
Integrals are puzzles!
Check out this dialogue between two calculus students (based
on a true story):
Devyn: Yo Riley, is it just me, or are integrals kind of fun?
Riley: I always feel accomplished when I finish one.
Devyn: I know! Also, even though antiderivatives are difficult, we
can always check our work by taking the derivative.
Riley: So awesome!
Devyn: But something is bothering me. When we are doing substitution, we have to find f and g such that
f (g(x)) g ® (x) dx =
f (g) dg.
How do we choose f and g?
Riley: Well, never ever pick g(x) = x, this doesn’t change anything!
Devyn: And never ever pick g(x) to be the entire integrand, this
doesn’t help either.
Riley: Somehow we must “see” one function “nested” inside of another.
Devyn: I’m not sure there’s an easy path to doing, this, I think it’s
gonna take practice.
Problem 2. Consider
sin5 (3x) cos(3x) dx =
if f (g) =
f (g(x)) g ® (x) dx
g5
, and
3
f (g(x)) g ® (x) dx =
f (g) dg.
what is g(x)?
Problem 3. In your own words, explain why Devyn and Riley
claim we should never pick g(x) = x or g(x) to be the entire
integrand.
In the problems that follow, we will be using the substitution
formula
f (g(x)) g ® (x) dx =
f (g) dg
While you may use a slightly different method to compute your
integrals, the skills developed by answering the problems below
will help you in your quest to conquer calculus.
Problem 1. Consider
sin5 (3x) cos(3x) dx =
f (g(x)) g ® (x) dx
if g(x) = 3x, and
f (g(x)) g ® (x) dx =
f (g) dg.
what is f (g)?
303
Working with substitution
Now
Dig-In:
36.2
Working with substitution
3
We begin by restating the substitution formula.
2
Theorem 62 (Integral Substitution Formula). If g is differentiable on the interval [a, b] and f is differentiable on the
interval [g(a), g(b)], and if they have continuous derivatives,
then
b
a
f ® (g(x))g ® (x) dx =
g(b)
g(a)
g(3)
1
dx =
x ln(x)
=
4
ln(2)
= ln(u)
f ® (u) du.
1
g(2) x
ln(3)
u
x du
1
du
u
5ln(3)
ln(2)
= ln(ln(3)) * ln(ln(2)).
We will work out a few more examples.
The next example requires a new technique.
Example 206. Compute:
Example 207. Compute:
3
2
Let
computing du, we find
1
dx
x ln(x)
u = ln(x),
1
du = dx
x
and solving for dx we find
dx = x du.
˘
x3 1 * x2 dx
Here it is not apparent that the chain rule is involved.
However, if it was involved, perhaps a good guess for g
would be
u = 1 * x2
and then
du = *2x dx,
*1
dx =
du.
2x
Now we consider the integral we are trying to compute
˘
x3 1 * x2 dx
and we substitute using our work above. Write with me
˘
x3 1 * x2 dx =
=
304
˘ ⇠ *1 ⇡
x3 u
du
2x
˘
*x2 u
du.
2
Working with substitution
However, we cannot continue until each x is replaced.
We know that
2
u=1*x
Ÿ
u * 1 = *x2
Ÿ
1 * u = x2
so now we may replace x2
˘
x 1 * x2 dx =
3
*
u = sec(y) + tan(y)
and we immediately see that
du = (sec(y) tan(y) + sec2 (y)) dy,
1
dy =
du.
sec(y) tan(y) + sec2 (y)
˘
(1 * u) u
2
du.
At this point, we are close to being done. Write
H ˘
˘
˘ I
(1 * u) u
u u
u
*
du =
*
du
2
2
2
˘
u
u3_2
=
du *
du
2
2
u5_2 u3_2
=
*
.
5
3
Now recall that u = 1 * x2 . Hence our final answer is
˘
(1 * x2 )5_2 (1 * x2 )3_2
x3 1 * x2 dx =
*
+ C.
5
3
Sometimes it is not obvious how a fraction could have been
obtained using the chain rule. A common trick though is to
substitute for the denominator of a fraction. Like all tricks,
this technique does not always work. Regardless the next two
examples present how this technique can be used.
Example 208. Compute:
sec(y) tan(y) + sec2 (y)
sec(y) + tan(y)
We substitute
dy
But this cancels perfectly with the numerator! So we
have that
sec(y) tan(y) + sec2 (y)
1
dy =
du
sec(y) + tan(y)
u
= ln u + C
= ln  sec(y) + tan(y) + C.
Notice that
sec(x) tan(x) + sec2 (x) sec(x)(tan(x) + sec(x))
=
= sec(x)
sec(x) + tan(x)
sec(x) + tan(x)
when sec(x) ë * tan(x). So in a very contrived way, we have
just proved
Theorem 63.
sec(x) dx = ln  sec(x) + tan(x) + C.
Notice the variable in this next example.
Example 209. Compute:
u
du
1 * u2
305
Working with substitution
We want to substitute for 1 * u2 . But the variable “u” has
already been used. . . OH NO! Never fear! We can substitute with whatever variable that we want. In particular,
let us use “w” for this problem. So we let
We begin by writing
sin(x)
cos(x)
sin(x) cos(x)
=
cos(x) cos(x)
sin(x) cos(x)
=
.
1 * sin2 (x)
tan(x) =
w = 1 * u2
and then
We then make the substitution
dw = *2u du,
*1
du =
dw.
2u
Thus
⇠
s = sin(x)
and so
⇡
u *1
dw
w 2u
1
1
=*
dw
2
w
1
= * ln w + C
2
1
= * ln 1 * u2  + C.
2
u
du =
1 * u2
ds = cos(x) dx,
1
dx =
ds.
cos(x)
Then
tan(x) dx =
=
Recall:
g(b)
g(a)
®
f (u) du =
g(b)
g(a)
®
f (w) dw =
g(b)
g(a)
®
f (s) ds.
So, we can call a variable u by any other name and “it would
smell as sweet".
Example 210. Compute:
tan(x) dx
=
sin(x) cos(x)
dx
1 * sin2 (x)
s cos(x)
1
ds
2
cos(x)
1*s
s
ds.
1 * s2
But this is the same problem as Example 209! And so
we know that
1
tan(x) dx = * ln 1 * s2  + C
2
1
= * ln 1 * sin2 (x) + C
2
1
= * ln  cos2 (x) + C
2
1
= ln  cos2 (x)* 2 + C
= ln  sec(x) + C.
306
Working with substitution
We have just proved
and
Theorem 64.
tan(x) dx = ln  sec(x) + C.
Note that in Example 210, we could have instead made the
substitution
u = 1 * sin2 (x).
This would have gotten us to the answer quicker and without
using Example 209. You are encouraged to work this out on
your own right now!
We end this section with two more difficult examples.
Example 211. Compute:
e2x
dx
1 * e2x
Maybe the biggest key to solving this problem is to recall
that
e2x = (ex )2 .
So we can rewrite the problem
(ex )2
dx.
1 * (ex )2
e2x
dx =
1 * e2x
Now, if we make the substitution u = ex , we have that
du = e dx,
1
dx = x du,
e
e2x
dx =
1 * e2x
=
=
(ex )2 1
du
1 * u2 ex
ex
du
1 * u2
u
du.
1 * u2
But now we are back to Example 209, and so we know
that
e2x
1
dx = * ln 1 * u2  + C
2
1 * e2x
1
= * ln 1 * e2x  + C.
2
Again, in the previous example we could have instead made
the substitution
u = 1 * e2x
and avoided using Example 209. In general, any time that you
make two successive substitutions in a problem, you could have
instead just made one substitution. This one substitution is the
composition of the two original substitutions. But sometimes
it may not be obvious to make one clever substitution, and so
two substitutions makes more sense. The next example helps
to demonstrate this.
Example 212. Compute:
16 t
x
0
4*
˘
x dx
While it is not obvious at all, let us try the substitution
˘
u = x.
307
Working with substitution
Then
So
1
du = ˘ dx,
2 x
˘
dx = 2 x du = 2u du,
and so
16 t
0
4*
˘
x dx =
=
g(16) ˘
g(0)
4
0
4 * u 2u du
˘
2u 4 * u du.
From here we now make the second (and more obvious)
substitution
w = 4 * u.
Then u = 4 * w, and
dw = * du,
du = * dw.
308
16 t
0
4*
˘
x dx =
=
4
0
w(4)
w(0)
0
=*
=
˘
2u 4 * u du
0
4
4
˘
2(4 * w) w(*1) dw
1
3
(8w 2 * 2w 2 ) dw
1
3
(8w 2 * 2w 2 ) dw
4 ⇠ ⇡
3
2
= 8
w2
3
⇠
16 32
=
(4) *
3
128 128
=
*
3
5
256
=
.
15
⇠ ⇡ 5 54
2
*2
w2
5
0
⇡
4 52
(4) * (0 * 0)
5
The Work-Energy Theorem
Dig-In:
36.3
The Work-Energy Theorem
In physics, we take measurable quantities from the real world,
and attempt to find meaningful relationships between them.
A basic example of this would be the physical ideal of force.
Force applied to an object changes the motion of an object.
Here’s the deal though, at a basic level
First we need to find the velocity at which the apple hits
the ground. Let a(t) = *9.8 represent acceleration at
time t and v(t) represent velocity. Since v(0) = 0, we
know that
v(t) =
=
t
0
0
t
a(x) dx
*9.8 dx
force = mass acceleration.
= *9.8t.
and while we can put a physical interpretation to this arithmetical definition, at the end of the day force is simply “mass
times acceleration.” The SI unit of force is a newton, which is
defined to be
1 N = 1 kg m_ s2 .
Now we need to know how long it takes for the apple to
hit the ground, after being dropped from a height of 1
meter. For this we’ll need a formula for position. Here
s(0) = 1, so we’ll need to use an indefinite integral:
Question 133. To get a feel for what a newton is, consider this:
if an apple has a mass of 0.1 kg, what force would an apple
exert on your hand due to the acceleration due to gravity?
In a similar way, the idea of kinetic energy, is “energy” objects
have from motion. It is defined by the formula
Ek =
m v2
.
2
The SI unit of energy is a joule, which is defined to be
1 J = 1 kg m2 _ s2 = 1 N m.
To get a feel for the “size” of a joule, consider this: if an apple
has a mass of 0.1 kg and it is dropped from a height of 1 m,
then approximately 1 joule of energy is released when it hits
the ground. Let’s see if we can explain why this is true.
s(t) =
v(t) dt
=
*9.8t dt
=
*9.8t2
+C
2
Since s(0) = 1, write with me
s(0) =
*9.8 0
+ C = 1,
2
hence C = 1 and s(t) =
*9.8t2
+1. Solving the equation
2
s(t) = 0
for t tells us the time the apple hits the ground. Write
Example 213. If an apple has a mass of 0.1 kg and it is
dropped from a height of 1 m, how much energy is released
when it hits the ground? Assume that the acceleration due to
gravity is *9.8 m_ s2 .
309
The Work-Energy Theorem
with me
Select All Correct Answers:
*9.8t2
2
s(t) = 0
(a) studying calculus
+1=0
(b) a car applying breaks to come to a stop over a distance of
100 f t
*9.8t2 = *2
2
t2 =
9.8
u
2
t=
.
9.8
u
2
So the apple hits the ground after
seconds. Finally,
9.8
the formula for kinetic energy is
m v2
2
0.1 (a t)2
=
2
0
t 12
2
0.1 (*9.8)
9.8
=
2
2
2
0.1 (9.8) 9.8
=
2
0.1 9.8 2
=
2
= 0.98.
Ek =
Ah! So the kinetic energy released by an apple dropped
from a height of 1 meter is approximately 1 joule.
Finally work is defined to be accumulated force over a distance.
Note, there must be some force in the direction (or opposite
direction) that the object is moving for it to be considered work.
Question 134. Which of the following are examples where
work of this kind is being done?
310
(c) a young mathematician climbing a mountain
(d) a young mathematician standing still, holding a 1000 page
calculus book for 10 minutes
(e) a young mathematician walking around with a 1000 page
calculus book
(f) a young mathematician picking up a 1000 page calculus
book
We can write the definition of work in the language of calculus
as,
W =
s1
s0
F (s) ds.
The SI unit of work is also a joule. To help understand this, 1
joule is approximately how much work is done when you raise
an apple one meter.
Let’s again see why this is true.
Example 214. If an apple has a mass of 0.1 kg, how much
work is required to lift this apple 1 meter? Assume that the
acceleration due to gravity is *9.8 m_ s2 .
Well, work is computed by
W =
s1
s0
F (s) ds.
Since force is mass times acceleration,
F (s) = 0.1 (*9.8)
= *0.98.
The Work-Energy Theorem
So, our integral becomes
1
0
• s(t0 ) represents the starting position, s0 ,
4
*0.98 ds =
• s(t1 ) represents the ending position, s1 ,
51
* 0.98s
• v(t0 ) represents the starting velocity, v0 ,
0
= *0.98.
Ah! So when lifting an apple 1 meter, requires *0.98
joules of work. The sign is negative since we are lifting
against the gravitational force.
Now we have a question:
Why do work and kinetic energy have the same
units?
• v(t1 ) represents the ending velocity, v1 .
Now write with me,
W =
W =
s1
s0
F (s) ds =
m v21
2
*
m v20
2
b
a
®
• s(t) represent position with respect to time,
• v(t) represent velocity with respect to time,
F (s) ds
®
g(b)
g(a)
f ® (g) dg
transforming from right to left, to see that
s(t1 )
s(t0 )
F (s) ds =
t1
t0
F (s(t))s® (t) dt
and we are now working with functions of time. Since
s® (t) = v(t), we may write
t1
t0
t1
F (s(t))s® (t) dt =
t0
F (s(t))v(t) dt
and now remember that F = m a, so
• a(s) represent acceleration with respect to position,
• t0 represent the starting time,
s(t0 )
f (g(x))g (x) dx =
.
First we need to get all of our symbolism out in the open.
Let:
s0
s(t1 )
F (s) ds =
here we are working with functions of distance. We will
use the substitution formula,
One way to answer this is via the Work-Energy Theorem.
Theorem 65 (Work-Energy Theorem). Suppose that an object
of mass m is moving along a straight line. If s0 and s1 are the
the starting and ending positions, v0 and v1 are the the starting
and ending velocities, and F (s) is the force acting on the object
for any given position, then
s1
t1
t0
F (s(t))v(t) dt =
t1
t0
m a(s(t))v(t) dt.
• t1 represent the ending time,
then we also have that
311
The Work-Energy Theorem
However, a(s(t)) = v® (t), so rearranging we have,
t1
t0
m a(s(t))v(t) dt = m
t1
t0
v(t)v® (t) dt.
Now we apply the substitution formula again, this time
we will transform left to right
b
a
®
g(b)
®
f (g(x))g (x) dx =
g(a)
f ® (g) dg
and so we see
m
t1
t0
v(t1 )
v(t)v® (t) dt = m
v(t0 )
v dv
and we are working with functions of velocity. At last,
setting v(t0 ) = v0 and v(t1 ) = v1 , we can evaluate this
integral,
4 2 5v1
v1
v
m
v dv = m
2 v
v0
=
m v21
2
0
*
m v20
2
.
The Work-Energy theorem says that:
s1
s0
F (s) ds =
m v21
2
*
m v20
2
This could be interpreted as:
The accumulated force over distance is
the change in kinetic energy.
312
Moreover, this answers our initial question of why work and
kinetic energy have the same units. In essence, energy powers
work.
Index
infinity, 65
line tangent to the curve, 90
negative infinity, 65
continuous on a half-closed interval, 43
continuous on an open interval, 43
cosecant, 26
cosine, 26, 43
cotangent, 26
critical point, 173
acceleration, 87
accumulation function, 278, 287
antiderivative, 235, 287
notation, 235
arccosecant, 29
arccosine, 27, 29
arccotangent, 29
arcsecant, 29
arcsine, 27, 29
arctangent, 29
area function, 287
asymptote
horizontal, 69
vertical, 66
average rate of change, 209
average value, 293
definite integral, 262, 288
degree, 21
derivative, 90
notation, 103
of arcsecant, 147
of arcsine, 146
of arctangent, 147
of cosine, 120
of ex , 107
of secant, 121
of sine, 109
of tangent, 120
of the natural logarithm, 137
derivative rules
chain, 118, 133
constant, 103
power, 104
product, 113
quotient, 114
sum, 105
determinate form, 59
differentiability implies continuity, 97
differentiable, 90
differential equation, 240, 243
displacement, 286, 291
distance, 291
domain, 8
base, 32
Binomial Theorem, 103
chain rule, 118, 133, 211
coefficients, 21
composition limit law, 49
concave down, 128
concave up, 128
concavity test, 128, 177
constant function, 43, 73
constant rule, 103
continuity of famous functions, 43, 73
continuity of polynomial functions, 48
continuity of rational functions, 49
continuous at a point, 42
continuous on a closed interval, 43
e, 107
Euler’s number, 107
313
The Work-Energy Theorem
even, 266
ex , 107
explicit function, 133
exponential function, 32, 43, 73
Extreme Value Theorem, 195
extremum
global, 194
local, 170
Fermat’s Theorem, 171
first derivative test, 176
First Fundamental Theorem of Calculus, 279
force, 309
function, 8
Fundamental Theorem of Algebra, 21
general solution, 241
global extremum, 194
global maximum, 194
global minimum, 194
greatest integer function, 9
grid points, 252
horizontal asymptote, 69
horizontal line test, 16
idenity function, 43
implicit differentiation, 133
indefinite integral, 235
indeterminate form, 59, 228
infinite limit, 65
infinity, 61
inflection point, 177
initial condition, 241
initial value problem, 241
instantaneous rate of change, 209
instantaneous velocity, 83
integral, 262
integral substitution formula, 298
Intermediate Value Theorem, 76
314
Inverse Function Theorem, 149
inverse trigonometric functions, 29, 73
inverses, 14
joule, 309, 310
kinetic energy, 309
L’Hôpital’s Rule, 228
left continuous, 43
limit, 37, 40
limit at infinity, 67
limit at negative infinity, 68
limit from the left, 40
limit from the right, 39
limit laws, 47
linear approximation, 203
local extremum, 170
local maximum, 170
local minimum, 170
logarithmic function, 32, 43, 73
marginal cost, 87
marginal revenue, 87
maximum/minimum
absolute, 194
local, 170
Mean Value Theorem, 198
Mean Value Theorem for integrals, 295
multiplying by the conjugate, 58
newton, 309
non-differentiable, 90
odd, 266
one-sided limit, 40
one-to-one, 15
optimization problem, 215
polynomial function, 21
polynomials, 73
The Work-Energy Theorem
position, 87, 291
power function, 43, 73
power rule, 104, 143
product rule, 113, 210
Pythagorean identities, 30
Pythagorean Theorem, 30
quotient rule, 114
range, 8
rational function, 23, 73
Riemann sum, 253
right continuous, 43
Rolle’s Theorem, 198
root, 21
sample point, 252
secant, 26
second derivative, 126
second derivative test, 178
Second Fundamental Theorem of Calculus, 284
sine, 26, 43
speed, 291
Squeeze Theorem, 51
sum rule, 105
tangent, 26
tangent line, 89
third derivative, 126
trigonometric function, 25, 26, 73
velocity, 87, 291
vertical asymptote, 66
vertical line test, 9, 16
work, 310
Work-Energy Theorem, 311
315