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MATH 221
FIRST SEMESTER
CALCULUS
fall 2009
Typeset:June 8, 2010
1
MATH 221 – 1st SEMESTER CALCULUS
LECTURE NOTES VERSION 2
...
The notes were written by Sigurd Angenent, starting
A
from an extensive collection of notes and problems compiled by Joel Robbin
...
math
...
edu/~angenent/Free-Lecture-Notes
They are meant to be freely available in the sense that “free software” is free
...
Angenent
...
2 or any later version published by the Free Software Foundation; with no Invariant
Sections, no Front-Cover Texts, and no Back-Cover Texts
...
Contents
Chapter 1
...
What is a number?
2
...
Functions
4
...
Exercises
3
...
5
...
7
...
9
...
Derivatives (1)
1
...
An example – tangent to a parabola
3
...
Rates of change
5
...
Exercises
21
21
22
25
25
27
27
29
32
33
34
35
36
36
38
Chapter 4
...
Derivatives Defined
2
...
Differentiable implies Continuous
4
...
Exercises
6
...
Differentiating powers of functions
8
...
Higher Derivatives
10
...
Differentiating Trigonometric functions
12
...
The Chain Rule
14
...
Implicit differentiation
16
...
Graph Sketching and Max-Min Problems
1
...
The Intermediate Value Theorem
11
...
13
...
15
...
Limits and Continuous Functions
1
...
The formal, authoritative, definition of limit
3
...
Variations on the limit theme
5
...
Examples of limit computations
7
...
What’s in a name?
9
...
Continuity
11
...
Exercises
13
...
Exercises
10
...
Exponentials and Logarithms (naturally)
1
...
Logarithms
3
...
Graphs of exponential functions and logarithms
5
...
Derivatives of Logarithms
7
...
Exponential growth and decay
9
...
The Integral
91
1
...
When f changes its sign
92
3
...
Exercises
94
5
...
Properties of the Integral
97
7
...
Method of substitution
99
9
...
Applications of the integral
105
1
...
Exercises
106
3
...
Examples of volumes of solids of revolution
109
5
...
Exercises
113
7
...
The length of a curve
116
9
...
Exercises
118
11
...
Work done by an electric current
119
Chapter 9
...
APPLICABILITY AND DEFINITIONS
2
...
COPYING IN QUANTITY
4
...
COMBINING DOCUMENTS
6
...
AGGREGATION WITH INDEPENDENT WORKS
8
...
TERMINATION
10
...
RELICENSING
125
125
125
125
126
126
126
126
126
126
126
4
CHAPTER 1
Numbers and Functions
The subject of this course is “functions of one real variable” so we begin by wondering what a real number
“really” is, and then, in the next section, what a function is
...
What is a number?
1
...
Different kinds of numbers
...
Together these form the integers or “whole numbers
...
These are the so called fractions or rational numbers such as
1 1 2 1 2 3 4
, , , , , , , ···
2 3 3 4 4 4 3
or
1
1
2
1
2
3
4
− , − , − , − , − , − , − , ···
2
3
3
4
4
4
3
By definition, any whole number is a rational number (in particular zero is a rational number
...
One day in middle school you were told that there are other numbers besides the rational numbers, and
the first example of such a number is the square root of two
...
e
...
e
...
n
x x2
Nevertheless, if you compute x2 for some values of x between 1 and 2, and check if you
get more or less than 2, then it looks like there should be some number x between 1
...
2 1
...
5 whose square is exactly 2
...
3 1
...
This raises several questions
...
4 1
...
4 and 1
...
5 2
...
6 2
...
It turns out to be rather difficult to give a precise
description of what a number is, and in this course we won’t try to get anywhere near the bottom of this
issue
...
One can represent certain fractions as decimal fractions, e
...
279
1116
=
= 11
...
25
100
5
Not all fractions can be represented as decimal fractions
...
333 333 333 333 333 · · ·
3
It is impossible to write the complete decimal expansion of 1 because it contains infinitely many digits
...
An electronic calculator, which always represents
numbers as finite decimal numbers, can never hold the number 1 exactly
...
If the decimal
expansion doesn’t end, then it must repeat
...
142857 142857 142857 142857
...
A real number is specified by a possibly unending decimal expansion
...
414 213 562 373 095 048 801 688 724 209 698 078 569 671 875 376 9
...
To give a precise description of a real number (such as 2) you have to
explain how you could in principle compute as many digits in the expansion as you would like
...
√
1
...
A reason to believe in 2
...
In
middle or high school you learned something similar to the following geometric construction
√
of a line segment whose length is 2
...
The figure you get
consists of 5 triangles of equal area and by counting triangles you see that the larger
square has exactly twice the area of the smaller square
...
Why are real numbers called real? All the numbers we will use in this first semester of calculus are
“real numbers
...
No real number has this property since the square of any real number is positive, so
it was decided to call this new imagined number “imaginary” and to refer to the numbers we already have
√
(rationals, 2-like things) as “real
...
3
...
It is customary to visualize the real numbers as points
on a straight line
...
We also
decide which direction we call “left” and hence which we call “right
...
”
To plot any real number x one marks off a distance x from the origin, to the right (up) if x > 0, to the
left (down) if x < 0
...
In particular, the
distance is never a negative number
...
To draw the half open interval [−1, 2) use a filled dot to mark the endpoint which is included
and an open dot for an excluded endpoint
...
To find
√
−1
0
1
√
2 2
2 on the real line you draw a square of sides 1 and drop the diagonal onto the real line
...
we write down in this course will be true for some
values of x but not for others
...
Below are some examples of sets of real numbers
...
The collection of all real numbers between two given real numbers form an interval
...
[a, b) is the set of all real numbers x which satisfy a ≤ x < b
...
[a, b] is the set of all real numbers x which satisfy a ≤ x ≤ b
...
E
...
(−∞, 2] is the interval of all real numbers
(both positive and negative) which are ≤ 2
...
4
...
A common way of describing a set is to say it is the collection of all real numbers
which satisfy a certain condition
...
(A,B,C,D,
...
e
...
This set consists of two parts: the interval (−∞, −1) and the interval (1, ∞)
...
Some sets can be very difficult to draw
...
In this course we will try to avoid such sets
...
Or the set
E = x | x3 − 4x2 + 1 = 0
which consists of the solutions of the equation x3 − 4x2 + 1 = 0
...
)
If A and B are two sets then the union of A and B is the set which contains all numbers that belong
either to A or to B
...
7
Similarly, the intersection of two sets A and B is the set of numbers which belong to both sets
...
2
...
What is the 2007th digit after the period in the expansion of 1 ?
7
4
...
Is it always true that
A ∩ B is an interval? How about A ∪ B?
2
...
3
25
15625
5
...
Are these sets the same?
6
...
3
...
Each of these
sets is the union of one or more intervals
...
Which of thee sets are finite?
Write the numbers
x = 0
...
,
y = 0
...
and z = 0
...
A = x | x2 − 3x + 2 ≤ 0
B = x | x2 − 3x + 2 ≥ 0
C = x | x2 − 3x > 3
D = x | x2 − 5 > 2x
E = t | t2 − 3t + 2 ≤ 0
F = α | α2 − 3α + 2 ≥ 0
G = (0, 1) ∪ (5, 7]
√
H = {1} ∪ {2, 3} ∩ (0, 2 2)
Q = θ | sin θ = 1
2
R = ϕ | cos ϕ > 0
as fractions (i
...
write them as
m
,
n
specifying m and n
...
A similar trick
works for y, but z is a little harder
...
Group Problem
...
e
...
99999999999999999
...
Functions
Wherein we meet the main characters of this semester
3
...
Definition
...
The set of numbers for which a function is defined is called its domain
...
The rule must be unambiguous: the same
xmust always lead to the same f (x)
...
Here the rule defining f is
“take the square root of whatever number you’re given”, and the function f will accept all nonnegative real
numbers
...
Most often it is a formula, as in
the square root example of the previous paragraph
...
Functions which are defined by different formulas on different intervals are sometimes called piecewise
defined functions
...
2
...
You get the graph of a function f by drawing all points whose coordinates are (x, y) where x must be in the domain of f and y = f (x)
...
The graph of a function f
...
m
P1
y1
1
y1 − y0
y0
P0
x1 − x0
n
x0
x1
Figure 4
...
The line is the graph of f (x) = mx + n
...
1
0
3
...
Linear functions
...
Its graph is a straight line
...
Conversely, any straight line which is not vertical (i
...
not
parallel to the y-axis) is the graph of a linear function
...
x1 − x0
This formula actually contains a theorem from Euclidean geometry, namely it says that the ratio (y1 − y0 ) :
(x1 − x0 ) is the same for every pair of points (x0 , y0 ) and (x1 , y1 ) that you could pick on the line
...
4
...
” In this course we will usually not be careful about
specifying the domain of the function
...
For instance, if we say that h is the
function
√
h(x) = x
9
y = x3 − x
Figure 5
...
”
The circle fails both tests
...
A systematic way of finding the domain and range of a function for which you are only given a formula is
as follows:
• The domain of f consists of all x for which f (x) is well-defined (“makes sense”)
• The range of f consists of all y for which you can solve the equation f (x) = y
...
5
...
The expression 1/x2 can be computed
for all real numbers x except x = 0 since this leads to division by zero
...
To find the range we ask “for which y can we solve the equation y = f (x) for x,” i
...
we for which y can you
solve y = 1/x2 for x?
If y = 1/x2 then we must have x2 = 1/y, so first of all, since we have to divide by y, y can’t be zero
...
On the other hand, if y > 0 then y = 1/x2 has a solution
be
(in fact two solutions), namely x = ±1/ y
...
3
...
Functions in “real life
...
If some
object is moving along a straight line, then you can define the following function: Let x(t) be the distance
from the object to a fixed marker on the line, at the time t
...
There are many examples of this kind
...
Here the domain is
the interval [0, T ], where T is the life time of the cell, and the rule that describes the function is
Given t, weigh the cell at time t
...
7
...
Generally speaking graphs of functions are curves in the plane but
they distinguish themselves from arbitrary curves by the way they intersect vertical lines: The graph of
a function cannot intersect a vertical line “x = constant” in more than one point
...
10
3
...
Examples
...
The collection of points determined by the equation x2 + y 2 = 1 is a circle
...
See Figure 6
...
Inverse functions and Implicit functions
For many functions the rule which tells you how to compute it is not an explicit formula, but instead an
equation which you still must solve
...
”
4
...
Example
...
In this example you can solve the equation for y,
3 − x2
...
y=
Here we have two definitions of the same function, namely
(i) “y = f (x) is defined by x2 + 2y − 3 = 0,” and
(ii) “f is defined by f (x) = (3 − x2 )/2
...
You see that with an “implicit function”
it isn’t the function itself, but rather the way it was defined that’s implicit
...
2
...
Define g by saying that for
any x the value y = g(x) is the solution of
x2 + xy − 3 = 0
...
x
Unlike the previous example this formula does not make sense when x = 0, and indeed, for x = 0 our rule for
g says that g(0) = y is the solution of
g(x) = y =
02 + 0 · y − 3 = 0, i
...
y is the solution of 3 = 0
...
√
y = + 1 − x2
x2 + y 2 = 1
√
y = − 1 − x2
Figure 6
...
11
4
...
Example: the equation alone does not determine the function
...
If x > 1 or x < −1 then x2 > 1 and there is no solution, so h(x) is at most defined when −1 ≤ x ≤ 1
...
The rule which defines a function must be unambiguous, and since we have not specified which of these two
solutions is h(x) the function is not defined for −1 < x < 1
...
Here are three possibilities:
h1 (x) = the nonnegative solution y of x2 + y 2 = 1
h2 (x) = the nonpositive solution y of x2 + y 2 = 1
h3 (x) =
h1 (x)
h2 (x)
when x < 0
when x ≥ 0
4
...
Why use implicit functions? In all the examples we have done so far we could replace the
implicit description of the function with an explicit formula
...
For instance, you can define a function f by
saying that y = f (x) if and only if
y 3 + 3y + 2x = 0
...
”
E
...
to compute f (0) you set x = 0 and solve y 3 + 3y = 0
...
To
compute f (1) you have to solve y 3 + 3y + 2 · 1 = 0, and if you’re lucky you see that y = −1 is the solution,
and f (1) = −1
...
Solving (1) is not easy
...
Here it is:
y = f (x) =
3
−x +
1 + x2 −
3
x+
1 + x2
...
4
...
Inverse functions
...
So to find y = f −1 (x) you solve the equation x = f (y)
...
1To see the solution and its history visit
http://www
...
org/~history/HistTopics/Quadratic_etc_equations
...
The graph of a function and its inverse are mirror images of each other
...
6
...
Consider the function f with f (x) = 2x + 3
...
2
So f −1 (x) is defined for all x, and it is given by f −1 (x) = (x − 3)/2
...
To see for which x the
inverse g −1 (x) is defined we try to solve the equation g(y) = x, i
...
we try to solve y 2 = x
...
But if x ≥ 0 then y = x does have a solution, namely y = x
...
4
...
Inverse trigonometric functions
...
x = f −1 (y)
y = f (x)
y = sin x
(−π/2 ≤ x ≤ π/2)
x = arcsin(y)
(−1 ≤ y ≤ 1)
y = cos x
(0 ≤ x ≤ π)
x = arccos(y)
(−1 ≤ y ≤ 1)
y = tan x
(−π/2 < x < π/2)
x = arctan(y)
The notations arcsin y = sin−1 y, arccos x = cos−1 x, and arctan u = tan−1 u are also commonly used for
the inverse trigonometric functions
...
Namely,
everybody writes the square of sin y as
2
sin y = sin2 y
...
Exercises
10
...
The functions f and g are defined by
2
2
y = f (x) ⇐⇒ x2 y − y = 6
...
Are f and g the same functions or are they different?
What is the domain of f ?
9
...
Let f be the function defined by y = f (x) ⇐⇒ y is
the largest solution of
y = f (x) ⇐⇒ x2 y + y = 7
...
What is the domain of f ?
13
Find a formula for f
...
Compute
12
...
(b) f (0)
(d) f (t)
y = f (x) ⇐⇒ 2x + 2xy + y 2 = 5 and y > −x
...
13
...
2) in three decimals where f is the implicitly defined function from §4
...
(There are (at least) two different ways of finding f (1
...
Does there exist a function f which satisfies
f (x2 ) = x + 1
14
...
(a) True or false:
for all x one has sin arcsin x = x?
(b) True or false:
for all x one has arcsin sin x = x?
for all real numbers x?
∗ ∗ ∗
The following exercises review precalculus material involving quadratic expressions ax2 + bx + c in one way or
another
...
On a graphing calculator plot the graphs of the following functions, and explain the results
...
)
f (x) = arcsin(sin x),
g(x) = arcsin(x) + arccos(x),
sin x
,
h(x) = arctan
cos x
cos x
k(x) = arctan
,
sin x
l(x) = arcsin(cos x),
m(x) = cos(arcsin x),
22
...
−2π ≤ x ≤ 2π
0≤x≤1
23
...
Find the inverse of the function f which is given by
f (x) = sin x and whose domain is π ≤ x ≤ 2π
...
m(x) = 1/(3 + 2x + x2 )
...
Group Problem
...
Find a number a such that the function f (x) =
sin(x + π/4) with domain a ≤ x ≤ a + π has an inverse
...
For each real number a we define a line
equation y = ax + a2
...
Draw the graph of the function h3 from §4
...
(a) Draw the lines corresponding to a
−2, −1, − 1 , 0, 1 , 1, 2
...
A function f is given which satisfies
for all real numbers x
...
where x and y are arbitrary real numbers
...
For which values of m and n does the graph of
f (x) = mx + n intersect the graph of g(x) = 1/x in
exactly one point and also contain the point (−1, 1)?
What are the range and domain of f ?
20
...
x+1
26
...
1
...
If you want to find the tangent to the
graph of f at some given point on the graph of f , how would you do that?
a secant
Q
tangent
P
Figure 1
...
If you are making a real paper and
ink drawing you would take a ruler, make sure it goes through P and then turn it until it doesn’t cross the
graph anywhere else
...
This line is called a
“secant,” and it is of course not the tangent that you’re looking for
...
15
So this is our recipe for constructing the tangent through P : pick another point Q on the graph, find the
line through P and Q, and see what happens to this line as you take Q closer and closer to P
...
We’ll write this in formulas in a moment, but first let’s worry about how close Q should be to P
...
If you choose Q different from P then you don’t get the tangent, but at best something that is “close”
to it
...
The concept of a limit is meant to solve this confusing problem
...
An example – tangent to a parabola
To make things more concrete, suppose that the function we had was f (x) = x2 , and that the point was
(1, 1)
...
Any line through the point P (1, 1) has equation
y − 1 = m(x − 1)
where m is the slope of the line
...
Let Q be the other point on the parabola, with coordinates (x, x2 )
...
Whatever x we choose, it must be
different from 1, for otherwise P and Q would be the same point
...
Now, as one changes x one thing stays the same, namely,
the secant still goes through P
...
By the “rise over run” formula, the slope of the secant line joining P and Q is
∆y
where ∆y = x2 − 1 and ∆x = x − 1
...
∆x
x−1
x−1
As x gets closer to 1, the slope mP Q , being x + 1, gets closer to the value 1 + 1 = 2
...
In symbols,
lim mP Q = 2,
Q→P
or, since Q approaching P is the same as x approaching 1,
(4)
lim mP Q = 2
...
e
...
A warning: you cannot substitute x = 1 in equation (3) to get (4) even though it looks like that’s what we
did
...
∆y
0
It is only after the algebra trick in (3) that setting x = 1 gives something that is well defined
...
We did a calculation which is valid for all x = 1, and later
looked at what happens if x gets “very close to 1
...
3
...
Here is
how it goes: When you are driving in your car the speedometer tells you how fast your are going, i
...
what
your velocity is
...
Namely, if it takes you two hours to cover 100 miles, then your
average velocity was
distance traveled
= 50 miles per hour
...
If the speedometer in your car tells you that you are driving 50mph, then that
should be your velocity at the moment that you look at your speedometer, i
...
“distance traveled over time
it took” at the moment you look at the speedometer
...
e
...
Your velocity at any moment is undefined
...
Let t be the time (in hours) that has
passed since we got onto the road, and let s(t) be the distance we have covered since then
...
We’ll write ∆t for the length of the time interval
...
A little later, at time t + ∆t we have traveled s(t + ∆t)
...
Our average velocity in that
time interval is therefore
s(t + ∆t) − s(t)
miles per hour
...
e
...
So we have the following formula (definition, really) for the velocity at time t
(5)
v(t) = lim
∆t→0
s(t + ∆t) − s(t)
...
Rates of change
The two previous examples have much in common
...
If you change
x to x + ∆x, then y will change from f (x) to f (x + ∆x)
...
∆x
∆x
17
This is the average rate of change of f over the interval from x to x + ∆x
...
If that happens then that “limiting number” is called the rate of change of f at x, or, the derivative of f at
x
...
∆x→0
∆x
Derivatives and what you can do with them are what the first half of this semester is about
...
In the
next chapter we’ll study limits so that we get a less vague understanding of formulas like (7)
...
Examples of rates of change
5
...
Acceleration as the rate at which velocity changes
...
Suppose v(t) is your velocity at time t (measured
in miles per hour)
...
You conclude that your
velocity increased by ∆v = v(t + ∆t) − v(t) during a time interval of length ∆t, and hence
average rate at which
your velocity changed
=
v(t + ∆t) − v(t)
∆v
=
...
Your
instantaneous acceleration at time t is the limit of your average acceleration as you make the time interval
shorter and shorter:
v(t + ∆t) − v(t)
...
e
...
Or, if you had measured distances in meters and time in seconds then velocities would be measured in meters
per second, and acceleration in meters per second per second, which is the same as meters per second2 , i
...
“meters per squared second
...
2
...
Think of a chemical reaction in which two substances A and B react to form
AB2 according to the reaction
A + 2B −→ AB2
...
Chemists write [A] for the amount of “A” in the chemical reactor (measured in
moles)
...
We’re mathematicians so we will write “[A](t)”
for the number of moles of A present at time t
...
e
...
∆t
This quantity is the rate of change of [A]
...
If
you open a paper on chemistry you will find that the derivative is written in Leibniz notation:
[A] (t) = lim
∆t→0
d[A]
dt
...
e
...
The law of mass-action
18
kinetics from chemistry states this more precisely
...
It’s a constant that you could try to measure by
timing how fast the reaction goes
...
Exercises
27
...
31
...
What is
the derivative of f (x) = x cos π at the points A and B
x
on the graph?
28
...
32
...
e
...
What units do the increments ∆y and ∆x, and the
derivative dy/dx have?
29
...
Should you trust your calculator?
Find the slope of the tangent to the parabola y = x2
at the point ( 1 , 1 ) (You have already done this: see
3 9
exercise 27)
...
A tank is filling with water
...
What
units does the derivative V (t) have?
Instead of doing the algebra you could try to compute
the slope by using a calculator
...
34
...
Compute
∆y
∆x
Let A(x) be the area of an equilateral triangle whose
sides measure x inches
...
1, 0
...
001, 10
(a) Show that
−12
, 10
...
Use at least 10 decimals
and organize your results in a table like this:
f (a)
...
...
...
...
...
∆y
...
...
...
(b) Which length does dA represent geometrically?
dx
[Hint: draw two equilateral triangles, one with side x and
another with side x + ∆x
...
]
∆y
∆x
∆x
0
...
01
0
...
...
...
35
...
Let A(x) be the area of a square with side x, and let
L(x) be the perimeter of the square (sum of the lengths
of all its sides)
...
2
Look carefully at the ratios ∆y/∆x
...
Give a geometric interpretation that explains why
∆A ≈ 1 L(x)∆x for small ∆x
...
Let A(r) be the area enclosed by a circle of radius
r, and let L(r) be the length of the circle
...
(Use the familiar formulas from geometry
for the area and perimeter of a circle
...
Simplify the algebraic expressions you get when you
compute ∆y and ∆y/∆x for the following functions
(a) y = x2 − 2x + 1
1
(b) y =
x
(c) y = 2x
37
...
Show that
V (r) = S(r)
...
)
19
CHAPTER 3
Limits and Continuous Functions
1
...
(See §2
...
1
...
Definition of limit (1st attempt)
...
” It means that if you choose values of x which are close but
not equal to a, then f (x) will be close to the value L; moreover, f (x) gets closer and closer to L as x gets
closer and closer to a
...
)
1
...
Example
...
1
...
Example: substituting numbers to guess a limit
...
We first try to substitute x = 2, but this leads to
22 − 2 · 2
0
=
22 − 4
0
which does not exist
...
Table 1 suggests that
f (x) approaches 0
...
f (2) =
x
3
...
500000
2
...
010000
2
...
600000
0
...
512195
0
...
500125
x
1
...
500000
0
...
010000
0
...
009990
1
...
009899
1
...
000000
Table 1
...
” (Values of f (x) and g(x) rounded to
six decimals
...
4
...
The previous example
shows that our first definition of “limit” is not very precise, because it says “x close to a,” but how close is
close enough? Suppose we had taken the function
g(x) =
101 000x
100 000x + 1
and we had asked for the limit limx→0 g(x)
...
000
...
2
...
” In
the end we don’t really know what they mean, although they are suggestive
...
e
...
Here is the definition
...
Go on to the next sections
...
2
...
Definition of limx→a f (x) = L
...
(2) for every ε > 0 one can find a δ > 0 such that for all x in the domain of f one has
(8)
|x − a| < δ implies |f (x) − L| < ε
...
The inequality |x − y| < δ says
that “the distance between x and y is less than δ,” or that “x and y are closer than δ
...
To prove that limx→a f (x) = L you must assume that
someone has given you an unknown ε > 0, and then find a postive δ for which (8) holds
...
2
...
Show that limx→5 2x + 1 = 11
...
So, if 2|x − a| < ε then we have |f (x) − L| < ε, i
...
1
if |x − a| < 2 ε then |f (x) − L| < ε
...
No matter what ε > 0 we are given our δ will also be positive, and if
2
|x − 5| < δ then we can guarantee |(2x + 1) − 11| < ε
...
22
y = f (x)
L+ε
L
How close must x be to a for f (x) to end up in this range?
L−ε
a
y = f (x)
L+ε
L
L−ε
For some x in this interval f (x) is not between L − ε and
L + ε
...
You need a smaller δ
...
Therefore the δ in this picture is small
enough for the given ε
...
3
...
We have f (x) = x2 , a = 1,
L = 1, and again the question is, “how small should |x − 1| be to guarantee |x2 − 1| < ε?”
We begin by estimating the difference |x2 − 1|
|x2 − 1| = |(x − 1)(x + 1)| = |x + 1| · |x − 1|
...
Here’s a more concrete situation in which ε and δ appear in exactly the same
roles:
Now you can ask the following question:
Suppose you want to know the area
with an error of at most ε,
then what is the largest error
that you can afford to make
when you measure the radius?
The answer will be something like this: if you want
the computed area to have an error of at most
|f (x) − A| < ε, then the error in your radius measurement should satisfy |x − R| < δ
...
You would expect that if your measured radius
x is close enough to the real value R, then your computed area f (x) = πx2 will be close to the real area
A
...
e how small you make ε) you can
always achieve that precision by making the error
in your radius measurement small enough (i
...
by
making δ sufficiently small)
...
You decide to
measure its radius, R, and then compute the area of
the circle by calculating
Area = πR2
...
When you measure the radius R you will make
an error, simply because you can never measure anything with infinite precision
...
Then the size of the error you made is
error in radius measurement = |x − R|
...
The error in your computed value
of the area is
error in area = |f (x) − f (R)| = |f (x) − A|
...
g
...
Here is a trick that allows you to replace the factor |x + 1| with a constant
...
If we do that, then we will always have
|x − 1| < δ ≤ 1, i
...
|x − 1| < 1,
and x will always be beween 0 and 2
...
If we now want to be sure that |x2 − 1| < ε, then this calculation shows that we should require 3|x − 1| < ε,
i
...
|x − 1| < 1 ε
...
We must also live up to our promise never to choose δ > 1, so
3
3
if we are handed an ε for which 1 ε > 1, then we choose δ = 1 instead of δ = 1 ε
...
3
We have shown that if you choose δ this way, then |x − 1| < δ implies |x2 − 1| < ε, no matter what ε > 0 is
...
We could
therefore say that in this problem we will choose δ to be
δ = min 1, 1 ε
...
4
...
Solution: We apply the definition with a = 4, L = 1/4 and
f (x) = 1/x
...
We begin by estimating |f (x) − 1 | in terms of |x − 4|:
4
|f (x) − 1/4| =
4−x
|x − 4|
1 1
1
=
=
−
=
|x − 4|
...
To achieve that we again agree not to take δ > 1
...
How large can 1/|4x| be
in this situation? Answer: the quantity 1/|4x| increases as you decrease x, so if 3 < x < 5 then it will never
1
be larger than 1/|4 · 3| = 12
...
< ε we could threfore require
1
12 |x
− 4| < ε,
i
...
|x − 4| < 12ε
...
Of course we have
to honor our agreement never to choose δ > 1, so our choice of δ is
δ = the smaller of 1 and 12ε = min 1, 12ε
...
Exercises
43
...
x→2
√
44
...
x→4
√
45
...
38
...
Joe offers to make square sheets of paper for Bruce
...
Bruce asks Joe for a square with
area 4 square foot
...
Bruce doesn’t mind
as long as the area of the square doesn’t differ more than
0
...
x→3
1+x
= 1
...
47
...
lim
= 1
...
lim
x→2
(a) What is the biggest error Joe can afford to make
when he marks off the length x?
49
...
However, she needs the error in the area to be less
than 0
...
(She’s paying)
...
Group Problem
...
) Joe is offering to build cubes of
side x
...
For instance, a cube with
2 foot sides has volume+area equal to 23 + 6 × 22 = 32
...
lim 2x − 4 = 6
If you ask Joe to build a cube whose volume plus
total surface area is 32 cubic feet with an error of at
most ε, then what error can he afford to make when he
measures the side of the cube he’s making?
x→1
2
40
...
x→2
41
...
Our definition of a derivative in (7) contains a limit
...
lim x3 = 27
x→3
4
...
” here we describe some possible variations on the concept of limit
...
1
...
When we let “x approach a” we allow x to be both larger or smaller than
a, as long as x gets close to a
...
All four notations are in use
...
The precise definition of right limits goes like this:
4
...
Definition of right-limits
...
Then
(9)
lim f (x) = L
...
The left-limit, i
...
the one-sided limit in which x approaches a through values less than a is defined in a
similar way
...
4
...
Theorem
...
x→a
In other words, if a function has both left- and right-limits at some x = a, then that function has a limit
at x = a if the left- and right-limits are equal
...
4
...
Instead of letting x approach some finite number, one can let x become “larger
and larger” and ask what happens to f (x)
...
x
∞
(“The limit for x going to infinity is L
...
5
...
The larger you choose x, the smaller its reciprocal 1/x becomes
...
x→∞ x
Here is the precise definition:
4
...
Definition of limit at ∞
...
If there is a number L such that for every ε > 0 one can find an A such that
x > A =⇒ |f (x) − L| < ε
for all x, then we say that the limit of f (x) for x → ∞ is L
...
Instead of δ which specifies how close
x should be to a, we now have a number A which says how large x should be, which is a way of saying “how
close x should be to infinity
...
7
...
To prove that limx→∞ 1/x = 0 we apply the definition to
f (x) = 1/x, L = 0
...
How do we choose A? A is not allowed to depend on x, but it may depend on ε
...
ε
This tells us how to choose A
...
Hence we have proved that limx→∞ 1/x = 0
...
Properties of the Limit
The precise definition of the limit is not easy to use, and fortunately we won’t use it very often in this
class
...
”
The following properties also apply to the variations on the limit from 4
...
e
...
Limits of constants and of x
...
x→a
Limits of sums, products and quotients
...
x→a
x→a
Then
(P3 )
(P4 )
(P5 )
lim F1 (x) + F2 (x) = L1 + L2 ,
x→a
lim F1 (x) − F2 (x) = L1 − L2 ,
x→a
lim F1 (x) · F2 (x) = L1 · L2
x→a
Finally, if limx→a F2 (x) = 0,
(P6 )
lim
x→a
F1 (x)
L1
=
...
One can prove these laws using the
definition of limit in §2 but we will not do this here
...
27
There are two more properties of limits which we will add to this list later on
...
6
...
1
...
One has
lim x2 = lim x · x
x→2
x→2
= lim x · lim x
x→2
by (P5 )
x→2
= 2 · 2 = 4
...
To apply (P6 ) we must check that the denominator (“L2 ”) is not zero
...
6
...
Try the examples 1
...
4 using the limit properties
...
x→2
x→2
to complete the computation we would like to apply the last property (P6 ) about quotients, but this would
give us
0
lim f (x) =
...
We have to do something else
...
For such functions there is an algebra trick which always allows you to compute the limit even
if you first get 0
...
In our case we have
0
x2 − 2x = (x − 2) · x,
x2 − 4 = (x − 2) · (x + 2)
so that
(x − 2) · x
x
= lim
...
) to compute
lim f (x) = lim
lim f (x) =
x→2
2
1
=
...
3
...
Of course, you would think that limx→2 x = 2 and you can
indeed prove this using δ & ε (See problem 44
...
However, if you assume that the limit exists then the limit properties allow us to find this limit
...
x→2
Then property (P5 ) implies that
√ √
x = lim x · x = lim x = 2
...
We can reject the latter because whatever x
does, its squareroot is always a positive number, and hence it can never “get close to” a negative number like
√
− 2
...
x gets close to
√
2
...
Find
√
√
x− 2
lim
x→2
x−2
assuming the result from the previous example
...
4
...
e
...
We use the same algebra trick as before, namely we factor numerator and denominator:
0
√
√
√
√
x− 2
x− 2
1
√ √
√ =√
√
...
x→2
x→2
x−2
4
x+ 2
2 2
6
...
Limit as x → ∞ of rational functions
...
bm xm + · · · + b1 x + b0
We have seen that
1
=0
x
We even proved this in example 4
...
Using this you can find the limit at ∞ for any rational function R(x) as
in (11)
...
”
lim
x→∞
To find limx→∞ R(x) divide numerator and denominator by xm (the highest power of x occurring in the
denominator)
...
+ 7x − 39
x→∞ 5x2
29
Remember the trick and divide top and bottom by x2 , and you get
3 + 3/x2
3x2 + 3
= lim
x→∞ 5 + 7/x − 39/x2
x→∞ 5x2 + 7x − 39
limx→∞ 3 + 3/x2
=
limx→∞ 5 + 7/x − 39/x2
3
=
5
Here we have used the limit properties (P∗ ) to break the limit down into little pieces like limx→∞ 39/x2
which we can compute as follows
lim
lim 39/x2 = lim 39 ·
x→∞
x→∞
1
x
2
=
lim 39 ·
x→∞
1
x→∞ x
lim
2
= 39 · 02 = 0
...
6
...
Compute
x
lim 3
...
This leads to
lim
x→∞ x3
x
1/x2
limx→∞ 1/x2
0
= lim
=
= = 0
...
6
...
When limits fail to exist
In the last couple of examples we worried about the possibility that a limit limx→a g(x) actually might
not exist
...
First let’s agree on what we will call a “failed limit
...
1
...
If there is no number L such that limx→a f (x) = L, then we say that the limit
limx→a f (x) does not exist
...
2
...
The “sign function1” is defined by
−1 for x < 0
sign(x) = 0
for x = 0
1
for x > 0
Note that “the sign of zero” is defined to be zero
...
e
...
But
for arbitrarily small negative values of x one has sign(x) = −1, so one would conclude that L = −1
...
e
...
lim sign(x) does not exist
...
If you think about this formula for a moment you’ll see that sign(x) = x/|x| for all x = 0
...
30
y = sign(x)
1
−1
Figure 1
...
In this example the one-sided limits do exist, namely,
lim sign(x) = 1 and lim sign(x) = −1
...
7
...
The example of the backward sine
...
e
...
x
When x = 0 the function f (x) = sin(π/x) is not defined, because its definition involves division by x
...
Then,
taking the sine, we see that sin(π/x) oscillates between +1 and −1 infinitely often as x → 0
...
B
1
2
D
1
2
3
y = sin π
x
E
C
Figure 2
...
Here again, the limit limx→0 f (x) does not exist
...
So the limit fails to exist in a stronger way than in the example
of the sign-function
...
In the present
example we see that even the one-sided limit
lim sin
x
0
does not exist
...
4
...
The expression 1/0 is not defined, but what about
lim
x→0
1
?
x
This limit also does not exist
...
“Common wisdom” is not always a reliable tool in mathematical proofs, so here is a better argument
...
Namely, suppose that
there were an number L such that
1
lim = L
...
· x = lim
x→0 x
x→0
x→0 x
1
On the other hand x · x = 1 so the above limit should be 1! A number can’t be both 0 and 1 at the same
time, so we have a contradiction
...
lim
7
...
Using limit properties to show a limit does not exist
...
Although it is perhaps not so obvious at first
sight, they also allow you to prove that certain limits do not exist
...
Here is another
...
You can turn this around and say that if limx→a g(x) + h(x) does not exist then either limx→a g(x) or
limx→a h(x) does not exist (or both limits fail to exist)
...
7
...
Limits at ∞ which don’t exist
...
x→∞
One can prove this from the limit definition (and see exercise 72)
...
x→∞
x+2
Once again we divide numerator and denominator by the highest power in the denominator (i
...
x)
L = lim
1
x+2− x
x→∞ 1 + 2/x
L = lim
Here the denominator has a limit (’tis 1), but the numerator does not, for if limx→∞ x + 2 −
since limx→∞ (2 − 1/x) = 2 exists,
lim x = lim
x→∞
x→∞
x+2−
would also have to exist, and limx→∞ x doesn’t exist
...
In this situation the limit L itself can never exist
...
8
...
”)
The difference between these two kinds of variables is this:
• if you replace a dummy variable in some formula consistently by some other variable then the value
of the formula does not change
...
• the value of the formula may depend on the value of the free variable
...
The limit is easy to compute:
lim 2x + 1 = 2a + 1
...
This computation says that if some number gets close to a then two times
that number plus one gets close to 2a + 1
...
But the result of our
computation shouldn’t depend on the name we choose, i
...
it doesn’t matter if we call it x or u
...
Some prefer to call x a
bound variable, meaning that in
lim 2x + 1
x→a
the x in the expression 2x + 1 is bound to the x written underneath the limit – you can’t change one without
changing the other
...
For instance, let’s try
setting x = 3 in our limit, i
...
what is
lim 2 · 3 + 1 ?
3→a
Of course 2 · 3 + 1 = 7, but what does 7 do when 3 gets closer and closer to the number a? That’s a silly
question, because 3 is a constant and it doesn’t “get closer” to some other number like a! If you ever see 3
get closer to another number then it’s time to take a vacation
...
For instance, if we set a = 3 (but leave x alone) then we get
lim 2x + 1
x→3
and there’s nothing strange about that (the limit is 2 · 3 + 1 = 7, no problem
...
In general you get 2a + 1
...
Limits and Inequalities
This section has two theorems which let you compare limits of different functions
...
(P6 ) from §5
...
e
...
The first theorem should not surprise you – all it says is that bigger functions have bigger limits
...
1
...
Let f and g be functions whose limits for x → a exist, and assume that f (x) ≤ g(x)
holds for all x
...
x→a
x→a
A useful special case arises when you set f (x) = 0
...
The statement may seem obvious, but it still needs a proof, starting from the ε-δ definition of limit
...
Here is the second theorem about limits and inequalities
...
2
...
Suppose that
f (x) ≤ g(x) ≤ h(x)
(for all x) and that
lim f (x) = lim h(x)
...
x→a
x→a
x→a
The theorem is useful when you want to know the limit of g, and when you can sandwich it between
two functions f and h whose limits are easier to compute
...
The inequalities f ≤ g ≤ h combined with the circumstance that f and h have the same
limit are enough to guarantee that the limit of g exists
...
Graphs of |x|, −|x| and x cos
34
π
x
for −1
...
2
9
...
Example: a Backward Cosine Sandwich
...
For example
1
−|x| ≤ x cos ≤ |x|
x
since the cosine is always between −1 and 1
...
x
Note that the limit limx→0 cos(1/x) does not exist, for the same reason that the “backward sine” did not
have a limit for x → 0 (see example 7
...
Multiplying with x changed that
...
Continuity
10
...
Definition
...
Note that when we say that a function is continuous on some interval it is understood that the domain
of the function includes that interval
...
10
...
Polynomials are continuous
...
To show that you have to prove that
lim P (x) = P (2),
x→2
i
...
lim x2 + 3x = 22 + 3 · 2
...
e
...
(P6 ) from §5 (just as good, and easier, even though it still takes a few lines to write it out – do both!)
P (x)
10
...
Rational functions are continuous
...
e
...
Then one has
P (x)
Q(x)
limx→a P (x)
=
limx→a Q(x)
P (a)
=
Q(a)
= R(a)
...
10
...
Some discontinuous functions
...
For instance, the sign function g(x) = sign(x) from example ?? is not continuous at x = 0
...
35
10
...
How to make functions discontinuous
...
In other words, we take a continuous function like g(x) = x2 , and change its value somewhere, e
...
at x = 3
...
x→3
The reason that the limit is 9 is that our new function f (x) coincides with our old continuous function g(x)
for all x except x = 3
...
10
...
Sandwich in a bow tie
...
Consider
f (x) =
x cos
0
1
x
for x = 0,
for x = 0
Then f is continuous at x = 0 by the Sandwich Theorem (see Example ??)
...
Since this limit is zero, f (0) = 0
is the only possible choice of f (0) which makes f continuous at x = 0
...
Substitution in Limits
Given two functions f and g one can consider their composition h(x) = f (g(x))
...
x→a
Suppose that you can find the limits
L = lim g(x) and lim f (u) = M
...
This is in fact a theorem:
11
...
Theorem
...
x→a
u→L
Another way to write this is
lim f g(x) = f lim g(x)
...
2
...
The given function is the composition of two functions, namely
√
x3 − 3x2 + 2 = u, with u = x3 − 3x2 + 2,
or, in function notation, we want to find limx→3 h(x) where
√
h(x) = f (g(x)), with g(x) = x3 − 3x2 + 2 and g(x) = x
...
You get the first limit from the limit properties (P1 )
...
The second limit says that taking the square
root is a continuous function, which it is
...
3
...
Normally, you write this whole argument as follows:
√
x3 − 3x2 + 2 = lim x3 − 3x2 + 2 = 2,
x→3
√
where you must point out that f (x) = x is a continuous function to justify the first step
...
12
...
52
...
54
...
56
...
65
...
lim
x→−7
t→0
lim (2x + 5)
lim (2x + 5)
x→−∞
x→−4
lim (x + 3)2007
69
...
lim (x + 3)2007
70
...
x→−4
x→−∞
59
...
True or
false?
x2 + 3
x2 + 4
72
...
1
In the text we proved that limx→∞ x = 0
...
Hint:
Suppose limx→∞ x = L for some number L
...
x
√
x−3
73
...
Hint: Multiply top and bottom
x→9 x − 9
√
by x + 3
...
lim
x +3
x2 + 4
lim
x2 + 1
x5 + 2
lim
(2x + 1)4
(3x2 + 1)2
x→∞
x→∞
x→∞
f (x)
71
...
t2 + t − 2
60
...
True or
false?
t2 + t − 2
t2 − 1
t→1
63
...
If limx→a f (x) exists then f is continuous at x = a
...
(2u + 1)4
(3u2 + 1)2
67
...
,
E in Figure 2 (the graph of y = sin π/x)
...
lim
61
...
C
The area of triangle OAC is
sin θ
tan θ
θ
1
2
sin θ
...
2
tan θ
...
O
A
74
...
x−2
1
√
x
−
77
...
x−2
76
...
75
...
is continuous
...
78
...
where a and b are constants
...
What are a and b?
is continuous for all x
...
Two Limits in Trigonometry
In this section we’ll derive a few limits involving the trigonometric functions
...
and
We will use these limits when we compute the derivatives of Sine, Cosine and Tangent
...
θ→0 θ
13
...
Theorem
...
The proof requires a few sandwiches and some geometry
...
Since the wedge OAC contains the triangle OAC its area must be larger
...
2
The Sandwich Theorem implies that
(14)
lim sin θ = 0
...
Next we compare the areas of the wedge OAC and the larger triangle OAB
...
Since tan θ = cos θ we can multiply with cos θ and divide by θ to get
2
θ
sin θ
π
for 0 < θ <
θ
2
If we go back to (15) and divide by θ, then we get
sin θ
cos θ <
<1
θ
The Sandwich Theorem can be used once again, and now it gives
sin θ
lim
= 1
...
To get the limit in which θ
0, you use that sin θ is an odd function
...
2
...
We will show that
(16)
lim
θ→0
1 − cos θ
1
=
...
Namely,
1 − cos θ
1
1 − cos2 θ
=
2
θ
1 + cos θ
θ2
2
sin θ
1
=
1 + cos θ θ2
2
1
sin θ
=
...
14
...
lim
sin(x2 )
...
lim
x(1 − cos x)
...
lim
Find each of the following limits or show that it does not
exist
...
sin(x2 )
...
lim
(two ways: with and without the double
α→0 sin α
angle formula!)
x→0
x→0
80
...
lim
θ→0
sin 3x
...
lim
x→π/2
tan θ
...
x− π
2
...
sin 2α
82
...
lim
x→0
84
...
lim
lim
cos x
...
lim
sin x
...
lim
sin x
...
x sin x
θ→π/2
x→π
1 − sin θ
θ − π/2
x→0
85
...
lim
x→0
π
) tan x
...
(x + 2)3
sin x
...
96
...
A = lim
x→∞
sin2 x
...
Is there a constant k such that the function
f (x) =
sin(1/x)
k
for x = 0
for x = 0
...
98
...
Compute limx→∞ x sin
substitute something)
...
(Hint:
x
100
...
(Geometry & Trig review ) Let An be the area of the
regular n-gon inscribed in the unit circle, and let Bn be
the area of the regular n-gon whose inscribed circle has
radius 1
...
On a historical note: Archimedes managed to compute A96 and B96 and by doing this got the most accurate
approximation for π that was known in his time
...
mcs
...
ac
...
html
(c) Compute limn→∞ An and limn→∞ Bn
...
gap-system
...
html
In chapter 2 we saw two mathematical problems which led to expressions of the form 0
...
After computing a few
derivatives using the definition we will spend most of this section developing the differential calculus, which is
a collection of rules that allow you to compute derivatives without always having to use basic definition
...
Derivatives Defined
1
...
Definition
...
The derivative of the function f at a is the value of the limit
f (x) − f (a)
...
(17)
f (a) = lim
f is called differentiable on the interval (c, d) if it is differentiable at every point a in (c, d)
...
2
...
One can substitute x = a + h in the limit (17) and let h → 0 instead of x → a
...
f (x) = lim
∆x→0
∆x
The interpretation is the same as in equation (6) from §4
...
If
you write y = f (x) then we can call the increase in f
(18)
f (a) = lim
∆y = f (x + ∆x) − f (x),
so that the derivative f (x) is
∆y
...
” The result of increasing x by this infinitely small quantity dx is that y = f (x)
increased by another infinitely small quantity dy
...
f (x) = lim
∆x→0
41
There are no “infinitely small real numbers,” and this makes Leibniz’ notation difficult to justify
...
This theory is called “non
standard analysis
...
Nonetheless, even though we won’t use infinitely small
numbers, Leibniz’ notation is very useful and we will use it
...
Direct computation of derivatives
2
...
Example – The derivative of f (x) = x2 is f (x) = 2x
...
The result was
f (x + h) − f (x)
(x + h)2 − x2
= lim
= lim (2x + h) = 2x
...
dx
2
...
The derivative of g(x) = x is g (x) = 1
...
h→0
h→0 h
h→0
h
h
g (x) = lim
In Leibniz’ notation:
dx
= 1
...
Nonetheless, this is not what is going on
...
2
...
The derivative of any constant function is zero
...
Then
we have
k(x + h) − k(x)
c−c
k (x) = lim
= lim
= lim 0 = 0
...
dx
2
...
Derivative of xn for n = 1, 2, 3,
...
x→a
x→a x − a
x−a
f (a) = lim
We need to simplify the fraction (xn − an )/(x − a)
...
x−a
For n = 1, 2, 3,
...
x−a
1But if you want to read more on this you should see Keisler’s calculus text at
http://www
...
wisc
...
html
I would not recommend using Keisler’s text and this text at the same time, but if you like math you should remember that it
exists, and look at it (later, say, after you pass 221
...
For instance, when n = 3 you would get
x × (x2 + xa + a2 ) = x3
−a × (x2 + xa + a2 ) =
(x − a) × (x2 + ax + a2 )
+ax2
−ax2
+a2 x
−a2 x
= x3
−a3
−a3
With formula (19) in hand we can now easily find the derivative of xn :
xn − an
x→a x − a
= lim xn−1 + xn−2 a + xn−3 a2 + · · · + xan−2 + an−1
f (a) = lim
x→a
= an−1 + an−2 a + an−3 a2 + · · · + a an−2 + an−1
...
One could also write this as f (x) = nxn−1 , or, in Leibniz’ notation
dxn
= nxn−1
...
3
...
1
...
If a function f is differentiable at some a in its domain, then f is also continuous at a
...
We are given that
f (x) − f (a)
x→a
x−a
lim
exists, and we must show that
lim f (x) = f (a)
...
4
...
1
...
Consider the function
f (x) = |x| =
x
−x
for x ≥ 0,
for x < 0
...
43
(more algebra)
(Limit Properties)
(f (a) exists)
To see this try to compute the derivative at 0,
|x|
|x| − |0|
f (0) = lim
= lim
= lim sign(x)
...
2)
If you look at the graph of f (x) = |x| then you see what is wrong: the graph has a corner at the origin
and it is not clear which line, if any, deserves to be called the tangent to the graph at the origin
...
The graph of y = |x| has no tangent at the origin
...
2
...
Another example of a function without a derivative at x = 0 is
f (x) =
|x|
...
Likewise, the limit from the left also does not exist (’tis “−∞)
...
That
observation does not give us a derivative, because the y-axis is vertical and hence has no slope
...
3
...
The previous two examples were about
functions which did not have a derivative at x = 0
...
It is easy to give examples of functions which are not differentiable
at more than one value of x, but here I would like to show you a function f which doesn’t have a derivative
anywhere in its domain
...
In this
graph you see a typical path of a Brownian motion, i
...
t is time, and x(t) is the position of a particle which
undergoes a Brownian motion – come to lecture for further explanation (see also the article on wikipedia)
...
yahoo
...
44
|x|
The tangent is vertical
y=
Figure 2
...
Exercises
101
...
For which value(s) is the function defined by
f (x) = x2 − 2x
1
g(x) =
x
k(x) = x3 − 17x
2
u(x) =
1+x
√
v(x) = x
1
w(x) = √
x
f (x) =
for x < 0
for x ≥ 0
differentiable at x = 0? Sketch the graph of the function
f for the values a and b you found
...
For which value(s) is the function defined by
f (x) =
ax2 + b
x − x2
for x < 1
for x ≥ 1
differentiable at x = 0? Sketch the graph of the function
f for the values a and b you found
...
105
...
Which of the following functions is differentiable at
x = 0?
ax2
x+b
for x < 2
for x ≥ 2
differentiable at x = 0? Sketch the graph of the function
f for the values a and b you found
...
x
f (x) = x|x|,
ax + b
x − x2
g(x) = x
106
...
True or false: If a function f is continuous at some
x = a then it must also be differentiable at x = a?
107
...
These formulas do not define k and at x = 0
...
True or false: If a function f is differentiable at
some x = a then it must also be continuous at x = a?
x(t)
t
Figure 3
...
Note how the graph doesn’t seem to have a a tangent anywhere at all
...
The Differentiation Rules
You could go on and compute more derivatives from the definition
...
This is fortunately not
necessary
...
In this section we’ll
look at rules which tell you how to differentiate a function which is either the sum, difference, product or
quotient of two other functions
...
The differentiation rules
The situation is analogous to that of the “limit-properties” (P1 )
...
6
...
Sum, product and quotient rules
...
The Differentiation Rules in function notation, and Leibniz notation, are
listed in figure 1
...
2
...
6
...
Proof of the Sum Rule
...
Then
f (x) − f (a)
f (a) = lim
(definition of f )
x→a
x−a
u(x) + v(x) − u(a) + v(a)
= lim
(use f = u + v)
x→a
x−a
u(x) − u(a) v(x) − v(a)
= lim
+
(algebra)
x→a
x−a
x−a
v(x) − v(a)
u(x) − u(a)
= lim
+ lim
(limit property)
x→a
x→a
x−a
x−a
= u (a) + v (a)
(definition of u , v )
6
...
Proof of the Product Rule
...
To find the derivative we must express the
change of f in terms of the changes of u and v
f (x) − f (a) = u(x)v(x) − u(a)v(a)
= u(x)v(x) − u(x)v(a) + u(x)v(a) − u(a)v(a)
= u(x) v(x) − v(a) + u(x) − u(a) v(a)
46
Now divide by x − a and let x → a:
lim
x→a
f (x) − f (a)
v(x) − v(a) u(x) − u(a)
= lim u(x)
+
v(a)
x→a
x−a
x−a
x−a
(use the limit properties)
u(x) − u(a)
v(x) − v(a)
+ lim
v(a)
x→a
x−a
x−a
= u(a)v (a) + u (a)v(a),
=
lim u(x)
lim
x→a
x→a
as claimed
...
6
...
Proof of the Quotient Rule
...
First we do the special
case where f (x) = 1/v(x), and then we use the product rule to differentiate
f (x) =
1
u(x)
= u(x) ·
...
We can express the change in f in terms of the change in v
f (x) − f (a) =
1
v(x) − v(a)
1
−
=
...
x−a
v(x)v(a)
x−a
Now we want to take the limit x → a
...
x→a
Therefore we find
1
v(x) − v(a)
v (a)
f (x) − f (a)
= lim
lim
=
...
In the second step we use the product rule to differentiate f = u/v
lim
f =
u
v
=
u·
1
v
=u ·
1
+u·
v
1
v
=
u
v
u v − uv
−u 2 =
...
5
...
The Quotient Rule
can be derived from the Product Rule as follows: if w = u/v then
w·v =u
(20)
By the product rule we have
w ·v+w·v =u,
so that
u −w·v
u − (u/v) · v
u ·v−u·v
=
=
...
4 above, this argument does not prove that w is differentiable if u and v are
...
w =
The trick which is used here, is a special case of a method called “implicit differentiation
...
47
6
...
Differentiating a constant multiple of a function
...
6
...
Picture of the Product Rule
...
If u and v are differentiable functions of x, then the changes ∆u and ∆v will be of the same order of magnitude
as ∆x, and thus one expects ∆u∆v to be much smaller
...
Leibniz would now divide by ∆x and replace ∆’s by d’s to get the product rule:
∆(uv)
∆v
∆u
=u
+v
...
The Product Rule
...
So the increase in area is approximately u∆v + v∆u, which explains why the product
rule says (uv) = uv + vu
...
Differentiating powers of functions
7
...
Product rule with more than one factor
...
e
...
)
After the first step you would get
f = u1 u2 · · · un + u1 u2 · · · un
...
This yields
f = u1 u2 · · · un + u1 u2 u3 · · · un + u2 (u3 · · · un )
= u1 u2 · · · un + u1 u2 u3 · · · un + u1 u2 u3 · · · un
...
7
...
The Power rule
...
dx
dx
(23)
7
...
The Power Rule for Negative Integer Exponents
...
The rule actually holds for all real exponents n, but the proof is harder
...
Suppose n = −m where
m is a positive integer
...
R
...
(um )2
Since m is a positive integer, we can use (23), so (um ) = mum−1 , and hence
(un ) = −
mum−1 · u
= −mu−m−1 · u = nun−1 u
...
4
...
So far we have proved that the power law holds if
the exponent n is an integer
...
The following derivation contains the trick called implicit differentiation which we will study in
more detail in Section 15
...
Assuming that both u and w are differentiable functions, we will show that
p
p
(24)
w (x) = u(x) q −1 u (x)
q
Raising both sides to the qth power gives
w(x)q = u(x)p
...
We get
qwq−1 · w = pup−1 · u
...
This proof is flawed because we did not show that w(x) = u(x)p/q is differentiable: we only showed what
the derivative should be, if it exists
...
5
...
If you choose the function u(x) in the Power Rule to be u(x) = x,
then u (x) = 1, and hence the derivative of f (x) = u(x)n = xn is
f (x) = nu(x)n−1 u (x) = nxn−1 · 1 = nxn−1
...
7
...
Example – differentiate a polynomial
...
For example, using the Sum Rule, the Power Rule
with u(x) = x, the rule (cu) = cu , the derivative of the polynomial
f (x) = 2x4 − x3 + 7
is
f (x) = 8x3 − 3x2
...
7
...
By the Quotient Rule the derivative of the
function
2x4 − x3 + 7
g(x) =
1 + x2
is
(8x3 − 3x2 )(1 + x2 ) − (2x4 − x3 + 7)2x
g (x) =
(1 + x2 )2
5
4
3
6x − x + 8x − 3x2 − 14x
...
√
7
...
Derivative of the square root
...
8
...
f (x) =
x
√
1+ x
118
...
Prove the statement in §6
...
e
...
1
(1 + x2 )2
117
...
Let f (x) = (1 + x2 )4
...
Are the answers the same?
ax + b
cx + d
116
...
Let f (x) = (x2 + 1)(x3 + 3)
...
Are your answers the same?
1−x
1+x
119
...
(try to simplify your answers)
3
x+
√
x
120
...
g(s) =
1−s
1+s
111
...
f (x) =
1+x
112
...
f (x) =
−1
122
...
Group Problem
...
(a) Find the derivative of f (x) = x4/3
...
Your answer should
have the form p/q where p and q are integers
...
]
√
(c) Approximate in the same way the numbers 143 and
√
145 (Hint: 12 × 12 = 144)
...
4 +7
2x
1 + x2
Note that f (x) = 1/g(x)
...
)
Instead of looking at the derivative of a function you can
look at the ratio of its derivative to the function itself, i
...
you can compute f /f
...
126
...
(a) Let x(t) = (1 − t2 )/(1 + t2 ),
y(t) = 2t/(1 + t2 ) and u(t) = y(t)/x(t)
...
(b) Now that you’ve done (a) there are two different ways
of finding du/dt
...
(a) Compute the logarithmic derivative of these functions (i
...
find f (x)/f (x))
k(x) = −x2 ,
u
u
125
...
Group Problem
...
Higher Derivatives
9
...
The derivative is a function
...
This function is called the derivative function of f , and it is denoted by f
...
The result, if it exists, is called the second derivative of f
...
The derivative of the second derivative is called the third derivative, written f , and so on
...
Thus
f (0) = f,
f (1) = f ,
f (2) = f ,
Leibniz’ notation for the nth derivative of y = f (x) is
dn y
= f (n) (x)
...
2
...
If f (x) = x2 − 2x + 3 then
f (x) = x2 − 2x + 3
f (x) = 2x − 2
f (x) = 2
f (3) (x) = 0
f (4) (x) = 0
...
...
51
f (3) = f ,
...
3
...
A common variation on Leibniz’ notation for derivatives is the so-called
operator notation, as in
d(x3 − x)
d 3
=
(x − x) = 3x2 − 1
...
Usually
d2 y
=
dx2
dy
dx
2
!!!!
10
...
Find f (x), f (x) and f (3) (x) if
127
...
Check
this
...
Group Problem
...
x+2
1
(b) Find the nth order derivative of f (x) =
(i
...
x+2
(n)
find a formula for f (x) which is valid for all n =
0, 1, 2, 3
...
x+2
Compute the third derivative of f (x) = 2x/(x2 − 1)
by using either the left or right hand side (your choice)
of (†)
...
Compute the first, second and third derivatives of the
following functions
f (x) = (x + 1)4
g(x) = x2 + 1
√
h(x) = x − 2
k(x) =
3
x2
x3
x4
x5
x6
+
+
+
+
...
(About notation
...
(b) Find du/dt and d2 u/dt2 if u = t/(t + 2)
...
d
x
d2
x
(c) Find
and
...
d
x
d
1
and
...
Find the derivatives of 10th order of the functions
1
f (x) = x12 + x8
g(x) =
x
12
x2
h(x) =
k(x) =
1−x
1−x
133
...
11
...
dx
dx
dx
cos2 x
Note the minus sign in the derivative of the cosine!
Proof
...
h
To simplify the numerator we use the trigonometric addition formula
sin (x) = lim
h→0
sin(α + β) = sin α cos β + cos α sin β
...
A similar computation leads to the stated derivative of cos x
...
cos x
g(x)
We get
tan (x) =
cos2 (x) + sin2 (x)
1
cos(x) sin (x) − sin(x) cos (x)
=
=
2 (x)
cos
cos2 (x)
cos2 (x)
as claimed
...
Exercises
Find the derivatives of the following functions (try to
simplify your answers)
is differentiable at x = π/4?
145
...
f (x) = sin(x) + cos(x)
f (x) =
135
...
f (x) = 3 sin(x) + 2 cos(x)
sin x
x
140
...
Group Problem
...
f (x) =
1 + sin x
cos x
143
...
sin x
144
...
If f is a given function, and you have another function
g which satisfies g(x) = f (x) + 12 for all x, then f and g
have the same derivatives
...
[Hint: it’s a short
proof – use the differentiation rules
...
f (x) = x cos(x) − sin x
141
...
f (x) = x sin(x) + cos(x)
139
...
With hindsight this was to be expected – why?
148
...
cos2 x
Hint: remember your trig to reduce work!
π
4
π
4
53
A depends on B depends on C depends on
...
Assuming
that the balloon is spherical you can say how large it
is by specifying its radius R
...
The volume of the balloon is a function of its radius, since the volume of a sphere of radius r is given
by
4
V = πr3
...
e
...
Schematically we can summarize this chain of
cause-and-effect relations as follows: you could either
say that V depends on r, and r depends on t,
volume
radius r
V
f (depends g (depends
time t −→
−→
on
on
time t)
radius
r)
r = f (t)
and the second tells you the volume of the balloon
given its radius
V = g(r)
...
A “real world example” of a composition of functions
...
The Chain Rule
13
...
Composition of functions
...
The notation for the composition is f ◦ g, and it is defined by the formula
f ◦ g(x) = f g(x)
...
For instance, if f (x) = x2 + x and g(x) = 2x + 1 then
f ◦ g(x) = f 2x + 1 = (2x + 1)2 + (2x + 1)
and g ◦ f (x) = g x2 + x = 2(x2 + x) + 1
Note that f ◦ g and g ◦ f are not the same fucntion in this example (they hardly ever are the same)
...
If a quantity z is a function of another quantity y, and if y itself depends on x,
then z depends on x via y
...
Give x one can compute y, and from y one can then compute z
...
One says that the composition of f and g is the result of subsituting g in f
...
2
...
If f and g are differentiable, so is the composition f ◦ g
...
The chain rule tells you how to find the derivative of the composition f ◦ g of two functions f and g
provided you now how to differentiate the two functions f and g
...
Suppose that y = g(x) and
z = f (y), then z = f ◦ g(x), and the derivative of z with respect to x is the derivative of the function f ◦ g
...
In short,
dz
= (f ◦ g) (x),
dx
dz
= f (y)
dy
and
dy
= g (x)
dx
so that the chain rule says
dz
dz dy
=
...
We first consider difference quotients
instead of derivatives, i
...
using the same notation as above, we consider the effect of an increase of x by an
amount ∆x on the quantity z
...
The ratio of the increase in z = f (g(x)) to the increase in x is
∆z
∆y
∆z
=
·
...
here are finite quantities, so this equation is just
algebra: you can cancel the two ∆ys
...
Proof of the chain rule
...
2 at some arbitrary value x = a, i
...
we will show that
(f ◦ g) (a) = f (g(a)) g (a)
...
x→a
x−a
x−a
The two derivatives on the right hand side are given by
(f ◦ g) (a) = lim
x→a
g (a) = lim
x→a
g(x) − g(a)
x−a
and
f (g(a)) = lim
y→a
f (y) − f (g(a))
...
So
we can substitute y = g(x) in the limit defining f (g(a))
(27)
f (g(a)) = lim
y→a
f (y) − f (g(a))
f (g(x)) − f (g(a))
= lim
...
There is one flaw in this proof, namely, we have divided by g(x) − g(a), which is not allowed when
g(x) − g(a) = 0
...
2
13
...
First example
...
The composition of these two functions is
z = f (g(x)) = (2x + 1)2 + (2x + 1) = 4x2 + 6x + 2
...
e
...
First, you simply differentiate the last formula we have:
dz
d(4x2 + 6x + 2)
=
= 8x + 6
...
dx
dx
Hence, by the chain rule one has
dz dy
dz
=
= (2y + 1) · 2 = 4y + 2
...
Once you remember that y = 2x + 1 you see that this is
indeed true:
y = 2x + 1 =⇒ 4y + 2 = 4(2x + 1) + 2 = 8x + 6
...
In this example there was no clear
advantage in using the chain rule
...
2 Briefly, you have to show that the function
h(y) =
{f (y) − f (g(a))}/(y − g(a))
f (g(a))
is continuous
...
4
...
We know what the derivative of sin x with
respect to x is, but none of the rules we have found so far tell us how to differentiate f (x) = sin(2x)
...
We know how to differentiate each of the two functions g and h:
g (u) = cos u,
h (x) = 2
...
Leibniz would have decomposed the relation y = sin 2x between y and x as
y = sin u,
u = 2x
and then computed the derivative of sin 2x with respect to x as follows
d sin 2x u=2x d sin u
d sin u
du
=
=
·
= cos u · 2 = 2 cos 2x
...
5
...
The Power Rule, which says that for any function f
and any rational number n one has
d
f (x)n = nf (x)n−1 f (x),
dx
is a special case of the Chain Rule, for one can regard y = f (x)n as the composition of two functions
y = g(u),
n
n−1
where g(u) = u
...
dx
du
dx
dx
= f (x) then gives you the Power Rule
...
6
...
Consider the “real world example” from page 53 again
...
The volume of this balloon is
4 3
4
πr = πf (t)3
...
V =
According to the chain rule the rate of change of the volume with time is now
dV
dV dr
=
dt
dr dt
i
...
it is the product of the rate of change of the volume with the radius of the balloon and the rate of change
of the balloon’s radius with time
...
dr
dt
For instance, if the radius of the balloon is growing at 0
...
0inch, then the
volume is growing at a rate of
dV
= 4π(3
...
5inch/sec ≈ 57inch3 /sec
...
7
...
Suppose you needed to find the derivative of
√
x+1
y = h(x) = √
( x + 1 + 1)2
We can write this function as a composition of two simpler functions, namely,
y = f (u),
with
f (u) =
u = g(x),
√
u
and g(x) = x + 1
...
2 x+1
Hence the derivative of the composition is
√
d
x+1
√
h (x) =
dx ( x + 1 + 1)2
= f (u)g (x) =
1
u−1
· √
...
2
3
dx ( x + 1 + 1)
( x + 1 + 1)
2 x+1
√
x + 1:
The last step (where you replace u by its definition in terms of x) is important because the problem was
presented to you with only x and y as variables while u was a variable you introduced yourself to do the
problem
...
” For instance, to compute
√
d 4 + 7 + x3
dx
you could set u = 7 + x3 , and compute
√
√
d 4 + u du
d 4 + 7 + x3
=
·
...
” You would then immediately write
d
(4 +
dx
1
7 + x3 ) = √
· 3x2
...
8
...
Often we have to apply the
Chain Rule more than once to compute a derivative
...
dx
du dv dx
In functional notation this is
(f ◦ g ◦ h) (x) = f (g(h(x)) · g (h(x)) · h (x)
...
Thus if b = h(a) and
c = g(b) the Chain Rule is
dy
dy
du
dv
=
·
·
...
=
·
·
=−
2 2 v
dx
du dv dx
(1 + u)
so
dy
dx
=
x=4
dy
du
·
u=6
du
dv
·
v=25
dv
dx
x=4
1 1
=− ·
· 8
...
Exercises
√
149
...
Say what plays the role of y = f (u) and u = g(x)
...
Find the derivative of f (x) = x cos
in Figure 3
...
Alice and Bob differentiated y = 1 + x3 with respect
√
to x differently
...
Assuming neither
one made a mistake, did they get the same answer?
p = f · g,
q = g · f
...
Group Problem
...
x
0
1
f (x)
1
3
g(x)
1
-4
f (x)
5
-1/3
g (x)
1/3
-8/3
Define
v(x) = f (g(x)),
w(x) = g(f (x)),
p(x) = f (x)g(x),
Compute the following derivatives
q(x) = g(x)f (x)
...
If there is insufficient information to answer the
question, so indicate
...
f (x) = sin 2x − cos 3x
π
155
...
f (x) = sin(cos 3x)
164
...
Find an equation for
the tangent line to the curve y = f (x2 ) at the point
(x, y) = (3, 7)
...
There is a function f whose second derivative satisfies
1 + x2
2
w = g ◦ f,
Find v(x), w(x), p(x), and q(x)
...
152
...
Find
dx
du
Express the former in terms of x and the latter in terms
of u
...
Suppose that f (x) = x, g(x) = 1 + x2 , v(x) =
f ◦ g(x), w(x) = g ◦ f (x)
...
158
...
Suppose that f (x) = x2 + 1, g(x) = x + 5, and
150
...
157
...
f (x) = cos x − cos x
f (x) = −64f (x)
...
Group Problem
...
At time t (seconds) the height of the water in the glass is h(t) (inch)
...
2 h2
(fluid ounces)
...
Group Problem
...
How
fast is Moe pouring water into the glass?
A cubical sponge, hereafter refered to as ‘Bob’, is
absorbing water, which causes him to expand
...
His volume is V (t)
...
e
...
If we measure
lengths in inches and time in minutes, then what units
do t, S(t), V (t), S (t) and V (t) have?
(c) Moe pours water at 1 ounce per second, and at
some moment the water level is going up at 0
...
What is the water level at that moment?
59
(c) What is the relation between S (t) and V (t)?
(d) At the moment that Bob’s volume is 8 cubic
inches, he is absorbing water at a rate of 2 cubic inch per
minute
...
Implicit differentiation
15
...
The recipe
...
We call this equation the defining equation for the function y = f (x)
...
Here is a recipe for computing the derivative of an implicitely defined function
...
dy
:
(3) Solve the equation in step 2 for
dx
dy
H(x, y)
=−
dx
G(x, y)
(31)
(4) If you also have an explicit description of the function (i
...
a formula expressing y = f (x) in terms
of x) then you can substitute y = f (x) in the expression (31) to get a formula for dy/dx in terms of
x only
...
In that case (31) is
as far as you can go
...
dy
dx
which contains both x
15
...
Dealing with equations of the form F1 (x, y) = F2 (x, y)
...
E
...
to deal with a function y = f (x) which satisfies
y 2 + x = xy
you rewrite this equation as
y 2 + x − xy = 0
and set F (x, y) = y 2 + x − xy
...
3
...
Consider the function
f (x) =
4
1 − x4 ,
−1 ≤ x ≤ 1
...
e
...
60
The direct approach goes like this:
d 1 − x4
f (x) =
dx
=
1
4
1 − x4
=
1
4
1 − x4
1/4
−3/4 d(1
−3/4
− x4 )
dx
−4x3
x3
=−
1 − x4
3/4
To find the derivative using implicit differentiation we must first find a nice implicit description of the
function
...
So our implicit description of the function
√
y = f (x) = 4 1 − x4 is
x4 + y 4 − 1 = 0;
The defining function is therefore F (x, y) = x4 + y 4 − 1
Differentiate both sides with respect to x (and remember that y = f (x), so y here is a function of x), and
you get
dx4
dy 4
d1
dy
+
−
= 0 =⇒ 4x3 + 4y 3
= 0
...
This last equation can be solved for dy/dx:
dy
x3
= − 3
...
To express dy/dx in terms of x
√
only, and remove the y dependency we use y = 4 1 − x4
...
15
...
Another example
...
e
...
For instance, if x = 2π then y = π, i
...
f (2π) = π
...
dx
dx
dx
dx dx
dx
Solve for
dy
dx
and you get
1
1
=
...
2 + cos π
2−1
If we were asked f (π/2), then all we would be able to say is
1
...
2
61
15
...
Derivatives of Arc Sine and Arc Tangent
...
15
...
Theorem
...
If y = arcsin x then x = sin y
...
You get
dy
,
1 = cos y
dx
and hence
1
dy
=
...
Therefore
1 − x2
...
This leaves us with
2
2
√
cos y = 1 − x2 , and hence
dy
1
...
16
...
State what
the expressions F (x, y), G(x, y) and H(x, y) from the
recipe in the beginning of this section are
...
169
...
x3 + xy + y 3 = 3
178
...
π
167
...
sin x + xy + y 5 = π
180
...
Then use this description to find the derivative
dy/dx
...
y = f (x) = 1 − x
1
2
168
...
x + y = xy
171
...
y = f (x) =
172
...
y = f (x) =
173
...
y = f (x) =
174
...
y = f (x) =
4
175
...
y = f (x) =
62
4
3
4
x + x2
√
1− x
√
x− x
√
2x + 1 − x2
x + x2
187
...
y = f (x) =
3
4
x−
√
200
...
The
bottom of the pole is pulled along the ground away from
the wall at the rate of 2 m/s
...
Group Problem
...
)
your simplification is valid
...
(a) sin arcsin x
A television camera is positioned 4000 ft from the
base of a rocket launching pad
...
(e) tan arctan z
(b) cos arcsin x
201
...
(f) tan arcsin θ
(c) arctan(tan θ)
(b) How fast is the camera’s angle of elevation changing at that same moment? (Assume that the television
camera points toward the rocket
...
What are they?
202
...
A 2-foot tall dog is walking away from a streetlight
which is on a 10-foot pole
...
How fast is the dog walking at that
moment?
190
...
f (x) = arcsin x
192
...
An isosceles triangle is changing its shape: the lengths
of the two equal sides remain fixed at 2 inch, but the
angle θ(t) between them changes
...
f (x) = sin arctan x
194
...
f (x) =
196
...
f (x) =
2
Let A(t) be the area of the triangle at time t
...
5inch2 /sec,
then how fast is the angle increasing or decreasing when
θ = 60◦ ?
1
1 + (arctan x)2
1 − (arcsin x)2
204
...
Its motion is parallel to the x-axis; its distance to the
x-axis is always 10 (feet)
...
We write θ for the angle between the positive
x-axis and the line segment from the origin to P
...
198
...
If the bottom of the pole is
8 feet away from the wall, and if it is sliding away from
the wall at 7 feet per second, then with what speed is the
top (A) going down?
(b) Where is the point when θ = π/3?
(c) Compute the rate of change of the angle θ at
the moment that θ = π
...
The point Q is moving on the line y = x with velocity 3
m/sec
...
pole of length 10 ft
206
...
Its x-coordinate is increasing at a constant rate
of 2 feet/minute
...
A pole 10 feet long rests against a vertical wall
...
(b) The ideal gas law turns out to be only approximately true
...
207
...
A certain amount of gas is trapped in a cylinder with
a piston
...
Suppose that the cylinder contains fictitious gas for
which one has a = 12 and b = 3
...
Then how fast is the pressure changing?
(a) If the pressure is 10psi (pounds per square inch),
if the volume is 25inch3 , and if the piston is moving so
64
CHAPTER 5
Graph Sketching and Max-Min Problems
The signs of the first and second derivatives of a function tell us something about the shape of its graph
...
1
...
The slope of the normal line to the graph is −1/m and thus one could write the equation for the normal as
x−a
...
In this case the equation for the
normal cannot be written as in (34), but instead one gets the simpler equation
y = f (a)
...
Both (35) and (34) are formulas that you shouldn’t try to remember
...
m
y
=
/1
m
=
m
un
/r
ise
r
=
run
e/
ris
−1
1
m
/
−1
m
x
Figure 1
...
”
slope of tangent
2
...
A more precise version of this statement is the Intermediate Value Theorem:
65
f (a)
y
f (b)
a c1
c2
c3
b
Figure 2
...
In this example there are three values of c for which f (c) = y
holds
...
1
...
If f is a continuous function on an interval a ≤ x ≤ b, and if y
is some number between f (a) and f (b), then there is a number c with a ≤ c ≤ b such that f (c) = y
...
2
...
Example – Square root of 2
...
Since f (1) < 2 and f (2) = 4 > 2
the intermediate value theorem with a = 1, b = 2, y = 2 tells us that there is a number c between 1 and 2
such that f (c) = 2, i
...
for which c2 = 2
...
2
...
Example – The equation θ +sin θ = π
...
It is a continuous
2
function at all x, so from f (0) = 0 and f (π) = π it follows that there is a number θ between 0 and π such
that f (θ) = π/2
...
Unlike the previous example, where we knew the solution was 2, there
is no simple formula for the solution to (36)
...
4
...
If we apply the intermediate value theorem to the function
f (x) = 1/x on the interval [a, b] = [−1, 1], then we see that for any y between f (a) = f (−1) = −1 and
f (b) = f (1) = 1 there is a number c in the interval [−1, 1] such that 1/c = y
...
But there is no such c, because 1/c is never zero! So we have done something wrong, and the mistake we
made is that we overlooked that our function f (x) = 1/x is not defined on the whole interval −1 ≤ x ≤ 1
because it is not defined at x = 0
...
Exercises
208
...
Does the graph of y = x4 −2x2 +2 have any horizontal
tangents? If so, where?
209
...
Where does the normal to the graph of y = x2 at the
point (a, a2 ) intersect the x-axis?
√
211
...
At some point (a, f (a)) on the graph of f (x) =
−1 + 2x − x2 the tangent to this graph goes through the
origin
...
True or False?
214
...
at the point
...
Find the equation for the tangents to the graph of the
Backward Sine at the points x = 1, x = 1 and at D (see
2
Figure 2 in §7
...
)
215
...
217
...
3)
The function
2
f (x) =
x + |x|
x
4
...
4
...
Theorem
...
Proof
...
If there were two such numbers then the intermediate value theorem would imply that
somewhere between x1 and x2 there was a c with f (c) = 0
...
4
...
Example
...
The zeros of f (i
...
the solutions of f (x) = 0) are − 1 , 1, 3
...
Theorem 4
...
For instance, f (x) has the
same sign for all x in the first interval (∞, − 1 )
...
g
...
Therefore f (x) > 0 for all x in the interval
(−∞, − 1 )
...
f (2) = (−1)(1) (5) < 0
f (4) = (1)(3) (9) > 0
If you know all the zeroes of a continuous function, then this method allows you to decide where the function
is positive or negative
...
For each of the factors x − 3, (x − 1)2 and (2x + 1)3 it is easy
to determine the sign, for any given x
...
Thus we have
• x − 3 is positive for x > 3 and negative for x < 3;
• (x − 1)2 is always positive (except at x = 1);
• (2x + 1)3 is positive for x > − 1 and negative for x < − 1
...
We can summarize this computation in the
following diagram:
67
−
−
−
−
−
+
+
+
+
+
−
−
+
+
+
−
+
−
−
−
+
+
+
+
+
+
+
+
−
−
−
−
+
1
+
+
+
−
−1/2
−
+
+
−
x−3
+
+
+
(x−1)
2
(2x+1)
+
3
f(x)
3
5
...
• A function is called increasing if a < b implies f (a) < f (b) for all numbers a and b in the domain
of f
...
• The function f is called non-decreasing if a < b implies f (a) ≤ f (b) for all numbers a and b in
the domain of f
...
You can summarize these definitions as follows:
f is
...
Increasing: a < b =⇒ f (a) < f (b)
Decreasing:
a < b =⇒ f (a) > f (b)
Non-increasing: a < b =⇒ f (a) ≥ f (b)
Non-decreasing:
a < b =⇒ f (a) ≤ f (b)
The sign of the derivaitve of f tells you if f is increasing or not
...
1
...
If a function is non-decreasing on an interval a < x < b then f (x) ≥ 0 for all x in that
interval
...
For instance, if f is non-decreasing, then for any given x and any positive ∆x one has f (x + ∆x) ≥ f (x)
and hence
f (x + ∆x) − f (x)
≥ 0
...
∆x
What about the converse, i
...
if you know the sign of f then what can you say about f ? For this we
have the following
5
...
Theorem
...
If f (x) > 0 for all a < x < b, then f is increasing
...
The proof is based on the Mean Value theorem which also finds use in many other situations:
68
a
c
b
Figure 3
...
This is true for any choice of a and b; c depends on a and b of course
...
3
...
If f is a differentiable function on the interval a ≤ x ≤ b, then
there is some number c, with a < c < b such that
f (c) =
f (b) − f (a)
...
2
...
Let x1 < x2 be
two numbers between a and b
...
Since we know that f (c) > 0 and x2 − x1 > 0 it follows that f (x2 ) − f (x1 ) > 0, i
...
f (x2 ) > f (x1 )
...
Examples
Armed with these theorems we can now split the graph of any function into increasing and decreasing
parts simply by computing the derivative f (x) and finding out where f (x) > 0 and where f (x) < 0 – i
...
we apply the method form the previous section to f rather than f
...
1
...
The familiar graph of f (x) = x2 consists of two parts, one
decreasing and one increasing
...
Therefore the function f (x) = x2 is decreasing for x < 0 and increasing for x > 0
...
2
...
The derivative of the function f (x) = 1/x = x−1 is
1
x2
which is always negative
...
But this isn’t true if you take a = −1 and b = 1:
f (x) = −
1
1
= −1 < 1 = !!
a
b
The problem is that we used theorem 5
...
The function in this example, f (x) = 1/x, is not defined on
the interval −1 < x < 1 because it isn’t defined at x = 0
...
a = −1 < 1 = b, but
On the other hand, the function is defined and differentiable on the interval 0 < x < ∞, so theorem 5
...
This means, that as long as x is positive, increasing x will
decrease 1/x
...
3
...
Consider the function
y = f (x) = x3 − x
...
We try to find out where f is positive, and where it is negative by factoring f (x)
√
√
1
f (x) = 3 x2 − 1 = 3 x + 1 3 x − 3 3
3
3
from which you see that
√
1
f (x) > 0 for x < − 3 3
√
√
1
f (x) < 0 for − 1 3 < x < 3 3
3
√
f (x) > 0 for x > 1 3
3
Therefore the function f is
√
√
√
√
increasing on −∞, − 1 3 , decreasing on − 1 3, 1 3 , increasing on 1 3, ∞
...
This leads us to the
3
following picture of the graph of f :
70
f (x) > 0
f (x) < 0
√
1
x = −3 3
f (x) > 0
x=
1
3
√
3
3
Figure 4
...
6
...
A function whose tangent turns up and down infinitely often near the origin
...
Somewhere in the mathematician’s zoo of curious functions the following will be on
exhibit
...
2
x
y = 1 x + x2
2
y = 1x
2
y = 1 x + x2 sin π
2
x
1
y = 2 x − x2
Figure 5
...
” The slopes at the intersection points alternate between 1 − π and 1 + π
...
This makes the function continuous
at x = 0
...
x→0
x→0 2
x−0
x
2
71
(To find the limit apply the sandwich theorem to −|x| ≤ x sin π ≤ |x|
...
e
...
) The point of this example is that
this turns out not to be true
...
It is given by
π
π
1
− π cos + 2x sin
...
of the graph with the line y = x/2
...
For larger and larger k the points Pk tend to the origin (the x coordinate is
The slope of the tangent at Pk is given by
1
k
which goes to 0 as k → ∞)
...
64159265358979
...
64159265358979
...
e
...
1
2
− π and
1
2
+ π,
In particular, the slope of the tangent at the odd intersection points is negative, and so you would expect
the function to be decreasing there
...
7
...
Global maxima are sometimes also called “absolute maxima
...
Every global maximum is a local maximum, but a local maximum doesn’t have to be a global maximum
...
1
...
Any x value for which f (x) = 0 is called a
stationary point for the function f
...
2
...
Suppose f is a differentiable function on some interval [a, b]
...
Proof
...
By assumption the
left and right hand limits
f (x) = lim
∆x
0
f (x + ∆x) − f (x)
f (x + ∆x) − f (x)
and f (x) = lim
∆x 0
∆x
∆x
both exist and they are equal
...
A function defined on an interval [a, b] with one interior absolute minimum, another interior
local minimum, an interior local maximum, and two local maxima on the boundary, one of which is in
fact an absolute maximum
...
In the first limit we
also have ∆x < 0, so that
f (x + ∆x) − f (x)
lim
≤0
∆x 0
∆x
Hence f (x) ≤ 0
...
Thus we have shown that f (x) ≤ 0 and f (x) ≥ 0 at the same time
...
7
...
How to tell if a stationary point is a maximum, a minimum, or neither
...
You can tell
what kind of stationary point c is by looking at the signs of f (x) for x near c
...
4
...
If in some small interval (c − δ, c + δ) you have f (x) < 0 for x < c and f (x) > 0 for
x > c then f has a local minimum at x = c
...
The reason is simple: if f increases to the left of c and decreases to the right of c
then it has a maximum at c
...
If in addition f (x) < 0 for x > c then f is decreasing for x between c and c + δ, so
that f (x) < f (c) for those x
...
73
7
...
Example – local maxima and minima of f (x) = x3 −x
...
3 we had found that the function
√
√
f (x) = x3 − x is decreasing when −∞ < x < − 1 3, and also when 1 3 < x < ∞, while it is increasing when
3
3
√
√
1
1√
− 1 3 < x < 3 3
...
Neither the local maximum nor the local minimum are global max or min since
lim f (x) = +∞ and lim f (x) = −∞
...
6
...
If you look for stationary
points of the function f (x) = x3 you find that there’s only one, namely x = 0
...
4 does not tell us anything
...
y = x3
8
...
2 is very useful since it tells you how to find (local) maxima and minima
...
It doesn’t say how to find maxima or minima, but it tells you
that they do exist, and hence that you are not wasting your time trying to find a maximum or minimum
...
1
...
Let f be continuous function defined on the closed interval a ≤ x ≤ b
...
In other words there exist real numbers c and
d such that
f (c) ≤ f (x) ≤ f (d)
whenever a ≤ x ≤ b
...
9
...
1 are
not met
...
The function on the left has no maximum, and the one on the right has no minimum
...
1
...
have a maximum on the interval 0 ≤ x ≤ 1?
Answer: No
...
On the other hand the function is never larger than 1
...
If you now search the interval for numbers a with f (a) = 1, then you notice that such an a does
not exist
...
What about Theorem 8
...
For at x = 1 one has
f (1) = 0 = 1 = lim f (x)
...
9
...
Question: Does the function
f (x) =
1
,
x2
1≤x<∞
have a maximum or minimum?
Answer: The function has a maximum at x = 1, but it has no minimum
...
Hence f (x) ≤ f (1) for all x in
the interval [1, ∞) and that is why f attains its maximum at x = 1
...
However, the equation f (a) = 0 has no solution – f does not attain its minimum
...
1 not apply? In this example the function f is continuous on the whole interval
[1, ∞), but this interval is not a closed interval, i
...
it is not of the form [a, b] (it does not include its endpoints)
...
General method for sketching the graph of a function
Given a differentiable function f defined on some interval a ≤ x ≤ b, you can find the increasing and
decreasing parts of the graph, as well as all the local maxima and minima by following this procedure:
(1) find all solutions of f (x) = 0 in the interval [a, b]: these are called the critical or stationary points
for f
...
Compute the function value f (x) at each stationary point
...
e
...
(5) the absolute maximum is attained at the stationary point or the boundary point with the highest
function value; the absolute minimum occurs at the boundary or stationary point with the smallest
function value
...
e
...
75
10
...
Example – the graph of a rational function
...
1 + x2
By looking at the signs of numerator and denominator we see that
f (x) =
f (x) > 0 for 0 < x < 1
f (x) < 0 for x < 0 and also for x > 1
...
Hence f (x) = 0 holds if and only if
1 − 2x − x2 = 0
√
and the solutions to this quadratic equation are −1 ± 2
...
To see if the derivative changes sign we factor the numerator and denominator
...
Therefore
< 0
f (x) > 0
<0
for x < A
for A < x < B
for x > B
It follows that f is decreasing on the interval (−∞, A), increasing on the interval (A, B) and decreasing again
on the interval (B, ∞)
...
Are these global maxima and minima?
abs
...
A
1
B
−1
abs
...
Figure 8
...
You
find
lim f (x) = lim f (x) = −1
...
Similarly, f is decreasing from B to +∞, so
−1 < f (x) ≤ f (−1 +
√
76
2) for B < x < ∞
...
If a graph is convex then all chords lie above the graph
...
Between the two stationary points the function is increasing, so
√
f (−1 − 2) ≤ f (x) ≤ f (B) for A ≤ x ≤ B
...
11
...
The function is called concave if the line segment connecting any pair of points on the graph lies below
the piece of the graph between those two points
...
Instead of “convex” and “concave” one often says “curved upwards” or “curved downwards
...
11
...
Theorem
...
11
...
Theorem
...
A proof using the Mean Value Theorem will be given in class
...
At an inflection point the tangent crosses the graph
...
3
...
The second derivative of the function f (x) =
x − x is
3
f (x) = 6x
which is positive for x > 0 and negative for x < 0
...
3, the origin is an inflection
point, and the piece of the graph where x > 0 is convex, while the piece where x < 0 is concave
...
4
...
In §7
...
There is another way of distinguishing
between local maxima and minima which involves computing the second derivative
...
5
...
If c is a stationary point for a function f , and if f (c) < 0 then f has a local maximum
at x = c
...
The theorem doesn’t say what happens when f (c) = 0
...
The basic reason why this theorem is true is that if c is a stationary point with f (c) > 0 then “f (x) is
increasing near x = c” and hence f (x) < 0 for x < c and f (x) > 0 for x > c
...
x
11
...
Example – that cubic function again
...
3 and
§11
...
We had found that this function has two stationary points, namely at x = ± 1 3
...
Instead
3
3
√
of looking at f (x) we could also have computed f (x) at x = ± 1 3 and applied the second derivative test
...
3
3
√
1√
Therefore f has a local maximum at − 3 3 and a local minimum at 1 3
...
7
...
Usually the second derivative test will work,
but sometimes a stationary point c has f (c) = 0
...
The figure below shows you the graphs of three functions, all three of which have a stationary point at
x = 0
...
As you can see, the stationary point can be a local maximum, a local minimum, or neither
...
Three functions for which the second derivative test doesn’t work
...
Proofs of some of the theorems
12
...
Proof of the Mean Value Theorem
...
e
...
This function is continuous (since f is continuous), and g attains its maximum and minimum at two numbers
cmin and cmax
...
Consider the first case: one of these two numbers is an interior point, i
...
if a < cmin < b or a < cmax < b,
then the derivative of g must vanish at cmin or cmax
...
e
...
The definition of m implies that one gets
f (b) − f (a)
...
b−a
We are left with the remaining case, in which both cmin and cmax are end points
...
Thus the maximal and minimal values of g are both zero! This means that g(x) = 0 for all x, and thus that
g (x) = 0 for all x
...
12
...
Proof of Theorem 5
...
If f is a non-increasing function and if it is differentiable at some interior
point a, then we must show that f (a) ≥ 0
...
Hence one also has
f (x) − f (a)
≥0
x−a
for all x > a
...
a
x−a
f (a) = lim
x
12
...
Proof of Theorem 5
...
Suppose f is a differentiable function on an interval a < x < b, and
suppose that f (x) ≥ 0 on that interval
...
e
...
To prove this we use the
Mean Value Theorem: given x1 and x2 the Mean Value Theorem hands us a number c with x1 < c < x2 , and
f (x2 ) − f (x1 )
...
Hence
f (x2 ) − f (x1 )
≥ 0
...
13
...
What does the Intermediate Value Theorem say?
228
...
What does the Mean Value Theorem say?
229
...
y = x4 + 27x
220
...
231
...
Show that this follows
from the Mean Value Theorem
...
Make a drawing of the situation,
then read the Mean Value Theorem again
...
y = x4 + 2x2 − 3
233
...
y = x5 + 16x
235
...
y =
x+1
x
237
...
What is a stationary point?
222
...
How can you tell if a local maximum is a global
maximum?
223
...
238
...
True or False?
x2
1 + x2
1 + x2
1+x
1
240
...
y = x −
x
239
...
What is an inflection point?
225
...
242
...
Group Problem
...
y = x3 + 2x2 − x
Draw four graphs of functions, one for each of the
following four combinations
244
...
y = x4 − 2x3 + 2x
f < 0 and f > 0
f < 0 and f < 0
246
...
y =
1 − x2
248
...
Group Problem
...
y =
Which of the following combinations are possible:
f (x) > 0 and f (x) = 0 for all x
4
1 + x2
1
1 + x4
The following functions are periodic, i
...
they satisfy f (x + L) = f (x) for all x, where the constant L is
called the period of the function
...
It therefore has infinitely many (local) minima and
maxima, and infinitely many inflections points
...
f (x) = 0 and f (x) > 0 for all x
Sketch the graph of the following functions
...
e
...
250
...
y = sin x + cos x
252
...
y = 2 sin x + sin2 x
80
254
...
In the following two problems it is not possible to solve
the equation f (x) = 0, but you can tell something from
the second derivative
...
y = 2 cos x + cos2 x
256
...
y = 2 + sin x
(a) Show that the function f (x) = x arctan x is
convex
...
(b) Show that the function g(x) = x arcsin x is
convex
...
2
Find the domain and sketch the graphs of each of
the following functions
259
...
f (x) =
10 < x < ∞
1 + x2
260
...
f (x) =
2 + x2
x3 − x
1
267
...
f (x) =
2 + x2
x3 − x
0
258
...
y = arctan(x2 )
262
...
y = 6 arcsin(x) − 10x2
14
...
If the function is continuous then according to theorem 8
...
If f is differentiable then we know what to do: any local maximum is either a stationary point or one of
the end points a and b
...
Usually there is only one global maximum, but sometimes there can be more
...
The
same recipe works (of course you should look for the smallest function value instead of the largest in step 4
...
Choosing which quantity to call x and finding the function f is half the job
...
1
...
Which rectangle has
the largest area, among all those rectangles for which the total length of the sides is 1?
Solution: If the sides of the rectangle have lengths x and y, then the total length of the sides is
L = x + x + y + y = 2(x + y)
and the area of the rectangle is
A = xy
...
The lengths of the sides can
also not be negative, so x and y must satisfy x ≥ 0, y ≥ 0
...
”
The quantity which we are asked to maximize is A, but it depends on two variables x and y instead of just
one variable
...
From this equation we get
1
2
L = 1 =⇒ y =
− x
...
e
...
So we end up with this problem:
Find the maximum of the function f (x) = x
1
2
− x on the interval 0 ≤ x ≤ 1
...
Therefore the theory guarantees that
2
there is a maximum and our recipe will show us where it is
...
The function value at this point is
1
1
f 1 ) = 1 ( 1 − 4 ) = 16
...
The area of such rectangles is zero, and so this is not the maximal value we are looking for
...
e
...
4
15
...
By definition, the perimeter of a rectangle is the sum
of the lengths of its four sides
...
Which rectangle of area 100in2 minimizes its height
plus two times its length?
�
������
271
...
Which
choice of θ and R will give you the wedge with the largest
area? Which choice leads to the smallest area?
The combined light intensity is the sum of the two
light intensities coming from both lamp posts
...
[A circular wedge is the figure consisting of two radii
of a circle and the arc connecting them
...
]
(b) What is the darkest spot between the two lights,
i
...
where is the combined light intensity the smallest?
272
...
(The lamp post problem)
273
...
If r is the
radius of the can and h its height, then which h and r
will give you the can with the largest volume?
In a street two lamp posts are 300 feet apart
...
(b) If instead of making a plain cylinder you replaced
the flat top and bottom of the cylinder with two spherical
82
caps, then (using the same 100in2 os sheet metal), then
which choice of radius and height of the cylinder give you
the container with the largest volume?
277
...
What are the smallest and largest values that
(cos x)(cos y) can have if x + y = π and if x and y
2
are both nonnegative?
279
...
A triangle has one vertex at the origin O(0, 0), another
at the point A(2a, 0) and the third at (a, a/(1 + a3 ))
...
Group Problem
...
The cost per hour of fuel to run a locomotive is v 2 /25
dollars, where v is speed (in miles per hour), and other
costs are $100 per hour regardless of speed
...
When she arrived on the north coast
of Africa (∼800BC) the locals allowed her to take as
much land as could be enclosed with the hide of one ox
...
Q
B
D
Josh is in need of coffee
...
He cuts out a wedge and glues the
two edges AC and BC together to make a conical filter
to hold the ground coffee
...
C
A
281
...
P
R
land
water
(a) If Dido wanted a rectangular region, then how
wide should she choose it to enclose as much area as
possible (the coastal edge of the boundary doesn’t count,
so in this problem the length AB + BC + CD is 100
yards
...
)
(a) Find the volume in terms of the angle θ
...
)
276
...
We know
x ≥ 1 and y ≥ 1
...
For the rest start at http://en
...
org/wiki/Dido
83
CHAPTER 6
Exponentials and Logarithms (naturally)
In this chapter we first recall some facts about exponentials (xy with x > 0 and y arbitrary): they should
be familiar from algebra, or “precalculus
...
g
...
718 281 828 459 045 235 360 287 471 352 662 497 757 247 093 699 95 · · ·
...
”
1
...
For any real number x and any positive integer n = 1, 2, 3,
...
xn
One defines x0 = 1 for any x = 0
...
One then defines
√
xp/q = q xp
...
One can define xa for irrational
a
numbers a by taking limits
...
e
...
4 =
14
10 ,
a3 = 1
...
g
...
e
...
414 = 1000 ,
···
...
Our definition of 2 2 then is
= lim 2an ,
n→∞
as the limit of the sequence of numbers
√
√
√
10
100
1000
2, 214 ,
2141 ,
21414 , · · ·
(See table 1
...
We will not go into these details in this course
...
One can show that these properties still hold if a and b are real numbers (not
necessarily fractions
...
85
x
1
...
4000000000
1
...
4140000000
1
...
4142100000
1
...
4142135000
...
...
000000000000
2
...
657371628193
2
...
665119088532
2
...
665143103798
2
...
...
√
Table 1
...
Note that as x √ closer to
gets
some number
...
√
2 the quantity 2x appears to converge to
Now instead of considering xa as a function of x we can pick a positive number a and consider the
function f (x) = ax
...
1
...
The trouble with powers of negative numbers
...
For instance 3 −8 = −2 because (−2)3 = −8
...
But there is a problem: since 2 = 1 you would think that (−8)2/6 = (−8)1/3
...
Another example:
√
(−4)1/2 = −4 is not defined
but, even though 1 = 2 ,
2
4
√
(−4)2/4 = 4 (−4)2 = 4 +16 = 2 is defined
...
The safest is just not to take fractional powers of negative numbers
...
For example, (−8)π is not defined1
...
Logarithms
Briefly, y = loga x is the inverse function to y = ax
...
In other words, loga x is the answer to the question “for which number y does one have x = ay ?” The number
loga x is called the logarithm with base a of x
...
For instance,
23 = 8,
21/2 =
so
log2 8 = 3,
log2
√
√
2,
2 =
1
,
2
2−1 =
log2
1
2
1
= −1
...
You will see this next semester if you take math 222
...
The graphs of y = 2x , 3x , (1/2)x , (0
...
The graphs are purposely not
labeled: can you figure out which is which?
Also:
log2 (−3) doesn’t exist
because there is no number y for which 2 = −3 (2y is always positive) and
y
log−3 2 doesn’t exist either
because y = log−3 2 would have to be some real number which satisfies (−3)y = 2, and we don’t take
non-integer powers of negative numbers
...
Properties of logarithms
In general one has
loga ax = x, and aloga x = x
...
Again, one finds the following formulas in precalculus texts:
loga xy = loga x + loga y
x
loga = loga x − loga y
y
loga xy = y loga x
(39)
loga x =
logb x
logb a
They follow from (38)
...
Graphs of exponential functions and logarithms
Figure 1 shows the graphs of some exponential functions y = ax with different values of a, and figure 2
shows the graphs of y = log2 x, y = log3 x, log1/2 x, log1/3 (x) and y = log10 x
...
)
From algebra/precalc recall:
If a > 1 then f (x) = ax is an increasing function
...
In other words, for a > 1 it follows from x1 < x2 that ax1 < ax2 ; if 0 < a < 1, then x1 < x2 implies ax1 > ax2
...
Graphs of some logarithms
...
Can you tell what a is for each graph?
5
...
= 2x lim
∆x→0
∆x
So if we assume that the limit
2∆x − 1
=C
∆x→0
∆x
lim
exists then we have
d2x
= C2x
...
693 147
...
Once we know (40) we can compute the derivative of ax for any other positive number a
...
By the chain rule we therefore get
d2x·log2 a
dax
=
dx
dx
dx · log2 a
dx
x·log2 a
= (C log2 a) 2
= C 2x·log2 a
= (C log2 a) ax
...
This is essentially our
formula for the derivative of ax , but one can make the formula look nicer by introducing a special number,
namely, we define
2∆x − 1
e = 21/C where C = lim
...
718 281 818 459 · · ·
This number is special because if you set a = e, then
1
= 1,
C log2 a = C log2 e = C log2 21/C = C ·
C
and therefore the derivative of the function y = ex is
dex
= ex
...
Thus we have
eln x = x
(42)
ln ex = x
where the second formula holds for all real numbers x but the first one only makes sense for x > 0
...
By the chain rule you then get
dax
= ax ln a
...
Derivatives of Logarithms
Since the natural logarithm is the inverse function of f (x) = ex we can find its derivative by implicit
differentiation
...
Then solve for f (x) to get
1
...
dx
x ln a
f (x) =
Finally we remember that af (x)
89
In particular, the natural logarithm has a very simple derivative, namely, since ln e = 1 we have
d ln x
1
=
...
Limits involving exponentials and logarithms
7
...
Theorem
...
Then, if a > 1,
lim xr a−x = 0,
x→∞
i
...
xr
= 0
...
For instance, as x → ∞ both
x1000 and (1
...
001)x
lim
so, in the long run, for very large x, 1
...
x
Proof when a = e
...
To do this consider the function f (x) =
e
...
dx
Therefore f (x) < 0 for x > r + 1, i
...
f (x) is decreasing for x > r + 1
...
e
...
Divide by x, abbreviate A = (r + 1)r+1 e−(r+1) , and we get
0 < xr e−x <
A
for all x > r + 1
...
Here are some related limits:
ax
= ∞ (D
...
E
...
To prove it you set x = et and then t = s/m, which leads to
lim
x→∞
ln x
xm
x=et
=
lim
t→∞
t
t=s/m
emt
=
1
s
lim
= 0
...
90
8
...
(45)
The constant X0 is the value of X(t) at time t = 0 (sometimes called “the initial value of X”)
...
e
...
dt
(46)
In words, for an exponentially growing quantity the rate of change is always proportional to the quantity itself
...
”
This property of exponential functions completely describes them, by which I mean that any function
which satisfies (46) automatically satisfies (45)
...
Then
de−kt
dX(t) −kt
dX(t)e−kt
= X(t)
+
e
dt
dt
dt
= −kX(t)e−kt + X (t)e−kt
= (X (t) − kX(t))e−kt
= 0
...
At t = 0 one has
X(t)e−kt = X(0)e0 = X0
and therefore we have
X(t)e−kt = X0 for all t
...
8
...
Half time and doubling time
...
In words, after time T goes by an exponentially growing (decaying) quantity changes by a factor ekT
...
If k < 0 then X(t) is decaying and one calls
T =
ln 2
−k
the half life because X(t) is reduced by a factor ekT = e− ln 2 =
91
1
2
every T time units
...
2
...
The general exponential growth/decay function (45) contains only two
constants, X0 and k, and if you know the values of X(t) at two different times then you can compute these
constants
...
Then we have
X0 ekt1 = X1 and X2 = X0 ekt2
in which t1 , t2 , X1 , X2 are given and k and X0 are unknown
...
t1 − t2
Once you have computed k you can find X0 from
X2
X1
X0 = kt1 = kt2
...
)
k=
9
...
298
...
)
1+x
1−x
(|x| < 1)
299
...
y = ln x2 − 3x + 2
(x > 2)
π
)
2
282
...
y = ln cos x
283
...
The function f (x) = e−x plays a central in statistics
and its graph is called the bell curve (because of its
shape)
...
(|x| <
2
284
...
y = e3x − 4ex
303
...
y =
1 + ex
287
...
2ex
1 + e2x
Find the limits
−x
288
...
y = xe−x/4
lim
x
290
...
A damped oscillation is a function of the form
291
...
y = ln x
f (x) = e−ax cos bx or f (x) = e−ax sin bx
where a and b are constants
...
e
...
294
...
y =
−1
ln x
296
...
y =
ln x
x
f (x)
and lim f (x)
x→∞
xn
where n can be any positive integer (hint: substitute
x =
...
y = ln
0
(0 < x < ∞, x = 1)
This function has many local maxima and minima
...
)
(x > 0)
(x > 0)
92
324
...
(a) If a sample has a mass of 200 mg find a formula for
the mass that remains after t days
...
(c) When will the mass be reduced to 10 mg?
(d) Sketch the graph of the mass as a function of time
...
Find the inflection points on the graph of f (x) =
(1 + x) ln x (x > 0)
...
(a) If x is large, which is bigger: 2x or x2 ?
(b) The graphs of f (x) = x2 and g(x) = 2x intersect at
x = 2 (since 22 = 22 )
...
Current agricultural experts believe that the world’s
farms can feed about 10 billion people
...
517 billion and the 1992 world population was 5
...
When can we expect to run out of
food?
Find the following limits
...
308
...
Group Problem
...
lim x
x→∞ 3 − 2x
310
...
One says that “when this baby is old enough to vote,
the world will have one billion new mouths to feed” and
the other says “in thirty six years, the world will have to
set eight billion places at the table
...
(Hint: 36 = 2 × 18
...
lim
x→∞ e2x + x
311
...
lim √ x
x→∞
e +1
314
...
The population of California grows exponentially at
an instantaneous rate of 2% per year
...
lim ln(1 + x) − ln x
x→∞
(a) Write a formula for the population N (t) of California t years after January 1, 2000
...
lim x ln x
315
...
At what rate is California consuming
pizzas t years after 1990?
x→0
317
...
lim √
x→0
ln x
x + ln x
(c) How many pizzas were consumed in California
from January 1, 2005 to January 1, 2009?
ln x
x + ln x
328
...
319
...
(a) If its population in the year 1980 was 1,980,000
and its population in the year 1990 was 1,990,000, what
is its population in the year 2000?
320
...
Find
d(xx )x
dxx dxx
,
, and
...
)
Hint: xx = e???
...
Group Problem
...
Find
du/dx
...
It will simplify the
calculation
...
)
329
...
cosh x
(a) Prove the following identities
sinh x =
323
...
(a) What is the half life of radon-222?
(b) How long would it take the sample to decay to 10%
of its original amount?
cosh2 x − sinh2 x = 1
cosh 2x = cosh2 x + sinh2 x
sinh 2x = 2 sinh x cosh x
...
d sinh x
= cosh x,
dx
d cosh x
= sinh x,
dx
d tanh x
1
=
...
The most common interpretation
of the integral is in terms of the area under the graph of the given function, so that is where we begin
...
Area under a Graph
y = f (x)
a
b
Let f be a function which is defined on some interval a ≤ x ≤ b and assume it is positive, i
...
assume
that its graph lies above the x axis
...
Look at figure
1 before you read on
...
e
...
These numbers split the interval [a, b] into n sub-intervals
[x0 , x1 ],
[x1 , x2 ],
...
,
∆xn = xn − xn−1
...
e
...
, and in the last interval we choose some number xn−1 ≤ cn ≤ xn
...
We then define n rectangles: the base of the k th rectangle is the interval [xk−1 , xk ] on the x-axis, while
its height is f (ck ) (here k can be any integer from 1 to n
...
e
...
Adding these we see that the total area of the rectangles is
(47)
R = f (c1 )∆x1 + f (c2 )∆x2 + · · · + f (cn )∆xn
...
If the partition is sufficiently fine then one would expect this sum, i
...
the total area of all rectangles
to be a good approximation of the area of the region under the graph
...
So you would expect the area to be
the limit of Riemann-sums like R “as the partition becomes finer and finer
...
TOP
...
In each of those intervals a point ci has been chosen at random, and the resulting rectangles
with heights f (c1 ),
...
The total area under the graph of the function is roughly
equal to the total area of the rectangles
...
Refining the partition
...
1
...
Definition
...
e
...
2
...
However the interpretation of the integral as “the area of the region between the graph and the x-axis” has a
twist to it
...
Illustrating a Riemann sum for a function whose sign changes
...
The Riemann-sum corresponding to this picture is the total
area of the rectangles above the x-axis minus the total area of the rectangles below the x-axis
...
When f can be positive or negative, then the terms in the Riemann sum can also be positive or negative
...
The Riemann sum is
therefore the area of the rectangles above the x-axis minus the area below the axis and above the graph
...
3
...
1
...
A function F is called an antiderivative of f on the interval [a, b] if one has F (x) =
f (x) for all x with a < x < b
...
2
3
...
Theorem
...
(48)
a
(a proof was given in lecture
...
We will abbreviate
def
F (b) − F (a) = F (x)
97
b
x=a
b
= F (x) a
...
3
...
In the integral
b
f (x) dx
a
the numbers a and b are called the bounds of the integral, the function f (x) which is being integrated is called
the integrand, and the variable x is integration variable
...
If you systematically replace it with another variable, the
resulting integral will still be the same
...
3
0
and if you replace x by ϕ you still get
1
0
Another way to appreciate that the integration variable is a dummy variable is to look at the Fundamental
Theorem again:
b
f (x) dx = F (b) − F (a)
...
4
...
f (x) = x4 − x2
330
...
Let f be the function f (x) = 1 − x2
...
f (x) = 1 + x +
Draw the graph of f (x) with 0 ≤ x ≤ 2
...
f (x) =
<2
x3
x4
x2
+
+
2
3
4
1
x
339
...
Draw the
corresponding rectangles (add them to your drawing of
the graph of f )
...
f (x) =
2
x
341
...
Make a new drawing of the graph of f and
include the rectangles corresponding to the right endpoint
Riemann-sum
...
f (x) =
1
2+x
ex − e−x
2
1
344
...
f (x) =
332
...
Look at figure 1 (top)
...
, c6 leads to the smallest Riemann sum?
Which choice would give you the largest Riemann-sum?
ex + e−x
2
1
346
...
f (x) =
(Note: in this problem you’re not allowed to change
the division points xi , only the points ci in between them
...
f (x) = sin x
Find an antiderivative F (x) for each of the following
functions f (x)
...
348
...
f (x) = cos x
333
...
f (x) = cos 2x
334
...
f (x) = sin(x − π/3)
2
335
...
f (x) = sin x + sin 2x
98
353
...
The region between the graph of y = 1/x and the
x-axis, and between x = a and x = b (here 0 < a < b
√
are constants, e
...
choose a = 1 and b = 2 if you have
something against either letter a or b
...
363
...
354
...
f (x) =
355
...
2
√
356
...
1
x
+ − 1
...
Compute
1
1 − x2 dx
0
√
without finding an antiderivative for 1 − x2 (you can
find such an antiderivative, but it’s not easy
...
The region above the x-axis and below the graph of
f (x) = x2 − x3
...
Group Problem
...
The region above the x-axis and below the graph of
f (x) = 4x2 − x4
...
359
...
1/2
1 − x2 dx
I=
0
360
...
1
|1 − x| dx
J=
−1
361
...
1
|2 − x| dx
K=
−1
5
...
In practice, much of the effort required to find an
integral goes into finding the antiderivative
...
For instance,
x2 dx =
1 3
x ,
3
1
sin 5x dx = − 5 cos 5x,
etc
...
It is called an indefinite
integral, as opposed to the integral in (49) which is called a definite integral
...
It is important to distinguish between the two kinds of integrals
...
b
a
f (x)dx is a number
...
“area under the graph of y = f (x)”, at
least when f (x) > 0
...
0
they are not equal
...
1
...
Suppose you want to find an antiderivative of a given
function f (x) and after a long and messy computation which you don’t really trust you get an “answer”,
F (x)
...
If
F (x) turns out to be equal to f (x), then your F (x) is indeed an antiderivative and your computation isn’t
important anymore
...
My cousin Louie says it might be F (x) = x ln x − x
...
dx
x
Who knows how Louie thought of this1, but it doesn’t matter: he’s right! We now know that ln xdx =
x ln x − x + C
...
2
...
Let f (x) be a function defined on some interval a ≤ x ≤ b
...
So one given function f (x) has many different antiderivatives, obtained by adding
different constants to one given antiderivative
...
3
...
If F1 (x) and F2 (x) are antiderivatives of the same function f (x) on some interval
a ≤ x ≤ b, then there is a constant C such that F1 (x) = F2 (x) + C
...
Consider the difference G(x) = F1 (x) − F2 (x)
...
Hence F1 (x) − F2 (x) = C for some constant
...
Two functions F1 (x) and F2 (x) can
both equal f (x) dx without equaling each other
...
This can sometimes lead to confusing situations, e
...
you can check that
2 sin x cos x dx = sin2 x
2 sin x cos x dx = − cos2 x
are both correct
...
To avoid this kind of confusion we will from
now on never forget to include the “arbitrary constant +C” in our answer when we
compute an antiderivative
...
cos x
2 1 − sin x
2
2
Table 1
...
All of these integrals should be familiar
1
from the differentiation rules we have learned so far, except for for the integrals of tan x and of cos x
...
6
...
Suppose we have two functions f (x) and g(x) with antiderivatives F (x) and G(x), respectively
...
1He took math 222 and learned to integrate by parts
...
These properties imply analogous properties for the definite integral
...
a
Definite integrals have one other property for which there is no analog in indefinite integrals: if you split
the interval of integration into two parts, then the integral over the whole is the sum of the integrals over the
parts
...
6
...
Theorem
...
a
c
Proof
...
Then
c
b
f (x)dx = F (c) − F (a) and
f (x)dx = F (b) − F (a),
a
c
so that
b
f (x)dx = F (b) − F (a)
a
= F (b) − F (c) + F (c) − F (a)
c
=
b
f (x)dx +
a
f (x)dx
...
The fundamental theorem
a
f (x)dx
...
xdx = −
1
0
7
...
I=
0
What does I depend on? To see this, you calculate the integral and you find
I=
1 3 x
3t 0
= 1 x3 − 1 0 3 = 1 x3
...
It does not depend on t, since t is a “dummy variable” (see §3
...
)
In this way you can use integrals to define new functions
...
Again, since t is a dummy variable
we can replace it by any other variable we like
...
The previous example does not define a new function (I(x) = x3 /3)
...
It is given by
def 2
erf(x) = √
π
(57)
x
2
e−t dt,
0
so erf(x) is the area of the shaded region in figure 3
...
Definition of the Error function
...
Since the integral in
(57) occurs very often in statistics (in relation with the so-called normal distribution) it has been given a
name, namely, “erf(x)”
...
e
...
a
A similar calculation gives you
d
dx
b
f (t) dt = −f (x)
...
π
erf (x) =
103
8
...
8
...
Example
...
It does not appear in the list of standard
d
antiderivatives we know by heart
...
So let’s call G(x) = x2 + 3, and
F (u) = − cos u, then
F (G(x)) = − cos(x2 + 3)
and
dF (G(x))
= sin(x2 + 3) · 2x = f (x),
dx
F (G(x))
G (x)
so that
2x sin(x2 + 3) dx = − cos(x2 + 3) + C
...
2
...
The most transparent way of computing an integral by
substitution is by following Leibniz and introduce new variables
...
Then we get
f (G(x))G (x) dx =
f (u) du = F (u) + C
...
As an example, let’s do the integral (58) using Leibniz’ notation
...
Then we
compute
dz = d x2 + 3 = 2x dx and sin(x2 + 3) = sin z,
so that
2x sin(x2 + 3) dx = sin z dz = − cos z + C
...
When we do integrals in this calculus class, we always get rid of the substitution variable because it is a
variable we invented, and which does not appear in the original problem
...
g
...
104
case you may want to skip the last step and leave the integral in terms of the (meaningful) substitution
variable
...
3
...
For definite integrals the chain rule
d
F (G(x)) = F (G(x))G (x) = f (G(x))G (x)
dx
implies
b
f (G(x))G (x) dx = F (G(b)) − F (G(a))
...
f (G(x))G (x) dx =
(59)
u=G(a)
x=a
8
...
Example of substitution in a definite integral
...
Since du = 2x dx, the associated indefinite integral is
1
1
x dx = 1
du
...
)
If x runs between x = 0 and x = 1, then u = G(x) = 1 + x2 runs between u = 1 + 02 = 1 and u = 1 + 12 = 2,
so the definite integral we must compute is
2
1
du,
u
1
2
ln u
1
x
dx =
1 + x2
0
which is in our list of memorable integrals
...
1
2
x
1
dx = 1
du,
2
2
1+x
u
x=0
u=1
to emphasize (and remind yourself) to which variable the bounds in the integral refer
...
Exercises
Compute these derivatives:
366
...
368
...
370
...
dt
0
372
...
Group Problem
...
wikipedia
...
How many inflection points does the graph of the
error function have?
s2 ds
1/t
(b) The graph of the error function on Wikipedia
shows that erf(x) is negative when x < 0
...
Is Wikipedia wrong? Explain
...
−4
x5 + 2 dx
395
...
1
− 7x dx
4
397
...
√
√
( t − 2/ t) dt
1
8
√
3
398
...
x
1
(x/a + a/x + x + a + ax) dx
1
r+ √
3
r
dr
0
√
377
...
(x + 1)3 dx
399
...
(3x − 5) dx
379
...
1
−2
9
4
380
...
(hm
...
t
−2
dt
403
...
x
−2
dt
√
5
x4
dx
dx
0
1
4
x5 +
2
x−1
√
dx
3
x2
404
...
2
(1 − 2x − 3x ) dx
383
...
(5y 4 − 6y 2 + 14) dy
407
...
√
x dx
409
...
2
384
...
(cos θ + sin 2θ) dθ
0
−3
π
1
386
...
π/2
π/3
√
3
0
1
388
...
1
2
1
1
− 4
t2
t
6
0
...
dt
0
1
2
391
...
x2 + 1
√
dx
x
413
...
(1/x) dx
ln 6
8ex dx
ln 3
9
2t dt
8
0
1
−e
√
√
u( u + 3 u) du
415
...
416
...
dx
√
1 − x2
4
2
392
...
cot x
dx
sin x
|x2 − 1| dx
−2
1
106
2
|x − x2 | dx
435
...
(x − 2|x|) dx
436
...
(x2 − |x − 1|) dx
437
...
417
...
−1
2
419
...
f (x) =
0
438
...
if 0 ≤ x < 1,
if 1 ≤ x ≤ 2
...
Find the area of the region included between the
parabola y 2 = x and the line x + y = 2
...
f (x) dx where
−π
f (x) =
x,
sin x,
if − π ≤ x ≤ 0,
if 0 < x ≤ π
...
Find the area of the region bounded by the curves
√
y = x and y = x
...
Group Problem
...
Compute
2
2x 1 + x2
I=
3
You asked your assistant Joe to produce graphs of a
function f (x), its derivative f (x) and an antiderivative
F (x) of f (x)
...
(b) Use the substitution u = 1 + x2
...
Identify which graph is
which and explain your answer
...
Compute
In =
2x 1 + x2
n
dx
...
If f (x) = x − 1/x2 and f (1) = 1/2 find f (x)
...
Sketch the graph of the curve y = x + 1 and determine the area of the region enclosed by the curve, the
x-axis and the lines x = 0, x = 4
...
Find the area under the curve y = 6x + 4 and above
the x-axis between x = 0 and x = 2
...
√
427
...
(Don’t
evaluate the integral, but compare with the area under
√
the graph of y = 1 − x2
...
Determine the area under the curve y = a2 − x2
and between the lines x = 0 and x = a
...
Graph the curve y = 2 9 − x2 and determine the
area enclosed between the curve and the x-axis
...
Group Problem
...
y = f (x)
c
a
2
430
...
Find the area of this region
...
Find the area bounded by the curve y = 4 − x2 and
the lines y = 0 and y = 3
...
Which among the following statements true?
432
...
(a) F (a) = F (c)
(b) F (b) = 0
(c) F (b) > F (c)
(d) The graph of y = F (x) has two inflection points?
433
...
434
...
Use a substitution to evaluate
the following integrals
...
x dx
√
x+1
453
...
sin 2x
dx
1 + sin x
5
s ds
√
3
s+2
455
...
443
...
0
445
...
0
2
447
...
450
...
1
, dr
r ln r
2
ξ(1 + 2ξ 2 )10 dξ
457
...
sin 2x
√
dx
1 + cos 2x
459
...
et
dt
t2
sin
2
1
π/3
sin2 θ cos θ dθ
451
...
In this chapter we will describe a number
of “things which are an integral
...
After letting the partition become arbitrarily fine we then find that the
quantity we are looking for is given by an integral
...
1
...
e
...
Then the area of the region between
the graphs of the two functions is
b
(61)
g(x) − f (x) dx
...
Choose a partition
a = x0 < x1 < · · · < xn = b of the interval [a, b]; choose a number ck in each interval [xk−1 , xk ]; form the
rectangles
xk−1 ≤ x ≤ xk ,
f (ck ) ≤ y ≤ g(ck )
...
Hence the combined area of the rectangles is
R = g(c1 ) − f (c1 ) ∆x1 + · · · + g(cn ) − f (cn ) ∆xn
which is just the Riemann-sum for the integral
b
(g(x) − f (x))dx
...
Finding the area between two graphs using Riemann-sums
...
2
...
Find the area of the region bounded by the parabola
y 2 = 4x and the line y = 2x
...
)
462
...
470
...
463
...
471
...
464
...
472
...
465
...
466
...
473
...
Then find the
4 16
area between these curves
...
Use integration to find the area of the triangular region
bounded by the lines y = 2x + 1, y = 3x + 1 and x = 4
...
Group Problem
...
Find this area
...
Hint: The part of a circle cut off by a line is a
circular sector with a triangle removed
...
Find the area bounded by the parabola x2 − 2 = y
and the line x + y = 0
...
Where do the graphs of f (x) = x2 and g(x) =
3/(2 + x2 ) intersect? Find the area of the region which
lies above the graph of g and below the graph of g
...
Cavalieri’s principle and volumes of solids
You can use integration to derive the formulas for volumes of spheres, cylinder, cones, and many many
more solid objects in a systematic way
...
”
3
...
Example – Volume of a pyramid
...
Our strategy will be to divide the pyramid into thin
horizontal slices whose volumes we can compute, and to add the volumes of the slices to get the volume of
the pyramid
...
e
...
The k th slice consists of those points on the pyramid whose height is between xk−1 and xk
...
Therefore the bottom of the k th slice is a square with side 1 − xk−1 , and its top is a
square with side 1 − xk
...
Thus the k th slice contains a block of height ∆xk whose base is a square with sides 1 − xk , and its volume
must therefore be larger than (1 − xk )2 ∆xk
...
The volume of the slice is therefore not more than
(1 − xk−1 )2 ∆xk
...
110
1
x + ∆x
x
0
1
1
Figure 2
...
Therefore there is some ck in the interval [xk−1 , xk ] such that
volume of k th slice = (1 − ck )2 ∆xk
...
The right hand side in this equation is a Riemann sum for the integral
1
(1 − x)2 dx
I=
0
and therefore we have
I = lim (1 − c1 )2 ∆x1 + · · · + (1 − cN )2 ∆xN = V
...
3
3
...
General case
...
The strategy always consists of dividing the solid into many thin (horizontal)
slices, compute their volumes, and recognize that the total volume of the slices is a Riemann sum for some
integral
...
To be more precise, let a and b be the heights of the lowest and highest points on the solid, and let
a = x0 < x1 < x2 <
...
Such a partition divides the
solid into N distinct slices, where slice number k consists of all points in the solid whose height is between
xk−1 and xk
...
If
A(x) = area of the intersection of the solid with the plane at height x
...
111
b
Area = A(x)
x
Slice at height x
a
Figure 3
...
The volume of one slice is approximately the product
of its thickness (∆x) and the area A(x) of its top
...
The total volume of all slices is therefore approximately
V ≈ A(c1 )∆x1 + · · · + A(cN )∆xN
...
···
On the other hand the sum on the right is a Riemann sum for the integral I =
exactly this integral
...
a
Figure 4
...
Both solids consist of a pile of horizontal slices
...
This operation does not affect the volumes of the slices, and hence both solids have the same volume
...
3
...
The formula (63) for the volume of a solid which we have just derived
shows that the volume only depends on the areas A(x) of the cross sections of the solid, and not on the
particular shape these cross sections may have
...
This principle is often illustrated by considering a stack of coins: If you put a number of coins on top of
each other then the total volume of the coins is just the sum of the volumes of the coins
...
e
...
112
r = f (x)
3
...
Solids of revolution
...
One class of solids for which the areas of the
cross sections are easy are the so-called “solids of revolution
...
A solid of revolution consists of all points in three-dimensional space whose distance r to the
x-axis satisfies r ≤ f (x)
...
More precisely, let f be a function which is defined on an interval [a, b] and which is always positive (f (x) > 0
for all x)
...
Yet another way of describing the solid of revolution is to say that the solid is the union of all discs which
meet the x-axis perpendicularly and whose radius is given by r = f (x)
...
Each slice is a disc of radius r = f (x) so that its area is A(x) = πr2 = πf (x)2
...
V =π
a
4
...
1
...
Consider the solid obtained by revolving the region
R = (x, y) | 0 ≤ x ≤ 2,
around the y-axis
...
Computing the volume of the solid you get when you revolve the region R around the y-axis
...
1
R
1
2
Solution: The region we have to revolve around the y-axis consists of all points above the parabola
y = (x − 1)2 but below the line y = 1
...
e
...
You can see it in the figure below: if you cut the region R horizontally
at height y you get the line segment AB, and if you rotate this segment around the y-axis you get the grey
ring region pictured below the graph
...
These radii are the two solutions of
y = (1 − r)2
so they are
rin = 1 −
√
y,
rout = 1 +
√
y
...
Rotation axis
SIDE VIEW:
y = (x − 1)2
1
A
B
xin 1
xout2
y
x
TOP VIEW:
rin
2
2
Area= πrout − πrin
rout
Problem 2: Rotate the line segment AB around
the vertical line x = −1 and you get a washer
...
The y-values which occur in the solid are 0 ≤ y ≤ 1 and hence the volume of the solid is given by
1
V =
1
A(y)dy = 4π
0
√
0
y dy = 4π ×
2
3
=
8π
...
2
...
Find the volume of the solid of revolution
obtained by revolving the same region R around the line x = −1
...
The height of each
plane will be called y
...
rin = 1 + xin = 2 − y,
The volume is therefore given by
1
1
2
2
πrout − πrin dy = π
V =
0
0
16π
√
8 y dy =
...
3
...
Compute the volume of the solid you get when
you revolve the same region R around the line y = 2
...
A typical slice is obtained by revolving the line segment AB about the line y = 2
...
rin = 1,
The area of the slice is therefore
A(x) = π 2 − (1 − x)2
2
− π12 = π 3 − 4(1 − x)2 + (1 − x)4
...
Volumes by cylindrical shells
Instead of slicing a solid with planes you can also try to decompose it into
cylindrical shells
...
Therefore the volume of a cylindrical shell of height
h, (inner) radius r and thickness ∆r is
r
∆r
h
πh(r + ∆r)2 − πhr2 = πh(2r + ∆r)∆r
≈ 2πhr∆r
...
By partitioning the interval a ≤ x ≤ b into many small intervals we can decompose the
solid into many thin shells
...
Adding the
volumes of the shells, and taking the limit over finer and finer partitions we arrive at the following formula
for the volume of the solid of revolution:
b
(65)
V = 2π
xf (x) dx
...
V = 2π
a
116
y
x
Figure 7
...
This particular tent is obtained
by rotating the graph of y = e−x , 0 ≤ x ≤ 1 around the y-axis
...
1
...
The region R from §4
...
The volume of the solid which we already computed in §4
...
1
...
Exercises
475
...
Find the volume of the solids you get by rotating
each of the following graphs around the x-axis:
What do the dots in “lim··· ” in equation (62) stand
for? (i
...
what approaches what in this limit?)
477
...
f (x) = 2 − x, 0 ≤ x ≤ 2
Draw and describe the solids whose
volume you are asked to compute
in the following problems:
479
...
f (x) = sin x, 0 ≤ x ≤ π
476
...
Here R and H are positive constants
...
481
...
f (x) = cos x, 0 ≤ x ≤ π
(!!)
483
...
484
...
485
...
486
...
(a) Find the volume of the solid obtained by rotating
the triangle T around the x-axis
...
Find the volume of the water in the bowl
...
(b) Water runs into a spherical bowl of radius 5 ft at
the rate of 0
...
How fast is the water level rising
when the water is 4 ft deep?
(c) Find the volume that results by rotating the
triangle T around the line x = −1
...
Distance from velocity, velocity from acceleration
7
...
Motion along a line
...
Therefore the position is an antiderivative of the velocity,
and the fundamental theorem of calculus says that
tb
v(t) dt = x(tb ) − x(ta ),
(66)
ta
or
tb
x(tb ) = x(ta ) +
v(t) dt
...
Equation (66) can also be obtained using Riemann sums
...
Let ∆sk be the distance travelled
during the time interval (tk−1 , tk )
...
During this time
interval the velocity v(t) need not be constant, but if the time interval is short enough then we can estimate
the velocity by v(ck ) where ck is some number between tk−1 and tk
...
e
...
The right hand side is again a Riemann sum for the integral in (66)
...
ta
The return of the dummy
...
e
...
For instance, you
might want to express the fact that the position x(t) is equal to the initial position x(0) plus the integral of
the velocity from 0 to t
...
The latter is a dummy
variable (see §8 and §3
...
To fix this formula we should choose a different letter or symbol for the integration
variable
...
So you can write
t
¯ ¯
v(t) dt
x(t) = x(0) +
0
118
7
...
Velocity from acceleration
...
v(t) = v(0) +
0
Conclusion: If you know the acceleration a(t) at all times t, and also the velocity v(0) at time t = 0, then you
can compute the velocity v(t) at all times by integrating
...
3
...
If you drop an object then it will fall, and as it falls
its velocity increases
...
This
constant is called g and is about 9
...
If we dsignate the upward direction as positive then
v(t) is the upward velocity of the object, and this velocity is actually decreasing
...
If you write h(t) for the height of the object at time t then its velocity is v(t) = h (t), and its acceleration
is h (t)
...
v(t) = v(0) +
0
Here v(0) is the velocity at time t = 0 (the “initial velocity”)
...
2
For instance, if you launch the object upwards with velocity 5ft/sec from a height of 10ft, then you have
h(0) = 10ft,
v(0) = +5ft/sec,
and thus
h(t) = 10 + 5t − 32t2 /2 = 10 + 5t − 16t2
...
To find that
5
height you compute h (t) = 5 − 32t and conclude that h(t) is maximal at t = 32 sec
...
64
7
...
Motion in the plane – parametric curves
...
This would give you two functions of t, namely,
x(t) and y(t), both of which are defined on the same interval t0 ≤ t ≤ t1 which describes the duration of the
motion you are describing
...
119
y(t)
x(t)
As an example, consider the motion described by
y(t) = sin t(0 ≤ t ≤ 2π)
...
As t increases from 0 to 2π the point (x(t), y(t)) goes around the unit circle exactly once, in the
counter-clockwise direction
...
Two motions in the plane
...
In another example one could consider
x(t) = t,
y(t) =
1 − t2 ,
(−1 ≤ t ≤ 1)
...
Unlike the previous example
we now always have y(t) ≥ 0 (since y(t) is the square root of something), and unlike the previous example the
motion is only defined for −1 ≤ t ≤ 1
...
120
y(t + ∆t)
7
...
The velocity of an object moving in the plane
...
If the object is allowed to move in the plane, so that its motion y(t)
∆x
is described by a parametric curve (x(t), y(t)), then we can differentiate both
x(t) and y(t), which gives us x (t) and y (t), and which leaves us with the
following question: At what speed is a particle moving if it is undergoing the
x(t)
x(t + ∆t)
motion (x(t), y(t)) (ta ≤ t ≤ tb ) ?
To answer this question we consider a short time interval (t, t+∆t)
...
Hence it has traveled a distance
∆s =
(∆x)2 + (∆y)2
where
∆x = x(t + ∆t) − x(t), and ∆y = y(t + ∆t) − y(t)
...
Letting ∆t → 0 you find the velocity at time t to be
(67)
2
dx
dt
v=
+
dy
dt
2
...
6
...
4
...
The velocity of this motion is therefore always the same; the point (cos t, sin t) moves along the unit circle
with constant velocity
...
4 we had x(t) = t, y(t) = 1 − t2 , so
dx
= 1,
dt
dy
−t
=√
dt
1 − t2
whence
t2
1
=√
...
v(t) =
12 +
8
...
1
...
Let (x(t), y(t)) be some parametric curve defined for ta ≤ t ≤ tb
...
At each moment in time the velocity v(t) of the
point is given by (67), and therefore the distance traveled should be
tb
(68)
s=
tb
x (t)2 + y (t)2 dt
...
Choose a partition
ta = t0 < t1 < · · · < tN = tb
P4
P3
∆y3
P2
of the interval [ta , tb ]
...
, PN (x(tN ), y(tN )), and after “connecting the dots” you get a polygon
...
The distance between
two consecutive points Pk−1 and Pk is
(∆xk )2 + (∆yk )2
∆sk =
2
∆xk
∆tk
=
∆yk
∆tk
+
2
∆tk
x (ck )2 + y (ck )2 ∆tk
≈
where we have approximated the difference quotients
∆yk
∆xk
and
∆tk
∆tk
by the derivatives x (ck ) and y (ck ) for some ck in the interval [tk−1 , tk ]
...
s=
ta
8
...
The length of the graph of a function
...
We will now find this length by representing the
graph as a parametric curve and applying the formula (68) from the previous section
...
This parametric curve traces the graph of y = f (x) from left to right as t increases from a to b
...
L=
a
The variable t in this integral is a dummy variable and we can replace it with any other variable we like, for
instance, x:
b
(69)
1 + f (x)2 dx
...
9
...
1
...
In §8 we parametrized the unit circle by
x(t) = cos t,
and computed
(0 ≤ t ≤ 2π)
y(t) = sin t,
x (t)2 + y (t)2 = 1
...
0
0
This cannot be a PROOF that the unit circle has length 2π since we have already used that fact to define
angles in radians, to define the trig functions Sine and Cosine, and to find their derivatives
...
9
...
Length of a parabola
...
While the area
under its graph was easy to compute ( 1 ), its length turns out to be much more complicated
...
dx =
0
To find this integral you would have to use one of the following (not at all obvious) substitutions1
√
1
1
x= 1 z−
(then 1 + 4x2 = (z + 1/z)2 so you can simplify the ·)
4
z
4
or (if you like hyperbolic functions)
x=
1
2
sinh w
(in which case
1 + 4x2 = cosh w
...
3
...
To compute the length of the curve given by
y = sin x, 0 ≤ x ≤ π you would have to compute this integral:
π
π
d sin x 2
dx =
1 + cos2 x dx
...
”) This happens very often with the integrals that you run into
when you try to compute the length of a curve
...
For example, since −1 ≤< cos x ≤ 1 we know that
√
√
1 ≤ 1 + cos2 x ≤ 1 + 1 = 2,
(70)
L=
1+
and therefore the length of the Sine graph is bounded by
π
π
0
i
...
π
1 + cos2 x dx ≤
1dx ≤
√
2 dx,
0
0
√
π ≤ L ≤ π 2
...
Exercises
487
...
2
488
...
One uses geometry (length of a circular
arc = radius × angle), the other uses an integral
...
Use both methods and check that you get the same
answer
...
Group Problem
...
123
Show that the Archimedean spiral, given by
Hint: you can set up integrals for both lengths
...
x(θ) = θ cos θ, y(θ) = θ sin θ, 0 ≤ θ ≤ π
has the same length as the parabola given by
y = 1 x2 ,
2
0 ≤ x ≤ π
...
Work done by a force
11
...
Work as an integral
...
e
...
If the force which acts is constant, then
the work done by this force is
Work = Force × Displacement
...
The amount of work you do is the product of the force you exert and
the length of the displacement
...
displacement
F
friction
F
push
The amount of work done by the friction is similarly the product of the friction force and the displacement
...
The work
done by the friction force is therefore negative
...
Suppose now that the force F (t) on the box is not constant, and that its motion is described by saying
that its position at time t is x(t)
...
To compute the work done by the varying force F (t) we choose a
partition of the time interval ta ≤ t ≤ tb into
ta = t0 < t1 < · · · < tN −1 < tN = tb
In each short time interval tk−1 ≤ t ≤ tk we assume the force is (almost) constant and we approximate it by
F (ck ) for some tk−1 ≤ ck ≤ tk
...
Therefore the work done by the force F during the time interval tk−1 ≤ t ≤ tk is
∆Wk = F (ck )v(ck )∆tk
...
Again we have a Riemann sum for an integral
...
124
11
...
Kinetic energy
...
If the position of the object at time t is x(t), then its velocity and acceleration are
v(t) = x (t) and a(t) = v (t) = x (t), and thus the total force acting on the object is
F (t) = ma(t) = m
dv
...
dt
Even though we have not assumed anything about the motion, so we don’t know anything about the velocity
v(t), we can still do this integral
...
dt
dt
(Remember that m is a constant
...
We get
tb
W =
m
ta
dv(t)
v(t) dt =
dt
tb
K (t) dt = K(tb ) − K(ta )
...
12
...
Therefore the total energy supplied during
a time interval t0 ≤ t ≤ t1 is the integral
t1
Energy supplied =
voltage = V(t)
I(t)V (t)dt
...
)
If a certain voltage is applied to a simple circuit (like a light bulb) then the current flowing through that
circuit is determined by the resistance R of that circuit by Ohm’s law 2 which says
I=
V
...
wikipedia
...
1
...
If the resistance of a light bulb is R = 200Ω, and if the voltage applied to it is
V (t) = 150 sin 2πf t
where f = 50sec
second?
−1
is the frequency, then how much energy does the current supply to the light bulb in one
To compute this we first find the current using Ohm’s law,
V (t)
150
I(t) =
=
sin 2πf t = 0
...
(Amp)
R
200
The energy supplied in one second is then
1 sec
I(t)V (t)dt
E=
0
1
(150 sin 2πf t) × (0
...
5
0
You can do this last integral by using the double angle formula for the cosine, to rewrite
sin2 (2πf t) =
1
2
1 − cos 4πf t =
1
2
−
1
2
cos 4πf t
...
5 ×
1
2
= 56
...
126
1
0
= 1,
2
CHAPTER 9
Answers and Hints
1 The decimal expansion of
34 (a) A(x) is an area so it has units square inch and x is
inch2
dA
is measured in
= inch
...
142857 142857 142857 · · ·
repeats after 6 digits
...
6 100x = 31
...
99
Similarly, y = 273
...
2 + 0
...
Now
very small triangle (ignore)
thin parallelogram
h
1000×0
...
4154154154 · · ·
= 15
...
0154154154 · · ·
= 15 2 + 0
...
0154154 · · · =
2
15 5
...
Ignore the area of the tiny triangle since the area of the
parallelogram will be much larger
...
∆A
h∆x
Conclusion:
≈
= h
...
From this you get
2
15 5
1076
=
...
Both are defined for all
real numbers, and both will square whatever number you
give them, so they are the same function
...
However, it is true that sin(arcsin x) = x for
all x in the interval [−1, 1]
...
However
the arcsine function always returns a number (angle) between −π/2 and π/2, so arcsin(sin x) = x can’t be true
when x > π/2 or x < −π/2
...
40 δ = min 1, 1 ε
6
41 |f (x) − (−7)| = |x2 − 7x + 10| = |x − 2| · |x − 5|
...
So, choosing δ ≤ 1 we always have |f (x) − L| < 4|x − 2|
and |f (x) − L| < ε will follow from |x − 2| < 1 ε
...
∆y = (x + ∆x)2 − 2(x + ∆x) + 1 − [x2 − 2x + 1]
42 f (x) = x3 , a = 3, L = 27
...
Therefore you can factor out x − 3 from x3 − 27 by doing
a long division
...
∆y
dy
32 ∆x : feet
...
∆x and dx are measured in
pounds per feet
...
|f (x) − L| = |x3 − 27| = |x2 + 3x + 9| · |x − 3|
...
Then |x − 3| < δ will imply
2 < x < 4 and therefore
|x2 + 3x + 9| ≤ 42 + 3 · 4 + 9 = 37
...
69 There are of course many examples
...
3)
1
Hence, if we choose δ = min 1, 37 ε then |x − 3| < δ
guarantees |x3 − 27| < ε
...
You have
√
√
√
( x − 2)( x + 2)
x−4
√
x−2=
= √
x+2
x+2
and therefore
1
|x − 4|
...
If we do that then for |x − 4| < δ we always have
3 < x < 5 and hence
1
1
√
< √
,
x+2
3+2
√
since 1/( x + 2) increases as you decrease x
...
√
A smarter solution: We can replace 1/( x + 2) by a
constant in (73), because for all x in the domain of f we
√
have x ≥ 0, which implies
1
1
√
≤
...
45 Hints:
√
x+6−9
x−3
x+6−3= √
= √
x+6+3
x+6+3
so
√
1
| x + 6 − 3| ≤ 3 |x − 3|
...
4+x
2
4+x
1
70 False! Here’s an example: f (x) = x and g(x) =
1
x − x
...
71 False again, as shown by the example f (x) = g(x) =
1
...
sin α
sin 2α
sin 2α
2α
2α
Other approach:
= sin α ·
...
80
3
...
Answer: the limit is 1
...
84 Hint: substitute θ =
−1
...
Answer:
90 Substitute θ = x − π/2 and remember that cos x =
cos(θ + π ) = − sin θ
...
θ→0 − sin θ
cos x
91 Similar to the previous problem, once you use tan x =
sin x
...
cos x
93 Substitute θ = x − π
...
θ→0
θ→0
x−π
θ
θ
Here you have to remember from trigonometry that
sin(π + θ) = − sin θ
...
Dividing by x then gives you a quantity which goes to zero
...
x
x
x
Since both −1/x and 1/x go to zero as x → ∞ the
function in the middle must also go to zero
...
7
4+x
5
Therefore
x−2
1
< 5 |x − 2|,
4+x
1
so if we want |f (x) − 2 | < ε then we must require
|x − 2| < 5ε
...
x
1
97 No
...
Therefore, no matter how you choose k, it will never be
1
true that limx→0 sin x = k, because the limit doesn’t
exist
...
51 The equation (7) already contains a function f , but
that is not the right function
...
2
67 A( 2 , −1); B( 2 , 1); C( 2 , −1); D(−1, 0); E(− 5 , −1)
...
e
...
This only happens if you choose
2x
A = 1
...
1
Since tan2 x = cos2 x − 1 one has g (x) = f (x) and
g (x) = f (x)
...
π
π
155 f (x) = − 2 cos
x
x
156 f (x)
=
cos(cos 3x) · (− sin 3x) · 3
−3 sin 3x cos cos 3x
...
If x = 0 then f (x) isn’t even
defined
...
This doesn’t contradict the IVT, because the function
isn’t continuous, in fact it isn’t even defined at x = 0, so
the IVT doesn’t have to apply
...
Consider the
example f (x) = x4 , and see the next problem
...
At such
a point you must have f (x) = 0, but by itself that it is
no enough
...
g
...
The second is impossible, since f is the derivative of f ,
so f (x) = 0 for all x implies that f (x) = 0 for all x
...
Only sign change at x = −1,
not at x = 0
...
No global max or min
...
3
No global max or min
...
231 mirror image of previous problem
...
Global min at x = 0; graph is convex, no
inflection points
...
234 Sign change at x = 0; function is always increasing
so no stationary points; inflection point at x = 0
...
inflection point at x = 0
...
For x > −1
sign change at x = 0, no stationary points, no inflection points (graph is concave)
...
For x < −1 no sign change , function is increasing and
convex, horizontal asymptote with limx→−∞ f (x) = 1
...
238 y = 0 at x = 0 but no sign changes anywhere; x = 0
is a global min; there’s no local or global max; two in√
flection points at x = ± 1 3; horizontal asymptotes at
3
height y = 1
...
For x > −1 the graph is
√
convex and has a minimum at x = −1 + 2; for x < −1
√
the graph is concave with a maximum at x = −1 − 2
...
240 Not def’d at x = 0
...
For x > 0 convex with minimum at x = 1, for
x < 0 concave with maximum at x = −1
...
At C one has x = − 2 ,
x
x
x
3
so cos π = 0 and sin π = −1
...
162 v(x) = f (g(x)) = (x + 5)2 + 1 = x2 + 10 x + 26
w(x) = g(f (x)) = (x2 + 1) + 5 = x2 + 6
p(x) = f (x)g(x) = (x2 + 1)(x + 5) = x3 + 5x2 + x + 5
q(x) = g(x)f (x) = f (x)g(x) = p(x)
...
e
...
So
sin 8x and sin(−8x) = − sin 8x are the two solutions you
find this way
...
All functions
of the form f (x) = A sin(8x + b) satisfy (†)
...
(b) S (t) is the rate with which Bob’s side grows with
time
...
Quantity
t
S(t)
V (t)
S (t)
V (t)
Units
minutes
inch
inch3
inch/minute
inch3 /minute
(c) Three versions of the same answer:
V (t) = f (S(t)) so the chain rule says V (t) =
f (S(t))S (t)
V (t) = S(t)3 so the chain rule says V (t) = 3S(t)2 S (t)
dV
dS
V = S 3 so the chain rule says
= 3S 2
...
Since V = S 3
we get S = 2
...
6
208 At x = 3
...
210 At x = a + 2a3
...
2
215 False
...
If x = 0 then this is the same as
x
x2 + |x| = 0, which has no solutions (both terms are
129
a line segment of length 1 and width 0, or the other way
2
around
...
241 Not def’d at x = 0
...
No stationary points
...
Non inflection points, no
horizontal asymptotes
...
Therefore the height plus twice the width
is f (x) = x + 2y = x + 2/x
...
e
...
This happens
0,
for x = 2
...
Inflection point at
x = −2/3
...
271 Perimeter is 2R + Rθ = 1 (given), so if you choose
the angle to be θ then the radius is R = 1/(2 + θ)
...
The smallest area arises when
you choose θ = 0
...
Hence the area will be A(R) = θR2 = R2 (1/R) − 2) =
R − 2R2
...
Again we note that this answer is
reasonable because values of θ > 2π don’t make sense,
but θ = 2 does
...
One zero
is obvious, namely at x = 0
...
The derivative is y = 4x3 − 3x2 − 1
...
So: one
stationary point at x = 1, which is a global minimum
The second derivative is y = 12x2 − 6x; there are two
1
inflection points, at x = 2 and at x = 0
...
Let’s call it
I(x)
...
Therefore
245 Again one obvious solution to y = 0, namely x = 0
...
The derivative is y = 4x3 − 6x2 + 2 which is also cubic,
but the coefficients add up to 0, so x = 1 is a root
...
There are three stationary points: local minima at x = 1,
√
√
x = − 1 − 1 3, local max at x = − 1 + 1 3
...
I(x) =
1000
125
+
x2
(1000 − x)2
(b) Find the minimum of I(x) for 0 < x < 1000
...
I (x) = 0 has one solution, namely, x = 1000
...
If you don’t like looking at signs, you could
instead look at the second derivative
246 Global min at x = 0, no other stationary points;
function is convex, no inflection points
...
247 The graph is the upper half of the unit circle
...
No horizontal asymptotes since
√
limx→±∞ 4 1 + x2 = ∞(DNE)
...
273 r =
√
50/3π, h = 100/(3πr) = 100/ 150π
...
Local min at x = 1 ln 2
...
limx→±∞ y = ∞, no asymptotes
...
285 dy/dx = 3e3x − 4ex
...
2
3
d2 y/dx2 = 9e3x − 4ex changes sign when e2x = 4 , i
...
at
9
x = 1 ln 4 = ln 2 = ln 2 − ln 3
...
limx→−∞ f (x) = 0 so negative x axis is a horizontal
asymptote
...
no asymptote there
...
Absolute max at x = π/4,
abs min at x = 5π/4
...
Note that sin x + cos x = 2 sin(x + π )
...
Global max at x = π/2, local max at x = 3π/2, global
min at x = 7π/6, 11π/6
...
Thus the area
2
enclosed is A(x) = x( 1 − x), and we’re only interested
2
1
in values of x between 0 and 2
...
) The minimal area occurs when either x = 0 or x = 1/2
...
Let’s figure out what sign
erf(−1) has (for instance)
...
You can’t change their
widths, but you can change their heights by changing
the ci
...
Since f appears to be decreasing,
the heights f (ci ) will be smallest when ci is as large as
possible
...
e
...
, c6 = x5
...
0
Note that in this integral the upper bound (−1) is less
than the lower bound (0)
...
π −1
The integral we have here is positive because it’s an integral of a positive function from a smaller number to
b
a larger number, i
...
it is of the form a f (x)dx with
f (x) ≥ 0 and with a < b
...
487 The answer is π/6
...
You get f (x) = −x/ 1 − x2 , so
1
...
2
373 (a) The first derivative of erf(x) is, by definition
2
2
erf (x) = √ e−x ,
π
so you get the second derivative by differentiating this:
2
−4x
erf (x) = √ e−x
...
131
GNU Free Documentation License
Version 1
...
http://fsf
...
Preamble
The “Title Page” means, for a printed book, the title page itself,
plus such following pages as are needed to hold, legibly, the material
this License requires to appear in the title page
...
The “publisher” means any person or entity that distributes copies
of the Document to the public
...
(Here
XYZ stands for a specific section name mentioned below, such
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We have designed this License in order to use it for manuals for free
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We recommend
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APPLICABILITY AND DEFINITIONS
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Add an appropriate copyright notice for your modifications adjacent to the other copyright notices
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Preserve all the Invariant Sections of the Document, unaltered in their text and in their titles
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In the combination, you must combine any sections Entitled “History” in the various original documents, forming one section Entitled “History”; likewise combine any sections Entitled “Acknowledgements”, and any sections Entitled “Dedications”
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FUTURE REVISIONS OF THIS LICENSE
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You may combine the Document with other documents released under this License, under the terms defined in section 4 above for modified versions, provided that you include in the combination all of
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If there are multiple Invariant Sections with the same name
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133
a version number of this License, you may choose any version ever
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