**Unformatted text preview: **Calculus I
Lecture Notes for
MATH 1013
(Fall 2015) Frederick Tsz-Ho Fong
Department of Mathematics
Hong Kong University of Science and Technology Contents 1 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Basic Notions 1 1.2 Compositions and Inverses 9 1.3 Logarithmic Functions 15 1.4 Inverse Trigonometric Functions 17 2 Limit and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.1 Limit of a Function 21 2.2 Continuity 31 2.3 Concept of Derivatives 35 3 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 3.1 Basic Derivatives 39 3.2 Differentiability 43 3.3 Differentiation Rules 47 3.4 Derivatives of Trigonometric Functions 49 3.5 Chain Rule 51 3.6 Implicit Differentiation 54 4 Applications of Differential Calculus . . . . . . . . . . . . . . . . . . . . . . . . . 61 4.1 Rate of Change 61 4.2 Numerical Methods 65 4.3 l’Hospital’s Rule 69 4.4 Derivatives and Graphs 72 4.5 Optimization 80 4.6 Mean Value Theorem 82 5 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 5.1 Anti-Derivatives 85 5.2 Definite Integrals 88 5.3 Fundamental Theorem of Calculus 96 5.4 Integration by Substitutions 101 1 — Functions 1.1 Basic Notions 1.1.1 What is a function? 12 |||| CHAPTER 1 FUNCTIONS AND MODELS It’s helpful to think of a function as a ma
the function f, then when x enters the machi
produces an output f &x' according to the ru
domain as the set of all possible inputs and t
FIGURE 2
Theafter
preprogrammed
A function can be regarded
as a machine
give it ƒan input, then
processing it functions in a calcu
Machine
diagram –foryou
a function
according to a defined rule, the machine gives you an output. We often
denote For
the example,
input by athe square root key o
machine.
variable x, and the output by f ( x ). For instance, we can write:
You press the key labeled s (or sx ) and e
domain
of this function; that is, x is not an a
f ( x ) = x2 + 1
cate an error. If x # 0, then an approximatio
your
calculator
to represent the “machine” that takes the input x and output the value
+ 1.onAs
such,
if we is not quite the sam
sxx2key
2
input the value 2 into this function, it will output the value 2 + 1, orby
equivalently,
f &x' ! sx 5.. In short,
we may simply write:
Another way to picture a function is by an
f (2) = 5.
ƒ
x
connects an element of D to an element of E
Likewise, it is easy to verify that fa(1) = 2, f (0) = 1, etc.
with x, f &a' is associated with a, and so on.
f(a)
The most common method for visualizing
Depending on the physical meaning, we can pick other suitable pair of letters to denote a
i
domain
D, then its graph is the set of ordere
function and its variable instead of the generic f and x. For instance,
x
(input) f ƒ
(output) • A(r ) = pr2 , where r means the radius of a circle and A means the area of this circle.
• h(t), where t means the time and fh means your height at time t. D !&x, f &x'' E (Notice that these are input-output pairs.) In FIGURE 3
1.1.2 Some terminology
points &x, y' in the coordinate plane such tha
Arrow
for set
ƒ of allowable inputs. For example, the function
1. The domain of a function
f ( x )diagram
means
the
The graph of a function f gives us a usef
p
defined by the expression f ( x ) = x has domain given by the interval:
a function. Since the y-coordinate of any poin [0, •), or equivalently, {x : x 0} the value of f &x' from the graph as being th Figure 4). The graph of f also allows us to p 1
which means the set of all non-negative numbers. The functionrange
g( x ) =
domain
on1the
as in Figure 5.
y-axis
x has
given by
( •, 1) [ (1, •)
y
{ x, ƒ}
which means the set of all real numbers except 1. r ƒ
f (2) f(1)
0 1 2 x x Functions 2
i If every real number is an allowable input for a function f ( x ), then we can denote
the domain of f ( x ) by ( •, •), or simply R. i Consider the function A(r ) = pr2 which is the area of a circle with radius r.
Although it is mathematically legitimate to input a negative r into A(r ), a circle
with negative or zero radius is not physically meaningful. Therefore, we can take
(0, •) to be the domain of A(r ), instead of R. i If D is the domain of a function f ( x ), then any (non-empty) subset E of D can also
taken to be the domain of f ( x ). When the domain of a function f ( x ) is not specified,
the domain is usually taken as the largest set in which f ( x ) is defined.
p
Take f ( x ) = x as an example. Since [0, •) can be taken to be the domain of f ( x ),
we can also declare that its subset [1, •)pto be the domain of f ( x ). However, without
any declaration, the domain of f ( x ) = x is taken to be [0, •) by default, since it is
the largest possible domain. 2. The codomain of a function f ( x ) is the set where the outputs belong to. In this course, the
p
output of a function is usually a real number, as in the examples f ( x ) = x, g( x ) = 1 1 x
and A(r ) = pr2 we have seen so far. We may simply say R, or ( •, •), is the codomain
of f ( x ), g( x ) and A(r ).
i The codomain of a function simply indicates what kind of objects the outputs are,
but it does not mean that everything in the codomain is a possible output.
Take the function h( x ) = x2 + 1 as an example. Given any input x, the output x2 + 1
is always positive (in fact at least 1). However, it is perfectly fine to say that the
codomain of h is R. 3. A function f ( x ) with domain D and codomain C is usually denoted by:
f :D!C
For example, we may indicate the domain and codomain of the function g( x ) =
writing:
g : ( •, 1) [ (1, •) ! R 1
1 x by 4. The range of a function f : D ! C is the set of all achievable outputs. It is usually denoted
by f [ D ] where D is the domain of f . In set notations, the range of f is usually defined by:
f [D] = { f (x) : x 2 D} For example, the range of h : R ! R defined by h( x ) = x2 + 1 is given by h[R ] = [1, •).
i However, if we declare the domain of h to be a smaller set such as [2, 3], then the
range of h will be smaller as it can only output values between 22 + 1 and 32 + 1.
Therefore, the range of h : [2, 3] ! R becomes [5, 10]. i There is no universal way to find the range of a given function, although it is easy
to do so in the above examples. One application of Calculus is to help us find the
range of a given function. Another way to find the range is to look at the graph of
the function which can be plotted by computer software. 1.1.3 Graph of a function
The graph of a function is a geometric way of representing a function. It is a curve on the
xy-coordinate plane which looks like the one in Figure 1.1.
To read off the value of f (1) from the graph, simply locate the point on the curve with
1 as the x-coordinate, then the y-coordinate of that point will be the value f (1). Similar for
f (2) and all other values f ( x ). Many computer softwares are able to plot the graph of a given
function. With the graph, it is easy to find (or at least to estimate) the range of the function
(see Example 1.1) FIGURE 3 points &x, y' in the coordinate plane such that y ! f &x' and x is in the domain of f .
The graph of a function f gives us a useful picture of the behavior or “life history
a function. Since the y-coordinate of any point &x, y' on the graph is y ! f &x', we can
the value of f &x' from the graph as being the height of the graph above the point x
Figure 4). The graph of f also allows us to picture the domain of f on the x-axis an
3
range on the y-axis as in Figure 5. Arrow diagram for ƒ 1.1 Basic Notions y y { x, ƒ} range ƒ
f (2) f (1)
0 1 2 x x 0 FIGURE
Figure4 1.1: graph of function
y domain x FIGURE 5 EXAMPLE 1 The graph of a function f is shown in Figure 6. ⌅ (a) possible)
Find the domain
values ofoffthe
and f &5'.defined by the expression:
&1' function
Example 1.1 Find the (largest
(b) What are the
s domain and range of f ?
x2 x 2
x
(a) We see from Figure36 that the point &1, 3' lies on the graph of f , so the value of SOLUTION f ( x ) = 1
0 y ! ƒ(x) 1 x at 1 is f &1' ! 3. (In other words, the point on the graph that lies above x ! 1 is 3 un By plotting the graph of f using computer softwares, estimate the range of f when the
abovepossible.
the x-axis.)
domain is taken to be the largest When x ! 5, the graph lies about 0.7 unit below the x-axis, so we estimate that
f &5' ( !0.7.
FIGURE 6 ⌅ Solution First we observe that x 3 is a divisor so we need to exclude 3 from the domain.
(b) We see that f &x' is defined when 0 "2 x " 7, so the domain of f is the closed int
Then, in order for the square root to be meaningful, we need to ensure x x x3 2 is non-negative,
so we solve the inequality: val $0, 7%. Notice that f takes on all values from !2 to 4, so the range of f is
x2 The notation for intervals is given in
Appendix A. N 2 x
x 3 " !y !2 " y " 4# ! $!2, 4% 0. It is standard arithmetics that the quotient of two numbers is non-negative if and only if both
numbers have the same sign (both 0, or both 0). Therefore, we can solve the inequality
by dividing it into two cases:
Case 1: x > 3
In this case the denominator x 3 is positive, so we need to have x2 x 2 0.
x2 x ( x + 1)( x
( 2 0 2) 0 (factorization) x+1 0
x 2 0
(
x
1
x 2
x or 2 ( x+1 0
x 20 or ( or x x 1
x2 As x > 3 in this case, and ( x > 3) and ( x 2 or x in this case is x > 3 .
Case 2: x < 3
In this case we need to have x2 x 2 0:
x2 x ( x + 1)( x 20 2) 0 1 1) simply means x > 3, the solution Functions 4
( x+1 0
x 2 0
(
x 1
x 2 or
or 1x2 (
( x+1 0
x 20
x
1
x2 Since the interval [ 1, 2] lies within x < 3, the solution in this case is simply
Overall, the domain of f is given by inequalities 1 x 2 or x > 3, i.e. 1x2. [ 1, 2] [ (3, •). 6 4 2 -2 2 4 6 8 10 From the graph of f , it suggests that the range of f should be in the form of [0, a] [ [b, •).
Later in the course, we will learn how to determine the exact values of a and b. In fact, the
range of f can be shown to be [0, 1] [ [3, •).
1.1.4 Well-definedness
When we define a function, it is important to keep in mind that:
Each input in the domain must have exactly one output.
It is the case for all examples we have seen so far. An expression with the property that every
input in the domain has exactly one output is said to be well-defined. The expressions below are
not well-defined and so they cannot be regarded as functions.
1. Let the domain be [0, •). If we define the output f ( x ) to be the number y such that
y2 = x, then such an f ( x ) is not well-defined. Here is the reason. If we let the input to be
1, then f (1) is the number y such that y2 = 1. There are two possibilities, namely y = 1
or y = 1. Therefore, there are two possible values for f (1), and so such an f is not
well-defined and cannot be regarded as a function.
p
2. Let the domain be Q, the set of all rational numbers. Given any input q where p and q
are integers, we define
✓ ◆
p
f
= p.
q
Given an input 12 , we have f ( 12 ) = 1, it seems that the output is simply 1. However, the
input 12 has many other forms, such as 24 that gives f ( 24 ) = 2 as output. The fractions 12
and 24 are regarded as the same input just expressed in two different ways. Now that this
input has at least two possible outputs, so f is not well-defined and cannot be regarded
as a function. t!x" ! 1
1
!
The
graph
of a function is a curve in the xy-plane. But the question arises: Which curves
x !x
x!x ! 1"
2 in the xy-plane are graphs of functions? This is answered by the following test. and division by 0 is not allowed, we see that t!x" is not defined when x ! 0 or x ! 1.
Thus the domain of t is
THE VERTICAL LINE TEST A curve in the xy-plane is the graph of a function of x if $ 1.1 Basic Notions
#x x " 0, xand
" only
1% if no vertical line intersects the curve more than once. 5 Given a curve on the xy-plane, one can determine whether or not it is the graph of a which could also be
writtenby
in so-called
interval notation
as reason
The
for the truth of the Vertical Line Test can be seen in Figure 13. If each verfunction
the vertical
line test:
tical line x ! a intersects a curve only once, at !a, b", then exactly one functional value
(Vertical Line Test) A curve on the xy-plane is the graph of a function of x if and only if no
M
!!$, 0" ! !0,
! !1,by$"f !a" ! b. But if a line x ! a intersects
is 1"
defined
the curve twice, at !a, b" and !a, c",
vertical line intersects the curve at more than one point (see Figures 1.2 and 1.3).
then the curve can’t represent a function because a function can’t assign two different valto a. But the question arises: Which curves
The graph of a function is a curve in the xyues
-plane. in the xy-plane are graphs of functions? This is answered by the following test.
y y x=a (a, c) x=a THE VERTICAL LINE TEST A curve in the xy-plane is the graph of a function of x if and only if no vertical line intersects the curve more than once. (a, b) (a, b) x
x
a 13. If each
a
0 be seen in Figure
The reason for the truth of the Vertical Line Test can
ver- 0
FIGURE 13
tical line x ! a intersects a curve
only once, at !a, b", then exactly one functional value
is defined by f !a" ! b. But if a line x ! a intersects the curve twice, at !a, b" and !a, c",
Figure 1.2: pass the vertical line test, hence2a graph of function
then the curve can’t represent a function because
function the
can’t
assignxtwo
val-in Figure 14(a) on the next page is not the
Fora example,
parabola
! ydifferent
! 2 shown
ues to a.
graph of a function of x because, as you can see, there are vertical lines that intersect the y parabola twice. The parabola, however, does contain the graphs of two functions of x.
y
Notice that
the equation x ! y 2 ! 2 implies y 2 ! x " 2, so y ! #sx " 2 . Thus the
x=a
upper and lower halves of the(a, c)
parabola are the graphs of the functions f !x" ! s x " 2
[from Example 6(a)] and t!x" ! !s x " 2 . [See Figures 14(b) and (c).] We observe that
if we reverse the roles of x and y, then the equation x ! h!y" ! y 2 ! 2 does define x as a
(a, b)
function of y (with y as the independent
variable and x as the dependent variable) and the
parabola now appears as the graph of the function h. x=a
(a, b) a 0 x 0 a x Figure 1.3: fail the vertical line test, hence not a graph of function For example, the parabola x ! y 2 ! 2 shown in Figure 14(a) on the next page is not the
graph of a function of x because, as you can see, there are vertical lines that intersect the
Although
each input
a function
haveofexactly
one output,ofit x
is. perfectly fine for a
parabola twice. The parabola,
however,
does ofcontain
themust
graphs
two functions
i
2 to give the same2output for two or more different inputs. For instance, consider the
function
Notice that the equation x ! y ! 2 implies y ! x " 2, so y ! #sx " 2 . Thus the
function g : R ! R defined by
upper and lower halves of the parabola are the graphs of the
functions f !x" ! s x " 2
g( x ) = x2 + 1.
[from Example 6(a)] and t!x"
[See
Figures
14(b)
and (c).] We observe that
!
!
x
"
2
.
Evidently,swe have both g(1) = 2 and g( 12) = 2. It is allowed.
if we reverse the roles of x and y, then the equation x ! h!y" ! y ! 2 does define x as a
function of y (with y as the independent variable and x as the dependent variable) and the
1.1.5
Piecewise
Defined
Functions
parabola now
appears
as the graph
of the
function h. Examples of functions we have seen so far are defined using a simple expression such as
f ( x ) = x2 + 1. However, in practice we sometimes define a function in a piecewise way, meaning
that the function can have different expressions on different part of the domain. Let’s look at an
example:
Example 1.2 The monthly bill of a smartphone data plan is calculated as follows: At a
monthly charge of HK$168, you get 1024MB of data usage for that month. If you go over
1024MB, you will be additionally charged at the rate of HK$0.5 per extra MB. Denote x to be
your monthly usage. Express your month bill amount as a function f ( x ).
⌅ Solution First the domain is clearly [0, •) as your usage cannot be negative. When
0 x 1024, you will be charged at the standard monthly rate of HK$168, so f ( x ) = 168 if
0 x 1024. When x > 1024, you have overused x 1024 (MB) and so you will be charged
by extra 0.5( x 1024) dollars. Therefore, f ( x ) = 168 + 0.5( x 1024) in HKD if x > 1024.
⌅ Functions 6
We can write down the expression of this function in the following way:
(
168
if 0 x 1024
f (x) =
168 + 0.5( x 1024) if x > 1024 A classic example of piecewise-defined function is the absolute value function, defined
piecewise by:
(
x
if x 0
|x| =
x if x < 0
For example, we can easily see that |1| = 1 and | 3| = ( 3). In short, this function turns
any negative number to its positive counterpart, and preserve all non-negative numbers. The
1 FUNCTIONS AND MODELS
graph of the absolute value function18f ( x||||
) = |CHAPTER
x | is shown
in Figure 1.4. EXAMPLE 8 Sketch the graph of the absolute valu y SOLUTION From the preceding discussion we know y=| x | $x$ !
0 x FIGURE
16
Figure
1.4: graph
of f ( x ) = | x |. • | xy| = | x | |y| and = x
if
$x if Using the same method as in Example 7, we see
line y ! x to the right of the y-axis and coincides
y-axis (see Figure 16). EXAMPLE 9 Find a formula for the function f gra It is useful to keep in mind that
x
y # |x|
;
|y| y • | x | = | x |;
• | x + y| | x | + |y| and | x y| | x | + |y|.
However, beware that generally we only have | x + y| | x | + |y|, and equality holds only when
x and y are of the same sign, i.e. both 0, or both 0.
FIGURE 17 Solving inequalities involving absolute values 1
0 1 To solve an inequality such as |2x 1| < 3, or |4 x | 2, etc., it is useful to keep in mind the
SOLUTION The line through !0, 0" and !1, 1" has slo
two facts:
equation is y ! x. Thus, for the part of the graph
If a is non-negative, we have:
• |y| a if and only if a y a.
f !x" ! x
if 0
• |y| a if and only if y a or y a.
If b is negative, we have:
The line through !1, 1" and !2, 0" has slope m !
• |y| b is always false!
Point-slope form of the equation of a line:
• |y| b is always true!
y $ y1 ! m!x $ x 1 "
y $ 0 ! !$1"!x $ 2"
N See Appendix B.
⌅ So we have Example 1.3 Determine the (largest possible) domain of the function: f (x) = ⌅ q |3x + 1| 2 Solution Again, we need to take the square root, and so we need: |3x + 1| 2
|3x + 1| 0
2 f !x" ! 2 $ x if We also see that the graph of f coincides with th
tion together, we have the following three-piece f # x
if
f !x" ! 2 $ x if
0
if EXAMPLE 10 In Example C at the beginning of t
of mailing a first-class letter with weight w. In ef because, from the table of values, we have
C
1 C!w" ! 0.39
0.63
0.87
1.11
!
!
! if
if
if
if 1.1 Basic Notions 7 From the rules mentioned above, we get: 3x + 1 x
The domain of f is ( •, 1 or 2 or 3x + 1 2, and so: 1
.
3 x 1] [ [ 13 , •). 1.1.6 Various Types of Functions Definition 1.1 — Odd and Even Functions. Let D be a subset of R which is symmetric about zero. A function f : D ! R is said to be:
• an even function if f ( x ) = f ( x ) for any x in D.
• an odd function if f ( x ) = f ( x ) for any x in D. The graph of an even function is reflexively symmetric about the y-axis (see Figure 1.5).
Important examples include:
cos x, x2 , | x | , . . . y SYMMETRY f(_x) If a function f satisfies f !!x"
" f !x"
SECTION
1.
even function. For instance, the func ƒ
y0 _x SECTION x x SYMMETRY f !!x"
If a function f satisfies f !!x" " f !x" fo
f(_x)
ƒ
even function. For instance, the functio
The geometric significance of an eve
_x
x
x
0
to the y-axis (see Figure 19).f This
!!x" me
"
Figure
1.5:19
graph of an even function
we obtain the entire graph simply by
FIGURE
If
f satisfies
f !!x" " of
!fan
!x"even
for fe
An even function
geometric
significance
The graph of an odd function is 180 -rotationally symmetric about The
the origin
(see Figure
theThis
function
1.6). Important examples include:
tofunction.
the y-axisFor
(seeexample,
Figure 19).
mean
y
FIGURE
19
sin x, tan x, An even...

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