стр. 1(всего 14)СОДЕРЖАНИЕ >>
Contents

1 Functions 2
1.1 The Concept of a Function . . . . . . . . . . . . . . . . . . . . 2
1.2 Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . 12
1.3 Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . 19
1.4 Logarithmic, Exponential and Hyperbolic Functions . . . . . . 26

2 Limits and Continuity 35
2.1 Intuitive treatment and deп¬Ѓnitions . . . . . . . . . . . . . . . 35
2.1.1 Introductory Examples . . . . . . . . . . . . . . . . . . 35
2.1.2 Limit: Formal Deп¬Ѓnitions . . . . . . . . . . . . . . . . 41
2.1.3 Continuity: Formal Deп¬Ѓnitions . . . . . . . . . . . . . 43
2.1.4 Continuity Examples . . . . . . . . . . . . . . . . . . . 48
2.2 Linear Function Approximations . . . . . . . . . . . . . . . . . 61
2.3 Limits and Sequences . . . . . . . . . . . . . . . . . . . . . . . 72
2.4 Properties of Continuous Functions . . . . . . . . . . . . . . . 84
2.5 Limits and Inп¬Ѓnity . . . . . . . . . . . . . . . . . . . . . . . . 94

3 Diп¬Ђerentiation 99
3.1 The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . 99
3.2 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 111
3.3 Diп¬Ђerentiation of Inverse Functions . . . . . . . . . . . . . . . 118
3.4 Implicit Diп¬Ђerentiation . . . . . . . . . . . . . . . . . . . . . . 130
3.5 Higher Order Derivatives . . . . . . . . . . . . . . . . . . . . . 137

4 Applications of Diп¬Ђerentiation 146
4.1 Mathematical Applications . . . . . . . . . . . . . . . . . . . . 146
4.2 Antidiп¬Ђerentiation . . . . . . . . . . . . . . . . . . . . . . . . 157
4.3 Linear First Order Diп¬Ђerential Equations . . . . . . . . . . . . 164

i
ii CONTENTS

4.4 Linear Second Order Homogeneous Diп¬Ђerential Equations . . . 169
4.5 Linear Non-Homogeneous Second Order Diп¬Ђerential Equations 179

5 The Deп¬Ѓnite Integral 183
5.1 Area Approximation . . . . . . . . . . . ...... . . . . . . 183
5.2 The Deп¬Ѓnite Integral . . . . . . . . . . . ...... . . . . . . 192
5.3 Integration by Substitution . . . . . . . . ...... . . . . . . 210
5.4 Integration by Parts . . . . . . . . . . . ...... . . . . . . 216
5.5 Logarithmic, Exponential and Hyperbolic Functions . . . . . . 230
5.6 The Riemann Integral . . . . . . . . . . ...... . . . . . . 242
5.7 Volumes of Revolution . . . . . . . . . . ...... . . . . . . 250
5.8 Arc Length and Surface Area . . . . . . ...... . . . . . . 260

6 Techniques of Integration 267
6.1 Integration by formulae . . . . . . . . . . . . . . . . . . . . . . 267
6.2 Integration by Substitution . . . . . . . . . . . . . . . . . . . . 273
6.3 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . 276
6.4 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . 280
6.5 Trigonometric Substitutions . . . . . . . . . . . . . . . . . . . 282
6.6 Integration by Partial Fractions . . . . . . . . . . . . . . . . . 288
6.7 Fractional Power Substitutions . . . . . . . . . . . . . . . . . . 289
6.8 Tangent x/2 Substitution . . . . . . . . . . . . . . . . . . . . 290
6.9 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . 291

7 Improper Integrals and Indeterminate Forms 294
7.1 Integrals over Unbounded Intervals . . . . . . . . . . . . . . . 294
7.2 Discontinuities at End Points . . . . . . . . . . . . . . . . . . 299
7.3 ......................... . . . . . . . . . 304
7.4 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . 314

8 Inп¬Ѓnite Series 315
8.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315
8.2 Monotone Sequences . . . . . . . . . . . . . . . . . . . . . . . 320
8.3 Inп¬Ѓnite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 323
8.4 Series with Positive Terms . . . . . . . . . . . . . . . . . . . . 327
8.5 Alternating Series . . . . . . . . . . . . . . . . . . . . . . . . . 341
8.6 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
8.7 Taylor Polynomials and Series . . . . . . . . . . . . . . . . . . 354
CONTENTS 1

8.8 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360

9 Analytic Geometry and Polar Coordinates 361
9.1 Parabola . . . . . . . . . . . ......... . . . . . . . . . . 361
9.2 Ellipse . . . . . . . . . . . . ......... . . . . . . . . . . 362
9.3 Hyperbola . . . . . . . . . . ......... . . . . . . . . . . 363
9.4 Second-Degree Equations . . ......... . . . . . . . . . . 363
9.5 Polar Coordinates . . . . . . ......... . . . . . . . . . . 364
9.6 Graphs in Polar Coordinates ......... . . . . . . . . . . 365
9.7 Areas in Polar Coordinates . ......... . . . . . . . . . . 366
9.8 Parametric Equations . . . . ......... . . . . . . . . . . 366
Chapter 1

Functions

In this chapter we review the basic concepts of functions, polynomial func-
tions, rational functions, trigonometric functions, logarithmic functions, ex-
ponential functions, hyperbolic functions, algebra of functions, composition
of functions and inverses of functions.

1.1 The Concept of a Function
Basically, a function f relates each element x of a set, say Df , with exactly
one element y of another set, say Rf . We say that Df is the domain of f and
Rf is the range of f and express the relationship by the equation y = f (x).
It is customary to say that the symbol x is an independent variable and the
symbol y is the dependent variable.

Example 1.1.1 Let Df = {a, b, c}, Rf = {1, 2, 3} and f (a) = 1, f (b) = 2
and f (c) = 3. Sketch the graph of f .

graph

Example 1.1.2 Sketch the graph of f (x) = |x|.
Let Df be the set of all real numbers and Rf be the set of all non-negative
real numbers. For each x in Df , let y = |x| in Rf . In this case, f (x) = |x|,

2
1.1. THE CONCEPT OF A FUNCTION 3

the absolute value of x. Recall that

x if x в‰Ґ 0
|x| =
в€’x if x < 0

We note that f (0) = 0, f (1) = 1 and f (в€’1) = 1.
If the domain Df and the range Rf of a function f are both subsets
of the set of all real numbers, then the graph of f is the set of all ordered
pairs (x, f (x)) such that x is in Df . This graph may be sketched in the xy-
coordinate plane, using y = f (x). The graph of the absolute value function
in Example 2 is sketched as follows:

graph

Example 1.1.3 Sketch the graph of
в€љ
x в€’ 4.
f (x) =

In order that the range of f contain real numbers only, we must impose
the restriction that x в‰Ґ 4. Thus, the domain Df contains the set of all real
numbers x such that x в‰Ґ 4. The range Rf will consist of all real numbers y
such that y в‰Ґ 0. The graph of f is sketched below.

graph

Example 1.1.4 A useful function in engineering is the unit step function,
u, deп¬Ѓned as follows:
0 if x < 0
u(x) =
1 if x в‰Ґ 0
The graph of u(x) has an upward jump at x = 0. Its graph is given below.
4 CHAPTER 1. FUNCTIONS

graph

Example 1.1.5 Sketch the graph of
x
f (x) = .
x2 в€’ 4
It is clear that Df consists of all real numbers x = В±2. The graph of f is
given below.

graph

We observe several things about the graph of this function. First of all,
the graph has three distinct pieces, separated by the dotted vertical lines
x = в€’2 and x = 2. These vertical lines, x = В±2, are called the vertical
asymptotes. Secondly, for large positive and negative values of x, f (x) tends
to zero. For this reason, the x-axis, with equation y = 0, is called a horizontal
asymptote.
Let f be a function whose domain Df and range Rf are sets of real
numbers. Then f is said to be even if f (x) = f (в€’x) for all x in Df . And
f is said to be odd if f (в€’x) = в€’f (x) for all x in Df . Also, f is said to be
one-to-one if f (x1 ) = f (x2 ) implies that x1 = x2 .

Example 1.1.6 Sketch the graph of f (x) = x4 в€’ x2 .
This function f is even because for all x we have

f (в€’x) = (в€’x)4 в€’ (в€’x)2 = x4 в€’ x2 = f (x).

The graph of f is symmetric to the y-axis because (x, f (x)) and (в€’x, f (x)) are
on the graph for every x. The graph of an even function is always symmetric
to the y-axis. The graph of f is given below.

graph
1.1. THE CONCEPT OF A FUNCTION 5

This function f is not one-to-one because f (в€’1) = f (1).

Example 1.1.7 Sketch the graph of g(x) = x3 в€’ 3x.
The function g is an odd function because for each x,
g(в€’x) = (в€’x)3 в€’ 3(в€’x) = в€’x3 + 3x = в€’(x3 в€’ 3x) = в€’g(x).
The graph of this function g is symmetric to the origin because (x, g(x))
and (в€’x, в€’g(x)) are on the graph for all x. The graph of an odd function is
always symmetric to the origin. The graph of g is given below.

graph

в€љ
в€љ
This function g is not one-to-one because g(0) = g( 3) = g(в€’ 3).
It can be shown that every function f can be written as the sum of an
even function and an odd function. Let
1
1
g(x) = (f (x) + f (в€’x)), h(x) = (f (x) в€’ f (в€’x)).
2 2
Then,
1
g(в€’x) = (f (в€’x) + f (x)) = g(x)
2
1
h(в€’x) = (f (в€’x) в€’ f (x)) = в€’h(x).
2
Furthermore
f (x) = g(x) + h(x).

Example 1.1.8 Express f as the sum of an even function and an odd func-
tion, where,
f (x) = x4 в€’ 2x3 + x2 в€’ 5x + 7.
We deп¬Ѓne
1
g(x) = (f (x) + f (в€’x))
2
1
= {(x4 в€’ 2x3 + x2 в€’ 5x + 7) + (x4 + 2x3 + x2 + 5x + 7)}
2
= x4 + x2 + 7
6 CHAPTER 1. FUNCTIONS

and
1
h(x) = (f (x) в€’ f (в€’x))
2
1
= {(x4 в€’ 2x3 + x2 в€’ 5x + 7) в€’ (x4 + 2x3 + x2 + 5x + 7)}
2
= в€’2x3 в€’ 5x.
Then clearly g(x) is even and h(x) is odd.
g(в€’x) = (в€’x)4 + (в€’x)2 + 7
= x4 + x2 + 7
= g(x)
h(в€’x) = в€’ 2(в€’x)3 в€’ 5(в€’x)
= 2x3 + 5x
= в€’h(x).
We note that
g(x) + h(x) = (x4 + x2 + 7) + (в€’2x3 в€’ 5x)
= x4 в€’ 2x3 + x2 в€’ 5x + 7
= f (x).
It is not always easy to tell whether a function is one-to-one. The graph-
ical test is that if no horizontal line crosses the graph of f more than once,
then f is one-to-one. To show that f is one-to-one mathematically, we need
to show that f (x1 ) = f (x2 ) implies x1 = x2 .

Example 1.1.9 Show that f (x) = x3 is a one-to-one function.
Suppose that f (x1 ) = f (x2 ). Then
0 = x3 в€’ x3
2
1
= (x1 в€’ x2 )(x2 + x1 x2 + x2 ) (By factoring)
2
1

If x1 = x2 , then x2 + x1 x2 + x2 = 0 and
2
1

x2 в€’ 4x2
в€’x2 В± 2 2
x1 =
2

в€’3x2
в€’x2 В± 2
= .
2
1.1. THE CONCEPT OF A FUNCTION 7

This is only possible if x1 is not a real number. This contradiction proves
that f (x1 ) = f (x2 ) if x1 = x2 and, hence, f is one-to-one. The graph of f is
given below.

graph

If a function f with domain Df and range Rf is one-to-one, then f has a
unique inverse function g with domain Rf and range Df such that for each
x in Df ,
g(f (x)) = x
and for such y in Rf ,
f (g(y)) = y.
This function g is also written as f в€’1 . It is not always easy to express g
explicitly but the following algorithm helps in computing g.
Step 1 Solve the equation y = f (x) for x in terms of y and make sure that there
exists exactly one solution for x.

Step 2 Write x = g(y), where g(y) is the unique solution obtained in Step 1.

Step 3 If it is desirable to have x represent the independent variable and y
represent the dependent variable, then exchange x and y in Step 2 and
write
y = g(x).

Remark 1 If y = f (x) and y = g(x) = f в€’1 (x) are graphed on the same
coordinate axes, then the graph of y = g(x) is a mirror image of the graph
of y = f (x) through the line y = x.

Example 1.1.10 Determine the inverse of f (x) = x3 .
We already know from Example 9 that f is one-to-one and, hence, it has
a unique inverse. We use the above algorithm to compute g = f в€’1 .

Step 1 We solve y = x3 for x and get x = y 1/3 , which is the unique solution.
8 CHAPTER 1. FUNCTIONS

Step 2 Then g(y) = y 1/3 and g(x) = x1/3 = f в€’1 (x).

Step 3 We plot y = x3 and y = x1/3 on the same coordinate axis and compare
their graphs.

graph

A polynomial function p of degree n has the general form

p(x) = a0 xn + a1 xnв€’1 + В· В· В· + anв€’1 x + an , a2 = 0.

The polynomial functions are some of the simplest functions to compute.
For this reason, in calculus we approximate other functions with polynomial
functions.
A rational function r has the form
p(x)
r(x) =
q(x)

where p(x) and q(x) are polynomial functions. We will assume that p(x) and
q(x) have no common non-constant factors. Then the domain of r(x) is the
set of all real numbers x such that q(x) = 0.

Exercises 1.1
1. Deп¬Ѓne each of the following in your own words.

(a) f is a function with domain Df and range Rf

(b) f is an even function

(c) f is an odd function

(d) The graph of f is symmetric to the y-axis

(e) The graph of f is symmetric to the origin.

(f) The function f is one-to-one and has inverse g.
1.1. THE CONCEPT OF A FUNCTION 9

2. Determine the domains of the following functions

x2
|x|
(a) f (x) = (b) f (x) = 3
x в€’ 27
x
в€љ x2 в€’ 1
(c) f (x) = x2 в€’ 9 (d) f (x) =
xв€’1

3. Sketch the graphs of the following functions and determine whether they
are even, odd or one-to-one. If they are one-to-one, compute their in-
verses and plot their inverses on the same set of axes as the functions.

(a) f (x) = x2 в€’ 1 (b) g(x) = x3 в€’ 1
в€љ
(d) k(x) = x2/3
9 в€’ x, x в‰Ґ 9
(c) h(x) =

4. If {(x1 , y1 ), (x2 , y2 ), . . . , (xn+1 , yn+1 )} is a list of discrete data points in
the plane, then there exists a unique nth degree polynomial that goes
through all of them. Joseph Lagrange found a simple way to express this
polynomial, called the Lagrange polynomial.
x в€’ x1
x в€’ x2
+ y2
For n = 2, P2 (x) = y1
x1 в€’ x2 x2 в€’ x1
(x в€’ x2 )(x в€’ x3 ) (x в€’ x1 )(x в€’ x3 )
For n = 3, P3 (x) = y1 + y2 +
(x1 в€’ x2 )(x1 в€’ x3 ) (x2 в€’ x1 )(x2 в€’ x3 )
(x в€’ x1 )(x в€’ x2 )
y3
(x3 в€’ x1 )(x3 в€’ x2 )

(x в€’ x2 )(x в€’ x3 )(x в€’ x4 ) (x в€’ x1 )(x в€’ x3 )(x в€’ x4 )
P4 (x) =y1 + y2 +
(x1 в€’ x2 )(x1 в€’ x3 )(x1 в€’ x4 ) (x2 в€’ x1 )(x2 в€’ x3 )(x2 в€’ x4 )

(x в€’ x1 )(x в€’ x2 )(x в€’ x4 ) (x в€’ x1 )(x в€’ x2 )(x в€’ x3 )
y3 + y4
(x3 в€’ x1 )(x3 в€’ x2 )(x3 в€’ x4 ) (x4 в€’ x1 )(x4 в€’ x2 )(x4 в€’ x3 )

Consider the data {(в€’2, 1), (в€’1, в€’2), (0, 0), (1, 1), (2, 3)}. Compute P2 (x),
P3 (x), and P4 (x); plot them and determine which data points they go
through. What can you say about Pn (x)?
10 CHAPTER 1. FUNCTIONS

5. A linear function has the form y = mx + b. The number m is called
the slope and the number b is called the y-intercept. The graph of this
function goes through the point (0, b) on the y-axis. In each of the
following determine the slope, y-intercept and sketch the graph of the
given linear function:

a) y = 3x в€’ 5 b) y = в€’2x + 4 c) y = 4x в€’ 3

d) y = 4 e) 2y + 5x = 10

6. A quadratic function has the form y = ax2 + bx + c, where a = 0. On
completing the square, this function can be expressed in the form
2
b2 в€’ 4ac
b
в€’ .
y=a x+
4a2
2a

b2 в€’ 4ac
b
The graph of this function is a parabola with vertex в€’ , в€’
2a 4a
в€’b
and line of symmetry axis being the vertical line with equation x = .
2a
The graph opens upward if a > 0 and downwards if a < 0. In each of
the following quadratic functions, determine the vertex, symmetry axis
and sketch the graph.

a) y = 4x2 в€’ 8 b) y = в€’4x2 + 16 c) y = x2 + 4x + 5

d) y = x2 в€’ 6x + 8 e) y = в€’x2 + 2x + 5 f) y = 2x2 в€’ 6x + 12

g) y = в€’2x2 в€’ 6x + 5 h) y = в€’2x2 + 6x + 10 i) 3y + 6x2 + 10 = 0

j) y = в€’x2 + 4x + 6 k) y = в€’x2 + 4x l) y = 4x2 в€’ 16x

7. Sketch the graph of the linear function deп¬Ѓned by each linear equation
and determine the x-intercept and y-intercept if any.

a) 3x в€’ y = 3 b) 2x в€’ y = 10 c) x = 4 в€’ 2y
1.1. THE CONCEPT OF A FUNCTION 11

d) 4x в€’ 3y = 12 f) 4x + 6y = в€’12
e) 3x + 4y = 12

g) 2x в€’ 3y = 6 h) 2x + 3y = 12 i) 3x + 5y = 15

8. Sketch the graph of each of the following functions:

b) y = в€’4|x|
a) y = 4|x|

c) y = 2|x| + |x в€’ 1| d) y = 3|x| + 2|x в€’ 2| в€’ 4|x + 3|

e) y = 2|x + 2| в€’ 3|x + 1|

9. Sketch the graph of each of the following piecewise functions.

x2 for x в‰¤ 0
if x в‰Ґ 0
2
b) y =
a) y =
в€’2 if x < 0 2x + 4 for x > 0

3x2
4x2 for x в‰¤ 1
if x в‰Ґ 0
d) y =
c) y =
3x3 4 for x > 1
x<0

e) y = n в€’ 1 for n в€’ 1 в‰¤ x < n, for each integer n.

f) y = n for n в€’ 1 < x в‰¤ n for each integer n.

10. The reп¬‚ection of the graph of y = f (x) is the graph of y = в€’f (x). In
each of the following, sketch the graph of f and the graph of its reп¬‚ection
on the same axis.

a) y = x3 b) y = x2 c) y = |x|

d) y = x3 в€’ 4x e) y = x2 в€’ 2x f) y = |x| + |x в€’ 1|

x2 + 1 for x в‰¤ 0
4 2
g) y = x в€’ 4x h) y = 3x в€’ 6 i) y =
x3 + 1 if x < 0
12 CHAPTER 1. FUNCTIONS

11. The graph of y = f (x) is said to be

(i) Symmetric with respect to the y-axis if (x, y) and (в€’x, y) are both
on the graph of f ;
(ii) Symmetric with respect to the origin if (x, y) and (в€’x, в€’y) are both
on the graph of f .

For the functions in problems 10 a) вЂ“ 10 i), determine the functions whose
graphs are (i) Symmetric with respect to y-axis or (ii) Symmetric with
respect to the origin.

12. Discuss the symmetry of the graph of each function and determine whether
the function is even, odd, or neither.

a) f (x) = x6 + 1 b) f (x) = x4 в€’ 3x2 + 4 c) f (x) = x3 в€’ x2

d) f (x) = 2x3 + 3x e) f (x) = (x в€’ 1)3 f) f (x) = (x + 1)4
в€љ
i) f (x) = (x2 + 1)3
g) f (x) = x2 + 4 h) f (x) = 4|x| + 2

в€љ
x2 в€’ 1
l) f (x) = x1/3
4 в€’ x2
j) f (x) = 2 k) f (x) =
x +1

1.2 Trigonometric Functions
The trigonometric functions are deп¬Ѓned by the points (x, y) on the unit circle
with the equation x2 + y 2 = 1.

graph

Consider the points A(0, 0), B(x, 0), C(x, y) where C(x, y) is a point on
the unit circle. Let Оё, read theta, represent the length of the arc joining
the points D(1, 0) and C(x, y). This length is the radian measure of the
angle CAB. Then we deп¬Ѓne the following six trigonometric functions of Оё as
1.2. TRIGONOMETRIC FUNCTIONS 13

follows:
y x y sin Оё
sin Оё = , cos Оё = , tan Оё = = ,
1 1 x cos Оё

1 1 1 1 x 1
csc Оё = = , sec Оё = = , cot Оё = = .
y sin Оё x cos Оё y tan Оё
Since each revolution of the circle has arc length 2ПЂ, sin Оё and cos Оё have
period 2ПЂ. That is,

sin(Оё + 2nПЂ) = sin Оё and cos(Оё + 2nПЂ) = cos Оё, n = 0, В±1, В±2, . . .

The function values of some of the common arguments are given below:

Оё 0 ПЂ/6 в€љПЂ/4 в€љПЂ/3 ПЂ/2 в€љ
2ПЂ/3 3ПЂ/4 5ПЂ/6 ПЂ
в€љ
sin Оё 0в€љ1/2 в€љ2/2 3/2 1 3/2 2/2 1/2 0
в€љ в€љ
в€’1/2 в€’ 2/2 в€’ 3/2 -1
cos Оё 1 3/2 2/2 1/2 0

Оё 7ПЂ/6 5ПЂ/4 4ПЂ/3 3ПЂ/2 5ПЂ/3 7ПЂ/4 11ПЂ/6 2ПЂ
в€љ в€љ в€љ в€љ
в€’1/2 в€’в€љ2/2 в€’ 3/2 в€’1 в€’ 3/2 в€’ 2/2 в€’1/2 0
sin Оё в€љ в€љ в€љ
в€’ 3/2 в€’ 2/2 в€’1/2
cos Оё 0 1/2 2/2 3/2 1

A function f is said to have period p if p is the smallest positive number
such that, for all x,

f (x + np) = f (x), n = 0, В±1, В±2, . . . .

Since csc Оё is the reciprocal of sin Оё and sec Оё is the reciprocal of cos(Оё), their
periods are also 2ПЂ. That is,

csc(Оё + 2nПЂ) = csc(Оё) and sec(Оё + 2nПЂ) = sec Оё, n = 0, В±1, В±2, . . . .

It turns out that tan Оё and cot Оё have period ПЂ. That is,

tan(Оё + nПЂ) = tan Оё and cot(Оё + nПЂ) = cot Оё, n = 0, В±1, В±2, . . . .

Geometrically, it is easy to see that cos Оё and sec Оё are the only even trigono-
metric functions. The functions sin Оё, cos Оё, tan Оё and cot Оё are all odd func-
tions. The functions sin Оё and cos Оё are deп¬Ѓned for all real numbers. The
14 CHAPTER 1. FUNCTIONS

functions csc Оё and cot Оё are not deп¬Ѓned for integer multiples of ПЂ, and sec Оё
and tan Оё are not deп¬Ѓned for odd integer multiples of ПЂ/2. The graphs of the
six trigonometric functions are sketched as follows:

graph

The dotted vertical lines represent the vertical asymptotes.
There are many useful trigonometric identities and reduction formulas.
For future reference, these are listed here.

sin2 Оё + cos2 Оё = 1 sin2 Оё = 1 в€’ cos2 Оё cos2 Оё = 1 в€’ sin2 Оё
tan2 Оё + 1 = sec2 Оё tan2 Оё = sec2 Оё в€’ 1 sec2 Оё в€’ tan2 Оё = 1
1 + cot2 Оё = csc2 Оё cot2 Оё = csc2 Оё в€’ 1 csc2 Оё в€’ cot2 Оё = 1

cos 2Оё = 1 + 2 sin2 Оё
cos 2Оё = 2 cos2 Оё в€’ 1
sin 2Оё = 2 sin Оё cos Оё
cos(x + y) = cos x cos y в€’ sin x sin y
sin(x + y) = sin x cos y + cos x sin y,
sin(x в€’ y) = sin x cos y в€’ cos x sin y, cos(x в€’ y) = cos x cos y + sin x sin y

tan x в€’ tan y
tan x + tan y
tan(x в€’ y) =
tan(x + y) =
1 в€’ tan x tan y 1 + tan x tan y

О±в€’ОІ
О±+ОІ
sin О± + sin ОІ = 2 sin cos
2 2

О±в€’ОІ
О±+ОІ
sin О± в€’ sin ОІ = 2 cos sin
2 2

О±в€’ОІ
О±+ОІ
cos О± + cos ОІ = 2 cos cos
2 2

О±в€’ОІ
О±+ОІ
cos О± в€’ cos ОІ = в€’2 sin sin
2 2
1.2. TRIGONOMETRIC FUNCTIONS 15

1
sin x cos y = (sin(x + y) + sin(x в€’ y))
2
1
cos x sin y = (sin(x + y) в€’ sin(x в€’ y))
2
1
cos x cos y = (cos(x в€’ y) + cos(x + y))
2
1
sin x sin y = (cos(x в€’ y) в€’ cos(x + y))
2
sin(ПЂ В± Оё) = sin Оё

cos(ПЂ В± Оё) = в€’ cos Оё

tan(ПЂ В± Оё) = В± tan Оё

cot(ПЂ В± Оё) = В± cot Оё

sec(ПЂ В± Оё) = в€’ sec Оё

csc(ПЂ В± Оё) = csc Оё

In applications of calculus to engineering problems, the graphs of y =
A sin(bx + c) and y = A cos(bx + c) play a signiп¬Ѓcant role. The п¬Ѓrst problem
has to do with converting expressions of the form A sin bx + B cos bx to one
of the above forms. Let us begin п¬Ѓrst with an example.

Example 1.2.1 Express y = 3 sin(2x)в€’4 cos(2x) in the form y = A sin(2xВ±
Оё) or y = A cos(2x В± Оё).
First of all, we make a right triangle with sides of length 3 and 4 and
compute the length of the hypotenuse, which is 5. We label one of the acute
angles as Оё and compute sin Оё, cos Оё and tan Оё. In our case,
4 3
3
, cos Оё = , and, tan Оё = .
sin Оё =
5 5 4

graph
16 CHAPTER 1. FUNCTIONS

Then,
y = 3 sin 2x в€’ 4 cos 2x
3 4
в€’ (cos(2x))
= 5 (sin(2x))
5 5
= 5[sin(2x) sin Оё в€’ cos(2x) cos Оё]
= в€’5[cos(2x) cos Оё в€’ sin(2x) sin Оё]
= в€’5[cos(2x + Оё)]
Thus, the problem is reduced to sketching a cosine function, ???
y = в€’5 cos(2x + Оё).
We can compute the radian measure of Оё from any of the equations
3 4 3
sin Оё = , cos Оё = or tan Оё = .
5 5 4

Example 1.2.2 Sketch the graph of y = 5 cos(2x + 1).
In order to sketch the graph, we п¬Ѓrst compute all of the zeros, relative
maxima, and relative minima. We can see that the maximum values will be
5 and minimum values are в€’5. For this reason the number 5 is called the
amplitude of the graph. We know that the cosine function has zeros at odd
integer multiples of ПЂ/2. Let
ПЂ1
ПЂ
2xn + 1 = (2n + 1) , xn = (2n + 1) в€’ , n = 0, В±1, В±2 . . . .
2 42
The max and min values of a cosine function occur halfway between the
consecutive zeros. With this information, we are able to sketch the graph of
1 1
the given function. The period is ПЂ, phase shift is and frequency is .
2 ПЂ

graph

For the functions of the form y = A sin(П‰t В± d) or y = A cos(П‰t В± d) we
make the following deп¬Ѓnitions:
1.2. TRIGONOMETRIC FUNCTIONS 17

2ПЂ 1 П‰
period = , frequency = = ,
П‰ period 2ПЂ
d
amplitude = |A|, and phase shift = .
П‰
The motion of a particle that follows the curves A sin(П‰tВ±d) or A cos(П‰tВ±d)
is called simple harmonic motion.

Exercises 1.2

1. Determine the amplitude, frequency, period and phase shift for each of
the following functions. Sketch their graphs.

(a) y = 2 sin(3t в€’ 2) (b) y = в€’2 cos(2t в€’ 1)
(d) y = 4 sin 2t в€’ 3 cos 2t
(c) y = 3 sin 2t + 4 cos 2t
sin x
(e) y =
x

2. Sketch the graphs of each of the following:

(a) y = tan(3x) (b) y = cot(5x) (c) y = x sin x
(d) y = sin(1/x) (e) y = x sin(1/x)

3. Express the following products as the sum or diп¬Ђerence of functions.

(a) sin(3x) cos(5x) (b) cos(2x) cos(4x) (c) cos(2x) sin(4x)
(d) sin(3x) sin(5x) (e) sin(4x) cos(4x)

4. Express each of the following as a product of functions:

(a) sin(x + h) в€’ sin x (b) cos(x + h) в€’ cos x (c) sin(5x) в€’ sin(3x)
(d) cos(4x) в€’ cos(2x) (e) sin(4x) + sin(2x) (f) cos(5x) + cos(3x)

в€’ПЂ ПЂ
в‰¤ x в‰¤ . Take the sample points
5. Consider the graph of y = sin x,
2 2

ПЂ ПЂ ПЂ ПЂ1 ПЂ
в€’ , в€’1 , в€’ ,в€’ , (0, 0), , , ,1 .
2 6 2 62 2
18 CHAPTER 1. FUNCTIONS

Compute the fourth degree Lagrange Polynomial that approximates and
agrees with y = sin x at these data points. This polynomial has the form
(x в€’ x2 )(x в€’ x3 )(x в€’ x4 )(x в€’ x5 )
P5 (x) = y1 +
(x1 в€’ x2 )(x1 в€’ x3 )(x1 в€’ x4 )(x1 в€’ x5 )
(x в€’ x1 )(x в€’ x3 )(x в€’ x4 )(x в€’ x5 )
+ В·В·В·
y2
(x2 в€’ x1 )(x2 в€’ x3 )(x2 в€’ x4 )(x2 в€’ x5 )
(x в€’ x1 )(x в€’ x2 )(x в€’ x3 )(x в€’ x4 )
+ y5 .
(x5 в€’ x1 )(x5 в€’ x2 )(x5 в€’ x3 )(x5 в€’ x4 )

6. Sketch the graphs of the following functions and compute the amplitude,
period, frequency and phase shift, as applicable.

a) y = 3 sin t b) y = 4 cos t c) y = 2 sin(3t)

ПЂ
d) y = в€’4 cos(2t) e) y = в€’3 sin(4t) f) y = 2 sin t + 6

ПЂ
i) y = в€’3 cos(2t в€’ ПЂ)
g) y = в€’2 sin t в€’ h) y = 3 cos(2t + ПЂ)
6

k) y = в€’2 cos(6t в€’ ПЂ)
j) y = 2 sin(4t + ПЂ) l) y = 3 sin(6t + ПЂ)

7. Sketch the graphs of the following functions over two periods.

b) y = в€’3 tan x
a) y = 2 sec x c) y = 2 cot x

ПЂ
d) y = 3 csc x e) y = tan(ПЂx) f) y = tan 2x + 3

ПЂ ПЂ
ПЂ
h) y = 3 sec 2x + i) y = 2 sin ПЂx +
g) y = 2 cot 3x + 2 3 6

8. Prove each of the following identities:

b) sin(3t) = 3 sin x в€’ 4 sin3 x
a) cos 3t = 3 cos t + 4 cos3 t

sin3 t в€’ cos3 t
4 4
c) sin t в€’ cos t = в€’ cos 2t d) = 1 + sin 2t
sin t в€’ cos t

sin(x + y) tan x + tan y
e) cos 4t cos 7t в€’ sin 7t sin 4t = cos 11t f) =
sin(x в€’ y) tan x в€’ tan y
1.3. INVERSE TRIGONOMETRIC FUNCTIONS 19

9. If f (x) = cos x, prove that

f (x + h) в€’ f (x) cos h в€’ 1 sin h
в€’ sin x
= cos x .
h h h

10. If f (x) = sin x, prove that

f (x + h) в€’ f (x) cos h в€’ 1 sin h
= sin x + cos x .
h h h

11. If f (x) = cos x, prove that

f (x) в€’ f (t) cos(x в€’ t) в€’ 1 sin(x в€’ t)
в€’ sin t
= cos t .
xв€’t xв€’t xв€’t

12. If f (x) = sin x, prove that

f (x) в€’ f (t) cos(x в€’ t) в€’ 1 sin(x в€’ t)
= sin t + cos t .
xв€’t xв€’t xв€’t

13. Prove that
1 в€’ tan2 t
.
cos(2t) =
1 + tan2 t

x
14. Prove that if y = tan , then
2
1 в€’ u2 2u
(a) cos x = (b) sin x =
1 + u2 1 + u2

1.3 Inverse Trigonometric Functions
None of the trigonometric functions are one-to-one since they are periodic.
In order to deп¬Ѓne inverses, it is customary to restrict the domains in which
the functions are one-to-one as follows.
20 CHAPTER 1. FUNCTIONS

ПЂ ПЂ
1. y = sin x, в€’ в‰¤ x в‰¤ , is one-to-one and covers the range в€’1 в‰¤ y в‰¤ 1.
2 2
Its inverse function is denoted arcsin x, and we deп¬Ѓne y = arcsin x, в€’1 в‰¤
ПЂ ПЂ
x в‰¤ 1, if and only if, x = sin y, в€’ в‰¤ y в‰¤ .
2 2

graph

2. y = cos x, 0 в‰¤ x в‰¤ ПЂ, is one-to-one and covers the range в€’1 в‰¤ y в‰¤ 1. Its
inverse function is denoted arccos x, and we deп¬Ѓne y = arccos x, в€’1 в‰¤
x в‰¤ 1, if and only if, x = cos y, 0 в‰¤ y в‰¤ ПЂ.

graph

в€’ПЂ ПЂ
< x < , is one-to-one and covers the range в€’в€ћ <
3. y = tan x,
2 2
y < в€ћ Its inverse function is denoted arctan x, and we deп¬Ѓne y =
в€’ПЂ ПЂ
arctan x, в€’в€ћ < x < в€ћ, if and only if, x = tan y, <y< .
2 2

graph

4. y = cot x, 0, x < ПЂ, is one-to-one and covers the range в€’в€ћ < y < в€ћ. Its
inverse function is denoted arccot x, and we deп¬Ѓne y = arccot x, в€’в€ћ <
x < в€ћ, if and only if x = cot y, 0 < y < ПЂ.

graph
1.3. INVERSE TRIGONOMETRIC FUNCTIONS 21

ПЂ ПЂ
5. y = sec x, 0 в‰¤ x в‰¤ or < x в‰¤ ПЂ is one-to-one and covers the range
2 2
в€’в€ћ < y в‰¤ в€’1 or 1 в‰¤ y < в€ћ. Its inverse function is denoted arcsec x,
and we deп¬Ѓne y = arcsec x, в€’в€ћ < x в‰¤ в€’1 or 1 в‰¤ x < в€ћ, if and only
ПЂ
ПЂ
if, x = sec y, 0 в‰¤ y < or < y в‰¤ ПЂ.
2 2

graph

в€’ПЂ ПЂ
в‰¤ x < 0 or 0 < x в‰¤ , is one-to-one and covers the
6. y = csc x,
2 2
range в€’в€ћ < y в‰¤ в€’1 or 1 в‰¤ y < в€ћ. Its inverse is denoted arccsc x and
we deп¬Ѓne y = arccsc x, в€’в€ћ < x в‰¤ в€’1 or 1 в‰¤ x < в€ћ, if and only if,
в€’ПЂ ПЂ
в‰¤ y < 0 or 0 < y в‰¤ .
x = csc y,
2 2

Example 1.3.1 Show that each of the following equations is valid.
ПЂ
(a) arcsin x + arccos x =
2
ПЂ
(b) arctan x + arccot x =
2
ПЂ
(c) arcsec x + arccsc x =
2
To verify equation (a), we let arcsin x = Оё.

graph

ПЂ
в€’ Оё = x, as shown in the triangle. It follows
Then x = sin Оё and cos
2
that
ПЂ
ПЂ
в€’ Оё = arccos x, = Оё + arccos x = arcsin x + arccos x.
2 2
The equations in parts (b) and (c) are veriп¬Ѓed in a similar way.
22 CHAPTER 1. FUNCTIONS

Example 1.3.2 If Оё = arcsin x, then compute cos Оё, tan Оё, cot Оё, sec Оё and
csc Оё.
ПЂ ПЂ
If Оё is в€’ , 0, or , then computations are easy.
2 2

graph

ПЂ ПЂ
Suppose that в€’ < x < 0 or 0 < x < . Then, from the triangle, we get
2 2
в€љ
в€љ 1 в€’ x2
x
в€љ
2,
cos Оё = 1 в€’ x tan Оё = , cot Оё = ,
x
1 в€’ x2
1 1
sec Оё = в€љ and csc Оё = .
x
1 в€’ x2

Example 1.3.3 Make the given substitutions to simplify the given radical
expression and compute all trigonometric functions of Оё.
в€љ
в€љ
(a) 4 в€’ x2 , x = 2 sin Оё (b) x2 в€’ 9, x = 3 sec Оё
(c) (4 + x2 )3/2 , x = 2 tan Оё
x
(a) For part (a), sin Оё = and we use the given triangle:
2

graph

Then
в€љ в€љ
4 в€’ x2 4 в€’ x2
x
, tan Оё = в€љ
cos Оё = , cot Оё = ,
2 x
2
4в€’x
2 2
sec Оё = в€љ , csc Оё = .
x
4 в€’ x2
в€љ
Furthermore, 4 в€’ x2 = 2 cos Оё and the radical sign is eliminated.
1.3. INVERSE TRIGONOMETRIC FUNCTIONS 23

x
(b) For part (b), sec Оё = and we use the given triangle:
3

graph

Then,
в€љ в€љ
x2 в€’ 4 x2 в€’ 4
3
sin Оё = , cos Оё = , tan Оё =
x x 3

x
3
cot Оё = в€љ , csc Оё = в€љ .
x2 в€’ 9 x2 в€’ 9
в€љ
Furthermore, x2 в€’ 9 = 3 tan Оё and the radical sign is eliminated.
x
(c) For part (c), tan Оё = and we use the given triangle:
2

graph

Then,
x 2 2
sin Оё = в€љ cos Оё = в€љ
, , cot Оё = ,
x
2+4 2+4
x x

в€љ в€љ
2+4 x2 + 4
x
sec Оё = , csc Оё = .
2 x
в€љ
Furthermore, x2 + 4 = 2 sec Оё and hence

(4 + x)3/2 = (2 sec Оё)3 = 8 sec3 Оё.
24 CHAPTER 1. FUNCTIONS

Remark 2 The three substitutions given in Example 15 are very useful in
calculus. In general, we use the following substitutions for the given radicals:
в€љ в€љ
a2 в€’ x2 , x = a sin Оё x2 в€’ a2 , x = a sec Оё
(a) (b)
в€љ
(c) a2 + x2 , x = a tan Оё.

Exercises 1.3

1. Evaluate each of the following:
в€љ
1 3
(a) 3 arcsin + 2 arccos
2 2
1 1
(b) 4 arctan в€љ в€љ
+ 5arccot
3 3
2
(c) 2arcsec (в€’2) + 3 arccos в€’ в€љ
3
(d) cos(2 arccos(x))
(e) sin(2 arccos(x))

2. Simplify each of the following expressions by eliminating the radical by
using an appropriate trigonometric substitution.

xв€’2
3+x
x
(a) в€љ (b) в€љ (c) в€љ
9 в€’ x2 x x2 в€’ 25
16 + x2
2 в€’ 2x
1+x
(d) в€љ (e) в€љ
x2 в€’ 2x в€’ 3
x2 + 2x + 2

(Hint: In parts (d) and (e), complete squares п¬Ѓrst.)

3. Some famous polynomials are the so-called Chebyshev polynomials, de-
п¬Ѓned by

Tn (x) = cos(n arccos x), в€’1 в‰¤ x в‰¤ 1, n = 0, 1, 2, . . . .
1.3. INVERSE TRIGONOMETRIC FUNCTIONS 25

(a) Prove the recurrence relation for Chebyshev polynomials:

Tn+1 (x) = 2xTn (x) в€’ Tnв€’1 (x) for each n в‰Ґ 1.

(b) Show that T0 (x) = 1, T1 (x) = x and generate T2 (x), T3 (x), T4 (x) and
T5 (x) using the recurrence relation in part (a).
(c) Determine the zeros of Tn (x) and determine where Tn (x) has its
absolute maximum or minimum values, n = 1, 2, 3, 4, ?.
(Hint: Let Оё = arccos x, x = cos Оё. Then Tn (x) = cos(nОё), Tn+1 (x) =
cos(nОё + Оё), Tnв€’1 (x) = cos(nОё в€’ Оё). Use the expansion formulas and
then make substitutions in part (a)).

4. Show that for all integers m and n,
1
[Tm+n (x) + T|mв€’n| (x)]
Tn (x)Tm (x) =
2
(Hint: use the expansion formulas as in problem 3.)

5. Find the exact value of y in each of the following
в€љ
в€љ
в€’1 3
a) y = arccos b) y = arcsin c) y = arctan(в€’ 3)
2 2

в€љ в€љ
в€љ
в€’ 33
d) y = arccot e) y = arcsec (в€’ 2) f) y = arccsc (в€’ 2)

2 2
g) y = arcsec в€’ в€љ3 h) y = arccsc в€’ в€љ3 i) y = arcsec (в€’2)

в€љ
в€’1
j) y = arccsc (в€’2) k) y = arctan l) y = arccot (в€’ 3)
в€љ
3

6. Solve the following equations for x in radians (all possible answers).

a) 2 sin4 x = sin2 x b) 2 cos2 x в€’ cos x в€’ 1 = 0

c) sin2 x + 2 sin x + 1 = 0 d) 4 sin2 x + 4 sin x + 1 = 0
26 CHAPTER 1. FUNCTIONS

e) 2 sin2 x + 5 sin x + 2 = 0 f) cot3 x в€’ 3 cot x = 0

g) sin 2x = cos x h) cos 2x = cos x

x
i) cos2 = cos x j) tan x + cot x = 1
2

7. If arctan t = x, compute sin x, cos x, tan x, cot x, sec x and csc x in
terms of t.

8. If arcsin t = x, compute sin x, cos x, tan x, cot x, sec x and csc x in terms
of t.

9. If arcsec t = x, compute sin x, cos x, tan x, cot x, sec x and csc x in
terms of t.

10. If arccos t = x, compute sin x, cos x, tan x, cot x, sec x and csc x in
terms of t.

Remark 3 Chebyshev polynomials are used extensively in approximating
functions due to their properties that minimize errors. These polynomials
are called equal ripple polynomials, since their maxima and minima alternate
between 1 and в€’1.

1.4 Logarithmic, Exponential and Hyperbolic
Functions
Most logarithmic tables have tables for log10 x, loge x, ex and eв€’x because
of their universal applications to scientiп¬Ѓc problems. The key relationship
between logarithmic functions and exponential functions, using the same
base, is that each one is an inverse of the other. For example, for base 10,
we have
N = 10x if and only if x = log10 N.
We get two very interesting relations, namely

x = log10 (10x ) and N = 10(log10 N ) .
1.4. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS27

For base e, we get
x = loge (ex ) and y = e(loge y) .
If b > 0 and b = 1, then b is an admissible base for a logarithm. For such an
admissible base b, we get

x = logb (bx ) and y = b(logb y) .

The Logarithmic function with base b, b > 0, b = 1, satisп¬Ѓes the following
important properties:
1. logb (b) = 1, logb (1) = 0, and logb (bx ) = x for all real x.

2. logb (xy) = logb x + logb y, x > 0, y > 0.

3. logb (x/y) = logb x в€’ logb y, x > 0, y > 0.

4. logb (xy ) = y logb x, x > 0, x = 1, for all real y.

5. (logb x)(loga b) = loga xb > 0, a > 0, b = 1, a = 1. Note that logb x =
loga x
.
loga b
This last equation (5) allows us to compute logarithms with respect to
any base b in terms of logarithms in a given base a.
The corresponding laws of exponents with respect to an admissible base
b, b > 0, b = 1 are as follows:
1. b0 = 1, b1 = b, and b(logb x) = x for x > 0.

2. bx Г— by = bx+y
bx
= bxв€’y
3. y
b
4. (bx )y = b(xy)
Notation: If b = e, then we will express

logb (x) as ln(x) or log(x).

The notation exp(x) = ex can be used when confusion may arise.
The graph of y = log x and y = ex are reп¬‚ections of each other through
the line y = x.
28 CHAPTER 1. FUNCTIONS

graph

In applications of calculus to science and engineering, the following six
functions, called hyperbolic functions, are very useful.

1x
(e в€’ eв€’x ) for all real x, read as hyperbolic sine of x.
1. sinh(x) =
2
1x
(e + eв€’x ), for all real x, read as hyperbolic cosine of x.
2. cosh(x) =
2

ex в€’ eв€’x
sinh(x)
3. tanh(x) = =x , for all real x, read as hyperbolic tangent
e + eв€’x
cosh(x)
of x.

ex + eв€’x
cosh(x)
=x , x = 0, read as hyperbolic cotangent of x.
4. coth(x) =
e в€’ eв€’x
sinh(x)

1 2
5. sech (x) = =x , for all real x, read as hyperbolic secant of
e + eв€’x
cosh x
x.
2
1
=x , x = 0, read as hyperbolic cosecant of x.
6. csch (x) =
e в€’ eв€’x
sinh(x)

The graphs of these functions are sketched as follows:

graph

Example 1.4.1 Eliminate quotients and exponents in the following equa-
tion by taking the natural logarithm of both sides.

(x + 1)3 (2x в€’ 3)3/4
y=
(1 + 7x)1/3 (2x + 3)3/2
1.4. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS29

(x + 1)3 (2x в€’ 3)3/4
ln(y) = ln
(1 + 7x)1/3 (2x + 3)3/2]
= ln[(x + 1)3 (2x в€’ 3)3/4 ] в€’ ln[(1 + 7x)1/3 (2x + 3)3/2 ]
= ln(x + 1)3 + ln(2x в€’ 3)3/4 в€’ {ln(1 + 7x)1/3 + ln(2x + 3)3/2 }
3 1 3
= 3 ln(x + 1) + ln(2x в€’ 3) в€’ ln(1 + 7x) в€’ ln(2x + 3)
4 3 2

Example 1.4.2 Solve the following equation for x:
log3 (x4 ) + log3 x3 в€’ 2 log3 x1/2 = 5.
Using logarithm properties, we get
4 log3 x + 3 log3 x в€’ log3 x = 5
6 log3 x = 5
5
log3 x =
6
5/6
x = (3) .

Example 1.4.3 Solve the following equation for x:
ex 1
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