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 de¬nitions . . . . . . . . . . . . . . . 35

2.1.1 Introductory Examples . . . . . . . . . . . . . . . . . . 35

2.1.2 Limit: Formal De¬nitions . . . . . . . . . . . . . . . . 41

2.1.3 Continuity: Formal De¬nitions . . . . . . . . . . . . . 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 In¬nity . . . . . . . . . . . . . . . . . . . . . . . . 94

3 Di¬erentiation 99

3.1 The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . 99

3.2 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 111

3.3 Di¬erentiation of Inverse Functions . . . . . . . . . . . . . . . 118

3.4 Implicit Di¬erentiation . . . . . . . . . . . . . . . . . . . . . . 130

3.5 Higher Order Derivatives . . . . . . . . . . . . . . . . . . . . . 137

4 Applications of Di¬erentiation 146

4.1 Mathematical Applications . . . . . . . . . . . . . . . . . . . . 146

4.2 Antidi¬erentiation . . . . . . . . . . . . . . . . . . . . . . . . 157

4.3 Linear First Order Di¬erential Equations . . . . . . . . . . . . 164

i

ii CONTENTS

4.4 Linear Second Order Homogeneous Di¬erential Equations . . . 169

4.5 Linear Non-Homogeneous Second Order Di¬erential Equations 179

5 The De¬nite Integral 183

5.1 Area Approximation . . . . . . . . . . . ...... . . . . . . 183

5.2 The De¬nite 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 In¬nite Series 315

8.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315

8.2 Monotone Sequences . . . . . . . . . . . . . . . . . . . . . . . 320

8.3 In¬nite 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 ≥ 0

|x| =

’x if x < 0

We note that f (0) = 0, f (1) = 1 and f (’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

√

x ’ 4.

f (x) =

In order that the range of f contain real numbers only, we must impose

the restriction that x ≥ 4. Thus, the domain Df contains the set of all real

numbers x such that x ≥ 4. The range Rf will consist of all real numbers y

such that y ≥ 0. The graph of f is sketched below.

graph

Example 1.1.4 A useful function in engineering is the unit step function,

u, de¬ned as follows:

0 if x < 0

u(x) =

1 if x ≥ 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 ’ 4

It is clear that Df consists of all real numbers x = ±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 = ’2 and x = 2. These vertical lines, x = ±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 (’x) for all x in Df . And

f is said to be odd if f (’x) = ’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 ’ x2 .

This function f is even because for all x we have

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

The graph of f is symmetric to the y-axis because (x, f (x)) and (’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 (’1) = f (1).

Example 1.1.7 Sketch the graph of g(x) = x3 ’ 3x.

The function g is an odd function because for each x,

g(’x) = (’x)3 ’ 3(’x) = ’x3 + 3x = ’(x3 ’ 3x) = ’g(x).

The graph of this function g is symmetric to the origin because (x, g(x))

and (’x, ’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

√

√

This function g is not one-to-one because g(0) = g( 3) = g(’ 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 (’x)), h(x) = (f (x) ’ f (’x)).

2 2

Then,

1

g(’x) = (f (’x) + f (x)) = g(x)

2

1

h(’x) = (f (’x) ’ f (x)) = ’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 ’ 2x3 + x2 ’ 5x + 7.

We de¬ne

1

g(x) = (f (x) + f (’x))

2

1

= {(x4 ’ 2x3 + x2 ’ 5x + 7) + (x4 + 2x3 + x2 + 5x + 7)}

2

= x4 + x2 + 7

6 CHAPTER 1. FUNCTIONS

and

1

h(x) = (f (x) ’ f (’x))

2

1

= {(x4 ’ 2x3 + x2 ’ 5x + 7) ’ (x4 + 2x3 + x2 + 5x + 7)}

2

= ’2x3 ’ 5x.

Then clearly g(x) is even and h(x) is odd.

g(’x) = (’x)4 + (’x)2 + 7

= x4 + x2 + 7

= g(x)

h(’x) = ’ 2(’x)3 ’ 5(’x)

= 2x3 + 5x

= ’h(x).

We note that

g(x) + h(x) = (x4 + x2 + 7) + (’2x3 ’ 5x)

= x4 ’ 2x3 + x2 ’ 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 ’ x3

2

1

= (x1 ’ x2 )(x2 + x1 x2 + x2 ) (By factoring)

2

1

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

2

1

x2 ’ 4x2

’x2 ± 2 2

x1 =

2

’3x2

’x2 ± 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 ’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 ’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 ’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 ’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 xn’1 + · · · + an’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. De¬ne 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 ’ 27

x

√ x2 ’ 1

(c) f (x) = x2 ’ 9 (d) f (x) =

x’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 ’ 1 (b) g(x) = x3 ’ 1

√

(d) k(x) = x2/3

9 ’ x, x ≥ 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 ’ x1

x ’ x2

+ y2

For n = 2, P2 (x) = y1

x1 ’ x2 x2 ’ x1

(x ’ x2 )(x ’ x3 ) (x ’ x1 )(x ’ x3 )

For n = 3, P3 (x) = y1 + y2 +

(x1 ’ x2 )(x1 ’ x3 ) (x2 ’ x1 )(x2 ’ x3 )

(x ’ x1 )(x ’ x2 )

y3

(x3 ’ x1 )(x3 ’ x2 )

(x ’ x2 )(x ’ x3 )(x ’ x4 ) (x ’ x1 )(x ’ x3 )(x ’ x4 )

P4 (x) =y1 + y2 +

(x1 ’ x2 )(x1 ’ x3 )(x1 ’ x4 ) (x2 ’ x1 )(x2 ’ x3 )(x2 ’ x4 )

(x ’ x1 )(x ’ x2 )(x ’ x4 ) (x ’ x1 )(x ’ x2 )(x ’ x3 )

y3 + y4

(x3 ’ x1 )(x3 ’ x2 )(x3 ’ x4 ) (x4 ’ x1 )(x4 ’ x2 )(x4 ’ x3 )

Consider the data {(’2, 1), (’1, ’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 ’ 5 b) y = ’2x + 4 c) y = 4x ’ 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 ’ 4ac

b

’ .

y=a x+

4a2

2a

b2 ’ 4ac

b

The graph of this function is a parabola with vertex ’ , ’

2a 4a

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

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

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

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

7. Sketch the graph of the linear function de¬ned by each linear equation

and determine the x-intercept and y-intercept if any.

a) 3x ’ y = 3 b) 2x ’ y = 10 c) x = 4 ’ 2y

1.1. THE CONCEPT OF A FUNCTION 11

d) 4x ’ 3y = 12 f) 4x + 6y = ’12

e) 3x + 4y = 12

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

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

b) y = ’4|x|

a) y = 4|x|

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

e) y = 2|x + 2| ’ 3|x + 1|

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

x2 for x ¤ 0

if x ≥ 0

2

b) y =

a) y =

’2 if x < 0 2x + 4 for x > 0

3x2

4x2 for x ¤ 1

if x ≥ 0

d) y =

c) y =

3x3 4 for x > 1

x<0

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

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

10. The re¬‚ection of the graph of y = f (x) is the graph of y = ’f (x). In

each of the following, sketch the graph of f and the graph of its re¬‚ection

on the same axis.

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

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

x2 + 1 for x ¤ 0

4 2

g) y = x ’ 4x h) y = 3x ’ 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 (’x, y) are both

on the graph of f ;

(ii) Symmetric with respect to the origin if (x, y) and (’x, ’y) are both

on the graph of f .

For the functions in problems 10 a) “ 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 ’ 3x2 + 4 c) f (x) = x3 ’ x2

d) f (x) = 2x3 + 3x e) f (x) = (x ’ 1)3 f) f (x) = (x + 1)4

√

i) f (x) = (x2 + 1)3

g) f (x) = x2 + 4 h) f (x) = 4|x| + 2

√

x2 ’ 1

l) f (x) = x1/3

4 ’ x2

j) f (x) = 2 k) f (x) =

x +1

1.2 Trigonometric Functions

The trigonometric functions are de¬ned 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 θ, 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 de¬ne the following six trigonometric functions of θ as

1.2. TRIGONOMETRIC FUNCTIONS 13

follows:

y x y sin θ

sin θ = , cos θ = , tan θ = = ,

1 1 x cos θ

1 1 1 1 x 1

csc θ = = , sec θ = = , cot θ = = .

y sin θ x cos θ y tan θ

Since each revolution of the circle has arc length 2π, sin θ and cos θ have

period 2π. That is,

sin(θ + 2nπ) = sin θ and cos(θ + 2nπ) = cos θ, n = 0, ±1, ±2, . . .

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

θ 0 π/6 √π/4 √π/3 π/2 √

2π/3 3π/4 5π/6 π

√

sin θ 0√1/2 √2/2 3/2 1 3/2 2/2 1/2 0

√ √

’1/2 ’ 2/2 ’ 3/2 -1

cos θ 1 3/2 2/2 1/2 0

θ 7π/6 5π/4 4π/3 3π/2 5π/3 7π/4 11π/6 2π

√ √ √ √

’1/2 ’√2/2 ’ 3/2 ’1 ’ 3/2 ’ 2/2 ’1/2 0

sin θ √ √ √

’ 3/2 ’ 2/2 ’1/2

cos θ 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, ±1, ±2, . . . .

Since csc θ is the reciprocal of sin θ and sec θ is the reciprocal of cos(θ), their

periods are also 2π. That is,

csc(θ + 2nπ) = csc(θ) and sec(θ + 2nπ) = sec θ, n = 0, ±1, ±2, . . . .

It turns out that tan θ and cot θ have period π. That is,

tan(θ + nπ) = tan θ and cot(θ + nπ) = cot θ, n = 0, ±1, ±2, . . . .

Geometrically, it is easy to see that cos θ and sec θ are the only even trigono-

metric functions. The functions sin θ, cos θ, tan θ and cot θ are all odd func-

tions. The functions sin θ and cos θ are de¬ned for all real numbers. The

14 CHAPTER 1. FUNCTIONS

functions csc θ and cot θ are not de¬ned for integer multiples of π, and sec θ

and tan θ are not de¬ned for odd integer multiples of π/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 θ + cos2 θ = 1 sin2 θ = 1 ’ cos2 θ cos2 θ = 1 ’ sin2 θ

tan2 θ + 1 = sec2 θ tan2 θ = sec2 θ ’ 1 sec2 θ ’ tan2 θ = 1

1 + cot2 θ = csc2 θ cot2 θ = csc2 θ ’ 1 csc2 θ ’ cot2 θ = 1

cos 2θ = 1 + 2 sin2 θ

cos 2θ = 2 cos2 θ ’ 1

sin 2θ = 2 sin θ cos θ

cos(x + y) = cos x cos y ’ sin x sin y

sin(x + y) = sin x cos y + cos x sin y,

sin(x ’ y) = sin x cos y ’ cos x sin y, cos(x ’ y) = cos x cos y + sin x sin y

tan x ’ tan y

tan x + tan y

tan(x ’ y) =

tan(x + y) =

1 ’ tan x tan y 1 + tan x tan y

±’β

±+β

sin ± + sin β = 2 sin cos

2 2

±’β

±+β

sin ± ’ sin β = 2 cos sin

2 2

±’β

±+β

cos ± + cos β = 2 cos cos

2 2

±’β

±+β

cos ± ’ cos β = ’2 sin sin

2 2

1.2. TRIGONOMETRIC FUNCTIONS 15

1

sin x cos y = (sin(x + y) + sin(x ’ y))

2

1

cos x sin y = (sin(x + y) ’ sin(x ’ y))

2

1

cos x cos y = (cos(x ’ y) + cos(x + y))

2

1

sin x sin y = (cos(x ’ y) ’ cos(x + y))

2

sin(π ± θ) = sin θ

cos(π ± θ) = ’ cos θ

tan(π ± θ) = ± tan θ

cot(π ± θ) = ± cot θ

sec(π ± θ) = ’ sec θ

csc(π ± θ) = csc θ

In applications of calculus to engineering problems, the graphs of y =

A sin(bx + c) and y = A cos(bx + c) play a signi¬cant role. The ¬rst 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 ¬rst with an example.

Example 1.2.1 Express y = 3 sin(2x)’4 cos(2x) in the form y = A sin(2x±

θ) or y = A cos(2x ± θ).

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 θ and compute sin θ, cos θ and tan θ. In our case,

4 3

3

, cos θ = , and, tan θ = .

sin θ =

5 5 4

graph

16 CHAPTER 1. FUNCTIONS

Then,

y = 3 sin 2x ’ 4 cos 2x

3 4

’ (cos(2x))

= 5 (sin(2x))

5 5

= 5[sin(2x) sin θ ’ cos(2x) cos θ]

= ’5[cos(2x) cos θ ’ sin(2x) sin θ]

= ’5[cos(2x + θ)]

Thus, the problem is reduced to sketching a cosine function, ???

y = ’5 cos(2x + θ).

We can compute the radian measure of θ from any of the equations

3 4 3

sin θ = , cos θ = or tan θ = .

5 5 4

Example 1.2.2 Sketch the graph of y = 5 cos(2x + 1).

In order to sketch the graph, we ¬rst compute all of the zeros, relative

maxima, and relative minima. We can see that the maximum values will be

5 and minimum values are ’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 π/2. Let

π1

π

2xn + 1 = (2n + 1) , xn = (2n + 1) ’ , n = 0, ±1, ±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 π, phase shift is and frequency is .

2 π

graph

For the functions of the form y = A sin(ωt ± d) or y = A cos(ωt ± d) we

make the following de¬nitions:

1.2. TRIGONOMETRIC FUNCTIONS 17

2π 1 ω

period = , frequency = = ,

ω period 2π

d

amplitude = |A|, and phase shift = .

ω

The motion of a particle that follows the curves A sin(ωt±d) or A cos(ωt±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 ’ 2) (b) y = ’2 cos(2t ’ 1)

(d) y = 4 sin 2t ’ 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 di¬erence 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) ’ sin x (b) cos(x + h) ’ cos x (c) sin(5x) ’ sin(3x)

(d) cos(4x) ’ cos(2x) (e) sin(4x) + sin(2x) (f) cos(5x) + cos(3x)

’π π

¤ x ¤ . Take the sample points

5. Consider the graph of y = sin x,

2 2

π π π π1 π

’ , ’1 , ’ ,’ , (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 ’ x2 )(x ’ x3 )(x ’ x4 )(x ’ x5 )

P5 (x) = y1 +

(x1 ’ x2 )(x1 ’ x3 )(x1 ’ x4 )(x1 ’ x5 )

(x ’ x1 )(x ’ x3 )(x ’ x4 )(x ’ x5 )

+ ···

y2

(x2 ’ x1 )(x2 ’ x3 )(x2 ’ x4 )(x2 ’ x5 )

(x ’ x1 )(x ’ x2 )(x ’ x3 )(x ’ x4 )

+ y5 .

(x5 ’ x1 )(x5 ’ x2 )(x5 ’ x3 )(x5 ’ 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)

π

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

π

i) y = ’3 cos(2t ’ π)

g) y = ’2 sin t ’ h) y = 3 cos(2t + π)

6

k) y = ’2 cos(6t ’ π)

j) y = 2 sin(4t + π) l) y = 3 sin(6t + π)

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

b) y = ’3 tan x

a) y = 2 sec x c) y = 2 cot x

π

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

π π

π

h) y = 3 sec 2x + i) y = 2 sin πx +

g) y = 2 cot 3x + 2 3 6

8. Prove each of the following identities:

b) sin(3t) = 3 sin x ’ 4 sin3 x

a) cos 3t = 3 cos t + 4 cos3 t

sin3 t ’ cos3 t

4 4

c) sin t ’ cos t = ’ cos 2t d) = 1 + sin 2t

sin t ’ cos t

sin(x + y) tan x + tan y

e) cos 4t cos 7t ’ sin 7t sin 4t = cos 11t f) =

sin(x ’ y) tan x ’ tan y

1.3. INVERSE TRIGONOMETRIC FUNCTIONS 19

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

f (x + h) ’ f (x) cos h ’ 1 sin h

’ sin x

= cos x .

h h h

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

f (x + h) ’ f (x) cos h ’ 1 sin h

= sin x + cos x .

h h h

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

f (x) ’ f (t) cos(x ’ t) ’ 1 sin(x ’ t)

’ sin t

= cos t .

x’t x’t x’t

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

f (x) ’ f (t) cos(x ’ t) ’ 1 sin(x ’ t)

= sin t + cos t .

x’t x’t x’t

13. Prove that

1 ’ tan2 t

.

cos(2t) =

1 + tan2 t

x

14. Prove that if y = tan , then

2

1 ’ 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 de¬ne inverses, it is customary to restrict the domains in which

the functions are one-to-one as follows.

20 CHAPTER 1. FUNCTIONS

π π

1. y = sin x, ’ ¤ x ¤ , is one-to-one and covers the range ’1 ¤ y ¤ 1.

2 2

Its inverse function is denoted arcsin x, and we de¬ne y = arcsin x, ’1 ¤

π π

x ¤ 1, if and only if, x = sin y, ’ ¤ y ¤ .

2 2

graph

2. y = cos x, 0 ¤ x ¤ π, is one-to-one and covers the range ’1 ¤ y ¤ 1. Its

inverse function is denoted arccos x, and we de¬ne y = arccos x, ’1 ¤

x ¤ 1, if and only if, x = cos y, 0 ¤ y ¤ π.

graph

’π π

< x < , is one-to-one and covers the range ’∞ <

3. y = tan x,

2 2

y < ∞ Its inverse function is denoted arctan x, and we de¬ne y =

’π π

arctan x, ’∞ < x < ∞, if and only if, x = tan y, <y< .

2 2

graph

4. y = cot x, 0, x < π, is one-to-one and covers the range ’∞ < y < ∞. Its

inverse function is denoted arccot x, and we de¬ne y = arccot x, ’∞ <

x < ∞, if and only if x = cot y, 0 < y < π.

graph

1.3. INVERSE TRIGONOMETRIC FUNCTIONS 21

π π

5. y = sec x, 0 ¤ x ¤ or < x ¤ π is one-to-one and covers the range

2 2

’∞ < y ¤ ’1 or 1 ¤ y < ∞. Its inverse function is denoted arcsec x,

and we de¬ne y = arcsec x, ’∞ < x ¤ ’1 or 1 ¤ x < ∞, if and only

π

π

if, x = sec y, 0 ¤ y < or < y ¤ π.

2 2

graph

’π π

¤ x < 0 or 0 < x ¤ , is one-to-one and covers the

6. y = csc x,

2 2

range ’∞ < y ¤ ’1 or 1 ¤ y < ∞. Its inverse is denoted arccsc x and

we de¬ne y = arccsc x, ’∞ < x ¤ ’1 or 1 ¤ x < ∞, if and only if,

’π π

¤ y < 0 or 0 < y ¤ .

x = csc y,

2 2

Example 1.3.1 Show that each of the following equations is valid.

π

(a) arcsin x + arccos x =

2

π

(b) arctan x + arccot x =

2

π

(c) arcsec x + arccsc x =

2

To verify equation (a), we let arcsin x = θ.

graph

π

’ θ = x, as shown in the triangle. It follows

Then x = sin θ and cos

2

that

π

π

’ θ = arccos x, = θ + arccos x = arcsin x + arccos x.

2 2

The equations in parts (b) and (c) are veri¬ed in a similar way.

22 CHAPTER 1. FUNCTIONS

Example 1.3.2 If θ = arcsin x, then compute cos θ, tan θ, cot θ, sec θ and

csc θ.

π π

If θ is ’ , 0, or , then computations are easy.

2 2

graph

π π

Suppose that ’ < x < 0 or 0 < x < . Then, from the triangle, we get

2 2

√

√ 1 ’ x2

x

√

2,

cos θ = 1 ’ x tan θ = , cot θ = ,

x

1 ’ x2

1 1

sec θ = √ and csc θ = .

x

1 ’ x2

Example 1.3.3 Make the given substitutions to simplify the given radical

expression and compute all trigonometric functions of θ.

√

√

(a) 4 ’ x2 , x = 2 sin θ (b) x2 ’ 9, x = 3 sec θ

(c) (4 + x2 )3/2 , x = 2 tan θ

x

(a) For part (a), sin θ = and we use the given triangle:

2

graph

Then

√ √

4 ’ x2 4 ’ x2

x

, tan θ = √

cos θ = , cot θ = ,

2 x

2

4’x

2 2

sec θ = √ , csc θ = .

x

4 ’ x2

√

Furthermore, 4 ’ x2 = 2 cos θ and the radical sign is eliminated.

1.3. INVERSE TRIGONOMETRIC FUNCTIONS 23

x

(b) For part (b), sec θ = and we use the given triangle:

3

graph

Then,

√ √

x2 ’ 4 x2 ’ 4

3

sin θ = , cos θ = , tan θ =

x x 3

x

3

cot θ = √ , csc θ = √ .

x2 ’ 9 x2 ’ 9

√

Furthermore, x2 ’ 9 = 3 tan θ and the radical sign is eliminated.

x

(c) For part (c), tan θ = and we use the given triangle:

2

graph

Then,

x 2 2

sin θ = √ cos θ = √

, , cot θ = ,

x

2+4 2+4

x x

√ √

2+4 x2 + 4

x

sec θ = , csc θ = .

2 x

√

Furthermore, x2 + 4 = 2 sec θ and hence

(4 + x)3/2 = (2 sec θ)3 = 8 sec3 θ.

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:

√ √

a2 ’ x2 , x = a sin θ x2 ’ a2 , x = a sec θ

(a) (b)

√

(c) a2 + x2 , x = a tan θ.

Exercises 1.3

1. Evaluate each of the following:

√

1 3

(a) 3 arcsin + 2 arccos

2 2

1 1

(b) 4 arctan √ √

+ 5arccot

3 3

2

(c) 2arcsec (’2) + 3 arccos ’ √

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.

x’2

3+x

x

(a) √ (b) √ (c) √

9 ’ x2 x x2 ’ 25

16 + x2

2 ’ 2x

1+x

(d) √ (e) √

x2 ’ 2x ’ 3

x2 + 2x + 2

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

3. Some famous polynomials are the so-called Chebyshev polynomials, de-

¬ned by

Tn (x) = cos(n arccos x), ’1 ¤ x ¤ 1, n = 0, 1, 2, . . . .

1.3. INVERSE TRIGONOMETRIC FUNCTIONS 25

(a) Prove the recurrence relation for Chebyshev polynomials:

Tn+1 (x) = 2xTn (x) ’ Tn’1 (x) for each n ≥ 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 θ = arccos x, x = cos θ. Then Tn (x) = cos(nθ), Tn+1 (x) =

cos(nθ + θ), Tn’1 (x) = cos(nθ ’ θ). 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|m’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

√

√

’1 3

a) y = arccos b) y = arcsin c) y = arctan(’ 3)

2 2

√ √

√

’ 33

d) y = arccot e) y = arcsec (’ 2) f) y = arccsc (’ 2)

2 2

g) y = arcsec ’ √3 h) y = arccsc ’ √3 i) y = arcsec (’2)

√

’1

j) y = arccsc (’2) k) y = arctan l) y = arccot (’ 3)

√

3

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

a) 2 sin4 x = sin2 x b) 2 cos2 x ’ cos x ’ 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 ’ 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 ’1.

1.4 Logarithmic, Exponential and Hyperbolic

Functions

Most logarithmic tables have tables for log10 x, loge x, ex and e’x because

of their universal applications to scienti¬c 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, satis¬es 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 ’ 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 — by = bx+y

bx

= bx’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 re¬‚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 ’ e’x ) for all real x, read as hyperbolic sine of x.

1. sinh(x) =

2

1x

(e + e’x ), for all real x, read as hyperbolic cosine of x.

2. cosh(x) =

2

ex ’ e’x

sinh(x)

3. tanh(x) = =x , for all real x, read as hyperbolic tangent

e + e’x

cosh(x)

of x.

ex + e’x

cosh(x)

=x , x = 0, read as hyperbolic cotangent of x.

4. coth(x) =

e ’ e’x

sinh(x)

1 2

5. sech (x) = =x , for all real x, read as hyperbolic secant of

e + e’x

cosh x

x.

2

1

=x , x = 0, read as hyperbolic cosecant of x.

6. csch (x) =

e ’ e’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 ’ 3)3/4

y=

(1 + 7x)1/3 (2x + 3)3/2

1.4. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS29

(x + 1)3 (2x ’ 3)3/4

ln(y) = ln

(1 + 7x)1/3 (2x + 3)3/2]

= ln[(x + 1)3 (2x ’ 3)3/4 ] ’ ln[(1 + 7x)1/3 (2x + 3)3/2 ]

= ln(x + 1)3 + ln(2x ’ 3)3/4 ’ {ln(1 + 7x)1/3 + ln(2x + 3)3/2 }

3 1 3

= 3 ln(x + 1) + ln(2x ’ 3) ’ ln(1 + 7x) ’ ln(2x + 3)

4 3 2

Example 1.4.2 Solve the following equation for x:

log3 (x4 ) + log3 x3 ’ 2 log3 x1/2 = 5.

Using logarithm properties, we get

4 log3 x + 3 log3 x ’ 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