Lectures in Calculus 1 - Lecture notes 1-45 PDF

Title Lectures in Calculus 1 - Lecture notes 1-45
Author Mohammed Althagafi
Course Mathematics Htam
Institution King Abdulaziz University
Pages 88
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Download Lectures in Calculus 1 - Lecture notes 1-45 PDF


Description

Lectur Lectures es in

Calculus (1) 101-1104

)1( ‫تفاضل وتكامل‬ second semester

1440 h

1

Chapter 1 Functions ELEMENTARY FUNCTIONS Definition: A function f : X → Y assigns to each element of the set X an element of Y. Picture a function as a machine,

The set X of inputs is called the domain of the function f. The set Y of all conceivable outputs is the codomain of the function f. The set of all outputs is the range of f. (The range is a subset of Y .) The most important criteria for a function is this: A function must assign to each input a unique output. Functions are often grouped into families, according to the form of their defining formulas or other common characteristics.

In this section we will discuss some of the most basic elementary functions.

2

LINEAR FUNCTIONS A function equation of the form y = m x + b is called linear function. It represents a line of slope

m and Y-intercept b . Example 1: Sketch the graph of the function f ( x) = 2 x + 3 and use the graph to find the domain and range. Solution

x

-4

-3

-2

-1

0

1

2

3

4

f(x)

-5

-3

-1

1

3

5

7

9

11

The domain is

and the range is

POWER FUNCTIONS A function of the form y = a x n , a  0 (constant), n 

is called a power function.

Example 2: Sketch the graph of the function y = x 2 and use the graph to find the domain and range.

3

Solution x

y

-4

16

-3

9

-2

4

-1

1

0

0

1

1

2

4

3

9

4

16 The domain is

and the range is [0, ) .

Example 3: −2 Sketch the graph of the function y = x and use the graph to find the domain and range.

Solution

x

y

-4

1/16

-3

1/9

-2

1/4

-1

1

1

1

2

1/4

3

1/9

4

1/16

f ( x) = x −2

and the range is (0, ) .

The domain is

4

1 n

POWER FUNCTIONS WITH NONINTEGER EXPONENTS y = x = n x Example 4: Sketch the graph of the function y = x and use the graph to find the domain and range. Solution

x

y

0

0

1

1

4

2

9

3

16

4 The domain is [0, ) and the range is [0, ) .

POLYNOMIALS A polynomial in x is a function that is expressible as a sum of finitely many terms of the form

c xn , where c is a constant and n is a nonnegative integer. Some examples of polynomials are 3 x +5, 3 x2 − x + 4, x3 + x −1, x4 + x2 − x + 2 The highest power of x that occurs with a nonzero coefficient is called the degree of the polynomial. Nonzero constant polynomials are considered to have degree 0, Polynomials of degree 1, 2, 3, 4,and 5 are described as linear, quadratic, cubic, quartic, and quintic, respectively.

5

Example 5: Sketch the graph of the function y = x2 + 2 x − 3 and use the graph to find the domain and range. Solution

x

y

-3

0

-2

-3

-1

-4

0

-3

1

0

2

5

3

12

4

21 The domain is

and the range is [ −4, ) .

RATIONAL FUNCTIONS A function that can be expressed as a ratio of two polynomials is called a rational function. If p( x) and q( x ) are polynomials, then the rational function f ( x ) =

Examples:

y=

p( x ) , q( x )  0 . q (x )

x 2x x 2 +1 f x ( ) = , , g (x ) = 4 3 2 x − 27 x +x 2 +1 x +1

Shifting a graph of a function Let us consider the most important transformation of the graph of the function

y = f (x ) :

(1) The graph f (x − a) , represents the initial graph shifted in the direction of the X-axis, by a quantity equal to a . (2) The graph f (x ) + b , represents the result of translating the initial graph about the Y-axis, by a quantity equal to b . 6

Remark: (Vertical and horizontal shifts) Let c be a positive constant number. Then we obtain the graph of (i) y = f ( x ) + c , by shifting the graph y = f (x ) a distance c units upward. (ii) y = f ( x ) − c , by shifting the graph y = f (x ) a distance c units downward. (iii) y = f ( x + c ) , by shifting the graph y = f (x ) a distance c units to the left. (iv) y = f ( x −c ) , by shifting the graph y = f (x ) a distance c units to the right

ALGEBRAIC FUNCTIONS Functions that can be constructed from polynomials by applying finitely many algebraic operations (addition, subtraction, multiplication, division, and root extraction) are called algebraic functions. Some examples are 3

f ( x) = x − x + 1, 2

x4 h( x) = x+ 3

g( x) = x ( x + 9),

Example 6: Sketch the graph of the function y = x − 2 by making a transformation of the original function

y= x Solution

y=

x y=

Domain = [2, ], Range = [0, ]

7

x −2

The absolute value function The absolute value of a real number x , is defined by the formula:

x x =  −x

x 0 x 0

if if

3 = 3, 0 = 0, − 5 = 5

For example:

More generally,

x − y x− y =  y − x

if

x y

if

xy

The absolute value function has the following properties: Properties of absolute values (1)

−a = a

2) a b = a b ,

(3) a + b  a + b

a a = b b

and

(the triangle inequality).

Example 7: Sketch the graph of the function f ( x ) = x and use the graph to find the domain and range. Solution:

x

y

-2

2

-1

1

0

0

The domain is

1

1

the range is [0, ).

2

2

8

and

Equations and Inequalities Involving Absolute Values

1) x = D  either x = D or x = −D 2) x  D  − D  x  D 3) x  D  − D  x  D 4) x  D  either x  D or x −D 5) x  D  either x  D or x  −D

More generally:

1) x− a = D  either x = a − D or x = a + D 2) x −a  D  a − D  x a + D 3) x− a  D  a− D x  a + D 4) x −a  D  either x a +D or x  a −D 5) x− a  D  either x  a + D or x  a − D

Example (8) Solve the equation : 2x Solution

2 x + 5 = 3  2x + 5 = 3 . So, either 2 x = −3 − 5 = −8 or 2 x = 3 − 5 = −2 . The solutions are x = −4 and

x = −1.

9

Example (9) Solve the inequality :

3x − 2  1

Solution:

3x − 2  1  −1  3x − 2  1. We solve this pair of inequalities: 3 x − 2  1  3 x  1 + 2  x 1 

and

 3 x − 2 −1   3 x  − 1+ 2  x1 3 

Thus the solutions lie in the interval [ 13 ,1]

Definition (Even and Odd functions) A function f is said to be an even function if f ( x ) = f (− x ) , and odd one if f ( x ) = − f (− x ) . Notice that, the graphs of even functions are symmetric about the y-axis, while graphs of odd functions are symmetric about the origin. Some examples of even functions 2 4 6 3 5 7 are x , x , x ; and some examples of odd functions are x , x , x .

10

THE TRIGONOMETRIC FUNCTIONS ANGLES There are two standard measurement systems for describing the size of an angle: degree measure 0

and radian measure. In degree measure, one degree (written 1 ) is the measure of an angle

1 of one revolution. Thus, there are 3600 in an angle of one revolution, 180◦ in generated by 360 an angle of one-half revolution, 900 in an angle of one-quarter revolution (a right angle). In radian measure, angles are measured by the length of the arc that the angle subtends on a circle of radius 1 when the vertex is at the center. One unit of arc on a circle of radius 1 is called one radian (written 1 radian or 1 rad) and hence the entire circumference of a circle of radius 1 is 2  radians. It follows that an angle of 3600 subtends an arc of 2  radians, an angle of 1800 subtends an arc of  radians, an angle of 90 0 subtends an arc of 2 radians.

The relationship between degree measure and radian measure for some important positive angles. DEGREES

300

45 0

600

900

1200

1350

1500

1800

2700

3600

RADIANS









2 3

3 4

5 6



3 2

2

6

4

3

2

From the fact that  radians correspond to 1800 , we obtain the following formulas, which are useful for converting from degrees to radians and conversely.

1 = 0

 rad  0.01745 rad 180

and

1 rad = (

Example 1: (a) Express the angle 3900 in radians. (b) Express the angle 3 in degrees. Solution 0 ( a) 390 = (390 

(b) 3 = (3 

 180

) rad =

13 rad 6

180 0 ) = 540 0



11

1800



)  570

Definition of (sin t and cos t) If the sides of the triangle are labeled "hyp" for hypotenuse, "adj" for the side adjacent to angle t, and ''opp'' for the side opposite angle t, then

cos t =

adj hyp

and

sin t =

opp hyp

Let C be the circle with center at the origin 0 and radius 1 , its equation is x2 + y2 = 1. Let A be the point (1,0) on C For any real number t , let Pt be the point on C . See the following Figure:

For any real t , the cosine of t (abbreviated cos t) and the sine of t (abbreviated sin t) are the X and Y-coordinates of the point P t

cos t = the X- coordinate of Pt ,

sin t = the Y- coordinate of Pt

12

Some Useful Identities: Many important properties of cos t and sin t follow from the fact that they are coordinates of the point

Pt on the circle C with equation x 2 + y 2 = 1.

− 1  cos t  1 and

(1)

−1  sin t  1

(2) cos 2 t + sin 2 t = 1

cos(t + 2 ) = cos t and sin(t + 2 ) = sin t (4) cos(2 − t)= sin t , and sin(2 − t)= cos t

(3)

sin( − t ) = sin t

(5)

cos( − t ) = − cos t

(6)

cos(−t ) = cos t and sin( −t ) = − sin t

(7)

cos(s + t ) = cos s cos t − sin s sin t

(8)

sin( s + t ) = sin s cos t + cos s sin t

(9)

cos(s − t) = cos s cos t + sin s sin t

and

(10)

sin( s − t ) = sin s cos t − cos s sin t

(11)

sin 2t = 2 sin t cos t

(12) cos 2t = cos2 t − sin2 t (13)

cos 2 t

(14) cos 2t = 1 − 2 sin2 t

(15) cos 2 t

2

2

13

cos and sin of special angles: Degrees

0

30

45

60

90

120

135

150

180

Radians

0









2 3

3 4



sin

1

1

3 2

1

3 2

1 2

cos

0

3 2

1 2 1 2

5 6 1

1

0

− 3 2

-1

6 2

4

3

2

2

The graph of sin x:

The graph of cos x :

14

−1

2

2

2

0

Definition of : (tan , cot , sec, cosec )

tan x = sin x , cot x = cos x , sec x = 1 , csc x = 1 cos x sin x cos x sin x Some Useful Identities:

1) 1

and

2) tan( x + y) =

tan x+ tan y 1− tan x tan y

3) tan( x − y ) =

tan x− tan y 1+ tan x tan y

1

The graph of the functions tan x, and cot x:

15

The graph of functions: sec x, and csc x:

16

The following table shows the Domain and the Range of the trigonometric functions : Function

sin x cos x tan x

Domain





sec x

[ −1,1] [ −1,1]

3 5 7  ,  ,  ,... 2 2 2

2

,  2 ,  3 ,...

cot x

Range





3 5 7  ,  ,  ,... 2 2 2

2

)

,  2 ,  3 ,...

csc x

)

Notice that, for all values of x , we have :

−1  sin x  1, and

−1  cos x  1,

or, in terms of absolute values :

| sin x |  1,

and

| cos x |  1,

The functions sin x and cos x have period 2 and the tangent function tan x has period 

i.e sin( x + 2 ) = sin x,

cos( x+ 2 )= cos x, tan( x + ) =tan x

17

Example 2 Evaluate the trigonometric functions of

( a)  = 45o (b )  = 30 o ( c)

 = 60 o

Solution

(a ) sin 45o = csc 45o = 2,

(b )

(c )

1 , 2

cos 45o =

1 , 2

sec 45o = 2,

cos 60o =

1 , 2

cot 45o = 1

3 , 2 2 csc30o = 2, sec 30o = , 3

sin 30o =

1 , 2

sec60o = 2,

tan 45o = 1

cos 30o =

3 , 2 2 csc 60o = , 3 sin 60o =

1

tan 30o =

3

cot 30o = 3 tan 60o =

3 1 3

cot 60o =

Example 3 Evaluate the trigonometric functions of

( a)  = (b )

=

 4



and  =

6

 3

Solution Since  =

 4

= 45 o,

=

 6

= 30 o and

=

 3

= 60 o this problem is equivalent to the above.

18

Example 4 If tan  =3 . Find the exact values of remaining five trigonometric functions. solution

1 3 3,  cot = , sin = 3 10

cos =

3

1 10 , sec  = 10, csc  = 3 10 1

Exponential functions x The function f ( x) = a , where a > 0 , is called an exponential function with base a . Some

examples are

( )

x f ( x) = 1 4 ,

x

f ( x) = 5 ,

f ( x) = e x where e

2

The domain of f(x) is (−, ) and the range is (0, ) If a and b are positive numbers and x,y are any real numbers, then

(1)

a x . a y = a x +y ,

(2)

ax = a x− y y a

(3)

(a )

(4)

( a. b ) x = a x . b x

x

y

=a

x .y

19

Example 5: Sketch the graph of the function y = 2 x . Solution x

y=f(x)

-3

1/8

-2

1/4

-1

1/2

0

1

1

2

2

4

3

8

The natural exponential function Among all possible bases for exponential functions there is one particular base that plays a special role in calculus. That base, denoted by the letter e, is a certain irrational number whose value to six decimal places is e  2.718282 f (x ) = e x is called the natural exponential function. To simplify typography, the natural exponential function is sometimes written as exp( x ) , in which case the relationship e x +y =ex . e y would be expressed as exp(x + y)= exp(x)exp(y)

The function

The graph of the function f ( x) = e

x

Logarithmic functions The function f (x ) = loga x is called the logarithmic function with base a

log a x = y  a y = x For example : log 232 = 5  2 5 = 32

and log 381 = 4  34 = 81

The Domain and Range of logarithmic function The domain of the logarithmic function is (0, ) and the range is

)

Remark: We write loga x, for a = 10 as log x and loga x for a = e as ln x.

Properties for logarithmic functions

(1) loga (a x ) = x , ln(e x ) = x  x  log x ln x (2) a a = x, e = x  x 

Remark: The logarithms with any base can be expressed in terms of the natural logarithm by

loga x =

ln x , ln a

For example: log 3 7 =

ln 7 ln 3

21

Example 6 Sketch the graph of the function y = log2 x Solution x

y

1/8

-3

1/4

-2

1/2

-1

1

0

2

1

4

2

8

3

Example 7 Sketch the graph of the function y = log( x ) Solution x

y

0.001

-3

0.01

-2

0.1

-1

1

0

10

1

100

2

1000

3

22

Note the relationship between the functions:

e x and ln(x )

Example 8 Find the value of

x in each equation :

(a ) log 3 x = −2 1 (b ) ln( ) = −3 x x (c ) 3 = 5 (d ) 3e 5x = 11 Solution (a) log 3 x = −2  3 x = 10 −2  x = 10 −6

1 1 (b ) ln( ) = −3  = e −3  x = e 3 x x (c ) 3 x = 5  ln 3 x = ln 5  x ln 3 = ln 5  x =

ln 5 ln 3

5x 11 11 1 11 (d ) 3e 5 x = 11  e 5x = 11 3  ln e = ln 3  5x = ln 3  x = 5 ln 3

23

Example 9

  813 x Expand the logarithm : log 3  2 3   ( x + 2) ( x − 1)  Solution

  813 x log 3  log 3(813 x ) − log 3(( x + 2)2 ( x − 1)3 )  ( x + 2) 2 ( x −1) 3  =   = log 3(81) + log 3( 3 x) − log 3(( x + 2)2 ) − log 3(( x −1)3 ) 1

= log 3(3 4) + log 3( x 3) − 2 log 3( x + 2) − 3log 3( x −1) 1 = 4 log3 (3) + log3 (x ) − 2 log3 (x + 2) − 3log3 (x − 1) 3 1 = 4 + log 3 ( x) − 2 log 3( x + 2) − 3log 3 ( x −1) 3 Example 10 Expand the logarithm

 5 x2 + 1 log   100( x2 −1)3 

   

Solution  5 x2 +1  log   = log( 5 x 2 + 1 ) − log(102 (x 2 − 1)3 )  100( x 2 − 1) 3    1


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