I.real Number

8/2/2019 I.real Number

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1.REAL NUMBERS



Calculus I (MA 1113) - Faculty of ScienceTelkom Institut of Technology

Schedule : Twice a week

1st lecture : Tuesday, 08.30-10.30 at A207A

2nd lecture : Thursday, 08.30-10.30 at A206A

Mark system

Middle Test 40%

Final Tet 40%

Quiz 20%




The Sillaby :

Chapter I Real numbers

Chapter II Function

Chapter III Limit and continuity

Chapter IV Differentiation

Chapter V Application of derivativeMidtest

Chapter VI Integral

Chapter VII Application of integral

Chapter VIII Transcendental functionChapter IX Techniques of integration

Chapter X Improrer integral

Final test




REFERENCES :

Purcell, dkk., Calculus with Geometry. Purcell, dkk., Kalkulus dan Geomteri Analitis Jilid I ,

terjemahan Penerbit Erlangga, Jakarta



Pre-test

Simplify:


. 4 3 6 13 2 5 9 ...a

1 2 1 1 1 2. ...3 5 2 3 5 9

b

2 2

2 2

6 2. . ...

1 5 6

x x x xc

x x x

3 3 3. 2 4 2 16 ...d

2

18 4 6. ...

33e

x x x x

2

2

3 4 3. ...5 5

1 3

x

x x x f

x x



Sistem bilangan

N : naturalnumber

Z : integerQ : rational number

R : real number

N : 1, 2, 3, ...

Z : …,-2,-1,0,1,2,..

0,,, b Z bab

aqQ :

R Q Irrational

,3,2

Irrational number :

Calculus I (MA 1113) - Faculty of Science Telkom Institut of Technology



Garis bilangan

0 1

The real number corresponding to a point on the line (real line)

-3

2

A line segment on a real line is said interval

Interval




Intervals

Set notation Interval notation

{ } a x x < a ,

{ } a x x ] a ,

{ } b x a x < < b a ,

{ } b x a x [ ] b a ,

{ } b x x > , b

{ } b x x [ , b

{ } x x ,

Geometric picture

a

a

a b

a b

b

b




Properties of Real Number

• Order properties :

Trichotomy

Let x and y are real numbers, then either x < y or x > y or

x = y Transitive

If x < y and y < z then x < z

Multiplication

Let z is a positive number and x < y then xz < yz.Let z is a negative number and x < y then xz > yz




Inequality

Inequality :

where A(x), B(x), D(x), E(x) are polynomialand B(x) ≠ 0, E(x) ≠ 0

x E

x D

x B

x A<




A solution of an inequality in an unknown x is avalue for x that makes the inequality a truestatement.

Steps to find solution of inequality :

1. Write inequality as :

0)(

)(<

xQ

xP




2. Factoring P(x) and Q(x) into a product linearfactors (ax+b) and/or irreducible quadraticfactors ( )

3. Find zero’s of linear factor and put these point

on real line

4. Use test points to find sign plus(+) or minus(-)

24 0

2

with bax bx c ac <




ExampleSolve

53213 x

352313 x

8216 x48 x

84 x

[ ]8,4Solution =

4 8

1




8462 < x

248 < x

248 > x

842 < x

22

1

< x

2,2

1

22

1

solution

2




03522 < x x

0312 < x x

Zero’s :21 x and 3 x

3

++ ++--

21

3

Solution =

3,

2

1




637642 x x x

x x 7642 6376 x xand

4672 x x and 6637 x x

4

109 x 010 xand

9

10 x 010 xand

9

10 x and 0 x




Solution= [

,09

10,

09

10

Thus

solution =

9

10,0




13

2

1

1

<

x x

0

13

2

1

1<

x x

0131

2213<

x x

x x

5.

0

1313 <

x x x

Zero : -1,1

3, 3

3

++ ++--

-1

--

1

3

Solution = 1

, 1 ,33




The Inequality with Absolute Value

Definition : The absolute value of realnumber x is defined by

<

0,

0,

x x

x x x

Geometrically : |x| is distance from x to original point (0)




Properties of absolute value :

y

x

y

x

2 x x

a xaaa x 0,

a xaa x 0, or a x y x

22 y x

6 The triangle inequality

y x y x

1

2

3

4

5

y x y x




Example

41 << x

Solve :

352 < x

Use second properties

3523 << x

53235 << x

822 << x

Solution = 4,11 4

1.




5432 x x2.

Use the fourth properties

22 5432 x x

254016912422 x x x x

01628122 x x

23 7 4 0 x x

3

4zero: , -1




Solution =4

[ , 1]

3

Plot zero on real line :

-13

4

++--++




<

22

22

2 x x

x x

x

<

11

111

x x

x x x

Real line is divided into 3 interval :

-1 2

I II III

1, [ 2,1 [ ,2

3. 2123 x x

Use definition of absolute value :




1< x

2123 x x

2123 x x

2136 x x

227 x

92 x

92 x

2

9 x

2

9,

I. For

or




1,2

9,

29-1

Solution1 =

Thus

Solution1 = 1,




21 < xII. For

2123 x x

2123 x x2136 x x

245 x

74 x

74 x

4

7 x

4

7,or




Solution2 = [ 2,14

7,

-1 24

7

thus

Solution2 =

4

7,1




Contoh : Menentukan HimpunanPenyelesaian

2 x

2123 x x

2123

x x2163 x x

272 x

52 x

III. For

2

5 x

,

2

5or




Contoh : Menentukan HimpunanPenyelesaian

Solution3 = [

,2,2

5

22

5

thus

Solution3 =

,

2

5




Solution = solution1 solution2 solution3

7 5

, 1 1, ,4 2

7 5, ,4 2




Problem

6. 2 3 4 5 x x Solve

5. 2 3 2 3 x x

21. 1

4 2

x x

x

2

2 1

2. 3

x x

x x

9. 3 2 x x

Calculus I (MA 1113) Faculty of Science Telkom Institut of Technology

47. 1 1

x <

2 18. 3

5

x

x

210.

1 1

x x

x x

2 53. 1

2 x x >

24. 2 3 0 x x >

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I.real Number