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Matematika Bisnis

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Introductory maths analysis chapter 00 official Presentation Transcript

  • 1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences ©2007 Pearson Education Asia Chapter 0Chapter 0 Review of AlgebraReview of Algebra
  • 2. ©2007 Pearson Education Asia INTRODUCTORY MATHEMATICAL ANALYSIS 0. Review of Algebra 1. Applications and More Algebra 2. Functions and Graphs 3. Lines, Parabolas, and Systems 4. Exponential and Logarithmic Functions 5. Mathematics of Finance 6. Matrix Algebra 7. Linear Programming 8. Introduction to Probability and Statistics
  • 3. ©2007 Pearson Education Asia 9. Additional Topics in Probability 10. Limits and Continuity 11. Differentiation 12. Additional Differentiation Topics 13. Curve Sketching 14. Integration 15. Methods and Applications of Integration 16. Continuous Random Variables 17. Multivariable Calculus INTRODUCTORY MATHEMATICAL ANALYSIS
  • 4. ©2007 Pearson Education Asia • To be familiar with sets, real numbers, real-number line. • To relate properties of real numbers in terms of their operations. • To review the procedure of rationalizing the denominator. • To perform operations of algebraic expressions. • To state basic rules for factoring. • To rationalize the denominator of a fraction. • To solve linear equations. • To solve quadratic equations. Chapter 0: Review of Algebra Chapter ObjectivesChapter Objectives
  • 5. ©2007 Pearson Education Asia Sets of Real Numbers Some Properties of Real Numbers Exponents and Radicals Operations with Algebraic Expressions Factoring Fractions Equations, in Particular Linear Equations Quadratic Equations Chapter 0: Review of Algebra Chapter OutlineChapter Outline 0.1) 0.2) 0.3) 0.4) 0.5) 0.6) 0.7) 0.8)
  • 6. ©2007 Pearson Education Asia • A set is a collection of objects. • An object in a set is called an element of that set. • Different type of integers: • The real-number line is shown as Chapter 0: Review of Algebra 0.1 Sets of Real Numbers0.1 Sets of Real Numbers { }...,3,2,1=integerspositiveofSet { }1,2,3..., −−−=integersnegativeofSet
  • 7. ©2007 Pearson Education Asia • Important properties of real numbers 1. The Transitive Property of Equality 2. The Closure Properties of Addition and Multiplication 3. The Commutative Properties of Addition and Multiplication Chapter 0: Review of Algebra 0.2 Some Properties of Real Numbers0.2 Some Properties of Real Numbers .then,andIf cacbba === .and numbersrealuniquearetherenumbers,realallFor abba + baababba =+=+ and
  • 8. ©2007 Pearson Education Asia 4. The Commutative Properties of Addition and Multiplication 5. The Identity Properties 6. The Inverse Properties 7. The Distributive Properties Chapter 0: Review of Algebra 0.2 Some Properties of Real Numbers ( ) ( ) ( ) ( )cabbcacbacba =++=++ and aaaa ==+ 1and0 ( ) 0=−+ aa 11 =⋅ − aa ( ) ( ) cabaacbacabcba +=++=+ and
  • 9. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.2 Some Properties of Real Numbers Example 1 – Applying Properties of Real Numbers Example 3 – Applying Properties of Real Numbers ( ) ( ) ( ) ( )354543b. 2323a. ⋅=⋅ +−=+− xwzywzyxSolution: a. Show that Solution: .0for ≠      = c c b a c ab ( )       =      ⋅=⋅= c b a c ba c ab c ab 11
  • 10. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.2 Some Properties of Real Numbers Example 3 – Applying Properties of Real Numbers b. Show that Solution: .0for c b ≠+= + c c a c ba ( ) c b c a c ba c ba 111 ⋅+⋅=+= + c b c a c ba c b c a c b c a += + +=⋅+⋅ 11
  • 11. ©2007 Pearson Education Asia • Properties: Chapter 0: Review of Algebra 0.3 Exponents and Radicals0.3 Exponents and Radicals 14. 1 3. 0for 11 2. 1. 0 = = ≠ ⋅⋅ == ⋅⋅= − − x x x x xxxxx x xxxxx n n factorsn n n factorsn n     n x exponent base
  • 12. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.3 Exponents and Radicals Example 1 – Exponents xx π = =−== == ==       =                        =      1 000 5 5- 5 5- 4 e. 1)5(,1,12d. 2433 3 1 c. 243 1 3 1 3b. 16 1 2 1 2 1 2 1 2 1 2 1 a.
  • 13. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.3 Exponents and Radicals • The symbol is called a radical. n is the index, x is the radicand, and is the radical sign. n x
  • 14. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.3 Exponents and Radicals Example 3 – Rationalizing Denominators Solution: Example 5 – Exponents ( ) x x x x xxx 3 32 3 32 3 2 3 2 3 2 b. 5 52 5 52 55 52 5 2 5 2 a. 6 55 6 566 5 1 6 1 6 5 6 1 2 1 2 1 2 1 2 1 2 1 === ⋅ = = ⋅ = ⋅ ⋅ == a. Eliminate negative exponents in and simplify. Solution: 11 −− + yx xy xy yx yx + =+=+ −− 1111
  • 15. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.3 Exponents and Radicals Example 5 – Exponents b. Simplify by using the distributive law. Solution: c. Eliminate negative exponents in Solution: ( )12/12/12/3 −=− xxxx 2/12/3 xx − ( ) .77 22 −− + xx ( ) ( ) 2222 22 49 17 7 17 77 xxxx xx +=+=+ −−
  • 16. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.3 Exponents and Radicals Example 5 – Exponents d. Eliminate negative exponents in Solution: e. Apply the distributive law to Solution: ( ) . 211 −−− − yx ( ) ( )2 222 22 211 11 xy yx xy xy xy xy yx yx − =      − =       − =      −=− −− −−− ( ).2 5 6 2 1 5 2 xyx + ( ) 5 8 2 1 5 2 5 6 2 1 5 2 22 xyxxyx +=+
  • 17. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.3 Exponents and Radicals Example 7 – Radicals a. Simplify Solution: b. Simplify Solution: 3233 33 323 46 )( yyxyyxyx =⋅⋅= 7 14 77 72 7 2 = ⋅ ⋅ = .3 46 yx . 7 2
  • 18. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.3 Exponents and Radicals Example 7 – Radicals c. Simplify Solution: d. If x is any real number, simplify Solution: Thus, and 210105 2152510521550250 += +−=+− .21550250 +− .2 x    <− ≥ = 0if 0if2 xx xx x 222 = ( ) .33 2 =−
  • 19. ©2007 Pearson Education Asia • If symbols are combined by any or all of the operations, the resulting expression is called an algebraic expression. • A polynomial in x is an algebraic expression of the form: where n = non-negative integer cn = constants Chapter 0: Review of Algebra 0.4 Operations with Algebraic Expressions0.4 Operations with Algebraic Expressions 01 1 1 cxcxcxc n n n n ++++ − − 
  • 20. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.4 Operations with Algebraic Expressions Example 1 – Algebraic Expressions a. is an algebraic expression in the variable x. b. is an algebraic expression in the variable y. c. is an algebraic expression in the variables x and y. 3 3 10 253 x xx − −− ( ) 2 3 + −+ y xyyx 2 7 5 310 y y + +−
  • 21. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.4 Operations with Algebraic Expressions Example 3 – Subtracting Algebraic Expressions Simplify Solution: ( ) ( ).364123 22 −+−+− xyxxyx ( ) ( ) ( ) ( ) 48 316243 )364()123( 364123 2 2 22 22 +−−= ++−−+−= +−−++−= −+−+− xyx xyx xyxxyx xyxxyx
  • 22. ©2007 Pearson Education Asia • A list of products may be obtained from the distributive property: Chapter 0: Review of Algebra 0.4 Operations with Algebraic Expressions
  • 23. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.4 Operations with Algebraic Expressions Example 5 – Special Products a. By Rule 2, b. By Rule 3, ( )( ) ( ) ( ) 103 5252 52 2 2 −−= −+−+= −+ xx xx xx ( )( ) ( ) 204721 45754373 4753 2 2 +−= ⋅+⋅+⋅+⋅= ++ zz zz zz
  • 24. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.4 Operations with Algebraic Expressions Example 5 – Special Products c. By Rule 5, d. By Rule 6, e. By Rule 7, ( ) ( ) 168 442 4 2 22 2 +−= +−= − xx xx x ( )( ) ( ) 8 31 3131 2 2 2 2 22 −= −+= −+++ y y yy ( ) ( ) ( )( ) ( ) ( ) ( ) 8365427 23233233 23 23 3223 3 +++= +++= + xxx xxx x
  • 25. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.4 Operations with Algebraic Expressions Example 7 – Dividing a Multinomial by a Monomial z zz z zzz x x xx 3 2 3 42 2 6384 b. 3 3 a. 2 23 2 3 −+−= −+− += +
  • 26. ©2007 Pearson Education Asia • If two or more expressions are multiplied together, the expressions are called the factors of the product. Chapter 0: Review of Algebra 0.5 Factoring0.5 Factoring
  • 27. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.5 Factoring Example 1 – Common Factors a. Factor completely. Solution: b. Factor completely. Solution: xkxk 322 93 + ( )kxxkxkxk 3393 2322 +=+ 224432325 268 zxybayzbayxa −− ( )24232232 224432325 342 268 xyzbazbyxaya zxybayzbayxa −−= −−
  • 28. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.5 Factoring Example 3 – Factoring ( ) ( )( ) ( )( ) ( ) ( )zzzz xxx yyyyyy xxxx xxx +=+ −=+− +−=−+ ++=++ +=++ 1e. 396d. 23231836c. 2313299b. 4168a. 4/14/54/1 22 23 2 42 ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )( ) ( )( )( )( )2222 333366 23 2222 3/13/13/13/2 24 j. 2428i. bbaah. 4145g. 1111f. yxyxyxyxyxyx yxyxyx xxxx bayxyxyxyx xxxx xxxx ++−+−+= −+=− ++−=− +−+=−+− −−=+− −++=−
  • 29. ©2007 Pearson Education Asia Simplifying Fractions • Allows us to multiply/divide the numerator and denominator by the same nonzero quantity. Multiplication and Division of Fractions • The rule for multiplying and dividing is Chapter 0: Review of Algebra 0.6 Fractions0.6 Fractions bd ac d c b a =            bc ad d c b a =÷
  • 30. ©2007 Pearson Education Asia Rationalizing the Denominator • For a denominator with square roots, it may be rationalized by multiplying an expression that makes the denominator a difference of two squares. Addition and Subtraction of Fractions • If we add two fractions having the same denominator, we get a fraction whose denominator is the common denominator. Chapter 0: Review of Algebra 0.6 Fractions
  • 31. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.6 Fractions Example 1 – Simplifying Fractions a. Simplify Solution: b. Simplify Solution: . 127 6 2 2 +− −− xx xx ( )( ) ( )( ) 4 2 43 23 127 6 2 2 − + = −− +− = +− −− x x xx xx xx xx . 448 862 2 2 xx xx −− −+ ( )( ) ( )( ) ( )22 4 214 412 448 862 2 2 + + = +− +− = −− −+ x x xx xx xx xx
  • 32. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.6 Fractions Example 3 – Dividing Fractions ( ) ( )( ) ( ) ( )( )41 2 82 1 1 4 1 82 1 4 c. 32 5 2 1 3 5 2 3 5 b. 32 5 3 5 25 3 2 a. 222 2 ++ = + − ⋅ − = − + − − − =⋅ − − =− − ++ − = + − ⋅ + = − + ÷ + xxxx x x x x xx x x xx x xx x x x x xx xx x x x x x x x x
  • 33. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.6 Fractions Example 5 – Adding and Subtracting Fractions ( ) ( ) 2 33 2 235 2 23 2 5 a. 2 2 2 − −+ = − ++− = − + + − − p pp p pp p p p p ( )( ) ( )( ) ( ) ( )( ) 3 4 32 2 31 41 65 2 32 45 b. 2 2 2 2 + −= ++ + − +− −− = ++ + − −+ +− x xx xx xx xx xx xx xx xx ( ) ( ) ( ) 1 7 7 7 425 149 84 7 2 7 5 c. 22 2 22 = − − = − −+−−−+ = +− +− + − − − − −+ x x x xxx xx x x x x xx
  • 34. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.6 Fractions Example 7 – Subtracting Fractions ( ) ( )( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )332 615 332 6512102 332 32322 92 2 96 2 2 2 2 22 222 −+ +− = −+ −−−+− = −+ ++−−− = − + − ++ − xx xx xx xxxx xx xxxx x x xx x
  • 35. ©2007 Pearson Education Asia Equations • An equation is a statement that two expressions are equal. • The two expressions that make up an equation are called its sides. • They are separated by the equality sign, =. Chapter 0: Review of Algebra 0.7 Equations, in Particular Linear Equations0.7 Equations, in Particular Linear Equations
  • 36. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.7 Equations, in Particular Linear Equations Example 1 – Examples of Equations zw y y xx x −= = − =++ =+ 7d. 6 4 c. 023b. 32a. 2 • A variable (e.g. x, y) is a symbol that can be replaced by any one of a set of different numbers.
  • 37. ©2007 Pearson Education Asia Equivalent Equations • Two equations are said to be equivalent if they have exactly the same solutions. • There are three operations that guarantee equivalence: 1. Adding/subtracting the same polynomial to/from both sides of an equation. 2. Multiplying/dividing both sides of an equation by the same nonzero constant. 3. Replacing either side of an equation by an equal expression. Chapter 0: Review of Algebra 0.7 Equations, in Particular Linear Equations
  • 38. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.7 Equations, in Particular Linear Equations Operations That May Not Produce Equivalent Equations 4. Multiplying both sides of an equation by an expression involving the variable. 5. Dividing both sides of an equation by an expression involving the variable. 6. Raising both sides of an equation to equal powers.
  • 39. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.7 Equations, in Particular Linear Equations Linear Equations • A linear equation in the variable x can be written in the form where a and b are constants and . • A linear equation is also called a first-degree equation or an equation of degree one. 0=+ bax 0≠a
  • 40. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.7 Equations, in Particular Linear Equations Example 3 – Solving a Linear Equation Solve Solution: .365 xx =− ( ) ( ) 3 2 6 2 2 62 60662 062 33365 365 = = = +=+− =− −+=−+− =− x x x x x xxxx xx
  • 41. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.7 Equations, in Particular Linear Equations Example 5 – Solving a Linear Equations Solve Solution: .6 4 89 2 37 = − − + xx ( ) ( ) ( ) 2 105 24145 2489372 64 4 89 2 37 4 = = =+ =−−+ =      − − + x x x xx xx
  • 42. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.7 Equations, in Particular Linear Equations Literal Equations • Equations where constants are not specified, but are represented as a, b, c, d, etc. are called literal equations. • The letters are called literal constants.
  • 43. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.7 Equations, in Particular Linear Equations Example 7 – Solving a Literal Equation Solve for x. Solution: ( ) ( ) ( ) ac a x aacx aaxxxcxax axxxca − = =− ++=++ +=++ 2 2 222 22 2 ( ) ( )22 axxxca +=++
  • 44. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.7 Equations, in Particular Linear Equations Example 9 – Solving a Fractional Equation Solve Solution: Fractional Equations • A fractional equation is an equation in which an unknown is in a denominator. . 3 6 4 5 − = − xx ( )( ) ( )( ) ( ) ( ) x xx x xx x xx = −=−       − −−=      − −− 9 4635 3 6 34 4 5 34
  • 45. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.7 Equations, in Particular Linear Equations Example 11 – Literal Equation If express u in terms of the remaining letters; that is, solve for u. Solution: , vau u s + = ( ) ( ) sa sv u svsau usvsau uvaus vau u s − = −=− =+ =+ + = 1 1
  • 46. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.7 Equations, in Particular Linear Equations Radical Equations • A radical equation is one in which an unknown occurs in a radicand. Example 13 – Solving a Radical Equation Solve Solution: .33 −=−− yy 4 2 126 963 33 = = = +−=− −=−− y y y yyy yy
  • 47. ©2007 Pearson Education Asia • A quadratic equation in the variable x is an equation that can be written in the form where a, b, and c are constants and • A quadratic equation is also called a second- degree equation or an equation of degree two. Chapter 0: Review of Algebra 0.8 Quadratic Equations0.8 Quadratic Equations 02 =++ cbxax .0≠a
  • 48. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.8 Quadratic Equations Example 1 – Solving a Quadratic Equation by Factoring a. Solve Solution: Factor the left side factor: Whenever the product of two or more quantities is zero, at least one of the quantities must be zero. .0122 =−+ xx ( )( ) 043 =+− xx ( ) ( ) 43 04or03 −== =+=− xx xx
  • 49. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.8 Quadratic Equations Example 1 – Solving a Quadratic Equation by Factoring b. Solve Solution: .56 2 ww = ( ) 6 5 or0 056 56 2 == =− = ww ww ww
  • 50. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.8 Quadratic Equations Example 3 – Solving a Higher-Degree Equation by Factoring a. Solve Solution: b. Solve Solution: .044 3 =− xx ( ) ( )( ) 1or1or0 0114 014 044 2 3 −=== =+− =− =− xxx xxx xx xx ( ) ( ) ( ) .0252 32 =++++ xxxxx ( ) ( ) ( ) ( ) ( ) ( )[ ] ( ) ( ) 7/2or2or0 0722 0252 0252 2 2 32 −=−== =++ =++++ =++++ xxx xxx xxxx xxxxx
  • 51. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.8 Quadratic Equations Example 5 – Solution by Factoring Solve Solution: .32 =x ( )( ) 3Thus, 3or3 033 03 3 2 2 ±= −== =+− =− = x xx xx x x
  • 52. ©2007 Pearson Education Asia Quadratic Formula • The roots of the quadratic equation can be given as Chapter 0: Review of Algebra 0.8 Quadratic Equations 02 =++ cbxax a acbb x 2 42 −±− =
  • 53. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.8 Quadratic Equations Example 7 – A Quadratic Equation with One Real Root Solve by the quadratic formula. Solution: Here a = 9, b = 6√2, and c = 2. The roots are 09262 2 =++ yy ( ) 3 2 18 026 or 3 2 18 026 92 026 −= −− =−= +− = ±− = yy y
  • 54. ©2007 Pearson Education Asia Quadratic-Form Equation • When a non-quadratic equation can be transformed into a quadratic equation by an appropriate substitution, the given equation is said to have quadratic-form. Chapter 0: Review of Algebra 0.8 Quadratic Equations
  • 55. ©2007 Pearson Education Asia Chapter 0: Review of Algebra 0.8 Quadratic Equations Example 9 –– Solving a Quadratic-Form Equation Solve Solution: This equation can be written as Substituting w =1/x3 , we have Thus, the roots are .08 91 36 =++ xx 08 1 9 1 3 2 3 =+      +      xx ( )( ) 1or8 018 0892 −=−= =++ =++ ww ww ww 1or 2 1 1 1 or8 1 33 −=−= −=−= xx xx