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# Introductory maths analysis chapter 13 official

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### Introductory maths analysis chapter 13 official

1. 1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences ©2007 Pearson Education Asia Chapter 13Chapter 13 Curve SketchingCurve Sketching
2. 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. 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. 4. ©2007 Pearson Education Asia • To find critical values, to locate relative maxima and relative minima of a curve. • To find extreme values on a closed interval. • To test a function for concavity and inflection points. • To locate relative extrema by applying the second-derivative test. • To sketch the graphs of functions having asymptotes. • To model situations involving maximizing or minimizing a quantity. Chapter 13: Curve Sketching Chapter ObjectivesChapter Objectives
5. 5. ©2007 Pearson Education Asia Relative Extrema Absolute Extrema on a Closed Interval Concavity The Second-Derivative Test Asymptotes Applied Maxima and Minima 13.1) 13.2) 13.3) Chapter 13: Curve Sketching Chapter OutlineChapter Outline 13.4) 13.5) 13.6)
6. 6. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.1 Relative Extrema13.1 Relative Extrema Increasing or Decreasing Nature of a Function • Increasing f(x) if x1 < x2 and f(x1) < f(x2). • Decreasing f(x) if x1 < x2 and f(x1) > f(x2).
7. 7. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.1 Relative Extrema Extrema
8. 8. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.1 Relative Extrema RULE 1 - Criteria for Increasing or Decreasing Function • f is increasing on (a, b) when f’(x) > 0 • f is decreasing on (a, b) when f’(x) < 0 RULE 2 - A Necessary Condition for Relative Extrema ( ) ( )     = ⇒    existnotdoes' or 0' at extremumrelative af af a implies
9. 9. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.1 Relative Extrema RULE 3 - Criteria for Relative Extrema 1. If f’(x) changes from +ve to –ve, then f has a relative maximum at a. 2. If f’(x) changes from -ve to +ve, then f has a relative minimum at a.
10. 10. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.1 Relative Extrema First-Derivative Test for Relative Extrema 1. Find f’(x). 2. Determine all critical values of f. 3. For each critical value a at which f is continuous, determine whether f’(x) changes sign as x increases through a. 4. For critical values a at which f is not continuous, analyze the situation by using the definitions of extrema directly.
11. 11. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.1 Relative Extrema Example 1 - First-Derivative Test If , use the first-derivative test to find where relative extrema occur. Solution: STEP 1 - STEP 2 - Setting f’(x) = 0 gives x = −3, 1. STEP 3 - Conclude that at−3, there is a relative maximum. STEP 4 – There are no critical values at which f is not ( ) 1for 1 4 ≠ + +== x x xxfy ( ) ( ) ( )( ) ( ) 1for 1 13 1 32 ' 22 2 −≠ + −+ = + −+ = x x xx x xx xf
12. 12. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.1 Relative Extrema Example 3 - Finding Relative Extrema Test for relative extrema. Solution: By product rule, Relative maximum when x = −2 Relative minimum when x = 0. ( ) x exxfy 2 == ( ) ( )2' += xxexf x
13. 13. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.2 Absolute Extrema on a Closed Interval13.2 Absolute Extrema on a Closed Interval Extreme-Value Theorem • If a function is continuous on a closed interval, then the function has a maximum value and a minimum value on that interval.
14. 14. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.2 Absolute Extrema on a Closed Interval Procedure to Find Absolute Extrema for a Function f That Is Continuous on [a, b] 1. Find the critical values of f . 2. Evaluate f(x) at the endpoints a and b and at the critical values in (a, b). 3. The maximum value of f is the greatest value found in step 2. The minimum value is the least value found in step 2.
15. 15. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.2 Absolute Extrema on a Closed Interval Example 1 - Finding Extreme Values on a Closed Interval Find absolute extrema for over the closed interval [1, 4]. Solution: Step 1: Step 2: Step 3: ( ) 542 +−= xxxf ( ) ( )2242' −=−= xxxxf ( ) ( ) endpointsatofvalues54 21 ff f = = ( ) ( )41,in2valuecriticalatofvalues12 ff = ( ) ( ) 12isminand54ismax == ff
16. 16. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.3 Concavity13.3 Concavity • Cases where curves concave upward: • Cases where curves concave downward:
17. 17. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.3 Concavity • f is said to be concave up on (a, b) if f is increasing on (a, b). • f is said to be concave down on (a, b) if f is decreasing on (a, b). • f has an inflection point at a if it is continuous at a and f changes concavity at a. Criteria for Concavity • If f’’(x) > 0, f is concave up on (a, b). • If f”(x) < 0, f is concave down on (a, b).
18. 18. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.3 Concavity Example 1 - Testing for Concavity Determine where the given function is concave up and where it is concave down. Solution: Applying the rule, Concave up when 6(x − 1) > 0 as x > 1. Concave down when 6(x − 1) < 0 as x < 1. ( ) ( ) 11a. 3 +−== xxfy ( ) ( )16'' 13' 2 −= −= xy xy
19. 19. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.3 Concavity Example 1 - Testing for Concavity Solution: Applying the rule, As y’’ is always positive, y = x2 is always concave up. 2 b. xy = 2'' 2' = = y xy
20. 20. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.3 Concavity Example 3 - A Change in Concavity with No Inflection Point Discuss concavity and find all inflection points for f(x) = 1/x. Solution: x > 0 f”(x) > 0 and x < 0  f”(x) < 0. f is concave up on (0,∞) and concave down on (−∞, 0) f is not continuous at 0  no inflection point ( ) ( ) 0for2'' 0for' 3 2 ≠= ≠−= − − xxxf xxxf
21. 21. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.4 The Second-Derivative Test13.4 The Second-Derivative Test • The test is used to test certain critical values for relative extrema. Suppose f’(a) = 0. • If f’’(a) < 0, then f has a relative maximum at a. • If f’’(a) > 0, then f has a relative minimum at a.
22. 22. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.4 The Second-Derivative Test Example 1 - Second-Derivative Test Test the following for relative maxima and minima. Use the second-derivative test, if possible. Solution: Relative minimum when x = −3. 3 3 2 18. xxya −= ( )( ) xy xxy 4'' 332' −= −+= 3havewe,0'When ±== xy ( ) ( ) 01234'',3When 01234'',3When >=−−=−= <−=−=+= yx yx
23. 23. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.4 The Second-Derivative Test Example 1 - Second-Derivative Test Solution: No maximum or minimum exists when x = 0. 186. 34 +−= xxyb ( ) xxy xxxxy 4872'' 1242424' 2 223 −= −=−= 1,0havewe,0'When == xy 0'',1When 0'',0When >= == yx yx
24. 24. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.5 Asymptotes13.5 Asymptotes Vertical Asymptotes • The line x = a is a vertical asymptote if at least one of the following is true: Vertical-Asymptote Rule for Rational Functions • P and Q are polynomial functions and the quotient is in lowest terms. ( ) ( ) ±∞= ±∞= − + → → xf xf ax ax lim lim ( ) ( ) ( )xQ xP xf =
25. 25. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.5 Asymptotes Example 1 - Finding Vertical Asymptotes Determine vertical asymptotes for the graph of Solution: Since f is a rational function, Denominator is 0 when x is 3 or 1. The lines x = 3 and x = 1 are vertical asymptotes. ( ) 34 4 2 2 +− − = xx xx xf ( ) ( )( )13 42 −− − = xx xx xf
26. 26. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.5 Asymptotes Horizontal and Oblique Asymptotes • The line y = b is a horizontal asymptote if at least one of the following is true: Nonvertical asymptote • The line y = mx +b is a nonvertical asymptote if at least one of the following is true: ( ) ( ) bxfbxf xx == −∞→∞→ limorlim ( ) ( )( ) ( ) ( )( ) 0limor0lim =+−=+− −∞→∞→ bmxxfbmxxf xx
27. 27. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.5 Asymptotes Example 3 - Finding an Oblique Asymptote Find the oblique asymptote for the graph of the rational function Solution: y = 2x + 1 is an oblique asymptote. ( ) 25 5910 2 + ++ == x xx xfy ( ) 25 3 12 25 5910 2 + ++= + ++ = x x x xx xf ( )( ) 0 25 3 lim12lim = + =+− ±∞→±∞→ x xxf xx
28. 28. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.5 Asymptotes Example 5 - Finding Horizontal and Vertical Asymptotes Find horizontal and vertical asymptotes for the graph Solution: Testing for horizontal asymptotes, The line y = −1 is a horizontal asymptote. ( ) 1−== x exfy ( ) ( ) 1101limlim1lim 1lim −=−=−=− ∞=− −∞→−∞→−∞→ ∞→ x x x x x x x ee e
29. 29. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.5 Asymptotes Example 7 - Curve Sketching Sketch the graph of . Solution: Intercepts (0, 0) is the only intercept. Symmetry There is only symmetry about the origin. Asymptotes Denominator ≠ 0  No vertical asymptote Since y = 0 is the only non-vertical asymptote Max and Min For , relative maximum is (1, 2). Concavity For , inflection points are (-√ 3, -√3), (0, 0), (√3, √3). 1 4 2 + = x x y 0 1 4 lim 2 = +∞→ x x x ( )( ) ( )22 1 114 ' + −+ = x xx y ( )( ) ( )32 1 338 '' + −+ = x xxx y
30. 30. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.5 Asymptotes Example 7 - Curve Sketching Solution: Graph
31. 31. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.6 Applied Maxima and Minima13.6 Applied Maxima and Minima Example 1 - Minimizing the Cost of a Fence • Use absolute maxima and minima to explain the endpoints of the domain of the function. A manufacturer plans to fence in a 10,800-ft2 rectangular storage area adjacent to a building by using the building as one side of the enclosed area. The fencing parallel to the building faces a highway and will cost \$3 per foot installed, whereas the fencing for the other two sides costs \$2 per foot installed. Find the amount of each type of fence so that the total cost of the fence will be a minimum. What is the minimum cost?
32. 32. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.6 Applied Maxima and Minima Example 1 - Minimizing the Cost of a Fence Solution: Cost function is Storage area is Analyzing the equations, Thus, yxCyyxC 43223 +=⇒++= x yxy 10800 800,10 =⇒= ( ) x x x xxC 43200 3 10800 43 +=      += 0since120 43200 30 2 >= −== xx xdx dC 0,120When 86400 2 2 32 2 >= = dx Cd x xdx Cd and Only critical value is 120. x =120 gives a relative minimum. ( ) 720 120 43200 3120 =+= xC
33. 33. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.6 Applied Maxima and Minima Example 3 - Minimizing Average Cost A manufacturer’s total-cost function is given by where c is the total cost of producing q units. At what level of output will average cost per unit be a minimum? What is this minimum? ( ) 4003 4 2 ++== q q qcc
34. 34. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.6 Applied Maxima and Minima Example 3 - Minimizing Average Cost Solution: Average-cost function is To find critical values, we set is positive when q = 40, which is the only relative extremum. The minimum average cost is ( ) q q q q q q c qcc 400 3 4 4003 4 2 ++= ++ === 0since40 4 1600 0 2 2 >=⇒ − == qq q q dq cd 32 2 800 qdq cd = ( ) 23 40 400 3 4 40 40 =++=c
35. 35. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.6 Applied Maxima and Minima Example 5 - Economic Lot Size A company annually produces and sells 10,000 units of a product. Sales are uniformly distributed throughout the year. The company wishes to determine the number of units to be manufactured in each production run in order to minimize total annual setup costs and carrying costs. The same number of units is produced in each run. This number is referred to as the economic lot size or economic order quantity. The production cost of each unit is \$20, and carrying costs (insurance, interest, storage, etc.) are estimated to be 10% of the value of the average inventory. Setup costs per production run are \$40. Find the economic lot size.
36. 36. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.6 Applied Maxima and Minima Example 5 - Economic Lot Size Solution: Let q be the number of units in a production run. Total of the annual carrying costs and setup is Setting dC/dq = 0, we get Since q > 0, there is an absolute minimum at q = 632.5. Number of production runs = 10,000/632.5 ≈ 15.8 16 lots  Economic size = 625 units ( ) 2 2 2 400000400000 1 4000010000 40 2 201.0 q q qdq dC q q q q C − =−= +=      +      = 5.632400000 400000 0 2 2 ≈= − == q q q dq dC
37. 37. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.6 Applied Maxima and Minima Example 7 - Maximizing the Number of Recipients of Health-Care Benefits An article in a sociology journal stated that if a particular health-care program for the elderly were initiated, then t years after its start, n thousand elderly people would receive direct benefits, where For what value of t does the maximum number receive benefits? 120326 3 2 3 ≤≤+−= ttt t n
38. 38. ©2007 Pearson Education Asia Chapter 13: Curve Sketching 13.6 Applied Maxima and Minima Example 7 - Maximizing the Number of Recipients of Health-Care Benefits Solution: Setting dn/dt = 0, we have Absolute maximum value of n must occur at t = 0, 4, 8, or 12: Absolute maximum occurs when t = 12. 120326 3 2 3 ≤≤+−= ttt t n 8or4 32120 2 == +−== tt tt dt dn ( ) ( ) ( ) ( ) 9612, 3 128 8, 3 160 4,00 ==== nnnn