1. Calculus Date: 12/17/13 Obj: SWBAT apply first derivative test http://youtu.be/PBKnttVMbV4http://youtu.be/PBKnttVMbV4 first derivative test inc. dec. Today – Cover the first derivative test In class: Start WS3-3A; complete for homework Tomorrow Cover the second derivative test (easier than 1 st ) Friday: With Christian complete any remaining worksheets, make sure You understand material from the test.,etc. Announcements: Break Packet online on Friday Merry Christmas if I don’t see you "Do not judge me by my successes, judge me by how many times I fell down and got back up again.“ Nelson Mandela

2. Increasing/Decreasing/Constant

5. Generic Example A similar Observation Applies at a Local Max. The First Derivative Test

6. The First Derivative Test leftright f (c) is a relative maximum f (c) is a relative minimum No changeNo relative extremum Determine the sign of the derivative of f to the left and right of the critical point. conclusion

7. The First Derivative Test Find all the relative extrema of 0 4 + 0 - 0 + Relative max. f (0) = 1 Relative min. f (4) = -31 Evaluate the derivative at points on either side of extrema to determine the sign.

8. The First Derivative Test Sketch of function based on estimates from the first derivative test.

9. The First Derivative Test Here is the actual function.

10. Another Example Find all the relative extrema of Stationary points: Singular points:

11. -1 0 1 + ND + 0 - ND - 0 + ND + Relative max. Relative min. Stationary points: Singular points: Evaluate the derivative at points on either side of extrema to determine the sign. Use Yvars for faster calculations ND – derivative not defined

12. Local max.Local min. Graph of -1 0 1 + ND + 0 - ND - 0 + ND +

13. Example: Graph There are roots at and. Set First derivative test: negative positive Possible extrema at. We can use a chart to organize our thoughts. y y=4 y=0

14. Example: Graph There are roots at and. Set First derivative test: maximum at minimum at Possible extreme at. y y=4 y=0

15. Example: Graph First derivative test: NOTE: On the AP Exam, it is not sufficient to simply draw the chart and write the answer. You must give a written explanation! There is a local maximum at (0,4) because for all x in and for all x in (0,2). There is a local minimum at (2,0) because for all x in (0,2) and for all x in.

16. Example: Graph First derivative test: There is a local maximum at (0,4) because for all x in and for all x in (0,2). There is a local minimum at (2,0) because for all x in (0,2) and for all x in.

17. Graph 1.Take the derivative f’(x) 2.Find the critical points f’(x) = 0; f’(x) = DNE 3.Make sign chart 1. Label critical points – put 0 or DNE on graph 2.Evaluate the derivative at points on either side of extrema to determine the sign. Use Yvars for faster calculations. Mark the chart as + or – in these areas 3.Evaluate f(x) at the critical points to determine actual values of extrema. 4.Add arrows to show increasing or decreasing regions in f(x) 4.Write out all extrema and use the value of the derivative and the appropriate interval to justify your answer.

18. Chapter 5 Applications of the Derivative Sections 5.1, 5.2, 5.3, and 5.4

19. Applications of the Derivative  Maxima and Minima  Applications of Maxima and Minima  The Second Derivative - Analyzing Graphs

20. Absolute Extrema Absolute Minimum Let f be a function defined on a domain D Absolute Maximum

21. The number f (c) is called the absolute maximum value of f in D A function f has an absolute (global) maximum at x = c if f (x)  f (c) for all x in the domain D of f. Absolute Maximum Absolute Extrema

22. Absolute Minimum Absolute Extrema A function f has an absolute (global) minimum at x = c if f (c)  f (x) for all x in the domain D of f. The number f (c) is called the absolute minimum value of f in D

23. Generic Example

26. Relative Extrema A function f has a relative (local) maximum at x  c if there exists an open interval (r, s) containing c such that f (x)  f (c) for all r  x  s. Relative Maxima

27. Relative Extrema A function f has a relative (local) minimum at x  c if there exists an open interval (r, s) containing c such that f (c)  f (x) for all r  x  s. Relative Minima

28. Generic Example The corresponding values of x are called Critical Points of f

29. Critical Points of f A critical number of a function f is a number c in the domain of f such that (stationary point) (singular point)

30. Candidates for Relative Extrema 1.Stationary points: any x such that x is in the domain of f and f ' (x)  0. 2.Singular points: any x such that x is in the domain of f and f ' (x)  undefined 3.Remark: notice that not every critical number correspond to a local maximum or local minimum. We use “local extrema” to refer to either a max or a min.

31. Fermat’s Theorem If a function f has a local maximum or minimum at c, then c is a critical number of f Notice that the theorem does not say that at every critical number the function has a local maximum or local minimum

32. Generic Example Two critical points of f that do not correspond to local extrema

33. Example Find all the critical numbers of Stationary points: Singular points:

34. Graph of Local max.Local min.

35. Extreme Value Theorem If a function f is continuous on a closed interval [a, b], then f attains an absolute maximum and absolute minimum on [a, b]. Each extremum occurs at a critical number or at an endpoint. a b Attains max. and min. Attains min. but no max. No min. and no max. Open IntervalNot continuous

36. Example Find the absolute extrema of Critical values of f inside the interval (-1/2,3) are x = 0, 2 Absolute Max.Absolute Min. Evaluate Absolute Max.

37. Example Find the absolute extrema of Critical values of f inside the interval (-1/2,3) are x = 0, 2 Absolute Min. Absolute Max.

38. Example Find the absolute extrema of Critical values of f inside the interval (-1/2,1) is x = 0 only Absolute Min. Absolute Max. Evaluate

39. Example Find the absolute extrema of Critical values of f inside the interval (-1/2,1) is x = 0 only Absolute Min. Absolute Max.

40. Start Here a. Reviewing Rolle’s and MVT b. Remember nDeriv and Yvars c. Increasing, Decreasing,Constant d. First Derivative Test

41. Finding absolute extrema on [a, b] 0. Verify function is continuous on the interval. Determine the function’s domain. 1.Find all critical numbers for f (x) in (a, b). 2.Evaluate f (x) for all critical numbers in (a, b). 3.Evaluate f (x) for the endpoints a and b of the interval [a, b]. 4.The largest value found in steps 2 and 3 is the absolute maximum for f on the interval [a, b], and the smallest value found is the absolute minimum for f on [a, b].

42. Rolle’s Theorem Given f(x) on closed interval [a, b] –Differentiable on open interval (a, b) If f(a) = f(b) … then –There exists at least one number a < c < b such that f ’(c) = 0 f(a) = f(b) a b c

43. The Mean Value Theorem (MVT) aka the ‘crooked’ Rolle’s Theorem If f is continuous on [a, b] and differentiable on (a, b) There is at least one number c on (a, b) at which a b f(a) f(b) c Conclusion: Slope of Secant Line Equals Slope of Tangent Line We can “tilt” the picture of Rolle’s Theorem Stipulating that f(a) ≠ f(b) How is Rolle’s Connected to MVT?

44. The Mean Value Theorem (MVT) aka the ‘crooked’ Rolle’s Theorem If f is continuous on [a, b] and differentiable on (a, b) There is at least one number c on (a, b) at which a b f(a) f(b) c Conclusion: The average rate of change equals the instantaneous rate of change evaluated at a point We can “tilt” the picture of Rolle’s Theorem Stipulating that f(a) ≠ f(b)

45. Finding c Given a function f(x) = 2x 3 – x 2 –Find all points on the interval [0, 2] where –Rolle’s? Strategy –Find slope of line from f(0) to f(2) –Find f ‘(x) –Set f ‘(x) equal to slope … solve for x

46. f(3) = 39 f(-2) = 64 For how many value(s) of c is f ‘ (c ) = -5? If, how many numbers on [-2, 3] satisfy the conclusion of the Mean Value Theorem. A. 0 B. 1 C. 2 D. 3 E. 4 CALCULATOR REQUIRED X XX

47. Given the graph of f(x) below, use the graph of f to estimate the numbers on [0, 3.5] which satisfy the conclusion of the Mean Value Theorem.

48. Relative Extrema Example: Find all the relative extrema of Stationary points: Singular points: None

49. First Derivative Test What if they are positive on both sides of the point in question? This is called an inflection point 

50. Domain Not a Closed Interval Example: Find the absolute extrema of Notice that the interval is not closed. Look graphically: Absolute Max. (3, 1)

51. Optimization Problems 1. Identify the unknown(s). Draw and label a diagram as needed. 2. Identify the objective function. The quantity to be minimized or maximized. 3. Identify the constraints. 4. State the optimization problem. 5. Eliminate extra variables. 6. Find the absolute maximum (minimum) of the objective function.

52. Optimization - Examples An open box is formed by cutting identical squares from each corner of a 4 in. by 4 in. sheet of paper. Find the dimensions of the box that will yield the maximum volume. x x x 4 – 2x x

53. Critical points: The dimensions are 8/3 in. by 8/3 in. by 2/3 in. giving a maximum box volume of V  4.74 in 3.

54. An metal can with volume 60 in 3 is to be constructed in the shape of a right circular cylinder. If the cost of the material for the side is $0.05/in. 2 and the cost of the material for the top and bottom is $0.03/in. 2 Find the dimensions of the can that will minimize the cost. top and bottom sidecost Optimization - Examples

55. So Sub. in for h

56. So with a radius ≈ 2.52 in. and height ≈ 3.02 in. the cost is minimized at ≈ $3.58. Graph of cost function to verify absolute minimum: 2.5

57. Second Derivative

58. Second Derivative - Example

59. Second Derivative

61. Concavity Let f be a differentiable function on (a, b). 1. f is concave upward on (a, b) if f ' is increasing on aa(a, b). That is f ''(x)  0 for each value of x in (a, b). concave upwardconcave downward 2. f is concave downward on (a, b) if f ' is decreasing on (a, b). That is f ''(x)  0 for each value of x in (a, b).

62. Inflection Point A point on the graph of f at which f is continuous and concavity changes is called an inflection point. To search for inflection points, find any point, c in the domain where f ''(x)  0 or f ''(x) is undefined. If f '' changes sign from the left to the right of c, then (c, f (c)) is an inflection point of f.

63. Example: Inflection Points Find all inflection points of

64. 2 - 0 + Inflection point at x  2

66. The Point of Diminishing Returns If the function represents the total sales of a particular object, t months after being introduced, find the point of diminishing returns. S concave up on S concave down on The point of diminishing returns is at 20 months (the rate at which units are sold starts to drop).

67. S concave up on S concave down on Inflection point The Point of Diminishing Returns