Lesson 5: Continuity

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continuity says that nearby points go to nearby values, and the Intermediate Value Theorem is an important property of continuous functions

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Lesson 5: Continuity

  1. 1. Section 1.5 Continuity V63.0121, Calculus I February 2–3, 2009 Announcements Quiz 1 is this week: Tuesday (Sections 11 and 12), Thursday (Section 4), Friday (Sections 13 and 14). 15 minutes, covers Sections 1.1–1.2 Fill your ALEKS pie by February 27, 11:59pm Congratulations to the Super Bowl XLIII Champion Pittsburgh Steelers!
  2. 2. Outline Warmup Grader Notes Continuity The Intermediate Value Theorem Back to the Questions
  3. 3. Hatsumon Here are some discussion questions to start. Were you ever exactly three feet tall?
  4. 4. Hatsumon Here are some discussion questions to start. Were you ever exactly three feet tall? Was your height (in inches) ever equal to your weight (in pounds)?
  5. 5. Hatsumon Here are some discussion questions to start. Were you ever exactly three feet tall? Was your height (in inches) ever equal to your weight (in pounds)? Is there a pair of points on opposite sides of the world at the same temperature at the same time?
  6. 6. Outline Warmup Grader Notes Continuity The Intermediate Value Theorem Back to the Questions
  7. 7. But first, a word from our friendly graders Please turn in neat problem sets: loose-leaf paper, stapled Label homework with Name, Section (10 or 4), Date, Problem Set number Do not turn in scratch work
  8. 8. Please label your graphs Example: Graph F (x) = |2x + 1|. y (0, 1) x (−1/2, 0) incomplete better
  9. 9. Example = explanation Problem If f is even and g is odd, what can you say about fg ? Let f (x) = x 2 (even) and g (x) = x 3 (odd). Then (fg )(x) = x 2 · x 3 = x 5 , and that’s odd. So the product of an even function and an odd function is an odd function. Dangerous!
  10. 10. The trouble with proof by example Problem Which odd numbers are prime? The numbers 3, 5, and 7 are odd, and all prime. So all odd numbers are prime. Fallacious!
  11. 11. Use the definitions Let f be even and g be odd. Then (fg )(−x) = f (−x)g (−x) = f (x)(−g (x)) = −f (x)g (x) = −(fg )(x) So fg is odd. Better
  12. 12. Outline Warmup Grader Notes Continuity The Intermediate Value Theorem Back to the Questions
  13. 13. Recall: Direct Substitution Property Theorem (The Direct Substitution Property) If f is a polynomial or a rational function and a is in the domain of f , then lim f (x) = f (a) x→a
  14. 14. Definition of Continuity Definition Let f be a function defined near a. We say that f is continuous at a if lim f (x) = f (a). x→a
  15. 15. Free Theorems Theorem (a) Any polynomial is continuous everywhere; that is, it is continuous on R = (−∞, ∞). (b) Any rational function is continuous wherever it is defined; that is, it is continuous on its domain.
  16. 16. Showing a function is continuous Example √ Let f (x) = 4x + 1. Show that f is continuous at 2.
  17. 17. Showing a function is continuous Example √ Let f (x) = 4x + 1. Show that f is continuous at 2. Solution We have √ lim f (x) = lim 4x + 1 x→a x→2 = lim (4x + 1) x→2 √ = 9 = 3. Each step comes from the limit laws.
  18. 18. Showing a function is continuous Example √ Let f (x) = 4x + 1. Show that f is continuous at 2. Solution We have √ lim f (x) = lim 4x + 1 x→a x→2 = lim (4x + 1) x→2 √ = 9 = 3. Each step comes from the limit laws. In fact, f is continuous on (−1/4, ∞).
  19. 19. Showing a function is continuous Example √ Let f (x) = 4x + 1. Show that f is continuous at 2. Solution We have √ lim f (x) = lim 4x + 1 x→a x→2 = lim (4x + 1) x→2 √ = 9 = 3. Each step comes from the limit laws. In fact, f is continuous on (−1/4, ∞). The function f is right continuous at the point −1/4.
  20. 20. The Limit Laws give Continuity Laws Theorem If f and g are continuous at a and c is a constant, then the following functions are also continuous at a: f +g f −g cf fg f (if g (a) = 0) g
  21. 21. Transcendental functions are continuous, too Theorem The following functions are continuous wherever they are defined: 1. sin, cos, tan, cot sec, csc 2. x → ax , loga , ln 3. sin−1 , tan−1 , sec−1
  22. 22. What could go wrong? In what ways could a function f fail to be continuous at a point a? Look again at the definition: lim f (x) = f (a) x→a
  23. 23. Pitfall #1 Example Let x2 if 0 ≤ x ≤ 1 f (x) = if 1 < x ≤ 2 2x At which points is f continuous?
  24. 24. Pitfall #1: The limit does not exist Example Let x2 if 0 ≤ x ≤ 1 f (x) = if 1 < x ≤ 2 2x At which points is f continuous? Solution At any point a in [0, 2] besides 1, lim f (x) = f (a) because f is x→a represented by a polynomial near a, and polynomials have the direct substitution property. However, lim f (x) = lim x 2 = 12 = 1 x→1− x→1− lim f (x) = lim+ 2x = 2(1) = 2 x→1+ x→1 So f has no limit at 1. Therefore f is not continuous at 1.
  25. 25. Graphical Illustration of Pitfall #1 y 4 3 2 1 x −1 1 2 −1
  26. 26. Pitfall #2 Example Let x 2 + 2x + 1 f (x) = x +1 At which points is f continuous?
  27. 27. Pitfall #2: The function has no value Example Let x 2 + 2x + 1 f (x) = x +1 At which points is f continuous? Solution Because f is rational, it is continuous on its whole domain. Note that −1 is not in the domain of f , so f is not continuous there.
  28. 28. Graphical Illustration of Pitfall #2 y 1 x −1 f cannot be continuous where it has no value.
  29. 29. Pitfall #3 Example Let 7 if x = 1 f (x) = π if x = 1 At which points is f continuous?
  30. 30. Pitfall #3: function value = limit Example Let 7 if x = 1 f (x) = π if x = 1 At which points is f continuous? Solution f is not continuous at 1 because f (1) = π but lim f (x) = 7. x→1
  31. 31. Graphical Illustration of Pitfall #3 y 7 π x 1
  32. 32. Special types of discontinuites removable discontinuity The limit lim f (x) exists, but f is not x→a defined at a or its value at a is not equal to the limit at a. jump discontinuity The limits lim f (x) and lim+ f (x) exist, but x→a− x→a are different. f (a) is one of these limits.
  33. 33. Graphical representations of discontinuities y 4 y 3 7 2 π 1 x x 1 −1 1 2 removable −1 jump
  34. 34. The greatest integer function y 3 2 1 x −2 −1 1 2 3 −1 −2
  35. 35. The greatest integer function y 3 2 1 x −2 −1 1 2 3 −1 −2 The greatest integer function f (x) = [[x]] has jump discontinuities.
  36. 36. Outline Warmup Grader Notes Continuity The Intermediate Value Theorem Back to the Questions
  37. 37. A Big Time Theorem Theorem (The Intermediate Value Theorem) Suppose that f is continuous on the closed interval [a, b] and let N be any number between f (a) and f (b), where f (a) = f (b). Then there exists a number c in (a, b) such that f (c) = N.
  38. 38. Illustrating the IVT f (x) x
  39. 39. Illustrating the IVT Suppose that f is continuous on the closed interval [a, b] f (x) x
  40. 40. Illustrating the IVT Suppose that f is continuous on the closed interval [a, b] f (x) f (b) f (a) x a b
  41. 41. Illustrating the IVT Suppose that f is continuous on the closed interval [a, b] and let N be any number between f (a) and f (b), where f (a) = f (b). f (x) f (b) N f (a) x a b
  42. 42. Illustrating the IVT Suppose that f is continuous on the closed interval [a, b] and let N be any number between f (a) and f (b), where f (a) = f (b). Then there exists a number c in (a, b) such that f (c) = N. f (x) f (b) N f (a) x a c b
  43. 43. Illustrating the IVT Suppose that f is continuous on the closed interval [a, b] and let N be any number between f (a) and f (b), where f (a) = f (b). Then there exists a number c in (a, b) such that f (c) = N. f (x) f (b) N f (a) x a b
  44. 44. Illustrating the IVT Suppose that f is continuous on the closed interval [a, b] and let N be any number between f (a) and f (b), where f (a) = f (b). Then there exists a number c in (a, b) such that f (c) = N. f (x) f (b) N f (a) x a c1 c2 c3 b
  45. 45. Using the IVT Example Prove that the square root of two exists.
  46. 46. Using the IVT Example Prove that the square root of two exists. Proof. Let f (x) = x 2 , a continuous function on [1, 2].
  47. 47. Using the IVT Example Prove that the square root of two exists. Proof. Let f (x) = x 2 , a continuous function on [1, 2]. Note f (1) = 1 and f (2) = 4. Since 2 is between 1 and 4, there exists a point c in (1, 2) such that f (c) = c 2 = 2.
  48. 48. Using the IVT Example Prove that the square root of two exists. Proof. Let f (x) = x 2 , a continuous function on [1, 2]. Note f (1) = 1 and f (2) = 4. Since 2 is between 1 and 4, there exists a point c in (1, 2) such that f (c) = c 2 = 2. In fact, we can “narrow in” on the square root of 2 by the method of bisections.
  49. 49. √ Finding 2 by bisections x f (x) = x 2 24 11
  50. 50. √ Finding 2 by bisections x f (x) = x 2 24 11
  51. 51. √ Finding 2 by bisections x f (x) = x 2 24 1.5 2.25 11
  52. 52. √ Finding 2 by bisections x f (x) = x 2 24 1.5 2.25 1.25 1.5625 11
  53. 53. √ Finding 2 by bisections x f (x) = x 2 24 1.5 2.25 1.375 1.890625 1.25 1.5625 11
  54. 54. √ Finding 2 by bisections x f (x) = x 2 24 1.5 2.25 1.4375 2.06640625 1.375 1.890625 1.25 1.5625 11
  55. 55. Using the IVT Example Let f (x) = x 3 − x − 1. Show that there is a zero for f . Solution f (1) = −1 and f (2) = 5. So there is a zero between 1 and 2.
  56. 56. Using the IVT Example Let f (x) = x 3 − x − 1. Show that there is a zero for f . Solution f (1) = −1 and f (2) = 5. So there is a zero between 1 and 2. (More careful analysis yields 1.32472.)
  57. 57. Outline Warmup Grader Notes Continuity The Intermediate Value Theorem Back to the Questions
  58. 58. Back to the Questions True or False At one point in your life you were exactly three feet tall.
  59. 59. Question 1: True! Let h(t) be height, which varies continuously over time. Then h(birth) < 3 ft and h(now) > 3 ft. So there is a point c in (birth, now) where h(c) = 3.
  60. 60. Back to the Questions True or False At one point in your life you were exactly three feet tall. True or False At one point in your life your height in inches equaled your weight in pounds.
  61. 61. Question 2: True! Let h(t) be height in inches and w (t) be weight in pounds, both varying continuously over time. Let f (t) = h(t) − w (t). For most of us (call your mom), f (birth) > 0 and f (now) < 0. So there is a point c in (birth, now) where f (c) = 0. In other words, h(c) − w (c) = 0 ⇐⇒ h(c) = w (c).
  62. 62. Back to the Questions True or False At one point in your life you were exactly three feet tall. True or False At one point in your life your height in inches equaled your weight in pounds. True or False Right now there are two points on opposite sides of the Earth with exactly the same temperature.
  63. 63. Question 3 Let T (θ) be the temperature at the point on the equator at longitude θ. How can you express the statement that the temperature on opposite sides is the same? How can you ensure this is true?

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