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# Lesson 5: Continuity (Section 21 slides)

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### Lesson 5: Continuity (Section 21 slides)

1. 1. Section 1.5 Continuity V63.0121.021, Calculus I New York University September 21, 2010Announcements Oﬃce Hours: Tuesday, Wednesday, 3pm–4pm TAs have oﬃce hours on website If you’re observing Sukkot, get me the homework earlier
2. 2. Announcements Oﬃce Hours: Tuesday, Wednesday, 3pm–4pm TAs have oﬃce hours on website If you’re observing Sukkot, get me the homework earlierV63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 2 / 47
3. 3. Grader’s corner HW Grades will be on blackboard this week, and the papers will be returned in recitation Remember units when computing slopes Not everything is linear. Remember to staple your papers—you have been warned. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 3 / 47
4. 4. Objectives Understand and apply the deﬁnition of continuity for a function at a point or on an interval. Given a piecewise deﬁned function, decide where it is continuous or discontinuous. State and understand the Intermediate Value Theorem. Use the IVT to show that a function takes a certain value, or that an equation has a solution V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 4 / 47
5. 5. Last timeDeﬁnitionWe write lim f (x) = L x→aand say “the limit of f (x), as x approaches a, equals L”if we can make the values of f (x) arbitrarily close to L (as close to L as welike) by taking x to be suﬃciently close to a (on either side of a) but notequal to a. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 5 / 47
6. 6. Limit Laws for arithmeticTheorem (Basic Limits) lim x = a x→a lim c = c x→aTheorem (Limit Laws)Let f and g be functions with limits at a point a. Then lim (f (x) + g (x)) = lim f (x) + lim g (x) x→a x→a x→a lim (f (x) − g (x)) = lim f (x) − lim g (x) x→a x→a x→a lim (f (x) · g (x)) = lim f (x) · lim g (x) x→a x→a x→a f (x) limx→a f (x) lim = if lim g (x) = 0 x→a g (x) limx→a g (x) x→a V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 6 / 47
7. 7. HatsumonHere are some discussion questions to start.True or FalseAt some point in your life you were exactly three feet tall. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 7 / 47
8. 8. HatsumonHere are some discussion questions to start.True or FalseAt some point in your life you were exactly three feet tall.True or FalseAt some point in your life your height (in inches) was equal to your weight(in pounds). V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 7 / 47
9. 9. HatsumonHere are some discussion questions to start.True or FalseAt some point in your life you were exactly three feet tall.True or FalseAt some point in your life your height (in inches) was equal to your weight(in pounds).True or FalseRight now there are a pair of points on opposite sides of the worldmeasuring the exact same temperature. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 7 / 47
10. 10. OutlineContinuityThe Intermediate Value TheoremBack to the Questions V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 8 / 47
11. 11. Recall: Direct Substitution PropertyTheorem (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→aThis property is so useful it’s worth naming. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 9 / 47
12. 12. Deﬁnition of ContinuityDeﬁnition Let f be a function deﬁned near a. We say that f is continuous at a if lim f (x) = f (a). x→a V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 10 / 47
13. 13. Deﬁnition of Continuity yDeﬁnition Let f be a function deﬁned near a. We say that f is f (a) continuous at a if lim f (x) = f (a). x→a A function f is continuous if it is continuous at every point in its domain. x a V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 10 / 47
14. 14. ScholiumDeﬁnitionLet f be a function deﬁned near a. We say that f is continuous at a if lim f (x) = f (a). x→aThere are three important parts to this deﬁnition. The function has to have a limit at a, the function has to have a value at a, and these values have to agree. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 11 / 47
15. 15. Free TheoremsTheorem(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. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 12 / 47
16. 16. Showing a function is continuousExample √Let f (x) = 4x + 1. Show that f is continuous at 2. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 13 / 47
17. 17. Showing a function is continuousExample √Let f (x) = 4x + 1. Show that f is continuous at 2.SolutionWe want to show that lim f (x) = f (2). We have x→2 √ √ lim f (x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f (2). x→a x→2 x→2Each step comes from the limit laws. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 13 / 47
18. 18. Showing a function is continuousExample √Let f (x) = 4x + 1. Show that f is continuous at 2.SolutionWe want to show that lim f (x) = f (2). We have x→2 √ √ lim f (x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f (2). x→a x→2 x→2Each step comes from the limit laws.QuestionAt which other points is f continuous? V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 13 / 47
19. 19. At which other points? √For reference: f (x) = 4x + 1 If a > −1/4, then lim (4x + 1) = 4a + 1 > 0, so x→a √ √ lim f (x) = lim 4x + 1 = lim (4x + 1) = 4a + 1 = f (a) x→a x→a x→a and f is continuous at a. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 14 / 47
20. 20. At which other points? √For reference: f (x) = 4x + 1 If a > −1/4, then lim (4x + 1) = 4a + 1 > 0, so x→a √ √ lim f (x) = lim 4x + 1 = lim (4x + 1) = 4a + 1 = f (a) x→a x→a x→a and f is continuous at a. √ If a = −1/4, then 4x + 1 < 0 to the left of a, which means 4x + 1 is undeﬁned. Still, √ √ lim+ f (x) = lim+ 4x + 1 = lim+ (4x + 1) = 0 = 0 = f (a) x→a x→a x→a so f is continuous on the right at a = −1/4. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 14 / 47
21. 21. Showing a function is continuousExample √Let f (x) = 4x + 1. Show that f is continuous at 2.SolutionWe want to show that lim f (x) = f (2). We have x→2 √ √ lim f (x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f (2). x→a x→2 x→2Each step comes from the limit laws.QuestionAt which other points is f continuous?AnswerThe function f is continuous on (−1/4, ∞). V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 15 / 47
22. 22. Showing a function is continuousExample √Let f (x) = 4x + 1. Show that f is continuous at 2.SolutionWe want to show that lim f (x) = f (2). We have x→2 √ √ lim f (x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f (2). x→a x→2 x→2Each step comes from the limit laws.QuestionAt which other points is f continuous?AnswerThe function f is continuous on (−1/4, ∞). It is right continuous at −1/4since lim f (x) = f (−1/4). V63.0121.021, 1/4+ x→−Calculus I (NYU) Section 1.5 Continuity September 21, 2010 15 / 47
23. 23. The Limit Laws give Continuity LawsTheoremIf f (x) and g (x) are continuous at a and c is a constant, then thefollowing functions are also continuous at a: (f + g )(x) (fg )(x) (f − g )(x) f (x) (if g (a) = 0) (cf )(x) g V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 16 / 47
24. 24. Why a sum of continuous functions is continuousWe want to show that lim (f + g )(x) = (f + g )(a). x→aWe just follow our nose: lim (f + g )(x) = lim [f (x) + g (x)] (def of f + g ) x→a x→a = lim f (x) + lim g (x) (if these limits exist) x→a x→a = f (a) + g (a) (they do; f and g are cts.) = (f + g )(a) (def of f + g again) V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 17 / 47
25. 25. Trigonometric functions are continuous sin and cos are continuous on R. sin V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 18 / 47
26. 26. Trigonometric functions are continuous sin and cos are continuous on R. cos sin V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 18 / 47
27. 27. Trigonometric functions are continuous tan sin and cos are continuous on R. sin 1 tan = and sec = are cos cos cos continuous on their domain, which is π sin R + kπ k ∈ Z . 2 V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 18 / 47
28. 28. Trigonometric functions are continuous tan sec sin and cos are continuous on R. sin 1 tan = and sec = are cos cos cos continuous on their domain, which is π sin R + kπ k ∈ Z . 2 V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 18 / 47
29. 29. Trigonometric functions are continuous tan sec sin and cos are continuous on R. sin 1 tan = and sec = are cos cos cos continuous on their domain, which is π sin R + kπ k ∈ Z . 2 cos 1 cot = and csc = are sin sin continuous on their domain, which is R { kπ | k ∈ Z }. cot V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 18 / 47
30. 30. Trigonometric functions are continuous tan sec sin and cos are continuous on R. sin 1 tan = and sec = are cos cos cos continuous on their domain, which is π sin R + kπ k ∈ Z . 2 cos 1 cot = and csc = are sin sin continuous on their domain, which is R { kπ | k ∈ Z }. cot csc V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 18 / 47
31. 31. Exponential and Logarithmic functions arecontinuousFor any base a > 1, ax the function x → ax is continuous on R V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 19 / 47
32. 32. Exponential and Logarithmic functions arecontinuousFor any base a > 1, ax the function x → ax is loga x continuous on R the function loga is continuous on its domain: (0, ∞) V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 19 / 47
33. 33. Exponential and Logarithmic functions arecontinuousFor any base a > 1, ax the function x → ax is loga x continuous on R the function loga is continuous on its domain: (0, ∞) In particular e x and ln = loge are continuous on their domains V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 19 / 47
34. 34. Inverse trigonometric functions are mostlycontinuous sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and right continuous at −1. π π/2 sin−1 V63.0121.021, Calculus I (NYU) −π/2 Section 1.5 Continuity September 21, 2010 20 / 47
35. 35. Inverse trigonometric functions are mostlycontinuous sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and right continuous at −1. π cos−1 π/2 sin−1 V63.0121.021, Calculus I (NYU) −π/2 Section 1.5 Continuity September 21, 2010 20 / 47
36. 36. Inverse trigonometric functions are mostlycontinuous sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and right continuous at −1. sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left continuous at −1, and right continuous at 1. π cos−1 sec−1 π/2 sin−1 V63.0121.021, Calculus I (NYU) −π/2 Section 1.5 Continuity September 21, 2010 20 / 47
37. 37. Inverse trigonometric functions are mostlycontinuous sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and right continuous at −1. sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left continuous at −1, and right continuous at 1. π cos−1 sec−1 π/2 csc−1 sin−1 V63.0121.021, Calculus I (NYU) −π/2 Section 1.5 Continuity September 21, 2010 20 / 47
38. 38. Inverse trigonometric functions are mostlycontinuous sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and right continuous at −1. sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left continuous at −1, and right continuous at 1. tan−1 and cot−1 are continuous on R. π cos−1 sec−1 π/2 tan−1 csc−1 sin−1 V63.0121.021, Calculus I (NYU) −π/2 Section 1.5 Continuity September 21, 2010 20 / 47
39. 39. Inverse trigonometric functions are mostlycontinuous sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and right continuous at −1. sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left continuous at −1, and right continuous at 1. tan−1 and cot−1 are continuous on R. πcot−1 cos−1 sec−1 π/2 tan−1 csc−1 sin−1 V63.0121.021, Calculus I (NYU) −π/2 Section 1.5 Continuity September 21, 2010 20 / 47
40. 40. What could go wrong?In what ways could a function f fail to be continuous at a point a? Lookagain at the deﬁnition: lim f (x) = f (a) x→a V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 21 / 47
41. 41. Continuity FAILExampleLet x2 if 0 ≤ x ≤ 1 f (x) = 2x if 1 < x ≤ 2At which points is f continuous? V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 22 / 47
42. 42. Continuity FAIL: The limit does not existExampleLet x2 if 0 ≤ x ≤ 1 f (x) = 2x if 1 < x ≤ 2At which points is f continuous?SolutionAt any point a in [0, 2] besides 1, lim f (x) = f (a) because f is represented by a x→apolynomial 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→1So f has no limit at 1. Therefore f is not continuous at 1. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 22 / 47
43. 43. Graphical Illustration of Pitfall #1 y 4 3 The function cannot be 2 continuous at a point if the function has no limit at that point. 1 x −1 1 2 −1 V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 23 / 47
44. 44. Continuity FAILExampleLet x 2 + 2x + 1 f (x) = x +1At which points is f continuous? V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 24 / 47
45. 45. Continuity FAIL: The function has no valueExampleLet x 2 + 2x + 1 f (x) = x +1At which points is f continuous?SolutionBecause f is rational, it is continuous on its whole domain. Note that −1is not in the domain of f , so f is not continuous there. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 24 / 47
46. 46. Graphical Illustration of Pitfall #2 y 1 The function cannot be continuous at a point outside its domain (that is, a point where it x has no value). −1 V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 25 / 47
47. 47. Continuity FAILExampleLet 7 if x = 1 f (x) = π if x = 1At which points is f continuous? V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 26 / 47
48. 48. Continuity FAIL: function value = limitExampleLet 7 if x = 1 f (x) = π if x = 1At which points is f continuous?Solutionf is not continuous at 1 because f (1) = π but lim f (x) = 7. x→1 V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 26 / 47
49. 49. Graphical Illustration of Pitfall #3 y 7 If the function has a limit and a value at a point the two must π still agree. x 1 V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 27 / 47
50. 50. Special types of discontinuitesremovable discontinuity The limit lim f (x) exists, but f is not deﬁned x→a 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 are x→a− x→a diﬀerent. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 28 / 47
51. 51. Graphical representations of discontinuities y y 7 2 π 1 x x 1 1 removable jump V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 29 / 47
52. 52. Graphical representations of discontinuities y y Presto! continuous! 7 2 π 1 x x 1 1 removable jump V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 29 / 47
53. 53. Graphical representations of discontinuities y y Presto! continuous! 7 2 π 1 continuous? x x 1 1 removable jump V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 29 / 47
54. 54. Graphical representations of discontinuities y y Presto! continuous! 7 2 continuous? π 1 x x 1 1 removable jump V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 29 / 47
55. 55. Graphical representations of discontinuities y y Presto! continuous! 7 2 continuous? π 1 x x 1 1 removable jump V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 29 / 47
56. 56. Special types of discontinuitesremovable discontinuity The limit lim f (x) exists, but f is not deﬁned x→a at a or its value at a is not equal to the limit at a. By re-deﬁning f (a) = lim f (x), f can be made continuous at a x→ajump discontinuity The limits lim f (x) and lim+ f (x) exist, but are x→a− x→a diﬀerent. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 30 / 47
57. 57. Special types of discontinuitesremovable discontinuity The limit lim f (x) exists, but f is not deﬁned x→a at a or its value at a is not equal to the limit at a. By re-deﬁning f (a) = lim f (x), f can be made continuous at a x→ajump discontinuity The limits lim f (x) and lim+ f (x) exist, but are x→a− x→a diﬀerent. The function cannot be made continuous by changing a single value. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 30 / 47
58. 58. The greatest integer function[[x]] is the greatest integer ≤ x. y 3 x [[x]] y = [[x]] 0 0 2 1 1 1.5 1 1 1.9 1 2.1 2 x −0.5 −1 −2 −1 1 2 3 −0.9 −1 −1 −1.1 −2 −2 V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 31 / 47
59. 59. The greatest integer function[[x]] is the greatest integer ≤ x. y 3 x [[x]] y = [[x]] 0 0 2 1 1 1.5 1 1 1.9 1 2.1 2 x −0.5 −1 −2 −1 1 2 3 −0.9 −1 −1 −1.1 −2 −2This function has a jump discontinuity at each integer. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 31 / 47
60. 60. OutlineContinuityThe Intermediate Value TheoremBack to the Questions V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 32 / 47
61. 61. A Big Time TheoremTheorem (The Intermediate Value Theorem)Suppose that f is continuous on the closed interval [a, b] and let N be anynumber between f (a) and f (b), where f (a) = f (b). Then there exists anumber c in (a, b) such that f (c) = N. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 33 / 47
62. 62. Illustrating the IVT f (x) x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 34 / 47
63. 63. Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] f (x) x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 34 / 47
64. 64. Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] f (x)f (b) f (a) a b x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 34 / 47
65. 65. Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let N be anynumber between f (a) and f (b), where f (a) = f (b). f (x)f (b) N f (a) a b x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 34 / 47
66. 66. Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let N be anynumber between f (a) and f (b), where f (a) = f (b). Then there exists anumber c in (a, b) such that f (c) = N. f (x)f (b) N f (a) a c b x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 34 / 47
67. 67. Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let N be anynumber between f (a) and f (b), where f (a) = f (b). Then there exists anumber c in (a, b) such that f (c) = N. f (x)f (b) N f (a) a b x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 34 / 47
68. 68. Illustrating the IVTSuppose that f is continuous on the closed interval [a, b] and let N be anynumber between f (a) and f (b), where f (a) = f (b). Then there exists anumber c in (a, b) such that f (c) = N. f (x)f (b) N f (a) a c1 c2 c3 b x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 34 / 47
69. 69. What the IVT does not sayThe Intermediate Value Theorem is an “existence” theorem. It does not say how many such c exist. It also does not say how to ﬁnd c.Still, it can be used in iteration or in conjunction with other theorems toanswer these questions. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 35 / 47
70. 70. Using the IVT to ﬁnd zeroesExampleLet f (x) = x 3 − x − 1. Show that there is a zero for f on the interval [1, 2]. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 36 / 47
71. 71. Using the IVT to ﬁnd zeroesExampleLet f (x) = x 3 − x − 1. Show that there is a zero for f on the interval [1, 2].Solutionf (1) = −1 and f (2) = 5. So there is a zero between 1 and 2.In fact, we can “narrow in” on the zero by the method of bisections. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 36 / 47
72. 72. Finding a zero by bisection y x f (x) x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 37 / 47
73. 73. Finding a zero by bisection y x f (x) 1 −1 x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 37 / 47
74. 74. Finding a zero by bisection y x f (x) 1 −1 2 5 x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 37 / 47
75. 75. Finding a zero by bisection y x f (x) 1 −1 1.5 0.875 2 5 x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 37 / 47
76. 76. Finding a zero by bisection y x f (x) 1 −1 1.25 − 0.296875 1.5 0.875 2 5 x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 37 / 47
77. 77. Finding a zero by bisection y x f (x) 1 −1 1.25 − 0.296875 1.375 0.224609 1.5 0.875 2 5 x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 37 / 47
78. 78. Finding a zero by bisection y x f (x) 1 −1 1.25 − 0.296875 1.3125 − 0.0515137 1.375 0.224609 1.5 0.875 2 5 x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 37 / 47
79. 79. Finding a zero by bisection y x f (x) 1 −1 1.25 − 0.296875 1.3125 − 0.0515137 1.375 0.224609 1.5 0.875 2 5 (More careful analysis yields x 1.32472.) V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 37 / 47
80. 80. Using the IVT to assert existence of numbersExampleSuppose we are unaware of the square root function and that it’scontinuous. Prove that the square root of two exists. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 38 / 47
81. 81. Using the IVT to assert existence of numbersExampleSuppose we are unaware of the square root function and that it’scontinuous. Prove that the square root of two exists.Proof.Let f (x) = x 2 , a continuous function on [1, 2]. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 38 / 47
82. 82. Using the IVT to assert existence of numbersExampleSuppose we are unaware of the square root function and that it’scontinuous. Prove that the square root of two exists.Proof.Let f (x) = x 2 , a continuous function on [1, 2]. Note f (1) = 1 andf (2) = 4. Since 2 is between 1 and 4, there exists a point c in (1, 2) suchthat f (c) = c 2 = 2. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 38 / 47
83. 83. OutlineContinuityThe Intermediate Value TheoremBack to the Questions V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 39 / 47
84. 84. Back to the QuestionsTrue or FalseAt one point in your life you were exactly three feet tall. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 40 / 47
85. 85. Question 1AnswerThe answer is TRUE. Let h(t) be height, which varies continuously over time. Then h(birth) < 3 ft and h(now) > 3 ft. So by the IVT there is a point c in (birth, now) where h(c) = 3. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 41 / 47
86. 86. Back to the QuestionsTrue or FalseAt one point in your life you were exactly three feet tall.True or FalseAt one point in your life your height in inches equaled your weight inpounds. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 42 / 47
87. 87. Question 2AnswerThe answer is 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 by the IVT there is a point c in (birth, now) where f (c) = 0. In other words, h(c) − w (c) = 0 ⇐⇒ h(c) = w (c). V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 43 / 47
88. 88. Back to the QuestionsTrue or FalseAt one point in your life you were exactly three feet tall.True or FalseAt one point in your life your height in inches equaled your weight inpounds.True or FalseRight now there are two points on opposite sides of the Earth with exactlythe same temperature. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 44 / 47
89. 89. Question 3AnswerThe answer is TRUE. 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? V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 45 / 47
90. 90. Question 3 Let f (θ) = T (θ) − T (θ + 180◦ ) Then f (0) = T (0) − T (180) while f (180) = T (180) − T (360) = −f (0) So somewhere between 0 and 180 there is a point θ where f (θ) = 0! V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 46 / 47
91. 91. What have we learned today? Deﬁnition: a function is continuous at a point if the limit of the function at that point agrees with the value of the function at that point. We often make a fundamental assumption that functions we meet in nature are continuous. The Intermediate Value Theorem is a basic property of real numbers that we need and use a lot. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 21, 2010 47 / 47