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# Lesson05 Continuity Slides+Notes

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### Lesson05 Continuity Slides+Notes

1. 1. Section 2.4 Continuity Math 1a October 3, 2007
2. 2. Questions True or False Right now there are two points on opposite sides of the Earth with exactly the same temperature.
3. 3. Questions True or False Right now there are two points on opposite sides of the Earth with exactly the same temperature. True or False At one point in your life your height in inches equaled your weight in pounds.
4. 4. Questions True or False Right now there are two points on opposite sides of the Earth with exactly the same temperature. True or False At one point in your life your height in inches equaled your weight in pounds. True or False At one point in your life you were exactly three feet tall.
5. 5. 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
6. 6. 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
7. 7. 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.
8. 8. Showing a function is continuous Example √ Let f (x) = 4x + 1. Show that f is continuous at 2.
9. 9. 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.
10. 10. 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 its whole domain, which is − 1 , ∞ . 4
11. 11. 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: 1. f + g 2. f − g 3. cf 4. fg f 5. (if g (a) = 0) g
12. 12. 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
13. 13. 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
14. 14. Pitfall #1 Example Let x2 if 0 ≤ x ≤ 1 f (x) = if 1 < x ≤ 2 2x At which points is f continuous?
15. 15. 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.
16. 16. Pitfall #2 Example Let x 2 + 2x + 1 f (x) = x +1 At which points is f continuous?
17. 17. 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.
18. 18. Pitfall #3 Example Let 46 if x = 1 f (x) = π if x = 1 At which points is f continuous?
19. 19. Pitfall #3: function value = limit Example Let 46 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) = 46. x→1
20. 20. 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.
21. 21. 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. The greatest integer function f (x) = [[x]] has jump discontinuities.
22. 22. 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.
23. 23. Illustrating the IVT f (x) x
24. 24. Illustrating the IVT Suppose that f is continuous on the closed interval [a, b] f (x) x
25. 25. Illustrating the IVT Suppose that f is continuous on the closed interval [a, b] f (x) f (b) f (a) x a b
26. 26. 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
27. 27. 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
28. 28. 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
29. 29. 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
30. 30. Using the IVT Example Prove that the square root of two exists.
31. 31. Using the IVT Example Prove that the square root of two exists. Proof. Let f (x) = x 2 , a continuous function on [1, 2].
32. 32. 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.
33. 33. 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.