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Section 1.4

Section 1.4. Definition of Continuity. Continuity at a Point: A function f is continuous at c if the following three conditions are met. 1. f ( c ) is defined.

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Section 1.4

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  1. Section 1.4

  2. Definition of Continuity • Continuity at a Point: A function f is continuous at c if the following three conditions are met. • 1. f (c) is defined.

  3. Continuity on an Open Interval: A function is continuous on an open interval (a, b) if it is continuous at each point in the interval. A function that is continuous on the entire real line (−∞, ∞) is continuouseverywhere. a b

  4. Discontinuity • Discontinuities fall into two categories: removable and nonremovable. • A discontinuity at c is called removable if f can be made continuous by appropriately defining (or redefining) f (c). Otherwise it is considered nonremovable.

  5. Removable Discontinuities To be a removable discontinuity there is a hole in the graph.

  6. Formally, a removable discontinuity is one at which the limit of the function exists but does not equal the value of the function at that point; this may be because the function does not exist at that point.

  7. Nonremovable Discontinuity Infinite Discontinuity Jump Discontinuity The limit does not exist for all nonremovable discontinuities.

  8. Formally, it is a discontinuity for which the limits from the left and right both exist but are not equal to each other. • 1. Jump Discontinuity is when the right-hand limit and the left-hand limit are unequal. • 2. Infinite Discontinuity is when at least one of the one-sided limits either does not exist or is infinite.

  9. Example 1 • Discuss the continuity of each function.

  10. The domain of f is all non zero real numbers. We can conclude that f is continuous at every value x in its domain. • However, at x = 0, f has a nonremovable discontinuitybecause the limit does not exist at x = 0.

  11. The domain of g is all real numbers except 1. The function is continuous at every x value in its domain. • It has a removable discontinuity at x = 1because there is a hole at x = 1.

  12. The domain for h is all real numbers. The function h is continuous on (−∞, 0) and (0, ∞), and because • h is continuous on the entire real line.

  13. Example 2 • Find the value of c which makes g(x) continuous for all x. • *Type of question on the Free Response Questions on the AP Exam.

  14. Remember to always find the one-sided limits for the x – value in question. You may also want to check your solution by graphing it.

  15. Definition of Continuity on a Closed Interval • A function f is continuous on the closed interval [a, b] if it is continuous on the open interval (a, b) and • The function f is continuous from the right at aand continuous from the left at b.

  16. (b, f(b)) (a, f(a)) f(x)

  17. The following types of functions are continuous at every point in their domain. • Polynomial functions • Rational functions • Radical functions • Trigonometric functions

  18. Intermediate Value Theorem • If f is continuous on the closed interval [a, b], f (a) ≠ f (b), and k is any number between f (a) and f (b), then there is at least one number c in [a, b] such that f (c) = k. • This is an existence theorem and will not provide a solution. It just tells us of the existence of a solution.

  19. The Intermediate Value Theorem does not help you find c. • It just guarantees the existence of at least one number c in the closed interval [a, b].

  20. If f is not continuous on [a, b], it may not exhibit the intermediate value property. k

  21. Example 3 • The function is continuous for −2 ≤ x ≤ 1. If f (−2) = −5 and f (1) = 4, decide if the following statements are true by the Intermediate Value Theorem?

  22. A. There exists c, where −2 ≤ x ≤ 1, such that f (c) ≥ f (x) for all x on the closed interval • −2 ≤ x ≤ 1. • True

  23. B. There exists c, −2 < c < 1, such that f (c) = 0. • True. Since f is continuous and f (−2) = −5 and f (1) = 4, then by the IVT there must be a c such that f(c) = 0

  24. C. There exists c, where −2 ≤ x ≤ 1, such that f (c) = 3. • True. Since f is continuous, by the IVT there must be a c such that f (c) = 3.

  25. Example 4 • Use Intermediate Value Theorem to show that there is a root of f (x) = x3 + 2x – 1 [0, 1] on the given interval.

  26. f (x) = x3 + 2x – 1 [0, 1] • Since f is a polynomial, we know that the function is continuous over the real number line. • f (0) = −1 and f (1) = 1 + 2 – 1 = 2 • Since f (0) < 0 and f (1) > 0, using the Intermediate Value Theorem you can conclude that there must be a c in the closed interval such that f (c) = 0.

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