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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. 2. lim f x exists. 3. lim f x f c xc xc 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 continuous everywhere. a b 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. Removable Discontinuities To be a removable discontinuity there is a hole in the graph. lim f x f c xc 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. Nonremovable Discontinuity Jump Discontinuity Infinite Discontinuity The limit does not exist for all nonremovable discontinuities. 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. Example 1 Discuss the continuity of each function. 2 1 x 1 a. f x b. g x x x 1 c. x 1, h x 2 x 1, x0 x0 a. 1 f x x 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 discontinuity because the limit does not exist at x = 0. b. x2 1 g x x 1 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 = 1 because there is a hole at x = 1. 4 2 -4 -2 2 -2 -4 4 c. x 1, h x 2 x 1, The domain for h is all real numbers. The function h is continuous on (−∞, 0) and (0, ∞), and because lim h x h 0 1 x0 h is continuous on the entire real line. x0 x0 Example 2 Find the value of c which makes g(x) continuous for all x. cx 1, x 3 g x 2 cx 1, x 3 *Type of question on the Free Response Questions on the AP Exam. lim g x g 3 cx 1, x 3 g x 2 cx 1, x 3 x3 g 3 3c 1 lim g x 3c 1 x3 lim+ g x 9c 1 x3 lim g x lim+ g x x3 x3 3c 1 9c 1 1 c 3 1 3 x 1, x 3 g x 1 x 2 1, x 3 3 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. 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 lim f x f a and lim f x f b . x a xb The function f is continuous from the right at a and continuous from the left at b. lim+ f x f a x a lim f x f b xb (b, f(b)) (a, f(a)) f(x) The following types of functions are continuous at every point in their domain. 1. Polynomial functions 2. Rational functions 3. Radical functions 4. Trigonometric functions 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. 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]. If f is not continuous on [a, b], it may not exhibit the intermediate value property. k 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? 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 y x 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 y x 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. y x 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. 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.