MATH CALCULUS I 1.5: Continuity

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1 MATH CALCULUS I 1.5: Continuity Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 12

2 Definition of Continuity Intuitively, a function is continuous at a point if the graph can be drawn through the point without lifting your pencil; that is, there are no holes or jumps in the graph at that point. This is not precise enough to accurately define continuity, however. We will use the following definition: Definition A function y = f (x) is continuous at x = a if f (a) is defined and lim f (x) = f (a). x a A consequence of this definition is that when x is near a, a small change in x results in a small change in y, and therefore no break or jump in the graph is possible. D.L. White (Kent State University) 2 / 12

3 Definition of Continuity We will use the following checklist to determine if a given function y = f (x) is continuous at a given point x = a: 1 Is f (a) defined? 2 Does lim x a f (x) exist? 3 Does lim x a f (x) = f (a)? If the answer to each of these is YES, then f is continuous at x = a. If the answer to any one of these is NO, then f is not continuous at x = a. D.L. White (Kent State University) 3 / 12

4 Example Determine whether the function f (x) = x 2 4 x 2 is continuous at x = 2 and justify your answer. Solution Notice that when x = 2, the denominator of the function is 0. Therefore f (2) is not defined, and so f is NOT continuous at x = 2. Note: Even though f (x) = (x 2)(x+2) x 2, we cannot cancel the x 2 factors in this situation. The function f (x) is undefined at x = 2 (and has a hole in its graph at x = 2), but the function g(x) = x + 2 has g(2) = 4. Cancelling changes the function; f (x) g(x). D.L. White (Kent State University) 4 / 12

5 Example Determine whether the function f (x) = x 2 4 x 2 if x 2 3 if x = 2 is continuous at x = 2 and justify your answer. D.L. White (Kent State University) 5 / 12

6 Solution 1 By the definition of the function, f (2) = 3, so f is defined at x = 2. 2 We have and so lim x 2 f (x) exists. lim f (x) = lim x 2 x 2 x 2 4, since x 2, x 2 (x 2)(x + 2) = lim x 2 x 2 = lim (x + 2), since x 2, x 2 = = 4, 3 Since f (2) = 3 4 = lim f (x), we have lim f (x) f (2), and so f is x 2 x 2 NOT continuous at x = 2. D.L. White (Kent State University) 6 / 12

7 Example Determine if f (x) = { x 2 + 3x + 1 if x < 1 2x 4 + 6x 2 3 if x 1 is continuous at x = 1 and justify your answer. D.L. White (Kent State University) 7 / 12

8 Solution 1 We have f (1) = 2(1 4 ) + 6(1 2 ) 3 = = 5, and so f (1) is defined. 2 To determine if lim f (x) exists, we compute the one-sided limits. and Since lim f (x) = lim (x 2 + 3x + 1) = = 5 lim f (x) = lim + +(2x 4 + 6x 2 3) = = 5. lim f (x) = lim f (x), lim f (x) does exist. + 3 Since lim f (x) = lim f (x) = 5, we have lim f (x) = 5 = f (1), and + so f is continuous at x = 1. D.L. White (Kent State University) 8 / 12

9 Example Determine if x 2 + 3x + 1 if x < 1 g(x) = 7 if x = 1 2x 4 + 6x 2 3 if x > 1 is continuous at x = 1 and justify your answer. D.L. White (Kent State University) 9 / 12

10 Solution 1 By the definition of the function, we have g(1) = 7, and so g(1) is defined. 2 To determine if lim g(x) exists, we compute the one-sided limits. and Since lim g(x) = lim (x 2 + 3x + 1) = = 5 lim g(x) = lim + +(2x 4 + 6x 2 3) = = 5. lim g(x) = lim g(x), lim g(x) does exist. + 3 Since lim g(x) = lim g(x) = 5, we have lim g(x) = 5 7 = g(1), + and so g is NOT continuous at x = 1. D.L. White (Kent State University) 10 / 12

11 Polynomials and Rational Functions Continuity for polynomials and rational functions is very easy to determine. From 1.4, we have Direct Substitution Property If f is a polynomial or rational function and a is in the domain of f, then lim f (x) = f (a). x a Notice that this condition is precisely the definition of continuity at the point x = a, and so we have Theorem If f is a polynomial or rational function and a is in the domain of f, then f is continuous at x = a. This means that if f is a polynomial, then f is continuous at x = a for every real number a, and if f is a rational function, then f is continuous at x = a if and only if the denominator of f is not 0. D.L. White (Kent State University) 11 / 12

12 Polynomials and Rational Functions Example Determine the values of x at which the function is not continuous. f (x) = x 2 + x 6 x 2 + 2x 3 Solution Since f is a rational function, it is discontinuous precisely at the values of x for which the denominator is 0. The denominator is x 2 + 2x 3 = (x + 3)(x 1), and so is 0 when x = 3 or x = 1. Therefore, f is discontinuous at x = 3 and at x = 1 and continuous at every other real number. D.L. White (Kent State University) 12 / 12