MATH 12002 - CALCULUS I 1.5: Continuity Professor Donald L. White - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I § 1.5: Continuity Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 12

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 x → a f ( x ) = f ( a ) . lim 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

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

Examples 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) , we cannot “cancel” the x − 2 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

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

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

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

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

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

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

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 x → a f ( x ) = f ( a ) . lim 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

Polynomials and Rational Functions Example Determine the values of x at which the function f ( x ) = x 2 + x − 6 x 2 + 2 x − 3 is not continuous. 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 + 2 x − 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