1.4 CONTINUITY AND ITS CONSEQUENCES

Transcription

1 Continuity: Informal Idea We say that a function is continuous on an interval if its graph on that t interval can be drawn without t interruption, ti that is, without lifting the pencil from the paper. Slide 1

2 DEFINITION 4.1 Slide 2

3 REMARK 4.1 For f to be continuous at x = a, the definition says that (i) f (a) must be defined, (ii) the limit must exist and (iii) the limit and the value of f at the point must be the same. Further, this says that a function is continuous at a point exactly when you can compute its limit at that point by simply substituting in. Slide 3

4 Examples of Points of Discontinuity Slide 4

5 EXAMPLE 4.1 Finding Where a Rational Function Is Continuous Slide 5

6 EXAMPLE 4.1 Solution Finding Where a Rational Function Is Continuous Slide 8

7 EXAMPLE 4.2 Removing a Hole in the Graph Extend the function from example 4.1 to make it continuous everywhere by redefining it at a single point. Slide 7

8 EXAMPLE 4.2 Removing a Hole in the Graph Solution Let for some real number a. If we let a = 4, Then and g(x) is continuous at x = 1. Slide 10

9 Removable Discontinuities When we can remove a discontinuity by redefining the function at that point, we call the discontinuity removable. Not all discontinuities are removable, however. Not removable Removable Slide 9

10 EXAMPLE 4.3 Functions That Cannot Be Extended Continuously Showthat (a) and (b) cannot be extended to a function that is continuous everywhere. Slide 10

11 EXAMPLE 4.3 Solution Functions That Cannot Be Extended Continuously Hence, no matter how we might define f (0), f will not be continuous at x = 0. Slide 11

12 EXAMPLE 4.3 Solution Functions That Cannot Be Extended Continuously Due to the endless oscillation, the limit does not exist, and there is no way to redefine df the function at x = 0 to make it continuous there. Slide 12

13 THEOREM 4.1 All polynomials are continuous everywhere. Additionally, sin x and cos x are continuous everywhere. is continuous for all x, when n is odd and for x > 0, when n is even. Slide 13

14 THEOREM 4.2 Suppose that f and g are continuous at x = a. Then all of the following are true: Simply put, Theorem 4.2 says that a sum, difference or product of continuous functions is continuous, while the quotient of two continuous functions is continuous at any point at which the denominator is nonzero. Slide 14

15 THEOREM 4.3 Suppose that and f is continuous at L. Then, Slide 15

16 COROLLARY 4.1 Slide 16

17 EXAMPLE 4.5 Continuity for a Composite Function Slide 17

18 EXAMPLE 4.5 Continuity for a Composite Function Solution Slide 18

19 DEFINITION 4.2 If f is continuous at every point on an open interval (a, b), we say that t f is continuous on (a, b). Following the figure, we say that f is continuous on the closed dinterval [a, b], if f is continuous on the open interval (a, b) and Slide 19

20 DEFINITION 4.2 Finally, if f is con nuous on all of (, ), we simply say that f is continuous. (That is, when we don t specify an interval, we mean continuous everywhere.) Slide 20

21 EXAMPLE 4.6 Continuity on a Closed Interval Slide 21

22 EXAMPLE 4.6 Continuity on a Closed Interval Solution Observe that f is defined only for 2 x 2. Note that f is the composition of two continuous functions and hence, is continuous for all x for which 4 x 2 > 0. Slide 22

23 EXAMPLE 4.6 Continuity on a Closed Interval Solution Since 4 x 2 > 0 for 2 < x < 2, we have that f is continuous for all x in the interval ( 2, 2), by Theorem 4.1 and Corollary 4.1. Slide 23

24 EXAMPLE 4.6 Continuity on a Closed Interval Solution Finally, we test the endpoints to see thatt and so that t f is continuous on the closed interval [ 2, 2]. Slide 24

25 THEOREM 4.4(Intermediate Value Theorem) Suppose that f is continuous on the closed interval [a, b] and W is any number between f (a) and f (b). Then, there is a number c [a, b] for which f (c) = W. Slide 25

26 Intermediate Value Theorem 4.4 Theorem 4.4 says that if f is continuous on [a, b], then f must take on every value between f (a) and f (b) at least once. That is, a continuous function cannot skip over any numbers between its values at the two endpoints. To do so, the graph would need to leap across the horizontal lline y = W, something that continuous functions cannot do. Slide 26

27 COROLLARY 4.2 Suppose that f is continuous on [a, b] and f (a) and f (b) have opposite signs [i.e., f (a) f (b) < 0]. Then, there is at least one number c (a, b) for which f (c) = 0. (Recall that c is then a zero of f.) Notice that Corollary 4.2 is simply pythe special case of the Intermediate Value Theorem where W = 0. Slide 27

28 EXAMPLE 4.8 Finding Zeros by the Method of Bisections (Since f is a polynomial of degree 5, we don t have any formulas for finding its zeros. The only alternative ti then, is to approximate the zeros.) Slide 28

29 EXAMPLE 4.8 Finding Zeros by the Method of Bisections Solution Corollary 4.2 suggests a simple yet effective method, called the method of bisections. Taking the midpoint of the interval [0, 1], since f (0.5) < 0 and f (0) = 3 > 0, there must be a zero between 0 and 0.5. Slide 29

30 EXAMPLE 4.8 Solution Next, the midpoint of [0, 0.5] is 0.25 and f (0.25) > 0, so that the zero is in the interval (0.25, 0.5). Finding Zeros by the Method of Bisections We continue in this way to narrow down the interval in which there s a zero Slide 30

31 EXAMPLE 4.8 Solution Finding Zeros by the Method of Bisections Slide 31

32 EXAMPLE 4.8 Finding Zeros by the Method of Bisections Solution Continuing this process through 20 more steps leads to the approximate zero x = , which is accurate to at least eight decimal places. The other zeros can be found in a similar fashion. Slide 32