## Xem mẫu

7.4 Continuity and the Intermediate and Extreme Value Theorems 271 F a c t 1 ( T h e I n t e r m e d i a t e V a l u e T h e o r e m ) If is continuous on the closed and bounded interval [, ] and , , then somewhere in the interval attains every value between and . In particular, if a continuous function changes sign on an interval, it must be zero somewhere on that interval. f A a b x B Figure 7.25 F a c t 2 ( T h e E x t r e m e V a l u e T h e o r e m ) If a function is continuous on a closed interval [, ], then takes on both a maximum (high) and a minimum (low) value on [, ].8 For to attain the maximum value of on [, ] means that there is a number in [, ] such that and for all in [, ]. Analogously, for to attain the minimum value of on [, ] means that there is a number in [, ] such that and for all in [, ]. high no highest value; no lowest value a no highest value; no lowest value b a b a b low a continuous function on a closed interval (i) a continuous function on a open interval (ii) a discontinuous function on a closed interval (iii) Figure 7.26 Studyinging parts (ii) and (iii) in Figure 7.26 should convince you that both conditions, the continuity of and the interval being closed, are necessary in order for the statement to hold. 8 For a a proof of either of these theorems, look in a more theoretical calculus book. 272 CHAPTER 7 The Theoretical Backbone: Limits and Continuity Note that even if , where is a constant, the statement holds. Given the denitions of maximum and minimum presented above, if on [, ], then for every [, ] is both a maximum value and a minimum value. Principles for Working with Limits and Their Implications for Derivatives The following general principles can be deduced from the denition of limits. Suppose lim 1 and lim 2, where 1 and 2 are nite. Then: 1. lim[] 1 2 2. lim[] 1 2 3. lim 2 , provided 2 0. The limit of a sum (difference) is the sum (difference) of the limits. The limit of a product is the product of the limits (in particular, may be constant: lim 1). The limit of a quotient is the quotient of the limits (provided the denominator has a nonzero limit). 4. If is continuous at 2, then lim 2lim . 5. If for all in the vicinity of (although not necessarily at ), then lim lim . WellalsosometimesusewhatisknownastheSandwichTheorem,ortheSqueezeTheorem. S a n d w i c h T h e o r e m If !for all in the vicinity of a (although not necessarily at ) and lim , then !. The idea behind this theorem is that the functions and act as a vise, as lower and upper bounds for ! in the vicinity of . As approaches the lower and upper jaws of the vise get arbitrarily close together, trapping lim ! between them. y g f j a x Figure 7.27 7.4 Continuity and the Intermediate and Extreme Value Theorems 273 EXERCISE 7.4 Usetheprinciplesforworkingwithlimits, alongwiththeconclusionsofExamples7.1, 7.2, and 7.3 and Exercises 7.1 and 7.2 to calculate the following. (a) lim 3 2 4 (b) lim 22 2 3 2 (c) lim 3 6 We can use principles (1) and (2) given above to prove two very useful properties of derivatives. Properties of Derivatives9 i. , where is any constant. Multiplying by a constant multiplies its derivative by . ii. [] The derivative of a sum is the sum of the derivatives. These are very important results. In the Exploratory Problems for this chapter you will be asked to prove these properties and to make sense out of them. Asanapplicationofsomeoftheprinciplesforworkingwithlimitsgiveninthissection, we will verify the following fact. T h e o r e m : D i f f e r e n t i a b i l i t y I m p l i e s C o n t i n u i t y If a function is differentiable at (that is, exists), then is continuous at . Although the line of reasoning in the proof is easier to follow than to come up with, the conclusion should make sense intuitively. If is differentiable at , then is locally linear at ; looks like a line near . It makes sense that must be continuous at . PROOF OF THEOREM Suppose that is a point in the domain of , where is dened on some open interval containing . We will assume that exists and show that is continuous at . According to our denition of continuity, is continuous at if lim . We will show that this is true provided exists. Showing that lim is equivalent to showing that lim[] 0. 9 Recall that """ means the derivative of """ 274 CHAPTER 7 The Theoretical Backbone: Limits and Continuity lim[] lim lim lim 0 0 We are multiplying by 1, since . The limit of a product is the product of the limits if both exist and are nite. The rst limit is , which exists by assumption, and the second equals zero. We have shown that if exists, then must be continuous at . Question: If is continuous at , is necessarily differentiable (locally linear) at ? Answer: No. Informally speaking, if has a sharp corner at , then it is not differ-entiable at because it is not locally linear there. A classic example of the latter situation is the function at 0. is continuous at 0 because lim0 00. However, as we saw in Example 7.13, 0 does not exist; is not differentiable at 0. f x Figure 7.28 You are now prepared to read Appendix C. There we offer proofs of facts about derivatives stated without proof in Chapter 5. Exploratory Problems for Chapter 7 275 Exploratory Problems for Chapter 7 Pushing the Limit 1. Let , where is a constant and is a differentiable function. (a) Use the principles of working with limits to show that . Begin with the limit denition of and then express in terms of . (b) Explain this result in graphical terms. Why is the slope of the tangent to at equal to times the slope of the tangent to at ? (c) Interpret this result in the case that is a position function and 2. More specically, consider the following exam-ple. Two women leave a Midwestern farmhouse and travel north on a straight road. One woman walks and the other woman runs. Suppose gives the distance between the walker and the farmhouse at time and 2 gives the distance between the runner and the farmhouse. Interpret the result 2in this context. Generalize to the case . 2. Let ! , where and are differentiable func-tions. (a) Use the principles of working with limits to show that ! . Begin with the limit denition of ! and then express ! in terms of and . (b) Explain this result in graphical terms. (c) Interpret this result in the following scenario. A teacher has put retirement money in two accounts, TIAA and CREF. Let be the value of his TIAA account at time and let be the value of his CREF account at time . Interpret the result ! in this context. 3. (a) Use the properties of limits and the results of the previous problems to differentiate 3 2 as ef-ciently as possible. (b) Alongtheway,youllneedtousethelimitdenitionofderiva-tive to differentiate 3. Explain why your answer makes sense graphically by looking at the graphs of 3 and 32. 4. Looking for Patterns: Use the results of Problem 3 above and formulas for derivatives of and 1 found in Chapter 5 to arrive at a formula for the derivative of for 0, 1, 2, 3, 1, and 1. Try your formula on another value of and see if it works. ... - tailieumienphi.vn