Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 30 docx

7.4 Continuity and the Intermediate and Extreme Value Theorems 271 Fact 1 (The Intermediate Value Theorem) If f is continuous on the closed and bounded interval [a, b] and f(a)=A, f(b)=B, then somewhere in the interval f attains every value between A and B. In particular, if a continuous function changes sign on an interval, it must be zero somewhere on that interval. f x ba B A Figure 7.25 Fact 2 (The Extreme Value Theorem) If a function f is continuous on a closed interval [a, b], then f takes on both a maximum (high) and a minimum (low) value on [a, b]. 8 For f to attain the maximum value of M on [a, b] means that there is a number c in [a, b] such that f(c)=Mand f(x)≤M for all x in [a, b]. Analogously, for f to attain the minimum value of m on [a, b] means that there is a number c in [a, b] such that f(c)=m and f(x)≥mfor all x in [a, b]. high aba a b 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) no highest value; no lowest value no highest value; no lowest value Figure 7.26 Studyinging parts (ii) and (iii) in Figure 7.26 should convince you that both conditions, the continuity of f 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 f(x)=k, where k is a constant, the statement holds. Given the definitions of maximum and minimum presented above, if f(x)=k on [a, b], then for every x ∈ [a, b] f(x)=kis 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 definition of limits. Suppose lim x→a f(x)=L 1 and lim x→a g(x) = L 2 , where L 1 and L 2 are finite. Then: 1. lim x→a [f(x)±g(x)] = L 1 ± L 2 The limit of a sum (difference) is the sum (difference) of the limits. 2. lim x→a [f (x)g(x)] = L 1 · L 2 The limit of a product is the product of the limits (in particular, g(x) may be constant: lim x→a kf (x) = kL 1 ). 3. lim x→a f(x) g(x) = L 1 L 2 , provided L 2 = 0. The limit of a quotient is the quotient of the limits (provided the denominator has a nonzero limit). 4. If h is continuous at L 2 , then lim x→a h(g(x)) = h(L 2 ) = h(lim x→a g(x)). 5. If f (x)<g(x)for all x in the vicinity of a (although not necessarily at x = a), then lim x→a f(x)≤lim x→a g(x). We’ll also sometimes use what is known as the Sandwich Theorem, or the Squeeze Theorem. Sandwich Theorem If f(x)≤j(x)≤g(x) for all x in the vicinity of a (although not necessarily at x = a) and lim x→a f(x)= lim x→a g(x) = L, then lim x→a j(x)=L. The idea behind this theorem is that the functions f and g act as a vise, as lower and upper bounds for j in the vicinity of a.Asxapproaches a the lower and upper jaws of the vise get arbitrarily close together, trapping lim x→a j(x) between them. a x y f g j Figure 7.27 7.4 Continuity and the Intermediate and Extreme Value Theorems 273 EXERCISE 7.4 Use the principles for working with limits, along with the conclusions of Examples 7.1, 7.2, and 7.3 and Exercises 7.1 and 7.2 to calculate the following. (a) lim x→∞ 3 · 2 −x + 4 (b) lim x→ √ 2 (2x 2 √ 2x + 3) (c) lim x→∞ 3 + 6x x We can use principles (1) and (2) given above to prove two very useful properties of derivatives. Properties of Derivatives 9 i. d dx kf (x) = k d dx f(x),where k is any constant. Multiplying f by a constant k multiplies its derivative by k. ii. d dx [f(x)+g(x)] = d dx f(x)+ d dx g(x) 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. As an application of some of the principles for working with limits given in this section, we will verify the following fact. Theorem: Differentiability Implies Continuity If a function is differentiable at x =a (that is, f  (a) exists), then f is continuous at x = a. Although the line of reasoning in the proof is easier to follow than to come up with, the conclusion should make sense intuitively. If f is differentiable at x = a, then f is locally linear at x =a; f looks like a line near x =a. It makes sense that f must be continuous at x = a. PROOF OF THEOREM Suppose that a is a point in the domain of f , where f is defined on some open interval containing a. We will assume that f  (a) exists and show that f is continuous at x = a. According to our definition of continuity, f is continuous at x = a if lim x→a f(x)=f(a). Wewill show that this is true provided f  (a) exists. Showing that lim x→a f(x)=f(a)is equivalent to showing that lim x→a [f(x)−f(a)]=0. 9 Recall that d dx means “the derivative of ” 274 CHAPTER 7 The Theoretical Backbone: Limits and Continuity lim x→a [f(x)−f(a)]= lim x→a  (f (x) − f(a)) x −a x − a  We are multiplying by 1, since x = a. =  lim x→a f(x)−f(a) x − a  ·  lim x→a (x − a)  The limit of a product is the product of the limits if both exist and are finite. = f  (a) · 0 The first limit is f  (a), which exists by assumption, and the second equals zero. = 0 We have shown that if f  (a) exists, then f(x)must be continuous at x = a. Question: If f is continuous at x = a,isf necessarily differentiable (locally linear) at x = a? Answer: No. Informally speaking, if f has a sharp corner at x = a, then it is not differ- entiable at x = a because it is not locally linear there. A classic example of the latter situation is the function f(x)=|x|at x = 0. f(x)=|x|is continuous at x = 0 because lim x→0 f(x)=f(0)=0. However, as we saw in Example 7.13, f  (0) does not exist; f is not differentiable at x = 0. x f 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 h(t) = kf (t), where k is a constant and f is a differentiable function. (a) Use the principles of working with limits to show that h  (t) = kf  (t). Begin with the limit definition of h  (t) and then express h in terms of f . (b) Explain this result in graphical terms. Why is the slope of the tangent to h at t equal to k times the slope of the tangent to f at t? (c) Interpret this result in the case that f(t)is a position function and k = 2. More specifically, 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 f(t) gives the distance between the walker and the farmhouse at time t and h(t) = 2f(t) gives the distance between the runner and the farmhouse. Interpret the result h  (t) = 2f  (t) in this context. Generalize to the case h  (t) = kf  (t). 2. Let j(t) =f(t)+g(t), where f and g are differentiable func- tions. (a) Use the principles of working with limits to show that j  (t) =f  (t) +g  (t). Begin with the limit definition of j  (t) and then express j in terms of f and g. (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 f(t) be the value of his TIAA account at time t and let g(t) be the value of his CREF account at time t. Interpret the result j  (t) = f  (t) + g  (t) in this context. 3. (a) Use the properties of limits and the results of the previous problems to differentiate f(x)=ax 3 +bx 2 + cx + d as effi- ciently as possible. (b) Along the way, you’ll need to use the limit definition of deriva- tive to differentiate x 3 . Explain why your answer makes sense graphically by looking at the graphs of x 3 and 3x 2 . 4. Looking for Patterns: Use the results of Problem 3 above and formulas for derivatives of √ x and 1 x found in Chapter 5 to arrive at a formula for the derivative of x n for n = 0, 1, 2, 3, −1, and 1 2 . Try your formula on another value of n and see if it works. 276 CHAPTER 7 The Theoretical Backbone: Limits and Continuity PROBLEMS FOR SECTION 7.4 1. (a) Find the following limits. Illustrate your answers with graphs. i. lim x→−∞ − 3 x ii. lim x→∞ − 3 x iii. lim x→∞ − 3 x − 3 iv. lim x→∞ x + 1 x v. lim x→∞ 2x + 3 x (b) In Section 7.1, Example 7.1, we showed lim x→∞ 1 x = 0. In Section 7.4 we listed some principles for working with limits. Show how your answers to all of the problems in part (a) can be deduced using lim x→∞ 1 x = 0 and lim x→∞ k =k, for any constant k and applying the principles listed in Section 7.4. 2. Find the following. (a) lim x→4 1 (x − 4) 2 (b) lim x→4 x + 3 (x − 4) 2 (c) lim x→4 x 2 + 16 (x − 4) 2 (d) lim x→4 + 1 (x − 4) (e) lim x→4 − 1 (x − 4) (f) lim x→4 x 2 − 16 (x − 4) 3. Find lim x→4 √ x−2 x−4 . 4. Suppose |h(x)|≤3for all x. Evaluate lim x→∞ h(x) x . Each of the functions in Problems 5 through 10 is either continuous on (−∞, ∞) or has a point of discontinuity at some point(s) x =a. Determine any point(s) of discontinuity. Is the point of discontinuity removable? In other words, can the function be made continuous by defining or redefining the function at the point of discontinuity? 5. f(x)= x 2 −4 x+2 6. f(x)=  x 3 , x =0, 3, x = 0. 7. f(x)= 1 x+2 8. f(x)= 1 x 2 +2 9. f(x)=  −x 2 −1, x>0, 5x −1, x<0. 10. f(x)=  −x 2 −x, x>0, 5x −1, x<0. 11. Let f(x)=  −x 2 +1, x>0, ax +b, x ≤0. What are the constraints on a and b in order for f to be continuous at x = 0? Exploratory Problems for Chapter 7 277 12. Let f(x)=  g(x), x ≥ a, h(x), x<a, where g is continuous on [a, ∞) and h is continuous on (−∞, a). What must be true about g and h in order for f to be continuous at x = a? 13. Let f(x)=  −1, if x is a rational number, 1, if x is an irrational number. (a) Is f(x)continuous at x = 1? (b) Is f(x)continuous at x = π? 14. In this problem we’ll look at a range of possible scenarios accompanying lim x→a f(x), where a is a finite number. Your job is to draw illustrations showing f in the vicinity of a with the specifications given. (a) Suppose lim x→2 f(x)=L,where L is a finite number. Draw three qualitatively different pictures of what f could look like in the vicinity of x = 2. The first picture should show a continuous function f . The second should show a function discontinuous at x = 2 but defined at x = 2. The third should show f undefined at x = 2. (b) Illustrate the following two scenarios. i. lim x→2 f(x)=∞ ii. lim x→2 f(x)=−∞ (c) Suppose lim x→2 f(x)is undefined. Draw three qualitatively different pictures of what f could look like in the vicinity of x = 2. The first picture should have f defined at 2. The second picture should have f undefined at 2 but the one-sided limits at 2 both finite. The third picture should have lim x→2 + f(x)=∞;lim x→2 − f(x) is left up to you. (Another possibility is that f(x)does not approach any single finite value but also does not increase or decrease without bound.) 2 15. Let f(x)=  x 2 , for x ≥ 0, −x 2 , for x<0. (a) Is f continuous at x = 0? (b) Is f differentiable at x = 0? If so, what is f  (0)? 16. Find the derivative of f(x)=kx 4 , where k is a constant. 17. Let g(x) =  x 2 , for x ≥ 0, x, for x<0. (a) Is g continuous at x = 0? (b) Is g differentiable at x = 0? If so, what is g  (0)? 278 CHAPTER 7 The Theoretical Backbone: Limits and Continuity 18. Let f(x)= 1 x . (a) Draw the graph of f(x)and f  (x). (b) Use the graphs you’ve drawn in part (a) to do the following. i. Find lim x→∞ f  (x). ii. Find lim x→0 + f  (x). iii. Find lim x→0 − f  (x). iv. Find lim x→−∞ f  (x). (c) Use the limit definition of derivative to find f  (x). Use your work to check your answers to parts (a) and (b). 19. The domain of a function f is all real numbers. The zeros of f(x)are x =−1, x = 2, and x = 6. There are no other x-values such that f(x)=0.Is it possible that f(3)>0 and f(4)<0?Explain. 20. The domain of a continuous function f is all real numbers. The zeros of f are x =−1, x = 2, and x = 6. There are no other x-values such that f(x)=0.Is it possible that f(3)>0and f(4)<0?Explain. 21. Sketch the graph of one function having all seven of the following characteristics. i. f(x)>0for all x, ii. lim x→4 f(x)=1, iii. f(4)=3, iv. lim x→∞ f(x)=1, v. lim x→−∞ f(x)=1, vi. lim x→0 + f(x)=5, vii. lim x→0 − f(x)=2. 22. Use the limit definition to differentiate f(x)= 1 x 2 . 8 CHAPTER Fruits of Our Labor: Derivatives and Local Linearity Revisited 8.1 LOCAL LINEARITY AND THE DERIVATIVE In Section 4.1 we discussed local linearity, but at that point we had not yet developed the concept of derivative. Therefore, in this section we revisit the idea. In Chapter 5 we pointed out that there are numerous forms in which the definition of derivative can be expressed, yet for the most part we’ve used f  (x) = lim h→0 f(x+h)−f(x) h . In this chapter, because we’ll look at approximations that are good for a small range of the independent variable, we’ll use notation that emphasizes the relative rates of change of the dependent and independent variables. With that in mind, we re-establish the following conventions. If y = f(x),then dy dx = lim x→0 y x , where y = f(x +x) − f(x). Equivalently, we can replace y by f , df dx = lim x→0 f x = lim x→0 f(x+x) − f(x) x . (Notice that x and h play the same role.) When we approximate a function near a point using local linearity, we might wonder whether the approximation is larger or smaller than the actual value. The answer depends on the second derivative, which was introduced in Chapter 6. There we saw that if f(t)gives position as a function of time, then f  (t) gives velocity, the rate of change of position with respect to time, and f  (t) gives acceleration, the rate of change of velocity with respect to time. Let’s now adopt a graphical perspective as well. If f  > 0, then f  is increasing and f is concave up. If f  < 0, then f  is decreasing and f is concave down. 279 280 CHAPTER 8 Fruits of Our Labor: Derivatives and Local Linearity Revisited f ″ > O => f ′ increasing f concave up f ″ < O => f ′ decreasing f concave down Figure 8.1 ◆ EXAMPLE 8.1 The weather pattern known as El Ni ˜ no brought extreme weather conditions throughout the globe in 1997 and 1998. Some areas experienced severe drought, while others were beset by flooding. In a certain town, the amount of water in the reservoir has been decreasing for the past few weeks and there is no indication that rain is to be expected any time soon. An awareness of the crisis is spreading and the rate at which water is being consumed is dropping slightly. At present there are G 0 gallons of water in the reservoir and the level is dropping at a rate of 115 gallons per day. (a) Let W(t) be the amount of water in the reservoir, where t is measured in days and we choose t = 0 to be today. Translate the information given above into statements about W , W  , and W  . (b) Approximate the amount of water in the reservoir two days from now. SOLUTION (a) Today there are G 0 gallons of water in the reservoir, so W(0)= G 0 . W(t) is decreasing and W  gives the rate of change of W , the amount of water in the reservoir, so W  (t) = dW dt < 0. dW dt    t=0 = W  (0) =−115 gallons/day t (days) W (gallons) W′ (0) = –115 G 0 Figure 8.2 The rate of water consumption is decreasing, so W  is becoming less negative as t increases. The graph of W is concave up. W  (t) ≥ 0. (b)  Amount of water in the reservoir 2 days from now  =  amount in the reservoir now  +  change in water in the next 2 days  . W(2)= G 0 +  change in water in the next two days  . . farmhouse and travel north on a straight road. One woman walks and the other woman runs. Suppose f(t) gives the distance between the walker and the farmhouse at time t and h(t) = 2f(t) gives the distance. definition of derivative to find f  (x). Use your work to check your answers to parts (a) and (b). 19. The domain of a function f is all real numbers. The zeros of f(x)are x =−1, x = 2, and x =. and formulas for derivatives of √ x and 1 x found in Chapter 5 to arrive at a formula for the derivative of x n for n = 0, 1, 2, 3, −1, and 1 2 . Try your formula on another value of n and see if it works. 276