Limits and Applying Differentiation Rules

Transcript

1. MATH 251 §2.3 redux, §2.4 February 4, 2013
2. Section 2.3: Limits
3. Idea of the Limit Let f(x) be any function, and let a and L represent numbers. Then the “mathematical phrase” lim x→a f(x) = L means that f(x) ≈ L if x ≈ a. Related phrases: lim x→a+ f(x) = L lim x→∞ f(x) = L lim x→a f(x) = ∞ lim x→a− f(x) = L lim x→−∞ f(x) = L lim x→a f(x) = −∞
4. Continuity Definition (Continuity) Let f be a function defined on an interval I containing x = a. If lim x→a f(x) = f(a) then f is continuous at the point x = a. If f is continuous at each point of I (including endpoints, if any), then f is continuous on the interval I.
5. Discontinuity This definition implies that a function is not continuous if one of the following occurs: f(a) is not defined; the limit does not exist; or f(a) is defined, the limit exists, and the two values are not equal.
6. Which functions are continuous? The following kinds of functions are continuous on their domains: Algebraic functions, like polynomials, roots, and rational functions Exponential and logarithmic functions Trigonometric functions All sums, differences, products, quotients, and compositions of these. For this reason, you have facts like lim x→4 √ x = 2.
7. Examples of continuous functions y = ex y = ex − e−x 2 y = sin(cos(x4 + x2 + 1)) Also, functions like y = 1 x + 1 are continuous on their domains, but y = 1 x + 1 is not continuous at x = −1 (division by zero).
8. Algebra with Limits Theorem 5 Suppose that lim x→a f(x) = L and lim x→a g(x) = M, where L and M are finite numbers. Let k be any constant. Then (i) lim x→a kf(x) = kL (ii) lim x→a (f(x) + g(x)) = L + M (iii) lim x→a (f(x) · g(x)) = LM (iv) lim x→a (f(x)/g(x)) = L/M if M = 0 This is why sums, products, and quotients of continuous functions are continuous on their domains.
9. Derivative example For example, if we look for the derivative of f(x) = √ x, we find f (x) = lim h→0 f(x + h) − f(x) h = lim h→0 √ x + h − √ x h √ x + h + √ x √ x + h + √ x = lim h→0   h   h √ x + h + √ x = 1 √ x + 0 + √ x = 1 2 √ x
10. Squeeze Theorem Example Let f(x) = sin 1 x . x 1 π 1 1.5π 1 2π 1 2.5π 1 3π 1 3.5π 1 4π 1 4.5π 1 5π f(x) 0 -1 0 1 0 -1 0 1 0
11. Failure of Limit Laws Examine the graph of x sin 1 x . Can’t use lim x→0 x sin 1 x = lim x→0 x lim x→0 sin 1 x
12. Squeeze Theorem Triumphs However, we can see that −|x| ≤ x sin 1 x ≤ |x| so this forces lim x→0 x sin 1 x = 0.
13. Squeeze Theorem Illustration
14. Precise Statement Squeeze Theorem Suppose that f(x) ≤ g(x) ≤ h(x) for all x in an interval containing x = a. If f(x) → L and h(x) → L as x → a, then g(x) → L as x → a.
15. Section 2.4: Using Derivative and Antiderivative Formulas
16. Stationary Points Recall that a stationary point of a function f is a number x such that f (x) = 0. Fact Let f be differentiable on an interval I. Maximum and minimum values of f can occur only at stationary points of f or at endpoints (if any) of I.
17. Example Find the maximum and minimum values of y = x(x + 1)(x − 2) on the interval 0 ≤ x ≤ 2. Can you estimate from the graph?
18. Exact values from the derivative We expand: f(x) = x(x + 1)(x − 2) = x3 − x2 − 2x Now we use our differentiation rules to find the derivative: f (x) = 3x2 − 2x − 2. We find the stationary points with the quadratic formula: x = 2 ± √ 4 + 24 6 = 1 3 ± √ 7 3
19. Using the fact The minimum and maximum values of f(x) = x(x + 1)(x − 2) on 0 ≤ x ≤ 2 occur at stationary points of f or at x = 0 or x = 2. Evaluate all three! x 0 1 3 + √ 7 3 2 f(x) 0 −14 √ 7−20 27 0 Hence, the greatest value y = 0 occurs at both endpoints x = 0 and x = 2, and the least value y = −14 √ 7 − 20 27 ≈ −2.11 occurs at the stationary point x = 1 3 + √ 7 3 ≈ 1.55.
20. Realistic Questions How do you fence in the largest rectangular area using the smallest amount of fence? How do you design a box to hold the most material? For a cylindrical aluminum can to be its strongest, the lid needs to be twice as thick as the sides. What should the dimensions of the can be?
21. Antiderivatives Given a function f(x), the derivative of f gives the rate of change of f(x) at any particular x. The antiderivative of f is a function whose derivative (i.e., rate of change) is f. For example, if f(t) denotes the values of a speedometer over time, one antiderivative would be the odometer reading at those times. Another would be the tripometer readings. There is never just one antiderivative.
22. Finding antiderivatives In the problems on the left, apply the power rule and the constant multiple rule. On the right, guess and check. f derivative x3 6x5 4 √ x derivative f x3 6x5 4 √ x
23. Finding antiderivatives In the problems on the left, apply the power rule and the constant multiple rule. On the right, guess and check. f derivative x3 3x2 6x5 4 √ x derivative f x3 6x5 4 √ x
24. Finding antiderivatives In the problems on the left, apply the power rule and the constant multiple rule. On the right, guess and check. f derivative x3 3x2 6x5 30x4 4 √ x derivative f x3 6x5 4 √ x
25. Finding antiderivatives In the problems on the left, apply the power rule and the constant multiple rule. On the right, guess and check. f derivative x3 3x2 6x5 30x4 4 √ x 2 √ x derivative f x3 6x5 4 √ x
26. Finding antiderivatives In the problems on the left, apply the power rule and the constant multiple rule. On the right, guess and check. f derivative x3 3x2 6x5 30x4 4 √ x 2 √ x derivative f x3 x4 4 6x5 4 √ x
27. Finding antiderivatives In the problems on the left, apply the power rule and the constant multiple rule. On the right, guess and check. f derivative x3 3x2 6x5 30x4 4 √ x 2 √ x derivative f x3 x4 4 6x5 x6 4 √ x
28. Finding antiderivatives In the problems on the left, apply the power rule and the constant multiple rule. On the right, guess and check. f derivative x3 3x2 6x5 30x4 4 √ x 2 √ x derivative f x3 x4 4 6x5 x6 4 √ x 8 3 x3/2
29. Finding antiderivatives In the problems on the left, apply the power rule and the constant multiple rule. On the right, guess and check. f derivative x3 3x2 6x5 30x4 4 √ x 2 √ x derivative f x3 x4 4 6x5 x6 4 √ x 8 3 x3/2 An antiderivative of xn is xn+1 n + 1 , provided n = −1.
30. Special Case The power rule does not work for taking the antiderivative of 1 x . d dx ln |x| = 1 x This means that an antiderivative of 1 x is ln |x|.
31. Graph
32. No “the” antiderivative Theorems 8 and 9 If F(x) is an antiderivative of f(x), then so is F(x) + C. These are all of the antiderivatives of f(x).1) For example, some antiderivativies of 3x2 are x3, x3 + 1, x3 − 3 2 , and so forth. The general antiderivative of 3x2 is x3 + C, where C is an arbitrary constant. 1Mostly. More discussion in Chapter 4.
33. Power Rule Theorem 10 Let k be any number besides −1. Then the antiderivatives of xk are all functions of the form xk+1 k+1 + C. Why are these all antiderivatives? Because d dx xk+1 k + 1 + C = 1 k + 1 d dx xk+1 + d dx C = xk.
34. Guess-and-Check What’s the general antiderivative of x2 + 8 3 √ x?
35. Guess-and-Check What’s the general antiderivative of x2 + 8 3 √ x? Rewrite as x2 + 8x1/3. Try applying the power rule to both pieces to get x3 3 + 8 · x4/3 4/3 = x3 3 + 6x4/3. Check using differentiation rules!
36. Some Antidifferentiation Rules Fact If F = f, G = g, and k is any constant, then (F + G) = f + g (F − G) = f − g (kF) = kf Thus, just as you can differentiate polynomials term-by-term, you can antidifferentiate polynomials term-by-term.
37. Custom-order curves Define piecewise functions which “hook up” smoothly. Define functions with prescribed stationary points, points of inflection, etc.