Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this document? Why not share!

1,223 views

Published on

No Downloads

Total views

1,223

On SlideShare

0

From Embeds

0

Number of Embeds

2

Shares

0

Downloads

57

Comments

0

Likes

1

No embeds

No notes for slide

- 1. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 Notes Section 1.5 Continuity V63.0121.021, Calculus I New York University September 20, 2010 Announcements Notes Oﬃce Hours: Tuesday, Wednesday, 3pm–4pm TAs have oﬃce hours on website V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 2 / 47 Grader’s corner Notes HW Grades will be on blackboard this week, and the papers will be returned in recitation Remember units when computing slopes Remember to staple your papers—you have been warned. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 3 / 47 1
- 2. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 Objectives Notes Understand and apply the deﬁnition of continuity for a function at a point or on an interval. Given a piecewise deﬁned function, decide where it is continuous or discontinuous. State and understand the Intermediate Value Theorem. Use the IVT to show that a function takes a certain value, or that an equation has a solution V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 4 / 47 Last time Notes Deﬁnition We write lim f (x) = L x→a and say “the limit of f (x), as x approaches a, equals L” if we can make the values of f (x) arbitrarily close to L (as close to L as we like) by taking x to be suﬃciently close to a (on either side of a) but not equal to a. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 5 / 47 Limit Laws for arithmetic Notes Theorem (Basic Limits) lim x = a x→a lim c = c x→a Theorem (Limit Laws) Let f and g be functions with limits at a point a. Then lim (f (x) + g (x)) = lim f (x) + lim g (x) x→a x→a x→a lim (f (x) − g (x)) = lim f (x) − lim g (x) x→a x→a x→a lim (f (x) · g (x)) = lim f (x) · lim g (x) x→a x→a x→a f (x) limx→a f (x) lim = if lim g (x) = 0 x→a g (x) limx→a g (x) x→a V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 6 / 47 2
- 3. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 Hatsumon Notes Here are some discussion questions to start. True or False At some point in your life you were exactly three feet tall. True or False At some point in your life your height (in inches) was equal to your weight (in pounds). True or False Right now there are a pair of points on opposite sides of the world measuring the exact same temperature. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 7 / 47 Outline Notes Continuity The Intermediate Value Theorem Back to the Questions V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 8 / 47 Recall: Direct Substitution Property Notes Theorem (The Direct Substitution Property) If f is a polynomial or a rational function and a is in the domain of f , then lim f (x) = f (a) x→a This property is so useful it’s worth naming. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 9 / 47 3
- 4. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 Deﬁnition of Continuity Notes y Deﬁnition Let f be a function deﬁned near a. We say that f is f (a) continuous at a if lim f (x) = f (a). x→a A function f is continuous if it is continuous at every point in its domain. x a V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 10 / 47 Scholium Notes Deﬁnition Let f be a function deﬁned near a. We say that f is continuous at a if lim f (x) = f (a). x→a There are three important parts to this deﬁnition. The function has to have a limit at a, the function has to have a value at a, and these values have to agree. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 11 / 47 Free Theorems Notes Theorem (a) Any polynomial is continuous everywhere; that is, it is continuous on R = (−∞, ∞). (b) Any rational function is continuous wherever it is deﬁned; that is, it is continuous on its domain. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 12 / 47 4
- 5. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 Showing a function is continuous Notes Example √ Let f (x) = 4x + 1. Show that f is continuous at 2. Solution We want to show that lim f (x) = f (2). We have x→2 √ √ lim f (x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f (2). x→a x→2 x→2 Each step comes from the limit laws. Question At which other points is f continuous? Answer The function f is continuous on (−1/4, ∞). It is right continuous at −1/4 since lim f (x) = f (−1/4). x→−1/4+ V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 13 / 47 At which other points? Notes √ For reference: f (x) = 4x + 1 If a > −1/4, then lim (4x + 1) = 4a + 1 > 0, so x→a √ √ lim f (x) = lim 4x + 1 = lim (4x + 1) = 4a + 1 = f (a) x→a x→a x→a and f is continuous at a. √ If a = −1/4, then 4x + 1 < 0 to the left of a, which means 4x + 1 is undeﬁned. Still, √ √ lim+ f (x) = lim+ 4x + 1 = lim+ (4x + 1) = 0 = 0 = f (a) x→a x→a x→a so f is continuous on the right at a = −1/4. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 14 / 47 Showing a function is continuous Notes Example √ Let f (x) = 4x + 1. Show that f is continuous at 2. Solution We want to show that lim f (x) = f (2). We have x→2 √ √ lim f (x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f (2). x→a x→2 x→2 Each step comes from the limit laws. Question At which other points is f continuous? Answer The function f is continuous on (−1/4, ∞). It is right continuous at −1/4 since lim f (x) = f (−1/4). V63.0121.021, 1/4+ x→−Calculus I (NYU) Section 1.5 Continuity September 20, 2010 15 / 47 5
- 6. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 The Limit Laws give Continuity Laws Notes Theorem If f (x) and g (x) are continuous at a and c is a constant, then the following functions are also continuous at a: (f + g )(x) (fg )(x) (f − g )(x) f (x) (if g (a) = 0) (cf )(x) g V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 16 / 47 Why a sum of continuous functions is continuous Notes We want to show that lim (f + g )(x) = (f + g )(a). x→a We just follow our nose: lim (f + g )(x) = lim [f (x) + g (x)] (def of f + g ) x→a x→a = lim f (x) + lim g (x) (if these limits exist) x→a x→a = f (a) + g (a) (they do; f and g are cts.) = (f + g )(a) (def of f + g again) V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 17 / 47 Trigonometric functions are continuous Notes tan sec sin and cos are continuous on R. sin 1 tan = and sec = are cos cos cos continuous on their domain, which is π sin R + kπ k ∈ Z . 2 cos 1 cot = and csc = are sin sin continuous on their domain, which is R { kπ | k ∈ Z }. cot csc V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 18 / 47 6
- 7. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 Exponential and Logarithmic functions are continuous Notes For any base a > 1, ax x loga x the function x → a is continuous on R the function loga is continuous on its domain: (0, ∞) In particular e x and ln = loge are continuous on their domains V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 19 / 47 Inverse trigonometric functions are mostly continuous Notes sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and right continuous at −1. sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left continuous at −1, and right continuous at 1. tan−1 and cot−1 are continuous on R. π cot−1 cos−1 sec−1 π/2 tan−1 csc−1 −1 sin V63.0121.021, Calculus I (NYU) −π/2 Section 1.5 Continuity September 20, 2010 20 / 47 −π What could go wrong? Notes In what ways could a function f fail to be continuous at a point a? Look again at the deﬁnition: lim f (x) = f (a) x→a V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 21 / 47 7
- 8. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 Continuity FAIL: The limit does not exist Notes Example Let x2 if 0 ≤ x ≤ 1 f (x) = 2x if 1 < x ≤ 2 At which points is f continuous? Solution At any point a in [0, 2] besides 1, lim f (x) = f (a) because f is represented by a x→a polynomial near a, and polynomials have the direct substitution property. However, lim f (x) = lim x 2 = 12 = 1 x→1− x→1− lim+ f (x) = lim+ 2x = 2(1) = 2 x→1 x→1 So f has no limit at 1. Therefore f is not continuous at 1. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 22 / 47 Graphical Illustration of Pitfall #1 Notes y 4 3 The function cannot be 2 continuous at a point if the function has no limit at that point. 1 x −1 1 2 −1 V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 23 / 47 Continuity FAIL: The function has no value Notes Example Let x 2 + 2x + 1 f (x) = x +1 At which points is f continuous? Solution Because f is rational, it is continuous on its whole domain. Note that −1 is not in the domain of f , so f is not continuous there. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 24 / 47 8
- 9. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 Graphical Illustration of Pitfall #2 Notes y 1 The function cannot be continuous at a point outside its domain (that is, a point where it x has no value). −1 V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 25 / 47 Continuity FAIL: function value = limit Notes Example Let 7 if x = 1 f (x) = π if x = 1 At which points is f continuous? Solution f is not continuous at 1 because f (1) = π but lim f (x) = 7. x→1 V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 26 / 47 Graphical Illustration of Pitfall #3 Notes y 7 If the function has a limit and a value at a point the two must π still agree. x 1 V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 27 / 47 9
- 10. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 Special types of discontinuites Notes removable discontinuity The limit lim f (x) exists, but f is not deﬁned x→a at a or its value at a is not equal to the limit at a. By re-deﬁning f (a) = lim f (x), f can be made continuous at a x→a jump discontinuity The limits lim f (x) and lim+ f (x) exist, but are x→a− x→a diﬀerent. The function cannot be made continuous by changing a single value. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 28 / 47 Graphical representations of discontinuities Notes y y Presto! continuous! 7 2 continuous? continuous? π 1 continuous? x x 1 1 removable jump V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 29 / 47 Special types of discontinuites Notes removable discontinuity The limit lim f (x) exists, but f is not deﬁned x→a at a or its value at a is not equal to the limit at a. By re-deﬁning f (a) = lim f (x), f can be made continuous at a x→a jump discontinuity The limits lim f (x) and lim+ f (x) exist, but are x→a− x→a diﬀerent. The function cannot be made continuous by changing a single value. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 30 / 47 10
- 11. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 The greatest integer function Notes [[x]] is the greatest integer ≤ x. y 3 x [[x]] y = [[x]] 0 0 2 1 1 1.5 1 1 1.9 1 2.1 2 x −0.5 −1 −2 −1 1 2 3 −0.9 −1 −1 −1.1 −2 −2 This function has a jump discontinuity at each integer. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 31 / 47 Outline Notes Continuity The Intermediate Value Theorem Back to the Questions V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 32 / 47 A Big Time Theorem Notes Theorem (The Intermediate Value Theorem) Suppose that f is continuous on the closed interval [a, b] and let N be any number between f (a) and f (b), where f (a) = f (b). Then there exists a number c in (a, b) such that f (c) = N. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 33 / 47 11
- 12. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 Illustrating the IVT Notes Suppose that f is continuous on the closed interval [a, b] and let N be any number between f (a) and f (b), where f (a) = f (b). Then there exists a number c in (a, b) such that f (c) = N. f (x) f (b) N f (a) a c1 c c2 c3 b x V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 34 / 47 What the IVT does not say Notes The Intermediate Value Theorem is an “existence” theorem. It does not say how many such c exist. It also does not say how to ﬁnd c. Still, it can be used in iteration or in conjunction with other theorems to answer these questions. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 35 / 47 Using the IVT to ﬁnd zeroes Notes Example Let f (x) = x 3 − x − 1. Show that there is a zero for f on the interval [1, 2]. Solution f (1) = −1 and f (2) = 5. So there is a zero between 1 and 2. In fact, we can “narrow in” on the zero by the method of bisections. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 36 / 47 12
- 13. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 Finding a zero by bisection Notes y x f (x) 1 −1 1.25 − 0.296875 1.3125 − 0.0515137 1.375 0.224609 1.5 0.875 2 5 (More careful analysis yields x 1.32472.) V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 37 / 47 Using the IVT to assert existence of numbers Notes Example Suppose we are unaware of the square root function and that it’s continuous. Prove that the square root of two exists. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 38 / 47 Outline Notes Continuity The Intermediate Value Theorem Back to the Questions V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 39 / 47 13
- 14. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 Back to the Questions Notes True or False At one point in your life you were exactly three feet tall. True or False At one point in your life your height in inches equaled your weight in pounds. True or False Right now there are two points on opposite sides of the Earth with exactly the same temperature. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 40 / 47 Question 1 Notes To be discussed in class! V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 41 / 47 Question 2 Notes V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 43 / 47 14
- 15. V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010 Question 3 Notes V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 45 / 47 Question 3 Notes V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 46 / 47 What have we learned today? Notes Deﬁnition: a function is continuous at a point if the limit of the function at that point agrees with the value of the function at that point. We often make a fundamental assumption that functions we meet in nature are continuous. The Intermediate Value Theorem is a basic property of real numbers that we need and use a lot. V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 47 / 47 15

No public clipboards found for this slide

Be the first to comment