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.

Like this presentation? Why not share!

- Lesson 19: The Mean Value Theorem by Matthew Leingang 3357 views
- Lar calc10 ch02_sec1 by ahmad 392 views
- 2D CAD Module by gonzalochris by Chris Gonzalo 1627 views
- Mean variance by odkoo 1893 views
- Ap calculus ftc applications frq by jaflint718 2177 views
- AP Calculus BC Integration By Parts... by jaflint718 4652 views

2,265 views

2,178 views

2,178 views

Published on

No Downloads

Total views

2,265

On SlideShare

0

From Embeds

0

Number of Embeds

0

Shares

0

Downloads

91

Comments

0

Likes

3

No embeds

No notes for slide

- 1. Section 4.2 The Mean Value Theorem V63.0121.002.2010Su, Calculus I New York University June 8, 2010 Announcements Exams not graded yet Assignment 4 is on the website Quiz 3 on Thursday covering 3.3, 3.4, 3.5, 3.7
- 2. Announcements Exams not graded yet Assignment 4 is on the website Quiz 3 on Thursday covering 3.3, 3.4, 3.5, 3.7 V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 2 / 28
- 3. Objectives Understand and be able to explain the statement of Rolle’s Theorem. Understand and be able to explain the statement of the Mean Value Theorem. V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 3 / 28
- 4. Outline Rolle’s Theorem The Mean Value Theorem Applications Why the MVT is the MITC Functions with derivatives that are zero MVT and diﬀerentiability V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 4 / 28
- 5. Heuristic Motivation for Rolle’s Theorem If you bike up a hill, then back down, at some point your elevation was stationary. Image credit: SpringSun V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 5 / 28
- 6. Mathematical Statement of Rolle’s Theorem Theorem (Rolle’s Theorem) Let f be continuous on [a, b] and diﬀerentiable on (a, b). Suppose f (a) = f (b). Then there exists a point c in (a, b) such that f (c) = 0. a b
- 7. Mathematical Statement of Rolle’s Theorem Theorem (Rolle’s Theorem) c Let f be continuous on [a, b] and diﬀerentiable on (a, b). Suppose f (a) = f (b). Then there exists a point c in (a, b) such that f (c) = 0. a b V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 6 / 28
- 8. Flowchart proof of Rolle’s Theorem endpoints Let c be Let d be are max the max pt the min pt and min f is is c an is d an yes yes constant endpoint? endpoint? on [a, b] no no f (x) ≡ 0 f (c) = 0 f (d) = 0 on (a, b) V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 8 / 28
- 9. Outline Rolle’s Theorem The Mean Value Theorem Applications Why the MVT is the MITC Functions with derivatives that are zero MVT and diﬀerentiability V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 9 / 28
- 10. Heuristic Motivation for The Mean Value Theorem If you drive between points A and B, at some time your speedometer reading was the same as your average speed over the drive. Image credit: ClintJCL V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 10 / 28
- 11. The Mean Value Theorem Theorem (The Mean Value Theorem) Let f be continuous on [a, b] and diﬀerentiable on (a, b). Then there exists a point c in (a, b) such that f (b) − f (a) b = f (c). b−a a
- 12. The Mean Value Theorem Theorem (The Mean Value Theorem) Let f be continuous on [a, b] and diﬀerentiable on (a, b). Then there exists a point c in (a, b) such that f (b) − f (a) b = f (c). b−a a
- 13. The Mean Value Theorem Theorem (The Mean Value Theorem) c Let f be continuous on [a, b] and diﬀerentiable on (a, b). Then there exists a point c in (a, b) such that f (b) − f (a) b = f (c). b−a a V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 11 / 28
- 14. Rolle vs. MVT f (b) − f (a) f (c) = 0 = f (c) b−a c c b a b a V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 12 / 28
- 15. Rolle vs. MVT f (b) − f (a) f (c) = 0 = f (c) b−a c c b a b a If the x-axis is skewed the pictures look the same. V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 12 / 28
- 16. Proof of the Mean Value Theorem Proof. The line connecting (a, f (a)) and (b, f (b)) has equation f (b) − f (a) y − f (a) = (x − a) b−a V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 13 / 28
- 17. Proof of the Mean Value Theorem Proof. The line connecting (a, f (a)) and (b, f (b)) has equation f (b) − f (a) y − f (a) = (x − a) b−a Apply Rolle’s Theorem to the function f (b) − f (a) g (x) = f (x) − f (a) − (x − a). b−a V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 13 / 28
- 18. Proof of the Mean Value Theorem Proof. The line connecting (a, f (a)) and (b, f (b)) has equation f (b) − f (a) y − f (a) = (x − a) b−a Apply Rolle’s Theorem to the function f (b) − f (a) g (x) = f (x) − f (a) − (x − a). b−a Then g is continuous on [a, b] and diﬀerentiable on (a, b) since f is. V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 13 / 28
- 19. Proof of the Mean Value Theorem Proof. The line connecting (a, f (a)) and (b, f (b)) has equation f (b) − f (a) y − f (a) = (x − a) b−a Apply Rolle’s Theorem to the function f (b) − f (a) g (x) = f (x) − f (a) − (x − a). b−a Then g is continuous on [a, b] and diﬀerentiable on (a, b) since f is. Also g (a) = 0 and g (b) = 0 (check both) V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 13 / 28
- 20. Proof of the Mean Value Theorem Proof. The line connecting (a, f (a)) and (b, f (b)) has equation f (b) − f (a) y − f (a) = (x − a) b−a Apply Rolle’s Theorem to the function f (b) − f (a) g (x) = f (x) − f (a) − (x − a). b−a Then g is continuous on [a, b] and diﬀerentiable on (a, b) since f is. Also g (a) = 0 and g (b) = 0 (check both) So by Rolle’s Theorem there exists a point c in (a, b) such that f (b) − f (a) 0 = g (c) = f (c) − . b−a V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 13 / 28
- 21. Using the MVT to count solutions Example Show that there is a unique solution to the equation x 3 − x = 100 in the interval [4, 5]. V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 14 / 28
- 22. Using the MVT to count solutions Example Show that there is a unique solution to the equation x 3 − x = 100 in the interval [4, 5]. Solution By the Intermediate Value Theorem, the function f (x) = x 3 − x must take the value 100 at some point on c in (4, 5). V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 14 / 28
- 23. Using the MVT to count solutions Example Show that there is a unique solution to the equation x 3 − x = 100 in the interval [4, 5]. Solution By the Intermediate Value Theorem, the function f (x) = x 3 − x must take the value 100 at some point on c in (4, 5). If there were two points c1 and c2 with f (c1 ) = f (c2 ) = 100, then somewhere between them would be a point c3 between them with f (c3 ) = 0. V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 14 / 28
- 24. Using the MVT to count solutions Example Show that there is a unique solution to the equation x 3 − x = 100 in the interval [4, 5]. Solution By the Intermediate Value Theorem, the function f (x) = x 3 − x must take the value 100 at some point on c in (4, 5). If there were two points c1 and c2 with f (c1 ) = f (c2 ) = 100, then somewhere between them would be a point c3 between them with f (c3 ) = 0. However, f (x) = 3x 2 − 1, which is positive all along (4, 5). So this is impossible. V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 14 / 28
- 25. Using the MVT to estimate Example We know that |sin x| ≤ 1 for all x. Show that |sin x| ≤ |x|. V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 15 / 28
- 26. Using the MVT to estimate Example We know that |sin x| ≤ 1 for all x. Show that |sin x| ≤ |x|. Solution Apply the MVT to the function f (t) = sin t on [0, x]. We get sin x − sin 0 = cos(c) x −0 for some c in (0, x). Since |cos(c)| ≤ 1, we get sin x ≤ 1 =⇒ |sin x| ≤ |x| x V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 15 / 28
- 27. Using the MVT to estimate II Example Let f be a diﬀerentiable function with f (1) = 3 and f (x) < 2 for all x in [0, 5]. Could f (4) ≥ 9? V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 16 / 28
- 28. Using the MVT to estimate II Example Let f be a diﬀerentiable function with f (1) = 3 and f (x) < 2 for all x in [0, 5]. Could f (4) ≥ 9? Solution y (4, 9) By MVT f (4) − f (1) (4, f (4)) = f (c) < 2 4−1 for some c in (1, 4). Therefore f (4) = f (1) + f (c)(3) < 3 + 2 · 3 = 9. (1, 3) So no, it is impossible that f (4) ≥ 9. x V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 16 / 28
- 29. Food for Thought Question A driver travels along the New Jersey Turnpike using E-ZPass. The system takes note of the time and place the driver enters and exits the Turnpike. A week after his trip, the driver gets a speeding ticket in the mail. Which of the following best describes the situation? (a) E-ZPass cannot prove that the driver was speeding (b) E-ZPass can prove that the driver was speeding (c) The driver’s actual maximum speed exceeds his ticketed speed (d) Both (b) and (c). V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 17 / 28
- 30. Food for Thought Question A driver travels along the New Jersey Turnpike using E-ZPass. The system takes note of the time and place the driver enters and exits the Turnpike. A week after his trip, the driver gets a speeding ticket in the mail. Which of the following best describes the situation? (a) E-ZPass cannot prove that the driver was speeding (b) E-ZPass can prove that the driver was speeding (c) The driver’s actual maximum speed exceeds his ticketed speed (d) Both (b) and (c). V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 17 / 28
- 31. Outline Rolle’s Theorem The Mean Value Theorem Applications Why the MVT is the MITC Functions with derivatives that are zero MVT and diﬀerentiability V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 18 / 28
- 32. Functions with derivatives that are zero Fact If f is constant on (a, b), then f (x) = 0 on (a, b). V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 19 / 28
- 33. Functions with derivatives that are zero Fact If f is constant on (a, b), then f (x) = 0 on (a, b). The limit of diﬀerence quotients must be 0 The tangent line to a line is that line, and a constant function’s graph is a horizontal line, which has slope 0. Implied by the power rule since c = cx 0 V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 19 / 28
- 34. Functions with derivatives that are zero Fact If f is constant on (a, b), then f (x) = 0 on (a, b). The limit of diﬀerence quotients must be 0 The tangent line to a line is that line, and a constant function’s graph is a horizontal line, which has slope 0. Implied by the power rule since c = cx 0 Question If f (x) = 0 is f necessarily a constant function? V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 19 / 28
- 35. Functions with derivatives that are zero Fact If f is constant on (a, b), then f (x) = 0 on (a, b). The limit of diﬀerence quotients must be 0 The tangent line to a line is that line, and a constant function’s graph is a horizontal line, which has slope 0. Implied by the power rule since c = cx 0 Question If f (x) = 0 is f necessarily a constant function? It seems true But so far no theorem (that we have proven) uses information about the derivative of a function to determine information about the function itself V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 19 / 28
- 36. Why the MVT is the MITC Most Important Theorem In Calculus! Theorem Let f = 0 on an interval (a, b). V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 20 / 28
- 37. Why the MVT is the MITC Most Important Theorem In Calculus! Theorem Let f = 0 on an interval (a, b). Then f is constant on (a, b). V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 20 / 28
- 38. Why the MVT is the MITC Most Important Theorem In Calculus! Theorem Let f = 0 on an interval (a, b). Then f is constant on (a, b). Proof. Pick any points x and y in (a, b) with x < y . Then f is continuous on [x, y ] and diﬀerentiable on (x, y ). By MVT there exists a point z in (x, y ) such that f (y ) − f (x) = f (z) = 0. y −x So f (y ) = f (x). Since this is true for all x and y in (a, b), then f is constant. V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 20 / 28
- 39. Functions with the same derivative Theorem Suppose f and g are two diﬀerentiable functions on (a, b) with f = g . Then f and g diﬀer by a constant. That is, there exists a constant C such that f (x) = g (x) + C . V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 21 / 28
- 40. Functions with the same derivative Theorem Suppose f and g are two diﬀerentiable functions on (a, b) with f = g . Then f and g diﬀer by a constant. That is, there exists a constant C such that f (x) = g (x) + C . Proof. Let h(x) = f (x) − g (x) Then h (x) = f (x) − g (x) = 0 on (a, b) So h(x) = C , a constant This means f (x) − g (x) = C on (a, b) V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 21 / 28
- 41. MVT and diﬀerentiability Example Let −x if x ≤ 0 f (x) = x2 if x ≥ 0 Is f diﬀerentiable at 0? V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 22 / 28
- 42. MVT and diﬀerentiability Example Let −x if x ≤ 0 f (x) = x2 if x ≥ 0 Is f diﬀerentiable at 0? Solution (from the deﬁnition) We have f (x) − f (0) −x lim = lim = −1 x→0− x −0 x→0 − x f (x) − f (0) x2 lim+ = lim+ = lim+ x = 0 x→0 x −0 x→0 x x→0 Since these limits disagree, f is not diﬀerentiable at 0. V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 22 / 28
- 43. MVT and diﬀerentiability Example Let −x if x ≤ 0 f (x) = x2 if x ≥ 0 Is f diﬀerentiable at 0? Solution (Sort of) If x < 0, then f (x) = −1. If x > 0, then f (x) = 2x. Since lim f (x) = 0 and lim f (x) = −1, x→0+ x→0− the limit lim f (x) does not exist and so f is not diﬀerentiable at 0. x→0 V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 22 / 28
- 44. Why only “sort of”? f (x) This solution is valid but less y f (x) direct. We seem to be using the following fact: If lim f (x) x→a does not exist, then f is not x diﬀerentiable at a. equivalently: If f is diﬀerentiable at a, then lim f (x) exists. x→a But this “fact” is not true! V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 23 / 28
- 45. Diﬀerentiable with discontinuous derivative It is possible for a function f to be diﬀerentiable at a even if lim f (x) x→a does not exist. Example x 2 sin(1/x) if x = 0 Let f (x) = . Then when x = 0, 0 if x = 0 f (x) = 2x sin(1/x) + x 2 cos(1/x)(−1/x 2 ) = 2x sin(1/x) − cos(1/x), which has no limit at 0. However, f (x) − f (0) x 2 sin(1/x) f (0) = lim = lim = lim x sin(1/x) = 0 x→0 x −0 x→0 x x→0 So f (0) = 0. Hence f is diﬀerentiable for all x, but f is not continuous at 0! V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 24 / 28
- 46. Diﬀerentiability FAIL f (x) f (x) x x This function is diﬀerentiable at But the derivative is not 0. continuous at 0! V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 25 / 28
- 47. MVT to the rescue Lemma Suppose f is continuous on [a, b] and lim+ f (x) = m. Then x→a f (x) − f (a) lim+ = m. x→a x −a V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 26 / 28
- 48. MVT to the rescue Lemma Suppose f is continuous on [a, b] and lim+ f (x) = m. Then x→a f (x) − f (a) lim+ = m. x→a x −a Proof. Choose x near a and greater than a. Then f (x) − f (a) = f (cx ) x −a for some cx where a < cx < x. As x → a, cx → a as well, so: f (x) − f (a) lim = lim+ f (cx ) = lim+ f (x) = m. x→a+ x −a x→a x→a V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 26 / 28
- 49. Theorem Suppose lim f (x) = m1 and lim+ f (x) = m2 x→a− x→a If m1 = m2 , then f is diﬀerentiable at a. If m1 = m2 , then f is not diﬀerentiable at a. V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 27 / 28
- 50. Theorem Suppose lim f (x) = m1 and lim+ f (x) = m2 x→a− x→a If m1 = m2 , then f is diﬀerentiable at a. If m1 = m2 , then f is not diﬀerentiable at a. Proof. We know by the lemma that f (x) − f (a) lim = lim f (x) x→a− x −a x→a− f (x) − f (a) lim+ = lim+ f (x) x→a x −a x→a The two-sided limit exists if (and only if) the two right-hand sides agree. V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 27 / 28
- 51. Summary Rolle’s Theorem: under suitable conditions, functions must have critical points. Mean Value Theorem: under suitable conditions, functions must have an instantaneous rate of change equal to the average rate of change. A function whose derivative is identically zero on an interval must be constant on that interval. E-ZPass is kinder than we realized. V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 28 / 28

No public clipboards found for this slide

Be the first to comment