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- 1. Sec on 4.2 The Mean Value Theorem V63.0121.011: Calculus I Professor Ma hew Leingang New York University April 6, 2011.
- 2. Announcements Quiz 4 on Sec ons 3.3, 3.4, 3.5, and 3.7 next week (April 14/15) Quiz 5 on Sec ons 4.1–4.4 April 28/29 Final Exam Thursday May 12, 2:00–3:50pm
- 3. Courant Lecture tomorrow Persi Diaconis (Stanford) “The Search for Randomness” (General Audience Lecture) Thursday, April 7, 2011, 3:30pm Warren Weaver Hall, room 109 Recep on to followVisit http://cims.nyu.edu/ for details and to RSVP
- 4. 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.
- 5. Outline Rolle’s Theorem The Mean Value Theorem Applica ons Why the MVT is the MITC Func ons with deriva ves that are zero MVT and diﬀeren ability
- 6. Heuristic Motivation for Rolle’s Theorem If you bike up a hill, then back down, at some point your eleva on was sta onary. .Image credit: SpringSun
- 7. Mathematical Statement of Rolle’sTheorem Theorem (Rolle’s Theorem) Let f be con nuous on [a, b] and diﬀeren able on (a, b). Suppose f(a) = f(b). Then there exists a point c in (a, b) such that f′ (c) = 0. . a b
- 8. Mathematical Statement of Rolle’sTheorem Theorem (Rolle’s Theorem) c Let f be con nuous on [a, b] and diﬀeren able on (a, b). Suppose f(a) = f(b). Then there exists a point c in (a, b) such that f′ (c) = 0. . a b
- 9. 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)
- 10. Outline Rolle’s Theorem The Mean Value Theorem Applica ons Why the MVT is the MITC Func ons with deriva ves that are zero MVT and diﬀeren ability
- 11. Heuristic Motivation for The Mean Value Theorem If you drive between points A and B, at some me your speedometer reading was the same as your average speed over the drive. .Image credit: ClintJCL
- 12. The Mean Value Theorem Theorem (The Mean Value Theorem) Let f be con nuous on [a, b] and diﬀeren able 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) Let f be con nuous on [a, b] and diﬀeren able on (a, b). Then there exists a point c in (a, b) such that f(b) − f(a) b = f′ (c). . b−a a
- 14. The Mean Value Theorem Theorem (The Mean Value Theorem) Let f be con nuous on c [a, b] and diﬀeren able on (a, b). Then there exists a point c in (a, b) such that f(b) − f(a) b = f′ (c). . b−a a
- 15. Rolle vs. MVT f(b) − f(a) f′ (c) = 0 = f′ (c) b−a c c b . . a b a
- 16. 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.
- 17. Proof of the Mean Value Theorem Proof. The line connec ng (a, f(a)) and (b, f(b)) has equa on f(b) − f(a) L(x) = f(a) + (x − a) b−a
- 18. Proof of the Mean Value Theorem Proof. The line connec ng (a, f(a)) and (b, f(b)) has equa on f(b) − f(a) L(x) = f(a) + (x − a) b−a Apply Rolle’s Theorem to the func on f(b) − f(a) g(x) = f(x) − L(x) = f(x) − f(a) − (x − a). b−a
- 19. Proof of the Mean Value Theorem Proof. The line connec ng (a, f(a)) and (b, f(b)) has equa on f(b) − f(a) L(x) = f(a) + (x − a) b−a Apply Rolle’s Theorem to the func on f(b) − f(a) g(x) = f(x) − L(x) = f(x) − f(a) − (x − a). b−a Then g is con nuous on [a, b] and diﬀeren able on (a, b) since f is.
- 20. Proof of the Mean Value Theorem Proof. The line connec ng (a, f(a)) and (b, f(b)) has equa on f(b) − f(a) L(x) = f(a) + (x − a) b−a Apply Rolle’s Theorem to the func on f(b) − f(a) g(x) = f(x) − L(x) = f(x) − f(a) − (x − a). b−a Then g is con nuous on [a, b] and diﬀeren able on (a, b) since f is. Also g(a) = 0 and g(b) = 0 (check both).
- 21. Proof of the Mean Value Theorem Proof. f(b) − f(a) g(x) = f(x) − L(x) = f(x) − f(a) − (x − a). b−a 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
- 22. Using the MVT to count solutions Example Show that there is a unique solu on to the equa on x3 − x = 100 in the interval [4, 5].
- 23. Using the MVT to count solutions Example Show that there is a unique solu on to the equa on x3 − x = 100 in the interval [4, 5]. Solu on By the Intermediate Value Theorem, the func on f(x) = x3 − x must take the value 100 at some point on c in (4, 5).
- 24. Using the MVT to count solutions Example Show that there is a unique solu on to the equa on x3 − x = 100 in the interval [4, 5]. Solu on By the Intermediate Value Theorem, the func on f(x) = x3 − 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.
- 25. Using the MVT to count solutions Example Show that there is a unique solu on to the equa on x3 − x = 100 in the interval [4, 5]. Solu on By the Intermediate Value Theorem, the func on f(x) = x3 − 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) = 3x2 − 1, which is posi ve all along (4, 5). So this is impossible.
- 26. Using the MVT to estimate Example We know that |sin x| ≤ 1 for all x, and that sin x ≈ x for small x. Show that |sin x| ≤ |x| for all x.
- 27. Using the MVT to estimate Example We know that |sin x| ≤ 1 for all x, and that sin x ≈ x for small x. Show that |sin x| ≤ |x| for all x. Solu on Apply the MVT to the func on f(t) = sin t on [0, x]. We get Since |cos(c)| ≤ 1, we get sin x − sin 0 sin x = cos(c) ≤ 1 =⇒ |sin x| ≤ |x| x−0 x for some c in (0, x).
- 28. Using the MVT to estimate II Example Let f be a diﬀeren able func on with f(1) = 3 and f′ (x) < 2 for all x in [0, 5]. Could f(4) ≥ 9?
- 29. Using the MVT to estimate II Solu on By MVT f(4) − f(1) = f′ (c) < 2 4−1 for some c in (1, 4). Therefore f(4) = f(1) + f′ (c)(3) < 3 + 2 · 3 = 9. So no, it is impossible that f(4) ≥ 9.
- 30. Using the MVT to estimate II Solu on By MVT y (4, 9) f(4) − f(1) = f′ (c) < 2 (4, f(4)) 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
- 31. Food for Thought Ques on A driver travels along the New Jersey Turnpike using E-ZPass. The system takes note of the me and place the driver enters and exits the Turnpike. A week a er his trip, the driver gets a speeding cket in the mail. Which of the following best describes the situa on? (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 cketed speed (d) Both (b) and (c).
- 32. Food for Thought Ques on A driver travels along the New Jersey Turnpike using E-ZPass. The system takes note of the me and place the driver enters and exits the Turnpike. A week a er his trip, the driver gets a speeding cket in the mail. Which of the following best describes the situa on? (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 cketed speed (d) Both (b) and (c).
- 33. Outline Rolle’s Theorem The Mean Value Theorem Applica ons Why the MVT is the MITC Func ons with deriva ves that are zero MVT and diﬀeren ability
- 34. Functions with derivatives that are zero Fact If f is constant on (a, b), then f′ (x) = 0 on (a, b).
- 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 quo ents must be 0 The tangent line to a line is that line, and a constant func on’s graph is a horizontal line, which has slope 0. Implied by the power rule since c = cx0
- 36. Functions with derivatives that are zero Ques on If f′ (x) = 0 is f necessarily a constant func on?
- 37. Functions with derivatives that are zero Ques on If f′ (x) = 0 is f necessarily a constant func on? It seems true But so far no theorem (that we have proven) uses informa on about the deriva ve of a func on to determine informa on about the func on itself
- 38. Why the MVT is the MITC(Most Important Theorem In Calculus!) Theorem Let f′ = 0 on an interval (a, b).
- 39. 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).
- 40. 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 con nuous on [x, y] and diﬀeren able 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.
- 41. Functions with the same derivative Theorem Suppose f and g are two diﬀeren able func ons 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.
- 42. Functions with the same derivative Theorem Suppose f and g are two diﬀeren able func ons 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.
- 43. Functions with the same derivative Theorem Suppose f and g are two diﬀeren able func ons 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)
- 44. Functions with the same derivative Theorem Suppose f and g are two diﬀeren able func ons 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)
- 45. Functions with the same derivative Theorem Suppose f and g are two diﬀeren able func ons 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
- 46. Functions with the same derivative Theorem Suppose f and g are two diﬀeren able func ons 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)
- 47. MVT and diﬀerentiabilityExampleLet { −x if x ≤ 0f(x) = x2 if x ≥ 0Is f diﬀeren able at 0?
- 48. MVT and diﬀerentiabilityExample Solu on (from the deﬁni on)Let We have { −x if x ≤ 0 f(x) − f(0) −xf(x) = lim− = lim− = −1 x2 if x ≥ 0 x→0 x−0 x→0 x f(x) − f(0) x2Is f diﬀeren able at 0? lim = lim+ = lim+ x = 0 x→0+ x−0 x→0 x x→0 Since these limits disagree, f is not diﬀeren able at 0.
- 49. MVT and diﬀerentiabilityExample Solu on (Sort of)Let If x < 0, then f′ (x) = −1. If x > 0, then { f′ (x) = 2x. Since −x if x ≤ 0f(x) = x2 if x ≥ 0 lim+ f′ (x) = 0 and lim− f′ (x) = −1, x→0 x→0Is f diﬀeren able at 0? the limit lim f′ (x) does not exist and so f is x→0 not diﬀeren able at 0.
- 50. Why only “sort of”? This solu on is valid but less f′ (x) direct. y f(x) We seem to be using the following fact: If lim f′ (x) does x→a not exist, then f is not . x diﬀeren able at a. equivalently: If f is diﬀeren able at a, then lim f′ (x) exists. x→a But this “fact” is not true!
- 51. Diﬀerentiable with discontinuous derivative It is possible for a func on f to be diﬀeren able at a even if lim f′ (x) x→a does not exist. Example { ′ x2 sin(1/x) if x ̸= 0 Let f (x) = . 0 if x = 0 Then when x ̸= 0, f′ (x) = 2x sin(1/x) + x2 cos(1/x)(−1/x2 ) = 2x sin(1/x) − cos(1/x), which has no limit at 0.
- 52. Diﬀerentiable with discontinuous derivative It is possible for a func on f to be diﬀeren able at a even if lim f′ (x) x→a does not exist. Example { ′ x2 sin(1/x) if x ̸= 0 Let f (x) = . 0 if x = 0 However, ′ f(x) − f(0) x2 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 diﬀeren able for all x, but f′ is not con nuous at 0!
- 53. Diﬀerentiability FAIL f(x) f′ (x) . x . x This func on is diﬀeren able But the deriva ve is not at 0. con nuous at 0!
- 54. MVT to the rescue Lemma Suppose f is con nuous on [a, b] and lim+ f′ (x) = m. Then x→a f(x) − f(a) lim+ = m. x→a x−a
- 55. MVT to the rescue 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
- 56. Using the MVT to ﬁnd limits Theorem Suppose lim f′ (x) = m1 and lim+ f′ (x) = m2 x→a− x→a If m1 = m2 , then f is diﬀeren able at a. If m1 ̸= m2 , then f is not diﬀeren able at a.
- 57. Using the MVT to ﬁnd limits 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.
- 58. Summary Rolle’s Theorem: under suitable condi ons, func ons must have cri cal points. Mean Value Theorem: under suitable condi ons, func ons must have an instantaneous rate of change equal to the average rate of change. A func on whose deriva ve is iden cally zero on an interval must be constant on that interval. E-ZPass is kinder than we realized.

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