Your SlideShare is downloading. ×
  • Like
Lesson 19: The Mean Value Theorem (slides)
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×

Now you can save presentations on your phone or tablet

Available for both IPhone and Android

Text the download link to your phone

Standard text messaging rates apply

Lesson 19: The Mean Value Theorem (slides)

  • 102 views
Published

The Mean Value Theorem is the most important theorem in calculus. It is the first theorem which allows us to infer information about a function from information about its derivative. From the MVT we …

The Mean Value Theorem is the most important theorem in calculus. It is the first theorem which allows us to infer information about a function from information about its derivative. From the MVT we can derive tests for the monotonicity (increase or decrease) and concavity of a function.

  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
    Be the first to like this
No Downloads

Views

Total Views
102
On SlideShare
0
From Embeds
0
Number of Embeds
0

Actions

Shares
Downloads
1
Comments
0
Likes
0

Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide

Transcript

  • 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.