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- 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

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