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# Lesson 20: The Mean Value Theorem

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The Mean Value Theorem is the most important theorem in calculus! It allows us to infer information about a function from information about its derivative. Such as: a function whose derivative is …

The Mean Value Theorem is the most important theorem in calculus! It allows us to infer information about a function from information about its derivative. Such as: a function whose derivative is zero must be a constant function.

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• 1. Section 4.2 The Mean Value Theorem V63.0121.034, Calculus I November 9, 2009 Announcements Quiz this week on §§3.1–3.5 . . Image credit: Jimmywayne22 . . . . . .
• 2. Outline Review: The Closed Interval Method Rolle’s Theorem The Mean Value Theorem Applications Why the MVT is the MITC . . . . . .
• 3. Flowchart for placing extrema Thanks to Fermat Suppose f is a continuous function on the closed, bounded interval [a, b], and c is a global maximum point. . . . c is a start local max . . . Is c an Is f diff’ble f is not n .o n .o endpoint? at c? diff at c y . es y . es . c = a or . f′ (c) = 0 c = b . . . . . .
• 4. The Closed Interval Method This means to ﬁnd the maximum value of f on [a, b], we need to: Evaluate f at the endpoints a and b Evaluate f at the critical points x where either f′ (x) = 0 or f is not differentiable at x. The points with the largest function value are the global maximum points The points with the smallest or most negative function value are the global minimum points. . . . . . .
• 5. Outline Review: The Closed Interval Method Rolle’s Theorem The Mean Value Theorem Applications Why the MVT is the MITC . . . . . .
• 6. 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 . . . . . .
• 7. 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 . . . . . . .
• 8. Mathematical Statement of Rolle’s Theorem c .. 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 . . . . . . .
• 9. Proof of Rolle’s Theorem Proof. By the Extreme Value Theorem f must achieve its maximum value at a point c in [a, b]. . . . . . .
• 10. Proof of Rolle’s Theorem Proof. By the Extreme Value Theorem f must achieve its maximum value at a point c in [a, b]. If c is in (a, b), great: it’s a local maximum and so by Fermat’s Theorem f′ (c) = 0. . . . . . .
• 11. Proof of Rolle’s Theorem Proof. By the Extreme Value Theorem f must achieve its maximum value at a point c in [a, b]. If c is in (a, b), great: it’s a local maximum and so by Fermat’s Theorem f′ (c) = 0. On the other hand, if c = a or c = b, try with the minimum. The minimum of f on [a, b] must be achieved at a point d in [a, b]. . . . . . .
• 12. Proof of Rolle’s Theorem Proof. By the Extreme Value Theorem f must achieve its maximum value at a point c in [a, b]. If c is in (a, b), great: it’s a local maximum and so by Fermat’s Theorem f′ (c) = 0. On the other hand, if c = a or c = b, try with the minimum. The minimum of f on [a, b] must be achieved at a point d in [a, b]. If d is in (a, b), great: it’s a local minimum and so by Fermat’s Theorem f′ (d) = 0. If not, d = a or d = b. . . . . . .
• 13. Proof of Rolle’s Theorem Proof. By the Extreme Value Theorem f must achieve its maximum value at a point c in [a, b]. If c is in (a, b), great: it’s a local maximum and so by Fermat’s Theorem f′ (c) = 0. On the other hand, if c = a or c = b, try with the minimum. The minimum of f on [a, b] must be achieved at a point d in [a, b]. If d is in (a, b), great: it’s a local minimum and so by Fermat’s Theorem f′ (d) = 0. If not, d = a or d = b. If we still haven’t found a point in the interior, we have that the maximum and minimum values of f on [a, b] occur at both endpoints. . . . . . .
• 14. Proof of Rolle’s Theorem Proof. By the Extreme Value Theorem f must achieve its maximum value at a point c in [a, b]. If c is in (a, b), great: it’s a local maximum and so by Fermat’s Theorem f′ (c) = 0. On the other hand, if c = a or c = b, try with the minimum. The minimum of f on [a, b] must be achieved at a point d in [a, b]. If d is in (a, b), great: it’s a local minimum and so by Fermat’s Theorem f′ (d) = 0. If not, d = a or d = b. If we still haven’t found a point in the interior, we have that the maximum and minimum values of f on [a, b] occur at both endpoints. But we already know that f(a) = f(b). . . . . . .
• 15. Proof of Rolle’s Theorem Proof. By the Extreme Value Theorem f must achieve its maximum value at a point c in [a, b]. If c is in (a, b), great: it’s a local maximum and so by Fermat’s Theorem f′ (c) = 0. On the other hand, if c = a or c = b, try with the minimum. The minimum of f on [a, b] must be achieved at a point d in [a, b]. If d is in (a, b), great: it’s a local minimum and so by Fermat’s Theorem f′ (d) = 0. If not, d = a or d = b. If we still haven’t found a point in the interior, we have that the maximum and minimum values of f on [a, b] occur at both endpoints. But we already know that f(a) = f(b). If these are the maximum and minimum values, f is constant on [a, b] and any point x in (a, b) will have f′ (x) = 0. . . . . . .
• 16. Flowchart proof of Rolle’s Theorem . . . Let c be Let d be endpoints . . . are max the max pt the min pt and min . . . f is is c. an y . es is d. an. y . es . constant endpoint? endpoint? on [a, b] n .o n .o . . . f′ (x) .≡ 0 . f (c) .= 0 ′ f (d) .= 0 ′ on (a, b) . . . . . .
• 17. Outline Review: The Closed Interval Method Rolle’s Theorem The Mean Value Theorem Applications Why the MVT is the MITC . . . . . .
• 18. 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 . . . . . .
• 19. 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 . b . f(b) − f(a) . . = f′ (c). a . b−a . . . . . .
• 20. 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 . b . f(b) − f(a) . . = f′ (c). a . b−a . . . . . .
• 21. 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 . b . f(b) − f(a) . . = f′ (c). a . b−a . . . . . .
• 22. Rolle vs. MVT f(b) − f(a) f′ (c) = 0 = f′ (c) b−a c .. c .. . b . . . . . . a . b . a . . . . . . .
• 23. 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. . . . . . .
• 24. 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 . . . . . .
• 25. 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 . . . . . .
• 26. 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. . . . . . .
• 27. 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) . . . . . .
• 28. 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 . . . . . .
• 29. Using the MVT to count solutions Example Show that there is a unique solution to the equation x3 − x = 100 in the interval [4, 5]. . . . . . .
• 30. Using the MVT to count solutions Example Show that there is a unique solution to the equation x3 − x = 100 in the interval [4, 5]. Solution By the Intermediate Value Theorem, the function f(x) = x3 − x must take the value 100 at some point on c in (4, 5). . . . . . .
• 31. Using the MVT to count solutions Example Show that there is a unique solution to the equation x3 − x = 100 in the interval [4, 5]. Solution By the Intermediate Value Theorem, the function 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. . . . . . .
• 32. Using the MVT to count solutions Example Show that there is a unique solution to the equation x3 − x = 100 in the interval [4, 5]. Solution By the Intermediate Value Theorem, the function 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 positive all along (4, 5). So this is impossible. . . . . . .
• 33. Example We know that |sin x| ≤ 1 for all x. Show that |sin x| ≤ |x|. . . . . . .
• 34. 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 . . . . . .
• 35. 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? . . . . . .
• 36. 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 . 4, 9 ) ( y . . By MVT f(4) − f(1) . 4, f(4)) ( = f′ (c) < 2 . 4−1 for some c in (1, 4). Therefore . 1, 3 ) ( . f(4) = f(1) + f′ (c)(3) < 3 + 2 · 3 = 9. So no, it is impossible that f(4) ≥ 9. . x . . . . . . .
• 37. Question A driver travels along the New Jersey Turnpike using EZ-Pass. 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) EZ-Pass cannot prove that the driver was speeding (b) EZ-Pass can prove that the driver was speeding (c) The driver’s actual maximum speed exceeds his ticketed speed (d) Both (b) and (c). Be prepared to justify your answer. . . . . . .
• 38. Question A driver travels along the New Jersey Turnpike using EZ-Pass. 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) EZ-Pass cannot prove that the driver was speeding (b) EZ-Pass can prove that the driver was speeding (c) The driver’s actual maximum speed exceeds his ticketed speed (d) Both (b) and (c). Be prepared to justify your answer. . . . . . .
• 39. Outline Review: The Closed Interval Method Rolle’s Theorem The Mean Value Theorem Applications Why the MVT is the MITC . . . . . .
• 40. Fact If f is constant on (a, b), then f′ (x) = 0 on (a, b). . . . . . .
• 41. 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 = cx0 . . . . . .
• 42. 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 = cx0 Question If f′ (x) = 0 is f necessarily a constant function? . . . . . .
• 43. 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 = cx0 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 . . . . . .
• 44. Why the MVT is the MITC Most Important Theorem In Calculus! Theorem Let f′ = 0 on an interval (a, b). . . . . . .
• 45. 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). . . . . . .
• 46. 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. . . . . . .
• 47. 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. . . . . . .
• 48. 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) . . . . . .
• 49. MVT and differentiability Example Let { −x if x ≤ 0 f(x) = x2 if x ≥ 0 Is f differentiable at 0? . . . . . .
• 50. 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. . . . . . .
• 51. 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 . . . . . .
• 52. This solution is valid but less direct. We seem to be using the following fact: If lim f′ (x) does not x→a exist, then f is not differentiable at a. equivalently: If f is differentiable at a, then lim f′ (x) exists. x→a But this “fact” is not true! . . . . . .
• 53. Differentiable with discontinuous derivative It is possible for a function f to be differentiable at a even if lim f′ (x) does not exist. x→a Example { x2 sin(1/x) if x ̸= 0 Let f′ (x) = . Then when x ̸= 0, 0 if 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. 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 differentiable for all x, but f′ is not continuous at 0! . . . . . .
• 54. 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 . . . . . .
• 55. 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 ′ (c x ) 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. 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. . . . . . .
• 57. 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. . . . . . .
• 58. What have we learned today? Rolle’s Theorem: there is a stationary point Mean Value Theorem: at some point the instantaneous rate of change equals the average rate of change (The Most Important Theorem in Calculus) Only constant functions have a derivative of zero. . . . . . .