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# Lesson 19: The Mean Value Theorem (handout)

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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.

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### Lesson 19: The Mean Value Theorem (handout)

1. 1. . V63.0121.001: Calculus I . Sec on 4.2: The Mean Value Theorem . April 6, 2011 Notes Sec on 4.2 The Mean Value Theorem V63.0121.001: Calculus I Professor Ma hew Leingang New York University April 6, 2011 . . Notes 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 . . Notes 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. . . . 1.
2. 2. . V63.0121.001: Calculus I . Sec on 4.2: The Mean Value Theorem . April 6, 2011 Notes 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 . . Notes 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 . Mathematical Statement of Rolle’s Notes Theorem 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 . . . 2.
3. 3. . V63.0121.001: Calculus I . Sec on 4.2: The Mean Value Theorem . April 6, 2011 Notes 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) . . Notes 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 . . Notes 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 . . 3.
4. 4. . V63.0121.001: Calculus I . Sec on 4.2: The Mean Value Theorem . April 6, 2011 Notes 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 . . Notes 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. . . Notes 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) y − f(a) = (x − a) b−a Apply Rolle’s Theorem to the func on f(b) − f(a) g(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) . . . 4.
5. 5. . V63.0121.001: Calculus I . Sec on 4.2: The Mean Value Theorem . April 6, 2011 Notes Proof of the Mean Value Theorem Proof. f(b) − f(a) g(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 . . Notes 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. . . Notes 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). . . . 5.
6. 6. . V63.0121.001: Calculus I . Sec on 4.2: The Mean Value Theorem . April 6, 2011 Notes 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? Solu on . . Notes 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). . . Notes 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.
7. 7. . V63.0121.001: Calculus I . Sec on 4.2: The Mean Value Theorem . April 6, 2011 Notes 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 . . Notes 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 . . Why the MVT is the MITC Notes (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. . . 7.
8. 8. . V63.0121.001: Calculus I . Sec on 4.2: The Mean Value Theorem . April 6, 2011 Notes Functions with the same derivative Theorem Suppose f and g are two diﬀeren able func ons on (a, b) with f′ = g′ . Proof. . . Notes MVT and diﬀerentiability Example Solu on (from the deﬁni on) Let We have { −x if x ≤ 0 f(x) − f(0) −x f(x) = lim = lim− = −1 x2 if x ≥ 0 x→0− x−0 x→0 x f(x) − f(0) x2 Is 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. . . Notes MVT and diﬀerentiability Example Solu on (Sort of) Let If x < 0, then f′ (x) = −1. If x > 0, then { f′ (x) = 2x. Since −x if x ≤ 0 f(x) = x2 if x ≥ 0 lim f′ (x) = 0 and lim− f′ (x) = −1, x→0+ x→0 Is f diﬀeren able at 0? ′ the limit lim f (x) does not exist and so f is x→0 not diﬀeren able at 0. . . . 8.
9. 9. . V63.0121.001: Calculus I . Sec on 4.2: The Mean Value Theorem . April 6, 2011 Notes 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! . . Notes 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. . . Notes 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! . . 9.
10. 10. . V63.0121.001: Calculus I . Sec on 4.2: The Mean Value Theorem . April 6, 2011 Notes 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! . . Notes 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 . . Notes 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 . . . 10.
11. 11. . V63.0121.001: Calculus I . Sec on 4.2: The Mean Value Theorem . April 6, 2011 Notes 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. . . Notes 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. . . Notes 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. . . . 11.