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Understanding Rolle’s and Mean Value Theorems in Calculus

Explore Rolle’s and Mean Value Theorems in calculus with examples and applications in finding zeroes and slopes of functions. Learn how these theorems connect derivatives and functions. Discover interesting problems and graphical interpretations.

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Understanding Rolle’s and Mean Value Theorems in Calculus

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  1. Calculus Date: 12/10/13 Obj: SWBAT apply Rolle’s and the Mean Value Thm http://youtu.be/4j2ZLtGoiLE derivatives of absolute values http://youtu.be/PBKnttVMbV4 first derivative test inc. dec. Do Now – HW Requests: pg 198 #11, 12, 13, 15, 17, 22, 23, 25, 29, 39, 41 SM pg 103 Read Section 5.2 • pg 183 #1-21 odds • In class: page SM 82/83 • HW: pg SM 84/85 • Step by Step Manual pg 4, 5, 8, 11-Friday • In class: SM 102 • Announcements: • No class next Wednesday • Step by Step Assignment Friday "Do not judge me by my successes, judge me by how many times I fell down and got back up again.“ Nelson Mandela

  2. I wonder how mean this theorem really is? The Mean Value Theorem Lesson 4.2

  3. Think About It • Consider a trip of two hours that is 120 miles in distance … • You have averaged 60 miles per hour • What reading on your speedometer would you have expected to see at least once? 60

  4. c Rolle’s Theorem • Given f(x) on closed interval [a, b] • Differentiable on open interval (a, b) • If f(a) = f(b) … then • There exists at least one numbera < c < b such that f ’(c) = 0 f(a) = f(b) b a

  5. Ex. Find the two x-intercepts of f(x) = x2 – 3x + 2 and show that f’(x) = 0 at some point between the two intercepts. f(x) = x2 – 3x + 2 0 = (x – 2)(x – 1) x-int. are 1 and 2 f’(x) = 2x - 3 0 = 2x - 3 x = 3/2 Rolles Theorem is satisfied as there is a point at x = 3/2 where f’(x) = 0.

  6. Let f(x) = x4 – 2x2 . Find all c in the interval (-2, 2) such that f’(x) = 0. Since f(-2) and f(2) = 8, we can use Rolle’s Theorem. f’(x) = 4x3 – 4x = 0 8 4x(x2 – 1) = 0 x = -1, 0, and 1 Thus, in the interval (-2, 2), the derivative is zero at each of these three x-values.

  7. c Mean Value Theorem • We can “tilt” the picture of Rolle’s Theorem • Stipulating that f(a) ≠ f(b) • Then there exists a c such that b a

  8. The Mean Value Theorem (MVT) aka the ‘crooked’ Rolle’s Theorem f(b) a c b f(a) • We can “tilt” the picture of Rolle’s Theorem • Stipulating that f(a) ≠ f(b) If f is continuous on [a, b] and differentiable on (a, b) There is at least one number c on (a, b) at which Conclusion: Slope of Secant Line Equals Slope of Tangent Line

  9. Finding c • Given a function f(x) = 2x3 – x2 • Find all points on the interval [0, 2] where • Strategy • Find slope of line from f(0) to f(2) • Find f ‘(x) • Set f ‘(x) equal to slope … solve for x

  10. Given f(x) = 5 – 4/x, find all c in the interval (1,4) such that the slope of the secant line = the slope of the tangent line. ? But in the interval of (1,4), only 2 works, so c = 2.

  11. Find the value(s) of c that satisfy the Mean Value Theorem for

  12. f(0) = -1 f(1) = 2

  13. If , how many numbers on [-2, 3] satisfy the conclusion of the Mean Value Theorem. A. 0 B. 1 C. 2 D. 3 E. 4 CALCULATOR REQUIRED f(3) = 39 f(-2) = 64 For how many value(s) of c is f ‘ (c ) = -5? X X X

  14. Find the value(s) of c that satisfy the Mean Value Theorem for Since has no real solution, there is no value of c on [-4, 4] which satisfies the Mean Value Theorem Note: The Mean Value Theorem requires the function to be continuous on [-4, 4] and differentiable on (-4, 4). Therefore, since f(x) is discontinuous at x = 0 which is on [-4, 4], there may be no value of c which satisfies the Mean Value Theorem

  15. Mean Value Theorem • Applied to a cubic equation Note Geogebera Example

  16. Find the value(s) of c which satisfy Rolle’s Theorem for on the interval [0, 1]. which is on [0, 1] Verify…..f(0) = 0 – 0 = 0 f(1) = 1 – 1 = 0

  17. Given the graph of f(x) below, use the graph of f to estimate the numbers on [0, 3.5] which satisfy the conclusion of the Mean Value Theorem.

  18. f(x) is continuous and differentiable on [-2, 2] On the interval [-2, 2], c = 0 satisfies the conclusion of MVT

  19. f(x) is continuous and differentiable on [-2, 1] On the interval [-2, 1], c = 0 satisfies the conclusion of MVT

  20. Since f(x) is discontinuous at x = 2, which is part of the interval [0, 4], the Mean Value Theorem does not apply

  21. f(x) is continuous and differentiable on [-1, 2] c = 1 satisfies the conclusion of MVT

  22. Modeling Problem • Two police cars are located at fixed points 6 miles apart on a long straight road. • The speed limit is 55 mph • A car passes the first point at 53 mph • Five minutes later he passes the second at 48 mph Yuk! Yuk! I think he was speeding, Enos We need to prove it, Rosco

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