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Calculus Date: 12/10/13 Obj: SWBAT apply Rolle’s and the Mean Value Thm

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

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Calculus Date: 12/10/13 Obj: SWBAT apply Rolle’s and the Mean Value Thm

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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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