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anonymous

  • 5 years ago

could someone explain Rolle theorem?

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  1. amistre64
    • 5 years ago
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    yes....

  2. amistre64
    • 5 years ago
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    it is the mean theorum but for a flat line.... really.....

  3. amistre64
    • 5 years ago
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    the mean theorum states that any slope between two points can be moved to a spot on the curve where it only touches at one point right?

  4. amistre64
    • 5 years ago
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    the Rolle theorum states that if the slope between two poits is a flat horizontal line, then there is a point on the curve that the flat line will only touch in one spot......

  5. anonymous
    • 5 years ago
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    ok thanks, how is it represented mathematically?

  6. amistre64
    • 5 years ago
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    ...that I dont know ;)

  7. amistre64
    • 5 years ago
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    if an interval I exists and is continuous between it end points and the interval I [a,b] has a slope of zero between f(a) and f(b) then there exists a point in the interval [a,b] called "c" such that the f'(c) = 0....matbe like that :)

  8. anonymous
    • 5 years ago
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    oh so in essence it's similiar to the mean value theorem?

  9. amistre64
    • 5 years ago
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    in essense...it IS the mean value theorum ;) only with a flat line. Why we need it to have its own name is a mystery to me....

  10. anonymous
    • 5 years ago
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    yeah I wonder also I am gonna send you a question and u solve it n show me the steps

  11. amistre64
    • 5 years ago
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    if I got the time, I gotta get to class in 30 minutes :)

  12. anonymous
    • 5 years ago
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    ok u gonna be back anytime soon?

  13. amistre64
    • 5 years ago
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    3 hours..if anything. post it up on the main side over there and see if anyone bites.

  14. amistre64
    • 5 years ago
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    or let me see it and ill let you know if I got the time :)

  15. anonymous
    • 5 years ago
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    ok

  16. anonymous
    • 5 years ago
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    \[let f:[-1,1]rightarrowR be such that\]

  17. anonymous
    • 5 years ago
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    let f:[−1,1]rightarrow R be such that

  18. amistre64
    • 5 years ago
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    let f(-1) to f(1) be.....

  19. anonymous
    • 5 years ago
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    \[f(x)= x^2 - 1/(1+x^2)\]

  20. amistre64
    • 5 years ago
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    plug in -1 and solve: f(-1) = 0 plug in 1 and solve: f(1) = 0

  21. anonymous
    • 5 years ago
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    yes and show that f satisfies all the conditions for the application of Rolle's theorem

  22. amistre64
    • 5 years ago
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    since the bottom of the function is never 0, we have no restrictions there and the function is continuous thruout the interval...

  23. amistre64
    • 5 years ago
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    if we can find the x = c such that f'(c) = 0 ..... then that shuold solve it.

  24. amistre64
    • 5 years ago
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    Rolles theorum doesnt tell you HOW to find "c", just that itlll exist :)

  25. amistre64
    • 5 years ago
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    are there any values of x in the interval [-1,1] that we should avoid using? I dont see any, so it is continuous right?

  26. anonymous
    • 5 years ago
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    yea it's continous

  27. amistre64
    • 5 years ago
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    what value of x will make the denominator go to zero? 1+x^2 = 0 .....none, nade, zip, ziltch.... its consinuous :) so there must be a c value in the interval [-1,1] that satisfies Rolles theorum

  28. anonymous
    • 5 years ago
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    so it's saying find all c which satisfy the conclusion of Rolle's them

  29. amistre64
    • 5 years ago
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    if you want to find it, yo could look for the hgihest or lowest value of f(x) and that would be it; or take the derivative and make it equal to 0 right?

  30. anonymous
    • 5 years ago
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    oh, so take the defferential of f(x) and equate it to zero

  31. anonymous
    • 5 years ago
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    n solve for x?

  32. amistre64
    • 5 years ago
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    yep, at f'(x) = 0 we have what they call "critical points" is is when the slope of the curve becomes "0"...a horizontal line

  33. amistre64
    • 5 years ago
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    yes.. (1+x^2)(2x) - (x^2 -1)(2x) = 0

  34. amistre64
    • 5 years ago
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    4x = 0....when x = 0

  35. amistre64
    • 5 years ago
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    f(0) = -1 :)

  36. anonymous
    • 5 years ago
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    ok thanx, have a productive class

  37. amistre64
    • 5 years ago
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    Ciao:)

  38. anonymous
    • 5 years ago
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    \[_{?}\]

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