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onegirl

Check the hypotheses of Rolle’s Theorem and the Mean Value Theorem and find a value of c that makes the appropriate conclusion true. Illustrate the conclusion with a graph. f(x) = x^2 + 1, [2,-2]

  • one year ago
  • one year ago

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  1. abb0t
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    Well, to satisfy rolle's theorem, you must satisfy the 3 conditions (you can find them in your book - if not, I can write them down for you if you don't have access to a book). Now to find the value of "c" using mean value theorem, you must start by finding the derivative of the function I'll leave you to find it's derivative as it is quite simple for this function. Basically, it must follow the first two conditions of rolle's theorem to apply the formula Then, plug it in to the formula: \[f'(c) = \frac{ f(b)-f(a) }{ b-a }\] on the given interval, which for your case it's -2<x<2 Now, set your derivative equal to that. and solve for "c" pick the value which matches the solution in the given interval. if the value falls outside of the given interval then the solution is excluded.

    • one year ago
  2. onegirl
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    so find the derivative of x^2 + 1?

    • one year ago
  3. abb0t
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    Yes. Find the derivative. Use the formula f'(c) = to solve for "c".

    • one year ago
  4. onegirl
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    okay got you thanks

    • one year ago
  5. abb0t
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    You don't need to check first two conditions since you know that your functions IS in fact continuous on the given interval and differentiable. So with that being said, best of luck.

    • one year ago
  6. onegirl
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    @zepdrix can u help?

    • one year ago
  7. onegirl
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    i did it but i think i got it wrong

    • one year ago
  8. onegirl
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    f'(c) = f(-2) - f(2)/ -2 - 2 so f(-2) -2^2 + 1 = -3, f(2) 2^2 + 1 = 5. So the slope will be 2 because -3 - 5/-2-2 = -8/-4 which is 2. So To c : f'(x) = 2x, f'(c) = 2c = 2 2c = 2 so c will equal one (1) but when i checked (a< c <b ) (2, 1 , -2) it doesn't make sense because 1 is not less than -2 :/

    • one year ago
  9. onegirl
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    @abb0t @zepdrix @experimentX ?

    • one year ago
  10. experimentX
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    http://en.wikipedia.org/wiki/Rolle's_theorem

    • one year ago
  11. experimentX
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    pick any two points ... on two sides of 0, you see that it satisfies the Rolle's theorem.

    • one year ago
  12. onegirl
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    Did i go wrong when i found f'x? and find the f prime of c?

    • one year ago
  13. experimentX
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    equate it to zero ... and you get x=0

    • one year ago
  14. experimentX
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    your graph is symmetric f(-2) - f(2) should be zero ... probably you made mistake somewhere.

    • one year ago
  15. onegirl
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    okay so c = to 0 right? i made a mistake in putting -2 and 2 ?

    • one year ago
  16. onegirl
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    @zepdrix are u there?

    • one year ago
  17. onegirl
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    @wio can u help

    • one year ago
  18. onegirl
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    okay so its 0?

    • one year ago
  19. experimentX
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    yes yes it is ..

    • one year ago
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