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Mean Value Theorem

  1. May 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Given the function f(x)= x(x^2-8)-5 satisfies the hypothesis of the Mean Value Thereom on the interval [1,4], find a number C in the interval (1,4) which satisfies this thereom.




    2. Relevant equations

    f'(c) = f(b)-f(a) / b-a

    3. The attempt at a solution

    1) Expand the equation first
    2) Find the first derivative.
    3) Equal the equation to 1

    Apparently, I got the wrong answer. What am I doing wrong?
    PLEASE HELP.
     
  2. jcsd
  3. May 5, 2008 #2

    Dick

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    That's a great strategy. Impossible to tell you how you got the wrong answer until you tell us what you got for (f(b)-f(a))/(b-a) and C.
     
  4. May 5, 2008 #3

    Defennder

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    What do you mean by "equal the equation to 1"? Don't you have to find both f(1) and f(4) then use the Mean Value theorem to find c?
     
  5. May 5, 2008 #4

    Dick

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    I thought the OP meant "equal the equation" to (f(b)-f(a))/(b-a). I may have been extrapolating on that.
     
  6. May 5, 2008 #5
    Hm.. I just realized it and I'm stuck. I don't know what to do or what I'm trying to get.. HAHA

    On the bright side, I do have the C value and the value for (f(b)-f(a))/(a-b):

    (f(b)-f(a))/(a-b)
    ( 27 + 12 )/(4-1) = 13

    C value = x^3-8x-5
    f' = 3x^2-8
    3x^2-8 = 1
    3x^2 = 9
    9 / 3 ^1/2
    = 3^1/2

    So, what to do next? Or what the heck am I suppose to get?
     
  7. May 5, 2008 #6
    I found out the tangent line is at (3^1/2, -13.66) which is parallel to the secant line through (1, -12) and (4, 27)

    Now, I don't even know if that helps.. but there it is. Lol!
     
  8. May 5, 2008 #7
    What I meant by equal the equation to one was that getting the derivative of the equation and equalling it to 1.

    1 = 3x^2 -8
     
  9. May 5, 2008 #8

    Dick

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    Don't "equal it to 1". Equal it to (f(b)-f(a))/(b-a)=13. Read the mean value theorem again.
     
  10. May 5, 2008 #9
    HAHA. Thanks. That's all I needed to know. You've solved one of my many problems, AGAIN! THANKS!
     
  11. May 5, 2008 #10

    Dick

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    Gotta admit, you resolve your own problems quickly. Hope this is a short lived phase of confusion.
     
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