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Mean value theorem and indefinate intgral

  1. May 11, 2008 #1
    1. The problem statement, all variables and given/known data

    http://img241.imageshack.us/img241/7753/scan0001io9.th.jpg [Broken]

    2. Relevant equations

    3. The attempt at a solution

    i completed the first part fine- knowing the function makes a u shape with min point being 0 and max being 1/2 at both +/- 1. i cant see how using the mean value theorem has much to do with part b) or how part a) helps.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. May 11, 2008 #2


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    For the mean value theorem, substitute f(c) for f'(c) in the statement, and let b=1 and a=-1. What do you observe? Also take note that by part (a), you have already found the maximum and minimum values of f(c) in the interval [-1,1].
  4. May 11, 2008 #3
    well from mvt f'(c)=0 which means there is a max/min between a and b - i know its a min, at (0,0). I cant see what any of this had to do with the integral or the inequality?
  5. May 11, 2008 #4


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    MVT doesn't say that f'(c) = 0. See here:

    Write out that expression, let b=1, a=-1. And then you should realise that f(c) is constrained less than the max value and greater than the min value which you have found in (a).
  6. May 11, 2008 #5
    but f(a) = f(b) so f'(c)=0?
  7. May 11, 2008 #6


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    What does f here refer to? Does it mean the f(x) in the question? If so, then what about F(x)? You didn't make use of that at all, which is required to solve the problem.
  8. May 11, 2008 #7


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    if you have to use an integral perhaps you should use the mvt for integrals?
  9. May 11, 2008 #8


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    No, that isn't required. The only thing required is this: [tex]f'(c) = \frac{f(b)-f(a)}{b-a}[/tex]. And the values of min and max as stated above.
  10. May 11, 2008 #9
    does this mean i have to integrate f(x)? if so how? i have tried substitution.
  11. May 11, 2008 #10


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    No you don't have to. In fact all you have to do is to consider what should f' and f be in the statement of the MVT in relation to this problem. You have to relate these 2 functions to F(x) and f(x) in the question.
  12. May 12, 2008 #11

    Gib Z

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    I might be easier to just integrate each part of the inequality produced in a) between -1 and 1.
  13. May 12, 2008 #12


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    It certainly is not necessary to actually integrate f.

    The mean value theorem says that there exist c between a and b such that
    [tex]\frac{F(b)- F(a)}{b- a}= F'(c)[/tex]

    You know that F'(x)= f(x) and from (a) you know the largest and smallest values of f on [-1, 1]. Put those in for F'(c).
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