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Differential calculus question (Mean value theorem)

  1. May 28, 2005 #1
    You are given the following information about the function f(x):
    i) There is an x-value x* approximately equal to 0.8 such that f(x*)=0
    ii) f(0.7) = C is negative
    iii) m1 < f'(x) < m2 for 0.7 < x < 0.9 where m1 and m2 are positive constants

    Apple the Mean Value Theorem to f(x) on the interval [0.7,x*] to find upper and lower bounds (in terms of m1, m2 and C) for x*


    I've been really struggling to understand the MVT and have been at this question for a while... i just can't seen to work it out though and end up going round in circles :frown: I have a test coming up so i really want to try to get my head around the MVT.
     
  2. jcsd
  3. May 28, 2005 #2

    Andrew Mason

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    By the Mean Value Theorem, there exists some w between .7 and x* such that:

    [tex]f'(w) = \frac{f(x*) - f(.7)}{x* - .7} = \frac{0 - C}{x* - .7}[/tex]

    Since m1 < f'(w) < m2,

    [tex]m1 < \frac{- C}{x* - .7} < m2[/tex]

    You should be able to work out the upper and lower bounds of x* from that.

    AM
     
  4. May 29, 2005 #3
    Great, that's what I did, only I keep getting the answer 0.7-C/m1 < x* < 0.7 - C/m2 but the actual answer is 0.7-C/m1 > x* > 0.7 - C/m2. I'm probably making some stupid mistake, but this is what I did:

    m1 < -C/(x*-0.7) < m2
    1/m1 > (x*-0.7)/-C > m2 (i changed the signs because i took the reciprocol)
    -C/m1 < (x*-0.7) < -C/m2 (i changed the signs because * by a negative number)
    0.7-C/m1 < x* < 0.7-C/m2

    Can you tell me where I fudged up? It's probably really obvious but I can't see it :(
     
  5. May 29, 2005 #4

    shmoe

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    -C is positive
     
  6. May 29, 2005 #5

    Andrew Mason

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    Take each inequality separately and do the algebra to separate x*.

    [tex]m1 < \frac{-C}{(x*-0.7)}[/tex]

    [tex]x* - .7 < \frac{-C}{m1}[/tex]

    [tex]x* < .7 - \frac{C}{m1}[/tex]

    Similarly:

    [tex]x* > .7 - \frac{C}{m2}[/tex]

    so:
    [tex].7 - \frac{C}{m1} > x* > .7 - \frac{C}{m2}[/tex]

    AM
     
  7. May 30, 2005 #6
    Thank you :)
     
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