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Rolle'n into a Theorems Proof

  1. Apr 28, 2010 #1
    Hey PF'ers :biggrin:

    I'm having trouble with a proof of Rolle's Theorem.

    I understand that by the Extreme Value theorem the function on the closed interval [a,b] must attain both a supremum and an infimum
    so if [tex]f(x) = f(a) = f(b) \forall x \in [a,b] and f'(c) = 0[/tex] then it must be a constant function.

    That is the trivial case.

    If I understand the proof correctly from here it goes as follows.

    By the Extreme Value Theorem applied to Rolle's theorem there must exist a point c ∈ [a,b] where the function changes from positive slope to negative slope.

    What we want to do is show that at the point f(c) the left and right limits, though both being of opposite sign, will be equal to the same point in the end.

    It's just that in the following link, http://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/node42.html" the proof has me confused.

    It's just the two cases they show in Rolle's Proof, the two if h>0, & if h<0, parts I need help understanding.

    1: Is the first one saying "if you go beyond the supremum c by a small amount h you'll end up below the value f(c) and because of this you'll end up with a value less than 0 (i.e. be negative) ???

    2: Also, in the h<0 part why does that become positive???
    If you're at the supremum of any function and you move either way you can only move down!

    I might be mixing something up or misunderstanding something about 2:.

    I think I may be most confused the the limits they use, they are both coming from the left side & shouldn't one be from the left and the other be from the right?

    If This is so, shouldn't the first limit be coming from the right (i.e. from the positive side of the graph toward the negative side)?
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Apr 28, 2010 #2
    hi,

    "
    2: Also, in the h<0 part why does that become positive???
    If you're at the supremum of any function and you move either way you can only move down!
    "

    f(c+h) - f(c) is negative and h is negative, so

    f(c+h) - f(c) / h is positive


    I think that you might be right about the limit sides
     
  4. Apr 28, 2010 #3
    1. If h>0, then f(c+h) - f(c) will be less than zero. This is because f(c+h) must be less than f(c) (as f(c) is the maximum). Because h is positive, dividing by it still gives a negative value.

    2. In the second part, f(c+h) - f(c) is again less than zero, but this time you are dividing by negative h, so the result is positive.
     
  5. Apr 28, 2010 #4
    Wow thanks for the quick responses guys.

    Yeah I think I understand it now.

    It's just confused me to have a positive result when you use h<0 and you're at the top of the function!

    As the limit is reached that positive number goes to zero.

    The positive result is the slope of the small increment below c that is striving up to it.
     
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