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Rolle's Theorem |
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| Nov11-05, 05:19 PM | #1 |
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Rolle's Theorem
Ok I think I can use Rolles theorem on F(x)=x^2+3x on the inteval [ 0,2]
because the derivative can be defined so then I think I use the formula [f(b)-f(a)] / [b-a] to find f'(c), then set f'(c) = F'(x) and solve is this process right?
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| Nov11-05, 06:03 PM | #2 |
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Recognitions:
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What are you even trying to prove?
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| Nov11-05, 09:13 PM | #3 |
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It seems you're using the mean value theorem, which is the one you listed. Rolle's theorem says that if f(a)=f(b) on some closed interval, then there must be some point c such that f'(c)=0.
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| Nov12-05, 11:50 AM | #4 |
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Recognitions:
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Rolle's Theorem
rolles theorem is a trivial consequence of the obvious (but deep) fact that a continuous function f which takes the same value twice must have a local extremum in between.
then rolle says if f is also differentiable, the derivative is zero there. the MVT is then a further trivial consequence of rolle. i.e. both rolle and MVT are trivial, but useful consequences of one deep result about continuity. my point is that the emphasis on these two as big time theorems is quite misplaced. even their statements take away something from the result, since just knowing the derivative is zero somewhere in between two points is decidedly weaker than knowing there is a local extremum. for instance the statement of rolles theorem does not imply that between two critical points of a continuously differentiable function there must be a flex, but the stronger result does. |
| Nov14-05, 03:57 PM | #5 |
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I had seen my error after my post I was using the mean value theorem instead of rolles theorem opps
thanks for the replys though
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