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731:= 0 and we started in x = 1, we'll never approach 0 - we'll get stuck approaching 0,5. Ok, for this example we can just pick smaller neighbourhood dx and start from smaller x, but we need to guarantee, that for an arbitrarilly small dx we'll be able to approach 0 (which might be wrong, if we have infinite number of such limitting points for the alpha(x,n) arbitrarilly close to x
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Note: "There exists x such that for all y, (A(x,y) or B(x,y))" is NOT the same as "There exists x such that for all y, A(x,y), or there exists x such that for all y, B(x,y)". The way it was previously worded, it just said that the derivative failed to vanish in some deleted neighbourhood. What you
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0 such that f is increasing on (c-r, c) and f is decreasing on (c, c+r). The standard counterexample is the function f:R→R defined by f(0)=0 and f(x) = -x^4*(2-sin(1/x)) for x ≠ 0. The function f is differentiable and has a local (indeed global) maximum at 0, but for every r:
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0 and f'(y)<0. Thus, f is not monotonic in any neighborhood of 0. A sufficient condition for the monotonicity of a function on each side of a local extremeum is that the function be continuously differentiable.
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0 upon small enough n. Then we would pass the points to the left of f(x-n) in the same way and show for them that the function there is even smaller than in f(x), thus if this sequence of points approaches
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This article needs revision. It is NOT true that if f is differentiable on an open interval I and f has a local maximum at c in I, then there exists r: -->
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It's very confusing like this, but I'm not familiar enough with calculus to want to edit it. Can someone step in please??
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735:- we'll be forced to make neighbourhood n smaller and smaller a la Zeno's paradox and never get infinitely close to x
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to delete these "External links modified" talk page sections if they want to de-clutter talk pages, but see the
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I can't figure out how to prove the test. Given that a continuous function within a certain area (x
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on
Knowledge. If you would like to participate, please visit the project page, where you can join
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https://web.archive.org/web/20060405222111/http://77.1911encyclopedia.org/S/SI/SIMPSON_THOMAS.htm
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differentiable at every poitn of an interval, then f' is
Darboux continuous. Hence, if
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Actually, one does not even have to appeal to
Darboux continuity. If
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want is that it is either entirely positive or entirely negative.
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Yeah, the implications in the description should be reversed: if
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When you have finished reviewing my changes, please set the
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for additional information. I made the following changes:
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But the problem is that this sequence might not approach x
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466:{\displaystyle f'|_{(x,x+r)}<0\,}
288:{\displaystyle f'|_{(x,x+r)}<0\,}
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385:{\displaystyle f'|_{(x-r,x)}: -->
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