Talk:Derivative test

Why is it necessery that ' f' has to be continuous '? If f differentiable at every poitn of an interval, then f' is Darboux continuous. Hence, if $f'|_{(x-r,x)}>0\,$ and $f'|_{(x,x+r)}<0\,$ (notations as in the article), then f' has to be zero at some point and this point can only be x. Mozó (talk) 08:12, 19 November 2007 (UTC)

I think you are right. — Emil J. 11:50, 17 December 2008 (UTC)

Actually, one does not even have to appeal to Darboux continuity. If $f'|_{(x-r,x)}>0\,$ then f is increasing on (x − r, x), and if $f'|_{(x,x+r)}<0\,$ then f is decreasing on (x, x + r), hence f has a maximum at x as long as it is continuous in x (which it is, being differentiable). — Emil J. 13:39, 17 December 2008 (UTC)

Conversion into Mathematics format

It's very confusing like this, but I'm not familiar enough with calculus to want to edit it. Can someone step in please?? Elakhna 17:40, 23 November 2008 (UTC)

Significant mathematical errors in this article

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>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>0 there exist x,y in (-r,r) such that f'(x)>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. Jalongi (talk) 04:59, 6 April 2009 (UTC)

Yeah, the implications in the description should be reversed: if f is increasing on (x − r, x) and decreasing on (x, x + r), then f has a local maximum in x, but not vice versa. — Emil J. 12:03, 6 April 2009 (UTC)

Another significant math error

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 want is that it is either entirely positive or entirely negative. 207.62.177.227 (talk) 04:22, 5 November 2009 (UTC)

Proof of the first derivative test

I can't figure out how to prove the test. Given that a continuous function within a certain area (x₀, x₀+r) (and symmetrically backwards) has f'(x₀) = 0 and for any x: x₀ < x < x₀ + r it has f'(x)>0 you still can't say that there is some dx: for all x: x₀ < x < x₀ + dx, f(x)>f(x₀). Here's my failed attempt of proof ad absurdum: suppose that for any dx < r there always exists at least one x:x₀ < x < x₀ + dx, f(x)<f(x₀). Then there is a certain neighbourhood n to the left from f(x), where f(x-n)<f(x) as f'(x-n)>0 and thus f(x)-f(x-n)=n * (f'(x-n) + alpha(x-n, n)), where alpha approaches zero upon decrease of n and thus f(x)-f(x-n)>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 x₀, it will be at least f(x)-f(x₀) below f(x₀), and this is in contrast with contiguity of function in x₀.

But the problem is that this sequence might not approach x₀. Potentially it might have another left limit. For instance, if alpha(x,n) = 1/(x-0,5)*n, x₀ = 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 x₀ - we'll be forced to make neighbourhood n smaller and smaller a la Zeno's paradox and never get infinitely close to x₀).

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