Points of inflection

May 2010
254
8
Hi there. I must find the points of inflection on the next function:

\(\displaystyle f(x)=\displaystyle\frac{x}{1+x^2}\)

I've calculated its second derivative (and the first ofcourse, I'll just post the second couse I have an exam tomorrow so I don't have much time).

\(\displaystyle f''(x)=\displaystyle\frac{-2x(3-x^2)}{(1+x^2)^3}\)

So, when I make f''(x)=0 I got \(\displaystyle {0,\sqrt[ ]{3}-\sqrt[ ]{3}}\) These three values. Now, I must analize which of them is an inflection point, and which is not. And for this I must "look" on a "ball" around the candidates. How may I do that?
 
Apr 2009
73
12
Ignore this.
 
Last edited:
May 2010
254
8
Hi. Well, no, not always that the second derivative is zero you got an inflection point. Actually, in the function I've posted the only inflection point is zero, but for the root of three you got that the second derivative is zero too, and theres no inflection point there.

And the inflection point is not when the first derivative is zero, its calculated using the second derivative.

Bye there.
 

skeeter

MHF Helper
Jun 2008
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check the value of f''(x) in each interval defined by the three critical values you found ... if the sign of f''(x) changes, then f(x) has an inflection point at that x-value.
 
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Apr 2009
73
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Hi. Well, no, not always that the second derivative is zero you got an inflection point. Actually, in the function I've posted the only inflection point is zero, but for the root of three you got that the second derivative is zero too, and theres no inflection point there.

And the inflection point is not when the first derivative is zero, its calculated using the second derivative.

Bye there.
Ah, yes, sorry, I was thinking of extrema.
 
May 2010
254
8
Np Diemo ;)

Thanks skeeter. I've asked because I just thought in some kind of "formal" demonstration. But I think I won't need it.

Bye there.
 

undefined

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Mar 2010
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Just to tie this in with the local ball idea: at any candidate inflection point x with f''(x) = 0, you can look at f''(x - h) and f''(x + h) for small h to see if the second derivative changes sign there; the interval [x-h, x+h] is a (closed) 2-dimensional ball centered at x of radius h. I personally would go with skeeter's approach, though.
 
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May 2010
254
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That was the kind of idea I was looking for. And actually what I wanted was a general argument for any h (I mean, not trying with smalls h's, but any h smaller as I want it to be). But anyway, it would be ok by following skeeters indications.
 

undefined

MHF Hall of Honor
Mar 2010
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Chicago
That was the kind of idea I was looking for. And actually what I wanted was a general argument for any h (I mean, not trying with smalls h's, but any h smaller as I want it to be). But anyway, it would be ok by following skeeters indications.
Well, I think we could formalise as follows: Given that f''(a) = 0 and f(x) is twice-differentiable, a is an inflection point if and only if there exists H > 0 such that for all h with 0 < h < H, (f''(a-h) < 0 and f''(a+h) > 0) or (f''(a-h) > 0 and f''(a+h) < 0).

But we know that inflection points can only exist where f''(x) = 0, so this reasoning is unnecessary, and it's much more straightforward to partition the real number line according to the candidates as skeeter suggested.
 
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May 2010
254
8
Yes, actually now I've found that thats the way the teacher did one example last class. I wasn't giving the right focus to the problem.

(Last class was one week ago here in argentina now we've got two hundred years!!)

Thanks.
 
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