# Points of inflection

#### Ulysses

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?

Ignore this.

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#### Ulysses

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
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.

Ulysses

#### Diemo

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.

#### Ulysses

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

MHF Hall of Honor
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.

Ulysses

#### Ulysses

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
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.

Ulysses

#### Ulysses

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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