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Consider $f(x)=x^{\frac {5}{7}}$, is it $x=0$ an inflection point? $$f'(x)=\frac {5}{7}x^{\frac {-2}{7}}$$ $$f''(x)=\frac {-10}{49}x^{\frac {-9}{7}}$$ As far as I know, the inflection point is the point in which $f''(x)=0$ or $f''(x)$ does not exist and $f''(x)$ also changes sign at that point.
However, here $f'(0)=+\infty $ from both sides, so it should have verical tangent at point $x=0$.
My calculation shows that $f_+''(0)=-\infty$ and $f_\_''(0)=-\infty$ so the $f''(x)$ changes sign around the point $x=0$. I'm not sure whether I'm correct, please tell me the safe steps to find the inflection point.

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  • $\begingroup$ $f$ is not real on the negative $x$-axis. Or do you mean its antisymmetric extension? $\endgroup$ – frog Oct 15 '14 at 6:42
  • $\begingroup$ Of course $f$ is real on negative $x$-axis, since the domain of $f$ is $R$. $\endgroup$ – FreeMind Oct 15 '14 at 6:44
  • $\begingroup$ This is no argument. But I guess I see what you mean: You mean probably $f(x):=\operatorname{sgn}(x)\exp\left( \frac{5}{7}\ln|x|\right)$... $\endgroup$ – frog Oct 15 '14 at 7:00
  • $\begingroup$ @frog Please, read my question carefully, what I'm simply asking is, whether $x=0$ is a point of inflection for $f$ or not and why? $\endgroup$ – FreeMind Oct 15 '14 at 7:33
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For an inflexion point, the only thing you need to find is whether concavity changes, i.e. if f"(x)=0, and you specifically test if there is a change in sign in f"(x)=0.

To test if it is a horizontal point of inflexion, f'(x) must also equal 0 at the point where there is an inflection.


This should be enough to test a point of inflexion.

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  • $\begingroup$ Could you please answer example? $\endgroup$ – FreeMind Oct 15 '14 at 6:45
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I think there is an issue with what is exactly meant by an inflection point. I think it is more convenient to work with the inverse function of $f$ since so the tangent at zero is simply the $x$-axis. (Inflection points are preserved under symmetries of the space).

So you look at $g(x)=x^{7/5}$ and you find $g'(0)=0$. But the second derivative is not defined in $0$. However, $g''(x)>0$ for $x>0$ and $g''(x)<0$ for $x<0$. Hence the curvature changes sign in $0$ but is undefined in $0$.

You can of course define such a point to be an inflection point too, but the usual definition says that $g''$ must vanish in an inflection point. Since it is not even defined here, I would say, $0$ is no inflection point.

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I think that some of the confusion comes from what exactly we call an inflection point. Unfortunately, I don't know that any of this terminology is standard. Here is how I would describe it.

By analogy, we call a critical point of a function $f(x)$ a point such that one of the following is satisfied

  • We have that $f'(c) = 0$
  • We have that the derivative of $f$ does not exist at $c$
  • We have that the function is not defined at $c$, but is on both sides.

Then if we are looking for extrema of a function, we search among all of these. However, a critical point may not be an extrema!

Now, for inflection points the case is a little different. The way I define inflection points (and I'm pretty sure that this is how the Stewart texts do) is much simpler. We say that $c$ is an inflection point if

  • $f''(x)$ exists on both sides of $c$ (although not necessarily at $c$!)
  • $f''(x)$ changes signs from one side to the other

So in your case, with the function $f(x) = x^{5/7}$ you get $f''(x) = -\frac{10}{49}x^{-9/7}$. This obviously doesn't exist at 0, but it is well defined away from zero. Moreover, the concavity changes signs from one side to the other. As such, I would call it an inflection point.

It should further be remarked that if $f''(x)$ is defined at $c$ and the concavity changes, then we would of course have $f''(c) = 0$. But this is not necessary, and may not occur if the derivative or second derivative does not exist at that point.

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