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$$f(x)=x-3x^{1/3}$$ $$f'(x)=1-x^{-2/3}$$ $$f''(x)=\frac{2}{3}x^{-5/3}$$ $f ''(x)=0$ has no solution. Does that mean f has no inflection point? If $x > 0$ then $f$ is concave up and if $x < 0$ then $f$ is concave down. Where $f$ changes concavity?

Inflection point is where the function changes concavity or where the second derivative is zero or ....?

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  • $\begingroup$ What is f``(x) at x = 0 ? $\endgroup$ Feb 24, 2014 at 13:39
  • $\begingroup$ Logically speaking, it's not defined, but in calculus it's infinite. $\endgroup$ Feb 24, 2014 at 13:42
  • $\begingroup$ At negative $x$--values $f(x)$ becomes complex. I'm not sure how convexity/concavity is defined for complex functions. $\endgroup$
    – JPi
    Feb 24, 2014 at 13:43
  • $\begingroup$ @JPi: The font-size is a little small, if you zoom you'll see that power is 1/3 not 1/2! $\endgroup$ Feb 24, 2014 at 13:47
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    $\begingroup$ @JPi negative reals do have real cube roots $\endgroup$ Feb 24, 2014 at 13:47

2 Answers 2

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If a curve $\gamma$ is presented as a graph of a $C^2$-function $f$ then the standard $f''(x)=0$ condition produces the inflection points. But in the case at hand the given function, while continuous at $x=0$, is not differentiable there. Therefore we have to construct a regular parametric representation of the given graph $\gamma$ as a preliminary step.

Now the definition of $x^{1/3}$ is open to debate when $x<0$. In accordance with widespread practice I interpret this expression as follows: $$x^{1/3}:={\rm sgn}(x)\>\root 3\of {|x|}\qquad(x\in{\mathbb R})\ .$$ It follows that $x(t):=t^3$ results in $x^{1/3}=t$, so that the graph $\gamma$ of $f$ can be written as $$\gamma:\quad t\mapsto\cases{x(t)=t^3\cr y(t)=t^3-3t\cr}\qquad(-\infty<t<\infty)\ .\tag{1}$$ One computes $$s'^2(t)=x'^2(t)+y'^2(t)=9(1-2t^2+2t^4)>0\qquad\forall t\ .$$ This shows that the representation $(1)$ of $\gamma$ is regular. The signed curvature $\kappa$ of $\gamma$ is given by $$\kappa(t)={x'(t)y''(t)-x''(t)y'(t)\over s'^3(t)}={18 t\over s'^3(t)}\ .$$ This shows that $\kappa$ changes sign at the origin and implies that we have an inflection point of $\gamma$ at $(0,0)$. See the following figure:

enter image description here

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  • $\begingroup$ Based on some of the other questions posted by the OP(precalculus level), I am going to go out on a limb and say that this probably isn't appropriate for the OP's level of study. $\endgroup$ Feb 24, 2014 at 22:10
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    $\begingroup$ @j4nbur53: If the curve $\gamma$ in the above figure is given by the equation $F(x,y):=(x-y)^3-27x=0$ then one can compute the curvature in individual points by means of a certain formula involving the partial derivatives of $F$. It will turn out that curvature at $(0,0)$ is zero, but a sign change of $\kappa$ will be more difficult to detect. $\endgroup$ Oct 15, 2016 at 10:07
  • $\begingroup$ @ChristianBlatter I feel your definition of $x^{1/3}$ is quite rigorous and ideal. I saw many calculus books or even analysis books that treat $x^{1/3}$ blurred and ambiguous when $x<0$, which leads a huge inconvenience. Can you tell me what books of calculus/analysis has explicitly stated the definition of such functions as you said? $\endgroup$
    – Eric
    Nov 1, 2017 at 5:18
  • $\begingroup$ @Eric: The applicable definition depends on the context. While the definition above is obviously fine for the problem at hand there are cases when one should insist on $x>0$, in particular when one wants to make free use of rules like $x^{\alpha+\beta}=x^\alpha\cdot x^\beta$, $\bigl(x^\alpha\bigr)^\beta=x^{\alpha\beta}$, etc. $\endgroup$ Nov 2, 2017 at 8:11
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An inflection point occurs at points of the domain at which the function changes concavity. These need not be points at which the second derivative (or even the first) exists. Now, if your function's second derivative is defined and continuous in a neighborhood of an inflection point, then it will be zero at that point.

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  • $\begingroup$ ""Now, if your function's second derivative is defined and continuous in a neighborhood of an inflection point, then it will be zero at that point."" Do you mean f''(0) is 0? $\endgroup$ Feb 24, 2014 at 13:45
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    $\begingroup$ @kptlronyttcn no, it is not defined there. $\endgroup$ Feb 24, 2014 at 13:53

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