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I am facing problems computing the inflection point

We have $$f(x) = (2x^2-x^3)^{1/3}$$

Let's assume $f$ is defined like: $f:\Bbb{R}\to \Bbb{R}$ (As far as I read on wikipedia, it's a matter of opinion. You could also say $f: (- \infty,2]\to \Bbb{R}$)

But we assume $f:\Bbb{R}\to\Bbb{R}$

$$f'(x) = \frac{4x-3x^2}{3(2x^2-x^3)^{\frac{2}{3}}}$$

$$f''(x) = \frac{-8x^2}{9(2x^2-x^3)^{\frac{5}{3}}}$$

Both derivatives are correct, I computed it two times and finally checked it online ( e.g. WolframAlpha).

I know that for an inflection point $f''(x) = 0$

BUT: $ -8x^2 =0 \Rightarrow x = 0$ That does not work, because $f''(0)$ would not be defined, since the denominator is $0$.

What am I doing wrong?

Thank you.

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  • $\begingroup$ What is \IR? Also, an inflection point happens when concavity switches. It is not necessary that $f''(x)=0$ Consider $f(x) = x^{1/3}$, which has an inflection point at $(0,0)$ even though $f$ is not differentiable at $0$. $\endgroup$ – Matthew Leingang Dec 29 '16 at 18:19
  • $\begingroup$ \IR = real numbers $\endgroup$ – Blnpwr Dec 29 '16 at 18:20
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    $\begingroup$ OK, more people will understand what you mean if you use LaTeX markup and write $\mathbb{R}$ $\endgroup$ – Matthew Leingang Dec 29 '16 at 18:21
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    $\begingroup$ Thank you, I am new to this format :) $\endgroup$ – Blnpwr Dec 29 '16 at 18:22
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    $\begingroup$ From Wlofram: "A necessary condition for x to be an inflection point is f''(x)=0" ... I can see why that seems contradictory. I interpret that to mean if f''(x) is defined then f''(x) must equal 0. But it's possible that f''(x) is undefined. In other words: if f''(x) != 0 then x is not an inflection point. But if f''(x) is undefined it might be. Use the for x < 0 f''(x) < 0 but for x > 0 f''(x) > 0. That is sufficient to show 0 must be an inflection point. $\endgroup$ – fleablood Dec 29 '16 at 18:37
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I'm not a fan of it, but some teachers call the solutions to $f"(x)=0$, PPI's. (Possible Point of Inflection.) It's an actual inflection point only if the concavity changes at that point.

But also, concavity can change at a point where the 2nd derivative fails to exist, which is the case in your problem. You can factor one $x$ out of the bottom of the first derivative, and you're left with a non-removable singularity at $x=0$.. So the first derivative doesn't exist at $x=0$. Therefore, neither does the second derivative.

When graphing functions, I have my students collect all the points where $f'$ and $f''$ are zero or discontinuous. Those are the points where increasing can change to decreasing and vv, and where concave up can change to concave down. (I don't worry about naming the points so much.) But finally, the answer to your question is that the concavity only may change at one of these points. In your example, it doesn't.

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  • $\begingroup$ So what I am supposed to do in order to check, whether a value is an inflection point or not? I have always been taught, f''(x) = 0 computing is enough. I guess: First, I have to compute f''(x) = 0 , and then I may get a value. I put this value into f''(x) and check, if there is a switch from - to + or + to - If so, this specific value was an inflection point. Is that right that far ? $\endgroup$ – Blnpwr Dec 29 '16 at 18:38
  • $\begingroup$ By collecting all the points where $f''$ is zero or undefined, you have found all the places where $f''$ might change sign. This divides the real line into finitely many intervals. On each interval, the sign of $f''$ can't change, so you just test one point from the interior of each interval and then you knwo the concavity on the whole interval. (The same technique is used for increasing/decreasing.) $\endgroup$ – B. Goddard Dec 29 '16 at 19:42
  • $\begingroup$ Ah, so $$ f''(x) = 0 $$ is only at x = 0 , but in the neighborhood of 0 there are no opposite signs , for example when I do $$ f(-1) $$ and $$ f(1) $$ both have the same sign,negative, so 0 can not be an inflection point, right ? $\endgroup$ – Blnpwr Dec 29 '16 at 20:05
  • $\begingroup$ Actually $f''(0)$ doesn't exist. You can factor an $x^2$ out of the bottom and cancel it, and you still have a small power of $x$ in the denominator. But the "neighborhood" observation is still correct. $\endgroup$ – B. Goddard Dec 29 '16 at 20:09
  • $\begingroup$ Well, I thought if $$ f''(0) $$ does not exist, it still COULD be an inflection point. Consider $$ f(x) = x^{1/3} $$. What about this? $\endgroup$ – Blnpwr Dec 29 '16 at 20:10
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We can verify B. Goddard and Matthew Leingang comments with the graph of $f(x) = (2x^2-x^3)^{1/3}$ (in blue):

enter image description here

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  • $\begingroup$ Is there an inflection point at x = 2 ? In the neighborhood of 2 , there are opposite signs, but $$ f''(2)$$ is not equal to 0 . So 2 can't be an inflection point, right? $\endgroup$ – Blnpwr Dec 29 '16 at 20:35
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It is neither necessary nor sufficient that $f''(c)$ exists and $f''(c) =0$ for the graph of $f$ to have an inflection point at $c$. Consider:

  • $f(x) = x^{1/3}$. The graph of $f$ has an inflection point at $(0,0)$, but $f$ is not even once differentiable at $0$:

  • $f(x) = x^4$. Then $f''(0) = 0$, but $f$ has a local minimum at $0$, not an inflection point.

Instead, the definition of inflection is a point at which concavity changes. It's harder to encode, but you can say it this way: $f$ has an inflection point at $c$ if there exist $a$ and $b$ with $a < c < b$ such that $f$ is concave up on $(a,c)$ and $f$ is concave down on $(c,b)$, or $f$ is concave down on $(a,c)$ and concave up on $(c,b)$.

It's not hard to show that if $f''(c) = 0$ and $f'''(c) \neq 0$, then $f$ has an inflection point at $c$. But that is not the case you have here. Instead, you would want to show that $f''(c)$ has one sign on one side of $0$, and the other sign on the other side.

Consider that $$ f(x) = -\frac{8x^2}{9(2x^2-x^3)^{5/3}} = -\frac{8x^2}{9[x^2(2-x)]^{5/3}} = -\frac{8x^2}{9x^{10/3}(2-x)^{5/3}} = -\frac{8}{9x^{4/3}(2-x)^{5/3}} $$ If $x$ is near zero, $(2-x)^{5/3}$ is positive, and $x^{4/3}$ is positive, too. So the quotient is negative on either side of $0$, meaning $0$ is not an inflection point.

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  • $\begingroup$ "It is neither necessary nor sufficient that f′′(c)=0" If f''(c) is defined, it is necessary (but not not sufficient) that f''(c) = 0. So if f''(c) != 0 but is defined then c is not an inflection point. Also I think the change of sign is sufficient but not necessary. $\endgroup$ – fleablood Dec 29 '16 at 18:53
  • $\begingroup$ I understand, but it is necessary that we have a value, to check if this value is an inflection point. And to get this value, one has to compute $$ f''(x) = 0 $$ And when we have $$ f(x) = x^{1/3} $$ , then , $$ f''(x) = \frac{-2}{9x^{5/3}} $$ So how to get the 0 ? So you have to look at the numerator. And the numerator here is -2. And $$f''(x) = 0$$ would be -2 = 0 , which is wrong. $\endgroup$ – Blnpwr Dec 29 '16 at 18:59
  • $\begingroup$ @fleablood: I agree on the first point; I've tried to reword my statement to make the distinction between “not $f''(c) = 0$” and “f''(c) \neq 0” more clear. As for the second, do you have an example? $\endgroup$ – Matthew Leingang Dec 29 '16 at 19:17
  • $\begingroup$ Well, This one I guess. As we can see by the graph x = 0 and x = 2 seem to be inflection points. x = 2 passes the sign change test but x = 0 doesn't. Anyway, I'm basically quoting wolfram: "The second derivative test is also useful. A necessary condition for x to be an inflection point is f^('')(x)=0. A sufficient condition requires f^('')(x+epsilon) and f^('')(x-epsilon) to have opposite signs in the neighborhood of x (Bronshtein and Semendyayev 2004, p. 231). " I interpret "necessary that f''(x) = 0" to mean if f''(x) exists. $\endgroup$ – fleablood Dec 29 '16 at 19:41

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