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I am a bit confused about the following statement.

Let $f$ be a real function of a real variable and $x_0 \in \mathbb{R}$. If $f$ is differentiable in a neighborhood $I$ of $x_0$, $f'(x_0) = 0$ and $f'(x_0) > 0$ for all $x \in I \setminus \{x_0\}$, then $x_0$ is an inflection point.

Is this statement true? Reading this question: Prove that $f$ has an inflection point at zero if $f$ is a function that satisfies a given set of hypotheses, I think my statement is false, but I cannot give a counterexample.

Can anyone help me out?

EDIT. (Thanks to @Zestylemonzi's comment) My definition of inflection point is as follows:

$x_0$ is an inflection point if $f$ is differentiable at $x_0$ and there exists a neighborhood $J$ of $x_0$ such that the function $d(x) = f(x) − f(x_0) − f'(x_0) (x − x_0)$ has the same sign as $x−x_0$ for all $x \in J \setminus \{ x_0 \}$, or $d(x) = f(x) − f(x_0) − f'(x_0) (x − x_0)$ and $x−x_0$ have opposite signs for all $x \in J \setminus \{ x_0 \}$

I think that according to this definition my statement is true. I am aware of another possible definition of inflection point:

$x_0$ is an inflection point if $f$ is differentiable at $x_0$ and there exists $\delta > 0$ such that $f$ is convex (respectively, concave) for $x \in (x_0 - \delta, x_0)$ and concave (respectively, convex) for $x \in (x_0, x_0 + \delta)$.

Is my statement true according to the second definition?

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    $\begingroup$ I think the crux of this lies with your definition of 'inflection point'. How are you defining it? $\endgroup$ – Zestylemonzi Jan 3 '17 at 22:13
  • $\begingroup$ @Zestylemonzi I am aware of two possibile definitions: $x_0$ is an inflection point if there exists a neighborhood $J$ of $x_0$ such that the function $d(x) = f(x) - f(x_0) - f'(x_0) (x - x_0)$ has the same sign as $x - x_0$ for all $x \in J \setminus \{x_0\}$, or $x_0$ is an inflection point if $f$ is convex for $x<x_0$ and concave for $x>x_0$, or vice versa. I think my statement is true according to the first definition; I don't know whether it's true according to the second one. $\endgroup$ – Paolo Jan 3 '17 at 22:32
  • $\begingroup$ Great, this makes a lot more sense now. Your comment is worth adding to your question, it adds clarity and context! $\endgroup$ – Zestylemonzi Jan 3 '17 at 22:36
  • $\begingroup$ I must say your definition is kind of weird because if $x_0$ is an inflection point for $f$ then it is not for $-f$. $\endgroup$ – Fimpellizieri Jan 4 '17 at 6:13
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Let's look at the question with the second definition; we will give an example of an $f$ for which the statement does not hold. For that, we will use a well-known lemma on the characterization of convex functions when differentiability is assumed.

Lemma: Let $I,J\subset \mathbb{R}$ be intervals and $g:I\longrightarrow J$ be differentiable. Then $g$ is convex (respectively, concave) if and only if $g'$ is non-decreasing (respectively, non-increasing).

By the lemma, it suffices to find an $f$ as in the question that, in addition, is such that $f'$ is never monotone near $x_0$.

In what follows we assume without loss of generality that $x_0=0$. One example is to take $f$ to be an antiderivative for $h:\mathbb{R}\longrightarrow\mathbb{R}$, given by

$$h(x)=x^2\cdot \exp\left({\sin\left(\frac1x\right)}\right),$$

where of course $h(0)=0$. Observe that $h>0$ when $x\neq 0$. Moreover, $h$ is continuous, so it is Riemann-integrable and its antiderivative is differentiable.

Finally, $h$ itself is differentiable on $\mathbb{R}\setminus\{0\}$ with

$$h'(x)=\left( 2x-\cos\left(\frac1x\right)\right)\cdot {\underbrace{\exp\left(\sin\left(\frac1x \right)\right)}_{>0}}\,\,\,,$$

so one need only check that $2x-\cos\left(\frac1x\right)$ changes sign infinitely often near $x=0$.

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