4
$\begingroup$

Let $f\colon (a,b) \to \mathbb{R}$ a non constant differentiable function.

Is the following statement true:

If $f$ has a local maximum and a local minimum then $f$ also does have an inflection point.

If so, how to prove it, if not, what would be a counterexample?

Remark

If there are $a_0,b_0 \in [a,b]$ with $a_0 < x_0 < b_0$ such that $f|_{(a_0,x_0)}$ is convex and $f|_{(x_0,b_0)}$ concave or vice versa, then $(x_0,f(x_0))$ is called inflection point of $f$ (Amann Escher Analysis I, p349).

Sometimes stricly concave (convex) is used in the definition. Does this change the theorem?

If the theorem is not true, does it hold if one allows only smooth functions or even more restrictive only polynomial ones (excluding linear functions)?

$\endgroup$
  • 1
    $\begingroup$ What is your exact definition of an inflection point? $\endgroup$ – Henning Makholm Dec 2 '14 at 19:19
  • $\begingroup$ If $X$ is a closed subset on the line, you can have $f$ linear which has maximum and minimum on the lateral limits of $X$. You might want to assume that $X$ is open and connected (this last one just to simplify things). $\endgroup$ – Marra Dec 2 '14 at 19:20
  • $\begingroup$ Also, assume that $f$ is not constant. $\endgroup$ – Marra Dec 2 '14 at 19:20
  • $\begingroup$ Limiting the interval does not makes the result true. Just take $(-3,3)\subset\mathbb{R}$ on the example of the answer below. $\endgroup$ – Marra Dec 2 '14 at 19:31
5
$\begingroup$

Consider this:

$$ f(x) = \begin{cases} (x+1)(x+3)-2 & \text{when }x\le -1 \\ 2x & \text{when } -1 < x < 1 \\ 2-(x-1)(x-3) & \text{when } x \ge 1 \end{cases} $$

(Graphed by Wolfram Alpha)

This has local minimum and maximum at $x=-2$ and $x=2$ respectively, but would you consider it to have an inflection point? Different defintions would give different answers.

In the definition you quote, every point in $[-1,1]$ would be an inflection point if you have "convex" and "concave"; but no points would be inflection point if you have "strictly convex/concave".


A counterexample to the claim in the non-strict case would be the indefinite integral of the Weierstrass function. It's easy to see that it must have local maxima and minima somewhere (e.g. you can find closed intervals where the endpoints are certainly not maxima, but a continuous function always attains its supremum over a closed interval). On the other hand the Weierstrass function itself is nowhere monotone, so its integral cannot be convex or concave on any interval, which again excludes the existence of inflection points.

Even requiring that $f$ is twice differentiable won't do; according to this anwswer there exist everywhere differentiable yet nowhere monotone functions, and (with an offset to make sure they cross the $x$ axis) their indefinite integrals would be counterexamples.

$\endgroup$
  • $\begingroup$ Now this is a very good answer. As I've seen on Calculus, an inflection point is where the concavity of the graphic changes. There is none to be found on this one despite the existence of points where the second derivative is zero (this happens on an interval however). $\endgroup$ – Marra Dec 2 '14 at 19:30
  • $\begingroup$ Thanks. But in the non strict case, is the theorem true or is there another counterexample? $\endgroup$ – Julia Dec 2 '14 at 20:04
  • $\begingroup$ @Julia: I think I've found a counterexample; see the edited answer. $\endgroup$ – Henning Makholm Dec 2 '14 at 20:36
  • $\begingroup$ In fact I suspect that there are $\mathcal C^\infty$ counterexamples. $\endgroup$ – Henning Makholm Dec 2 '14 at 21:27
1
$\begingroup$

Consider the bump function $$\phi(x) = \begin{cases} e^{-1/(1-x^2)}, & -1 < x < 1 \\ 0, & \text{else.} \end{cases}$$ It's a standard exercise to verify that $\phi$ is $C^\infty$ and has a local maximum at $x=0$. Set $g(x) = \phi(x-2) - \phi(x+2)$, so that $g$ is $C^\infty$, strictly negative on $(-3,-1)$, strictly positive on $(1,3)$, and zero on $[-1,1]$. In particular, there is no point at which $g(x)$ changes from strictly positive to strictly negative or vice versa.

Let $f(x) = \int_0^x g(t)\,dt - c$ for some sufficiently small positive constant $c$ so that $f$ has exactly one zero in $(-3,-1)$ and another in $(1,3)$. Then let $F(x) = \int_0^x f(t)\,dt$. Now $F$ is $C^\infty$ and has a local maximum in $(-3,-1)$ and a local minimum in $(1,3)$. But $F'' = g$ so there is no "strict inflection point" at which $F$ changes from strictly concave to strictly convex or vice versa.

For polynomials this cannot happen. I will try to write more later, but in short the issue is that a polynomial can only be zero at finitely many points.

$\endgroup$
0
$\begingroup$

Definition of inflection point: when $ f''(x)=0. $

Between two extrema with second derivatives of opposite sign there will always be at least one inflection point. The graph of second derivative must pass through zero as consequence of Rolle's theorem.

Non-strict concave/convex situation is when $ f'(x)=0, f''(x)=0. $ It does not change the theorem.

A point of inflection (PI) exists in all the four situations:

maximum,PI,minimum real root,PI,maximum

maximum,PI,minimum imaginary root,PI,maximum

minimum,PI,maximum real root,PI,minimum

minimum,PI,maximum imaginary root,PI,minimum

$\endgroup$
  • 1
    $\begingroup$ This is not the definition of inflection point. It is just a neccesary condition if $f$ is twice differentiable. It would be a sufficient condition for example if $f$ is three times differentiable and additionally $f'''(x) \neq 0$. $\endgroup$ – Julia Dec 2 '14 at 19:44
  • 1
    $\begingroup$ You sure about that definition? With $y=x^4$, the second derivative is $0$ at the origin, but it's a minimum, not an inflection point. $\endgroup$ – Akiva Weinberger Dec 2 '14 at 19:46
  • $\begingroup$ Furthermore $x$ beeing a local extremum doesn't imply that $f''(x) \neq 0$ (only that $f'(x)=0$). $\endgroup$ – Julia Dec 2 '14 at 20:08
  • $\begingroup$ @columbus8myhw $y=x^4$ to be minimum at x=0 its second derivative should be positive, not 0. $\endgroup$ – Narasimham Dec 2 '14 at 20:16
  • $\begingroup$ @Narasimham Look at a graph. Also, for it to be a minimum, we need $f'$ to be increasing, which implies that $f''\ge0$—not $f''>0$. $\endgroup$ – Akiva Weinberger Dec 2 '14 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.