I'm trying to solve this problem

Let $f:[a,b] \to \mathbb R$ a differentiable function with continuous derivative.

Suppose further that $f$ is twice differentiable on $(a,b)$.

Prove that if $f(a)=f(b)$ and $f'(a)=f'(b)=0$, then exist $x_1, x_2 \in (a,b)$ with $x_1 \neq x_2$ such that $f''(x_1) = f''(x_2)$.

I'm trying to solve this problem graphically intuitively ; I tried, and likely the problem is real: if $f(a) = f(b)$ and $f'(a) = f'(b) = 0$, the points $a$ and $b$ are the points of maximum and minimum for the function. but how can you prove that there are $x_1, x_2 \in (a,b)$ with $x_1 \neq x_2$ such that $f''(x_1) = f''(x_2)$?

  • $\begingroup$ The last paragraph of your question doesn't make much sense. You could consider fixing it. $\endgroup$ – user21436 Dec 28 '11 at 17:29
  • $\begingroup$ There's this weird spacing when you write $[a,b]$ and $(a,b)$ before the second bracket/parenthesis that I can't get rid of by editing. Weird. $\endgroup$ – Patrick Da Silva Dec 28 '11 at 17:30
  • $\begingroup$ @Patrick: Done. $\endgroup$ – Did Dec 28 '11 at 17:46
  • $\begingroup$ @Didier : What was the problem exactly? I looked at the edit but saw nothing, just you replacing some characters by the exact same characters. $\endgroup$ – Patrick Da Silva Dec 28 '11 at 18:27
  • $\begingroup$ @Patrick: What we saw as b plus a space was in fact a compound of the letter b and of two invisible pseudo-characters attached to it and translated by the system as a space. That is, an encoding gone awry (but please do not ask me for more precise details...). $\endgroup$ – Did Dec 29 '11 at 9:16

If the function is constant, everything is easy.

If the function is not constant, it attains a maximum and a minimum value in our interval, and these values are distinct.

The max and min cannot both occur at endpoints, since $f(a)=f(b)$. So there is a local extremum in $(a,b)$, and therefore a point $c\in(a,b)$ such that $f'(c)=0$. Now you should be able to use Rolle's Theorem to show that there exist points $x_1\in(a,c)$ and $x_2\in(c,b)$ such that $f''(x_1)=f''(x_2)=0$.

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  • $\begingroup$ A nitpick: Why do you essentially prove Rolle's Theorem and then appeal to it? $\endgroup$ – David Mitra Dec 28 '11 at 18:02
  • $\begingroup$ @David Mitra: Of course you are right. I wanted to connect the finding of $c$ with something familiar, to make the geometry more concrete-seeming. $\endgroup$ – André Nicolas Dec 28 '11 at 18:07

Hint: Apply the Mean Value Theorem three times (or Rolle's Theorem if you prefer).

Applying it once to $f$ over $[a,b]$ gives you a point $c$ with $a<c<b$ where $f'(c)=0$.

Now apply the Mean Value Theorem to $f'$ on each of the intervals $[a,c]$ and $[c,b]$.

(Note that the Mean Value Theorem gives you a point $c$ strictly between the endpoints of the interval that you're working over.)

It's not true that $a$ and $b$ necessarily give the minimum and maximum of the function. They could in fact give neither. For instance, the graph could resemble a $\sin$ wave over $[0,2\pi]$ that's been "smoothed out" at the endpoints (so that the derivative is zero at the endpoints).

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