2
$\begingroup$

I have the following exercise:

Let $f$ be a twice differentiable function on $\mathbb R$ such that:

$f(-1)<0$

$f(1)>0$

$\lim_{x\to \pm \infty} f(x) =0$

Prove that $f''(x)= 0$ for at least three values of $x.$

I managed to find one such point reasoning about local maximum/minimum, and I know that the function has at least one zero. But I don't know how to proceed further.

  • 0
    @SimpleArt I don't want to sound childish but we didn't actually studied this at the course (however I was thinking myself about such a relationship but I didn't manage to formalize it rigorously) so if you can explain it to me or send me some source about I would be very glad :) – 2017-01-03

1 Answers 1

2

Because $f:\mathbb R\to \mathbb R$ is continuous, $f(-1)<0,$ and $\lim_{x\to\pm \infty}f(x) = 0,$ $f$ assumes a negative minimum value at some $a\in \mathbb R.$ Similary $f$ assumes a positive maximum value at some $b\in \mathbb R.$

WLOG, $a

Because the maximum of $f$ occurs at $b,$ we have $f''(b)\le 0.$ If $f''(b) = 0,$ we have found a point in $[b,\infty)$ where $f'' = 0,$ so we can relax.

If $f''(b) < 0,$ then because $f'(b) =0,$ we have $f'(d) < 0$ for some $d>b.$ There must then exist a point in $(b,\infty)$ where $f''=0.$ Otherwise, by Darboux, $f''<0$ on $(b,\infty).$ This would imply $f'(x) < f'(d)$ for $x\in [d,\infty).$ But that forces $f(x) \to -\infty$ as $x\to \infty,$ contradiction.

So we've shown there exists a point in $[b,\infty)$ where $f''=0.$ The same ideas show there exists a point in $(-\infty,a]$ where $f''=0.$ That gives us three distinct points where $f''=0.$