Exercise on a twice differentiable function 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.
 A: 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, $a<b.$ (Otherwise consider $f(-x).$) Then $f'(a) = f'(b)=0.$ By Rolle, $f''(c) = 0$ for some $c\in (a,b).$
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 $f''=0.$
