If $f$ has a zero and $|f''|\leq M$, then $f$ is monotone on $(-h,h)$, where $h=\sqrt{2|f(0)|/3M}$ 
Let $f$ be twice differentiable on $\mathbb R$ and let $M$ be a bound
  of $f''$, $|f''|\leq M$ on $\mathbb R$. Assume $f(0)\neq0$ and define
  $h=\sqrt{\frac{2|f(0)|}{3M}}$. Prove that if $f$ has a zero in $(-h,h)$, it's
  monotone in $(-h,h)$.

So let's assume there's a zero $a\in (-h,h)$ such that $f(a)=0$ and also there's $b\in (-h,h)$ such that $f'(b)=0,\ f''(b)\neq0$. We need to get a contradiction.
This question is under the Taylor Expansion chapter although I can't really get anything out of $f(x)=f(0)+f'(0)x+\frac{f''(\xi_x)}{2}x^2$ nor by the expansions near $a,b$. I get that in $(-h,h)$ we have $\frac{M}{2}x^2<\frac{|f(0)|}{3}$ which may be related to the 2nd derivative term though.

This is not true.
Consider $ f(x) = exp (-x^2) -0.1$
It's not monotone on $(-h,h)$ for any given $h>0$, yet it's second derivative is bounded by a finite $M$ and $f(0)=1$, and it has a root.
