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 $\frac{M}{2}x^2&lt;\frac{|f(0)|}{3}$ which may be related to the 2nd derivative term though.
Suppose that $f'(x_0)=0$ with $|x_0|&lt;h.$ Then $|f'(x)|\leq M|x-x_0|&lt; M(|x|+h)$ for all $|x|&lt;h.$ In particular for $0&lt;x&lt;h$ we have $$|f(x)-f(0)|\leq \int_0^h |f'(x)|dx\leq \int_0^h M(x+h)dx&lt;\tfrac32 Mh^2&lt;|f(0)|$$ with the same bound holding for $-h&lt;x&lt;0.$ This means that $f$ cannot have a zero in $(-h,h).$
This is not true.
Consider $ f(x) = exp (-x^2) -0.1$

It's not monotone on $(-h,h)$ for any given $h&gt;0$, yet it's second derivative is bounded by a finite $M$ and $f(0)=1$, and it has a root.