Prove using the formule of Taylor Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a $4$ times differentiable function and let $f(0) = 0$, $f'(0) = 0$, $f''(0) = 0$, $f'''(0) = 6$. Prove that there exists an interval $I = [a,b]$ such that $0 \in (a,b)$ and for all $x \in I$:
$$ f''(x) > 0 $$
for $x >0$,
$$f''(x) < 0$$
for $x < 0$.
I tried using the formula of Taylor. For every $x \in I\setminus \{0\}$ there is a $c \in (0,x)$ such that 
$$f(x) = \frac{f(0)}{0!} x^0 + \frac{f'(0)}{1!}x^1 + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f''''(c)}{4!}x^4 = x^3 + \frac{f''''(c)}{4!}x^4$$
From this follows:
$$f'(x) = 3x^2 + \frac{f''''(c)}{3!}x^3,$$
and 
$$f''(x) = 6x + \frac{f''''(c)}{2!}x^2. $$
I am not sure if I can use this in my proof.
 A: Well, I'll give it a shot. Use Taylor for $f''$ centered at 0, hence, for $h >0$:
$$f''(h) = f''(0) + hf'''(0) + \frac{h^2 f''''(0)}{2} + o(h^2)$$
$$\therefore$$
$$\frac{f''(h)}{h^2} = \frac{6h}{h^2} + \frac{h^2 f''''(0)}{2h^2} + \frac{o(h^2)}{h^2}$$
So, there exists a small enough $h$ such that $\frac{6h}{h^2} > \frac{h^2 f''''(0)}{2h^2} + \frac{o(h^2)}{h^2} \implies f''(h) >0$. 
Finally, apply the same for $-h$:
$$f''(-h) = f''(0) - hf'''(0) + \frac{h^2 f''''(0)}{2} + o(h^2)=-6h + \frac{h^2 f''''(0)}{2} + o(h^2)$$
$$\therefore$$
$$\frac{f''(-h)}{h^2} = -\frac{6h}{h^2} + \frac{h^2 f''''(0)}{2h^2} + \frac{o(h^2)}{h^2}$$
Again, make $h$ small enough, and we have $f''(-h)<0$.
Since the interval is centered at 0, then $f(0) = 0$ belongs to it.
A: The problem is far simpler than it appears. What a relief!
Let $g(x) =f''(x) $. Then we are given $g(0)=0,g'(0)=6>0$. This means that $$\lim_{h\to 0}\frac{g(h)-g(0)}{h}>0$$ It follows via a simple application of definiton of limit that there is a $\delta>0$ such that $$\frac{g(h)}{h}>0$$ for all $0<|h|<\delta$ (think it as trivial: if a function tends to a positive limit then its values being close to the limit must themselves be positive). In other words $g(h) $ has same sign as that of $h$ for all non-zero $h$ lying in interval $(-\delta, \delta) $. Thus $I=[-\delta/2, \delta/2] $ is one such desired interval which meets the requirements of the question.

The other data given in question like $f(0)=0,f'(0)=0$ are meant to draw some more conclusions. In particular note that $f''$ is positive in $(0,\delta /2]$ and negative in $[-\delta/2,0)$ and therefore $f' $ decreases strictly in $[-\delta/2,0]$ and increases strictly in $[0,\delta/2]$. Since $f'(0)=0$ it follows that $f'$ is positive in all of $I$ except at $0$. And finally this means that $f$ is strictly increasing in $I$. 
