show that $f^{(3)}(c) \ge 3$ for $c\in(-1,1)$ Let $f:I\rightarrow \Bbb{R}$, differetiable three times on the open interval $I$ which contains $[-1,1]$. Also: $f(0) = f(-1) = f'(0) = 0$ and $f(1)=1$.  
Show that there's a point $c \in (-1, 1)$ such that $f^{(3)}(c) \ge 3$
I'd be glad to get a guidace here how to start.

You can use argument contradiction to show. Suppose $f'''(x)<3$ for $\forall x\in(-1,1)$. In [-1,0], by MVT, we have
$$ f(0)-f(-1)=f'(c_1) $$
where $c_1\in(-1,0)$. So $f(c_1)=0$. Similarly in $[c_1,0]$, we have
$$ f'(0)-f'(c_1)=f''(c_2) $$
where $c_2\in(c_1,0)$. So $f''(c_2)=0$. In [0,1], we have
$$ f(1)-f(0)=f'(b_1) $$
where $b_1\in(0,1)$. So $f(b_1)=1$. In $[0,b_1]$, we have
$$ f'(b_1)-f'(0)=f''(b_2)b_1 $$
where $b_2\in(0,b_1)$. So $f''(b_2)=\frac{1}{b_1}>1$.
Since $f'''(x)$ exists in (0,1), we have that $f''(x)$ is contiuous in $(0,1)$ and hence there exists $a\in[c_2,b_2]$ such that $f'':[c_2,a]\to [0,1]$ is an onto map. Choose $c_2\in(c_1,0)$ such that $c_2$ is the largest. Thus in $ (c_2,a)$, $f''>0$ and hence $f''(x)>0$ in $[0,a]$.
For $x\in[0,1]$, we have
$$ f''(x)-f''(0)=f'''(c_3)(x-0) \tag{1} $$
where $c_3\in(0,x)$. (1) implies $$ f''(x)<3x+1 \tag{2}. $$
Integrating (2) twice from (0,x) and using $f(0)=f'(0)=0$, we have
$$ f(x)<\frac{1}{2}x^3+\frac{1}{2}x^2. \tag{3} $$
Letting $x=1$ in (3), we obtain 
$$ f(1)<\frac{1}{2}+\frac{1}{2}=1$$
which is against $f(1)=1$.
