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.

This is only a partial answer. I will try to prove this for a polynomial function $f(x)$. The degree of this polynomial is at least 3. Zero is a double root and -1 is a single root:
$$f(x)=c(x)x^2(x+1)$$
The first thing I want to prove is that there exists an $a<0$ with $f''(a)=0$ and a $b>0$ with $f''(b)>1$. Remember that $f^{(3)}$ exists, so $f''(x)$ is continuous. As a result of continuity, $f''(x)$ reaches all the values between $0$ and $1$ in the interval $(a,b)$. Choose the largest such $a$ and the smallest such $b$, so that $f''(x)$ reaches in $(a,b)$ every value $y \in (0,1)$ exactly once. The image of $f''$ for $x \in (a,b)$ is thus $(0,1)$. 
(1) Proof that there is an $a \in (-1,0)$ with $f''(a)=0$: 
$f(-1)=0$ and $f(0)=0$ so by Rolle's theorem there is a $c \in (-1,0)$ with $f'(c)=0$.
$f'(c)=0$ and $f'(0)=0$ so by Rolle's theorem there is an $a \in (c,0)$ with $f''(a)=0$.
(2) Proof that there is a $b \in (0,1)$ with $f''(b)>1$:
$f(0)=0$ and $f(1)=1$ so by Lagrange's theorem there is a $d \in (0,1)$ with $f'(d)=1$.
$f'(0)=0$ and $f'(d)=1$ so by Lagrange's theorem there is an $b \in (0,d)$ with $1=f''(b)d$. Since $0<d<1$, $f''(b)>1$. 
Equipped with this information, I want to prove that there exists an $x \in (0,b)$ with $f^{(3)}(x)>3$:
$$f^{(3)}(x)=c'''(x)(x^3+x)+3c''(x)(3x^2+1)+3c'(x)(6x)+6c(x)$$
Remember that 


*

*$\forall x \in (0,b)$: $f(x) > 0$ and the same must be true for $c(x)$. You can proof this by contradiction. 

*$\forall x \in (0,b)$: $f'(x) > 0$ and the same must be true for $c'(x)$ (not completely sure how to proof this).

*As we already know: $\forall x \in (0,b)$: $f''(x) > 0$ and the same must be true for $c''(x)$.

*$\forall x \in (0,b)$: $f^{(3)}(x) \geq 0$ (because $f''(x)$ is monotone increasing in $(a,b)$) and the same must be true for $c''(x)$.


If those statements are true, then $\forall x \in (0,b): f^{(3)}(x)>3(3x^2+1)>3$.
I know this is not an incomplete proof, but nevertheless I hope it helps you. 
