Third derivative at point is greater than 3 f : $I \to \Bbb R$ is differentiable 3 times in open interval $I$ which contains the closed interval [-1,1]. $f(0)=f(-1)=f'(0)=0$ and $f(1)=1$
show that exists a point $c \in (-1,1) s.t. f^{(3)} (c) \ge 3$
What I did:
I used Rolle's theorem to prove that there are points in the derivatives where they equal 0. Don't really know how to get to 3... maybe using the intermediate value somehow- but don't really have an idea...
Thanks in advance
 A: A stronger statement holds: there exists $c$ in $(-1,1)$ such that $f^{(3)}(c)=3$.
Observe that the degree $3$ polynomial
$$
g(x)=\frac{1}{2}x^2(x+1)
$$
has exactly the same properties as your function $f$. And note that $g^{(3)}(x)=3$.
Now $h=:f-g$ satisfies
$$
h(-1)=h(0)=h(1)=h'(0)=0.
$$
By Rolle, there exist $-1<a<0<b<1$ such that $h'(a)=h'(0)=h'(b)=0$.
Two more applications of Rolle yield $-1<c<1$ such that
$$
0=h^{(3)}(c)=f^{(3)}(c)-g^{(3)}(c)=f^{(3)}(c)-3.
$$
A: I have a sketch of an answer that might help:
I believe that since $[-1,1]$ is closed, the function $f'''$ must reach a maximum $m$ on the interval. 
So we have: $f'''(x) \leq m$
Then you can integrate both sides over the range $[0,x]$, i.e.
$\int^x_0 f'''(t) dt \leq mx$
$f''(x) - f''(0) \leq mx$
$f''(x)  \leq mx +  f''(0)$
Trouble is, we are not given $f''(0)$ so leave as is for now. 
Now repeat the process:
$\int^x_0 f''(t) dt \leq \frac{1}{2}mx^2 +  f''(0)x$
$f'(x) - f'(0) \leq \frac{1}{2}mx^2 +  f''(0)x$
$f'(x) \leq \frac{1}{2}mx^2 +  f''(0)x$
and again:
$\int^x_0 f'(t) dt \leq \frac{1}{6}mx^3 +  \frac{1}{2}f''(0)x^2$
$f(x) - f(0) \leq \frac{1}{6}mx^3 +  \frac{1}{2}f''(0)x^2$
$f(x) \leq \frac{1}{6}mx^3 +  \frac{1}{2}f''(0)x^2$
Setting $x=1$ :
$1 \leq \frac{1}{6}m + \frac{1}{2}f''(0)$
So you get an inequality involving $m$ and $f''(0)$. 
Now you can repeat the whole process again but this time use the interval $[-x,0]$. You should then get another inequality involving $m$ and $f''(0)$. 
Using substitution you should be able to derive an inequality involving just $m$ that should yield your answer. 
