TITLE: Third derivative at point is greater than 3
QUESTION [5 upvotes]: 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

REPLY [4 votes]: 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
$$
0=h^{(3)}(c)=f^{(3)}(c)-g^{(3)}(c)=f^{(3)}(c)-3.
$$