Proving second value of second derivative from function and first derivative $  Let \ f: [-1,1] \rightarrow \mathbb{R} $ be a function, which is continuous on [-1,1] and twice differentiable on (-1,1) with: $$ f(-1)=f(1)= 0 \ and \ f(0)=-3 $$
Show that there exists $ c \ within \ (-1,1) $ s.t. $ f''(c) > 3 $
My intuition was to use Rolle's theorem to show that there must be a K within (-1,1) s.t. f'(k) = 0, then use prove by contradiction and use mean value theorem to show that there is a f''(c) which cant be $ \leq 3$ however, i am having serious trouble with this. I think i am overseeing something simple, I also figure i need to use the fact that $ f(0)=-3 $ somehow but can't work out where. Some hints would be greatly appreciated! 
 A: Hint: Instead of using the mean value theorem (in the form of Rolle's theorem) for $f$ on all of $[-1,1]$, use it separately on each of the intervals $[-1,0]$ and $[0,1]$.
A: One can even show that $f''(x) \ge 6$ for some $c \in (-1, 1)$
under the given conditions.
(The function
$f(x) =3(x^2-1)$ shows that this is the best possible bound.)
Assume that $f''(x) < 6$ for all $x \in (-1, 1)$.
Let $x_0 \in (-1, 1)$ be a point where $f$ assumes it's minimum, then
$f(x_0) \le -3$ and $f'(x_0) = 0$. Because of the symmetry of the problem we
can assume that $x_0 \ge 0$.
Taylor's formula applied to the interval $[x_0, 1]$ gives
$$
f(1) = f(x_0) + f'(x_0)(1-x_0) + \frac{f''(\xi)}{2}(1-x_0)^2
$$
for some $\xi \in (x_0, 1)$, and therefore
$$
 0 = f(1) < -3 + \frac 62 (1-x_0)^2 \le -3 + 3 = 0
$$
which is a contradiction.

An alternative (mimicking the proof of Taylor's theorem) is to consider
the function
$$
 g(x) = f(x) +  (x^2 -1) \cdot f(0)\, .
$$
Then $g(-1) = g(0) = g(1)$. Repeated application of Rolle's theorem
shows that $g'$ has two distinct zeros, and therefore
$g''(c) = 0$ for some $c \in (-1, 1)$. So
$$
 0 = g''(c) = f''(c) + 2f(0) \implies f''(c) = -2f(0) = 6 \, .
$$
A: $-3=\frac{f(-1)-f(0)}{-1\ -0}=f'(a)$ for some $a\in(-1,0)$ and $3=\frac{f(1)-f(0)}{1-0}=f'(b)$ for some $b\in(0,1)$ . Now $\frac{f'(b)-f'(a)}{b-a}=f''(c)$ for some $c\in(a,b)$, also $0<b-a<2$.Therefore $f''(c)>3$
A: Note that by Taylor's theorem we have two numbers $c_1,c_2$ with $-1<c_1<0<c_2<1$ such that $$f(-1)=f(0)-f'(0)+\frac{f''(c_1)}{2},\,f(1)=f(0)+f'(0)+\frac{f''(c_2)}{2}$$ Adding these and noting that $f(0)=-3, f(-1)=f(1)=0$ we get $$\frac{f''(c_1)+f''(c_2)}{2}=6$$ It now follows via intermediate value property of derivatives that there is a $c\in[c_1,c_2]$ such that $f''(c) =6$. 
