Got this question. Struggled with it for some time and recieved the hint to fit a polynomial through the given points and then use Rolle's Theorem. I think I have the proof for this question now. The full question is below. Please let me know if it makes sense, or if perhaps there is another simpler way for my own knowledge!
Given $f$ continuous on $[-1, 1]$ and twice differentiable on $(-1, 1)$ and $f(-1) = 0, f(0) = 0, f(1) = 1$. Prove $f''(c) = 1$ for some $c \in (-1, 1)$.
First, let us fit a function $g(x) = ax^2 + bx+c$ so that
$g(0) = c = 0$
$g(-1) = a - b = 0, a=b$
$g(1) = a + b = 1, a = \frac{1}{2}$.
Then we get $g(x) = \frac{x^2}{2} + \frac{x}{2}$. Then let us construct another function $h(x) = f(x) - g(x)$. Note that since $f$ is continuous on $[-1, 1]$ and twice differentiable on $(-1, 1)$ and since $g$ is a polynomial then $h(x)$ is also continuous on $[-1, 1]$ and twice differentiable on $(-1, 1)$.
Also note that $h''(x) = f''(x) - 1$.
Therefore, since $h(-1) = h(0) = 0$ and $h(0) = h(1) = 0$ then by Rolle's Theorem there exists a $c_1 \in (-1, 0), c_2 \in (0, 1)$ so that $h'(c_1) = h'(c_2) = 0$.
Now, since $h'(c_1) = h'(c_2) =0$ applying Rolle's Theorem again, there exists a $c \in (c_1, c_2) \subset (-1, 1)$ so that $h''(0) = 0.$
Hence, we have some $c \in (-1, 1)$ so that $h''(c) = f''(c) - 1 = 0$ and so $f''(c) = 1.$