Using Rolle's Theorem and MVT for twice differentiable functions Let $f$ be a function that is twice differentiable on $\mathbb{R}$. Given that $f'(0)=f'(1)=0$, prove that there exists a $c \in(0,1)$ such that $|f''(c)|\geq 4|f(1)-f(0)|$ .
Yes, I know that many questions are already posted about this problem, but do hear me out.
Many other theorems use Definite  Integrals; however, I need to do this question using only Rolle's Theorem and the Mean Value Theorem.
My understanding so far:

*

*Use MVT to get $f'(c_0)=f(1)-f(0)$

*Use Rolle's theorem to get $f''(c_1)=0.$ Since $f'(0)=f'(1)=0$
I would sincerely appreciate it if you guys could help me with this question and not close it because it is a duplicate.
Thank you, Sirs and Madames.
 A: If $f(0) = f(1)$ the result is obvious, otherwise we can (by scaling and translating $f$ if necessary) assume wlog that $f(0) = 0$ and $f(1) = 1$, so that we need to prove $\vert f''(c) \vert \geq 4$ for some $c \in (0,1)$.
Assume towards a contradiction that for all $x \in (0, 1)$ we have $\vert f''(x)\vert  < 4$,
then the MVT gives us
$$f'(x) = f'(x) - f'(0) < 4x.$$
If we consider the function
\begin{alignedat}{2}
g : \mathbb{R} &\to \mathbb{R}\\ 
  x  &\mapsto f(x) - 2x^2,
\end{alignedat}
for all $x \in (0, 1)$ we get
$$ g'(x) = f'(x) - 4x <0$$
so $g$ is decreasing on $[0, 1]$ and therefore
\begin{alignedat}{2}
 0 = g(0) &> g\left(\frac{1}{2}\right) = f\left(\frac{1}{2}\right) - \frac{1}{2}\\
\implies f\left(\frac{1}{2}\right) &< \frac{1}{2}.
\end{alignedat}
Finally consider the function
\begin{alignedat}{2}
h : \mathbb{R} &\to \mathbb{R}\\ 
  x  &\mapsto 1 - f(1-x),
\end{alignedat}
which has many of the same properties as $f$:
$$
h(0) = 0,\\ 
h(1) = 1,\\
h'(0) = h'(1) = 0,\\
$$
and for all $x \in (0, 1)$ we still have
$$\vert h''(x) \vert = \vert -f''(1-x)\vert <4.$$
Therefore just like before we get
\begin{alignedat}{2}
&&h\left(\frac{1}{2}\right) &< \frac{1}{2}\\
&\implies 
&1 - f\left(\frac{1}{2}\right) &< \frac{1}{2}\\
&\implies 
&f\left(\frac{1}{2}\right) &> \frac{1}{2},
\end{alignedat}
which is a contradiction.
