Let $f(x)$ be two times differentiable in $[a, b]$. The Mean Value Theorem tells us that there's such a point $a\lt c\lt b$, such that ${f}'(c)=\frac{f(b)-f(a)}{b-a}$.
If $\forall x \in [a, b], {f}''(x)\gt 0$ and $f''(x)$ is strictly increasing then $c\gt \frac{a+b}{2}$.
I'm having a hard time figuring this one out. More so, I can't seem to understand why this is true, intuitively.
I tried the following:
Write $f(x)$ Taylor's expansion of degree 1, around $a$. Then, for $x=b$ we get: $f(b)-f(a)=f'(a)(b-a)+\frac{1}{2}f''(\xi)(b-a)^2 \implies f'(c)=f'(a)+\frac{1}{2}f''(\xi)(b-a)$
Now I still haven't used the fact that $f''(x)$ is positive and strictly increasing in $[a, b]$ (which btw also implies that $f'(x)$ is strictly increasing). I tried playing with the expression above but I just don't see which direction to go.
Please don't give a full solution. Hints and intuition are appreciated.
Best Answer
Try sketching a graph, since you are looking for intuition only.
A sketch of the f' will be easier as then you can use the Intermediate Value Theorem too.
Say we want to see that the MVT holds for the case of $f(x)$ in [a,b]
We need to prove that the slope $\frac{f(b)-f(a)}{b-a}$ is inside $[f'(a),f'(b)]$ This is not too difficult to prove.