Suppose that $f:[a,b]\to\mathbb{R}$ be a twice-differentiable function, and that there exists $c\in(a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$.
Show that if $f''(x)>0$ for all $x\in[a,b]$ and $f''$ is strictly increasing on $[a,b]$, then $c>\frac{a+b}{2}$.
Using Taylor's theorem, i solved the problem.
But, i'd like to prove that using convexity.
Give some comments or hints. Thank you!
As indicated in Kavi Rama Murthy’s comment, convexity is not enough. Consider $g(x)=f(a+b-x)$. Then $g$ satisfies the same conditions as $f$ except that $g’’$ is decreasing, while $$ \frac{g(b)-g(a)}{b-a}=g’(a+b-c), $$ with $a+b-c<(a+b)/2$.