For any increasing, differentiable, strictly convex real function $f: [a,b] \to \mathbb{R},\ $ by the mean value theorem and the nature of $f,$ there exists a unique $\ \mu\in (a,b)\ $ such that $f'(\mu) = \frac{f(b) - f(a)}{b - a}.$
I have an example of a increasing, differentiable, strictly convex real function $f: [a,b] \to \mathbb{R},\ $ that also has the property that $f'(\mu) = \frac{f(b) - f(a)}{b - a}\implies \mu \in \left(a, \frac{a+b}{2}\right),\ $ namely $f(x) = x^{3/2}$ on $[0,100].$
My question is, is there an example of a increasing, differentiable, strictly convex real function $f: [a,b] \to \mathbb{R},\ $ that also has the property that $f'(\eta) = \frac{f(b) - f(\mu)}{b - \mu}\implies \eta \in \left(a, \frac{a+b}{2}\right),\ $ where $\mu$ is defined as the point such that $f'(\mu) = \frac{f(b) - f(a)}{b - a}\ ?$
If the answer is no, then I guess Darboux Theorem is relevant to prove that no such function exists (but I have no idea how to do this). I imagine a (counter)-example does exist and I am not being creative enough in my attempts to come up with one.
$f$ defined by $g(x)=1+9e^{-x}$ is strictly convex, differientiable, not increasing but $f(x)=9x+g(x)$ is and seems to be an example with $a=0$, $b=10$. I find $\mu=\ln(10)-\ln(-e^{-10})\approx 2.3$ and $\eta$ such that $f'(\eta)=\frac{f(b)-f(\mu)}{b-\mu}$ is less than $\frac{a+b}{2}$ as $f'(\eta)<f'(5)$ which is less computation to check as it stands to $\mu>10+e^{-5}+\frac{e^{-5}-e^{5}}{10}$ the right member being approx. $-4.8$