Does there exist a strictly increasing function $f$ satisfying $c=a$ or $c=b$? The mean value theorem:

Theorem: Let $f:[a,b]→[a,b]$ be a continuous function on the closed interval $[a,b]$, and differentiable on the open interval $(a,b)$, where $a<b$. Then there exists some $c\in(a,b)$ such that
$$f′(c)=\frac{f(b)- f(a)}{b-a}.$$

Generally, $c$ is included strictly in the open interval $(a,b)$, i.e., $c≠a$  and $c≠b$.
My question is: Does there exist a strictly increasing function $f$ (with a strictly increasing derivative $f'$) for which  "$c=a$"  or "$c=b$", i.e.
$$f′(a) = \frac{f(b)- f(a)}{b-a} \;\;\text{ or } \;\;f′(b)= \frac{f(b)- f(a)}{b-a}$$
?
 A: The theorem says that a point $c\in(a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$ exists. It says nothing about $f'(a)$ or $f'(b)$. Under the theorem's hypothesis, they may even not exist.
But if you are looking for a function $f$ such that $f'(a)$ exists and
$$f'(a)=\frac{f(b)-f(a)}{b-a}$$
you can take for example a stright line.
If you also want the derivative $f'$ to be strictly increasing, then it is impossible, since the theorem says that a point $c\in(a,b)$ exists, and $f'$ is injective, that is, there is no $c_1\neq c$ such that $f'(c_1)=f'(c)$.
A: The answer is no, because of the theorem you cited!
Let us call define the constant $D$ as
$$D = \frac{f(b)-f(a)}{b-a}.$$
Then the mean value theorem says that there exists some $c \in (a,b)$ such that
$$f'(c) = D.$$
Now, I believe your question asks if there exists a function $f$ with both $f$ and $f'$ are strictly increasing, such that
$$f'(a) = D \;\;\;\text{   or   } \;\;\;f'(b) = D.$$
Let's assume that the first one is true, i.e. $f'(a) = D$. Then $f'(a) = D = f'(c)$, but since $c \in (a,b)$ you have that $a < c$. Then $f'$ is not strictly increasing anymore! The same argument holds if $f'(b) = D$.
