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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

$$f′(c)=\frac{f(b)- f(a)}{b-a}.$$

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. $$f′(a) = \frac{f(b)- f(a)}{b-a} \;\;\text{ or } \;\;f′(b)= \frac{f(b)- f(a)}{b-a}$$ ?

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    If $c\in (a,b)$ then $c \neq a$. What do you then mean with *"verifying $c = a$"*? – 2017-01-02
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    @Therkel: I am asking about the possibility for this to happened – 2017-01-02
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    If $c > a$ then the statement $c = a$ is false. There does not exist a real number $c$ such that $c = a$ and $c\neq a$. Perhaps you are asking if $f'(c) = a$? – 2017-01-02
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    @Therkel: My **question** is: Does there exist a strictely increasing function $f$ (with a strictely increasing derivative) verifying $c=a$ or $c=b$? – 2017-01-02
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    Sorry, I must be missing something here. If $c\in (a,b)$ surely $c\neq a$? Then asking if it could be that $c = a$ is absurd. – 2017-01-02
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    @Therkel I believe he means that $f'(a)$ or $f'(b)$ equal $(f(b) - f(a)) / (b-a)$. – 2017-01-02
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    If the derivative $f'$ is strictly increasing,then there is a ***unique*** $c$ with $f'(c)=(f(b)-f(a))/(b-a)$ and it is in $(a,b)$, right? – 2017-01-02
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    The mean value theorem is usually stated for functions $f:[a,b]\to\mathbb R.$ For your purposes, it is important that $f:[a,b]\to[a,b]$? – 2017-01-02
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    @bof: Yes, right – 2017-01-02

2 Answers 2

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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 $f'(b) = D$.

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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 $f'(c_1)=f'(c)$.