Is a function $f$ defined on a closed interval $[a,b]$ constant, if $f'(x)=0$ for all $x∈(a,b)$ but $f'(a)$ or $f'(b)$ are nonzero real numbers?

Question

Suppose we have a function $f$ that is defined on a closed interval $[a,b]$.
The following can be proven from the Mean Value Theorem:

If

• $f$ is continuous on the interval $[a,b]$
• $f'(x)=0$ for all $x∈(a,b)$
  

then $f$ is constant in $[a,b]$

The problem I have with the above theorem is the following:
If there exists a function $f$ such that all of the above conditions for the theorem hold true and additionally, $f'(a)$ or $f'(b)$ are non-zero real numbers, then, according to the theorem, $f$ would be constant for all $x∈[a,b]$.
Obviously, $f(a)=c$ for all $x∈[a,b]$ and $f'(a)≠0$ can't both be true at the same time. Where is the problem here?

Answer 1

The point of the Mean Value Theorem is that the function need only be continuous on $[a,b]$. It need not even be differentiable at $a$ or $b$ for the conclusion to hold.

However, if in fact $f$ is differentiable at $a$ (or $b$), then the derivative must be $0$. This is because of a theorem which states that the derivative function always has the intermediate value property; thus, it cannot have jump discontinuities. If $f'(a)$ exists and $\lim\limits_{x\to a^+} f'(x)=\ell$, then $f'(a)=\ell$.

EDIT: Spurred on by @MartinR's comment, here's a much more direct proof. Suppose the right-hand derivative $f'_+(a)$ exists. Since we've established that by continuity $f$ must be constant on $[a,b]$, then we have $$f'(a)=\lim_{x\to a^+} \frac{f(x)-f(a)}{x-a} = \lim_{x\to a^+}\frac 0{x-a} = 0.$$

Answer 2

In the given situation, there are only two possibilities:

1. $f$ is differentiable at $a$ (resp. $b$) and its derivative must be equal to $0$.
2. $f$ is not differentiable at $a$ (resp. $b$)

So a function that is constant on $[a,b]$, has $f'(x)=0$ for all $x \in (a,b)$ and has non-zero derivative at $a$ and $b$ does not exist. The reason why we say it must have a zero-derivative on $(a,b)$ is just because the derivative in $a$ or $b$ need not exist.