Why does the Mean Value Theorem require a closed interval for continuity and an open interval for differentiability?

Why does the Mean Value Theorem assume a closed interval for continuity and an open interval for differentiability?

The MVT says: Let $f$ be a continuous function on $[a,b]$ that is differentiable on $(a,b)$, then....

Is there any example where one of them isn't true so that the MVT is not true?

Solution 1:

Relax the first constraint: Let $f:[0,1] \to \mathbb R$ so that $f(0) = 1,f(x) = 0$ for $x \in \left]0,1\right]$. Then $(f(1) - f(0))/(1-0) = -1$ but $f'(x) = 0$ on $]0,1[$.

Relax the second contraint: Let $f(x) = |x|$ on $[-1,1]$, then $(f(1)-f(-1))/(1-(-1)) = 0$ but $f'(x) = 0$ nowhere.

Solution 2:

The MVT is a consequence of Rolle's Theorem. you need continuity at $[a,b] $ to be sure that the function is bounded. if its extremum is attained at $x=c\in (a,b) $ you use differentiability at $(a,b) $ to get $f'(c)=0.$