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?
Best Answer
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.