# 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?

• Because it isn't necessarily true if these conditions are relaxed. – carmichael561 Jun 11 '17 at 0:50
• You may also wonder why we assume differentiability on the open interval $(a,b)$ rather than (perhaps more naturally) assuming that "$f$ is differentiable on $[a,b]$" The reason is that functions like $f(x) = \sqrt{x}$ defined on $[0,1]$ are not differentiable at an endpoint, but nonetheless the MVT holds. – MathematicsStudent1122 Jun 11 '17 at 0:59
• @MathematicsStudent ..sorry but I didn't understand why derivative doesn't exist at endpoint..are u talking about 0??but RHD is $+\infty$..so if derivative is infinite can we say function is not differentiable at that point – Believer Jan 8 at 11:13

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.
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 $f'(c)=0.$