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?

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    $\begingroup$ Because it isn't necessarily true if these conditions are relaxed. $\endgroup$ Jun 11, 2017 at 0:50
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    $\begingroup$ 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. $\endgroup$ Jun 11, 2017 at 0:59
  • $\begingroup$ @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 $\endgroup$
    – Believer
    Jan 8, 2019 at 11:13
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    $\begingroup$ @Believer Saying that the derivative of $\sqrt{x}$ at $0$ is $+\infty$ is the same thing as saying that the derivative doesn't exist or that the graph of $\sqrt{x}$ is completely vertical there. MathematicsStudent1122 must have been talking about the endpoint at $0$ since the derivative at the other endpoint, $1$, is $1$. The point is that it's a counterexample to the natural assumption that you could define the MVT with functions differentiable on $[a,b]$. $\endgroup$
    – Jam
    Jul 17, 2019 at 14:32

2 Answers 2


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 $f'(c)=0.$

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    $\begingroup$ You need continuity for more than just boundedness. Henricus W.'s first example is bounded, but still fails. $\endgroup$ Jun 11, 2017 at 1:36
  • $\begingroup$ @PaulSinclair and to attain extrema. $\endgroup$ Jun 11, 2017 at 1:44
  • $\begingroup$ Without continuity at the end-points, the MVT is obviously false, isn't it? It's nothing to do with boundedness. $\endgroup$
    – TonyK
    Jun 11, 2017 at 11:20
  • $\begingroup$ @TonyK In the counterexample above, the function is not continuous but still bounded. $\endgroup$ Jun 11, 2017 at 11:45

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