The usual formulation of the mean value theorem in a real analysis course is something like this:

Let $f\colon [a,b] \to \mathbb{R}$ be continous on $[a,b]$ and differentiable on $]a,b[$. Then there is a $\xi \in ]a,b[$ such that

$$ f(b) - f(a) = f'(\xi)(b-a) $$

Since differentiability implies continuity one could impose the slightly less general condition that $f$ should be just be differentiable on $[a,b]$.

Are there any "non exotic" cases or any theorems with are proven using the mean value theorem where one really need the more general form above?

Is it correct that I need only the less general version for deriving the following standard calculus theorems:

  • characterizing monotonic and constant function with the derivative
  • the inverse function theorem
  • the second derivative test for minima/maxima and the change of sign test

2 Answers 2


Derivatives of functions defined on an open interval have the intermediate value property. There is nothing "exotic" here; consider the semicircle defined by $$f(x) = \sqrt{1 - x^2},\qquad x \in [-1,1].$$ This function is differentiable at neither endpoint but this form of the MVT says it has a horizontal tangent in $[-1,1]$.

  • $\begingroup$ However if you know only the less general version you could say, that $f$ has the MVT on every closed sub-interval of $]-1,1[$. What about the theorems (see question above). $\endgroup$
    – Julia
    Dec 15, 2014 at 18:41

Consider the example $f(x) = \sqrt{x}$. Then $f$ is not differentiable at $0$, but it is continuous.


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