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Why in Rolle's theorem the function is given to be continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a,b)$? What if we take open interval for continuity as well? Please answer if anyone knows.

Did
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Waqar Ali Shah
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    Then the condition f(a)=f(b) would be empty hence no conclusion could be drawn from the hypothesis. – Did Jun 28 '15 at 09:43

2 Answers2

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Consider the function $f : [0, 1] \to \mathbb{R}$, given by

$$\begin{cases} x & 0 \leq x < 1\\ 0 & x = 1. \end{cases}$$

Note that $f$ is continuous and differentiable on $(0, 1)$, and $f(0) = f(1)$, but $f'(x) = 1$ for all $x \in (0, 1)$; in particular, there is no $c \in (0, 1)$ such that $f'(c) = 0$.

As the above example shows, continuity on the open interval only is insufficient for Rolle's Theorem.

Michael Albanese
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    I was typing, got there a little later than you did. Qe thought of the same exact example. Funny! – Lonidard Jun 28 '15 at 09:56
  • @MichaelAlbanese, what if we take the closed interval [a,b] for differentiability? Will it create some problem? – Waqar Ali Shah Jun 29 '15 at 18:01
  • Excellent counterexample there. – ireallydonknow Jun 29 '15 at 18:05
  • @WaqarAliShah: You would have to first define what it means for a function to be differentiable on $[a, b]$. The problem here is what it means to be differentiable at the endpoints. – Michael Albanese Jun 29 '15 at 18:11
  • @MichaelAlbanese, Given the closed interval [a,b], Left-hand derivative at "b" and right -hand derivative at "a" exist. I have also seen some theorems and examples where the function has been given to be differentiable on the closed interval. – Waqar Ali Shah Jun 29 '15 at 18:30
  • OK. Note that differentiable on $[a, b]$ implies differentiable on $(a, b)$, so what is your question exactly? – Michael Albanese Jun 29 '15 at 18:57
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The theorem does not hold. For instance, consider the function $f: [0,1] \to \mathbb{R}$

$$ f\left(x\right) = \left\{ \begin{array}{lr} x & \text{if} \ x \in (0,1) \\ 0 & \text{otherwise} \end{array} \right.\\ $$

All of the hypotheses of the theorem hold (except continuity on the extremal points), but $f'(x)=1$ for every $x \in (0,1)$.

Lonidard
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  • Of course! Otherwise, it would of course be continuous in the closed interval. Thank you for the remark, I'll fix it immediately. – Lonidard Jun 28 '15 at 10:57