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Mean Value Theorem. If $f:[a,b] \rightarrow R$ is continuous on [a,b] and differentiable on (a,b), then there exists a point $c \in (a,b)$ where

$$f'(c) = \frac{f(b)-f(a)}{b-a} $$

I am having a hard time understanding why $f$ must be continuous on a closed interval and differential on open interval. What is the significance of having such a requirement? What if it was continuous on an open interval and differential on closed interval? Also why is it that there exists a point $c \in (a,b)$ and not $c \in [a,b]$?

I don't know if a similar question has been asked before as I found no satisfactory post. If one has been asked previously, a link to the post would be much appreciated.

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5 Answers 5

Let's go through the possible variations one by one

  • Continuous on the open internval, differentiable on the closed one. Differentiable at $x$ implies continuous at $x$. Thus, the range of values where $f$ is continuous is always at least as large as the range of values where it is differentiable. So having $f$ continuous on the open interval but differentiable on the closed on doesn't make sense.

  • Continuous and Differentiable on the open interval. The theorem fails in this case. Let $f(x) = 1$ if $x > 0$ and $f(0) = 0$. Since $f$ is constant on $(0,1)$, $f$ is clearly continuous and differentiable. But the derivative is identically zero, so $0 = f'(c) \neq \frac{f(1) - f(0)}{1-0} = 1$ for all $c \in (0,1)$, contradicting the theorem.

  • Continuous on the closed interval, differentiable on less than the open interval. The theorem fails in this case too. Let $f(x) = x$ if $x < \frac{1}{2}$ and $f(x) = 1$ if $x \geq \frac{1}{2}$. $f'$ exists on $(0,1)\setminus\left\{\frac{1}{2}\right\}$, and takes only the values $2$ (on $\left(0,\frac{1}{2}\right)$) and $0$ (on $\left(\frac{1}{2},1\right)$). Yet again $\frac{f(1) - f(0)}{1-0} = 1$, so $\frac{f(1) - f(0)}{1-0} \neq f(c)$ for all $c$ where $f'$ exists.

  • Continuous and Differentiable on the closed interval. This is strictly weaker than the original theorem, and therefore is of course true. But since the theorem also holds if we require differentiability only on the open interval, why state the weakened form, if we might just as well state the stronger one? The function $f(x) = (x-x^2)\sin \frac{1}{x-x^2}$ for $x=[0,1]$, where $f(0)=f(1)=0$ wittnesses that this version is actually strictly weaker. The function itself is continuous on $[0,1]$, since the factor $(x^2-x)$ ensures that it goes to zero as $x$ approaches $0$ respectively $1$. But the same isn't true for the derivative, and therefore the derivative exists only on $(0,1)$.

Thus the theorem is stated so that it is as general as possible while still being true.

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Basically, we need $f$ to be continuous on $[a,b]$ (closed), since if not we can have scenarios where $f(a),f(b)$, "blow up", violating the differentiability around these points, and making our quotient $\frac{f(b)-f(a)}{b-a}$, plausibly not well defined.

And we need $f$ to be differentiable on $(a,b)$ (open) since, differentiation is not well defined on a closed interval, because we need to know what is happening at each side of a point, i.e. if we say $f$ is differentiable at some point $z$, then we have to know how the function behaves on the "left" and "right" of $z$.

This cannot happen on a closed interval, i.e $[a,b]$, since we do not know how the function is behaving on the "left" of $a$ or the "right" of $b$.

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The proof of the mean value theorem uses Rolle's theorem. In order to apply Rolle's theorem, $f$ has to be continuous on $[a,b]$, since the proof of Rolle's theorem uses the fact that a continuous function on a compact interval attains it's maximum. This is obviously false for non-compact intervals.

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Well, of course $f$ has to be continuous on $[a,b]$ because that is what the theorem tells us. But my question is simple. Why? –  Eames Cobb Apr 12 at 20:34
    
@EamesCobb See my edit –  Vincent Boelens Apr 12 at 20:37

Notice that $f'(c)$ is the slope of the tangent line of the curve at the point $(c,f(c))$ so the geometric interpretation of the mean value theorem: there's $c$ in which the tangent line of the curve is parallel to the segment joining the point $(a,f(a))$ and $(b,f(b))$.

enter image description here

If $f$ is not continuous on $[a,b]$ then we can find easily a counterexample as shown in the figure below

enter image description here

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If $f$ is differentiable le on the closed interval then it is also continuous on the closed interval. However, $f$ can be continuous on the closed interval, differentiable on the open interval and not differentiable on the closed interval.

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