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The mean value theorem is stated as follows

Let there be a function $f$ which is continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists $c$ belonging to $(a,b)$ such that $f′(c)=\frac{f(b)−f(a)}{b−a}$.

Now, here it is assumed that the function is continuous on a closed interval. What I don't understand is the use of making the function continuous on a closed interval. I understand that then $f(a)$ and $f(b)$ will not be defined if we use rather an open interval (a,b), but then we can take into account the limits as x approaches a and b of the function (sometimes one-sided limits) instead of the values of the function at $a$ and $b$.

I know this would be tedious and would make the theorem complicated, but what I want to ask is that whether there is any other reason for proposing a closed continuous function rather than an open or half-open one.

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  • $\begingroup$ It's also important to notice that the proof of the MVT (or the proof of Rolle's Theorem by proxy) employs the Extreme Value Theorem which demands a continuous function on a closed interval to say anything. $\endgroup$ – Robert Wolfe May 30 '18 at 0:19
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If you just consider continuity of $f(x)$ on $(a, b)$, then the limits at endpoints could be infinity or do not exist at all.

For example $f(x)= \tan (x)$ on $(-\pi /2, \pi /2)$.

In that case the statement of the theorem does not make sense as it stands.

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Because in the particular cases you evoke, it is always possible to bring you back to that form of the theorem.

For example if $f$ has only a left limit in $a$, you would define the function $g$ as $$ g(x) := f(x)\quad \text{if} \quad x>a $$ and $$ g(a) := \lim_{x\rightarrow a^+}f(x) $$ Then you can apply the theorem to $g$.

But it will much more complicated to state all the particular cases in the hypothesis, with no real gain, since they can be easily deduced with simple procedures as above.

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Let $f$ be a differentiable function on $(a,b)$. If the right side limit at $a$ and the left side limit at $b$ exist and are finite, then you can (uniquely) extend $f$ to a continuous function $\widetilde{f}$ on $[a,b]$ and apply the mean value theorem for $\widetilde{f}$ to get the desired result for $f$. But, in general, the side limits could be infinite, or they could not exist at all.

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If $f$ is not continuous at the endpoints, then how could knowing the values at the endpoints possibly tell you anything about its behavior elsewhere? Suppose we have the function $f: x \to x^2$ on the interval $(0,1)$. If $f$ is continuous, then $f(0) = 0$, $f(1) = 1$, so the MVT says that there is some place where the derivative of $f$ is $\frac{1-0}{1-0} = 1$. And indeed, $f'(0.5) = 1$. But if $f$ is not continuous, then $f(0)$ and $f(1)$ could be anything. For instance, $f(0)$ could be $-100$ and $f(1)$ could be $100$. Then the MVT would require there be a point in the interval such that the derivative is $200$, which clearly is not the case.

Yes, you could take the limit at $f(a)$ instead of $f(a)$, but then you have to worry about whether that limit exists, and if it does exist, then you're basically creating a new function with the value at a replaced by the limit at $a$. So why have a theorem that says "Given any function $f_1$, if $f_1$ can be turned into a continuous function $f_2$, then this holds for $f_2$", when you can just say " This holds for a continuous function"?

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