Difficulties in stating mean value theorem for functions which are not continuous on a closed interval.

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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"?