# Does this contradict the fact that we cannot apply the mean value theorem?

Mean value theorem: Let $f:[a,b]\to \mathbb {R}$ be a continuous function on the closed interval $[a,b]$, and differentiable on the open interval $(a,b)$, where $a<b$. Then there exists some $c$ in $(a,b)$ such that

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

Contrary, consider the case where the function $f$ is not defined in a point $d$ in $[a,b]$, i.e., $lim_{x→d⁻}f(x)=+∞$ and $lim_{x→d⁺}f(x)=-∞$. Then it is not possible to apply the above theorem to find certain $c$. Now, for $a<d$ and $b>d$. Assume that a direct calculation gives certain values of $c$. Does this contradict the fact that we cannot apply the mean value theorem.

• If $f$ is continuous on $[a,b]$ then it must actually be defined on $[a,b]$ (else what does "continuous" mean?). – lulu Dec 2 '16 at 15:16
• I see the edit, but then I don't understand the question. The Mean Value Theorem does not apply to discontinuous functions. Of course, the conclusion of that Theorem might happen to be true for some discontinuous function but you can't prove it by invoking the Theorem. – lulu Dec 2 '16 at 15:23
• You said "Then it is not possible to apply the above theorem to find certain c". The Mean value theorem doesn't give you the value of c, only guarantees the existence – user261263 Dec 2 '16 at 15:24
• As a, perhaps, simpler illustration of the (logical) point: The Intermediate value theorem tells us, among other things, that for continuous $f$, $f(a)>0,f(b)<0\implies \exists c\in [a,b]$ such that $f(c)=0$. But obviously we can have a continuous $f$ such that both $f(a)$ and $f(b)$ are greater than $0$ but nevertheless there is some $c\in [a,b]$ such that $f(c)=0$. – lulu Dec 2 '16 at 15:27
• From a logical point of view "$P\implies Q$" doesn't implies that $\lnot P \land Q$ is false. – user251257 Dec 2 '16 at 15:47

## 1 Answer

My understanding is that the mean value theorem is set up like:

1. $f:[a,b]\rightarrow \mathbb{R}$ is continuous on $[a,b]$
2. $f'$ is continuous on $(a,b)$

$(1$ and $2) \Rightarrow \exists c\in(a,b):f'(c)=\frac{f(b)-f(a)}{b-a}$

Which has the logical form of $A\Rightarrow B$. You will note that this does not imply that $\neg A \Rightarrow \neg B$. Instead it has the form of $\neg B \Rightarrow \neg A$. You will also note that the mean value theorem does not use the only if language which would also be a pathway to asserting what you assert.