Alternative to Rolle's Theorem? I sincerely hope this is not a dumb question. I was doing some reading through an analysis book and was looking at Rolle's Theorem. That is, if a function $f:[a,b]:\rightarrow \mathbb{R}$ is continuous on [a,b] and differentiable on (a,b), and that $f(a)=f(b)$. Then $\exists c\in(a,b)$ such that $f'(c)=0$.
Now, it seems to me that the intuition here is that if the "end points" of an function's value are equal (and the function is continuous), then the function's maximum or minimum has occurred somewhere between the endpoints. However, this intuition is obviously much more general than theorem. It doesn't seem to require differentiability(or does it?). 
Hence, my question is this: Is there a analogous theorem that captures the complete intuition without differentiability as a requirement? Why or why not?
As a possibly less important question:if we allow positive and negative infinity can we likewise get rid of the the continuity condition? Why or why not? Or why might this be a unimportant question?
Please let me know if I'm not being entirely clear somewhere, so I can clarify what I mean.
EDIT: I'm sorry I didn't use an example to maybe explain my question. The absolute value function has a minimum, but it is not differentiable there, but I would still like to speak about that point. 
Thanks for your time and help in advance!
 A: From the Extreme Value Theorem:

For any continuous $f$ defined on $[a,b]$, there exists $c_1,c_2\in[a,b]$ such that for all $x\in[a,b]$ $f(c_1)\leq f(x)\leq f(c_2)$.

We get a continuous version of Rolle:

If $f$ is continuous on $[a,b]$ and $f(a)=f(b)$ then there exists an $c_0\in(a,b)$ such that $f(c_0)$ is either the minimum or the maximum value of $f$ on $[a,b]$.

Proof: If $f(x)$ is constant on $[a,b]$ then you can pick any $x_0\in(a,b)$.
Otherwise, assume $f(c)>f(a)$ for some $c\in(a,b)$. (The second case, $f(c)<f(a)$, is the same.) By the Extreme Value Theorem, there exists $c_0$ such that $f(c_0)\geq f(x)$ for all $x\in[a,b]$. Since $f(c_0)\geq f(c)>f(a)=f(b)$, we know that $c_0\neq a,b$, so $c_0\in(a,b)$.

Rolle's theorem follows from this by the proof that, if $f$ is differentiable on $(a,b)$ then a maximum or minimum on $(a,b)$ must have derivative zero.
A: There is a theroem that states something pretty close to what you want :
Let $a, b \in \mathbb{R}, a<b$ and $f:[a,b]\to\mathbb{R}$ continous over $[a,b]$. 
Then, $\exists m, M \in [a, b]$ such that $f([a, b])=[m, M]$ and hence :
$$\forall x \in [a, b], m \leq f(x) \leq M$$
However, from the moment $f$ is not continuous, this is not necessarily true. e.g. $f(x) = \frac{1}{x^2}, f(-2) = f(2)$ but it has no maximum on $[-2, 2]$.
Note also that Rolle's theorem is different, it does not say that there is a maximum but more specifically that there is a value for which the derivative is $0$ (which is why it needs the function do be differentiable). 
If you allow $±\infty$ to be maximum/minimum values, this won't change anything. They can be infimum and supremum but not maximum and minimum because your function will never reach it.
Another example is for instance the function $1-e^{-x}$. It will never reach one, so one is not a maximum.
