Why is $f$ continuous on $[a,b]$ and differentiable on $(a,b)$ in this proof of the mean value theorem?

Question

I stopped understanding the proof at "since $f$ is". I don't understand why $f$ is necessarily continuous on $[a,b]$ and differentiable on $(a,b)$. After all, $f$ is some kind of abstract function, it can be anything.

Answer 1

Regarding your comment about how you stopped understanding the proof at "since $f$ is," the continuity of $g$ follows from the theorems (often called) algebraic continuity theorem (a.c.t.):

Let $[a,b]$ be a closed interval of some field (if your field if just $\mathbb{R}$ you will of course be fine). Assume your hypotheses ($f$ is continuous on $[a,b]$ and differentiable on the open interval $(a,b)$ ). Since $fa,$ $fb,$ $a,$ and $b$ are all constants(scalars) with $a\neq b$, $\text{ }\text{ }\frac{fb-fa}{b-a}\text{ }\text{ }$ is a (defined) constant. Recall that the identity function is continuous. Then by a.c.t, the functions (of variable $x$) defined by $\text{ }\text{ }\text{ }x-a\text{ }\text{ }\text{ }$ and $\text{ }\text{ }\text{ }\frac{fb-fa}{b-a}(x-a)\text{ }\text{ }\text{ }$ are continuous on $[a,b]$. Hence, again because $fa$ is just a constant, the function defined by $\frac{fb-fa}{b-a}(x-a)+fa$ is continuous on $[a,b]$. Therefore, by assumption of continuity of $f$ and the a.c.t, $g$ defined by $\text{ }\text{ }\text{ } fx-\frac{fb-fa}{b-a}(x-a)+fa\text{ }\text{ }\text{ }$ is continuous on $[a,b].\blacksquare$

The differentiability of $g$ follows similarly.

Answer 2

No, it cannot be anything, since the statement of the Mean Value Theorem assumes that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$.

Answer 3

To apply Rolle's theorem of course $g(x)$ needs to be continuous on $[a,b]$ and differentiable on $(a,b)$ therefore also $f(x)$ needs to be continuous on $[a,b]$ and differentiable on $(a,b)$.