This paper presents an elementary proof of Lagrange's form of the remainder for the Taylor series of a real function.
The following is proved.
Proposition. Suppose $f$ has an $(n+1)^\text{th}$ derivative in $[a,b]$ and that $m\leq f^{(n+1)}(x)\leq M$ for all $a<x<b$. Then for all $a\leq x\leq b$ we have $$\frac{m}{(n+1)!}(x-a)^{n+1}\leq R_nf(x)\leq \frac{m}{(n+1)!}(x-a)^{n+1}.$$
The authors then proceed to formulate and prove the Lagrange form as follows.
Theorem. Suppose $f$ has an $(n+1)^\text{th}$ derivative in $[a,b]$. Then there is some $a\leq c\leq b$ such that $R_nf(b)=\frac{f^{(n+1)}(c)}{(n+1)!}(b-a)^{n+1}$.
Proof. Choose $m=\inf_{[a,b]}f^{(n+1)},M=\sup_{[a,b]}f^{(n+1)}$ (if $f^{(n+1)}$ is unbounded we allow $m,M=\pm \infty$). Thus by the proposition, $R_nf(b)=\frac{k}{(n+1)!}(b-a)^{n+1}$ for some $m\leq k\leq M$. If one of the equalities holds, then the result is immediate from the proposition. Otherwise it follows directly from Darboux's intermediate value theorem.
If neither equality holds, i.e $m<k<M$, then I see Darboux's theorem takes care of things. However, I don't understand the preceding sentence.
Question. If one of the equalities holds, i.e $k\in \left\{ m,M \right\}$, then why is the assertion immediate from the proposition?
At first I thought perhaps a bounded derivative attains its extrema, but this answer gives a counterexample.
In the paper there is a 4th lemma that shows that $k$ is $M$ (or $m$) iff the $(n+1)$-th derivative is constant.