Further insights into the intermediate value theorem and Rolle's theorem

The intermediate value theorem states: If $f$ is a continuous function on a closed interval $\lbrack a,b\rbrack$, and if $y_{0}$ is any value strictly between $f(a)$ and $f(b)$, that is $min\{ f(a),f(b)\} < y_{0} < max\{ f(a),f(b)\}$, then $y_{0} = f(c)$ for some $c$ in $(a,b)$.
The first insight this essay would reveal is:
if $f(b) > f(a)$, then there is some $c$ in $(a,b)$ such that $y_{0} = f(c)$ and $f^{'}(c) \geq 0$ if $f$ is differentiable at $c$, if we do not impose $f$ is differentiable at $c$, we say $f$ is non-decreasing at $c$, and even further the satisfying $c$ could be the greatest or the least number in set $\{ c \mid f(c) = y_{0}\}$.

In other words, to go continuously from $f(a)$ to $f(b)$, and if $f(a) < f(b)$, $f$ must be non-decreasingly reaches each value strictly between $f(a)$ and $f(b)$ at some point in $(a,b),$ though $f$ may not be a monotonic increasing function. For $f(a) > f(b)$, there is a similar conclusion¹.

The proof as follows. According to the intermediate value theorem, there is a set $C =$ $\left\{ c \mid f(c) = y_{0} \right\}$, since $b$ is an upper bound for $C$, there should be a least upper bound $M$ for $C$. if $C$ is a finite set, then $M \in C$, while if $C$ is an infinite set², then for every positive real number $\varepsilon_{n}$, there is a $c_{n}$ in $C$ such that $M - \varepsilon_{n} < c_{n} \leq M$, so that one can find a sequence $\left\{ c_{n} \right\}$ with the limit $M$, because of the continuity of $f(x)$, therefore $\lim_{n \rightarrow \infty}\mspace{2mu} f\left( c_{n} \right) = f(M) = y_{0}$, thus $M$ is the greatest one in $C$ whether $C$ is infinite or not. Now supposing $f$ is decreasing at $M$ in case of $f(a) < f(b)$, then there is a $d$ in some right neighborhood of $M$ such that $f(M) = y_{0} > f(d)$, then according to the intermediate value theorem there should be another $y_{0}$ in the range of $f$ such that $f(d) < y_{0} < f(b)$ on $(d,b)$, which is contradictory to the definition of $M$, thus $f$ is non-decreasing at $M$, i.e. $f^{'}(M) \geq 0$ if $f$ is differentiable at $M$. Similar arguments for the lower bound $a$ and the case $f(a) > f(b)$ totally complete the proof.

By the conclusions above, we can also extend Rolle's theorem, which states “suppose that $y = f(x)$ is continuous over the closed interval $\lbrack a,b\rbrack$ and differentiable at every point of its interior $(a,b)$. If $f(a) = f(b)$, then there is at least one number $c$ in $(a,b)$ at which $f^{'}(c) = 0$.” to $f^{'}(c) \geq 0$³ if $f(b) > f(a)$ and $f^{'}(c) \leq 0$ if $f(b) < f(a)$, where the $c$, in addition to that determined by the mean value theorem $f^{'}(c) = \frac{f(b) - f(a)}{b - a}$, could be some in $(a,b)$ with $min\{ f(a),f(b)\} < f(c) < max\{ f(a),f(b)\}$.

1. For $f(b) = f(a)$, we could not draw such a conclusion, because $y_{0} = f(b) = f(a)$ for $c$ in $(a,b)$ is not always possible.??

2. Think about $y = xsin\frac{1}{x}$ or $f\left( \pm \frac{1}{2^{n}} \right) = {( - \frac{1}{2})}^{n}$ with the value 0 at $x = 0$.??

3. Think about $y = x^{3}$ around $x = 0$.??