Examine if Rolle's Theorem is applicable to the function $f(x) = [x]$ for $x \in [5, 9]$. Can you say something about the converse of Rolle's Theorem from this example?

(N/A) According to Rolle's Theorem,for a function $f: [a, b] \to \mathbb{R}$,if:
$1)$ $f$ is continuous on $[a, b]$
$2)$ $f$ is differentiable on $(a, b)$
$3)$ $f(a) = f(b)$
Then,there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.
For the function $f(x) = [x]$ on $[5, 9]$:
$1)$ The greatest integer function $[x]$ is discontinuous at all integral points. Since $5, 6, 7, 8, 9$ are integers in $[5, 9]$,$f(x)$ is not continuous on $[5, 9]$.
$2)$ $f(5) = [5] = 5$ and $f(9) = [9] = 9$. Thus,$f(5) \neq f(9)$.
$3)$ Since $f(x)$ is discontinuous at integral points,it is also not differentiable at these points in $(5, 9)$.
Since the conditions of Rolle's Theorem are not satisfied,the theorem is not applicable to $f(x) = [x]$ on $[5, 9]$.
Regarding the converse: The converse of Rolle's Theorem states that if there exists $c \in (a, b)$ such that $f'(c) = 0$,then $f(a) = f(b)$. This is not necessarily true. For example,if $f(x) = x^2$,$f'(x) = 2x$. Setting $f'(c) = 0$ gives $c = 0$. However,for any interval $[a, b]$ not containing $0$,the converse does not hold.