Examine if Rolle's Theorem is applicable to the following function. Can you say something about the converse of Rolle's Theorem from this example?
$f(x) = [x]$ for $x \in [-2, 2]$

(N/A) According to Rolle's Theorem,for a function $f: [a, b] \rightarrow \mathbb{R}$,if:
$a)$ $f$ is continuous on $[a, b]$
$b)$ $f$ is differentiable on $(a, b)$
$c)$ $f(a) = f(b)$
Then,there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.
For the function $f(x) = [x]$ on the interval $[-2, 2]$:
$1.$ The greatest integer function $[x]$ is discontinuous at all integral points. Since the interval $[-2, 2]$ contains integers like $-1, 0, 1$,the function $f(x)$ is not continuous on $[-2, 2]$.
$2.$ The function is also not differentiable at integral points in the interval $(-2, 2)$.
$3.$ $f(-2) = [-2] = -2$ and $f(2) = [2] = 2$. Thus,$f(-2) \neq f(2)$.
Since the function fails to satisfy all the conditions of Rolle's Theorem,the theorem is not applicable to $f(x) = [x]$ on $[-2, 2]$.