ચકાસો કે આપેલ વિધેયમાં રોલનું પ્રમેય લગાડી શકાય કે નહિ : $f(x)=[x],$ $x \in[-2,2]$

By Rolle's Theorem, for a function $f:[a, b] \rightarrow R,$ if

a) $f$ is continuous on $[a, b]$

b) $f$ is continuous on $(a, b)$

c) $f(a)=f(b)$

Then, there exists some $c \in(a, b)$ such that $f^{\prime}(c)=0$

Therefore, Rolle's Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

$f(x)=[x]$ for $x \in[-2,2]$

It is evident that the given function $f(x)$ is not continuous at every integral point.

In particular, $f(x)$ is not continuous at $x=-2$ and $x=2$

$\Rightarrow f=(x)$ is not continuous in $[-2,2]$

Also, $f(-2)=[2]=-2$ and $f(2)=[2]=2$

$\therefore f(-2) \neq f(2)$

The differentiability of in $(-2,2)$ is checked as follows.

Let $\mathrm{n}$ be an integer such that $n \in(-2,2)$

The left hand limit of $f$ at $x=\mathrm{n}$ is,

$\mathop {\lim }\limits_{h \to {0^\prime }} \frac{{f(n + h) - f(n)}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} \frac{{[n + h] - [n]}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} \frac{{n - 1 - n}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} \frac{{ - 1}}{h} = \infty $

The right hand limit of $f$ at $x=n$ is,

$\mathop {\lim }\limits_{h \to {0^\prime }} \frac{{f(n + h) - f(n)}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} \frac{{[n + h] - [n]}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} \frac{{n - n}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} 0 = 0$

Since the left and right hand limits of $f$ at $x=n$ are not equal, $f$ is not differentiable at $x=n$

$\therefore f$ is not continuous in $(-2,2).$

It is observed that $f$ does not satisfy all the conditions of the hypothesis of Rolle's Theorem.

Hence, Roller's Theorem is not applicable for $f(x)=[x]$ for $x \in[-2,2]$