 # Examine if Rolle's theorem is applicable to any one of the following functions. `
Question:

Examine if Rolle's theorem is applicable to any one of the following functions.

(i) $f(x)=[x]$ for $x \in[5,9]$

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

Can you say something about the converse of Rolle's Theorem from these functions?

Solution:

By Rolle's theorem, for a function $f:[a, \mathrm{~b}] \rightarrow \mathbf{R}$, if

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

(b) $f$ is differentiable on $(a, b)$ and

(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.

(i) $f(x)=[x]$ for $x \in[5,9]$

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

In particular, f(x) is not continuous at = 5 and = 9.

Thus, f (x) is not continuous on [5, 9].

Also, $f(5)==5$ and $f(9)==9$

$\therefore f(5) \neq f(9)$

The differentiability of f on (5, 9) is checked in the following way.

Let be an integer such that n ∈ (5, 9).

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

$\lim _{h \rightarrow 0^{-}} \frac{f(n+h)-f(n)}{h}=\lim _{h \rightarrow 0^{-}} \frac{[n+h]-[n]}{h}=\lim _{h \rightarrow 0^{-}} \frac{n-1-n}{h}=\lim _{h \rightarrow 0^{-}} \frac{-1}{h}=\infty$

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

$\lim _{h \rightarrow 0^{+}} \frac{f(n+h)-f(n)}{h}=\lim _{h \rightarrow 0^{+}} \frac{[n+h]-[n]}{h}=\lim _{h \rightarrow 0^{+}} \frac{n-n}{h}=\lim _{h \rightarrow 0^{+}} 0=0$

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

Thus, is not differentiable on (5, 9).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.

Hence, Rolle’s theorem is not applicable on  $f(x)=[x]$ for $x \in[5,9]$

(ii) $f(x)=[x]$ 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 = −2 and = 2.

Thus, f (x) is not continuous on [−2, 2].

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

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

The differentiability of f on (−2, 2) is checked in the following way.

Let be an integer such that n ∈ (−2, 2).

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

$\lim _{h \rightarrow 0^{-}} \frac{f(n+h)-f(n)}{h}=\lim _{h \rightarrow 0^{-}} \frac{[n+h]-[n]}{h}=\lim _{h \rightarrow 0^{-}} \frac{n-1-n}{h}=\lim _{h \rightarrow 0^{-}} \frac{-1}{h}=\infty$

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

$\lim _{h \rightarrow 0^{+}} \frac{f(n+h)-f(n)}{h}=\lim _{h \rightarrow 0^{+}} \frac{[n+h]-[n]}{h}=\lim _{h \rightarrow 0^{+}} \frac{n-n}{h}=\lim _{h \rightarrow 0^{+}} 0=0$

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

Thus, is not differentiable on (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.

Hence, Rolle’s theorem is not applicable on $f(x)=[x]$ for $x \in[-2,2]$