# Examine if Rolle’s Theorem is applicable to any of the following functions.

Question:

Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s Theorem from these examples?

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

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

(iii) $f(x)=x^{2}-1$ for $x \in[1,2]$

Solution:

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

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

(b) $f$ is differentiable 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.

(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

$\Rightarrow f(x)$ is not continuous in $[5,9] .$

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

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

The differentiability of $f$ in $(5,9)$ is checked as follows.

Let $n$ be an integer such that $n \in(5,9)$.

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,

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

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

is not differentiable in (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 for $f(x)=[x]$ for $x \in[5,9]$.

(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 = −2 and = 2

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

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

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

The differentiability of f in (−2, 2) is checked as follows.

Let $n$ be an integer such that $n \in(-2,2)$.

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 right hand limits of f at x = n are not equal, f is not differentiable at x = n

is not differentiable in (−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 for $f(x)=[x]$ for $x \in[-2,2]$.

(iii) $f(x)=x^{2}-1$ for $x \in[1,2]$

It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).

$f(1)=(1)^{2}-1=0$

$f(2)=(2)^{2}-1=3$

$\therefore f(1) \neq f(2)$

It is observed that f does not satisfy a condition of the hypothesis of Rolle’s Theorem.

Hence, Rolle's Theorem is not applicable for $f(x)=x^{2}-1$ for $x \in[1,2]$.