Question

Examine if Rolles theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolles theorem from these example ?
(i) $f\left(x\right)=\left[x\right]$ for $x\in [5,9]$
(ii) $f\left(x\right)=\left[x\right]$ for $x\in [-2,2]$
(iii) $f\left(x\right)=x$ ${}^2-1$ for $x\in [1,2]$

Answer 1

Rolle's theorem holds for a function $f:[a,b]\to R$, if following three conditions holds-

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

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

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

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

(i)
Given $f\left(x\right)=\left[x\right]$ for $x\in$ [5, 9]
Since, the greatest integer function is not continuous at integral values.
So, $f\left(x\right)$ is not continuous at $x=5,6,7,8,9$
So, $f\left(x\right)$ is not continuous in $[5,9]$.

Since, condition (1) does not holds ,so need to check the other conditions.

Hence, Rolle's theorem is not applicable on given function

(ii)
Given function $f\left(x\right)=\left[x\right]$ for $x\in [-2,2]$
Since, the greatest integer function is not continuous at integral points.
So, $f\left(x\right)$ is not continuous at $x=-2,-1,0,1,2$
$f\left(x\right)$ is not continuous in $[-2,2]$.

(iii)

$f\left(x\right)=x^2-1$ for $x\in [1,2]$
It is evident that $f$, being a polynomial function, is continuous in $[1,2]$ and is differentiable in $(1,2).$
Also $f\left(1\right)=(1{)}^2-1=0$
and $f\left(2\right)=(2{)}^2-1=3$
$\therefore f\left(1\right)\ne f\left(2\right)$
It is observed that $f$ does not satisfy a condition of the hypothesis of Rolles Theorem.
Hence, Rolles Theorem is not applicable for $f\left(x\right)=x^2-1$ for $x\in [1,2].$