Q

# 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 example? f (x) = [x] for x Î [– 2, 2]

Q2 (2)   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 example?

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

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According to Rolle's theorem function must be
a )  continuous in given closed interval say [x,y]
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then their exist a $c \ \epsilon \ (x,y)$  such that   $f^{'}(c)= 0$
If all these conditions are satisfies then we can verify  Rolle's theorem
Given function is
$f (x) = [x]$
It is clear that Given function $f (x) = [x]$ is not  continuous for each and every point in [-2,2]
Now, lets check differentiability of f(x)
L.H.L. at x = n , $n \ \epsilon \ [-2,2]$
$\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$
$( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)
Now,
R.H.L. at x = n , $n \ \epsilon \ [-2,2]$
$\lim_{h\rightarrow 0^+}\frac{f(n+h)-f(n)}{h} = \lim_{h\rightarrow 0^+}\frac{[n+h]-[n]}{h} = \lim_{h\rightarrow 0^+}\frac{n-n}{h} = \lim_{h\rightarrow 0^-}\frac{0}{h}=0$
$( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)$
We can clearly see that R.H.L. is not equal to L.H.L.
Therefore, function is not differential in (-2,2)
Hence, Rolle's theorem is not applicable for given function   $f (x) = [x]$    ,  $x \ \epsilon \ [-2,2]$

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