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]

Answers (1)

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)<n)
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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