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According to Rolle's theorem function must be
such that 
![f (x) = [x]](https://learn.careers360.com/latex-image/?f%20%28x%29%20%3D%20%5Bx%5D)
![f (x) = [x]](https://learn.careers360.com/latex-image/?f%20%28x%29%20%3D%20%5Bx%5D)
is not continuous for each and every point in [-2,2]![n \ \epsilon \ [-2,2]](https://learn.careers360.com/latex-image/?n%20%5C%20%5Cepsilon%20%5C%20%5B-2%2C2%5D)
![\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}](https://learn.careers360.com/latex-image/?%5Clim_%7Bh%5Crightarrow%200%5E-%7D%5Cfrac%7Bf%28n+h%29-f%28n%29%7D%7Bh%7D%20%3D%20%5Clim_%7Bh%5Crightarrow%200%5E-%7D%5Cfrac%7B%5Bn+h%5D-%5Bn%5D%7D%7Bh%7D%20%3D%20%5Clim_%7Bh%5Crightarrow%200%5E-%7D%5Cfrac%7Bn-1-n%7D%7Bh%7D%20%3D%20%5Clim_%7Bh%5Crightarrow%200%5E-%7D%5Cfrac%7B-1%7D%7Bh%7D)
![( [n+h]=n-1 \because h < 0 \ therefore \ (n+h)<n)](https://learn.careers360.com/latex-image/?%28%20%5Bn+h%5D%3Dn-1%20%5Cbecause%20h%20%3C%200%20%5C%20therefore%20%5C%20%28n+h%29%3Cn%29)
![n \ \epsilon \ [-2,2]](https://learn.careers360.com/latex-image/?n%20%5C%20%5Cepsilon%20%5C%20%5B-2%2C2%5D)
![\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](https://learn.careers360.com/latex-image/?%5Clim_%7Bh%5Crightarrow%200%5E+%7D%5Cfrac%7Bf%28n+h%29-f%28n%29%7D%7Bh%7D%20%3D%20%5Clim_%7Bh%5Crightarrow%200%5E+%7D%5Cfrac%7B%5Bn+h%5D-%5Bn%5D%7D%7Bh%7D%20%3D%20%5Clim_%7Bh%5Crightarrow%200%5E+%7D%5Cfrac%7Bn-n%7D%7Bh%7D%20%3D%20%5Clim_%7Bh%5Crightarrow%200%5E-%7D%5Cfrac%7B0%7D%7Bh%7D%3D0)
![( [n+h]=n \because h > 0 \ therefore \ (n+h)>n)](https://learn.careers360.com/latex-image/?%28%20%5Bn+h%5D%3Dn%20%5Cbecause%20h%20%3E%200%20%5C%20therefore%20%5C%20%28n+h%29%3En%29)
![f (x) = [x]](https://learn.careers360.com/latex-image/?f%20%28x%29%20%3D%20%5Bx%5D)
, ![x \ \epsilon \ [-2,2]](https://learn.careers360.com/latex-image/?x%20%5C%20%5Cepsilon%20%5C%20%5B-2%2C2%5D)
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
If all these conditions are satisfies then we can verify Rolle's theorem
Given function is
It is clear that Given function
Now, lets check differentiability of f(x)
L.H.L. at x = n ,
Now,
R.H.L. at x = 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