Q2 (3)   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^2 - 1 \: \:for \: \: x \epsilon [ 1,2]


 

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 there exist a c \ \epsilon \ (x,y)  such that   f^{'}(c)= 0
If all these conditions are satisfied then we can verify  Rolle's theorem 
Given function is 
 f (x) = x^2-1 
Now, being a polynomial , function f (x) = x^2-1 is continuous in [1,2] and differentiable in(1,2)
Now,
f(1)=1^2-1 = 1-1 = 0
And
f(2)=2^2-1 = 4-1 = 3
Therefore,  f(1)\neq f(2)
Therefore, All conditions are not satisfied
Hence, Rolle's theorem is not applicable for given function   f (x) = [x]    ,  x \ \epsilon \ [-2,2]

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