Answer :

First, let us write the conditions for the applicability of Rolle’s theorem:


For a Real valued function ‘f’:


a) The function ‘f’ needs to be continuous in the closed interval [a,b].


b) The function ‘f’ needs differentiable on the open interval (a,b).


c) f(a) = f(b)


Then there exists at least one c in the open interval (a,b) such that f’(c) = 0.


Given function is:


f(x) = log(x2 + 2) – log3 on [ – 1,1]


We know that logarithmic function is continuous and differentiable in its own domain.


We check the values of the function at the extremum,


f( – 1) = log(( – 1)2 + 2) – log3


f( – 1) = log(1 + 2) – log3


f( – 1) = log3 – log3


f( – 1) = 0


f(1) = log(12 + 2) – log3


f(1) = log(1 + 2) – log3


f(1) = log3 – log3


f(1) = 0


We have got f( – 1) = f(1). So, there exists a c such that cϵ( – 1,1) such that f’(c) = 0.


Let’s find the derivative of the function f,





We have f’(c) = 0



2c = 0


c = 0ϵ( – 1,1)


Rolle’s theorem is verified.


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