Answer :

Given that f is continuous and differentiable in the interval [ – 5,5].


It is also given that f’(x) doesn’t vanish anywhere.


According to Rolle’s theorem for a differentiable function on [a,b] will have atleast one cϵ(a,b) such that f’(c) = 0, if the following condition had satisfied:


f(a) = f(b).


According to the problem it is given for any value of x, say r the values never equals to zero.


f’(r)≠0


This is possible when Rolle’s theorem is not applicable.


Let us Recap the 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.


First, two conditions are satisfied according to the problem, so the only condition that cannot be satisfied is (c).


So, we can clearly say that f( – 5)≠f(5).


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