Answer :

Condition (1):


Since, f(x)= x2-5x+6 is a polynomial and we know every polynomial function is continuous for all xϵR.


f(x)= x2-5x+6 is continuous on [-3,6].


Condition (2):


Here, f’(x)=2x-5 which exist in [-3,6].


So, f(x)= x2-5x+6 is differentiable on (-3,6).


Condition (3):


Here, f(-3)=(-3)2-5×(-3)+6=30


And f(6)= 62-5×6+6=12


i.e. f(-3)≠f(6)


Conditions (3) of Rolle’s theorem is not satisfied.


So, Rolle’s theorem is not applicable.


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