Answer :

Condition (1):


Since, f(x)=x(x-5)2 is a polynomial and we know every polynomial function is continuous for all xϵR.


f(x)= x(x-5)2 is continuous on [0,5].


Condition (2):


Here, f’(x)= (x-5)2+ 2x(x-5) which exist in [0,5].


So, f(x)= x(x-5)2 is differentiable on (0,5).


Condition (3):


Here, f(0)= 0(0-5)2=0


And f(5)= 5(5-5)2=0


i.e. f(0)=f(5)


Conditions of Rolle’s theorem are satisfied.


Hence, there exist at least one cϵ(0,5) such that f’(c)=0


i.e. (c-5)2+ 2c(c-5)=0


i.e.(c-5)(3c-5)=0


i.e. or c=5


Value of


Thus, Rolle’s theorem is satisfied.


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