Answer :

Condition (1):


Since, f(x)=x(x-4)2 is a polynomial and we know every polynomial function is continuous for all xϵR.


f(x)= x(x-4)2 is continuous on [0,4].


Condition (2):


Here, f’(x)= (x-4)2+2x(x-4) which exist in [0,4].


So, f(x)= x(x-4)2 is differentiable on (0,4).


Condition (3):


Here, f(0)=0(0-4)2=0


And f(4)= 4(4-4)2=0


i.e. f(0)=f(4)


Conditions of Rolle’s theorem are satisfied.


Hence, there exist at least one cϵ(0,4) such that f’(c)=0


i.e. (c-4)2+2c(c-4)=0


i.e. (c-4)(3c-4)=0


i.e. c=4 or c=3÷4


Value of


Thus, Rolle’s theorem is satisfied.


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