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. or c=5

Value of

Thus, Rolle’s theorem is satisfied.

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