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.