Condition (1):
Since, f(x) = x(x - 5)2 is a polynomial and we know every polynomial function is continuous for all x ϵ R.
⇒ 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. c = 5/3 or c = 5
Value of c = 5/3 ∈ (0, 5)
Thus, Rolle’s theorem is satisfied.