Given, f(x) = x3(x - 1)2
f'(x) = 3x2(x - 1)2 + x3 2(x - 1) = x2(x - 1) (3(x - 1) + 2x)
or, f(x) = x2(x - 1)(5x - 3)
Clearly, f(x) is differentiable and continuous for all n.
So, (ii) f(x) is differentiable in (0,1) also
(iii) f(0) = 0 and f(1) = 0
∴ f(0) = f(1)
Hence, all the condition of Rolle's theorem are satisfied for f(x) in [0,1].
So, f'(C) = 0
c2(c - 1)(5c - 3) = 0
c = 0, 1, 3/5
But 0 < c < 1
∴ c = 3/5
Thus, there exist at least one c, c = 3/5 between 0 & 1 such that f'(C) = 0.
Hence, Rolle's theorem has been verified