Verify Rolle’s theorem for each of the following functions on the indicated intervals :

f(x) = x(x – 1)2 on [0, 1]

First let us write the conditions for the applicability of Rolle’s theorem:


For a Real valued function ‘f’:


a) The function ‘f’ needs to be continuous in the closed interval [a,b].


b) The function ‘f’ needs differentiable on the open interval (a,b).


c) f(a) = f(b)


Then there exists at least one c in the open interval (a,b) such that f’(c) = 0.


Given function is:


f(x) = x(x – 1)2 on [0,1]


Since, given function f is a polynomial it is continuous and differentiable everywhere i.e, on R.


Let us find the values at extremums:


f(0) = 0(0 – 1)2


f(0) = 0


f(1) = 1(1 – 1)2


f(1) = 02


f(1) = 0


f(0) = f(1), Rolle’s theorem applicable for function ‘f’ on [0,1].


Let’s find the derivative of f(x):



Differentiating using UV rule,



f’(x) = ((x – 1)2×1) + (x×2×(x – 1))


f’(x) = (x – 1)2 + 2(x2 – x)


f’(x) = x2 – 2x + 1 + 2x2 – 2x


f’(x) = 3x2 – 4x + 1


We have f’(c) = 0 cϵ(0,1), from the definition given above.


f’(c) = 0


3c2 – 4c + 1 = 0






ϵ(0,1)


Rolle’s theorem is verified.


2