Q. 655.0( 1 Vote )

Verify the Rolle’s theorem for each of the functions

f(x) = x (x – 1)2 in [0, 1].

Answer :

Given: f(x) = x(x – 1)2

f(x) = x (x2 + 1 – 2x)


f(x) = x3 + x – 2x2 in [0,1]


Now, we have to show that f(x) verify the Rolle’s Theorem


First of all, Conditions of Rolle’s theorem are:


a) f(x) is continuous at (a,b)


b) f(x) is derivable at (a,b)


c) f(a) = f(b)


If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0


Condition 1:


On expanding f(x) = x(x – 1)2, we get f(x) = x3 + x – 2x2


Since, f(x) is a polynomial and we know that, every polynomial function is continuous for all x R


f(x) = x3 + x – 2x2 is continuous at x [0,1]


Hence, condition 1 is satisfied.


Condition 2:


f(x) = x3 + x – 2x2


Since, f(x) is a polynomial and every polynomial function is differentiable for all x R


f(x) is differentiable at [0,1]


Hence, condition 2 is satisfied.


Condition 3:


f(x) = x3 + x – 2x2


f(0) = 0


f(1) = (1)3 + (1) – 2(1)2 = 1 + 1 – 2 = 0


Hence, f(0) = f(1)


Hence, condition 3 is also satisfied.


Now, let us show that c (0,1) such that f’(c) = 0


f(x) = x3 + x – 2x2


On differentiating above with respect to x, we get


f’(x) = 3x2 + 1 – 4x


Put x = c in above equation, we get


f’(c) = 3c2 + 1 – 4c


, all the three conditions of Rolle’s theorem are satisfied


f’(c) = 0


3c2 + 1 – 4c = 0


On factorising, we get


3c2 – 3c – c + 1 = 0


3c(c – 1) – 1(c – 1) = 0


(3c – 1) (c – 1) = 0


(3c – 1) = 0 or (c – 1) = 0



So, value of


Thus, Rolle’s theorem is verified.


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