Verify the Rolle’s theorem for each of the functions
f(x) = x (x – 1)2 in [0, 1].
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.