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.