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 – 1)(x – 2)² on [1,2]

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(1) = (1 – 1)(1 – 2)²

⇒ f(1) = 0(1)²

⇒ f(1) = 0

⇒ f(2) = (2 – 1)(2 – 2)²

⇒ f(2) = 1.0²

⇒ f(2) = 0

∴ f(1) = f(2), Rolle’s theorem applicable for function ‘f’ on [1,2].

Let’s find the derivative of f(x):

⇒

Differentiating using UV rule

⇒

⇒ f’(x) = ((x – 2)²×1) + ((x – 1)×2×(x – 2))

⇒ f’(x) = x² – 4x + 4 + 2(x² – 3x + 2)

⇒ f’(x) = 3x² – 10x + 8

We have f’(c) = 0 cϵ(1,2), from the definition given above.

⇒ f’(c) = 0

⇒ 3c² – 10c + 8 = 0

⇒

⇒

⇒

⇒

⇒ ϵ(1,2) (neglecting the value 2)

∴ Rolle’s theorem is verified.