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) = (x² – 1)(x – 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)² – 1)( – 1 – 2)

⇒ f( – 1) = (1 – 1)( – 3)

⇒ f( – 1) = (0)( – 3)

⇒ f( – 1) = 0

⇒ f(2) = (2² – 1)(2 – 2)

⇒ f(2) = (4 – 1)(0)

⇒ 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)×2x) + ((x² – 1)×1)

⇒ f’(x) = 2x² – 4x + x² – 1

⇒ f’(x) = 2x² – 4x – 1

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

⇒ f’(c) = 0

⇒ 2c² – 4c – 1 = 0

⇒

⇒

⇒

⇒

⇒

⇒ ϵ( – 1,2)

∴ Rolle’s theorem is verified.