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 – 4x + 3 on [1,3]

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) = 1² – 4(1) + 3

⇒ f(1) = 1 – 4 + 3

⇒ f(1) = 0

⇒ f(3) = 3² – 4(3) + 3

⇒ f(3) = 9 – 12 + 3

⇒ f(3) = 0

∴ f(1) = f(3), Rolle’s theorem applicable for function ‘f’ on [1,3].

Let’s find the derivative of f(x):

⇒

⇒

⇒

⇒ f’(x) = 2x – 4

We have f’(c) = 0 cϵ(1,3), from the definition given above.

⇒ f’(c) = 0

⇒ 2c – 4 = 0

⇒ 2c = 4

⇒

⇒ C = 2ϵ(1,3)

∴ Rolle’s theorem is verified.