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 – 5x + 4 on [1,4]

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 – 5(1) + 4

⇒ f(1) = 1 – 5 + 4

⇒ f(1) = 0

⇒ f(4) = 4² – 5(4) + 4

⇒ f(4) = 16 – 20 + 4

⇒ f(4) = 0

∴ We got f(1) = f(4). So, there exists a cϵ(1,4) such that f’(c) = 0.

Let’s find the derivative of f(x):

⇒

⇒

⇒

⇒ f’(x) = 2x – 5

We have f’(c) = 0

⇒ f’(c) = 0

⇒ 2c – 5 = 0

⇒ 2c = 5

⇒

⇒ C = 2.5ϵ(1,4)

∴ Rolle’s theorem is verified.