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(x – 4)² on [0,4]

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(0) = 0(0 – 4)²

⇒ f(0) = 0

⇒ f(4) = 4(4 – 4)²

⇒ f(4) = 4(0)²

⇒ f(4) = 0

∴ f(0) = f(4), Rolle’s theorem applicable for function ‘f’ on [0,4].

Let’s find the derivative of f(x):

⇒

Differentiating using UV rule,

⇒

⇒ f’(x) = ((x – 4)²×1) + (x×2×(x – 4))

⇒ f’(x) = (x – 4)² + 2(x² – 4x)

⇒ f’(x) = x² – 8x + 16 + 2x² – 8x

⇒ f’(x) = 3x² – 16x + 16

We have f’(c) = 0 cϵ(0,4), from the definition given above.

⇒ f’(c) = 0

⇒ 3c² – 16c + 16 = 0

⇒

⇒

⇒

⇒

⇒ ϵ(0,4)

∴ Rolle’s theorem is verified.