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) = log(x² + 2) – log3 on [ – 1,1]

We know that logarithmic function is continuous and differentiable in its own domain.

We check the values of the function at the extremum,

⇒ f( – 1) = log(( – 1)² + 2) – log3

⇒ f( – 1) = log(1 + 2) – log3

⇒ f( – 1) = log3 – log3

⇒ f( – 1) = 0

⇒ f(1) = log(1² + 2) – log3

⇒ f(1) = log(1 + 2) – log3

⇒ f(1) = log3 – log3

⇒ f(1) = 0

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

Let’s find the derivative of the function f,

⇒

⇒

⇒

We have f’(c) = 0

⇒

⇒ 2c = 0

⇒ c = 0ϵ( – 1,1)

∴ Rolle’s theorem is verified.