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 c∈(– 1,1) such that f’(c) = 0.
Let’s find the derivative of the function f,
∴ Rolle’s theorem is verified.