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 – 1)(x – 2) on [– 1, 2]
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) = (( – 1)2 – 1)( – 1 – 2)
⇒ f(– 1) = (1 – 1)(– 3)
⇒ f(– 1) = (0)(– 3)
⇒ f(– 1) = 0
⇒ f(2) = (22 – 1)(2 – 2)
⇒ f(2) = (4 – 1)(0)
⇒ f(2) = 0
∴ f(– 1) = f(2), Rolle’s theorem applicable for function ‘f’ on [– 1, 2].
Let’s find the derivative of f(x):
\(\Rightarrow f'(x) = \frac{d((x^2-1)(x-2))}{dx}\)
Differentiating using UV rule,
We have f’(c) = 0 c∈(– 1,2), from the definition given above.
∴ Rolle’s theorem is verified.