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Verify Rolle’s theorem for each of the following functions on the indicated intervals :

f(x) = (x2 – 1)(x – 2) on [– 1, 2]

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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.

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