Answer :

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) = sinx – sin2x on [0,]

We know that sine function is continuous and differentiable over R.

Let’s check the values of the function ‘f’ at the extremums.

f(0) = sin(0)–sin2(0)

f(0) = 0 – sin(0)

f(0) = 0

f() = sin() – sin2()

f() = 0 – sin(2)

f() = 0

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

Let’s find the derivative of the function ‘f’

f’(x) = cosx – 2(2cos2x – 1)

f’(x) = cosx – 4cos2x + 2

We have f’(c) = 0

cosc – 4cos2c + 2 = 0

We can see that cϵ(0,)

Rolle’s theorem is verified.

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