Question

Verify Rolle's theorem for the following function on the indicated interval f(x) = sin x − sin 2x on [0, π]?

Answer 1

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, 1]

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(π) = sin(π) − sin2(π)

⇒ f(π) = 0 − sin(2π)

⇒ f(π) = 0

We got f(0) = f(π). So, there exists a ce(0,m) such that f'(c) = 0.

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

⇒ f' (x) = `(s(sinx - sin2x))/dx`

⇒ f' (x) = cosx − cos2x `(d(2x))/dx`

⇒ f' (x) = cosx − 2cos2x

⇒ f' (x) = cosx − 4cos2x + 2

We have f' (c) = 0

⇒ cosc − 4cos2c + 2 = 0

⇒ cosc = `(-1±sqrt((1)^2 - (4 xx -4 xx 2)))/(2 xx -4)`

⇒ cosc = `(-1±sqrt(1 + 33))/(-8)`

⇒ c = `cos^-1 ((-1 ± sqrt33)/(-8))`

We can see that C∈ (0, π)

∴ Rolle's theorem is verified.