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