Given as the function is f (x) = 2sinx + sin2x on [0, π]
As we know that sine function continuous and differentiable over R.
Let us check the values of function f at the extremes
⇒ f (0) = 2sin(0) + sin2(0)
⇒ f (0) = 2(0) + 0
⇒ f (0) = 0
⇒ f (π) = 2sin(π) + sin2(π)
⇒ f (π) = 2(0) + 0
⇒ f (π) = 0
f(0) = f(π), therefore there exist a c belongs to (0, π) such that f’(c) = 0.
Let us find the derivative of function f.
f'(x) = d(2sin x + sin 2x)/dx
f'(x) = 2cos x + cos 2x(d(2x)/dx)
⇒ f’(x) = 2cosx + 2cos2x
⇒ f’(x) = 2cosx + 2(2cos2x – 1)
⇒ f’(x) = 4 cos2x + 2 cos x – 2
f'(c) = 0, from definition
⇒ 4cos2c + 2 cos c – 2 = 0
⇒ 2cos2c + cos c – 1 = 0
⇒ 2cos2c + 2 cos c – cos c – 1 = 0
⇒ 2 cos c (cos c + 1) – 1 (cos c + 1) = 0
⇒ (2cos c – 1) (cos c + 1) = 0
cos c = 1/2 or cos c = -1
c = (π/3) ∈ (0,π)
Thus, Rolle'* theorem is verified.
