Verify the Rolle’s Theorem for functions on the indicated intervals:

1.	Verify the Rolle’s Theorem for functions on the indicated intervals: f(x) = 2 sin x + sin 2x on [0, π]
Answer» 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's theorem is verified.

Verify the Rolle’s Theorem for functions on the indicated intervals: f(x) = 2 sin x + sin 2x on [0, π]

Answer»

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's theorem is verified.