582. Verify Rolle's Theorem for the function x² + 2 x - 8 in [-4, 2]

 f(x)=x² + 2 x - 8
This is polynomial in x.
(i) Sisce every polynomial in x is continuous function for all x.
∴ f is continuous in [-2, 2]
(ii) f'(x) = 2 x + 2, which exists in (-4, 2)
∴ f is differentiable in (-4, 2)
(iii) f(-4) = (-4)² + 2(-4)-8 = 16 - 8 - 8 = 0
f(2) = (2)² + 2 (2) - 8 = 4 + 4 - 8 = 0
∴ f(-4) = f(2).
∴ f satisfies all the conditions of Rolle's Theorem.
∴ there exists at least one value c of x such that f'(c) = 0, where - 4 < c < 2.
Now    f'(c) = 0 gives us 2 c + 2 = 0 or 2 c = -2
∴ c = - 1 ∈ (- 4, 2)
∴ Rolle's Theorem is verified.