Verify Rolle’s theorem for the function f (x) = x2 + 2x – 8, x ∈ [– 4, 2].
The given function is f (x) = x2 + 2x – 8 and x ∈ [-4, 2].
By Rolle’s Theorem, for a function f : [a, b] → R, if
(a) f is continuous on [a, b]
(b) f is differentiable on (a, b)
(c) f(a) = f(b)
Then there exists some c in (a, b) such that f '(c) = 0.
As f(x) = x2 + 2x – 8 is a polynomial function,
(a) f(x) is continuous in [-4, 2]
(b) f'(x) = 2x + 2
So, f(x) is differentiable in (-4, 2).
(c) f(a) = f(-4) = (-4)2 + 2(-4) – 8 = 16 – 8 – 8 =16 – 16 = 0
f(b) = f(2) = (2)2 + 2(2) – 8 = 4 + 4 – 8 = 8 – 8 = 0
Hence, f(a) = f(b).
∴ There is a point c ∈ (-4, 2) where f'(c) = 0.
f(x) = x2 +2x – 8
f'(x) = 2x + 2
f'(c) = 0
⇒ f'(c) = 2c + 2 = 0
⇒2c = -2
⇒ c = -2 /2
⇒ c = -1 where c = -1 ∈ (-4, 2)
Hence, Rolle’s Theorem is verified.