# 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.

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