Question

Verify Rolle’s theorem for the function f (x) = x² + 2x – 8, x ∈ [– 4, 2].

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Let's check all conditions :

as f(x) = x² + 2x - 8 is a polynomial function.

f(x) is continuous for all real value of x

hence, f(x) is continuous in [ -4 , 2]

we know, every polynomial function are differentiable. e.g., f'(x) = 2x + 2

hence, f(x) is differentiable in [-4, 2]

now, f(-4) = (-4)² + 2(-4) -8

= 16 - 8 - 8 = 16 - 16 = 0

f(2) = (2)² + 2(2) - 8

= 4 + 4 - 8 = 8 - 8 = 0

hence, f(-4) = f(2)

hence, a point c exists in (-4,2) in such that f'(c) = 0.

because f'(x) = 2x + 2

put x = c , f'(c) = 2c + 2 = 0

c = -1

hope it helps you...