Question

Verify Rolle’s Theorem for the function f(x)=x^2 +2x-8, x∈[-4,2]

Answer 1

$\textbf{\underline{according to Rolle's theorem,}}\\\textbf{for a function} f:[a,b]\rightarrow\mathbb{R}\\\textbf{if}\\\textbf{(a) f is continuous in [a,b]}\\\textbf{(b) f is differentiable in (a,b)}\\\textbf{f(a)=f(b)}\\\textbf{Then there exists some c in}\\\textbf{ (a, b) such that f '(c) = 0.}$


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
$\textbf{\underline{Rolle's theorem is verified}}$

Answer 2

hope this will help you