Given as f(x) = x³ – 2x² – x + 3 on [0, 1]

The every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. Therefore it is continuous in [0, 1] and differentiable in (0, 1). Therefore both the necessary conditions of Lagrange’s mean value theorem is satisfied.

Therefore, there exist a point c ∈ (0, 1) such that:

f'(c) = (f(1) - f(0))/1 - 0

f'(c) = (f(1) - f(0))/1

f (x) = x³ – 2x² – x + 3

Differentiate with respect to x

f’(x) = 3x² – 2(2x) – 1

= 3x² – 4x – 1

For the f’(c), put the value of x = c in f’(x)

f’(c)= 3c² – 4c – 1

For the f(1), put the value of x = 1 in f(x)

f (1)= (1)³ – 2(1)² – (1) + 3

= 1 – 2 – 1 + 3

= 1

For the f(0), put the value of x = 0 in f(x)

f (0)= (0)³ – 2(0)² – (0) + 3

= 0 – 0 – 0 + 3

= 3

∴ f’(c) = f(1) – f(0)

⇒ 3c² – 4c – 1 = 1 – 3

⇒ 3c² – 4c = 1 + 1 – 3

⇒ 3c² – 4c = – 1

⇒ 3c² – 4c + 1 = 0

⇒ 3c² – 3c – c + 1 = 0

⇒ 3c(c – 1) – 1(c – 1) = 0

⇒ (3c – 1) (c – 1) = 0

c = 1/3, 1

c = (1/3) ∈ (0,1)

Thus, lagrange's mean value theorem is verified.