Given: f(x) = x² – 4x – 3 in the interval [1, 4]

Mean Value Theorem states that for a function f : [a, b] → R, if

(a)f is continuous on [a, b]

(b)f is differentiable on (a, b)

Then there exists some c ∈ (a, b) such that

As f(x) is a polynomial function,

(a) f(x) is continuous in [1, 4]

(b) f'(x) = 2x – 4

So, f(x) is differentiable in (1, 4).

∴

f(4) = 4² – 4(4) – 3 = 16 – 16 – 3 = -3

f(1) = 1² – 4(1) – 3 = 1 – 4 – 3 = -6

∴ There is a point c ∈ (1, 4) such that f'(c) = 1

⇒ f'(c) = 1

⇒ 2c – 4 = 1

⇒ 2c = 1+4 =5

⇒ c = 5/2 where c ∈ (1,4)

The Mean Value Theorem is verified for the given f(x).