Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

f(x) = (x – 1) (x – 2)(x – 3) on [0, 4]

= (x² – x – 2x + 3) (x – 3)

= (x² – 3x + 3) (x – 3)

= x³ – 3x² + 3x – 3x² + 9x – 9

= x³ – 6x² + 12x – 9 on [0, 4]

Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.

Here, f(x) is a polynomial function. So it is continuous in [0, 4] and differentiable in (0, 4). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x³ – 6x² + 12x – 9

Differentiating with respect to x:

f’(x) = 3x² – 6(2x) + 12

= 3x² – 12x + 12

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

f’(c)= 3c² – 12c + 12

For f(4), put the value of x=4 in f(x):

f(4)= (4)³ – 6(4)² + 12(4) – 9

= 64 – 96 + 48 – 9

= 7

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

f(0)= (0)³ – 6(0)² + 12(0) – 9

= 0 – 0 + 0 – 9

= – 9

⇒ 3c² – 12c + 12 = 4

⇒ 3c² – 12c + 8 = 0

For quadratic equation, ax² + bx + c = 0

Hence, Lagrange’s mean value theorem is verified.