Verify mean value theorem for f(x) = (x – 3)(x – 6)(x – 9) in [3, 5].

Given: f(x) = (x – 3)(x – 6)(x – 9) in [3, 5]

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 – 3)(x – 6)(x – 9) on [3, 5]

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

Here, f(x) is a polynomial function. So it is continuous in [3, 5] and differentiable in (3, 5). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = (x – 3)(x – 6)(x – 9) = x³ – 18x² + 99x – 162

Differentiating with respect to x:

f’(x) = 3x² – 2(18x) + 99

= 3x² – 36x + 99

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

f’(c)= 3c² – 36c + 99

For f(3), put the value of x=3 in f(x):

f(3)= (3 – 3)(3 – 6)(3 – 9)

= 0

For f(5), put the value of x=5 in f(x):

f(5)= (5 – 3)(5 – 6)(5 – 9)

= (2)(-1)(-4)

= 6

⇒ 3c² – 36c + 99 = 3

⇒ 3(c² – 12c + 33) = 3

⇒ c² – 12c + 33 = 3

⇒ c² – 12c + 33 – 1 = 0

⇒ c² – 12c + 32 = 0

⇒ c² – 8c – 4c + 32 = 0

⇒ c(c – 8) – 4(c – 8) = 0

⇒ (c – 8) (c – 4) = 0

⇒ c = 4 or 8

As, 3 < c < 5

⇒ c = 4

Hence, Lagrange’s mean value theorem is verified.