Question

Verify Lagrange’s mean value theorem for the following functions : f(x) = (x – 1)(x – 2)(x – 3) on [0, 4].

Answer 1

The function f given as f(x) = (x – 1)(x – 2)(x – 3)
= (x – 1)(x² – 5x + 6)
= x³ – 5x² + 6x – x2 + 5x – 6
= x³ – 6x² + 11x – 6 is a polynomial function.
Hence, it is continuous on [0, 4] and differentiable on (0, 4).
Thus, the function f satisfies the conditions of Lagrange’s mean value theorem.
∴ there exists c ∈ (0, 4) such that

f'(c) = `(f(4) - f(0))/(4 - 0)`              ...(1)

Now, f(x) = (x – 1)(x – 2)(x – 3)
∴ f(0)
= (0 – 1)(0 – 2)(0 – 3)
= (– 1)(– 2)(– 3)
= – 6
and
f(4)
= (4 – 1)(4 – 2)(4 – 3)
= (3)(2)(1)
=  6

Also, f'(x) = `d/dx(x^3- 6x^2 + 11x - 6)`

= 3x² – 6 x  2x + 11 x 1 – 0
= 3x² – 12x + 11
∴ f'(c) = 3c² – 12x + 11

∴ from (1), 3c² – 12c + 11 = `(6 - (-6))/(4)`

∴ 3c² – 12c + 11 = 3
∴ 3c² – 12c + 8 = 0.

∴ c = `(12 ± sqrt(144 - 4(3)(8)))/(2(3)`

∴ c = `(12 ± sqrt(48))/(6)`

= `(12 ± 4sqrt(3))/(6)`

∴ c = `2 ± (2)/sqrt(3) ∈ (0, 4)`

Hence, Lagrange's mean value theorem is verified.