Question

verify lagrange's Mean Value Theorem for the function :

f(x) = x (x - 1)(x - 2) in the interval [0,1/2]​

Answer 1

Answer:

We have f(x) = x (x – 1) (x – 2) = x3 – 3 x2 + 2x .

(1) f(x) is polynomial function and is continuous in [0, 1/2] .

(2) f’(x) = 3x2 – 6x + 2 , which exists and hence is differentiable in (0, 1/2) .

Now f’(c) = 3c2 – 6c + 2 , f(0) = 0 , f(1/2) = 1/2(1/2 – 1) (1/2 – 2) = 3/8 .

[f(b) – f(a)]/(b – a) = f’(c)

Or, (3/8 – 0)/(1/2 – 0)= 3c2 – 6c + 2

Or, 3/4 = 3 c2 – 6c + 2

Or, 12 c2 – 24 c + 5 = 0

Or, c = [24 ± √(576 – 240)] /24 = [24 ± √ 336]/24 = 1 ± [(√21)/6] .

Or, c = 1 – √21 /6 = 0.236 (approx).

And c = 1 + √21/6 > 1/2 , is not acceptable.