Given as f(x) = x (x – 1) on [1, 2]
= x2 – x
The every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. Therefore it is continuous in [1, 2] and differentiable in (1, 2). Therefore both the necessary conditions of Lagrange’* mean value theorem is satisfied.
So, there exist a point c ∈ (1, 2) such that:
f'(c) = (f(2) - f(1))/(2 - 1)
f'(c) = (f(2) - f(1))/1
f (x) = x2 – x
Differentiate with respect to x
f’(x) = 2x – 1
For the f’(c), put the value of x=c in f’(x):
f’(c)= 2c – 1
For the f(2), put the value of x = 2 in f(x)
f (2) = (2)2 – 2
= 4 – 2
= 2
For the f(1), put the value of x = 1 in f(x):
f (1) = (1)2 – 1
= 1 – 1
= 0
∴ f’(c) = f(2) – f(1)
⇒ 2c – 1 = 2 – 0
⇒ 2c = 2 + 1
⇒ 2c = 3
c = (3/2) ∈ (1,2)
Thus, lagrange'* mean value theorem is verified.
