Verify Lagrange’s mean value theorem for the following functions on the indicated intervals. In each find a point ‘c’ in the indicated interval as stated by the Lagrange’s mean value theorem :
f(x) = x (x – 1) on [1, 2]
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 (x – 1) on [1, 2]
= x2 – x
Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.
Here, f(x) is a polynomial function. So it is continuous in [1, 2] and differentiable in (1, 2). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.
f(x) = x2 – x
Differentiating with respect to x:
f’(x) = 2x – 1
For f’(c), put the value of x=c in f’(x):
f’(c)= 2c – 1
For f(2), put the value of x=2 in f(x):
f(2)= (2)2 – 2
= 4 – 2
= 2
For 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
Hence, Lagrange’s mean value theorem is verified.