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) = 2x – x2 on [0, 1]
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) = 2x – x2 on [0, 1]
Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments.
Here, f(x) is a polynomial function. So it is continuous in [0, 1] and differentiable in (0, 1). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.
⇒ f’(c) = f(1) – f(0)
f(x) = 2x – x2
Differentiating with respect to x:
f’(x) = 2 – 2x
For f’(c), put the value of x=c in f’(x):
f’(c)= 2 – 2c
For f(1), put the value of x=1 in f(x):
f(1)= 2(1) – (1)2
= 2 – 1
= 1
For f(0), put the value of x=0 in f(x):
f(0) = 2(0) – (0)2
= 0 – 0
= 0
f’(c) = f(1) – f(0)
⇒ 2 – 2c = 1 – 0
⇒ – 2c = 1 – 2
⇒ – 2c = – 1
Hence, Lagrange’s mean value theorem is verified.