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)

\(\Rightarrow f'(c)=\frac{f(b) - f(a)}{b-a}\)

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.

Therefore, there exists a point c∈(0, 1) such that:

\(f'(c)=\frac{f(1) - f(0)}{1-0}\)

⇒ 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

\(\Rightarrow c=\frac{-1}{-2}=\frac{1}{2}∈(0, 1)\)

Hence, Lagrange’s mean value theorem is verified.