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) = 2x² – 3x + 1 on [1, 3]

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, 3] and differentiable in (1, 3). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = 2x² – 3x + 1

Differentiating with respect to x:

f’(x) = 2(2x) – 3

= 4x – 3

For f’(c), put the value of x=c in f’(x):

f’(c)= 4c – 3

For f(3), put the value of x=3 in f(x):

f(3)= 2(3)² – 3(3) + 1

= 2(9) – 9 + 1

= 18 – 8 = 10

For f(1), put the value of x=1 in f(x):

f(1) = 2(1)² – 3(1) + 1

= 2(1) – 3 + 1

= 2 – 2 = 0

⇒ 4c = 5 + 3

⇒ 4c= 8

Hence, Lagrange’s mean value theorem is verified.