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) = x² – 2x + 4 on [1, 5]

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, 5] and differentiable in (1, 5). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x² – 2x + 4

Differentiating with respect to x:

f’(x) = 2x – 2

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

f’(c)= 2c – 2

For f(5), put the value of x=5 in f(x):

f(5)= (5)² – 2(5) + 4

= 25 – 10 + 4

= 19

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

f(1) = (1)² – 2(1) + 4

= 1 – 2 + 4

= 3

⇒ 2c = 4 + 2

⇒ 2c= 6

Hence, Lagrange’s mean value theorem is verified.