Answer :

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]

= x^{2} – 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) = x^{2} – 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.

Rate this question :

Verify the Rolle’Mathematics - Exemplar

The value of c inMathematics - Exemplar

For the function Mathematics - Exemplar

Verify the Rolle’Mathematics - Exemplar

Verify the Rolle’Mathematics - Exemplar

Discuss theRD Sharma - Volume 1

Using Rolle’s theMathematics - Exemplar

Verify the Rolle’Mathematics - Exemplar

State True Mathematics - Exemplar

Discuss the appliMathematics - Exemplar