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# 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) = x^{2} + x – 1 on [0, 4]

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^{2} + x – 1 on [0, 4]

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, 4] and differentiable in (0, 4). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

f(x) = x^{2} + x – 1

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(4), put the value of x=4 in f(x):

f(4)= (4)^{2} + 4 – 1

= 16 + 4 – 1

= 19

For f(0), put the value of x=0 in f(x):

f(0) = (0)^{2} + 0 – 1

= 0 + 0 – 1

= – 1

⇒ 2c + 1 = 5

⇒ 2c = 5 – 1

⇒ 2c = 4

Hence, Lagrange’s mean value theorem is verified.

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