Q. 1 C5.0( 1 Vote )

Verify Lagrange’s

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]


= x2 – 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) = x2 – 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.


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