Q. 1 G5.0( 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) = 2x – x2 on [0, 1]


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, 1] and differentiable in (0, 1). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.




f’(c) = f(1) – f(0)


f(x) = 2x – x2


Differentiating with respect to x:


f’(x) = 2 – 2x


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


f’(c)= 2 – 2c


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


f(1)= 2(1) – (1)2


= 2 – 1


= 1


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


f(0) = 2(0) – (0)2


= 0 – 0


= 0


f’(c) = f(1) – f(0)


2 – 2c = 1 – 0


– 2c = 1 – 2


– 2c = – 1



Hence, Lagrange’s mean value theorem is verified.


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