Q. 1 E5.0( 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) = 2x2 – 3x + 1 on [1, 3]


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, 3] and differentiable in (1, 3). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.





f(x) = 2x2 – 3x + 1


Differentiating with respect to x:


f’(x) = 2(2x) – 3


= 4x – 3


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


f’(c)= 4c – 3


For f(3), put the value of x=3 in f(x):


f(3)= 2(3)2 – 3(3) + 1


= 2(9) – 9 + 1


= 18 – 8 = 10


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


f(1) = 2(1)2 – 3(1) + 1


= 2(1) – 3 + 1


= 2 – 2 = 0





4c = 5 + 3


4c= 8



Hence, Lagrange’s mean value theorem is verified.


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