Q. 1 D5.0( 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) = x2 – 3x + 2 on [ – 1, 2]


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 – 3x + 2


Differentiating with respect to x:


f’(x) = 2x – 3


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


f’(c)= 2c – 3


For f(2), put the value of x=2 in f(x):


f(2)= (2)2 – 3(2) + 2


= 4 – 6 + 2


= 0


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


f( – 1) = ( – 1)2 – 3( – 1) + 2


= 1 + 3 + 2


= 6





2c = – 2 + 3


2c= – 1



Hence, Lagrange’s mean value theorem is verified.


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