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) = x3 – 2x2 – x + 3 on [0, 1]

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) = x3 – 2x2 – x + 3 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(x) = x3 – 2x2 – x + 3


Differentiating with respect to x:


f’(x) = 3x2 – 2(2x) – 1


= 3x2 – 4x – 1


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


f’(c)= 3c2 – 4c – 1


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


f(1)= (1)3 – 2(1)2 – (1) + 3


= 1 – 2 – 1 + 3


= 1


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


f(0)= (0)3 – 2(0)2 – (0) + 3


= 0 – 0 – 0 + 3


= 3


f’(c) = f(1) – f(0)


3c2 – 4c – 1 = 1 – 3


3c2 – 4c = 1 + 1 – 3


3c2 – 4c = – 1


3c2 – 4c + 1 = 0


3c2 – 3c – c + 1 = 0


3c(c – 1) – 1(c – 1) = 0


(3c – 1) (c – 1) = 0




Hence, Lagrange’s mean value theorem is verified.


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