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^{2} – 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) = x^{2} – 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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