Question

Verify Lagrange’s mean value theorem for the functions on the indicated intervals. Find a point ‘c’ in the indicated interval as stated by the Lagrange’s mean value theorem:

f(x) = x³ – 5x² – 3x on [1, 3]

Answer 1

Given as f(x) = x³ – 5x² – 3x on [1, 3]

The every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Since, f(x) is a polynomial function. Therefore it is continuous in [1, 3] and differentiable in (1, 3). Therefore both the necessary conditions of Lagrange’s mean value theorem is satisfied.

So, there exist a point c ∈ (1, 3) such that:

f'(c) = (f(3) - f(1))/(3 - 1)

f'(c) = (f(3) - f(1))/2

f(x) = x³ – 5x² – 3x

Differentiate with respect to x:

f’(x) = 3x² – 5(2x) – 3

= 3x² – 10x – 3

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

f’(c)= 3c² – 10c – 3

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

f (3) = (3)³ – 5(3)² – 3(3)

= 27 – 45 – 9

= – 27

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

f (1)= (1)³ – 5 (1)² – 3 (1)

= 1 – 5 – 3

= – 7

Thus, lagrange's mean value theorem is verified.