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 – 5x2 – 3x on [1, 3]
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 – 5x2 – 3x 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) = x3 – 5x2 – 3x
Differentiating with respect to x:
f’(x) = 3x2 – 5(2x) – 3
= 3x2 – 10x – 3
For f’(c), put the value of x=c in f’(x):
f’(c)= 3c2 – 10c – 3
For 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 f(1), put the value of x=1 in f(x):
f(1)= (1)3 – 5(1)2 – 3(1)
= 1 – 5 – 3
= – 7
⇒ 3c2 – 10c – 3 = – 10
⇒ 3c2 – 10c – 3 + 10 = 0
⇒ 3c2 – 10c + 7 = 0
⇒ 3c2 – 7c – 3c + 7 = 0
⇒ c(3c – 7) – 1(3c – 7) = 0
⇒ (3c – 7) (c – 1) = 0
Hence, Lagrange’s mean value theorem is verified.