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) = x (x + 4)2 on [0, 4]
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 (x + 4)2 on [0, 4]
= x [(x)2+2(4)(x)+(4)2]
= x(x2 + 8x + 16)
= x3 + 8x2 + 16x on [0, 4]
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, 4] and differentiable in (0, 4). So both the necessary conditions of Lagrange’s mean value theorem is satisfied.
f(x) = x3 + 8x2 + 16x
Differentiating with respect to x:
f’(x) = 3x2 + 8(2x) + 16
= 3x2 + 16x + 16
For f’(c), put the value of x=c in f’(x):
f’(c)= 3c2 + 16c + 16
For f(4), put the value of x=4 in f(x):
f(4)= (4)3 + 8(4)2 + 16(4)
= 64 + 128 + 64
= 256
For f(0), put the value of x=0 in f(x):
f(0)= (0)3 + 8(0)2 + 16(0)
= 0 + 0 + 0
= 0
⇒ 3c2 + 16c + 16 = 64
⇒ 3c2 + 16c + 16 – 64 = 0
⇒ 3c2 + 16c – 48 = 0
For quadratic equation, ax2 + bx + c = 0
Hence, Lagrange’s mean value theorem is verified.