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# Verify Lagrange’s

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 (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.

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