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.

Here,

⇒ x² – 4 >0

⇒ x² > 4

⇒ f(x) exists for all values expect ( – 2, 2)

∴ f(x) is continuous in [2, 4]

Differentiating with respect to x:

Here also,

⇒ f’(x) exists for all values of x except (2, – 2)

∴ f(x) is differentiable in (2, 4)

So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

On differentiating with respect to x:

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

For f(4), put the value of x=4 in f(x):

For f(2), put the value of x=2 in f(x):

⇒ f(2) = 0

Squaring both sides:

⇒ c² = 3(c² – 4)

⇒ c² = 3c² – 12

⇒ – 2c² = – 12

⇒ c² = 6

Hence, Lagrange’s mean value theorem is verified.