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,

⇒ 25 – x² >0

⇒ x² < 25

⇒ – 5 < x < 5

∴ f(x) is continuous in [ – 3, 4]

Differentiating with respect to x:

Here also,

⇒ – 5 < x < 5

∴ f(x) is differentiable in ( – 3, 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):

⇒ f(4) = 3

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

⇒ f( – 3) = 4

Squaring both sides:

⇒ 49c² = 25 – c²

⇒ 50c² = 25

Hence, Lagrange’s mean value theorem is verified.