f(x) = \(\sqrt{x^2 - 4}\)
f(x) will exist, if
x2 – 4 ≥ 0
x2 ≥ 4
\(\therefore\) x ≤ -2 or x ≥ 2
\(\because\) for each x Є [2, 3], the function f(x) has a unique definite value, f(x) is continuous on (2, 3).
Exists for all x Є (2, 3).
So, f(x) is differentiable on (2,3).
Hence, both the conditions of Lagrange’s Theorem are satisfied.
Consequently, there exists c Є (2, 3) such that,
Squaring both sides,
Hence, c = \(\sqrt{5}\) Є (2, 3) such that f'(c) = \(\frac{f(3) - f(2)}{3 - 2}\)
Hence the above explanation verifies the Lagrange’s Theorem.