f(x) will exist, if

x² – 4 ≥ 0

x² ≥ 4

x ≤ -2 or x ≥ 2

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,

0

Squaring both sides,

5x² – 20 – x² = 0

4x² = 20

x² = 5

x = ± √5

Hence, c = √5 Є (2, 3) such that

Hence the above explanation verifies the Lagrange’s Theorem.