If the function \(\sqrt {{X^2} - 4} \) in [2, 4] satisfies the Lagrange’s mean value theorem, then there exists some c ∈ [2, 4]. The value of c is

Concept:

Let f(x) is a function define on [a ,b] such that, 

1. f(x) is a Continuous on [a , b]
2. f(x) is Differentiable on [a , b]

Then, there exist a real number C ∈ (a , b) such that, According to Lagrangian Mean Value Theorem,

f'(c) =  \(\frac{{f\left( b \right) ~- ~f\left( a \right)}}{{b ~-~ a}}\) 

Calculation:

Given:

f(x) = \(\sqrt {{X^2} - 4} \)    , and f'(x) = \(\frac{1}{{2\sqrt {{X^2}~ -~ 4} }} \times 2X\)

The function satisfies Lagrange's Mean Value Theorem that means it satisfies two condition given above 1 and 2

Therefore for the value of C we can write down above formula 

f'(c) = \(\frac{{f\left( b \right)~ -~ f\left( a \right)}}{{b ~- ~a}}\)  = \(\frac{{\sqrt {16~ -~4} ~-~ \sqrt {4~ -~ 4} }}{{4~ -~ 2}}\)   = \(\frac{{\sqrt {12} }}{2}\)  = √3 

\(\frac{{{c}}}{{\sqrt {c^2~-~4}}}\) = √3

Squaring on both sides,

3c² - 12 = c² 

2c² = 12

c² = 6

c = √6

Additional Information

• (a, b) means an open interval.
• The value of C is obtained in the open interval (a, b) which means between a and b.