Question

If the function f(x) = \(\sqrt{x^2 - 4}\)  satisfies the Lagrange’s Mean Value Theorem on \([2, 4]\), then the value of \( C \) is} 

• A. \( 2\sqrt{3} \)
• B. \( -2\sqrt{3} \)
• C. \( \sqrt{6} \)
• D. \( -\sqrt{6} \)

Hint:

To apply the Mean Value Theorem, first verify that the function is continuous and differentiable on the given interval. Then, compute the average rate of change and set it equal to the derivative to find \( c \).

Answer 1

The Correct Option is C

Step 1: Verify the applicability of the Mean Value Theorem 
Lagrange’s Mean Value Theorem states that if a function \( f(x) \) is continuous and differentiable on \( [a, b] \), then there exists some \( c \in (a, b) \) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a}. \] The given function is: \[ f(x) = \sqrt{x^2 - 4}. \] It is continuous and differentiable on \( (2,4] \), satisfying the conditions of the Mean Value Theorem. 
Step 2: Compute \( f(2) \) and \( f(4) \) 
\[ f(2) = \sqrt{2^2 - 4} = \sqrt{0} = 0. \] \[ f(4) = \sqrt{4^2 - 4} = \sqrt{12} = 2\sqrt{3}. \] 
Step 3: Compute the Average Rate of Change 
\[ \frac{f(4) - f(2)}{4 - 2} = \frac{2\sqrt{3} - 0}{2} = \sqrt{3}. \] 
Step 4: Compute \( f'(x) \) 
Using the chain rule, we differentiate \( f(x) \): \[ f'(x) = \frac{1}{2\sqrt{x^2 - 4}} \cdot 2x = \frac{x}{\sqrt{x^2 - 4}}. \] 
Step 5: Solve for \( c \) 
By the Mean Value Theorem, \[ \frac{c}{\sqrt{c^2 - 4}} = \sqrt{3}. \] Squaring both sides, \[ \frac{c^2}{c^2 - 4} = 3. \] Cross-multiplying, \[ c^2 = 3(c^2 - 4). \] \[ c^2 = 3c^2 - 12. \] \[ -2c^2 = -12. \] \[ c^2 = 6. \] \[ c = \pm \sqrt{6}. \] Since \( c \) is in \( (2,4) \), we take the positive root: \[ c = \sqrt{6}. \]
Final Answer: \( \boxed{\sqrt{6}} \).