The value of c prescribed by Largrange's mean value . Theorem, when `f(x)=sqrt(x^(2)-4),a=2 and b=3` is

Options

A. `2.5`
B. `sqrt(5)`
C. `sqrt(3)`
D. `sqrt(3)+1`

Explanation

To find the value of \( c \) prescribed by Lagrange's Mean Value Theorem for the function \( f(x) = \sqrt{x^2 - 4} \) over the interval \( [2, 3] \), we will follow these steps:

### Step 1: Verify the conditions of the Mean Value Theorem
The Mean Value Theorem states that if \( f(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
Here, \( a = 2 \) and \( b = 3 \).

### Step 2: Calculate \( f(a) \) and \( f(b) \)
- Calculate \( f(2) \):
\[ f(2) = \sqrt{2^2 - 4} = \sqrt{4 - 4} = \sqrt{0} = 0 \]
- Calculate \( f(3) \):
\[ f(3) = \sqrt{3^2 - 4} = \sqrt{9 - 4} = \sqrt{5} \]

### Step 3: Apply the Mean Value Theorem
Now, we can find the average rate of change of \( f(x) \) over the interval \([2, 3]\):
\[ \frac{f(b) - f(a)}{b - a} = \frac{f(3) - f(2)}{3 - 2} = \frac{\sqrt{5} - 0}{1} = \sqrt{5} \]

### Step 4: Find the derivative \( f'(x) \)
To find \( f'(x) \), we first differentiate \( f(x) \):
\[ f(x) = \sqrt{x^2 - 4} \]
Using the chain rule:
\[ f'(x) = \frac{1}{2\sqrt{x^2 - 4}} \cdot (2x) = \frac{x}{\sqrt{x^2 - 4}} \]

### Step 5: Set \( f'(c) \) equal to the average rate of change
According to the Mean Value Theorem:
\[ f'(c) = \sqrt{5} \]
Substituting the expression for \( f'(x) \):
\[ \frac{c}{\sqrt{c^2 - 4}} = \sqrt{5} \]

### Step 6: Solve for \( c \)
Cross-multiply to eliminate the fraction:
\[ c = \sqrt{5} \cdot \sqrt{c^2 - 4} \]
Square both sides:
\[ c^2 = 5(c^2 - 4) \]
Expanding the right side:
\[ c^2 = 5c^2 - 20 \]
Rearranging gives:
\[ 20 = 5c^2 - c^2 \implies 20 = 4c^2 \implies c^2 = 5 \]
Taking the square root:
\[ c = \sqrt{5} \quad \text{(we discard the negative root since } c \text{ must be in } (2, 3)) \]

### Conclusion
Thus, the value of \( c \) prescribed by Lagrange's Mean Value Theorem is:
\[ c = \sqrt{5} \]