Step 1: Understanding the Concept:
Lagrange's Mean Value Theorem (LMVT) is a cornerstone of differential calculus.
It states that if a function \( f(x) \) 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 the interval \((a, b)\) such that the derivative of the function at that point is equal to the average rate of change of the function over the entire interval.
Geometrically, this means there is a point on the curve where the tangent line is parallel to the secant line passing through the points \((a, f(a))\) and \((b, f(b))\).
In this problem, we are dealing with a semi-circular function \( f(x) = \sqrt{25 - x^2} \), which represents the upper half of a circle centered at the origin with a radius of 5.
The theorem is applicable here because the function is well-defined and smooth within the requested domain.
Step 2: Key Formula or Approach:
The mathematical formula for finding \( c \) is given by:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
Where:
- \( a = 1 \) and \( b = 5 \).
- \( f'(x) \) is the first derivative of the function.
- \( c \) is the value we need to find such that \( 1<c<5 \).
Step 3: Detailed Explanation:
1. First, let us verify the conditions of LMVT.
The function \( f(x) = \sqrt{25 - x^2} \) is defined for \( 25 - x^2 \ge 0 \), which implies \( x \in [-5, 5] \).
Since our interval is \([1, 5]\), the function is continuous.
The derivative is \( f'(x) = \frac{-x}{\sqrt{25 - x^2}} \), which is defined for \( x \in (-5, 5) \).
Thus, the function is differentiable on \((1, 5)\).
2. Calculate the values at the boundaries:
\( f(a) = f(1) = \sqrt{25 - 1^2} = \sqrt{24} = 2\sqrt{6} \).
\( f(b) = f(5) = \sqrt{25 - 5^2} = \sqrt{0} = 0 \).
3. Calculate the slope of the secant line:
\[ \text{Slope} = \frac{f(5) - f(1)}{5 - 1} = \frac{0 - \sqrt{24}}{4} = \frac{-2\sqrt{6}}{4} = -\frac{\sqrt{6}}{2} \]
4. Set the derivative \( f'(c) \) equal to this slope:
\[ \frac{-c}{\sqrt{25 - c^2}} = -\frac{\sqrt{6}}{2} \]
Squaring both sides to eliminate the radicals:
\[ \frac{c^2}{25 - c^2} = \frac{6}{4} = \frac{3}{2} \]
Cross-multiplying:
\[ 2c^2 = 3(25 - c^2) \]
\[ 2c^2 = 75 - 3c^2 \implies 5c^2 = 75 \implies c^2 = 15 \]
This gives \( c = \pm \sqrt{15} \).
Step 4: Final Answer:
Since LMVT requires \( c \) to be in the open interval \((1, 5)\), we reject \( -\sqrt{15} \) because it is negative.
The value \( c = \sqrt{15} \approx 3.87 \) lies within the interval \((1, 5)\).
Therefore, the required value is \( \sqrt{15} \).