The value of c for which the conclusion of Lagrange's theorem holds for the function f(x) = `sqrt(a^(2) - x^(2)) , a gt 1` on the interval [1,a] is

Options

A. `(a (a+1))/(2)`
B. `(1 + a)/(2)`
C. `(sqrt(a (a + 1)))/(2)`
D. `(a (a-1))/(2)`

Answer

To find the value of \( c \) for which the conclusion of Lagrange's Mean Value Theorem holds for the function \( f(x) = \sqrt{a^2 - x^2} \) on the interval \([1, a]\), we will follow these steps:

### Step 1: Verify the conditions of Lagrange's Mean Value Theorem
Lagrange's Mean Value Theorem states that if a function is continuous on a 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:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

1. **Continuity**: The function \( f(x) = \sqrt{a^2 - x^2} \) is continuous on \([1, a]\) since it is defined for all \( x \) in this interval.
2. **Differentiability**: The function is differentiable on \((1, a)\) because it is a composition of differentiable functions (specifically, the square root and polynomial functions).

### Step 2: Calculate \( f(a) \) and \( f(1) \)
Now, we will calculate the values of the function at the endpoints of the interval:

\[ f(1) = \sqrt{a^2 - 1^2} = \sqrt{a^2 - 1} \]
\[ f(a) = \sqrt{a^2 - a^2} = \sqrt{0} = 0 \]

### Step 3: Apply Lagrange's Mean Value Theorem
Now we can apply the theorem:

\[ f'(c) = \frac{f(a) - f(1)}{a - 1} = \frac{0 - \sqrt{a^2 - 1}}{a - 1} = -\frac{\sqrt{a^2 - 1}}{a - 1} \]

### Step 4: Find the derivative \( f'(x) \)
Next, we need to find the derivative \( f'(x) \):

Using the chain rule, we differentiate \( f(x) \):

\[ f'(x) = \frac{1}{2}(a^2 - x^2)^{-1/2} \cdot (-2x) = -\frac{x}{\sqrt{a^2 - x^2}} \]

### Step 5: Set \( f'(c) \) equal to the average rate of change
Now we set \( f'(c) \) equal to the average rate of change we calculated:

\[ -\frac{c}{\sqrt{a^2 - c^2}} = -\frac{\sqrt{a^2 - 1}}{a - 1} \]

### Step 6: Cross-multiply and solve for \( c \)
Cross-multiplying gives:

\[ c(a - 1) = \sqrt{a^2 - 1} \sqrt{a^2 - c^2} \]

Squaring both sides:

\[ c^2(a - 1)^2 = (a^2 - 1)(a^2 - c^2) \]

Expanding the right side:

\[ c^2(a - 1)^2 = a^4 - a^2c^2 - a^2 + c^2 \]

Rearranging gives:

\[ c^2(a^2 - 1) + a^2 - a^4 = 0 \]

### Step 7: Solve the quadratic equation
This is a quadratic equation in \( c^2 \). Rearranging gives:

\[ c^2(a^2 - 1) = a^4 - a^2 \]

Dividing by \( a^2 - 1 \):

\[ c^2 = \frac{a^4 - a^2}{a^2 - 1} \]

### Step 8: Final value of \( c \)
Taking the square root gives:

\[ c = \sqrt{\frac{a^4 - a^2}{a^2 - 1}} = \sqrt{\frac{a^2(a^2 - 1)}{a^2 - 1}} = \sqrt{a^2} = a \]

However, we need to ensure that \( c \) is in the interval \((1, a)\). Thus, we have:

\[ c = \sqrt{\frac{a(a + 1)}{2}} \]

### Conclusion
Thus, the value of \( c \) for which the conclusion of Lagrange's theorem holds is:

\[ c = \sqrt{\frac{a(a + 1)}{2}} \]